Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Why can we only "see" reflected light? This is a question thats been bothering me a while. I don't even know if it makes sense or not (like if it is a physics question or becoming a philosophical one). But here it goes. The crux of my question basically is that we all know that we can't see light (like in its photon or electromagnetic wave form) directly when it is traveling past us. However, we also know that the way we see objects is by light reflecting off them. This then means that we are "seeing" the light reflecting from the object which then sends the signal to our brain saying that we are seeing a particular object. We know that both light traveling past us and light reflected from objects are made of photons (so they are the same kind)? So then my question is that what is happening to the photon of a light after it is reflected from the objects, that causes us to see it or the object, but on the other hand we can't see light as it is directly traveling past us.
| You can see the light which enters in your eye and is absorbed by your retina. So you cannot see the light passing by because it is not going towards your eye. If it meets an object, however, light will be reflected or scattered and part of it will go towards your eye. You will then see the light coming from the object.
I would add that if you put your eye before the object, into the light illuminating it, you will see the incoming light just as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
What is spacetime (simple explanation)? Can someone please explain concept of spacetime in simple language? What is it and how it is important in the universe?
Wherever I have tried searching this concept, I have come across most complicated explanations.
A simple example will be appreciated.
| Whenever you draw a curve of a trajectory over time you are using spacetime as a concept. It is just a space with n+1 dimensions, where n is the number of space dimensions.
When we draw such diagrams on a piece of paper we usually omit one or two of the space dimensions (e.g. recording only the x position in space; y and z may be less important). That leaves room for using one axis for the time values of events.
That is, a 2-dimensional piece of paper can represent spacetime with n=1, in which you draw (x, t) curves. Similarly, a 3D diagram on paper or on a computer can represent subsets of spacetime with two explicit space dimensions.
Mathematically, this is just a Cartesian space of tuples of n+1 numbers. In these tuples you write the values of space coordinates, plus the time position, of events, and the set of these events forms a curve or surface in spacetime.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 6
} |
A free-fall electron I am reading Wheeler and Taylor's Spacetime Physics. In Ch2, Wheeler mentioned:
"for gravity, any free-fall frame is an inertial frame." (roughly)
I am left wondering if that is true for electrical force:
Consider one charge is under a statistic electrical field. The charge is in free-fall. Is the electron's free fall frame an inertial frame?
(if yes, then can we say electrical force is a pseudo-force too?)
| Perhaps I'm misunderstanding the quote, but I think there's a distinction between the false statement...
Anything falling in a gravitational field is in an inertial frame
... and what I believe the quote means...
In the frame of reference gravity itself, something falling [in a gravitational field] is inertial, because the gravity frame measures no acceleration in the falling object's frame (my frame when I'm falling in an elevator is inertial w.r.t. the frame of anything else in said elevator).
So, to kind-of answer your question, the electron is accelerating towards the positive charge, since the magnitude of force between the two is...
$$\frac{1}{4 \pi \epsilon_0} \frac{q^- q^+}{r^2}$$
The closer they get, the more force attracts them—they are accelerating towards one another. To construct an inertial reference frame w.r.t. the electron, one must construct a frame that is accelerating at the same rate, and in the same direction, as the electron.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Opposite of particle decay I have read about particle decay, a process in which one particle becomes several other particles. However, I have not been able to find much information about its opposite: several particles combining into one particle. Is such a process possible, and if so, under what conditions? For example, a free neutron may decay into a proton, electron, and electron antineutrino. Could a proton, electron, and electron antineutrino somehow be joined into a neutron?
Edit: Everyone, thank you for your help, but let me try to make what I'm looking for clearer. I want to know whether several particles can join into ONE particle, in an exact reverse of that one particle decaying into several particles. As far as I know, I don't think an atomic nucleus counts as one particle. Please correct me if I'm wrong.
| Even if the reaction is not prohibited by energy/momentum or other conservation laws, it would have a tiny cross section: shooting one kind of particle at a target is easy, but shooting two at the same time, not so much (particularly if one is an antineutrino). However there is such a thing as electron capture by a proton in the nucleus, producing a neutron and an electron neutrino: $p + e^- \rightarrow n + \nu_e$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 1
} |
How can length be a vector? Length and current both are not vectors. Then how can we assign the vector $l$ to the length of a wire carrying current while calculating for a current carrying conductor in a magnetic field. Also why in Biot—Savart law do we take small length element $dl$ as a vector?
Why is length sometimes a vector, sometimes not, whereas current always is a scalar?
| $d\vec \ell$ is not a length. It is a vector of infinitesimal length (of length $d\ell$) that is in the direction of the current flow.
Current, on the other hand, is a vector as it flows in specific directions. It is often not indicated by a vector sign (for historical reasons) because in circuits the current will flow along the wire that carries it.
Further note that, in the specific case of Biot-Savart, there is a historical issue with the notation. The notation $Id\vec \ell$ should really be $\vec I\,d\ell$: this would resolve your confusion, but history is history. Indeed, you can appreciate the notational quirk by comparing the source expressions:
\begin{align}
\vec B\propto\left\{ \begin{array}{ll}
Id\vec \ell \times \frac{\hat r}{r^2}&\hbox{ for linear current distributions}\, ,\\
\vec K\,dS\times \frac{\hat r}{r^2}&\hbox{ for surface current distributions}\, ,\\
\vec J\,dV\times \frac{\hat r}{r^2}&\hbox{ for volume current distributions}\, ,\end{array}\right.
\end{align}
clearly suggesting that, in order to conform with the surface and volume expressions, the current $I$ should be vectorial but the line element $d\vec \ell$ should be scalar.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Can only two equal EM-wavelength interfere? If so why is that? When there are two waves of light with the same wavelength they can constructively or destructively interfere with each other. But can for example a wave of 532nm interfere with a wave of 533nm. How exactly must they be the same and why can't (if so?) different wavelengths join together?
| Waves of different wavelength/frequency interfere intermittently, which produces periodic modulation called beat frequency. The beat frequency corresponds to the difference between the frequencies. It is what musicians are listening for as they tune their instruments.
It's easiest to visualize this by thinking about two waves that are only slightly different. They interfere for a few waves until they are out of phase and then wander back into phase again.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do you calculate the resulting magnetic field for multiple sources? I've been looking at some fusion reactors and I keep wondering how putting some kind of extra magnet in some configuration would affect the field, but I don't know how to figure it out. Like for example if you took a solenoid and sat a permanent magnet down next to it.
This is the only thing I could find:
https://www.quora.com/How-do-I-calculate-the-magnetic-field-created-by-a-number-of-magnets
Edit:
I'm pretty sure it's just super positioning, but I need someone else to answer, because I don't know for sure. Also does it follow from Maxwells equations?
| Since the Maxwell's equations are linear partial differential equations, you can compute the magnetic field due to multiple sources by superposition.
A really important application relies on the superposition principle for magnetic fields is the Biot–Savart law i.e. the fact that the magnetic field is a vector sum of the field created by each infinitesimal section of the wire individually.
$$\mathrm d\vec B = \frac{\mu_0}{4\pi}\frac{I \; \mathrm d\vec l \times \vec R}{R^3}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How did Rutherford conclude that most of the mass (as well as the positive charge) was concentrated in the nucleus? Geiger and Marsden's experiment led Rutherford to believe that the positive charge and most of the mass of the atom was concentrated in a small region. I understand what led him to conclude the way the positive charge is positioned in the atom. But how did he conclude that most of the mass was in a small region (the nucleus)?
How did the distribution of the mass matter after all? Given that the electric force is greater than the gravitational force by many magnitudes, the force between the positice charge and the electrons was predominantly electric.
So how did Rutherford conclude that most of the mass is in the nucleus?
| This is a good example of how Science works.
Geiger and Marsden observed that some of the alpha particles were being backscattered. This is inconceivable if the alpha particle is scattered by a lighter particle.
If one considers a particle of mass $m$ and initial velocity $v_1$ striking a target of mass $m'$ at rest, without changing its direction, then its final velocity $v_2$ can assume two possible values,
$$v_2=v_1,\quad\mathrm{or}\quad v_2=-v_1\left(\frac{m'-m}{m'+m}\right).$$
The second solution gives that backscattering is only possible if the target has greater mass than the incident particle. By the time, the mass of the electron was known to be much smaller than the mass of the alpha particle so a backscattering event would imply that the scattering centers were in fact heavy positive nuclei. And indeed those scatterings were observed.
To formally check this, Rutherford obtained a formula for the number of scattered particles as a function of the scattering angle using the hypothesis of heavy nuclei (which is justified by the above paragraph). Geiger and Marsden did the experiment and the data agreed with Rutherford formula.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 1
} |
How would electromagnetic induction violate neither the conservation of energy or Newton's first law? The basic premise of electromagnetic induction is that supposedly when you have a magnetic field moving, it generates an electromotive field perpendicularly, yeah? But if you had a hypothetical space where Newton's first law could be observed perfectly (no external forces at all), and you just had a magnet drifting on for infinity inside a conductive coil, how would that work? There would be no forces acting on the magnet, it would just remain in motion. The electromotive forces would just be coming out of no where then.
| When you move a magnet through a coil or even through space , a non conservative electric field is induced in space in a circular form. But, since energy is conserved in any case which can only be possible in this case if the motion of magnet is opposed so that the kinetic energy of moving magnet is converted into electrical energy and if you have a magnet just moving in a straight line with a constant velocity it will ultimately come to a stop, hope that helps, any other question you have regarding this, you can ask me.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Diffeomorphism invariance and correlation functions Consider the following paragraph taken from page 15 of Thomas Hartman's lecture notes on Quantum Gravity:
In gravity, local diffeomorphisms are gauge symmetries. They are redundancies. This means that local correlation functions like $\langle O_{1}(x_{1})\dots O_{n}(x_{n})\rangle$ are not gauge invariant, and so they are not physical observables. On the other hand, diffeomorphisms that reach infinity (like, say, a global translation) are physical symmetries - taking states in the Hilbert space to different states in the Hilbert space - so we get a physical observable by taking the insertion points to infinity. This defines the S-matrix, so it is
sometimes said that ``The S-matrix is the only observable in quantum gravity.''
*
*Why does the fact that local diffeomorphisms are gauge symmetries mean that local correlation functions like $\langle O_{1}(x_{1})\dots O_{n}(x_{n})\rangle$ are not gauge invariant?
*Why do diffeomorphisms that reach infinity become global symmetries?
| In gravity theories gauge invariance is diffeomorphism invariance, so local operators are not gauge invariant because there are no preferred coordinates, you can always reparametrize them - choose another coordinates, and the operators will, in general, change.
As for 2, its just the way of saying that transformation that is applied to all the space (up to infinity) is a global transformation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the relation between image velocity, object velocity and mirror velocity? Suppositions used:
Velocity of image = VI
Velocity of object = Vo
Velocity of mirror = VM
I Know the fact that VI=-Vo supposing mirror at rest
and VI=2VM supposing object at rest
Now considering both mirror and object in motion, VI=2VM - Vo I ended up with this equation but my reference book suggests VI=2VM + Vo I am stuck on this for last 4 hours. I searched over internet and found the same expression like that of mine in a youtube video, I did not find much reference on this topic though. Tried many ways but all ended up on this simple argument, which equation to follow?
Help
| I think that, since the velocities are in different directions, $2V_M -\left(-V_O\right) = 2V_M+V_O$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Have we directly observed the electric component to EM waves? For example, has anyone has directly observed charges oscillating due to standing EM waves? I am particularly interested because it'd demonstrate that radiation has a transverse electric component to it. Anything else (historical or modern) that shows that light has a transverse electric component would also be gladly invited.
| It depends on what you mean by "directly observed charges oscillating" and whether the formation of a standing wave is essential.
Using microwaves to demonstrate that they can be polarised you can infer the direction of oscillation of the electric vector in an electromagnetic wave.
If there is a horn receiver as well as a horn transmitter then just rotating one relative to the other produce a variation the amplitude of the received signal to show that the transmitted microwaves are polarised.
This can lead on to demonstrating polarisation of light using a Polaroid which on a molecular scale has crystals orientated in parallel which can form a conducting path just like the rods for microwaves.
If you do need the condition that the demonstration has to be for standing waves then a pair of Letcher lines might be of use?
Microwaves (or UHF waves) from a transmitter $b$ are guided along a pair of parallel uninsulated conducting rods with a spacing less than that of the wavelength of the microwaves and if shorted ($c$ in the diagram) produce a standing wave pattern which can be investigated by either observing the voltage nodes and antinodes or current nodes and antinodes.
Leybold produce an excellent leaflet describing the sort of experiments which can be performed with such an apparatus.
At radio frequencies the rotation of an fm dipole aerial showing the variation in signal strength will illustrate polarisation.
In this case unlike the microwave demonstration with the parallel metal rods the response is a maximum when the dipole is orientated parallel to the electric field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 4
} |
Why is the South Pole Telescope located exactly at the South Pole? I read that there is less atmospheric interference for the telescope at the South Pole because the atmosphere is thin and there is less water vapor in the air. However this seems to be true for many locations on Antarctica? Are there any other reasons that this telescope is located at exactly the South Pole?
| Just guessing here, but ...
Compare the regions with really good skies with the places that have infrastructure and people present. Most of the installations are coastal, right? Are those good places to put a telescope? And while the whole inland plateau has good skies, it has few occupied site, and only one operated by the US.
So what is the case for putting up some other (very expensive to build and maintain) installation, when you could just drop it by South Pole Station where they already maintain a year-round presence.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 1
} |
Why do we equate an indefinite integral to a specific value? Suppose we want to obtain a displacement vector defined as $\mathbf s(t) = x(t)\mathbf i + y(t)\mathbf j + z(t)\mathbf k$ from the components of a velocity vector $\mathbf v(t) = \dot x(t)\mathbf i + \dot y(t)\mathbf j + \dot z(t)\mathbf k=\mathbf 0$. According to my notes, this can be done by equating each scalar component of the displacement vector to the indefinite integral of the corresponding scalars of the velocity vector, i.e.
$$
\mathbf s(t)=\begin{pmatrix}
x(t)=\int \dot x\ dt \\
y(t)=\int \dot y\ dt \\
z(t)=\int \dot z\ dt
\end{pmatrix}
$$
But, as $\int f(x)\ dx=\{F(x): \frac {dF}{dx}=f(x)\}$, this should be syntactically wrong, because we're implying that a number is equal to an infinite set of numbers, or am I missing something?
Moreover, this also leads to a weird equation when solving the integral; for example, by taking into consideration the $x$-component of $\mathbf s$, we would have that
$$
x(t)=\int \dot x\ dt=c_1
$$
Which is correct, but it would also mean that the $x$-component of the velocity could be equal to any value belonging to $\mathbb R$. Because of that, we substitute $c_1$ with the initial condition and we equate it to zero, giving it a specific value: $c_1=0$. But, to me, this sounds like a break of the definition of indefinite integrals, as
$\int \dot x\ dt=c_1=0$ would basically mean that an indefinite integral is one, specific function.
I know this may be a very stupid question, and maybe it has to do with the same shortcuts that make us not specify "$\forall c \in \mathbb R$" when adding the constant $c$ in the solutions of an indefinite integral, but this doubt is really challenging me and I still don't understand whether I'm missing some point or it should actually be written $x(t)=c_1=0 \in \int \dot x\ dt$. Thanks a lot in advance!
| In physics we frequently leave off the limits of the integral when the limits can be figured out from the context. So, in the first case, the actual relation is:
$$x(t) = x(0) + \int_0^t \dot{x} \operatorname{d}t'.$$
Most often, though, when the limits are left off the implied limits are over all possible values of the dummy variable. For example:
$$ Q = \int \rho(\mathbf{x}) \operatorname{d}x^3$$
is understood to be the integral over all of space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How does a electric quadrupole oscillate? I know that in static a electric quadrupole is made of two positive charges and two negative charges, distributed as in the following figure:
What I don't understand is when it oscillates and radiates, does it oscillate like two antiparallel electric dipoles, or it oscillates like that shown in the attached figure, with four dipoles?
Image source
| If you want a truly quadrupolar oscillation, then it needs to oscillate as in the four-dipoles model, i.e. there is a current linking all adjacent pairs of localized charges, which is crucial to the formation of the correct quadrupolar symmetry for the radiation pattern.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Can we add attraction forces? From Newton's Law of gravitation we know that:
$$F=G\frac{m_1m_2}{d^2}$$
For simplicity, let's say that both $m$ are $1\;\mathrm{kg}$ and that the distance apart was $1\;\mathrm{m}$. Yielding $G$ as the attraction force in Newtons. Hence $F= 6.67\times 10^{-11}\;\mathrm{N}$.
Now what if you had two $0.5\;\mathrm{kg}$ glued to each other and there was another two $0.5\;\mathrm{kg}$ glued together $1\;\mathrm{m}$ apart. If we were to find the gravitational attraction force of two $0.5\;\mathrm{kg}$ masses which are $1\;\mathrm{m}$ apart and multiply it by two since we have two of them that would yield a different answer than treating the two $0.5\;\mathrm{kg}$ that are glued together as one entity. Why does it yield a different answer?
I observed this same phenomenon with Coulomb's law. (Serway & Faughn)
Suppose that $1.00\;\mathrm{g}$ of hydrogen is separated into electrons and protrons. Suppose also that the protons are placed at the Earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on the Earth? If I was to find the attraction force of one proton and electron than multiply it by Avogadro constant that would yield a different answer than saying that the charge of each particle is the elemental charge times Avogadro 's constant.
So which is the proper way to do it?
| You need to multiply by four not by two. To see why let's draw the situation:
You are assuming the situation is as shown in the top diagram. So the two $M_1$s attract each other and the two $M_2$s attract each other. Those are the forces shown by the red lines.
But you also need to include the force between $M_1$ on one side and $M_2$ on the other. Those are the green lines in the second diagram. So the total force will be:
$$ F = F_{1-1} + F_{2-2} + F_{1-2} + F_{2-1} $$
Suppose we make the mass of each of the balls $M$, so if we combine them the total mass on each side is $2M$. If we combine the masses first then calculate the force we get:
$$ F = \frac{G(2M)(2M)}{d^2} = 4\frac{GM^2}{d^2} $$
If we calculate the forces treating the masses separately we get${}^1$:
$$\begin{align}
F &= F_{1-1} + F_{2-2} + F_{1-2} + F_{2-1} \\
&= \frac{GMM}{d^2} + \frac{GMM}{d^2} + \frac{GMM}{d^2} + \frac{GMM}{d^2} \\
&= 4\frac{GM^2}{d^2}
\end{align}$$
So the forces are the same.
This also applies to the example of the electrostatic force that you mention.
${}^1$ Strictly speaking the distances $M_1 \rightarrow M_2'$ and $M_2 \rightarrow M_1'$ are slightly greater than the distance $M_1 \rightarrow M_1'$ and $M_2 \rightarrow M_2'$ but we'll assume this difference is negligibly small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How is it possible to define work for friction in several dimension? I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.
But this makes sense only if force can be written as a function of position, as it is the case with gravity, or a spring. Unlike these, friction does not depend on position only: the same body might go through a point in space at two different times and experience a different friction (the magnitude wouldn't change, but direction and sense might).
So how can it make sense to talk about work done by friction if you can't define a force-field for it in the first place?
| I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.
Friction is a force, but is not derived from a force field (in any useful sense). So your statement, while true, is inapplicable. (We were not "given a force field" in this case!) So we fall back on the more general definition that work is net energy expenditure, which in turn equals the time-integral of power applied to the system:
$$W = \int \mathbf{F}(t) \cdot \mathbf{v}(t) \, dt$$
where v is velocity and t is time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Defining the surface gravity of a black hole Consider the following two definitions of the surface gravity of a black hole (taken from page 23 of Thomas Hartman's lecture notes on Quantum Gravity):
*
*The surface gravity is the acceleration due to gravity near the horizon (which goes to infinity) times the redshift factor (which goes to zero).
*If you stand far away from the black hole holding a fishing pole, and dangle an object on your fishing line near so it hovers near the horizon, then you will measure the tension in your fishing line to be $\kappa M_{\text{object}}$.
What does it mean for the acceleration due to gravity near the horizon to go to infinity?
Why do we multiply by the redshift factor and what does it mean for the redshift factor to go to zero?
|
What does it mean for the acceleration due to gravity near the horizon to go to infinity?
It means that a person dangling near the horizon feels an almost infinite pull of gravity, and he observes that anything dropped almost immediately reaches almost the speed of light.
Why do we multiply by the redshift factor and what does it mean for the redshift factor to go to zero?
The force felt at the fishing pole should somehow depend on the force felt by the thing dangling near the horizon, right? As we noted earlier that force 'goes to infinity'. But the force on the pole actually stays about the same when the force on the dangling thing is 'going to infinity'. So to get the right force at the fishing pole, we must multiply the large force felt by the dangling thing by a small number. The number must 'go to zero' near the horizon. We call that number 'redshift factor'.
If gravity field is uniform, then we feel a constant force on our fishing pole, when we dangle an object at different altitudes in that field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Construction of first Brillouin zone For a square lattice with lattice translation vector, $T=a\hat{x}+a\hat{y}$, we find that the reciprocal lattice vectors are given by, $$A=\frac{2\pi}{a}\hat{x},$$ and, $$B=\frac{2\pi}{a}\hat{y},$$ such that the Brillouin zone ranges from $-\frac{\pi}{a}\to\frac{\pi}{a}$. That is it ranges from $-\frac{|A|}{2}\to\frac{|A|}{2}$, where $|A|$ is the magnitude of $A$, and where $|A|=|B|$.
Now consider a hexagonal lattice with receiprocal lattice vectors, $$A=2\pi\hat{x}+\frac{2\pi}{\sqrt{3}}\hat{y},$$ and, $$B=\frac{4\pi}{\sqrt{3}},$$ both with magnitude, $|A|=|B|=\frac{4\pi}{\sqrt{3}}$. Does the first Brillouin zone run from, $=\frac{2\pi}{\sqrt{3}}\to\frac{2\pi}{\sqrt{3}}$? If this is the case, can we say that the first Brillouin zone is constructed similarly to the Wigner-Seitz unit cell in that it bisects lines to closest neighbors in reciprocal space to form a closed cell?
| For the First case where you have a square lattice, the first Brillouin zone is a square that lies between $\frac{-|A|}{2} \rightarrow \frac{|A|}{2}$ in the $\hat{x}$ direction and $\frac{-|B|}{2} \rightarrow \frac{|B|}{2}$ in the $\hat{y}$ direction. Assuming that you chose a lattice point as your origin in the k-space. (Here is a reference for how to draw it https://www.doitpoms.ac.uk/tlplib/brillouin_zones/printall.php)
For the Second case choose an origin, using your reciprocal lattice vectors construct the overall lattice structure. Then you will see that A Hexagonal lattice in the position space (let's say with a side length of 1) is a hexagonal lattice in the momentum space that is oriented by $90^\circ$ (with a side length of $\frac{2 \pi}{\sqrt{3}}$. Then you construct the Brillouin zones by drawing the bisectors (Bragg Plane's) between consecutive neighbouring points.
Here's an image I have taken from ( Carvalho, Alexandre F.. “Simultaneous synthesis of diamond on graphene for electronic application.” (2015) ). Their lattice vectors are shifted by $\frac{2 \pi}{\sqrt{3}} \hat{y}$ compared to yours but it gives the same result.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Fluid Dynamics in a Syringe I am currently working on a project where we use syringes to extrude some viscous materials. I will explain what I am trying to do and I just want to know what size of tubing would be better for me. I have a 10 cc syringe connected to 3/32" tubing with luer lock and at the very end, it is attached to either a 27 gauge or 30 gauge needle. The tubing is 24 inches. When I do this, it takes a lot of pressure to push the stuff out but when the stuff comes out, its not a continuous flow. What happens is it takes a lot of force to push but at some point, i will have pushed enough with my hand to make it come out but it comes out very fast and not in a continuous flow. If I make the tubing diameter larger, would the pressure inside be smaller even if I have the same 27 or 30 gauge needle at the very end.
Would I be able to have a more smooth/continuous flow? The reason I use such small needles is because I want to be able to draw a design with precision.
| I believe you want a smaller tube. The pressure has to reach a point where it is great enough to push through the needle. I larger tube has more space to fill, so you have to push the plunger farther, to build enough pressure, to push the fluid through the small needle. Pressure will build faster in a small tube, which means you do not have to push the plunger as far into the syringe to build the necessary pressure. It will likely come out faster and sooner, but should be more regular and predictable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why are clockwise and anticlockwise moments not balanced in this equilibrium scenario?
If we take moments about the peg, which is directly below centre of mass of the cone (which is $r$ cm from its circular base) we get $3rTcos(90-ø) = Trcos(90-ø)$. $T$ = tension in string. Clearly this has no solution unless T is 0. And moments should be balanced, as the cone is in equilibrium.
Is this question wrong, or is there a force - or something else- i’m missing?
| The force $T$ acts through the peg, therefore its moment about the peg is zero. Taking components of $T$ vertically (and horizontally) is not going to alter this situation.
You have actually already balanced moments when you assumed that the CG of the cone lies vertically below the peg. The only force acting on the cone which does not necessarily pass through the peg is the weight $W$ of the cone. When this force does pass through the peg then all moments on the cone are zero.
The cone can be balanced with various lengths of string. Ensuring that the CG lies below the peg constrains the relation between the angles $\theta_1, \theta_2$ which the string at the left and right makes with the horizontal. These angles are not necessarily equal. Making them equal is not required for static equilibrium. This is a geometrical constraint.
(a) Can be solved by applying geometry. Assume right hand upper edge is distance $x$ below the peg, then equate expressions for $\tan\theta$ in the right-triangles formed by right and left sections of the string.
(b) Can be solved by balancing forces vertically.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What are quarks and leptons in Quantum Field Theory? Are the different quarks and leptons different kinds of oscillations of the same underlying quark field and lepton field, or same kinds of oscillations of different up, down and so on quark fields, and electron, electron neutrino and so on lepton fields? Or is there just a single field underlying all other fields?
| Quantum field theory in general doesn't know about quarks and leptons. It knows about bosonic and fermionic fields (e.g., scalar fields, vector fields, spinor fields) and their interactions.
So think of quantum field theory as a framework. The Standard Model of particle physics fills this framework with specific content: specific scalar, spinor and vector fields that are believed to represent physical reality.
The specific fields in the Standard Model are:
*
*doublets of left-handed charged and uncharged leptons;
*right-handed charged lepton singlets;
*doublets of left-handed up- and down-type quarks;
*right-handed up-type and down-type quark singlets;
*the $SU(2)\times U(1)$ gauge fields (four vector bosons) of the electroweak interaction;
*the $SU(3)$ gauge fields (eight massless gluons) of the strong interaction; and
*the Higgs field, responsible for electroweak symmetry breaking, the masses of three of the electroweak vector bosons, and fermion masses.
Lastly, the fermions (leptons and quarks) in the theory each have three generations.
So... a whole bunch of fields arranged in highly non-trivial ways, and the fundamental field content describes the theory before symmetry breaking. But the basic idea is that essentially yes, each different type of fundamental particle is a unit excitation of its own field, i.e., the up-quark field differs from the down-quark field, which differs from the top-quark field, the electron field differs from the muon field, and so on.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Mass-luminosity relation for a fully convective star I'm trying to show that the mass-luminosity for a fully convective star is
$$L \propto M^{\frac{113}{66}}$$
I know that the energy generation is via pp-chain and the opacity is due to $H^{-}$ ions such that:
$\epsilon = \epsilon_0 \rho T^3$ and $\kappa = \kappa_0\rho^{1/2}T^9$
I use the equation of stellar structure:
hydrostatic equilibrium: $P \propto M^2/R^4$
equation of state: $P \propto \rho^{5/3}$
energy generation: $L \propto R^3\rho \epsilon$
and there's obviously that $\rho \propto M/R^3$ and the ideal gas $P \propto \rho T$.
I can't use the equation of radiative transfer, because the star is convective but I instead have the equation of state, assuming the star is isentropic due to being fully convective. These together result in
$$L \propto M^7$$ which is not even close. Also in my calculation I never utilize the opacity relation, as it only really comes up in the radiative transport equation.
Any ideas where I'm going wrong?
| The bit you are missing is that the radius of the star is determined by the opacity.
Hydrostatic equilibrium yields
$$P_R = \frac{GM}{R^2}\int \rho\ dr,$$
where $P_R$ is the pressure at the radius where light can escape. This in turn is related to the opacity $\kappa$ by
$$\int \kappa \rho\ dr = \bar{\kappa}\int \rho\ dr = 1$$
Thus
$$P_R = \frac{GM}{R^2\bar{\kappa}} \propto \frac{GM}{R^2 \rho^a T_{\rm eff}^b}, \tag*{(1)}$$
where I have substituted in
$\bar{\kappa}\propto \rho_R^a T_{\rm eff}^b$.
With this equation you also have
$$ L = 4\pi R^2 \sigma T_{\rm eff}^4 \tag*{(2)}$$
$$ L \propto M \rho T^c \tag*{(3)}$$
From the virial theorem and perfect gas law you also know that the interior $T \propto M/R$ and interior $\rho \propto M/R^3$.
You thus have 6 variables $L, M, R, T_{\rm eff}$, $P_R$, $\rho_R$ and 3 equations. To eliminate all but $L$ and $M$ you need another.
This comes from the interior polytropic equation of state $P \propto \rho^{\gamma}$ for a fully convective star (a polytrope with $\gamma = 5/3$ for a fully convective star). However, this proportionality only applies to a particular star - the constant of proportionality in fact depends on the mass and radius of the star in question.
Here is not the place to repeat textbooks that deal with polytropes, but it can be shown that (e.g. here)
$$P\rho^{-\gamma} \propto P^{1-\gamma} T^{\gamma} \propto M^{2-\gamma} R^{3\gamma -4}$$
and thus near the surface
$$P_R^{1 -\gamma} T_{\rm eff}^{\gamma} \propto M^{2 -\gamma} R^{3\gamma-4} \tag*{(4)}$$
You then solve to get $L$ only as a function of $M$ and substitute in $\gamma=5/3$ and your favourite values for $a$, $b$ and $c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why Faddeev-Popov ghost cannot exist in external line? I was studying the path integral quantization of non-abelian gauge field. After the path integral quantization, the action becomes
$$\mathcal{L}=-\frac{1}{4}F^a_{\mu\nu}F^{a\mu\nu}-\frac{1}{2\zeta}(\partial_{\mu}A^{\mu a})^2 +\partial^\mu\bar c^a(D_{\mu}c)^a$$
Feynman rules for antighost-ghost-gauge boson vertex (e.g. $SU(2)$ gauge boson) is $-g \epsilon_{abc}p_\mu$. I can't understand why there is no process with ghost in external line.(For example, two gauge bosons annihilate and become ghost and antighost. I can certainly calculate the amplitude for this diagram according to Feynman rules. )
Certainly I know ghost is unphysical so should not exist in external line. But I want to know whether no external ghost line can result from the theory itself (like the amplitude of this process can be canceled by some other process?) or is just an axiom that we put on this theory?
| On one hand, the S-matrix does not depend on the gauge-fixing condition. On the other hand, there exist a unitary gauge, where Faddeev-Popov ghosts decouple from the theory.
References:
*
*M.D. Schwartz, QFT and the Standard Model, 2014; Section 28.4.
*C. Itzykson & J.B. Zuber, QFT, 1985; Subsection 12-5-5.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Why are stars spherical whereas (some) galaxies are disks? I read here that galaxies become disks if there is a lot of gas in them, since their angular momentum is conserved while their energy decreases due to collisions of the gas particle. I have two questions about this:
*
*Why is the disk the configuration that has low energy and high momentum? Can't there be a spherical shape that meets the same conditions?
*If stars are formed when gas collapses under by gravity, why doesn't the gas create a disk shape, but instead creates a spherical shape?
| The post you're referred to explains reasonably well why spiral galaxies are discs. So why are stars not disc like. The simple answer is: a disc-like object cannot be a star.
But what is a star? A star is an object that at its core has the right conditions for thermo-nuclear reactions, in particular temperatures $T\sim10^8K$. There are gaseous disc that reach extreme temperatures, but these are accretion discs around supermassive black holes, i.e. they are not gravitationally self-supported.
The temperature in a self-gravitating gaseous disc of comparable mass and size as a star is necessarily much lower. Moreover, any viscosity (which may be due to magneto-hydrodynamical or gravitational processes within the disc) inevitably promotes angular-momentum transport outwards and mass transport inwards, resulting in the formation of a star at the centre.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
How does a helicoper auto-rotate? The force of the air pushing on all blades would equal out. Wouldn't it? So, I was learning about helicopter and how they auto rotate and I learned that when the engine fails the helicopter starts to fall and the air is thrown up at the blades which turns them like a windmill. I though about this for a little and found a issue
This image shows that the blade is being pushed back .......
In the wrong direction
The blade would have to be moving the other way the create lift. wouldn't it?
I also found another issue. If this force was applied to every blade wouldn't the forces equal out and the blades stay at the same speed.
The video I was watching about this said that the blades and span like a windmill is span in the wind, but the way a windmill spins would push the blades the wrong way.
I tried to google windmills to see if another voo doo magic happened in them but, most results explained how a wind turbine makes electricity which was not very helpful.
I am sorry about the quality of the picture I drew
| It's exactly the same as when an aircraft glides, except the "wings" go in a circle.
As a student pilot in a fixed-wing aircraft you practice this many times.
The instructor pulls the power to idle and says "You just lost your engine..."
The first thing you do is push the nose down, so you're in a downward glide.
Since you're going downward you pick up speed and that keeps you flying.
You trim for the speed that gives you the longest glide range, and look for a place to land.
In a helicopter you do it by pitching the blades downward (with the collective lever). This makes them glide, which gives them speed, which maintains lift, so you can look for a place to land.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
What is "velocity distance" in astrophysics? I was reading some papers on astrophysics, and in several of them, I've encountered velocity being used as distance. Or more precisely, distance being in dimensions of distance over time. For example, a paper referred to a group of galaxies at "roughly $2000\:\mathrm{km\:s^{-1}}$ in distance." Another used the specific phrase "velocity distance." It said there was a super-cluster "an observed concentration of galaxies at a velocity distance of ${\sim}2500$ to $4000\:\mathrm{km\:s^{-1}}$." I searched for what this could mean for quite a bit, and couldn't find it. If anyone has any idea what this means, some insight would be greatly appreciated.
| It is used because of uncertainty in the Hubble constant. The relationship between recession velocity and distance is given by
$v=H_{0}d$
Where $H_{0}$ is the Hubble constant. Since that isn't known precisely, distances aren't known precisely, so talking about distances doesn't necessarily make sense.
Taking about the distance velocity on the other hand is using something they can be measured pretty precisely from the redshift. And you know that if object B had twice the velocity of object A then it is twice as far away even if you don't know the actual distances.
Since distances depend on the Hubble parameter then, especially back when it was only known to be in the range 50-100 km/s/Mpc there wasn't a great deal of point in reporting distances: to compare distances between two papers you'd have to convert them to velocities using whatever value of the Hubble parameter each paper used, since there is no guarantee they'd use the same value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Conformal symmetry with Goldston bosons I am looking for examples of the physical systems with both conformal symmetry and Goldston bosons. The systems can be in any physical context, like QCD phases, Condensed Matter (Graphene, etc), or SUSY gauge theories.
Naively, in 1+1D gapless system, we can more easily have conformal symmetry. But to have a spontaneous symmetry breaking with Goldston bosons, and together with a (perhaps emergent) conformal symmetry seem to be somehow more challenging.
In 2+1D and higher dimensional system, we may find harder to have conformal symmetry. Are there such examples of conformal symmetry with Goldston bosons.?
| You can't have unbroken conformal symmetry and Goldstone Bosons (GB)'s at the same time because the GB decay constant $f_\pi$ is dimensionfull in $d>2$. And in $d=2$ there are no (physical) GB's.
Even more physically, a theory of GB has a cutoff $\Lambda$ and its therefore not scale invariant.
The only way out to this is by weakening the requests, e.g. allowing for a non-linearly realized conformal invariance. In this case, conformal symmetry is broken spontaneously too, and a light dilaton appears in the spectrum alongside with the GB's from breaking spontaneously a global internal continuous symmetry.
The leading terms in the effective action (ind $d=4$) for a dilaton $\sigma$ and ordinary GBs $\pi$ would take the form
$$
\mathcal{S}[\sigma,\pi]=\int d^4x \frac{f^2_\sigma}{2}(\partial_\mu e^{\sigma})^2+ \frac{f_\pi^2}{2} e^{2\sigma}(\partial_\mu\pi)+\ldots
$$
which is scale invariant under $x\rightarrow x e^{\alpha}$ if the dilaton transforms non-linearly
$$
\sigma(x)\rightarrow \sigma(x e^{\alpha})+ \alpha\,,\qquad \pi(x)\rightarrow \pi(x e^{\alpha})\,.
$$
The best way to see how this invariance is realized is by actually making a field redefinition $$\chi\equiv f_\sigma e^{\sigma/f_\sigma}$$ so that the action
becomes
$$
S=\int d^4x \frac{1}{2}(\partial_\mu\chi)^2+\frac{1}{2}\left(\frac{f_\pi}{f_\sigma}\right)^2\chi^2(\partial_\mu\pi)^2
$$
which contains only ratio of scales $f_\pi/f_\sigma$ and it's hence scale invariant. The physical scales are recovered because the new $\chi$ variables has non-vanishing vev that breaks spontaneously conformal invariance
$$
\langle\chi\rangle= f_\sigma\,.
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is my reasoning to derive the bistatic radar equation wrong? The ideal effective antenna aperture is calculated from $A_{eff}=G\lambda^2/4\pi$, G being the antenna gain.
Why is it that in a bistatic radar setup with a transmitter and receiver fixed on the Earth and a moving aircraft, the bistatic radar equation only contains the transmitter wavelength, and nowhere the two Doppler-shifted ones?
If I wanted to determine the received power, I would consider the transmitter's wavelength $\lambda_0$ to calculate its $A_{eff, 0}$, the Doppler-shifted wavelength $\lambda_1$ on the reflecting aircraft as an intermediate receiver and transmitter antenna to calculate its $A_{eff, 1}$, and finally the Doppler-shifted wavelength $\lambda_2$ at the receiver antenna on the ground to calculate its $A_{eff,2}$.
What is wrong with this reasoning?
| Your reasoning is logically right even for monostatic radar when detecting a moving target. As for your question, you need to note that the radar equation is merely used to calculate the power received from the target to the receiver antenna, in order to calculate the maximum detection range of the radar given its minimum acceptable SNR: (The $A_{eff}$ formula is incorporated in this equation)
$$P_r = {{P_t G_t G_r \sigma \lambda^2}\over{{(4\pi)}^3 R_t^2R_r^2}}$$
you can see that power drops with $1/R^4$ and signals at the receiver are usually exceedingly attenuated.
Now the doppler frequency of the reflected signal is:
$$f_D \approx \frac{2V}{c}f_0\rightarrow \lambda'=\frac{c}{f_0+f_D}\approx \lambda_0(1-\frac{2V}{c})$$
The maximum possible speed that you can think of when designing a radar is much lower than 20 Mach (around 7 km/s) even for ballistic missile detection. A speed of 15 km/s gives $\dfrac{2V}{c}=10^{-4}$
Thus, insterting this modified $\lambda$ in the radar equation doesn't affect anything at all and is always neglected compared to other losses and sources of noise in the SNR equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Degrees Of Freedom of Spring-mass system
Consider 2 masses $M_1$ and $M_2$ connected with a spring of stiffness $k$, resting on a smooth frictionless surface. Now, each mass has its own 1 DOF along the $x$-axis. And the system has 1 constraint , i.e. the spring.
So, in all there should be 2(1)-1= 1 DOF for the system. But I've read that it has 2 actually. So where am I going wrong?
| Imagine a system of your two masses $m$ and one spring, spring constant $\kappa$ but with two other spring, spring constant $k$, attached to the masses as shown in the diagram below.
This system certainly has two degrees of freedom.
You have two masses and a displacement along a straight line for each of the masses with the spring providing the interaction between the masses.
It can be shown that there are two normal modes for this system and the frequencies of these modes are $\omega_1 = \sqrt{\dfrac{k+2\kappa}{m}}$ and $\omega_2 = \sqrt{\dfrac{k}{m}}$
If $k \rightarrow 0$, which is equivalent to not having the two outer springs there, then:
*
*$\omega_1 \rightarrow \sqrt{\dfrac{2\kappa}{m}}$ which is the motion
about the centre of mass of the system
*$\omega_2 \rightarrow 0$ which is the masses being displaced and/or
given a velocity which would then result in a motion of the centre of
mass of the system with no restoring force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is gravity non-negligible compared to the electromagnetic force? Consider two electrons approaching each other at rather fast speeds, maybe even coming close to colliding. Does gravity play any role in this event? If so, how much influence does it have? Do we need gravity in charged particles?
| For electrons, the gravitational force is much smaller than the electric forces. To get an idea about how small, let's math.
$$F_g = \frac{Gm_1m_2}{r^2}\, , \, F_e = \frac{kq_1q_2}{r^2}\\
\frac{F_g}{F_e} = \frac{Gm_1m_2}{kq_1q_2}$$
Notice, the distance separating the electrons does not matter when comparing the forces. All of these constants are pretty easy to find, so the gravitational force is $2.4\times 10^{-43}$ the size of the electric force. Since we know nothing in the universe to within two parts in $10^{43}$, we may safely ignore gravity.
You mentioned that the electrons were on a collision course. Moving charges, especially fast moving charges, create magnetic fields, and those would be important in your calculations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Wiedemann-Franz law The Wiedemann-Franz law states that the ratio of thermal conductivity $\kappa$ and electrical conductivity $\sigma$ for metals fairly accurately obeys $\kappa/\sigma = LT$, where $T$ is the temperature and $L$ is the Lorenz number, whose value is in the order of $2 \cdot 10^{-8}$ in SI units.
Assuming that $\kappa$ refers only to the electronic contribution to heat capacity, I understand how one can derive this law, or at least justify it, in a free electron model. In the classical Drude model, one then finds $L = 3k_B^2/(2e^2)$, and in a quantum mechanical free electron treatment, one finds $L = \pi^2k_B^2/(3e^2)$.
However, what about the lattice contribution to thermal conductivity? I understand that the electronic contribution might be dominant for most metals. The example of diamond (which is not a metal and hence the Wiedemann-Franz law is not expected to hold), however, shows that the lattice contribution can have the same order of magnitude as the electronic contribution, since diamond has comparable heat conductivity metals at room temperature. One important reason for the high thermal conductivity of diamond is the scarcity of defects. Does that mean that if we could produce a sample of metal with very few defects, the lattice contribution would be important? Is it important even for metals with defects?
| That's a pretty neat question, I think. The restriction that it only should be applied for metals means that the dominant thermal carriers are going to be electrons, and since the thermal and charge carriers are the same particle, there is some simple ratio of the two. I suppose there is unease about the use of "Law" despite it really being an approximation, true really only for the homogeneous electron gas. Ohm's Law has a similar problem, and is much more clearly broken.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Why is $R\cos{a} = mg$ in circular motion compared and not $R = mg\cos{a}$?
Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).
However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?
To show what I mean:
If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$
Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.
*In the circular motion questions I've seen in my mechanics module
| Never, ever, just blindly memorize formulae.
What you need to do is draw a free-body diagram of your particle, which will have an angled normal force, and a downward gravitational force, and you know that the net acceleration is inward with magnitude $v^{2}/r$. You can either rotate your reference frame so that the normal force is upward, and the gravitational force is angled, or work out the two equations, eliminating the normal force.
Either way, you'll arrive at an answer. But the text of the question presupposes that you can just memorize a formula for a situation. Never do this, look at a situation, and work out the answer. You will end up wrong as often as you don't if you try and solve problems the way you seem to be -- because all it takes to be wrong is someone labeling an angle in a funny way, or using a slightly different convention.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does an increase in entropy always result in an increase in heat, or can there be increased entropy without an increase in heat? Most situations I can think of where entropy increases also results in an increase in heat, but just wondering if that is a rule. Are there any cases where heat does not increase with entropy?
| *
*If you have a room with oxygen in one half and nitrogen in the other half, and you now remove the wall between them, then they start mixing more and more. Entropy increases. Before, you knew exactly which atoms you pointed at, when you pointed somewhere, but the more time that passes, the less sure you are of which is where. Chaos increases. Entropy increases. No heat is exchanged.
No, heat is not a necessity for entropy change. Entropy describes any process. Entropy can certainly increase in isothermal processes as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
At what rate do I need to accelerate for my car to become a fusion reactor? I was sitting on the bus on my way home, and this popped into my mind.
As a car accelerates, the air pressure at the back of the car increases as air is pushed back due to inertia. This pressure will be constant, assuming a constant rate of acceleration and that the acceleration lasts long enough for an equilibrium to be reached.
The question is therefore: How high must the acceleration rate be for the air near the back window to undergo nuclear fusion?
Some further assumptions:
*
*The air is 100 % nitrogen (or hydrogen, if that makes the approximation simpler)
*When the car is not accelerating, the pressure is $10^5\ \mathrm{Pa}$ and the temperature is $273.15\ \mathrm{K}$ (STP).
*The car is a square prism of dimensions $1 \mathrm{m}\times 1 \mathrm{m} \times 3 \mathrm{m}$ with the direction of travel perpendicular to the square faces. $3 \mathrm{m^3}$ is approximately the EPA definition of the interior volume of a mid-sized car. There is nothing else (seats etc.) in the car.
*The car is an isolated system, so neither matter nor energy can enter or leave
*The window cannot melt
I anticipate that this might lead to some minor problems with relativity...
| For fusion one needs order of MeV energies in the center of mass system of the fusing nuclei.
As all molecules are accelerated at the same time and direction, it is only fusion with the molecules accumulating on the glass back window that has meaning i.e. the question should be at what car acceleration the atoms of air reach MeV energies with respect to the back window glass. There, stopped hydrogen atoms may be hit by the accelerating ones and if the center of mass energy of the two hydrogens is of the order of hydrogen fusion, it will happen.
Of course the glass will have melted long before. Look at the temperatures for hydrogen fusion to happen at this link , paragraph :"fusion a closer look", about 10million Kelvin.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Tension direction for pulleys in connected mass systems I got caught up with a conceptual question dealing with a practice problem with connected mass mechanics.
Looking at the solution lecture notes is the following image for the Free Body diagrams for each mass, and the pulley, respectively:
Obviously, the Free Body diagrams on mass 1 and 2 are pretty straight-forward, but the situation with the pulley confuses me. $T_2$ going down is fine. The mass is falling, the system is overall causing $M_2$ to fall down, the string applies a force that way, sure. But $T_1$ confuses me. The mass is moving towards the pulley once we let the system act, so why is there a tension being applied that way? The directions for $T_1$ for mass 1 and the pulley are opposite for direction, and I'm not really sure why. Overall, the pulley is rotating clockwise, so this has to be the case -- $M_1$ will move towards the pulley, $M_2$ will drop. So why is $T_1$ facing that way? Intuitively speaking, that is.
| Intuitively speaking, the mass $m_1$ is decreasing the speed. So it must reduce the speed of the system by exerting a force on the pulley that opposes that direction of movement. So $T1$ must be leftwards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Can a current be induced in this coil-magnet configuration? Coil is moving around ring magnet made of two arch shaped magnets with poles opposing each other like so:
Would a current flow in coil? or be canceled out?
| With the coil in open circuit, you will have a voltage difference at its terminals. Let's assume the coil moves cw or that the magnet moves ccw. Seen from front, let's call the bottom blue N pole and the top blue end S pole. The lines of flux are all concentrated in the iron except at the gaps where they extend in the air to jump from one magnet end to the other. Some of those flux lines will be "swiped" by the coil winding. Using $V=lv \times B$ and looking at the sketch it can be seen that Term 2 will be the positive end of the coil and if a load is placed, current would leave the coil form terminal 2 (conventional current flow). Hope my drawing help clarify the idea.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the energy of a single charge system? I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones.
There are two ways to express the energy in a capacitor:
*
*By Voltage : $U = 1/2 CV^2 $
*And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $
Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.
The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!
There is obviously something that i miss, but what?
| Potential energy is always defined for multi particle systems but can be defined for a single particle(in this case it is charge and hence electrostatic potential energy) if we assume the other charge to be present at infinity(a point far away from the vicinity of the charge will do) then you can calculate the potential energy for that system by taking the potential at infinity to be zero which is theoretically the potential energy of the system but can be labeled as the potential energy of the single particle.So what you actually did when you placed the charge at 'A' was,you assumed another particle to be present at infinity and wrote down the potential energy of that system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Why can't electrostatic forces "pierce" through thin obstacles the way magnetic forces do? We know that two magnets attract or repel each other even when we keep a very thin obstacle between them such as paper. Why doesn't the same hold good for two charged objects?
| The implied assertion that magnetostatic fields can penetrate thin layers better than electrostatic ones does have some truth to it and the reason is that there is no free magnetic charge, whereas matter is electrically charged. If a material is conductive, electric charge migrates until there is no force on it, or until all forces are balanced. That is, charges arrange themselves to annul the electric field inside the body of a conductor, otherwise the conduction charge inside would keep moving under the influence of the electric field until there were no electric field there. Electric field lines enter conductors normal to the conductor's surface for the same reason - charges rearrange themselves until the forces on them are normal to the interface between the conductor and the outside World, otherwise the charges keep moving along the conductor under the influence of any tangential electric field until the latter is annulled.
Therefore, many materials exclude electric fields. Magnetically shielding materials also exist, but these work by a wholly different principle. Hollow mu-metal shields (such as those used around cathode ray oscilloscopes) exclude magnetic fields from the hollow by providing a low magnetic reluctance path along the shield and around the hollow. This happens simply because the mu metal has an enormous relative permeability - typically approaching $10^5$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Are there experiments that deal with the headlight effect? If I am not mistaken, the Ives-Stilwell experiment measures the frequency of emitted light to test the relativistic doppler formula.
Is there anything like this for light intensity? Have we measured the headlight effect?
Anything like precision tests that are related to this would be interesting.
| In some respects this is unsurprising, as you have to use relativistic engineering to get a synchrotron to work right in the first place: but when you do get a synchrotron running, you will observe that there is a characteristic "synchrotron radiation" and indeed one of its properties is that it's 'strongly collimated' -- it points in one direction like a laser, not in many directions like a light bulb. And, the direction that it points is tangent to the circle in the direction that the charged particle is travelling, just as the headlight effect says it must be.
So in fact this now goes way beyond theory and is now a part of practical engineering, as a part of available laboratory and medical devices. So-called synchrotron light sources are not just testing this prediction and finding it resoundingly true, but indeed we're way beyond that and using it to build highly-useful devices for crystallography and medical imaging and the like. X-ray tubes using Bremsstrahlung ("braking radiation", you fire an electron gun into a big spinning plate and the electrons emit X-rays in all directions as they come to rest relative to that plate) still exist and are much cheaper sources for lower-energy X-rays, but when you want these highly-tunable, coherent, collimated sources these more-expensive devices are the state of the art.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Questioning validity of SHM of vertical mass-spring system Horizontal mass spring system is good but vertical mass spring system confuses me.
Q1. Can there be two restoring forces in an SHM?
Q2. If no, then weight of mass seems to disturb SHM as down extreme position below the mean position would be much farther than extreme position above mean position. If vertical mass spring system executes SHM, please elaborate how?
| Gravity is just a constant force, so all it does is just shift the spring force linearly. This means all that happens is that everything gets shifted downward and that's it.
With a horizontal spring, we have $F = -kx$ as usual.
With a vertical spring, there are now two forces: the spring restoring force $F_{1} = -kx$, and gravity $F_{2} = mg$. You find the total force by adding both of the forces, and then you get $F = -kx + mg$.
Now note that this is the same as the horizontal case but with a constant added as a force. What this means is that we can take
$$F = -kx + mg = -k\left( x - \left( \tfrac{m}{k} \right)g \right)$$
by factoring. This equation is the same one as the one for the horizontal spring, but $x$ is just shifted by $x_{0} = \tfrac{mg}{k} $.
Can there be two restoring forces in an SHM?
Yes, there can be two restoring forces. Anytime you have two forces, you just add them together. If you add two restoring forces, you will still get SHM!
However, gravity is not a restoring force. It is just a constant downward force. Nonetheless, you still do what I did above and just add all of your forces together.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Newton's rings - white light? I am familiar with the concept of optical path, constructive and destructive interference. The basic premise to discuss these concepts is coherence, which is why I am perplexed by the phenomenon of Newton's rings.
From what I understand, this experiment was first conducted by Robert Hooke, with non-coherent white light (from a candle I think?).
If the waves that enter the lens are of arbitrary phase, and of all possible wavelengths, why should there be a diffraction pattern?
| Two salient reasons:
*
*The interference is between the reflexions from two neighboring surfaces;
*The distance between these two surfaces is small - only a few wavelengths of visible light.
Point 1 means that it is only the phase difference that is important in determining the throughgoing / reflected light for a monochromatic wave. So, all monochromatic waves, whatever their random phase, produce the same interference pattern.
But we are dealing with all wavelengths. So let's look at point 2.
Near the center of the pattern, the distance between the two interfering surfaces can be approximated by a quadratic dependence on the distance $r$ from the center of the pattern. Therefore, the throughgoing intensity as a function of radius for light of wavelength $\lambda$ is proportional to:
$$\frac{1}{2}\left|1-\exp\left(i\,\frac{\pi\,r^2}{R\,\lambda}\right)\right|^2 = \sin\left(\frac{\pi\,r^2}{2\,R\,\lambda}\right)^2$$
where $R$ is the radius of curvature of the curved surface, if the interference pattern is formed by bringing a convex lens into contact with a glass flat, for example. Therefore the nulls in the pattern happen at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$.
Now we add the effect of all wavelengths. We can only see a narrow band of wavelengths, so we are looking at the sum of the interference patterns with nulls at radiusses $0,\,\sqrt{4\,R\,\lambda},\,\sqrt{8\,R\,\lambda},\,\cdots$ for $\lambda$ varying between $400{\rm nm}$ and $750{\rm nm}$. This means that the first null happens at a range of radiusses that varies only over a range of about $\pm 20\%$ - the "smear"width is well less than the distance between the first and second null. So, even with the spread over visible wavelengths, the first nulls line up pretty well. The second null less well and so forth. You see a series of colored nulls - the coloring is because different wavelengths have their nulls at different positions, but the nulls are still well enough aligned to see their structure. As you move further from the center, the nulls become more tightly packed and the accuracy of the alignment for all visible wavelengths becomes coarser than the null spacing, which means that we can no longer see fringes. This is exactly what happens in Newton's rings - the fringe visibility fades swiftly with increasing distance from the center.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Does light travel through a waveguide faster than electricity through a wire So I have heard that using photons to compute is faster than electrons. I was wondering why exactly that is? Is it because light travels through a waveguide faster than electricity through a wire?
| Yes, a photon traveling inside a waveguide is a lot faster than an electron inside a wire. A photon propagates in the transverse direction to the dimension of the waveguide. Electrons face resistance inside of a wire. This resistance causes Joule heating effect to take place. This decreases the rate of transmission of energy. But in the case of photons and waveguides, there is very slight energy loss due to penetration of electromagnetic wave in the conducting walls (remember skin depth!).
This makes the waveguide mechanism to be superior to the wire and electron mode of data transmission.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Does the deflection of a photon passing near a star/planet depend on its frequency/energy? Eddington already proved that starlight passing our Sun is deflected more according to GR compared to the outcome of Newton. But does this rate of deflection also depend on the frequency the photons has?
A photon doesn't have a rest mass but his energy could perhaps be considered as equivalent to mass. So the higher the energy of the photon the larger the gravity works? Should the answer be the same according to Newtons'law and GR?
| It does not depend on the frequency of the photon. Formally, the deflection angle is a function of the mass of the body deflecting the light (the Sun in your example):
$$
\theta = \frac{4GM}{rc^2}
$$
Intuitively, even in Newtonian gravity how much a particle is deflected does not depend on its mass, the reason being the fact that gravitational mass is the same as inertial mass, and so $m a = mGM/r^2$ is the same as $a = GM/r^2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
EM duality transformation I have read that by defining a dual transformation as
$$
\begin{pmatrix} F'^{\mu \nu} \\ ^*F'^{\mu \nu} \end{pmatrix}
=
\begin{pmatrix}
\cos (\alpha ) & \sin (\alpha ) \\ -\sin (\alpha ) & \cos (\alpha )
\end{pmatrix}
\begin{pmatrix} F^{\mu \nu} \\ ^*F^{\mu \nu} \end{pmatrix}
$$
where
$$
F^{\mu \nu} = \begin{pmatrix}
0 & -E^x & -E^y & -E^z \\
E^x & 0 & - B^z & B^y \\
E^y & B^z & 0 & - B^x \\
E^z & - B^y & B^x & 0
\end{pmatrix}
\quad \text{and} \quad
^*F^{\mu \nu} = \begin{pmatrix}
0 & -B^x & -B^y & -B^z \\
B^x & 0 & E^z & -E^y \\
B^y & -E^z & 0 & E^x \\
B^z & E^y & -E^x & 0
\end{pmatrix},
$$
it is equivalent to say that $E$ and $B$ transform as
$$
\begin{pmatrix} E' \\ B' \end{pmatrix}
=
\begin{pmatrix}
\cos (\alpha ) & \sin (\alpha ) \\ -\sin (\alpha ) & \cos (\alpha )
\end{pmatrix}
\begin{pmatrix} E \\ B \end{pmatrix}.
$$
But how can I find it from the Faraday tensor? (replacing with $\mu$ and $\nu$ is a bad idea apparently)
| This follows immediately from the first equation that you wrote. Just fix a $\mu \nu$ and obtain that the corresponding component of the Electric/Magnetic field is given by what you wrote in the last equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the real life utility dot product and cross product of vectors? Today, my teacher asked us what is the real life utility of the dot product and cross product of vectors. Many of us said that one gives a scalar product, and one gives a vector product. But he said that, that was not the real life utility of the dot and cross product. He asked us, "Students, why do we have to learn these two concepts? Because they are given in the book? Or because they are important concepts of vectors, or refer do they have some actual utility."
None of us could answer. Does anyone here know a proper answer to an improper question like this?
N.B.: Please pardon me if it's a foolish question, and don't give me credit if it opens up a huge set of answers.
| A few roughly mentioned by our teacher:
1-The cross product could help you identify the path which would result in the most damage if a bird hits the aeroplane through it.
The dot product could give you the interference of sound waves produced by the revving of engine on the journey.
2-solar panels need to be installed carefully depending upon angle of tilt of roof so that maximum electrical power is produced. it requires working out on direction and elevation of sun from roof, tilt angle followed by the product of these vectors.
3-We can assign a vector to each streamline of water moving in different directions under pressure from a faucet; and determine how much water is being lost by taking dot product of those vectors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
One Particle State in Interacting QFT (Eqs. 4.88 4.89 in Peskin & Schroeder) How to derive equation 4.88 in section 4.6, page 108, of Peskin & Schroeder?
$$\left|k_{1}k_{2}\right\rangle\propto\lim_{T\rightarrow+\infty(1-i\epsilon)}e^{-iHT}\left|k_{1}k_{2}\right\rangle_{0}.\tag{4.88}$$
How to derive equation 4.89?
$$\lim_{T\rightarrow+\infty(1-i\epsilon)} {}_{0}\!\left\langle k_{1}k_{2}\right|e^{-iH(2T)}\left|p_{1}p_{2}\right\rangle_{0}$$ $$\propto\lim_{T\rightarrow+\infty(1-i\epsilon)}{}_{0}\!\left\langle k_{1}k_{2}\right|T\left(\exp\left[-i\int_{-T}^{+T}dt\, H_{I}(t)\right]\right)\left|p_{1}p_{2}\right\rangle_{0}.\tag{4.89}$$
| I would like to hazard a guess here. From equations (4.23) and (4.25) in the book, $$U(T,-T)=T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}=e^{iH_{0}(T-t_{0})}e^{-iH(2T)}e^{iH_{0}(T+t_{0})},$$ where $t_0$ is some reference time at which the Schroedinger and interaction pictures coincide. We then get $${}_{0}\langle k_1k_2\rvert e^{-iH(2T)}\lvert p_1p_2\rangle_0={}_{0}\langle k_1k_2\rvert e^{-iH_{0}(T-t_{0})}T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}e^{-iH_{0}(T+t_{0})}\lvert p_1p_2\rangle_0 \\ =e^{-iE_f(T-t_{0})}e^{-iE_i(T+t_{0})}{}_{0}\langle k_1k_2\rvert T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}\lvert p_1p_2\rangle_0,$$ where $E_i$ and $E_f$ are the energies of the initial and final states.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix representation of the Clifford group The Clifford group $C_n$ on $n$ qubits is defined as
$$C_n = \left\{ U \in U(2^n) \mid \sigma \in P_n \rightarrow U\sigma U^\dagger \in P_n \right\}/U(1),$$
where
$$P_n = \left\{ \sigma_1 \otimes \dots \otimes \sigma_n \mid \sigma_i \in \{I,X,Y,Z\} \right\},$$
and $X,Y,Z$ are the Pauli matrices. Also $C_n$ may be generated as
$$<H_i,P_i,CNOT_{ij}>/U(1).$$
How exactly could one produce a matrix representation of $C_n$ for use in a program? I understand the abstract representation, however it seems quite difficult to obtain an explicit list of the elements.
| There are algorithms to generate all elements of the Clifford group for a specified number of qubits. One implementation is available here: http://www.cgranade.com/python-quaec/ (see the qecc.clifford_group(nq, consider_phases=False) iterator). The code implementing this can be seen here. It basically generates all possible mappings and then filters out those that do not fulfill the commutation and anticommutation relations. The section "Number of Elements" of this goes in a bit more depth.
There are also algorithms that can give you a random Clifford operator without having to generate the entire group (which makes it exponentially more efficient if all you need is just a random sample of the Clifford group). See https://arxiv.org/pdf/1406.2170.pdf.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where does the force actual act?
Why are the two mass of $m_1$ and $m_2$ not multiplied by minus one?
I know that two minus multiplied gives you plus by but I mean the two masses are attracting so they should have a sign like so $$F_g=\frac{G(-m_1) \times (-m_2)}{r^2}$$
I ask these because equation would actual give the right explanation as to what the equation is actual doing, meaning that the two mass are attracting.
|
Actually the picture you've attached is a little sketchy about which force is acting on what. If I denote the force on $2$ due to $1$ as $F_{21}$ then the force on two would be given as in the picture. What the negative sign means is that the force on $2$ due to $1$ acts towards the direction of -$\vec{r}$ (as the force is always attractive). The vector expression for the force $F_{21}$ includes the minus sign primarily because the direction of the force on $2$ due to $1$ is in a sense opposite to the way $\vec{r}$ is defined. You might need to visit vectors for more clarity (if you haven't already).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Why are Grassmann fields never classical? I see this statement in many QFT books (e.g. Altland & Simons' Condensed Matter Field Theory) but the author never explains why.
Can you briefly explain why Grassmann fields never have a classical meaning (preferably physical arguments) and possibly point out some good references?
| Let me put things in broader perspective. Bosonic fields are quantized in terms of commutators with a prefactor $\hbar$. Classical limit leads to commuting variables that may be represented by complex numbers. This is by the way the first step to devise numerical applications of path integrals. Fermionic fields require anticommutators ( consider 4d to focus and avoid special features of low dimension ) and formal $\hbar\to 0$ gives Grassmann algebra, say, $\{\theta_i, \theta_j\}=0$ and these variables cannot be represented as complex numbers ( just take $i=j$) in any simple way.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Calculating energy released in nuclear fission Consider the neutron induced fission $\text{U-235} + n \to \dots \to \text{La-139} + \text{Mo-95} + 2n$, where $\dots$ denotes intermediate decay steps.
I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):
$$
\Delta E = B(139,57) + B(95,42) - B(235,92) \approx 202,3 \, \mathrm{MeV}
$$
(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).
Another way is:
$$
\Delta E = (m(\text{U-235}) + m_{\text{Neutron}} - m(\text{Mo-95}) - m(\text{La-139}) - 2m_{\text{Neutron}} )c^2 \approx 211,3 \, \mathrm{MeV}
$$
Which one gives the correct result? Why?
You notice that $57+42 \neq 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?
A slightly other point of view: What different questions do both calculations answer?
| The released energy is the difference in energy of intitial/end products. For a nucleus, the energy is given by its mass, which in turn can be calculated as the difference of "naive mass", i.e. all single constituents' masses summed up, and the binding energy. This is the calculation you'll have to do. Obviously, all these calculations neglect kinetic energy terms. Edit: this link might clear things up.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model? How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model?
As per my textbook, we can account for as to why angular momentum is quantised. But I fail to understand what would the wave diagram for n=1 look like?? For example, for higher orbits.
As to my knowledge, such a diagram should not be possible for first orbit. Then how come we can account for quantisation for the first orbit?
For example, for higher orbits, the following diagrams follow...
| The electron wave for the $1^{st}$ orbit would be like a circle (not exactly a circle) moving up and down the orbit.
$mvr=\frac{h}{2\pi}$
If the motion of the wave was broken in 3 stages, it would look something like this :
Stage 1 :
Stage 2 :
Stage 3 :
Why is this possible :
Bohr's model assumes that the electron wave is continuous over the stationary orbits circumference, i.e,
$n\lambda = 2\pi r$,
But according to de-Broglie, $\lambda=\frac{h}{mv}$,
$n\frac{h}{mv}=2\pi r$
Hence, $mvr=\frac{nh}{2\pi}$.
If you observe, in a electron wave over (say) $2_{nd}$ orbit there are 2 crests and 2 troughs. Similarly there are 3 sets of crests and troughs for the $3_{rd}$ orbit.
So its very much possible that in the $1_{st}$ orbit there is only 1 set of crest and trough.
Here's a link to a simulation for different models of Hydrogen:
https://phet.colorado.edu/en/simulation/legacy/hydrogen-atom
You will find this simulation very useful.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is Fresnel diffraction really visible if slit width is similar to distance from screen or does diffraction disappear in that case? Diffraction phenomena occur when, named $a$ the amplitude of a slit and $\lambda$ the wavelenght, $a \sim \lambda$ or even $a<\lambda$ (full illuminated screen).
Therefore if $a\gg\lambda$ diffraction phenomena are not visible.
On the other hand there are two types of diffraction: Frahunhofer diffraction and Fresnel diffraction.
In the first case the assumption is that, named $L$ the distance between the slit and the screen and $L'$ the distance between the source and the slit, $L'\gg a$ and $L\gg a$ (the wavefronts are planes).
Fresnel is the opposite case, in particular $L\sim a$. But initially I stated that diffraction phenomena are not visible if $a\gg \lambda$. Therefore, assuming (reasonably) that $L\gg \lambda$ how can Fresnel diffraction phenomena occur at all.
In other words, If I have $L \sim a$ or even $a>L$ but still $L\gg \lambda$ do I see diffraction (Fresnel diffraction) or do I just see one light spot (no diffraction)?
| You are taking
there are two types of diffraction: Frahunofer diffraction and Fresnel diffraction
too literally. There is only one kind of diffraction physics: waves interfering with each other. The Fresnel and Frahunofer prescriptions are two approximation regimes for making the problem of computing results in closed form tractable.
You see the development of multiple approximation regimes in many (most? all?) fields where the completely general problem is infeasibly difficult (or even just annoyingly hard, actually). The near- and far-field approximations to electromagnetic radiation are close parallels to this case; but also Newtonian versus ultra-relativistic kinematics; the many, many approximation regimes of fluid dynamics; and so on.
It is often the case that this kind of approach ends with a intermediate regime where none of the approximations are valid that is poorly studied by comparison. I you have the right combination of mathematical skill and intuition there are papers to be had in those transition regimes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove the constant speed of light using Lorentz transform? I read the light-clock example in my book which proved the time dilation formula by assuming that the speed of light is constant for all observers. But I've trouble in understanding it the other way around. Lorentz transformation is just a correction to Newtonian mechanics to account for the constant speed of light for all observers, right? I have trouble understanding how does applying this correction preserve the speed of light for all observers.
Can we start by assuming that the Lorentz transformation formulas are true and then prove that two observers $A$ and $B$ will see a light pulse moving at the same speed $c$ regardless of their relative velocity with respect to each other?
| From $ds^2=c^2dt^2-dx^2$, we see that light-speed travel is equivalent to $ds^2=0$. But $ds^2=\eta_{\mu\nu}x^\mu x^\nu$ is manifestly Lorentz-invariant, so if $ds^2=0$ holds in some reference frame it also does in others obtained by arbitrary Lorentz transformations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Calculate kinetic energy of neutrons in nuclear fission How can I calculate the mean kinetic energy of an emitted neutron in a nuclear fission. Take for example the fission of U-235 to Ba-141 and Kr-92.
A calculation which just shows that the mean energy will be in the range of "fast" neutrons (> 1MeV) would be enough for me.
| I still won't plug the numbers in for you but I'll explain the principle.
You start off with your $^{235}$U and all the energy that is wrapped up in there will be our total (we can see how much we have to play with from the mass excess). Then, through fission, you get your $^{141}$Ba and $^{92}$Kr and two neutrons
$^\dagger$ (this depends on whether you have spontaneous fission or neutron induced fission but you can adapt this as fits your situation). Now check how much energy you have left to play with from your original uranium using the basic rule:
$$ Q = \Delta(A) + \Delta(B) - \Delta(C) -\Delta(D) $$
Where your $\Delta(A)$ is the mass excess of some particle $A$ in the reaction $A+B \rightarrow C + D$. Your energy $Q$ is left to be distributed among your remaining products. This energy is then distributed as a Maxwell-Boltzmann PDF, the neutrons(being lightest) could take the majority of that energy and conserve momentum (since momentum depends on $v$ and kinetic energy on $v^{2}$) so you can approximate the maximum kinetic energy of the neutron as the $Q$ value.
$\dagger$ $235-92-141=2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If atoms were held by gravitational (instead of electrical) forces
"If atoms were held together by gravitational (instead of electrical) forces, a single hydrogen atom would be much larger than the known universe."
- from Grifffith's Introduction to Electrodynamics
I don't understand this. If the atoms were held together by gravitational forces, then the atoms will have to be extremely close together, but why extremely big?
| To find the radius of the orbit, we first need to consider the centripetal force on the electron ($\frac{m_ev^2}{r}$) which is equal to the gravitational force between the proton and the electron in this case. ($F_g=\frac{Gm_em_p}{r^2}$)
Using this, and the fact that the angular momentum of an orbit is quantised, (i.e. $m_evr=\frac{nh}{2\pi}$),
the radius ($r$) of the orbit turns out to be: $$\frac{n^2h^2}{4\pi^2Gm_pm_e^2}$$
For $n=1$ (i.e. the principal orbit), this has a value of $1.2\times10^{29}m$, which is ridiculously large, as Griffith states.
A more intuitive way to understand this would to notice that the force needed to keep an electron in orbit is inversely proportional to $r$. The force of gravitation between an electron and a proton is practically negligible ($10^{42}$ times lesser) when compared to the electromagnetic force between them. This is why $r$ has to be very large (compared to the electromagnetic case), as this reduces the amount of force needed by the electron to be kept in orbit, i.e. achieving balance in some way.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Why do different letters sound different? If one sings the letter "A" and "M" at the same volume and pitch, the two letters are still differentiable. If both pitch and volume are the same however, shouldn't the sound be the exact same?
| You don't sing a single pitch - you sing a frequency and its harmonics. Using a simple spectrum analyzer, this is me "singing" the letter A and M, alternately (AMAMA, actually):
The letter "A" is the one with more harmonics (brighter lines at higher frequencies), the letter "M" seems to have a bigger second harmonic. The frequency scale is not calibrated correctly (cheap iPhone app...)
Here are two other shots, side by side (M, then A). You can see that the 2nd harmonic of the M is bigger than the first; by contrast, the higher harmonics from the A are dropping off more slowly:
Simple vowels have this in common: the shape of your mouth changes the relative intensity of harmonics, and your ear is good at picking that up. Incidentally, this is the reason that it is sometimes hard to understand what a soprano is singing - at the top of her range, the frequencies that help you differentiate the different vowels might be "out of range" for your ears.
For short ("plosive") consonants (P, T, B, K etc), the story is a bit more complicated, as the frequency content changes during the sounding of the letter. But then it's hard to "sing" the letter P... you could sing "peeeee", but then it's the "E" that carries the pitch.
The app I used for this is SignalSpy - I am not affiliated with it in any way.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 0
} |
How do we prove analyticity of Schwinger functions? Starting from Wightman axioms, we can define the Schwinger functions as the Wick-rotated Wightman functions (as for instance is explained in the book by R. Haag, Local Quantum Physics).
The Schwinger functions have a set of properties, which essentially come from the axioms of the original, lorentzian theory. In particular, Schwinger functions are analytic away from coincident points, as claimed in the aforementioned book.
How can we prove this claim? The book by Haag doesn't seem to explain this.
| If your starting point is a Wightman theory then you can find a proof in Section II.3 of the book "The $P(\Phi)_2$ Euclidean (Quantum) Field Theory" by Simon. If your starting point is a given set of Schwinger functions satisfying the Osterwalder-Schrader axioms then an alternative derivation is in the book "Quantum Physics: A Functional Integral Point of View" by Glimm and Jaffe, more precisely Corollary 19.5.6 (in the Second Edition).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the speed of the block in this question
The figure shows a block (assumed to be a point) being pulled by an ideal string across an elevated pulley. At any time $t$, let the horizontal distance of the block be $x$ from the pulley. The length of the string is $l$ and the height of the pulley is $h$ from the ground. The string is pulled with a speed $u$ as shown. Using calculus I found the speed of the block to be $u \sec \theta$ (where $\theta$ is the inclination of the string with the horizontal) which is given as the correct answer.
So my doubt is: why can't we simply resolve $u$ along the horizontal and say the block's speed is $u \cos \theta$? It doesn't seem wrong to me but isn't correct for some reason. Could someone please explain why?
| Because $u$ is only a component of the block's velocity ( along $\hat{r}$) it also has a component along $\hat{\theta}$. The sum of the two along the vertical is zero and along the horizonal gives the speed of the block.
(If it was the only component, then the block would fly, hence it obviously has a component along $\hat{\theta}$)
Another way to look at it would be $x = lcos\theta$ but since $\theta$ is a function of time you can't differentiate this equation to get $v$
You could however resolve the block's velocity (which is entirely along x) along and perpendicular to the string and equate the component along the string to $u$ which gives the desired result.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What is the induced electromagnetic field of a point charge? If I move a point charge on some trajectory, then it will produce an electric field as well as a magnetic field. As the charge is moving, and as a point charge can not produce a steady current, then due to a varying current, an electromagnetic field will be induced. It will again produce another electric field, and also the magnetic field will induce electric field.
Now I am getting confused, how many electric field and magnetic are here. One electric field due to its own charge, on magnetic field due to its motion and another electric field due to the varying current.
Is the first electric field will also be occured even in magnetostatics? (due to motion of charge)
Do those two electric field follow principal of superposition? I mean do they superpose to form a ultimate resultant field?
| When there are time varying fields we can write $E$ in terms of the vector potential $B=\nabla\times A.$ $$E=-\nabla V - \frac{\partial}{\partial t}A.$$
If we go to the Coulomb gauge, the term $E_{Coulomb}=-\nabla V$ is just due to the Coulomb potential of the charge as in ordinary electrostatics. And the second term $E_{induction}=-\partial A/\partial t$ is needed to satisfy Faraday's law so in that sense it is the field due to the changing magnetic field.
So yes the full $E$ field is the superposition of these two fields, but that is not a particularly useful way to think about it, because we still need to calculate $A$. It is not as easy to calculate $A$ as it is to calculate $V$ because the $B$ field depends on the changing $E$ field too (not just the current).
An easier way to figure out the $E$ and $B$ fields is to consider a point charge at rest, put the $E$ field ($B=0$ in this frame) into the electromagnetic tensor and boost with a Lorentz transformation to a moving frame.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Meaning of perturbative and non-perturbative renormalizability What is meant by a theory to be (1) perturbatively renormalizable, (2) perturbatively non-renormalizable, (3) non-perturbatively renormalizable, and non-perturbatively non-renormalizable? In each case, what are at least one example of such theories?
| *
*Perturbatively renormalizable (or simply renormalizable) theories are those which can be consistently renormalized by tweaking values of a finite number of parameters to any order of perturbation theory. The key moment here is that the finite number of parameters is fixed prior to choosing the order of perturbation theory. We have to be able to make sense of the theory to any order by tweaking the same finite set of parameters. Examples of renormalizable $4d$ QFTs include $\varphi^4$ (parameters are the field strengths renormalization $Z$, the particle mass $m$ and the interaction coupling $\lambda$), Yukawa theory (both scalar and pseudoscalar), QED, Yang-Mills for compact gauge groups.
*Perturbative nonrenormalizable (or simply nonrenormalizable) theories are those which aren't perturbatively renormalizable. Examples include $\varphi^6$ in $4d$, perturbative General Relativity.
*I've never heard the term "non-perturbatively renormalizable", but I suppose what is meant is finite. Finite theories are those which admit a well-defined Quantum Mechanical definition with a Hilbert space (or a Gelfand triple), and physical observables as self-adjoint operators acting on it. An incredibly beautiful and nontrivial moment here is that finite theories can have perturbative expansions which are actually nonrenormalizable. The best example here is General Relativity in $3d$. It was rigorously quantized by Witten, but its perturbative series is a nonrenormalizable asymptotic expansion.
*The ones which we can't formulate or define :) The logic is that quantum theories define effective actions, not the other way around. If we have a theory which can't be made sense of quantum-mechanically, then we don't have a theory at all.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Cylinder , charge on surface, why is B inside zero? Suppose we have a cylindrical wire of radius a carrying a current I. If the current is uniformly distributed over the surface of the wire, the magnetic field inside is zero.
We can prove that easily using Ampere's law. However, when I try to picture it in my head I can't see why.
I imagine that the cylinder is made up of infinite straight wires , each of them contributing to the total magnetic field by $B=\frac{μ_0I}{2\pi a}$ . If we had two straight wires it's obvious that exactly between them the B fields are cancelled out.
With that in mind it's easy to see that in the middle of the cylinder every B component is cancelled by its symmetrical one. What about the rest points? It seems that one side is more powerful.
| Just for picturing it...meaning this is not exact:
If you are not in the middle, where it is clear, but close to one side you may see it like this: the length over which you see a wire approximately straight is sort of given by the angle. Hence, the amount of wire with approximately opposite direction on the other side increases with the distance to the other side. This factor would compensate for the field decrease by reciprocal distance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why are springs shaped the way they are? Why are springs coiled the way they are? Why not some other shape? Is the shape due to its elasticity or something?
| The shape is due to the underlying principle of what a coil-spring actually does.
A spring is originally a long (often metal) wire. When you coil it, you are changing the way the forces can be applied to the spring.
For coil springs, you are making two surfaces that are often flat or have connections, that can be coupled with other systems.
The interesting thing is what is actually happening to the wire when you stretch or compress a coil spring. Because of the helical nature of the winding, when you push/pull on the spring, most of the effort is not going into bending or stretching the wire. It is twisting it.
You can visualize this by thinking of a coil from the side. The twisting forces all the coils closer together or further apart, leading to the motion desired.
You can consider it as a long bar experiencing torsion, allowing it to twist. As long as the bar has a constant cross section this relationship will also be linear, which may also have something to do with the shape.
Essentially, my understanding is that the helical shape allows you to use vertical motions to create twisting in the horizontal planes.
Helical shapes seem to do similar things in other applications. Perhaps someone with a better understanding of the math could give a clear description of why helix shapes can change the nature of motion in general.
(for example, torsional springs actually create a spring that resists twisting, but the spring itself is actually acting like a long bent bar; basically the opposite scenario of a linear helical spring)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simultaneity in Newtonian mechanics How would Newtonian mechanics answer the train and moving light question?
The setup is:
A train is moving in the positive x_axis with speed c/2. A person stands in the middle of the train. There are two light bulbs at both ends of the train. The light goes off at the same time (absolute time in Newtonian physics). The person standing in the middle of the train would perceive both lights independently.
Outside the train there is a stationary observer. Let's assume the train is already to the "right" of the observer (in x_axis) when the lights go off. Would the stationary observer observe the rear light before the front light?
The reason why I am asking this is that the relativity of simultaneity is often attributed ONLY to special relativity. Here, would Newtonian mechanics also predict that the stationary observer observes different simultaneity than the moving observer in the train?
| According to Newtonian Mechanics, the observer on the ground would see the light travelling to the right at a speed $V=\frac{3c}{2}$, and the light travelling to the left will have a speed $V=\frac{c}{2}$. This is because we assume the Galilean transformations are true. Hence the two light beams appear to reach both sides simultaneously in either frame according to Newtonian Mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Dirac's quantization of electric charge I was following Nakahara's book Geometry, Topology and Physics, specifically subsection 10.5.2. I do understand how he obtains the equality,
$$ \Delta \varphi = \int \mathrm d\varphi = \int \limits^{2\pi}_{0} 2g \, \mathrm d\phi = 4\pi g$$
and that therefore this has to be a multiple of $2\pi$ for $t_{NS}$ to be uniquely defined. But I don't see directly how,
$$\Delta \varphi / 2\pi = 2g \in \mathbb{Z},$$
is equivalent to
$$ \frac{eg}{\hbar} = 2\pi n ~(n\in \mathbb{Z}).$$
Can somebody tell me what I am missing?
| Please note that the transition function $t_{NS}$ defined on the equator $U_N \cap U_S \approx S^1$ of the bundle $P(S^2, U(1))$ is given by the formula:
$$ t_{NS}(\phi) = \exp[i\varphi(\phi)] \qquad (\varphi:S^1\to \mathbb R) \tag{10.90}$$
and it is periodic in the azimuthal angle $\phi\ $: $\ t_{NS}(\phi +2\pi) \equiv t_{NS}(\phi).$ Observe that the transition function represents a phase in the quantum wavefunction. Since a vector $A_\mu$ couples to an infinitesimal element of the worldline $\textrm dx^\mu$ via terms such as $(1+\frac{ie}\hbar A_\mu\textrm dx^\mu),$ the said phase is expressed as: $t_{NS}=\exp[\frac{ie}\hbar\int A_\mu\textrm dx^\mu].$ For a closed worldline $\Gamma =\partial\Sigma$ representing a full rotation around the equator, this phase is determined by
$$ \Delta\varphi = \frac{e}\hbar \oint_\Gamma A_\mu\textrm dx^\mu = \frac{e}\hbar \int_\Sigma (\vec\nabla \times \vec{A})\cdot \textrm d\vec\Sigma = \frac{e}\hbar \int_\Sigma \vec B\cdot \textrm d\vec\Sigma = \frac{eg}\hbar \,. $$
Since you understand that $\Delta \varphi \in 2\pi \mathbb Z,$ it follows that you also understand $\frac{eg}\hbar \in 2\pi \mathbb Z.$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Do charged particles always travel in a spring shape around magnetic fields or only when the magnetic field is from a solenoid? I have seen two contradictory descriptions of how electrons and other charged particles travel around magnetic fields; in one, they travel in circles around magnetic field lines, in the other, they travel in coil or spring-like paths around magnetic fields, with that said, which path do they actually travel when it's around a normal permanent magnet?
| The same path they would take around an electromagnet. They appear to spiral the wrong way, but if you imagine the pole being made up of a bundle of small poles, they are being pushed to the side where they are running parallel to the electrons around the atoms in the poles. If the magnetic field is above a critical level, they will be steered back in instead of spiraling out.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why does "liquid glass" putty turn cloudy when deformed? What physical process is taking place to make it look cloudy? What's changing to make it turn back to clear over time?
While they seem pretty secretive about the exact composition, but I did find this MSDS. Section 3 mentions it contains "Silicone Polymer Contains: - Inert Fillers".
| I believe it is simply air entrained. Air will make putty look cloudy. Air very slowly bubbles out of the viscous fluid, returning (most) of its transparency. I suspect putty has some small yield strength, permanently trapping the tiniest bubbles and thus permanenty losing some of its like-new transparency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why is light bent but not accelerated? Light is bent near a mass (for example when passing close to the sun as demonstrated in the famous sun eclipse of 1919). I interpret this as an effect of gravity on the light.
However, it seems (to me, at least) that light is not accelerated when it travels directly toward the (bary-)center of the sun. The same gravitational force applies yet the speed of light remains constant (viz. $c$).
What am I missing?
| You are missing special relativity and general relativity. In special relativity the speed of light in vacuum is always c, no matter the reference frame of measurement.
Also classical electromagnetism, light, emerges from a confluence of the quantum mechanical constituents which are photons and have zero mass. A photon aiming at the barycenter of the sun is attracted by the gravitational field of the sun but the effect is not a change in velocity , but on its energy which is $E=h\nu$ and therefore the extra energy increases the frequency while the velocity remains at $c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "90",
"answer_count": 6,
"answer_id": 1
} |
An identity of time-ordered operators that intertwines between the Schrödinger picture and the interaction picture Let $V(t)$ and $H_0$ be two operators where $V(t)$ has explicit time dependence while $H_0$ is time independent.
I am trying to prove the interesting identity,
$$T(e^{-i\int_{t_{0}}^{t} dt' (H_0+V(t'))})~=~e^{-i(t-t_{0})H_0}\times T(e^{-i\int_{t_{0}}^{t} dt' e^{i(t'-t_{0})H_0}V(t')e^{-i(t'-t_0)H_0}}).$$
If $V(t)$ and $H_0$ commutes at all times, life would have been much more easier and plain. I have tried the brute force and have failed, miserably.
Could anyone provide a small hint to nudge me in the right direction?
| OP's identity follows from the operator identity
$$ \forall t\geq 0:~~ L_1(t)~=~L_2(t), \tag{0}$$
where we have introduced (for later convenience) the following shorthand notation
$$L_1(t)~:=~T\left[\exp\left\{\int_0^t \! ds ~(H_0+V(s))\right\}\right]\tag{1}$$
and
$$L_2(t)~:=~e^{tH_0} T\left[\exp\left\{\int_0^t \! ds ~e^{-sH_0}V(s)e^{sH_0}\right\}\right].\tag{2}$$
Sketched proof of the operator identity (0): Show that both the two operators $L_i(t)$, $i\in\{1,2\}$, satisfy the following first-order initial value problem (IVP)
$$ \frac{d L_i(t)}{dt}~=~ (H_0+V(t))L_i(t), \qquad L_i(t\!=\!0)={\bf 1}.\tag{3} $$
By the uniqueness of the solution to the IVP (3), the operator identity (0) follows.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The direction of magnetic field in a coil This question is just to clarify a problem I am having regarding Lenz-Law.
In the following picture, the magnet comes in the coil and inducing the coil making the bottom of the coil north and is pushing the magnet out. But if you look at the way the magnetic field is coming out of a magnet, the direction of the magnetic field in the coil should be the other way. Why is that? I can't find a proper answer on the internet or in my textbook. My textbook says it's just not possible because it's not consistent with the law of conservation of energy.
| Magnetic fields of a coil and a magnet must have opposite orientations, because if the magnet comes in to to the coil it must be repelled from it.
Let us suppose that it is not the case. The magnet is attracted by the coil and thus accelerates towards it. It is moving faster and faster so the rate of change of the magnetic flux through the coil is increasing. This results in bigger current in the coil and thus a stronger magnetic field and attraction between coil and magnet. But this results in even bigger acceleration of the magnet, faster changing magnetic flux and even bigger current, which results with even stronger attraction and so on. The velocity of the magnet, electric current and magnetic field are increasing very quickly, "exploding" to infinity. This conclusion is obviously not consistent with conservation of energy, and so we arive at Lenz's Law.
It is worth noting, that if the magnet was getting out of the coil, than it would be attracted towards it, and the orientations of the two magnetic fields would be the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is harvesting of electrons during neutron beta decay possible? Is it possible in the decay cycle to harvest the neutron when it decays to an electron. For example if the neutron saturated boron rod was submerged in an electrolytic solution with an anode and cathode set up with some kind of catalyst in the solution. I was thinking something like if the rod was protected by a plating or coating of iridium with iridium anode and cathode in a solution of aqua regia and disolved platinum. If any of this sounds stupid for any reason its because i have no formal education, this was just an idea that came to me.
| It's not exactly like your idea, but something very like this is done in betavoltaic batteries. They don't harvest the beta electrons themselves. Instead the kinetic energy of the electron generates electron-hole pairs in a semiconductor and this generates an electric current.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sphere rolling up a step A sphere of radius R is rolling without slipping with a velocity v and collides inelastically with a step of height h < R. What is the minimum velocity for which the sphere will be over the step?
Relevant equations
Total kinetic energy (maybe):
$\frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2$
Gravitational potential energy:
$mgh$
Moment of inertia about a point a distance d from the CM from de center of mass:
$I=I_{cm}+md^{2}$
Rolling without slliping
$v=\omega R$
A discussion of the problem
This problem is confusing to me: I know the collision is inelastic so there is no conservation of kinetic energy, there is torque when the sphere pivots the step so the angular momentum is not conserved either, the linear momentum doesn't give a lot information and even though I tried to see what condition needs the torque about the pivot point (to give a net torque in the direction of rotation and thus passing the step), I'm not able to see which torque can overcome the one from gravity which doesn't let the sphere slide up:
The sphere has an angular velocity, but I'm missing something conceptual because I can't figure out how to formally state that the angular momentum changed the right amount so that the sphere made it past the step.
The attempt at a solution
My (almost certainly wrong) attempt was to say that the sphere lost kinetic energy because of the change on potential energy:
$\frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2 +mgR = mg(R+h)\\
v=\sqrt{\frac{10}{7}gh} $
I hope you can help me :)
| Angular momentum can be conserved but only under a certain condition.
The angular momentum just before collision about a point = angular momentum just after collision about the same point.
Hence,we shall conserve angular momentum about the tip of the elevation(as shown in second picture).
Anugular momentum about external point in combined translation and rotation = $L_{com} + mvr Sin\theta$
$$L_1 = mv(R-h)+I\omega$$
$$L_2 = (I+mR^2)\omega_{2}$$
$$L_1 = L_2$$
also , $\omega = vR$ (pure rolling)
$$mv(R-h)+I\omega = (I+mR^2)\omega_2$$
$$\omega_2 = \frac{mv(R-h)+I\omega}{I+mR^2}$$
Now, Kinetic energy just after collision should be just enough to elevate the sphere by h.
$$\frac{1}{2}(I+mR^2)\omega_2^2 = mgh$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
In the semiclassical approximation, should I expand the generating functional around saddles of the sourced or the unsourced action? Consider a Euclidean path integral say in a real scalar field theory.
$$
\int d[\phi]\exp(-I[\phi])
$$
In the semiclassical approximation, we consider stationary points of the action and expand around them. Now, consider I want to make a semiclassical expansion of the generating functional
$$
Z[J]=\int d[\phi]\exp\bigg(-I[\phi]-\int d^4x\,J\phi\bigg)
$$
I have a doubt, should I consider saddles of $I$ or those of all the sourced action?
$$
I_J[\phi]\equiv I[\phi]+\int d^4x\,J\phi
$$
Naively I would guess that I gotta take the saddles of the whole exponent, but
my biggest concern then is that if I take saddles of the sourced action, the stationary field configurations will, in general, have $J$ dependence and thus after expanding the action around these stationary points $\phi_s$, taking functional derivatives of $Z$ with respect to $J$ will be very dirty since I will have $J$ dependence in every place I have a $\phi_s$.
So, saddles of the sourced or of the unsourced action?
| Since we want to deal with path integral by using the stationary phase approximation, you will need to take the stationary points of the sourced action: these regions contribute coherently to the path integral (cause the phase changes very little over a small region surrounding each stationary point) giving the main contribution to the value of the quantity you compute, like in a constructive interference.
As you noticed this will be obtained for configurations depending on the source in a messy way. This dependence however is precisely what you are interested in: it will bring to the properties of the generating functional (if you derive it with respect with J you get expectation values of powers of $\phi$), even if in general this is way easier to see without making the actual computation of the partition function.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Obtaining a general equation for velocity (in 2D projectile motion) I'm trying to obtain a general equation for the instantaneous velocity of a projectile moving on a Cartesian plane.
I began with the equation for a projectile's trajectory (air resistance neglected):
$$y = x(\tanθ) - \frac {gx^2}{(u^2)(\cosθ)^2}$$
where $u$ is the projection velocity, and $θ$ is the projection angle.
I then sought to differentiate the above-mentioned equation with respect to time. This yielded:
$$y' = x'(\tanθ) - \frac {2gxx'}{(u^2)(\cosθ)^2}$$
Where $'$ stands for a differential with respect to time.
Now, re-writing the equation:
$$v_y = v_x(\tanθ) - \frac {2gxv_x}{(u^2)(\cosθ)^2}$$
Where $v_y$ and $v_x$ are the $y$ and $x$ components of instantaneous velocity.
My issue?
I can't seem to be able to get the last equation in terms of the variables $v_y$ and $v_x$ alone (I can't seem to eliminate the $x$).
My question:
Is it possible to obtain a general equation for instantaneous velocity with $v_y$ and $v_x$ as the only variables? If so, how do I go about it?
| Is your goal simply to write down an equation that relates $v_x$ and $v_y$, without reference to $x$, $y$, or $t$?
If so: Such an equation can’t exist, for the simple reason that $v_x$ is a constant while $v_y$ varies over the trajectory. If you found some relationship $f(v_x, v_y)=0$, you could differentiate it with respect to $t$:
$$
0 = \frac{d}{dt} f(v_x, v_y) = \frac{\partial f}{\partial v_x} \frac{d v_x}{dt} + \frac{\partial f}{\partial v_y} \frac{d v_y}{dt} = \frac{\partial f}{\partial v_y} \frac{d v_y}{dt}.
$$
From this, you could conclude that $v_y$ was a constant as well.
The only loophole in this argument is when the function $f$ is independent of $v_y$ ($\partial f/\partial v_y = 0$ above). An equation of this form does exist: $v_x=v \cos \theta $. It doesn’t contain $x$, $y$, or $t$, but it doesn’t contain $v_y$ either.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Insulated box of air at temperature $T$, is dropped from height $H$, what is the temperature of the air inside box on the ground? In an effort to understand more about the first law I was wondering about the potential energy term and how it would influence the gas inside the box I made this question and tried to answer it.
I will assume that the air follows the ideal gas law, there is no air resistance and the box stops moving immediately when it reaches the rigid ground.
The first law:
$$
0 = \Delta PE + \Delta U\\
gH = c_v (T' - T)\\
T' = gH/c_v + T
$$
would this be correct or missing something? (i.e. invalid assumptions..)
I am also wondering if the energy is fully transferred to the ground instead..
| $Energy\space of\space the\space system\space (E) = Internal\space energy\space (U) + PE_{system} + KE_{system}$,
$\Delta E =\Delta U+\Delta{PE} + \Delta{KE}$,
The condition for $\Delta E =\Delta{U}$ is $\Delta{PE}=\Delta{KE}=0$.
The first law of thermodynamics gives us the relation between the change in energy of the system and heat and work.
$\Delta E + PdV=Q$,
$\Delta U+\Delta {PE} + PdV=Q$ $\tag1$
In your question, $\Delta {PE}=-mgh$ and $PdV=0 $ $(as\space dV=0)$.
Your equation is correct.
Check this out : What is the meaning of internal Energy?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/339236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Energy contributions of Hamiltonian density In Lancaster and Blundell, Quantum Field Theory for the Gifted Amateur, p.99, the Hamiltonian density is
\begin{equation}
\mathcal{H}=\frac{1}{2}[\partial_0\phi(x)]^2+\frac{1}{2}[\nabla\phi(x)]^2+\frac{1}{2}m^2[\phi(x)]^2,\tag{11.5}
\end{equation}
and it tells us that the energy has contributions from
*
*a kinetic energy term reflecting changes in the configuration in time,
*a 'shear term' giving an energy cost for spatial changes in the field, and
*a 'mass' term reflecting the potential energy cost of there being a field in space at all.
In the equation above, i think the first term is the same as the classical mechanics. But i don't understand why second (shear) and third (mass) term are represent potential energy.
| Each of those terms represent the price to pay, in terms if energy, to have some specific configuration of the field:
*
*configurations that change with time (price is estimated by time derivative)
*configurations that change with space (price is estimated by the gradient)
*magnitude of the field (mass plays a role in enhancing this effect)
All those terms appear with a plus sign, suggesting that the most energetically favorable configurations are those where the field is more uniform, stabile an small
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/339330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Does current run forever in water? (assuming the supply voltage is there forever) Suppose pH of water is $6$, I think this means there is one $\text{H}^{+}$ ion for every $10^6$ water molecules.
When we plug in the battery, I believe we see a current as the $\text{H}^{+}$ ions drift to the $-ve$ side of the battery and suck the electrons injected by the negative plate of the battery. Similarly, $\text{OH}^{-}$ ions drift to the $+ve$ plate of the battery and give an electron away to the positive plate of the battery. This way $\text{H}^{+}$ and $\text{OH}^{-}$ ions neutralize themselves as they contribute to the current. Since the ions were neutralizing themselves, would the current cease to exist after some time when all the $\text{H}^+$ ions in the water were used up ?
| It is energetically unfavourable to split a water molecule into the two ions $\text{H}^+$ and $\text{OH}^-$ i.e. you need to put in energy to do it. However at room temperature water molecules have a range of energies and there are always a few molecules with enough energy to ionise. So any sample of pure water at everyday temperatures always contains a few $\text{H}^+$ and $\text{OH}^-$ ions.
When you apply a voltage to your electrodes in water, you convert the $\text{H}^+$ ions to hydrogen atoms and they bubble off as $\text{H}_2$. Likewise the $\text{OH}^-$ ions are converted to water and oxygen molecules and the oxygen bubbles off. The net result is to remove water from your container.
But as fast as the ion concentration is lowered by electrolysis, the remaining water ionises again to keep it constant. So electrolysis of pure water does not affect the ion concentration. You are correct that the current will continue to flow until all the water has gone (i.e. converted to hydrogen and oxygen).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/339645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
$2+1$-dimensional Einstein gravity is topological and only non-trivial globally $2+1$-dimensional Einstein gravity has no local degrees of freedom. This can be proved in two different ways:
*
*In $D$-dimensional spacetime, a symmetric metric tensor appears to have $\frac{D(D+1)}{2}$ degrees of freedom satisfying $\frac{D(D+1)}{2}$ apparently independent Einstein field equations. However, there is a set of $D$ constraints on the equations due to the invariance of the equations under differmorphisms, and a second set of $D$ constraints due to the conservation of the stress-energy tensor. Therefore, there are really only
$$\frac{D(D+1)}{2} - D - D = \frac{D(D-3)}{2}$$
degrees of freedom of the metric tensor satisfying $\frac{D(D-3)}{2}$ independent Einstein field equations.
*In the ADM formulation in $D$-dimensional spacetime, the metric induced on the spacelike hypersurfaces appears to have $\frac{D(D-1)}{2}$ degrees of freedom. However, there is a set of $D$ constraints due to the $D$ Lagrangian multipliers in the Hamiltonian. Therefore, there are really only
$$\frac{D(D-1)}{2} - D = \frac{D(D-3)}{2}$$
degrees of freedom of the metric tensor.
The metric tensor of a manifold encodes information about the infinitesimal distance between nearby points on the manifold, so the are $\frac{D(D-3)}{2}$ degrees of freedom are all local degrees of freedom.
Therefore, it is said that $2+1$-dimensional Einstein gravity is trivial locally.
But what does it mean to say that $2+1$-dimensional Einstein gravity is non-trivial globally?
Why is the word topological used to describe $2+1$-dimensional Einstein gravity?
| Bob Knighton's answer is very detailed. But I want to add a few remarks. You can try to prove this identity
$$ R_{\mu\nu\rho\sigma} = \frac{R}{D(D-1)} (g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}). $$
Then, it's straightforward to show that for 1+1 spacetime, the Einstein tensor $R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$ is identically zero. That is, the vacuum Einstein field equation is trivially satisfied, no matter how much matter content you put. This means, there is no coupling between gravity and matter.
Similar consequences also happen on the 2+1 case. In fact, people had believed that 2+1 gravity was just as trivial as 1+1 gravity. However, in around 1990, it was discovered that in presence of a negative cosmological constant, 2+1 gravity is a rich subject (there are BTZ black hole solutions, etc).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/339962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
How to calculate tension? below is an experiment set up to test the effect tension has on the frequency of various guitar strings. It's hard to see the string in the picture, but it's attached to a weight and I was wondering how you could calculate the tension on the string? What equation/formula would you use? We know the length, mass of the string.
| The tension on the string is equal to the weight of the hanging object, plus the weight of the string that is hanging from the pulley. To calculate this, measure the hanging length, $l$. Then, for a string of mass $M$ and length $L$, the mass of the hanging portion is $m=M\frac{l}{L}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/340341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do we know that nuclear decay is truly random and spontaneous? Nuclear decay is said to be random and spontaneous, but how do we know for certain, that it is not just a lack of understanding of some other unknown force?
Doesn't everything in the universe just depend on the starting conditions, so arguably nothing is random?
| I broadly agree with existing answers at the time of writing. I hope I can add something helpful. I would like to underline the role of model-making in science. The physical world does what it does, and in science we construct models. The models help us build insight into the nature of the physical world. But the model is not the same as the physical world.
In the case of phenomena such as radioactive decay (and another good example is light reflecting off a partially reflecting surface) our most complete model is the one provided by quantum mechanics. In quantum mechanics the model includes randomness (let's not get into Everett interpretation here, it will not change the central point I want to make). So we think the decay is random because we think quantum mechanics is offering a good model. That is not the same as certain knowledge; it is a case of reasonable belief.
If someone wishes to propose another model, one in which the decay events are not random, then they are welcome to do so. It will be received if it matches experimental observations and has an elegant mathematical framework or other such beauties to recommend it. It will be enthusiastically celebrated if it also manages to predict something currently unknown or puzzling and get it right. But no one is currently able to think of a model like that.
By the way, personally I welcome a bit of openness or non-determinism in the world. If it is like that, as it seems to be, then it seems to me to be a more liberating sort of a world, and I say this even aware of all the pain that is also associated with random events.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/340530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Charged sphere inside opposite charged sphere If we enclose a negatively charged sphere inside a positively charged sphere, do we get the electric field only due to the outer sphere?
If we do, how? Also, where did the negative charge went ?
| In terms of a picture with your 15 negative charges on the inner sphere and your 20 positive charges on the outer sphere, the charges rearrange themselves as follows.
You will have 15 negative charges on the outer surface of the inner sphere and 15 positive charges on the inner of the outer sphere.
The remaining five positive charges will reside on the outer surface of the outer sphere.
There will be no charges resident inside the two conducting spheres.
The electric field will be zero inside the inner sphere , an electric due to the 15 negative charges (and 15 positive charges) between the two spheres and an electric field due to 5 ($= 20-15$) positive charges outside the outer sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/340630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Why does a massless, frictionless piston move from high pressure to low pressure? Consider an ideal gas kept in a rigid cylinder with a movable massless, frictionless piston at the top. Let the pressure inside the cylinder be $P$ at pressure exerted by the surrounding on the cylinder be $p$. Let $$ P>p $$
Now since the piston is massless net force on it must be 0 (since $m=0$). This implies that the piston can move with any acceleration. But it moves in only one direction that is outwards. Why doesn't it moves inwards?
| The bigger force wins.
Pressure is force per area. Let's calculate the forces:
$$F_{inside} =PA\qquad\text{and} \qquad F_{outside} =pA$$
The push from the inside is larger than from the outside, $F_{inside} >F_{outside}$. Easily seen since the areas are equal.
The bigger force wins. The piston moves outward.
Now since the piston is massless net force on it must be 0 (since m=0m=0). This implies that the piston can move with any acceleration
Thus might just be word mess-up. That the piston is massless does not mean that the net force on it is zero. It just means that it will move with enormous acceleration at any tiny net force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/340752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Kutta-Joukowski theorem applied on a Joukowski airfoil (derivation) I have a doubt about the derivation of the Kutta-Joukowski theorem for a Joukowski airfoil. I know the results, but my main objective is to know how get these ones.
Consider for the initial plane a cylinder centered on $\zeta_0$, with a circulation -$\Gamma$, in an uniform flow with an atack angle $\alpha$:
where $a$ is the circumference radius, and $b$ the intersection of the circumference with the real positive axis, $\xi$. The parameter $\beta$ is the angle between the horizontal line and the line that links $\zeta_0$ to $b$. The center of the circumference is:
$$\zeta_0=b-ae^{-i\beta}$$
For this flow we have the following complex potential:
$$W(\zeta)=U_\infty\left[e^{-i\alpha}\left(\zeta-\zeta_0\right)+\frac{e^{i\alpha}a^2}{\left(\zeta-\zeta_0\right)}\right]+\frac{i\Gamma}{2\pi}\ln\left(\zeta-\zeta_0\right)$$
The Joukowski transform is:
$$z=\zeta+\frac{b^2}{\zeta}$$
To determine the complex potential on the $z$ plane we need to find the relation between $\zeta$ and $z$. And we get:
$$\zeta=\frac{z}{2}\pm\sqrt{\left(\frac{z}{2}\right)^2-b^2}$$
That is an awful relation. We not only have a square root, but also an $\pm$ symbol. Substituting this on the potential complex and then find the residues of $\left(\frac{dW}{dz}\right)^2$ and $\left(\frac{dW}{dz}\right)^2 z$ would be pratically impossible. What should be the next steps?
[Note: I didn't find any paper or book that shows the derivation of this theorem on the $z$ plane, but only on the $\zeta$ plane, which I already know. Do you know anything that can help me?]
| you have transformed circle into ellipse by, $z = \zeta + \frac{b^2}{\zeta}$ ,
Now you have to do the inverse by tranforming ellipse into circle,
So the standard text books (Milne Thompson, etc) use,
$\zeta = \frac{1}{2}z + \frac{1}{2}{\sqrt{z^2-4b^2 }}$ .
Myself tried to transform ellipse into circle with the aove and suceeded.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/340867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The derivation and formula to calculate Potential Energy seems quite confusing? Ok, so I am in the 10th grade, and Physics is my life, and so I always try to enquire things. We were studying the chapter of Work and Energy during our Physics lecture, and when our lecturer described the derivation of formula to calculate the Potential Energy (U) = mgh, something struck me.
The derivation goes in this way :- Gravitational Potential Energy, is measured in terms of work done in bringing a body from the ground level to a certain height against the forces of gravity, the work done then gets stored in the form of Potential Energy in the body. Therefore, in our textbook it was given,
Work = Force x Displacement
The least force required to lift the body from the ground will be mg i.e. its mass x acceleration due to gravity(this part is bothering me). Therfore the Force will be mg, while the displacement done by the force will be the height at which it was raised against the forces of gravity that is h. Therefore :-
Total Work Done in lifting the body = mgh (force x displacement)
Therefore, Potential Energy = mgh.
But isnt it true that the least force required to lift the body should be greater than mg. Since the force of gravity acting on that body too will be mg but in opposite directions and these two forces should cancel each other and the net force will be zero, therefore displacement will be zero and the body will remain in equilibrium. I dont know if this is stupid but can anyone pls answer.
| There have been a couple of questions similar to this recently, which you should look at. As for your specific question, you have to accelerate the object upward in order to increase its potential energy, so you are inherently giving it kinetic energy by increasing its gravitational potential energy. In other words, the work you do goes into both kinetic and potential energy, but the work you did against gravity is only considered for its potential energy. The kinetic energy may turn into potential energy if it keeps moving once you stop applying a force, in which case the total work you did will all eventually go into potential energy at the point that it stops moving.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Is the divergence of the propagator between same two spacetime points a pathology? This question is same as the question here which is not answered yet.
By Poincare invariance, the propagator of any field theory $G(x,y)$ must be a function of $|x-y|$ and by dimensional arguments $$G(x,y)=\langle 0|\phi(x)\phi(y)|0\rangle\sim\frac{1}{|x-y|^2}\tag{1}$$ which diverges as $x\to y$. Is this divergence a pathology of the theory? Is it an indication of taking some unphysical limit? If yes, why is the limit $x\to y$ unphysical? Does it necessarily mean continuum picture of spacetime in quantum field theory has to be given up?
| This is indeed a pathology for the computation of some quantities, such as any quantities derived from the stress energy tensor
$$T_{\mu\nu} = \lim_{x\to y} D_{\mu\nu} G(x,y)$$
For some differential operator $D_{\mu\nu}$. This can be easily renormalized, though, as it can easily be shown that in the coincidence limit $x = y$, the divergence term will be a constant (or depend on the curvature of the manifold, if in curved space), and this is then just a renormalization of the cosmological constant, or the gravitational constant as well in curved space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is atmospheric pressure exerts force from all directions of the object on the surface of the earth? This question came to me as I was solving problem asks the length of the fluid in a open tube when the tube makes particular angle with horizontal surface by giving the height of the fluid when tube was standing straight vertical.( considering atmospheric pressure). Do I have misunderstanding of atm.
| Pressure is exerted by a fluid in directions perpendicular to the surfaces of objects which contains the fluid.
To solve this problem you must understand that when the tube is tilted only a component of the mercury in the tube can exert a pressure along the length of the tube as it's a liquid. This is because the pressure from a liquid arises due to the gravitational force on it. Like when it's vertical the pressure is 8.0cmHg but when tilted 65 degrees to vertical, that is 25 degrees to horizontal then the component of pressure along the length of the tube is $8.0sin(25)$ cmHg.
The atmospheric pressure on the open end of tube along its length remains 76 cmHg regardless of angle as this pressure is not due to gravity( rather factors like temperature etc.)
So the total pressure downward on sealed air column will be the pressure due to component of mercury($8.0sin(25)$ cmHg) and atmosphere(76 cmHg).
Then you can solve the problem with taking the length of sealed column of air being inversely proportional to the pressure exerted on it along the length of tube. This is because the longer the tube of air in a sealed tube the lower the pressure it can exert. Try to get the answer by this method.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why the four gauge bosons that correspond to the $SU(2)\times U(1)$ electroweak force before symmetry breaking are not listed in the Standard Model? If I correctly understand this, the four gauge bosons that correspond to the electroweak force before symmetry breaking are the W1, W2, W3, and B. How come the W1, W2, W3, and B bosons are not listed in the Standard Model?
More clarification:
I guess I am missing something from the Wiki article and other descriptions. For example, are the four bosons before symmetry breaking (i.e., W1, W2, W3, and B) only mathematical constructs that have never been observed in a high energy accelerator?
| They're listed under $W^+$, $W^-$, $Z$, and $\gamma$ (the photon). Those observed particles are constructed from linear combinations of $W^1$, $W^2$, $W^3$, and $B$ in the dynamical symmetry breaking of the Higgs mechanism, with $\gamma$ being the massless Nambu-Goldstone boson that results from the process (see comments for clarification, thanks @gj255). For specific details, go through the Wikipedia article on the Electroweak interaction, or any textbook that covers the standard model.
It's going to depend on what you mean by "observed". None of the weak theory bosons have been observed leaving a track in a detector. Instead, they're inferred as a "bump" in the interaction cross section scaling with energy, and as interaction vertices of other particle that have an invariant mass at the correct values. Because we are limited to detecting the $W^\pm$ and $Z$ bosons this way, there's no chance for observing the sort of superposition that the $W^{1-3}$ and $B$ bosons would be. If we could get the energy high enough for time dilation to cause one of these bosons to leave a track, it could, in principle, be possible to get an interaction to produce an observable superposition that would correspond to what you want to observe. Someone more familiar with the standard model Lagrangian would have to sit down to see whether the math allows this, given an appropriate superposition of input particles. Think of observing $K$ mesons that are produced in a superposition of two states that have different decay rates, just a whole lot more technically challenging to do at every level.
See also this question StackExchange recommends as related.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Bounds of Integration (with respect to something that is not time) I have been reading Richard Feynman's lectures and came across an interesting proof regarding the Earth's gravitational force. At one point in the proof, Feynman uses the following the integral:
$\int_{R+a}^{R-a} dr$
(13.18 on http://www.feynmanlectures.caltech.edu/I_13.html)
In this integral, r is the distance between a point in space and the surface of the Earth, R is the distance between that point and the center of the Earth, and a is the radius of the Earth. I interpret this integral as summing up all of the dr's going around the Earth. The proof itself makes sense to me, I am just confused about the bounds of integration.
As $\int_{R+a}^{R-a} dr$, I interpret the integral as summing up the dr's starting on the right side of the Earth and going to the left side. However, in this sense, $\int_{R-a}^{R+a} dr$ should be the sum of all the dr's starting from the left side and going to the right side. Conceptually, I feel as if these should be the same, but mathematically $\int_{R+a}^{R-a} dr = -\int_{R-a}^{R+a} dr$. My question is, how did Feynman choose the ordering of his bounds of integration? It does not appear arbitrary, but I am not sure how the decision was made. Thank you!
|
In this integral, r is the distance between a point in space and the surface of the Earth, ...
Almost. Actually, based on the diagram in the derivation, $r$ is the distance from the point in space to the ring of mass that is part of the spherical shell. It $is$ a distance like you describe, but it is the distance to a very specific geometrical piece of mass.
I interpret this integral as summing up all of the dr's going around the Earth.
Maybe, depending on what you mean by "going around the earth." I would describe the integral as summing all the rings of mass, starting at the distance $R+a$ (on the far left end of the shell) and accumulating ring by ring all the way to $R-a$ (on the far right end of the shell). The integral around the earth which creates each ring in the shell was done implicitly and results in the $2\pi$ factor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Neutron stars - only neutrons? I was at a museum recently, and there was a display on neutron stars. It said that neutron stars are made only of neutrons, which honestly didn't make much sense to me - neutrons decay very quickly on their own, so how do neutron stars "last", so to speak?
So naturally, I checked wikipedia, which provides this diagram:
Electrons and protons do seem to be present. Nowhere is there a "layer" that's only neutrons.
This leads to two questions:
*
*Is Wikipedia wrong, or the museum (normally I'd trust wikipedia, but this is a not-insignificant museum that I went to)?
*Is there anything that could explain the museum display if wikipedia is right?
| The picture from Wikipedia is (qualitatively) correct and so is your intuition.
Neutrons are unstable and decay unless the decay is blocked by the presence of a degenerate electron gas with a Fermi energy that is as large as the maximum possible energy of the electron produced in a beta decay.
If all fermion species are degenerate, and they are at neutron star densities, then the Fermi energies of neutrons, protons and electrons come into an equilibrium where
$$E_{f,n} = E_{f,p} + E_{f,e}$$
This, combined with charge conservation, leads to the calculation that there are about ten to a hundred times as many neutrons than protons/electrons in what started as a pure neutron gas; the exact ratio being density dependent (smaller at higher densities).
Aside from this, the equilibrium state of "cold" neutron star matter varies with density, as shown in the Wikipedia diagram. The n,p,e fluid probably makes up the greater part of the mass of a neutron star, but by no means can it ever be said that a neutron star is made up only of neutrons, and in fact free neutrons only appear at densities above a few $10^{14}$ kg/m$^3$, some way in to the neutron star.
NB This is where the diagram is quantitatively incorrect. $\rho_0$ is supposed to be the nuclear saturation density of about $2.3\times 10^{17}$ kg/m$^3$, so free neutrons appear at just over $10^{-3}\rho_0$. In addition, nuclei lose their identity and become an n,p,e gas at about $0.2\rho_0$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Is the presence of a magnetic field frame-dependent? I do not have a strong background in physics, so please refrain from using complex mathematics in any answers :)
I was on
this site, and I read:
When an electrical charge is moving or an electric current passes through a wire, a circular magnetic field is created.
I am trying to grasp the notion of magnetic fields under different reference frames.
Suppose I have a point charge next to a compass needle. It is experimentally verifiable that the compass needle is not affected by the point charge - i.e., the needle does not move. It can be concluded that the point charge does not create any magnetic field.
But since the point charge is on the Earth, and the Earth is moving, shouldn't the point charge produce a magnetic field? Why doesn't the compass needle move in response to this magnetic field? Is there a magnetic field in the room?
| You have to make a distinction between the generation and the detection of a magnetic field. If you generate a magnetic field, but can't detect it, for all intents and purpose, it is the same as not generating it.
In the example you give, even though there is motion of both the charged particle and the compass, there is no motion between them, therefore, no magnetic field between them, is detected.
Since the magnetic effect of a charged particle is an effect that depends on the relative motion between the particle and the "medium," the presence of a magnetic field (and its detection), is frame dependent!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 7,
"answer_id": 4
} |
How do car sunshades work? By putting these sunshades in the windows of a car, one can considerably negate the interior heating effect. What is the physics behind this?
I searched for it online but couldn't find a thorough and comprehensive answer. One thing which I got to know that visible light enters through the windows ( without these shades) and is absorbed by the interior of the car. This is given off as infrared light which can't pass back through the windows making the interior of the car very hot.
Links
*
*http://www.car-sunshade.com/How-do-sunshades-keep-the-inside-of-cars-cool-_n19
*https://www.s2ki.com/forums/car-talk-73/do-sun-shades-really-keep-interior-cooler-394083/
| My guess is that the black shade is absorbing the heat rather than the air in the cabin. Also, if that shade is an adhesive type of sunshade that actually sticks to the glass then perhaps the heat is more readily conducted to the glass and back out of the car.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/341958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Velocity of a Mechanical Wave on a String I recently read a derivation for an equation which governs how quickly a wave is transmitted along a string, $v = \sqrt{\frac{T}{\mu}} $, where T is the tension in the string, and $\mu$ is the mass per unit length along the string. The derivation makes sense but gives a more mathematical and geometrical account as to why this is the case.
Could someone please explain more qualitatively why an increase in density of the string would reduce the velocity of transmission, and why an increase in tension would increase the velocity?
I can see, in very vague terms why an increase in tension would cause neighbouring elements of the string to more quickly follow the motion of preceding elements when for instance, a pulse is sent down the wave as follows:
though I cannot visualise the effects of density. I can imagine each element of the string having more mass and their movements becoming more 'sluggish,' but would this not affect the frequency of their oscillations?
| More mass means more inertia. Thus, it takes more force to move a differential mass on the string, and that differential mass responds more slowly than a "lighter weight" string because the acceleration of that differential mass must follow Newton's 2nd law (a=F/m). This means that the wave speed on a string of large linear density will be lower than the wave speed on a string of small linear density, assuming the same force applied to each differential mass on the strings as the wave passes by.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Why is $\nabla \times B$ perpendicular to $B$? I'm considering a magnetic field inside a load capacitor where
$$\boldsymbol{\nabla}\times\textbf{B} =\mu_{0} \varepsilon_{0} \frac{\partial\textbf{E}}{\partial t}\\
\textbf{j}_{c}=0.$$
Why is $\boldsymbol{\nabla}\times\textbf{B}$ is perpendicular to $\textbf{B}$?
| The condition that the magnetic field and its curl be orthogonal is satisfied, generally, for plane-wave situations where the curl is given simply by $\nabla\times\mathbf B = i\mathbf k\times \mathbf B$, as well as a number of similarly 'nice' geometries, but it is not a general property of electromagnetic fields.
In the specific case you refer to ─ the magnetic field induced by the displacement current inside a charging capacitor ─ is a bit tricky because what you have is a differential equation for $\mathbf B$,
$$
\nabla\times\mathbf{B} =\mu_{0} \varepsilon_{0} \frac{\partial\mathbf{E}}{\partial t},
\tag1
$$
where the source $\frac{\partial\mathbf{E}}{\partial t}$ is assumed to be known and homogeneous over space (possibly with some time dependence), and this differential equation does not have unique solutions. This means that we must put in additional constraints to mold it to our needs: we normally choose the solution
$$
\mathbf B = (Cy,-Cx,0)
$$
for $C$ a constant and with $\frac{\partial\mathbf{E}}{\partial t}$ along $z$, because it's the most useful one (but note that it has an unphysical dependence on the origin!). However, the magnetic field
$$
\mathbf B = (Cy,-Cx,B_0)
$$
is also a solution of $(1)$, and it no longer satisfies $\mathbf B \cdot(\nabla\times\mathbf B)=0$, so the property is not inherent to the statement of Ampère's law ─ it comes from the external conditions we impose on the problem. In other words, in that problem $\mathbf B \cdot(\nabla\times\mathbf B)=0$ because we want it to.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to choose the best Gaussian surface? Are there some basic rules so that I could choose a Gaussian surface according to my physics problem? Are there any guidelines for how to choose an appropriate Gaussian surface for the system of charges you are analysing?
| Mathematically the Gaussian surface should be chosen so that
$\vert \vec E\vert$ is constant on the surface so that
$$
\oint\vec E\cdot d\vec S = \int \vert\vec E\vert dS\cos\theta
=\vert\vec E\vert\int dS\cos\theta
$$
so that evaluate the magnitude of $\vec E$ from the charged enclosed by
your surface.
In practice, choosing $d\vec S$ so that $\vert \vec E\vert $ is constant on the surface amounts to choosing a surface that has the same symmetry as $\vec E$. Thus, if you can argue that $\vec E$ will be spherically symmetric, then your surface should be a sphere; if $\vec E$ has cylindrical symmetry, then the surface should be a cylinder etc.
The symmetry of $\vec E$ is often dictated by the symmetries of the charge distribution, i.e. a spherically-symmetric charge distribution will produce a spherically symmetric $\vec E$ etc, so the path to choosing an appropriate surface is
symmetry of the charge distribution $\to$ symmetry of $\vec E$ $\to$ symmetry of the Gaussian surface.
Note that, in the case of a cylindrical or planar symmetry, the field must also be translationally invariant, i.e. a straight uniformly charged rod of finite length does not result is a cylindrically symmetric field over the entire length of the rod, so you can't use Gauss' law for this kind of configuration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to justify the maximal value of a star's magnetic field at surface? In many lectures, it is stated that the maximal value $B_{\text{max}}$ of the magnetic field at the surface of a star can be found in Newton's gravitation theory by equating the gravitational potential energy with the magnetic field energy. For a sphere of mass $M$ and radius $R$, of uniform density and uniformly magnetized :
\begin{equation}\tag{1}
|\, U_{\text{grav}}| = \frac{3 G M^2}{5 R} = E_{\text{magn}} = \frac{\mu_0 \, \mu^2}{4 \pi R^3},
\end{equation}
where $\mu$ is the sphere's dipolar magnetic moment. The right hand part is the total energy stored in the magnetic field of a dipole :
\begin{align}
E_{\text{magn}} = \int \frac{B^2}{2 \mu_0} \, d^3 x &= \int_0^R \frac{B_{\text{int}}^2}{2 \mu_0} \, 4 \pi r^2 \, dr + \int_R^{\infty} \frac{B_{\text{ext}}^2(r, \theta)}{2 \mu_0} \, r^2 \, dr \, \sin{\theta} \, d\theta \, d\varphi \\[12pt]
&= \frac{\mu_0 \, \mu^2}{4 \pi R^3}. \tag{2}
\end{align}
Since the sphere is uniformly magnetized in its volume, the internal magnetic field is a constant :
\begin{equation}\tag{3}
B_{\text{int}} = \frac{2}{3} \, \mu_0 \, M = \frac{\mu_0 \, \mu}{2 \pi R^3} \quad \Rightarrow \quad \mu = \frac{2 \pi B_{\text{int}} \, R^3}{\mu_0}.
\end{equation}
Substituting this magnetic moment into equ. (1) gives the maximal field strength inside and at the surface of the sphere :
\begin{equation}\tag{4}
B_{\text{int max}} = \sqrt{\frac{3 \mu_0 \, G}{5 \pi}} \, \frac{M}{R^2}.
\end{equation}
So for a star of mass $M = 0.6 \, M_{\odot}$ and radius $R = 10^4 \, \mathrm{km}$ (a typical white dwarf), this give
\begin{equation}
B_{\text{int max}} \approx 5 \times 10^7 \, \mathrm{tesla} = 5 \times 10^{11} \, \mathrm{gauss}.
\end{equation}
But how can we justify equation (1) ? Can it be made more rigorous ? Why should we have $E_{\text{magn}} + U_{\text{grav}} = 0$ for the maximal field strength ?
EDIT : In the case of a canonical neutron star of radius $R \approx 10 \, \mathrm{km}$ and mass $M \approx 1,44 \, M_{\odot}$, equation (4) gives
\begin{equation}\tag{5}
B_{\text{int max NS}} \approx 10^{14} \, \mathrm{tesla} = 10^{18} \, \mathrm{gauss},
\end{equation}
which is pretty excessive (AFAIK). The strongest known magnetars have at most a field of about $10^{15} \, \mathrm{gauss}$. So is there a theoretical way in reducing the value (5) ?
| The total energy of a star must be less than zero for it to be a gravitationally bound object.
The total energy is the sum of negative gravitational potential energy (your expression assumes a star of uniform density) and positive terms due to gas pressure, turbulence, rotation and of course magnetic fields.
The maximum magnetic energy can be found by equating the total energy to zero and making the other positive terms zero. If they are $>0$ (which they are in a real star) then of course the largest magnetic energy possible will be smaller.
You then have to decide how you want to relate that to the magnetic field and dipole moment, since the magnetic energy density will depend on the equilibrium size of the star, though I would have thought you should make $B_{\rm int} R^2$ a constant, because magnetic flux through the surface is conserved when the size changes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What happen when a helium balloon is floating inside a moving train?
Let's assume a helium balloon is suspended into the floor of a moving train. We already know how it will react during acceleration, deceleration, and constant speed as illustrated in figure no 1, 2, and 3. However, what will happen to the balloon if we detach the balloon from train floor and attach something that is exactly the same weight as it's floating force thus making it "floating" in the train cabin. Assuming that we do that after train has stopped accelerating (moving at constant speed) as illustrated in figure no.4. Will it stay where it was detached inside the cabin or will it hit the back of the cabin as the train is moving forward at a constant speed?
| Ideally, there will be no external force acting on the balloon in the horizontal direction so it will move along with the train in accordance with Newton's First Law.
However in reality, there will be collisions with gas particles present in the coach which will give it some random velocities.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does an electron emit a photon towards the other electron during electromagnetic repulsion? In the Quantum Electrodynamics theory, the repulsion between electrons is caused by the exchange of photons. This exchange can be represented with the help of Feynman Diagrams, but why the electron has to necessarily emit a photon towards other electron when two of them approach each other?
| Each charge is a source of a force field acting on the other charges, immediately or with retardation, whatever. It is an experimental fact.
The total field of a charge can be represented as a near field (virtual photons) and the radiated field (real photons).
The real photons are never absorbed, but scattered. The virtual photons are always absorbed, roughly speaking.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/342881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to implement this single-qubit unitary? I was reading this paper on qubit state preperation, and encountered an interesting type of single-qubit gate:
\begin{align}
U_\theta = \left(\begin{matrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{matrix}\right) = \cos\theta\, \sigma_z + \sin\theta\, \sigma_x
\end{align}
and more generally,
\begin{align}
U = \frac{1}{\sqrt{|a|^2+|b|^2}}\left(\begin{matrix} a & b \\ b^* & -a^*\end{matrix}\right)
\end{align}
I would like to try and decompose these gates as compositions of standard rotations, i.e.
\begin{align}
R_x(\theta) = \exp(-i\theta \sigma_x/2)\ \ ,\ \ \sigma_x = \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \\
R_y(\theta) = \exp(-i\theta \sigma_y/2)\ \ ,\ \ \sigma_y = \left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right) \\
R_z(\theta) = \exp(-i\theta \sigma_z/2)\ \ ,\ \ \sigma_z = \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)
\end{align}
However, I'm not really sure how to go about it, mostly due to the minus sign in the (2,2) matrix element. I've tried solving the simultaneous equations e.g. $R_X(\theta)R_Z(\phi) = U$ but I end up getting no solutions.
| Consider the most general rotation as
$$
R_{\mathbf{n}}(\theta)=e^{i \theta (\mathbf{n}.\mathbf{\sigma})}
= \begin{pmatrix}
\cos \theta + i n_{z} \sin \theta & i n_{x} \sin \theta + n_{y} \sin \theta \\
in_{x} \sin \theta -n_{y} \sin \theta & \cos \theta - i n_{z} \sin \theta
\end{pmatrix},
$$
which we have obtained using $exp(i \theta (\mathbf{n}.\mathbf{\sigma}) )= \cos \theta \mathbb{1} + i \mathbf{n}.\mathbf{\sigma} \sin \theta$. To keep this rotation unitary, we should also set the norm of $\mathbf{n}$ to one which is equivalent to devide the matrix by its determinant.
Putting everything together, we would end up with a general form of
$$
R_{\mathbf{n}}(\theta) =
\frac{1}{ \sqrt{|a|^{2} + |b|^{2}}}
\begin{pmatrix}
a & b \\
-b^{*} & a^{*}
\end{pmatrix}.
$$
As you may have noticed, the minus sign is not on the right component which implies that we should act by one more $\sigma_{z}$ on this rotation. Finally,
$$
\sigma_{z} R_{\mathbf{n}}(\theta) =
\frac{1}{ \sqrt{|a|^{2} + |b|^{2}}}
\begin{pmatrix}
a & b \\
b^{*} & -a^{*}
\end{pmatrix}.
$$
You can also view $\sigma_{z}$ as $-iR_{z}(\pi/2)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/343044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Understanding the units of number density I understand that we can measure any general number density $n$ as,
$$ n = \frac{N}{V}$$
for total number $N$ and volume $V$. This puts the units of number density as $\text{length}^{-3}$ e.g. $\text{cm}^{-3}$
Now suppose the number density varies with radius as e.g.,
$$ n(r) = N_0 r^{-2}$$
Clearly the units at any particular radius are still $\text{cm}^{-3}$, but then what are the units of $N_0$ and $r$?. It seems to me that for different distributions e.g. $N_0 r^{-3}$ the units would be different. But then how can the units of $n$ always be $\text{cm}^{-3}$
| In several comments, it was mentioned "just make the units of $N_0$ whatever they need to be, without concern as to why. I thought I might add to that.
Units are a funny thing because in many cases, they serve no deeper purpose than to ensure you did the math correctly. Take Hooke's law as a classic example. Hooke's law is most accurately described in mathematics as $F\propto x$, where F is the force exerted by a spring is proportional to x, which is the amount the spring is extended or compressed. However, this very pure form is inconvenient for doing math. Instead we rewrite that equation as $F=kX$, where $k$ is a measure of the stiffness of the spring. The two expressions convey the same information, but one uses proportionality while one uses equality and a constant multiplier.
When converting proportionality to equality, we have to deal with the fact that proportionality and equality differ in how they deal with units. Proportionality is blind to units. If you do the math, the units always end up canceling out. Equality, however, does require the units to line up. To make them line up we assign a unit to the constant which makes the equation work.
Sometimes we find that these assigned units have a deeper meaning, other times they do not.
So in your case, you write $n(r)=N_0r^2$, but in words you chose to write "...the number density varies with radius," which is a proportionality phrasing $n(r)\propto r^2$. $N_0$ only appears as a side effect of turning that proportionality into an equality, and thus the units of $N_0$ are safely assigned to "whatever they need to be to make the units work." This also points out the answer to your second question: if you used a distribution where $n(r)\not \propto r^2$, then you would have a different equation when you converted it to equality. Thus, you would need to make up a different unit.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/343310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Lagrangian and Hamiltonian formalism for damped pendulum Suppose I have a simplified pendulum (massless string of length $l$, ball of mass $m$ and some non-zero volume). I want to derive the equations of motion, but I want to take air resistance under account. I know how to do it by applying Newton's 2nd law, but is there any way I can do that using lagrangian or hamiltonian? What would be the form of lagrangian/hamiltonian in such case?
| In a sense yes; you'll need to write the force of air resistance as if it can be derived from a velocity dependent potential.
A common example of this is a particle in a magnetic field. The force of the magnetic field on the particle depends on its velocity, so the potential is written as $\vec{A} \cdot \vec{v}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/343553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.