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Two boys pull a rope in the two edges with the force 50 N each. Will the rope rip if it can handle only a pull of 80 N? I think we should make the sum the two forces 50+50=100 N and say it can't handle? Can you explain me using the 3rd Newton Law?
| This question comes up again and again, in different guises.
If you pull on a rope with a force of 50N, then the tension in the rope is 50N, not 100 N. It's no different whether the other end of the rope has a weight of 50N on it, or whether it's attached to an anchor point in the wall, or to another boy.
Newton 3 talks about the fact that A exerts a force F on B, then A must feel that force too. A cannot feel more or less than the force he is exerting - that would be a short path to violation of conservation of momentum (if A and B are both floating in space, then $F\Delta t$ for A must be equal and opposite to $F\Delta t$ for B so that their total momentum is unchanged; and since they feel the force for the same amount of time, the direction of the forces felt must be equal and opposite).
Nothing in the above doubles the force.
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Parallel Universe Moving Backwards in time We know that different parallel universes- if they truly exist - are governed by different sets of laws. But, Could there be a parallel universe that is moving back in time(?) - in different direction the flow of time in our universe?
| We know that different parallel universes- if they truly exist - are governed by different sets of laws.
No we don't. If parallel universes exist, we know nothing about them. They could have the same laws, different laws, or no laws.
in different direction the flow of time in our universe - ?
The only thing that really defines the "direction" of time in our universe (rather than the rate) is entropy, which increases over time. So a universe could exist where entropy decreases over time under a different set of laws, until it reaches 0 entropy.
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Temperature of electroweak phase transition How does one estimate the temperature at which electroweak phase transition (EWPT) occurred? Somewhere I have read it is around 100GeV but the reason was not explained.
| Let's define $T_{EW}$ the temperature where the coefficient $m^2_H(T)$ of the operator $H^2$ in the SM lagrangian vanishes:
$$
m_H^2(T=T_{EW})=0\,.
$$
For $T>T_{EW}$ the Higgs vev is vanishing, the EW symmetry in unbroken, and the elementary particles are massless. For $T<T_{EW}$ the the vev is non-vanishing, $v_T\propto -m_H^2/\lambda\neq 0$, the EW is broken spontaneously, and the various elementary particles acquire masses $m_i$.
Let's now determine $T_{EW}$ quantitatively. Since $m_i\simeq 0$ for $T\simeq T_{EW}$ is OK to make an expansion in small $m/T$. In this regime the 1-loop corrections to the Higgs potential from the thermal propagators are, expanded at leading order in $m/T$, given by
\begin{equation}
m^2(\tilde{T}_{EW})=m^2_{T=0}+\tilde{T}_{EW}^2\left[\frac{y_t^2}{4}+\frac{\lambda}{2}+\frac{g^{\prime\,2}}{16}+\frac{3g_2^2}{16}\right]
\end{equation}
where $m^2_{T=0}=m_H(T=0)=-\lambda v_{T=0}^2$, with $v_{T=0}=246$ GeV and $2\lambda v_{T=0}=m_h^2=(125\mathrm{GeV})^2$. The negative mass-squared receive positive thermal mass contributions form the loops of the particles the Higgs couples to.
The leading contribution to $m_H^2(T)$ is, non-sorprisingly, coming from the largest couploing, top quark yukawa $y_t=\sqrt{2}m_t/v_{T=0}$. Neglecting the gauge couplings and the higg-self coupling, and solving $m^2(\tilde{T}_{EW})=0$ for $T_{EW}$ we get
\begin{equation}
\tilde{T}^2_{EW}\simeq m_h^2 v^2/m_t^2\simeq (178\mathrm{GeV})^2\,.
\end{equation}
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What are galactic speeds measured against? The Earth moves through space at 67,000 MPH. The Milky Way travels through a local group at 2,237,000 MPH.
Wouldn't you need a fixed point to be able to measure velocity against? After all, compared to the total speed of our Milky Way, the Earth isn't moving through space. What fixed points do we compare against? When we say the speed of light is $c$, what is that relative to?
| There are a number of different frames of references.
For the velocities of celestial objects we use: (i) The geocentric frame: This is a velocity measured with respect to the Earth's centre. Obviously this is quite useful for artificial satellites, but also for things like meteors. (ii) The heliocentric frame: this is the velocity as seen from the centre of mass of the Sun. Heliocentric corrections to measured velocities from the Earth correct for the Earth's rotation and the motion of the Earth relative to the Sun (corrections of order tens of km/s). This is often the frame used for stellar velocities. (iii) The barycentric frame: This is similar to the heliocentric frame, but now referred to the centre of mass of the solar system. The difference between the two is only of order 10m/s, but this is important when discussing the velocities of stars when looking for doppler shifts due to exoplanets. Also crucial when looking at timing analysis from pulsars. (iv) The local standard of rest: this is set to follow the mean motion of objects in the vicinity of the Sun. The Sun actually moves at about 20 km/s with respect to the defined LSR. This frame is often used for discussing the motions of objects in our Galaxy (orbits of stars around the centre etc.).
The speeds of extragalactic objects are usually heliocentric velocities - precision is not usually an issue. However if one wished to convert heliocentric/barycentric velocities to the frame of rest of the cosmic microwave background, then the solar-system baycentre moves at $368\pm 2$ km/s in the direction $l=263.85 \pm 0.10$, $b=48.25\pm 0.04$, where $l,b$ are the Galactic latitude and longitude in degrees.
Your last sentence is answered in the comments. The speed of light is a defined quantity and is the same in all frames of reference.
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Does spacetime have a "mass" value? or What is "Spacetime" made out of? When measuring the MASS within the Universe, does "space" or "spacetime" have a value?
I only ask, because when speaking of expansion, space is expanding. Could it be possible, to reverse the process, by "uncreating" space?
For if it can be created, surely it can be destroyed. And if it is "created", it must be created out of something. What is this "something", and is it only gravity that can manipulate it.
| Given that all existing matter and space had a common origin, it seems reasonable to assume that there is a possibility that if you could calculate the net mass of all the matter in the universe this value might be matched by a negative mass of what remains (i.e. space). If this was true then space itself would have a negative mass, and given this mass would be spread out amongst all existing space the negative mass would be very small when compared to a matching volume of matter. Whilst this might sound like strange thinking you have to factor in that if this was true then large amounts of space (e.g. the space between galaxies) might go some way to explaining the cause of why the expansion of our universe is actually accelerating rather slowing down.
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Does lunar module need the same amount of fuel for landing and take off? Let's assume there is no atmosphere and let's assume there is no change in weight due to fuel consumption, will reactive rocket need the same amount of fuel for landing on a planet as for take off?
In theory - I think - you need the same escape velocity to get rocket to orbit as you need to break it to 0 speed after free fall from the orbit, but this changes if the descend is slower than free fall, is that right?
What is the real world(moon) situation in case of lunar module? (extrapolating for fact that the Apollo Lunar Module leaves the descend stage behind)
| Alternative - just because it's fun to think about these things.
Moving away a bit from the traditional rocket science, in principle you could land with very little fuel and a good set of wheels/brakes: apply a small fuel burn to change your orbit just enough to make a glancing pass at the surface of the moon, then apply the brakes as you touch down. You dissipate the energy as heat with very little fuel burn (passing the momentum to the moon instead of to the burnt fuel). But you can't easily regain momentum without using a source of energy (regenerative braking?). This would make the take-off much more costly.
Note that the orbital velocity at the surface of the moon required for this maneuver is approximately 1500 m/s - as I said you would need a very fine set of wheels / suspension...
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Light in a box with positron walls instead of electrons I understand that with a hollow cube with the inner walls covered in mirrors given a light source briefly, the light would eventually be absorbed. This is due to electron excitation I believe. So suppose the cube had walls of positrons instead of mirrors, since there are no electrons what would happen?
| I'm imagining a box made completely out of anti-matter so that your situation is realistic.
Positrons are the antiparticle of the electron (i.e., the anti-matter equivalent). Meaning, in this case, they're identical to electrons except for charge.
Photons, though, have no charge. So don't give a hoot whether a charged particle it's interacting with is positive or negative; the resulting interaction will be the same regardless.
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What do units like joule * seconds imply? I can easily understand what divisive units imply, but not what multiplicative units imply.
What I mean is, when I read "$12 \:\mathrm{eggs/carton}$", I mentally convert it to, "There are 12 eggs for each carton". I get that.
But, when I see units like joules * seconds, $\:\mathrm{J \cdot s}$, it really bothers me for some reason. I'm not sure what it's saying. It's not saying the amount of joules expended over a certain time-frame, otherwise it would be joules/second, wouldn't it? So, when I read units like joules * second, I try to relate it to my egg-carton analogy, and it comes out like this:
"There are 12 eggs for each reciprocal of a carton."
It just doesn't make any sense for me. Is this a problem for anybody else? Should this be on math SE? Is this even a valid question or am I just asking why $2 + 2 = 4$?
| Building on the answer that Johannes gave:
Imagine you have a broad band source of energy, transmitting on multiple frequencies. It might be the sun, or it might be a RF transmitter, or ...
Now we are going to measure the energy transmitted in each wave band for a certain time, and plot this as a function of the frequency observed. This leads to a plot with Hz along the X axis and J/Hz (the energy density per unit frequency, that is the energy measured divided by the bandwidth of my receiver) along the Y axis.
And boom - out of nowhere I have a unit J/Hz (or J sec) which makes physical sense...
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How does airborne laser system deflect the laser beam? In the Airborne Laser system, the laser beam is directed from the nose of a Boeing airplane towards the target. In this system, the angle of laser beam with respect to airplane body can be changed at the nose of the plane.
What kind of material can deflect such high energy beam? Can a missile get protected from the laser by deflecting the light, if the same material is coated around the missile body?
| The ABL beam director system is, or at least was, available in schematic form somewhere on the LockheedMartin website.
To answer your specific question: the mirrors in the system are highly developed to withstand high-energy laser beams without damage. The coatings are quite specialized and the substrates are designed for efficient heat-sinking. Even so, there are sensors to correct not only atmospheric-induced beam aberrations but aberrations due to the internal optics, platform (i.e. the aircraft) vibrations, and so on.
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What can we learn from a band structure diagram? Other than the band gap and its magnitude, what are the things that we can immediately learn about the properties of the material just by glancing at its band structure?
Can we say something about the bonding? Can we see if the material exhibits some exotic properties like whether it is a topological insulator or not? Etc, etc..
| Depending on what is shown in the band diagram, you can see if it's a topological insulator. Most band diagrams just show what's going on in the bulk material -- what you'd get with an infinitely large chunk of material with not surface. However, to be a topological insulator, you need surface states that conduct. These can be (and sometimes are) drawn on top of the bulk band structure. Since the surface states conduct, they have no band gap. See the first figure in the topological insulators wikipedia article.
As @engineer noted, you can also see if the material has a direct or indirect bad gap. The final thing I can think of is that by looking at the curvature of the bands near their extrema, you can get the effective mass of the electrons and holes.
I'm sure there are other uses as well.
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Using infrared temperature sensor to measure water surface temperature Can I use infrared temperature sensor (such as TS118-3) to measure water surface temperature?
I'm afraid some effects such as reflection of infrared waves from water surface and blinking because of the water ripple will make it impossible.
| Yes, water is the ideal material to measure because water (specifically ice water at 0 °C) is the material used to calibrate IR sensors for temperature readings. You do not have to worry about "reflection" as IR sensors do not themselves emit any of the IR (see page 3) used in the measurement, they just receive the IR emitted by the target.
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Close or open circuit? A electric source provides a non electrostatic influence on the charges inside the source which pushes the positive charges from the negative terminal to the positive one. Does this happen when the source is part of a close circuit or of an open one?
| I know it is the back electromotive force in a circuit carrying coil for which such case occurs. Most electrical components carry coil , that's why back emf occurs in open or close loop circuits because every coil discharging current must produce magnetic field and such fields comes with polarity and will likely attract or repel neighbouring fields.
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Is there a commonly used unit of measure (other than temperature units) that is not absolute? I live in a country where we use Degree-Celsius(°C) to measure the temperature.
Sometimes from one day to the other, the temperature rises from 10°C to 20°C and I hear people say, "Wow! Today is twice as hot as yesterday!".
I try to explain that today is not two times hotter than yesterday because Celsius(°C) is not an absolute unit of measure, that is 0° does not mean the absence of temperature.
If I convert Celsius(°C) to Fahrenheit(°F) or Kelvin(K) it gets clear, but I wish I could provide another example of non-absolute unit of measure to clarify things.
In short, do you know any other unit of measurement (except for temperature ones) where 0 does not mean the absence of the physical phenomenon that is being measured?
| Fahrenheit is not "absolute" either, by your definition. 0 degrees Fahrenheit is equivalent to -17.8 degrees Celsius; absolute zero is -460 degrees Fahrenheit.
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Is it possible to "focus" a radio wave to target an area much smaller than its wavelength? Recently I was reading about a technology that uses radio waves to stimulate neurons to fire. The radio waves have the advantage of being able to pass through the skull (hence being non-invasive) but they are very coarse. Would it be possible to target individual neurons with radio waves (for example by having a bunch of low-powered radio waves converge on a focal point only nanometers large without diffusing to surrounding tissue)?
http://sfari.org/news-and-opinion/toolbox/2012/radio-waves-turn-on-gene-expression
| something like the microwave generator in the microwave oven can shoot out energy waves of different spectrum. and its is relatively focused like the laser beam.
human brain wave of thoughts present in brain as small area patterns. mind control/manipulation and detection hence achieved.
| {
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What is the initial angular momentum of a rigid body given an offset impulsed force? What is the imparted angular momentum to a rigid body if the impulse force is offset by a distance $h$ from the center of mass and the imparted momentum from the center of mass is $mv$?
For a homogeneous sphere I said the imparted angular momentum is $L=mvh$, but I am not sure if that is correct.
| When you have an impulse $F\Delta t$ (I prefer that notation over $m\Delta v$ because it allows impulse to be imparted without worrying about the mass of the thing giving the impulse), then
*
*The momentum of the center of mass changes as though the impulse was applied there, so $$m\Delta v = F\Delta t$$
*The angular momentum changes according to the torque imparted $$\Delta \vec{L} = \vec{F}\Delta t \times \vec{h} $$
So yes, you got it right.
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Why are some elemental materials grey? How does grey occur in elemental materials such as metals? I believe that grey arises from the simultaneous reflection and absorption of all colors of the spectrum (in different atoms of course), as mixing opposite colors on the spectrum would do. How would this occur in a one-element material? How does this occur on an atomic level?
| A clean un-oxidized metal surface is not usually grey but rather reflective like a mirror.
In fact, mirrors are constructed by coating a sheet of glass with a thin layer of metal atoms.
The grey color of a slightly oxidized surface, which includes almost metal surfaces you encounter in daily life, look grey because some of the ambient white light is absorbed.
How would this occur in a one-element material? How does this occur on an atomic level?
Metals have relatively mobile electrons.
Therefore, when an electromagnetic wave comes in, the electrons can move around easily under the influence of that wave's electric field.
That means that an incoming wave causes the electrons to oscillate at the same frequency as the wave itself.
Therefore, the electrons in the metal surface emit radiation at exactly the same frequency as the incoming wave.
In other words, the light is reflected at the same color as it had before.
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How can the thrust due to radiation pressure be amplified in photonic laser thruster? The thrust is amplified due to repeated bouncing of photons between two mirrors as shown in the diagram in this: Why does repeated bouncing of photons produce amplified thrust when the answer in 'Mirror problem of radiation pressure' indicates that the radiation pressure will only be doubled?
| The answer is misleading. If you consider one photon it is giving each mirror a kick every T seconds. There is no "continuous pressure". Now multiply that by N photons.
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Does a positive or negative charge attract a neutral object?
Three objects are brought close to each other, two at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that:
(a) objects A and C possess charges of the same sign.
(b) objects A and C possess charges of opposite sign.
(c) all three of the objects possess charges of the same sign.
(d) one of the objects is neutral.
(e) we need to perform additional experiments to determine information about the charges on the objects.
From this question, I thought we can create two scenarios:
scenario 1: | This is the phenomenon of electrostatic induction.
Since "neutral" objects are made out of many positive and negative charges in equal measure, some of which can move, the presence of an electric field from a charged object will move these charges, and result in a region of opposite (to the object creating the field) charge where the neutral object is nearest to the charged object, and this will indeed result in an attraction between the formerly neutral object and the charged object.
Therefore, you cannot conclude from the attraction of two conducting objects that they must have the opposite charge - one of them may well be uncharged.
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Selection of system in Conservation of momentum I came across a question in which a cart is moving and having sand. Suddenly the sand valve malfunctioned and the sand starts falling from the cart. So momentum of which system will remain conserved.
| First, define clearly what is in The System; it can be anything you want. This definition is then fixed.
If any part of this system experiences a force from outside the system in a particular direction, then the total momentum of the system in that direction is not conserved.
If all the forces on any part of the system are from other parts of the system, then the momentum of the system is conserved.
If you define the system as just the cart, and the sand falls vertically out through the bottom of the cart, then the sand exerts no horizontal force on the cart, and the momentum of this system is conserved.
If you define the system as the cart plus original amount of sand, then, as the sand falls out and hits the ground, the ground exerts a horizontal force on part of the system, and the momentum of this system is not conserved.
Edited to address comment:
Suppose you define the system to be the cart and all the sand in it at a particular instant, $t_i$, and allow a small amount of this sand to dribble out the valve and start to fall to the ground. Let's consider such a small amount of sand that the first part of the dribbling sand doesn't have time to reach the ground before the last part of the small amount leaves the valve
There are no external horizontal forces acting on this system, so the total momentum is conserved. The falling sand has no horizontal forces acting on it, so that part of the system has no change in momentum. So the cart and the remaining sand (after the dribble is over) also has its momentum conserved.
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Total angular momentum in QFT
Can anyone show explicitly how the QFT total angular momentum operator $$\hat{\vec{J}} = - i \int \frac{d^3p}{(2 \pi)^3} \hat{a}^{\dagger}_{\vec{p}} ( \vec{p} \times \nabla_{\vec{p}}) \hat{a}_{\vec{p}}$$ gives $$\hat{\vec{J}} |\vec{0} \, \rangle = 0~?$$
Derivation of all the above is here. The question is essentially the same but I can't really implement the existing answer mathematically.
$|\vec{0} \, \rangle$ being a momentum eigenstate.
| If the state you're using is just the vacuum, then my comment to your question applies. If otherwise $|\mathbf 0\rangle = a_{\mathbf 0}^\dagger|0\rangle$, then just use that $[a_{\mathbf p},a_{\mathbf q}^\dagger] = \delta_{\mathbf p,\mathbf q}1$, with $\mathbf q=\mathbf 0$ to fix the integral at the term with $\mathbf p = \mathbf 0$ through the Dirac delta $\delta_{\mathbf p,\mathbf 0}$.
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Diagonalizability as a measure of uncertainty (discrete case) I have seen two characterizations of the problem in measuring a discrete variable of a state ψ exactly with each of two non-commuting Hermitian operators A and B:
(1) that the product of the standard deviations ( = √(<ψ|A2|ψ>-<ψ|A|ψ>2), & ditto for B) ≥ 1
(2) that one cannot simultaneously diagonalize the matrix representations of A and B
(i.e., if A = U†CU and B = V†DV, for unitary U and V and diagonal C and D, with † denoting the adjoint, then U≠V.
Where is the link between these two?
| If you have a situation where $[A,B] = iC$, and $\omega$ is any state of the C*-algebra containing $A,B,C$, then you have the relation
$$\Delta_\omega A\Delta_\omega B\geq \frac12|\omega(C)|$$
which is an answer to your first question, provided that, e.g., $[A,B] = i2\cdot\mathbf 1$.
Part two is basically the fact that, unless $A$ and $B$ commute, their product is not self-adjoint, and therefore it won't diagonalise. This violates the fact that, if they were diagonalisable, then their product would be diagonal.
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What is dimension and how many types of dimensions are there in the universe? What is dimension and how many types of dimensions are there in the universe? I mean how many total dimensions are there? I have only heard about 2d and 3d. Other than these two, are there any other dimensions. If yes than please explain each of them clearly.
| In the mathematics used for physics dimensions are independent mathematical fields on which a variable can be assigned . Independent means that in the algebra used each projected on the other gives zero, is orthogonal. The example is the three dimensions we live in which are assigned orthogonal directions and the field is the real numbers on the axis.
So classical physics uses these three dimensions to model mathematically all observations on macroscopic scales, and in this formulation time is a parameter.
For very large energies it was experimentally found that the mathematics was most efficient by assigning a dimension to time and treating it as the fourth dimension whose mathematical field is an imaginary number, i.e. the real mathematical field multiplied by the square root of minus one. Thus there are four verified space dimensions as far as physics goes.
In the quantum regime, where the size of the items under observation is very small, new theories are proposed in trying to formulate an overall theory. These may have many extra space dimensions and some even time dimensions. String theories as an example. Until they are validated experimentally it is a moot question whether their extra dimensions are observable in our physical reality.
Thus, as far as physics goes, there are four dimensions over all at the moment, three of space and one of time.
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What is antimatter? Can you give a visual example of what is antimatter?
With the re-opening of Large Haldron Collider scheduled in Mar 2015, I'm reading that they smash two particles together to try to re-create particles that might have been there are the beginning of the Big Bang, and this includes antimatter?
Is antimatter something we can see, or it is some invisible field. I'm trying to get my arms around this concept.
| Antimatter is simply matter with an opposite charge in each particle. It will (probably) look and behave exactly the same as regular matter.
However, no one has made enough antimatter to actually test that it does look the same. The major obstacles are the energy cost and the handling problems. Antimatter has to be built up from pure energy, and the stuff has the nasty habit of exploding violently when it contacts regular matter. The Hiroshima bomb turned about 700 milligrams of matter into energy, you could do the same thing with 350 milligrams of antimatter - that's a ball of anti-iron about 5mm across. Even the tiniest leak in your containment device will result in your lab, and anything else within a kilometer, simply disappearing.
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Who is doing the normalization of wave function in the time evolution of wave function? In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
| Nobody is "doing the normalization".
Normalization is not even necessary. We often normalize for convenience, since that means that the Born rule for $\lvert \psi \rangle$ being the state $\lvert \phi \rangle$ reads
$$ P(\psi,\phi) = \lvert \langle\psi\vert\phi\rangle \rvert ^2$$
which is certainly easier to recall/write than
$$ P(\psi,\phi) = \frac{\lvert \langle\psi\vert\phi\rangle \rvert ^2}{\lvert \langle\phi\vert\phi\rangle \rvert \lvert \langle\psi\vert\psi\rangle \rvert }$$
but nothing in the formalism forces normalisation. The basic principle says that states are rays in the Hilbert space, so that $\lvert \psi \rangle$ and $c\lvert \psi \rangle$ represent the same state for all $c \in \mathbb{C}$, and are, for all purposes, fully equivalent representants of the same state. (This, by the way, means that if we want a space where every element corresponds to a distinct quantum state, we should look at the projective Hilbert space instead)
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Why is kinetic energy defined as $(1/2)m v^2$? What is special about $(1/2)m v^2$ that makes physicists believe that it is a representation of kinetic energy?
| It's the work done to accelerate a particle from rest to a final speed $v$, i.e. the energy "put in" to initiate motion.
Work $W \equiv \int_0 ^{v} \,F \, dr$, with $F = ma = m\frac{dv}{dt} = m \frac{dv}{dr}\frac{dr}{dt} = mv\frac{dv}{dr}$, so:
$$W = \int_0 ^{v} \,F \, dr=\int_0 ^{v} \, mv'\frac{dv'}{dr} dr = \int_0 ^{v} \, mv'\,dv' = \frac{1}{2}mv^2$$
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Sign of the totally anti-symmetric Levi-Civita tensor $\varepsilon^{\mu_1 \ldots}$ when raising indices I am confused with the sign we get when we want to raise or lower all indices of the totally anti-symmetric tensor of any rank. Take the metric to be mostly plus ($-+\ldots+$). Then is it
$$\varepsilon^{ijk}=\varepsilon_{ijk}$$
or
$$\varepsilon^{ijk}=-\varepsilon_{ijk}?$$
so I am confused to as which one is true.
And if we consider higher rank does something change? For example
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}$$
or
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}?$$
| Sean Carroll's Spacetime and Geometry has a thorough discussion of this, and, even better, this discussion is in the lecture notes that turned into the book (see Chapter 2: Manifolds).
In full generality (at least for any right-handed coordinate system), we start off with the symbol $\tilde{\epsilon}_{\mu_1\cdots\mu_n} \equiv [\mu_1\,\cdots\,\mu_n]$, which is $0$ or $\pm1$ depending on the sign of $\mu_1\,\cdots\,\mu_n$ as a permutation of $0\,\cdots\,(n-1)$. Then the tensor with lower indices obeys
$$ \epsilon_{\mu_1\cdots\mu_n} = \sqrt{\lvert g \rvert} \tilde{\epsilon}_{\mu_1\cdots\mu_n} = \sqrt{\lvert g \rvert}\ [\mu_1\,\cdots\,\mu_n], $$
where $g$ is the determinant of the metric.
Some people define a symbol with upper indices that is the same as that with lower indices, but on the other hand some people include an extra factor of $\operatorname{sgn}(g)$. In fact, this is ambiguous enough that Carroll states one version in the linked notes, and the other version in the final published book.
In any event, the tensor with upper indices is unambiguously
$$ \epsilon^{\mu_1\cdots\mu_n} = \frac{1}{g} \epsilon_{\mu_1\cdots\mu_n} = \frac{\operatorname{sgn}(g)}{\sqrt{\lvert g \rvert}} [\mu_1\,\cdots\,\mu_n]. $$
In the end, you should be using the full tensors in calculations, at least if everything else in the formula is a true tensor. In flat spacetime, the metric is $\operatorname{diag}(-1,+1,+1,+1)$, so $g = -1$ and we have
\begin{align}
\epsilon_{\mu_1\cdots\mu_n} & = \phantom{-}[\mu_1\,\cdots\,\mu_n], \\
\epsilon^{\mu_1\cdots\mu_n} & = -[\mu_1\,\cdots\,\mu_n].
\end{align}
| {
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Are there new physics scenarios that predict low lying hadrons? There is a significant ongoing experimental effort to search for new hadrons with masses in the GeV range. This is used to find the spectra of QCD bound states, with a particular emphasis on finding exotic resonances such as the tetraquark. To my knowledge, they have not found any state whose mass is in contradiction with the theoretical prediction using lattice QCD thus far (though e.g., there are a few tetraquark candidates, such objects are not in conflict with lattice predictions).
These searches are clearly very important as they confirm our understanding of QCD and in particular, they verify the validity of lattice QCD, which can subsequently be used to study new phenomena. But my question is, are there any mainstream new physics scenarios which predict a deviation in the spectra of QCD and could be found at for example, LHCb?
EDIT: I'm interested in changes to the low energy ($\sim$ GeV) bound state spectrum measured by these experiments
| There are none. The reason is that the mainstream extensions of the Standard Model leave its gauge group intact at LHC energies and only add matter to the spectrum.
Such a modification will not lead to new hadronic states. Exotic particles with masses comparable to quark masses are mostly excluded (unless there is some intricate hiding scheme, e.g. for light stops).
In general mingling with the ingredients of the low energy bound spectrum is both "dangerous" and not very fruitful from a model building point of view - the low energies is not where the SM has its troubles; on the contrary it is the low energy bound spectrum where the Standard Model shines.
tl;dr:
Since there is no need (and little room) to build models affecting the low energy spectrum nobody does.
| {
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Defining Reference Directions for Voltage and Power (sign convention)
My professor decided to use the above reference directions when calculating power in circuits. He says that when power > 0, power is consumed. When p < 0, power is generated.
This definition is counter intuitive to what I would have assigned--I would have said negative means power is consumed and positive power means power is generated.
So my question is this: Are these sign conventions actually what is happening physically? In other words, when you look at the diagrams above, what determines (or what are the formal conditions based on the parameters above) that a system is generating or consuming power?
Do we understand it like this: On the left diagram, charge is flowing from "high" to "low" potential so charge is being used. And on the right, charge is flowing in its "opposite" direction so power is being generated?
| When we discuss power in the general physics sense, we mean the rate at which energy is transferred from one system to another, or possibly one form to another. The sign convention your professor specified is very common, and yields the power consumed by the device, taking energy out of the charge flow system and putting it somewhere else at a certain rate. This somewhere else might be internal energy of a resistor, or energy of the EM field of a transformer or capacitor or inductor, etc. In this convention, a negative number for the power means that energy is being transferred into the charge flow system.
The nice thing about allowing for negative power consumption is that for closed circuit systems (i.e., all sources and sinks are shown in the circuit diagram), the total power consumed is zero, so you can check your solutions of voltages and currents.
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Is gravity a force? If gravity just emerges from the curvature of spacetime, is it actually a force?
Why is it one of the 4 fundamental forces of nature?
| It could be or it could not be. It's really not clear. You still need some sort of means to communicate with space-time and tell it to curve. This is where gravitons could potentially come into play as the force mediator, and that would technically make gravity a fundamental force. It is certainly expected to be quite different from the other 3 forces. But I'm certainly not an expert in GR and could be mistaken.
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If wave speed is dependent on medium only, then how to reconcile $v\propto f$? I have read and learnt in many places that velocity of a wave depends only on the medium through which it travels. It is clear from this that the velocity of a wave doesn't depend on the frequency of the wave because both the sound of a roaring lion and crying baby reaches our ear with the same speed. But we also know that $\text{speed} = \frac{\text{distance}}{\text{time}}\ \Rightarrow\ v=\frac{\lambda}{T}\ \Rightarrow\ v=\lambda f=\text{wavelength}\times\text{frequency}$. In this derivation, velocity is found to be dependent on frequency.
Can anyone please explain this contradiction? Is there any fault in my perception of the concept?
| When pitch of a voice is changed, both the wavelength and frequency change. For example, a higher pitch will have a higher frequency (of course) but a smaller wavelength.
Okay, that being said, the independence of the propagation speed of a wave to the properties of the wave itself it an model to use in many circumstances, but not all. I don't know about acoustics, but one example that comes to my mind is in the field of optics; you might find dispersion a useful search term.
And for a more general advice, be careful when using equations involving more than two variables. Think about what is varying and what isn't; the equation alone won't tell you.
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Would a rotating magnet emit photons? If so, what causes the torque that gradually slows the rotation? If a magnet is rotating around an axis perpendicular to the axis north-south axis of the magnet (which I assume to be cylindrically symmetric), in space (so no-gravity/freefall or friction), should it still slow down because it emits electromagnetic radiation/photons?
I would think so, due to conservation of energy. The power output of the oscillating magnetic field should mean a decrease of the rotational energy of the magnet.
But what causes the torque that gradually slows the magnet's rotation? One way of looking at it would be conservation of (angular)momentum and the fact that photons have momentum. But how would you express the torque in terms of electromagnetism/Maxwell's equations?
| Yes, a rotating magnet emits radiation, exactly like a pulsar does. The solution to the Maxwell equations is known (but not very well known, actually). The analytical expressions for the electromagnetic fields $\mathbf{E}(t, \mathbf{r})$ and $\mathbf{B}(t, \mathbf{r})$ are very complicated. About the total angular momentum, lets introduce the rotating magnetic dipole vector $\boldsymbol{\mu}(t)$. Then, the angular momentum lost in the form of radiation is this (you can find this formula in Landau-Lifchitz book, and probably in Jackson too):
$$\tag{1}
\frac{d \mathbf{L}}{dt} = -\, \frac{2}{3} \, \frac{\mu_0}{4 \pi c^3} \, \dot{\boldsymbol{\mu}} \times \ddot{\boldsymbol{\mu}},
$$
where the dot is a time derivative. If the dipole is just rotating, you get this:
$$\tag{2}
\frac{dL}{dt} = -\, \frac{2}{3} \, \frac{\mu_0 \mu^2 \omega^3}{4 \pi c^3} \sin^2{\alpha},
$$
where $\alpha$ is the tilt of the dipole, relative to the rotation axis.
If there's no external agent to sustain the rotation, then the dipole will lose angular velocity (its angular momentum is radiated away).
You could substitue $L = I \omega$ in the left member of (2), to get the same equation as the one you get from the power radiated away.
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How can I prove that $\langle\Omega\vert \phi(x) \vert\Omega\rangle \langle\Omega\vert\phi(y)\vert\Omega\rangle=0$ for a scalar field? From Peskin-Schroeder, p.212:
The term
$$ \langle \Omega | \phi(x) | \Omega \rangle \langle \Omega |\phi(y) | \Omega \rangle$$
is usually zero by symmetry; for higher-spin fields, it is zero by Lorentz invariance.
Where $|\Omega \rangle$ is the ground state and $\phi(x)$ a scalar field.
How can I prove this?
| It's a bit more involved than creation and annihilation operators. What is meant is that the field has no vacuum expectation value, and in a way this is by construction. You may think about why $\langle x \rangle = 0$ for a harmonic oscillator (or better yet, an an harmonic one). It's because $x\to -x$ is a symmetry of the Hamiltonian. It's not true for cubic potentials, though, but those theories can be sick.
In many situations, when we quantize a field, we perform perturbation theory around it's classical solution, and the dynamical field is the total field minus is vev, I.e. we write $\Phi = \langle \Phi \rangle + \phi$, and then work in terms of $\phi$. Thus the vev of the dynamical field vanishes by construction.
Edit:
To clarify, the reason you can't rely on creation and annihilation operators is because $\left\langle \Omega | a |\Omega\right\rangle$ is only zero if $\Phi$ has a zero vev, (circular argument). Imagine trying to naively quantize the Higgs field, which has a "Mexican hat" style potential where the minimum is located away from $\Phi = 0$. In that case, you do not have $\left\langle \Omega | \Phi |\Omega\right\rangle$ until you redefine the field to $\Phi = \langle \Phi \rangle + \phi$.
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Is the shell theorem only an approximation? I've read the shell theorem during gravitation lectures, i.e. I know it states that the net gravitational field inside a 3D spherical shell or a uniform 2D ring is zero.
Now, assume a thin spherical shell. If I put a particle inside the shell, so that it was infinitesimally close to one of the regions of the shell, shouldn't the particle move towards the shell and touch the portion of the shell it was closest to? (Since as the distance goes to zero, the magnitude of the field between the particle and that portion of the shell should be very high, when compared to the field from other regions.)
But in the same case if I apply the shell theorem, the particle shouldn't move at all! Since it states the net gravitational field inside the shell is zero.
Can anybody explain this difference, or if there isn't any, how am I wrong?
| The shell theorem assumes a continuous distribution of matter in the shell.
If you came infinitesimally close to a real, physical shell you would discover that it, too, is made of particles. As you passed through the shell one of two things would happen:
*
*You could crash into one of the particles and experience a non-gravitational force.
*You could pass through a hole in the shell to the outside.
The second case is interesting.
For a continuous shell there is a discontinuity between the force inside (zero) and the force outside (equivalent to a point mass at the shell's center).
For a particle passing through a small hole in a spherical shell that discontinuity gets smoothed out.
If the particle doesn't pass through the center of the hole, the smoothed-out gravitational force will include some component parallel to the surface of the shell; the details depend on exactly how "clumpy" the shell is.
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Supermassive Black holes at centers of galaxies Why there is super-massive black holes at the center of our galaxy and other galaxies ?
| Because at the centre of the galaxy, there is much higher star density, and thus, much higher chances for having huge stars.
These stars will inevitably collapse, forming black holes. These black holes if located at the center of a galaxy, are able to easily gain more mass from nearby stars. Increasing their mass further and further.
So finally, we have super-massive black holes.
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Baryogenesis via Leptogenesis Baryon number is directly violated through electroweak anomaly and so does the Lepton number, for each transition from one vacuum to another. The two violations are of equal amount $\Delta B=\Delta L=N_f[N_{CS}(t_i)-N_{CS}(t_f)]$ since $(B-L)$ is anomaly free. Both the violations (i.e., $\Delta B,\Delta L$) will occur simultaneously for each transition. But in this manner $B$ is violated directly and I do not think that $L$ violation is inducing $B$ violation. Is this true? When standard model is extended with right-handed neutrinos, then decays of heavy Majorana neutrinos give rise additional to $L-$violation. How can this leptogenesis induce baryogenesis?
If the question is not clear enough I can clarify it further.
| When right handed neutrinos are introduced, they imply $L$ violation through their Majorana nature. Their decay into lepton and Higgs doublets
$$N\,\longrightarrow l\,\Phi^\dagger$$
violates the lepton number (since $N$ has zero lepton number being Majorana). These decays take place out of thermal equilibrium, therefore a net lepton asymmetry is produced. However, in the Early Universe ($T\gtrsim 10^9\,\mbox{GeV}$) several other particle physics interactions are efficient. Among them, the so-called sphalerons, that violate $B$ and $L$, while still preserving $B-L$.
By considering all these additional processes which are in equilibrium at high temperatures, it is possible to show that the lepton asymmetry produced by the decay of $N$ (i.e. by the leptogenesis mechanism) is partly converted into a baryon asymmetry. Therefore, leptogenesis can be a viable mechanism to produce the baryon asymmetry of the Universe because the Standard Model itself provides the necessary way to link lepton and baryon asymmetry.
| {
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Which Symmetry class and what kind of topological invariant for $2D -p+ip$? What kind of topological invariants are there for $2D-p+ip$ topological superconductor and to which symmetry class it belongs to?
| All topological insulators can be classified according to their symmetry classes. There is time reversal symmetry ($T$) , charge conjugation symmetry ($C$) and the combination $S=T*C$ symmetry. The $T$ and $C$ symmetries can be either positive or negative, i.e the energy spectrum may change sign under the symmetry operation.
The $p+ip$ superconductor is a $p$-vave superconductor and has positive $C$ symmetry and therefore falls into symmetry class $D$. From the table if the follows that it has an $\mathbb Z$ valued topological invariant.
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What would happen to matter if it was squeezed indefinitely? I hope that this is a fun question for you physicists to answer.
Say you had a perfect piston - its infinitely strong, infinitely dense, has infinite compression ... you get the idea. Then you fill it with some type of matter, like water or dirt or something. What would happen to the matter as you compressed it indefinitely?
Edit:
I'm getting some responses that it would form a black hole. For this question I was looking for something a little deeper, if you don't mind. Like if water kept getting compressed would it eventually turn into a solid, then some sort of energy fireball cloud? I'm not as concerned about the end result, black hole, as I am in the sequence.
| Presumably, it's going to be similar to the Big Bang in reverse, which is also what happens at the singularity of a black hole or any situation where matter is continuously compressed.
(Beyond a certain point, we don't really know.)
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General relativity: is curvature of spacetime really required or just a convenient representation? I'm not really far into the general theory of relativity but already have an important question: are there formulations that can do without spacetime curvature and describe the general theory of relativity/all associated gravitational effects in global cartesian coordinates? Idea: Einstein chose spacetime curvature so that one had not to build gravitational effects into the rest of physics equations like Maxwell's equations. I assume that spacetime curvature provides a convenient way to apply physics laws not related to gravity unmodified locally because at small distances we may use special relativity as an approximation.
Please correct me everywhere I am wrong.
| There are formulations of classical general relativity that do not utilize the concept of the curvature of spacetime, which are equivalent to the traditional formulation that interprets gravity as the curvature of spacetime.
As with anything else, the framework that one chooses to do calculations in is a matter of circumstance, or philosophy, etc... sometimes both. But, history of science has shown time and time again, that it is well worth it to explore alternative formulations of theories that are known to be successful.
Quoted from the abstract of the paper above, "In these notes we discuss two alternative, though equivalent, formulations of General Relativity in flat spacetimes, in which gravity is fully ascribed either to torsion or to non-metricity, thus putting forward the existence of three seemingly unrelated representations of the same underlying theory."
This seems pretty clear that these operate in flat spacetime, i.e. as you say, "global cartesian coordinates." I can write a more thorough explanation if required by the OP.
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Changing of spin for electron? Can we change the spin of electron by applying magnetic field from $\uparrow$ to $\downarrow$ configuration?
| A magnetic field consists of photons. Photons are spin $1$ particles, which means for a given $z$ axis a measurement of spin can yield $+1$ or $-1$. If a photon collides with an electron, we know that spin must be conserved. Let's assume the electron is in a spin up state $\uparrow_e= + \frac{1}{2}$ and the photon in a spin down state $\downarrow_P= -1$. There are two possibilities if these two meet:
$\uparrow_e \rightarrow \uparrow_e$ and $\downarrow_P \rightarrow \downarrow_P$ which means no spin flip
or
$\uparrow_e \rightarrow \downarrow_e$ and $\downarrow_P \rightarrow \uparrow_P$, which is possible because $ + \frac{1}{2} - 1 = -\frac{1}{2} $
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Moving electric charges I just wanted to double-check these three statements, as I'm not entirely sure I understood them completely:
1) A stationary electric charge (let's say a proton) produces electric field.
2) A moving non-accelerating proton produces also magnetic field.
3) An accelerating proton produces electromagnetic waves.
Is that true? Does an accelerating proton also produce magnetic field or only electromagnetic waves? What is the rationale behind the third sentence? Is it because some of a proton's energy gets lost in the process of changing velocity and transfers into the electromagnetic wave energy?
| For all of these points it is important that you ask 'relative to what' because the idea of an electric and a magnetic field was shown by Maxwell to be a different way of looking at the same thing. So if we were in the frame of reference of a charged particle moving past another charged particle (which would look like a charged particle moving past us in the opposite direction!) we could see a magnetic field rather than an electric one which could be the opposite of what the other particle sees
So to answer your points, number 1 and 2 are true and for point 3: There are two separate questions here. 1) The idea of a field in quantum field theory is that it can be thought of as an area of effect caused by waves from a source. So whenever you think about a field you can think about things in the field being effected by EM radiation and 2) You might be talking about 'synchrotron radiation' which is, as you say, generated by forcing charged particles to accelerate...so yes to point 3 too.
Cheers
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Gravity and bottomless pits Assuming that someone is theoretically able to make a hole through the center of a large planet, and then jumps down the hole, what will happen? Given my understanding of gravity and energy, my estimate is that in the absence of any resistive forces, the person would fall right through to the other side, with at least part of their body coming through, and would continue oscillating like this. If there were resistive forces, such as air resistance, the person would oscillate similar to a spring winding down, eventually stopping in the middle of the planet. Is this what would happen, or are my calculations off somewhere?
| You are correct about what you believe would happen.
Here's an important fact:
Consider yourself positioned at a random location within a solid planetary sphere. Let that point define a radius from the center of the sphere to you. Any matter that lies beyond that radius has a ZERO net gravitational pull on you. In other words, that mass does not matter at all to you. Only the matter within that radius has a net gravitational pull on you.
Also, if the object was a spherical shell and you were ANYwhere inside that shell, you would just float around. Your location within that shell wouldn't matter at all. Yes, you could be closer to one section than another, but, the closeness to the smaller amount of mass would be exactly equalized in force by the much greater amount of mass further away from you. Calculus easily proves this with either double or triple integrals.
Please don't jump in the hole. Your velocity would be tremendous as you reached the center. One brush against the side of the hole and you'd get a heck of a burn.
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What kind of object is the Landau--Lifshitz pseudotensor? I understand that it's called a pseudo-tensor because it's not a tensor. Wikipedia says
most pseudo-tensors are sections of jet bundles, which are perfectly valid objects in GR.
Refer Here
Is the Landau-Lifshitz pseudo-tensor a section of a jet bundle?
| Because the Landau-Lifshitz pseudotensor behaves as a tensor only with respect to restricted coordinate transformations, it would be considered as a part of the jet bundle within the manifold when used to this end only.
| {
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Why are Majorana fields usually used to introduce gravity in the Rarita-Schwinger Lagrangian? When first introducing the gravitational interaction for a spin-3/2 Rarita-Schwinger field, Majorana fields are usually used (see for example here at chapter 4, or in Ramond, (6.4.112) ).
Why is this? What are the advantaged of imposing a Majorana condition in this context?
| In this context, the word "Majorana" doesn't mean that it is the "real one-half" of the ordinary simple Dirac spinor. It means that it is any half-integer field with a reality (Majorana) condition!
For example, in the Chapter 4 of the DAMTP lectures, the field introduced is a standard spin-3/2 field with one spinor index and one vector index, as required for a gravitino. The gravitino's spin has to be 3/2 because it differs by 1/2 from $j=2$ of the gravitons, and $j=5/2$ is already too much and would require too large gauge invariance.
The Majorana reality condition is imposed on the gravitinos simply because the metric tensor is naturally a real (or, quantum mechanically, Hermitian) field which is why it must be possible to naturally impose the reality condition on the superpartner field, the gravitino field, too.
The minimum $N=1$ supersymmetry is generated by 4 real supercharges which may be imagined to combine to one Weyl (complex chiral) spinor or one Majorana (real non-chiral) spinor. Using the latter approach, the gravitino superpartners of the real graviton by the real supercharges are real, too.
Analogously, we usually describe the gauginos, superpartners of (naturally real) gauge bosons, as Majorana fermions while the superpartners of (complex but not only complex) scalar fields are naturally Weyl fermions – even though the information in one Weyl fermion and one Majorana fermion is really "the same".
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What does 'channel' mean? I see many plots like the following that graph counts per channel,
I know what a 'count' is, but I don't know what a 'channel' is. Could somebody please explain to me? My guess is that it is that each channel represents a bin in the histogram, and so channel 453 would be the 453$^{\textrm{rd}}$ bin in the histogram starting from the first bin which would most likely correspond with the, in the case of energy measurements, lowest energy accepted into the histogram.
[Sorry for my ignorance]
| In this case a "channel" is a separate register in the data acquisition (assuming the data comes from a MCA or other ADC driven collection device).
Each channel represents a discrete range in some input to the data acquisition system (generally time or charge) which is linked through the physics of the detector to some physics quantity of interest.
In this case the physics quantity appears to be the energy of a gamma ray (which has presumably been converted to light in a scintillator and the light converted to charge in a PMT or similar detector element).
Or to talk only in terms of your graphs, each of the data points on the graph represents the number of counts that fell into a particle energy range. Those ranges are called channels.
It is, perhaps, worth emphasizing that the word channel gets applied in many different contexts and has very different meanings there. In a context like "t-channel scattering" they refer to ways of drawing the fundamental Feynman-diagrams that can give the same observables. In the context of particle physics data reduction they might refer to different observables that can indicate the same underlying processes or to different processes which can generate the same observable. And I know the word is widely used in other sub-fields as well, but I won't expose my ignorance by trying to give any examples.
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When is a unitary operator a quantum gate? Quantum gates we use like X, Y, Z, H, CNOT, etc. are all unitary. When can an arbitrary unitary operator be considered as a quantum gate?
| Quantum gates are all unitary transformations on a state of qubits. Any unitary transformation can be considered a "gate", although the ones you mention are primitive ones from which others can be constructed. More complex ones are usually referred to as circuits. The two qubit gates, $\mathrm{H}$, $\frac{\pi}{8}$ and $\mathrm{CNOT}$ are considered universal gates, because any gate set can be constructed out of those.
You may want to take a look at these lecture notes, in particular, Lemma 12. I would also suggest getting a hold of the textbook by Nielsen & Chuang.
Addendum:
I said something incorrect about the Toffoli gate, which is universal for classical computation, but as Peter Shor pointed out in the comments, will not give you complex entries (both Hadamard and Toffoli are real).
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Why does the surface structure of a metal make it hydrophobic? I was just reading this article from phys.org describing water-repellant surfaces.
However the article doesn't go into enough details of explaining why a particular structure repels the water.
Can someone please explain why the water molecules react to a particular pattern on the surface of a metal in this way?
| The article Gowtham linked to seems to be the one you want. Basically (if I understood correctly), if the material is already hydrophobic (water is more attracted to itself than the material), the surface tension of the water prevents it from filling small empty spaces in the surface which will remain filled with air. Thus the contact surface and attractive force between water droplet and the surface will be very small.
The pattern itself is quite simple. It is simply very small empty spaces the walls of which are also full of even smaller empty spaces. This together with the material means that barring some outside pressure forcing the water into the spaces, surface tension will keep it out. Gravity will not be enough.
And in fact if you force water into the surface it will try to bounce back as long as water is less attracted to the original matter than it is to water. Water trapped in the small spaces will be more attracted to the rest of the drop than it is to the surface and if the drop is small enough and the space is small enough, this attraction wins over the gravity trying to force water to the surface.
Hope this is in the correct direction AND at least somewhat understandable.
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How does "pushing-start" a dead-battery manual car work? A few days ago the battery of my car went (almost) dead. As it is a manual car, my father once told me that the way to get it going without jumper cables was to push it or let it roll down a hill, sink the clutch, shift to 2nd gear and then let go the clutch.
After the joy of being able to 'revive' the car, I got to wonder of the reason for why this works. All I could think of was Electromagnetic Induction, however I couldn't find anything on the web to support this. I'm sure that there might be all sorts of engineering details, so I'm only looking for the physical principles involved and the basic process that make this work.
|
I'm sure that there might be all sorts of engineering details,
Not really. An electric starter motor requires a large amount of power to turn the engine over. If the battery is nearly exhausted, it cannot provide the power to rotate the electric starter motor.
However, as long as there enough power from the battery to run the car's electronic engine controller and ignition system, the engine can be turned over by non-electric means and started.
For older designs without electronic ignition and controllers, no battery is required at all.
I was recently reading about the Coffman starter which used what looked like shotgun shells (cartridges) to start aircraft piston engines. If you've ever watched the movie "Flight of the Phoenix", there is a dramatic scene involving a Coffman starter.
In early automobiles, the engines were often started with a hand crank.
| {
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Limit as $x_1 \to x_0$ for the propagator of the harmonic oscillator
Consider a non-relativistic particle of mass $m$, moving along the $x$-axis in a potential $V(x) = m\omega^2x^2/2$. use path-integral methods to find the probability to find the particle between $x_1$ and $x_1 + dx_1$ if the particle is at $x_0$ at time $t = 0$.
One finds the propagator to be $$P(x_1, t_1; x_0, 0) = \sqrt{{{m\omega}\over{2\pi i\hbar \sin \omega t_1}}}e^{{{im\omega}\over{2\hbar \sin\omega t_1}}((x_0^2 + x_1^2)\cos\omega t_1 - 2x_0x_1)}$$and the requisite probability to be $$\left|\int P(x_1, t_1; x_0, 0)\psi(x_1)\,dx_1\right|^2$$for some wavepacket $\psi$ localized between $x_1$, $x_1 + dx_1$. In the limit $x_1 \to x_0$, we have$$\lim_{(x_1 - x_0) \to 0} P(x_1, t_1; x_0, 0) = \sqrt{{m\omega}\over{2\pi i\hbar \sin\omega T}}e^{{{im\omega}\over{\hbar\sin\omega T}}(x_0^2\cos\omega T - x_0^2)}.$$
My question is, what is the physical significance of taking this limit $x_1 \to x_0$?
| Ideologically speaking, the absolute value of the propagator squared
$$\tag{1} |K(x_f,t_f;x_i,t_i)|^2 \mathrm{d}x_f ~=~ \frac{m\omega}{2\pi\hbar\sin\omega \Delta t}\mathrm{d}x_f, \qquad \Delta t~:=~t_f-t_i~>~0,$$
is the probability that a harmonic oscillator starting at $t_i$ in position $x_i$ will finish within the position interval $[x_f,x_f+\mathrm{d}x_f]$ at time $t_f$.
In particular OP can study the case $x_i=x_f$, i.e. the probability of returning to the same position in a given time $\Delta t$.
Counter-intuitively, according to eq. (1), the probability does not depend on the start and end positions $x_i$ and $x_f$ at all! This foresees the fact that the notion of absolute (as opposed to relative) probabilities of the Feynman kernel $K(x_f,t_f;x_i,t_i)$ cannot be maintained on a non-compact position space, cf. e.g. Ref. 1 and this Phys.SE post.
In general, the probabilistic interpretation of eq. (1) only holds for short times $\Delta t\ll \tau$, where $\tau$ is some characteristic time scale of the system.
References:
*
*R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.
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Sending information faster than light If I could ever send my friend any information faster than light it would violate causality. If he just guesses the information and acts on it before he could ever receive it, everything is fine. What is different here?
I can understand that nothing can ever move faster than light but I can't understand why causality would be violated if something did. Or does it really have to? Since wormholes are mathematically sound, is it only a question of traversing faster than light? Is it ok if I find a way of transferring information faster than light as long as I don't move anything faster than light?
|
What is different here?
In some reference frames, your friend guesses the information and acts before you send it and in others, he guesses and acts after you send it.
But there is no causality problem since his action is caused by his guess rather than the received information.
In all reference frames, the guess precedes the action.
Now consider the actual effect of receiving the information. Perhaps a tone is sounded or a light is activated.
If the information propagates faster than $c$ from the transmitter to the receiver in some frames of reference, there are other frames of reference in which the event that the tone is sounded occurs before the event the information is transmitted, i.e., the cause and effect are reversed.
This is clear if you draw a spacetime diagram of two events, E1 and E2, with spacelike interval and note that, in some reference frames, t(E1) < t(E2) and, in others, t'(E2) < t'(E1).
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How do I solve this Gaussian path integral? Suppose
$$
Z = \int \mathcal D[\phi^*] \mathcal D[\phi] \exp(\phi^*A\phi + \phi B\phi)
$$
where $A$ and $B$ are operators. I know how to solve a Gaussian path integral involving only $\phi^* A \phi$ but I don't know how to handle the other quadratic term.
| You just divide $\phi$ to the real and imaginary part, to make things clear:
$$\phi = f + ig, \quad \phi^* = f-ig, \quad f,g\in{\mathbb R}$$
Up to some totally universal normalization factor, the integration measure is simply
$$\int {\mathcal D} f \,\,{\mathcal D} g $$
and the exponent in the exponential may be written as
$$[(f-ig) A + (f+ig) B] (f+ig) $$
Writing the column $(f,g)^T$ as $h$, the bilinear expression above is nothing else than
$$ h M h $$
where the matrix $M$ is, in a block-diagonal form,
$$ M = \left(\begin{array}{cc}A+B&-iA+iB\\iA+iB&A-B\end{array}\right) $$
Now, I assume you may calculate the integral
$$\int {\mathcal D} h\,\exp(hMh) $$
which is completely analogous to the $\exp(\phi^* A \phi)$ integral. However, with the matrix $M$ enough, the integral is infinity because $M$ is singular (infinity, due to flat directions) because the second row (of blocks) is $i$ times the first. However, you will get a nonsingular result if the exponent will also contain the Hermitian conjugate $\phi^* B^\dagger \phi^*$ or something like that.
If there are algebraic mistakes above, it should be possible to fix them.
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What are the dimensions, width and length, of a photon? Everyone is always talking about photon's wavelength. But what about its dimensions?
What is length and width of it?
And does it even have a point to think about such things? Or those dimensions are non-existent in such cases?
| The photon can be experimentally shown to not be point-like. The Young's slits experiment involves the interference of a photon with itself (the photon behaves in some ways like a particle, in some ways like a wave and in some ways like a probability distribution. in reality these are convenient models that we apply to it - in reality it is none of these - it is a photon). The interference patterns shown in the Young's slit experiment remain even if the stream of photons is reduced to the extent that photons go through 1 at a time. The length difference in the 2 paths can be varied to determine the coherence length. I believe this turns out to be in the order of 1m.
Further experimental proof that the photon must have a significant length is that the spread in its frequency is minimal. if it reduced to 0 amplitude over (e.g.) only 3 wavelengths then this would give it a significant spread in frequencies
| {
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What do quarks look like? I've heard everything from zero-dimensional points, to squares, and I would love to know what they really look like, or if they have any physical shape.
| I've not studied this at all but I have a theory that they are shaped like a triangular pyramid, all four sides being equilateral. It is the least amount of lines to create a three dimensional object, and being the simplest form of matter, it would be simple... But like I've said, this is just an unstudied guess
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If the solar system is a non-inertial frame, why can Newton's Laws predict motion? Since there is no object in the universe that doesn't move, and the solar system likely accelerates through space, how did Newton's Laws work so well? Didn't he assume that the sun is the acceleration-less center of the universe? Shouldn't there be many psuedo-forces to account for planetary motion?
| There's no doubt the solar system is accelerating. The milky way galaxy rotates, and we're quite on the outside. Hence, there's a permanent acceleration vector pointing to the center.
However, this is a phenomenally small acceleration. If you'd try to measure it here on earth, you run into all kind of practical problems when you try to isolate it. For instance, the earth's gravity isn't really that constant, at this scale. The tides move ocean water around, in reaction to the moon's gravity.
So, in practice, when Newton's law is a sufficiently good approximation (relativistic effects small enough), the Sun can be considered to stand still.
| {
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How do you determine the "phase" of a hydrogen eigenfunction? I've been reading the wikipedia article on the atomic orbitals of hydrogen. They have a nice collection of diagrams, such as this one for n,l,m = 3,1,1
This is apparently showing the wavefunction, not the probability density, and the blue area represents positive phase, red represents negative.
My problem is this: this particular wavefunction contains a term $\exp(+i\phi)$, so how has this graph been drawn taking into account the complex part? And what are they referring to when they talk about phase?
Edit, this is the 311 wavefunction containing the $\exp(+i\phi)$ term
| If you are referring to a global phase factor $e^{i\phi_0}$ (with $\phi_0 \in \mathbb{R}$ a real number, NOT the azimuthal angle $\phi$ upon which the hydrogenic wavefunction depends) multiplying the whole wavefunction, then that has probably been simply taken to be 1, as this choice does not affect any of the physics. This is because the probability density of finding a particle in some location is given the square modulus of the wavefunction, and with such an operation the phase factor vanishes.
You could equally well take the phase factor to be $−1$, and then red and blue in the picture would be exchanged. Indeed, you should keep in mind that the only thing that matters is the difference in sign between the two regions, not that one is "positive" and the other "negative".
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Bra-ket of products I was trying to solve the following problem.
(Lifted from Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund)
I came across across a solution for the problem as follows
(Lifted from http://theochemlab.asu.edu/teaching/chm598/ch2soln.pdf)
The 4th line in the solution doesn't seem to make sense to me. From what I understand, if we let:
$\chi$ be an $M \times O$ matrix
$\psi$ be an $M \times N$ matrix
$\alpha$ (and $\beta$) be an $N \times O$ matrix
then $\langle \psi_i^\alpha | \psi_j^\beta \rangle$ is an $N \times N$ matrix and $\langle \alpha | \beta \rangle$ is an $O \times O$ matrix. This means that we cannot multiply the 2 resulting matrices, right? Also, how was $\langle \chi_{2i-1} | \chi_{2i} \rangle$ expanded into $\langle \psi_i^\alpha | \psi_j^\beta \rangle \langle \alpha | \beta \rangle$?
| I will consider non-relativistic quantum mechanics in this answer. The wave function of a particle with spin can be thought of as living in the tensor product space $L^2(\mathbb R^3,\mathbb C)\otimes V$, where $V$ is the spin space (usually just $\mathbb C^2$). This tensor product vector space becomes a Hilbert space (after a possible quotient and completion in the general case) through the extension of the inner product on elementary tensors
$$(\psi\otimes\chi,\phi\otimes\eta):=(\psi,\phi)(\chi,\eta),\qquad\forall\psi,\phi\in L^2(\mathbb R^3,\mathbb C),\chi,\eta\in\mathbb C^2$$
to the whole tensor product space.
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What was Einstein's 1923 Nature paper "The Theory of the Affine Field" about? After his divorce with Mileva, Einstein published a paper in Nature entitled "The Theory of the Affine Field." Allegedly it confused renowned scientists. Why? What did the paper accomplish? What was it about, really?
| It was Einstein's reaction to Eddington's ideas about a unified field theory based on a symmetric affine connection. Ultimately, it did not go anywhere.
Cf. Hubert F. M. Goenner: On the History of Unified Field Theories. There's also a second part.
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What is a reasonably accurate but simple model of the Milky Way's gravitational field? I am putting together a toy program which shows how stars move around in the galaxy.
To run the simulation I need to know strength of the Milky Way's gravitational field at any location in it. I'm looking for a model (e.g. a collection of uniformly dense planes/rods) rather than a database of potentials.
Where can I get such a model?
I could simply construct an infinite plane of uniform density, but is that good enough? This is only a toy so I'm looking for something which preserves integrity of the overall shape and statistics of the galaxy, rather than worrying about the specific location of any particular star.
| Note first that there are three different sources of gravitational potential: the disk, the bulge, and the dark halo.
There are a few different models of the gravitational field of the disk, two of the more common potentials are:
*
*Kuzmin model:
$$\Phi(r,z)=-\frac{GM}{\sqrt{r^2+(a+|z|)^2}}$$
*Miyamoto-Nagai model:
$$\Phi(r,z)=-\frac{GM}{\sqrt{r^2+(a+\sqrt{z^2+b^2})^2}}$$
where $a$ and $b$ are scale lengths.
For the bulge, you can use spherically symmetric potentials such as
*
*Plummer model:
$$\Phi(r)=−\frac{GM}{\sqrt{r^2+a^2}}$$
*Jaffe model:
$$\Phi(r)=\frac{GM}{a}\ln\left(\frac{r}{r+a}\right)$$
where $a$ also is a scale length and not necessarily the same as those for the disk.
The dark halo takes a spherical form,
$$
\Phi(r)=\frac12V_h^2\ln\left(r^2+a^2\right)
$$
where $V_h$ is the radial velocity of the galaxy at far distances ($\sim200$ km/s) and $a$ another scale length that isn't necessarily the same as above.
See also
*
*http://www.ifa.hawaii.edu/~barnes/ast626_97/gp.html
*http://www.astro.utu.fi/~cflynn/galdyn/lecture4.html
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Why do we call a white led with high color temperature "cool"? one can buy LED bulbs with defined color temperature.
why cool white = many kelvins (= high temperature?)
why warm white = few kelvins (= low temperature?)
| The designation of different colors of light as warm and cool, in contradiction to their actual color temperatures, has to do with the items that we experience in ordinary life.
Most objects hat we encounter are not warm enough to emit detectable levels of visible light. As they warm up, they begin to emit more visible light, principally at the red end of the spectrum. We associate this appearance of red light with dangerous heat: stove elements, candle flames, red-hot pokers... We seldom encounter items hot enough to emit predominantly blue light...
On the other hand, the presence of snow and ice can give a decidedly blue tint to natural light, especially if it is reflecting a lot of blue sky light. So we associate this blue color with the coldness of the climate.
I remember on an eclipse cruise a long time ago, coming out of a lecture on Cool Red Giants and Hot White Dwarves, and staring in confusion at the red and blue color coded taps in the washroom! Add to this the fact that the sink was several meters below sea level. There is such a thing as knowing too much physics...
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Confusion with potential in simple pendulum I'm a maths student taking a course in classical mechanics and I'm having some confusion with the definition of a potential.
If we consider a simple pendulum then the forces acting on the end are $mg$ and $T$. Now I know that the potential is defined such that $F = -\nabla V$. Now I also know that the total energy of this system is $$\frac{1}{2}m \dot{\vec{x}}^2 + mgz.$$ Now if we take the gradient of the potential we have $(0,0,mg)$. My question is, why doesn't the potential involve the tension in the pendulum?
| So you've seen that $\frac{1}{2} m \dot{\vec{x}}^2+ mgz$ doesn't describe everything. Now, you'll learn about how to go from a formula almost like this to equations of motion later, but let's ignore that for a bit. Let's say there's a potential $U(\vec{x})$ which is zero on the circle, and off the circle is equal to a constant $k$ times the shortest distance to the circle, squared.
This is like a rigid pendulum. Imagine $k$ getting larger and larger. Due to conservation of energy, the point in question is constrained to only move within a certain distance of the desired curve. As $k$ gets larger and larger, this distance of constraint shrinks. As $k\to \infty$, the distance shrinks to $0$.
Now in this idealization when the particle is on the path, $U(x)$ is - in this limiting sense - zero on the circle and infinite everywhere else. There are mathematical proofs that this works (or discussions of it) in Arnold, Mathematical Methods of Classical Mechanics.
The physical intuition behind this is that you can have a lot of force without any energy. If you compress water as hard as you can, when released you won't notice anything. If you compress air as hard as you can, when released there can be a big explosion.
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Why does the mathematical constant $e$ enter into quantum mechanics so much? In A. Zee's book Quantum Field Theory in a Nutshell, he mentions on pages 11-12 the following formula which he assumes reader had encountered before:
\begin{equation}
\langle q | p \rangle ~=~ \frac{e^{iqp/\hbar}}{\sqrt{2\pi\hbar}}.
\end{equation}
I keep seeing $e$ come up in quantum mechanics. For instance, in that same chapter he writes $e^{-iH\delta t}$.
Why is $e$ useful in quantum mechanics? What does it signify in this context relative to the inner product of the bra and ket vector on the left?
| One reason is because harmonic oscillation is so common in nature. An oscillation is called harmonic if the force is proportional to how far the system is from equilibrium. E.g. a mass attached to a spring in a uniform gravitational field. If you lift the mass a bit, gravity takes over and there's a force in the down direction. If you pull the mass down past the equilibrium point, the force from the spring wants to pull it up. If you let it go, a harmonic oscillation starts.
If you solve the equations for harmonic oscillators you find they are sine (or cosine) functions. A lot of the maths becomes much easier if you use the Euler equation connecting sin and cos,
$$e^{i\phi}=\cos\phi + i\sin\phi.$$
| {
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Do we know where newly formed quark-antiquark pairs come from in the process of hadronization? The only explanations I have found are very vague, such as "spontaneously created from the vacuum" and because "it is more energetically favorable".
| Hadronization - the process by which colored objects form uncolored hadrons - is poorly understood, but we know the basics. Hadronization is a long-distance process, because, contrary to e.g. gravity, the strong force gets stronger at large distances.
When a quark-antiquark pair is created from a high-energy collision, they are always connected by a sort of "web" of gluons (the carriers of the strong force). Think of it as lots of strings connecting the two particles.
Some of the gluons connecting the pair might emit more gluons, and those gluons might decay into extra quark-antiquark pairs. The new quark-antiquark pairs ultimately form colorless hadrons with the original quarks. These hadrons are energetically favored, because the quarks bound into hadrons have a smaller mass than their free constituents.
| {
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Is a black hole really a hole in space? What if when a supernova occurs, instead of it condensing into a singularity it creates enough force to tear a hole into the fabric of space? Is a black hole just what is sounds like, a hole in space?
| No. Spacetime is smooth and continuous everywhere except at the central singularity. If you jumped into a big enough black hole you would cross the event horizon without noticing anything special happening.
The requirement for it to be a big black hole is because the tidal forces at the event horizon decrease with increasing black hole size. Jump into a small black hole and you will be spaghettified before you reach the horizon. But even with a small black hole spacetime remains smooth and continuous - you just wouldn't be alive to appreciate it.
If you're interested there are some animations of a trip into a black hole on this web site.
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Tensor product of two different Pauli matrices $\sigma_2\otimes\eta_1 $ I'm solving problem 3.D in H. Georgi Lie Algebra etc for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces.
Defining the basis in our four dimensional tensor product space
$$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\
\left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\
\left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\
\left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$
Now we know that when we multiply representations, the generators add in the sense of
$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements).
Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by
$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix}
0 & \mathbf{1} & -i & 0 \\
1 & 0 & 0 & -i \\
i & 0 & 0 & 1 \\
0 & i & 1 & 0 \end{pmatrix}$$
(The bold $\mathbf{1}$ is just notation, see below!)
I am not asking you to redo the calculations for me but does $(3)$ make sense?
Appendix.
My calculations were done in the following fashion [using equation $(2)$]:
$$\tag{4}\langle 1| \sigma_2\otimes \eta_1 |1\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=1\rangle
\\
=
[\sigma_2]_{11}\delta_{11}+\delta_{11}[\eta_1]_{11}
\\
= 0.$$
Similarly for eg
$$\tag{5}
\langle 1| \sigma_2\otimes \eta_1 |2\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=2\rangle
\\
=
[\sigma_2]_{11}\delta_{12}+\delta_{11}[\eta_1]_{12}
\\
= 1.
$$
This is how the bold $\mathbf{1}$ was obtained.
So are my calculations $(4), (5)$ totally wrong?
The Pauli matrices
$$\begin{align}
\sigma_1 &=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix} \\
\sigma_2 &=
\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix} \\
\sigma_3 &=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} \,.
\end{align}
$$
| Your equation (2) is right, in principle: it is the standard coproduct of Lie algebras,
but it is irrelevant, and should have never been used for anything here. The language confused you. It should read
$$ \boldsymbol{J^a} = \boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a} .$$
If you wished to apply it to two doublet reps, you should have used the same Pauli matrix $\sigma^a$ for both $j^a$s, and multiplying by the same angle and exponentiating you would have seen how nicely the group elements tensor-factor in the respective subspaces 1 and 2: $\exp (i\theta^a \boldsymbol{J}^a)=$ $\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a}))=\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1)) \exp(i\theta^a (1\!\!1\otimes \boldsymbol{j^a}))= \exp(i\theta^a \boldsymbol{j^a}) \otimes\exp(i\theta^a\boldsymbol{j^a} )$.
But, instead, your assignment asked you to simply mechanically evaluate the tensor product of two different matrices, to see if you understand the rules @jabirali correctly applied to get the correct answer you were meant to find. So, your equation (3) is magnificently wrong: you evaluated $\boldsymbol{\sigma_2} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{\sigma_1} $. jabirali is actually using your conventions, basis (1), exactly.
As a further exploratory excursion, you might use his and your rules, "right matrix into entries of left matrix", "left-coarse, right-fine" to compute (2) for a common matrix, e.g. $\sigma^2$, and then C-G rotate/reduce the 4x4 matrix to find $J_2$ in the triplet representation (3x3 block) and a singlet (0! in the remaining 1x1 block).
| {
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Is $ds^2$ just a number or is it actually a quantity squared? I originally thought $ds^2$ was the square of some number we call the spacetime interval. I thought this because Taylor and Wheeler treat it like the square of a quantity in their book Spacetime Physics. But I have also heard $ds^2$ its just a notational device of some sort and doesn't actually represent the square of anything. It is just a number and that the square sign is simply conventional.
Which is true?
| It is a mnemonic notation that indicates that $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ is the object whose square root is to be used as the infinitesimal line element, traditonally denoted $\mathrm{d}s$, when determining the lengths of worldlines $x : [a,b] \to \mathcal{M}$ by integrating the line element along them as
$$ \begin{align*}
L[\gamma] & = \int^b_a \lvert\lvert x'(t) \rvert\rvert \mathrm{d}t = \int_a^b \sqrt{\lvert g_{\mu\nu} x'^\mu(t) x'^\nu(t)\rvert}\mathrm{d}t = \int_a^b \sqrt{\lvert g_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t}\rvert}\mathrm{d}t \\ & = \int_x\sqrt{\lvert g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu\rvert}\frac{\mathrm{d}t}{\mathrm{d}t} = \int_x\sqrt{\lvert g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu\rvert} = \int_x \mathrm{d}s
\end{align*} $$
where the left hand side defines the length functional and the right hand side is obtained by scetchy manipulation of differentials, which is why you should not take $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ too seriously.
| {
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If a Killing vector field is timelike, can it be set to $\partial/\partial t$? If one has a Killing vector that turned out to be a timelike Killing vector field because of negative norm. Can we set this Killing vector field equal to $\partial/\partial t$?
| Comments to the question (v3):
*
*Given a manifold $M$, if a smooth vector field $X\in \Gamma(TM)$ does not vanish in a point $p\in M$, then one may choose a local coordinate neighborhood $U\subseteq M$ of $p$, with local coordinates $(x^1, \ldots, x^n)$, so that $X=\frac{\partial}{\partial x^1}$. This procedure is sometimes called stratification or straightening out of a vector field. It is a special case of Frobenius theorem.
*A timelike vector field $X$ does not vanish by definition, so it can be locally stratified $X=\frac{\partial}{\partial x^1}$, cf. 1. Since $X$ is timelike, one would call $x^1$ a time-coordinate.
*The Killing vector field property is irrelevant for local stratification.
| {
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Is the method of images applicable to gravity? It is well known that the method of images is a useful tool for solving electrostatics problems. I was wondering why this technique is not applied when considering newtonian gravity?
Obviously there is no "negative mass" to correspond to a negative charge in electromagnetism, but surely could the unphysical nature of negative mass be ignored and considered a mathematical trick to solve a given problem?
The classic example for the method of images is the point charge near an infinite conducting plane, is there a way to apply a similar method to calculate the gravitational field between a point mass and an infinite thin plane?
Some preliminary research online has resulted in no resources on this idea so any references for/against this would be great.
| While rare, there are a few uses of the method of images to gravitational problems. As lurscher says, the problem is finding equipotential surfaces. In most problems, such a surface doesn't exist, and hence the scare use of the method of images in GR.
One class of problems for which it does applies are the so-called Dirichlet problems. Suppose one was interested in solving for the metric in some region, with specified boundary conditions on the boundary surface. This is not usually what is done--usually the entire spacetime is solved for. For the case of Dirichlet boundary conditions (requiring the metric to approach some specified value on the boundary surface), image charges can be useful. In this case the image charges could correspond to image black holes, for example.
However, this is somewhat of an exotic problem, and I've only seen perhaps one or two examples where image charges have been used.
| {
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What physical evidence is there that subatomic particles pop in and out of existence? What physical evidence shows that subatomic particles pop in and out of existence?
| My current understanding is that the physical reality of vacuum fluctuations, particle-antiparticle pairs being created and then annihilating, is disputed. The Casimir effect is often cited as physical evidence but there's a few authors which have come to dispute that the Casimir effect is convincing evidence for the reality of vacuum fluctuations, as they argue that the same results can be extracted from treating the effect as a result of retarded van der Waals forces, not vacuum fluctuations.
See this paper: http://arxiv.org/abs/hep-th/0503158
and
this summary of the situation http://orbi.ulg.ac.be/bitstream/2268/137507/1/238.pdf
Maybe read the summary first, it's easy and quick to read :) As far as I know aside from the Casimir effect we have no other evidence for the physical reality of vacuum fluctuations. If you want to delve deeper, a good start is the papers cited by the two above.
| {
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Calculating electrostatic potential A continuous charge distribution is spherically symmetric and has a volume charge density
$$\rho(r) = \rho_oe^{−\alpha r}$$
I need to find the potential as a function of '$r$' i.e. $V(r)$.
It seems fairly straight forward. I can easily find the Electric field as a function of $r$ using Gauss's Law, and then integrate the negative of the field from infinity to $r$ to get the potential. But it turns out that calculating the integrals is a tough task. Is there a simpler way to solve this problem?
| No, that is the simplest way to solve the problem. As mentioned in the comments, this is in the absolute scale of things a very easy problem: the spherical symmetry allows you to even have an integral to calculate, and the exponential is not only exactly integrable, but easily so. If you allow general spherically symmetric charge densities
$$\rho(\mathbf r)=\rho(r),$$
and you require only that $\rho(r)$ be an elementary function, then in general you can find the radial electric field as the integral
$$
E_r(r)=\frac{1}{\epsilon_0 r^2}\int_0^r\rho(r')r'^2\mathrm d r'
$$
but this will not in general be expressible as an elementary function. (For more details look up e.g. the Risch algorithm.) For the functions $\rho(r)$ that come up in practical, interesting problems, the integrals in question are generally doable but typically much harder than for an exponential. As you progress into electromagnetism, be prepared to work out much harder integrals.
I should note at this point that there is indeed an equivalent procedure, based not on integration but on differential equations, and it's correspondingly based on the differential form of Gauss's law,
$$
\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2 E_r\right]=\nabla\cdot\mathbf E =\rho/\epsilon_0,
$$
or in terms of the potential
$$\nabla^2 V=-\rho/\epsilon_0.$$
This may or may not be more useful, depending on your tastes, but it should be immediately obvious that it's essentially the same as the integral version. There's no getting around the calculation.
| {
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Why is the bottom part of a candle flame blue? What’s the explanation behind the bottom part of a candle flame being blue? I googled hard in vain. I read this. I don’t understand how it’s explained by the emission of excited molecular radicals in the flame. I read that a radical is a molecule or atom which has one unpaired electron. That made me more confused. I want a more detailed, clearer explanation.
| In the book "Physics of the Plasma Universe" Dr. Anthony Peratt puts candle flames near the bottom of "energy in electronvolts" portion of the 'plasma spectrum'. If you look at the chart below, you'll see candles flames about midway (ok, cosmologically) between the ends:
*
*solar bodies and laser radiation
*terrestrial flames
*interstellar charged gases
I don't have the studies to do a discursive rendering of the chemical energetics, but you can see from this spectrum that most plasmas have fuel sources that are nearly-inexhaustible by comparison. Terrestrial flames have a very tight length-scales (relative to humans) before the dynamics become unsuitable (in terms of electrons per $cm^3$) to sustaining a plasma reaction — which produces the higher-energetic blues you see at the base of the flame.
You can see this more clearly when looking at a candle flame burning in space. The absence of gravity allows convection to very-evenly mix the soot and other combustibles, leading to a very clean plasma.
| {
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Spacelike to timelike four vectors First at all, let me just say that I'm not a Physicist, I study mathematics. So, I have this question. If you have a spacelike four vector, is there any transformation that could change it to be a timelike four vector? I mean, I know that every Lorentz Transformation (LT) preserves this properties (timelike $\rightarrow$ timelike, spacelike $\rightarrow$ spacelike, etc.), but I was thinking in another frame $S'$, different from the former $S$, where a spacelike four-vector (in $S$) will be timelike (in $S'$). If it is possible to have this other frame then, the way to relate events between frames is not a LT? or I'm missing something?
| The proper time, $\Delta\tau$, between two events is a conserved quantity in special relativity i.e. all observers will agree on its value. Since the definition of timelike is $(\Delta\tau)^2 \gt 0$, and the definition of spacelike is $(\Delta\tau)^2 \lt 0$, there can be no coordinate transformation that interconverts spacelike and timelike vectors.
| {
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$\mathrm{\rho^0}$ meson decay via the weak interaction? Of course, the $\mathrm{\rho^0}$ meson can decay in $\mathrm{\pi^{+}\ \pi^{-}}$ through the strong interaction. Using Feynman diagrams, I cannot understand why the same decay couldn't happen through the weak interaction. I attach the diagram I've drawn.
Strong decay:
Weak decay:
| Assume for the sake of argument that this decay channel happens.
Now, ask yourself how you are going to prove it?
*
*How about we compute the rate for each channel and see if the real rate is the sum? This works when the added channel is a reasonable fraction of the dominate channel, but you're talking about comparing a strong decay to a weak one. This is hugely impractical.
*OK, the weak interaction respects fewer symmetries than the strong. We can take advantage of that. Now you're on the right track. This is how the weak interaction form-factors of the proton have been measured: by observing the parity violation rate in $\vec{e}(p,e'p)$ and similar channels. But that is a comparison between a electromagnetic process and weak one and it still requires around $10^{13}$ separate events to extract the results to useful precision. It's going to take more than that to pull a weak signal out from under a strong one. Still impractical, I'm afraid.
So, long story short: this is an experimentally inaccessible channel.
| {
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What does $v=c$ in the Lorentz transformation for time tell us? For the simpler cases as boost in the x-direction, the time dilation formula following the Lorentz transformation for time is $$\Delta t'=\gamma(\Delta t-v\frac{\Delta x}{c^2})$$Now, we observe that as $v\to c$, $\gamma\to \infty$. And we also observe that if we increase $v$, the term in brackets beside $\gamma$ decreases more rapidly than the denominator of $\gamma$ i.e., $\sqrt{1-\frac{v^2}{c^2}}$. But if we put $v=c$ suddenly, $\gamma$ becomes undefined and the denominator of $\gamma$ and the term in brackets beside $\gamma$ becomes zero mathematically at the same time while the term in brackets was decreasing more rapidly. But while increasing $v$ to $c$ we see that $\Delta t'$ decreases and approaches zero and as we know $c$ is the maximum speed why don't we get $\Delta t'=0$ on putting $v=c$? Why do we get an indeterminate form? What is the explanation for this?
| Also, the fact that $\Delta t' \rightarrow 0$ if we formally let $v \rightarrow c$ can be interpreted as saying that no time at all passes for a particle moving at the speed of light. Photons cannot "age" or in any other way change over time.
| {
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Cosmology: what is a quantity that is called "$h$" in regard to angular size of a galaxy? I am trying to solve a Cosmology problem, but a certain quantity $h$ appears in it, of which I do not know the definition (I have never seen it mentioned anywhere before). So I thought maybe someone here could tell me what this $h$ is?
The problem goes as follows:
If $a$ is the scale factor in a FLRW-universe and $1+z=1/a$ is the redshift, then the luminosity distance of a far away object that is emitting light is given by $d_L=a_0r(1+z)$ (where $a_0$ is the scale factor today and can be scaled to $a_0=1$). Knowing the luminosity distance, we can get the angular diameter distance $d_A=l/\theta=d_L/(1+z)^2$, where $l$ is the proper size of the source and $\theta$ the apparent size. Knowing $l$, what is the minimum angular size of the radiating object? Be sure to first express your finding in terms of $h$, and then use $h=0.7$ to get a numerical value.
So, the problem asks to find the minimum angular size (which is $\theta$ I assume), and I can do that. But I have no idea what $h$ is supposed to be! Can someone clarify? Thanks for any suggestion!
| This goes back to the history of the Hubble constant. It's easy enough to write most cosmological formulas in terms of this value, but measuring it was something of a challenge for a while.
For several decades, we were confident it was between $50$ and $100\ \mathrm{km/s/Mpc}$. Many results scale with this value (to some power), so what people did was write
$$ H_0 = h \times 100\ \mathrm{km/s/Mpc}. $$
This way, you could put any value of $h$ you personally believed into the final result, but since this factor is of order unity you can also remove it from the result (replace it with $1$) to get a quick sense of what the answer is. Today we know $h = 0.7$ well enough that this practice of leaving $h$ in equations is disappearing... slowly. It's still a useful exercise, though, if you are comparing results that assume different values of $H_0$, in order to see how the change in $H_0$ affects the results.
| {
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Why doesn't the speed of the wind have an effect on the apparent frequency? A boy is standing in front of stationary train. The train blows a horn of $400Hz$ frequency . If the wind is blowing from train to boy at speed at $30m/s$, the apparent frequency of sound heard by the boy will be?
The answer: The frequency remains the same at $400Hz$
MY QUESTION:
Why doesn't the speed of the wind have an effect on the apparent frequency?
| The problem is equivalent with considering stationary air and both train and boy moving 30m/s relative to the ground (and air). As they move in the same direction relative to the air, there is no relative motion between them (the observer and source are neither approaching nor receding). So there is no reason to expect a Doppler shift.
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Are the Wigner and Husimi transforms injective? I am wondering if the Wigner function is injective. By injective I mean, that, for every density matrix $\rho$, there is a different Wigner distribution. The same question applies to the Husimi distribution.
If the dynamical group is $\mathrm{SU}(2)$, the Husimi function does not recognize all states as being different, whereas the Wigner function does. How far can I "trust" the Wigner function, in general? Is there some proof of a one-to-one correspondence between density matrices and phase space distributions?
| The answer is a resounding "yes", cf Ref. 1, provided by Groenewold in 1946, op cit, and countless emulators since.
The Husimi is completely equivalent, so, injective, to the Wigner d.f., and so the answer is ipso facto "yes" here too.
I do not understand your particular SU(2)-blindness attributed to the Husimi, but I trust it is just an artifact of some breezy specific implementation implicitly used. At least in "regular" physics of the "street", (i.e. outside the business of manufacturing instructive freak counterexamples) They are all representation changes w.r.t. each other.
References:
*
*Thomas L. Curtright, David B. Fairlie, & Cosmas K. Zachos, A Concise Treatise on Quantum Mechanics in Phase Space, World Scientific, 2014. The PDF file is available here.
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Pumping charged particles (of same charge) into a blackhole Would would happen if you started pumping charged particles of same charge into a black hole? Let's assume that you have an infinite number of those charged particles. What will happen to the event horizon and the singularity? Please give both perspectives - that of the charged particle's falling in and that of an external observer's.
| Well if you add charge gradually to a black hole, it will become more charged. If you started with a neutral Schwarzschild black hole, it will then become a Reissner Nordstrom one, and each time you throw in another charged particle the charge will increase.
As the charge of a black hole is increased (keeping it's mass fixed), it becomes colder (if you are unfamiliar with the fact that black holes have a temperature, see this Wikipedia article: http://en.wikipedia.org/wiki/Black_hole_thermodynamics). So eventually the black hole will become very cold and then reach a zero-temperature extremal black hole.
Except that's not physically realizable, because all the charged particles we have at our disposal also have mass. So electrons are thrown in, the mass of the black hole is increasing and so is the charge. It turns out that this fact means that the black hole will never reach zero temperature, but will just get colder and colder with each new electron. From a more ordinary thermodynamics point of view, this is pleasing, since you should never be able to reach $T=0$ in a finite number of steps.
| {
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Is there a formal proof for the superposition theorem? I was just wondering whether there is a formal prove for the superposition theorem in electric circuits? I tried searching it online but couldn't find anything sufficient. Most of the sources assume it follows from the definition of linear systems, but then how do you prove linearity without using the superposition theorem? Any input will be much appreciated!
| There are many proofs, the problem is that they become too mathematical and hence end up distracting electrical engineering students from core electrical engineering.
Very rigorous proofs involve assuming your circuits can be represented as planar graphs. Additional assumptions including no short circuits, independent current and voltage sources only. Proofs involving dependent sources become complicated.
In any case, to do the full proof, you lay down KVL and KCL (node analysis, mesh analysis) and you end up with linear system of equations needed to solve the circuit.
Of course, you can just solve the linear equation using elementary linear algebra techniques, but if you pay careful attention you see that the equations are always of the form:
linear combination of node voltages = linear combination of independent voltage sources and current sources
(OR, of course, linear combination of branch currents ) = ...
(pay attention to the fact that the coefficients for the voltage sources and current sources in the above equations are simply algebraic functions of resistors ONLY)
... from here it should be easy too see that we can solve circuits by simply flipping on one source at a time (and keeping all other sources set to zero), and then adding up the "effects"
so to sum it up: what we are doing is assuming a linear system of equations exist, that this is system has as many unknowns as there are equations (not over/under specified) -- but we are not going to write this system down and solve it, instead, we set all V's and I's to zero, and flip back one V or I at a time and solve for currents/node voltages at a time.
| {
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How could a cord withstand a force greater than its breaking strength? How could a 100 N object be lowered from a roof using a cord with a breaking strength of 80 N without breaking the cord?
My attempt to answer this question is that we could use a counter weight. But I don't really understand the concept behind counterweights so I hope someone can clear that up for me and if there is a better answer I'll love to know it.
| I would not be surprised if a cord with a given breaking strength of 80N held 100N once. It should be regarded as trash after that though.
You see, they set the breaking strength as guaranteed to hold under the worst-case setup. Select a low-reduction knot (the rating not is the double figure-eight which is common but not the optimal choice) and apply the load smoothly, and you will find the real breaking strength is higher than rated.
It would be unwise to depend on this though.
EDIT: The double figure eight's coefficient is .75 which means the straight line breaking strength is 106N. If we assume 0 margin (as we do in homework problems) this solution requires a coefficient of .943. This means long splice is the only "knot" with a good enough coefficient (the coefficient of the long splice is not known but estimated at 1). While we can make a loop with a long splice, this is really suboptimal.
| {
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What is the expectation value of the position times momentum operator? Should I write the expectation of the position times momentum operator as:
$$\langle xp\rangle = \langle \psi|x (-i\hbar \partial_x) |\psi \rangle$$
or
$$\langle xp\rangle = \langle \psi| (-i\hbar \partial_x x) |\psi\rangle$$
| in short the first way, but you can see this via using the position basis,
$$
<x|\psi> = \psi(x) , \ 1 = \int |x><x| \ dx, \ p|x> = |x> (-i \hbar \partial_x), \ <x|x'> =\delta(x-x') \ \rightarrow \\
<\psi|xp|\psi> = \int dx \ dx' <\psi|x><x| x p | x'><x'|\psi> = \int dx \ dx' \psi^*(x) x <x|p | x'> \psi(x') \\
=\int dx \ dx' \psi^*(x) x <x|x'>(-i \hbar \partial_{x'}) \psi(x') = -i \hbar \int dx \ dx' \psi^*(x) x \ \delta(x-x')\partial_{x'}\psi(x') \\
= -i \hbar \int dx \ \psi^*(x) \ (x \ \partial_{x}\psi(x) ) = <xp>_\psi
$$
typically you don't have to be this explicit but hopefully this helps. If this is a bit too much you can find more information in many introductory texts to QM such as Sakurai. Hope this helps
| {
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What is basically the difference between static pressure and dynamic pressure? What is basically the difference between static pressure and dynamic pressure?
While studying Bernoulli's theorem, I came before these terms. The law says:
When the fluid flows through a small area, its pressure energy decreases & kinetic energy increases and vice versa.
Now that's wierd as I know due to having KE, ie. having momentum, one can impart pressure. Then why distinction ? What is then pressure energy??
In order to understand that I went to wikipedia & quora & others; there I found fluid exerts two pressure: Static & dynamic. But really nothing could be understood more than that. What are they actually?
| The quantity $\frac{1}{2}\rho v^2$ is called dynamic pressure for two reasons: because it arises from the motion of the fluid, and because it has the dimensions of a pressure.
It is not really a pressure at all: it is simply a convenient name for the quantity (half the density times the velocity squared), which represents the decrease in the pressure due to the velocity of the fluid.
| {
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Why the electromagnetic wave transmits momentum in the direction of the movement? Ordinarily a transverse wave does not transmit momentum, as it is the case of the well-known sea wave phenomenon. But the electromagnetic radiation has both fields transversal to the movement of the wave. Anyway, the electromagnetic wave transmits momentum in the direction of the wave.
| It's possible to prove that while the fields $E$ and $B$ oscillations are transverse to the direction of propagation of the e.m. wave, the momentum carried by the electromagnetic field moves along the direction of propagation of the field(in vacuum) since it is defined in units of area and time by the relation $$\vec{S}=\frac{\vec{E}\times\vec{B}}{\mu_0}$$ where $\vec{S}$ is the Poynting vector and the transferred momentum is $$\vec{P}=\frac{\vec{S}}{c}$$ that's a mathematical property of classical electrodynamics. Obviously a more precise explanation would require the concept of electromagnetic quanta, the photon which carries momentum $|p|=\frac{E=h\nu}{c }$ (from the special relativistic dispersion relation for massless particles $E=|p|c$) clearly along the direction of it's propagation and it's polarisation is transverse respect to it's direction of propagation since it is a massless particle.
| {
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Does all the theoretical work of astrophysicists have to be confirmed by the observations of astronomers? I am a chemist an I have some doubts about the work of astrophysicists.
I know that astrophysicists do a lot of theoretical calculations based in other theoretical work and also based in real measurements, which are normally made by astronomers.
But I don´t know if all the theoretical work of astrophysicists has to be confirmed by the measurements of the astronomers.
There is a lot more theoretical work done than there is equipment (telescopes, spectrometers, etc) necessary to measure and otherwise confirm their work, so do astrophysicists have to wait years in order for the observatories to make measurements confirming/contradicting their theories?
| Theoretical physics, in general, does not have to be confirmed by observations. Theories are proposed as an effort to explain observations, so some consistency with observations is expected. However, it's not necessary to wait for observations. A theoretical astrophysicist can propose work that is consistent with current observations and build off of it, making new predictions that observational tests have yet to confirm. But rather than waiting to see if tests confirm the predictions, they can continue working on the theory to see what further predictions or interpretations are available. Alternatively, they can also develop competing theories, use the unconfirmed theories as a basis for new theories or start working on a different topic altogether. If the observational data simply isn't available, a theoretical physicist won't sit on their hands and wait. They can operate as if the theory is confirmed until the data from tests is received. That speeds up the process if they were correct and changes nothing if they were wrong. Others will likely have developed different theories and, more than likely, one theory at least will fit the new data as well. Then the work done on that theory is accepted and everyone refocuses to that framework. Wash, rinse, and repeat.
| {
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If any, what would be the ideal modulation frequency for a phase-sensitive light sensing experiment in a non-dark environment? I have constructed some equipment for sensing a beam of light. The equipment is enclosed in a rudimentary darkbox to block out the majority of the ambient room lighting, but is still partially exposed. In order to isolate the beam from the ambient lighting we are modulating the beam with an optical chopper and sensing it with a phase-sensitive detector (lock-in amplifier).
I'm trying to identify the ideal modulation frequency of the chopper in order to avoid sources of noise in the environment. So far I have tried to avoid multiples of 60 Hz to account for the electrical lines and fluorescent lighting in the room. I have tried using 1039 Hz on the basis of it being a prime number, but the ambient lighting is clearly still having some influence, because the reading changes significantly depending on how open/closed the darkbox is. This may not be an issue of modulation frequency at all; maybe it has something to do with the dynamic reserve of the lock-in amplifier.
I'm hoping someone around here might have worked with a similar set up before and can inform from their experience.
| From your description, it seems that your detection system is operating in a nonlinear region. Nonlinearities in the detection system will complicate things, so making sure the detector is operating in its linear range will be very important. If the detection is linear, separation of the ambient and signal beam effects should be fairly straightforward. A useful approach can be to directly measure the ambient light that leaks into your box. You can then find a way to separate the effects of your light beam and the ambient light.
| {
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Are electrons miniature black holes? For something to be a blackhole, it must have gravity and the radius must be smaller than the schwarzschild radius for its mass.
-Electrons have gravity
-Electron are theoretically believed to be infinitely small points
Since it has gravity it is capable of being a black hole. Since its radius is infinitely small, it must have a schwarzschild radius and thus be a black hole.
| There is no universally accepted quantum theory of gravity.
Quantumly, the "shape" of a fundamental particle is a very fuzzy notion - we know that states are often not localized, so it is wholly unclear what it means to say "the electron is pointlike". The proper formal interpretation of a "pointlike particle" is simply a particle that is not composite - has no substructure we know of.
But the electron is not believed to be a "miniature black hole" because no one expects gravity to work exactly as in GR at the quantum level. Your question is meaningless at the current state of knowledge because black holes are classical macroscopic objects, and the "pointlike structure" of the electron is not meant in the sense that it is a classical point particle.
| {
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Solution to Schrödinger equation I'm trying to solve the Schrödinger equation for a given potential. With some assumptions I end up with:
$$\frac{\hbar^2}{2M}\frac{d^2u(r)}{dr^2} = - \left(E - V(r)\right)u(r)$$
Since it's a square well potential I'm looking at, I have for the first region ($r \leq r_0$) that $V(r) = V_0$. So if I plug that into the above equation and solve that differential equation, I end up with:
$$u(r) = A\sin(kr) + B\cos(kr),$$
where $k = \sqrt{2M(E+V_0)}/\hbar$. I can then make some conditions and stuff to reduce it even further, but that's not my problem.
The next region ($r > r_0$) I get that $V(r) = 0$, which mean I have to solve:
$$\frac{\hbar^2}{2M}\frac{d^2u(r)}{dr^2} = - E\,u(r)$$
In my books, and websites I've seen, the solution to this is the same as above (With different constants) and the expontential functions instead of cosine and sine. And this is what I don't understand, why is that ? Whenever I try to solve that on my computer, I still get cosine and sine, but no exponential functions - which I think makes sense, since the only difference is some constant for the potential.
So what am I missing ? Am I solving it the wrong way, or is there some trick I don't know ?
| What you're missing is that you're interested in $E < 0$. Such states are "bound" in the square well. The equation:
$ \frac{d^2 f}{dx^2} ~=~ - k^2 ~ f(x)$
is solved by $f(x) ~=~ A e^{i k x} + B e^{-i k x}$. If $k^2 < 0$ then you get another $i$ in this picture and you switch from sinusoidal behavior to plus/minus exponential behavior.
| {
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How eddy current brakes function Take the following example:
where a rectangular sheet of metal is entering a constant magnetic field at $v \dfrac{m}{s}$. Due to Faraday's law of induction + Lenz's law, we can state that an eddy current will be generated to oppose the increase of magnetic flux through the sheet of metal, so as to produce a magnetic field coming out of the page (represented by the red dots). Intuitively, I believe that this induced magnetic field should act as a 'brake' on the metal plate, as Lenz's law implies that the induced current should always in some way act against the motion, but I don't see how to calculate this 'retarding' force that would act to reduce the plate's speed?
| Part of the issue you're forgetting is the dynamics of the situation. The surface fields are not perpendicular because of the movement and the physical delay in setting up the induced eddy current to produce a counter field. If the velocity is very low or static then the forces cancel, which is why eddy breaking doesn't work well at low speeds.
With regards to calculating the breaking force, this is very complicated. The force that you get from magnets moving near copper or aluminum structures depends on many factors, including:
*
*The strength of the magnetic field in the metal and the magnitude of
the change in field strength. This is affected by the size and
strength of the magnet the position of the magnet(s) relative to the
metal part, which relates to the field strength the shape, thickness
and geometry of the metal.
*And Velocity of the magnet/metal as faster speeds yield more force
(up to a point).
Most measurements are done empirically but estimations of the force can be done with three-dimensional FEA (Finite Element Analysis).
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Proof that a traceless strain tensor is pure shear deformation How can i proove that the traceless part of linear strain tensor $e$ in the Euler description:
$$e_{i,j}={ 1 \over 2 } \left({ \partial u_i \over \partial x_j}+{ \partial u_j \over \partial x_i} \right)$$
is alway a pure shear deformation i.e. it does conserve volume. In case it is not clear this is the Euler strain tensor with the assumption $ { \partial u_i \over \partial x_j} << 1$, which means the part $\sum \limits_k{ \partial u_k \over \partial x_j} { \partial u_k \over \partial x_i}$ is neglected.
This apperently always has to hold when decomposed into traceless part $ e^t$ and a generally not traceless part $e^l$:
$$u= \underbrace{{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^l} +\left( \underbrace{e -{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^t} \right) $$
Even just the direction of how this can be done would be appreciated.
EDIT NOTE: the notation in my script is confusing so i changed it
EDIT UPDATE:
One hint might be that if we look at small local deformations the strain tensor can probably be approximated as equal in all directions in first order. So we can write $e= \mathbb{I} \epsilon+...$. The first term must be nonzero for small deformations, and so the trace in that order does not vanish, but ist there a better more general Argument? This seems very hand waving.
| Consider a volume element $dV$ at a certain point $\vec{x}$. Let the strain tensor at it be given by
\begin{equation}
e_{ij} = \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}
\end{equation}
Let us diagonalize the strain tensor at this point and let its diagonal entries be
$u^{(i)}$. Since the trace of tensor is invariant,
\begin{equation}
e_{ii} = u^{(1)} + u^{(2)} + u^{(3)}
\end{equation}
In terms of the diagonalized strain tensor, the volume element $dV = dx_1 dx_2 dx_3$ is deformed as $dV^\prime = dx_1(1 + u^{(1)})dx_2(1 + u^{(2)})dx_3(1 + u^{(3)})$. Ignoring higher than linear terms, $dV^\prime - dV = u^{(1)} + u^{(2)} + u^{(3)}$ Clearly, $dV = dV^\prime$ means that $u^{(1)} + u^{(2)} + u^{(3)}$ is zero, which is equivalent to vanishing of trace of the strain tensor.
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If a black body is a perfect absorber, why does it emit anything? I'm trying to start understanding quantum mechanics, and the first thing I've come across that needs to be understood are black bodies. But I've hit a roadblock at the very first paragraphs. :( According to Wikipedia:
A black body (also, blackbody) is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.
OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it. An object that always looks totally black, no matter under what light you view it. Good. But then it follows with:
A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation.
Say what? Which part of "absorbs" does this go with? How can it absorb anything if it just spits it right back out, even if modified? That's not a black body, that's a pretty white body if you ask me. Or a colored one, depending on how it transforms the incoming waves.
What am I missing here?
| Any body above 0K emits radiation. Same is the case with black body, it absorbs radiation and it also emits, now the rate at which it absorbs /emits depends upon the surrounding.
| {
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My physics teacher gave us this equation $v= -3 +3t$ She asked us if the body was accelerating or slowing down, and I immediately said that it was accelerating (because the $a=3>0$). Then she said that I was wrong because the direction of the acceleration vector was the opposite of the direction of initial speed($v_0=-3$). I do not understand why it slows down, because with the passage of time the body moves faster. Can someone give me an explanation?
|
She asked us if the body was accelerating or slowing down
Acceleration is defined as the time rate of change of velocity and, in this example, the acceleration is constant and positive.
So, the full answer is: the velocity of the body is always increasing while the speed is decreasing for $t<1$ and increasing for $t>1$.
In this plot, the velocity $v$ is blue and the speed $|v|$ is red
$$v = -3 + 3t$$
$$|v| = |-3 + 3t| $$
| {
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Proof of Lorentz contraction? The measurement of the flux of muons at the Earth's surface shows that many more muons are detected than would be expected, based on their mean half-lifetime of 2,2 microseconds. This is a good proof for the time dilatation as predicted by the special relativity theory.
QUESTION: does there exist a similar proof for the Lorentz contraction?
| Imagine for each observer there is a grid in space that maps how far light travels in an amount of time.
You can measure an object at rest in this grid by how long it takes light to travel its length and back again (this is important to measure it both ways).
Since light speed is a constant, the object needs to change shape to keep its 'light length/duration' the same.
It is hard to measure length and duration contraction directly, but you can sort of get evidence through secondary effects, but these could possibly be from other causes.
| {
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Neutrino Reaction: Is the Following Reaction Allowed? Is the following reaction allowed and why?
$$
\nu_e \to e^- + \mu^+ + \nu_{\mu}
$$
I would say it is allowed since individual lepton number and charge are conserved.
| It is by lepton number and charge, but you can't get energy/momentum to balance. In the $\nu_e$ rest frame there isn't enough energy to make the products. If there is a nucleus around, you can imagine the $\nu_e$ emitting a virtual $W^+$ making the $e^-$, the $W^+$ scattering electromagnetically off a nucleus to deal with the momentum, then decaying into $\mu^++\nu_\mu$
| {
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Proving and demonstrating vacuum in container without breaking it Let there be a hollow container made of glass or some other transparent material, roughly the size and shape of an apple. Let the walls be of sufficient thickness for the container to be safely evacuated to some reasonable degree, perhaps around $10^{-8}$ mbar, and then hermetically and evenly sealed.
*
*How could one prove that the container is evacuated, and with what accuracy? Would x-ray crystallography, laser scattering, light absorption or emission or ultrasound be possible ways?
*Is there a simple way of demonstrating that the container is evacuated, perhaps by placing something inside it before sealing which behaves in a very specific and obvious way in a vacuum, without actually removing said vacuum? Aside from the obvious feather, which would fall without air resistance. I was thinking of something along the lines of a small quantity of cesium, but that wouldn't be distinguishable from an inert gas atmosphere.
Thanks!
Edit: There is no reference container to compare with and the container itself isn't standardized, so density/weight considerations are out, if I am right. Actually, so is light refraction, probably, as the container walls aren't really level enough, considering the small refraction delta mentioned in Andrew S.'s reply.
| Why don't you just put a balloon inside? It will enlarge because the proportion of the pressure inside and outside the balloon will change.
| {
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Why aren't all black holes the same "size"? The center of a black hole is a singularity. By definition, a singularity has infinite density. So how can a black hole with a different mass or density be described?
| The singularity probably does not exist, as GR likely breaks down at those size / energy scales. When we have a full quantum description of gravity we may know what's really there.
By the way, the part of the black hole we fully understand is actually the vacuum solution - the Schwarzschild metric - which includes the event horizon but not the source mass. GR is agnostic about what happens on the inside and only cares about the total energy content.
| {
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Is it possible to have a black hole in empty space? If the escape velocity of two very massive objects is near the speed of light, and those objects are orbiting each other (let's ignore the Roche limit for this exercise), is it possible that the combined mass of these two objects is great enough that their center of mass, while in the empty space between, is effectively a black hole? Is it possible that this is already the case with black holes we've observed, but we're unable to tell, due to the information capturing nature of black holes?
Update
The reason I asked this question is that the the recent discovery of a very massive black hole made me wonder whether it was possible that this black hole was in fact a collection of nearly coalesced stars (perhaps from many galaxies), and its combined mass was so great that, while locally there was information due to its accumulation of space time, it appeared to our quiet part of space as one large black hole. I didn't initially mention this because I didn't want to distract from the fundamental idea on which my question was predicated.
| No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other surfaces, they would still, by construction, be just larger than the combined Schwarzschild radius. If you wanted to compress them the go below, then you would need to compress the individual bodies to go below their individual Schwarzschild radii.
Also, anything that forms a black hole will collapse to a point, keeping no memory of its former constituents (like the number of objects used to create it). Only mass, charge, and angular momentum is remembered.
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Time dilation as an effect of energy density Has any relation been observed or postulated to exist between the energy-density (or the surrounding space) of an object and time dilation?
i.e. Higher energy density==>Slower rate of time?
| The formula for gravitational time dilation1 is
$$\frac{t_0}{t_f}=\sqrt{1-\frac{2GM}{rc^2}}$$
For a sphere,
$$M=V \rho = \frac{4}{3} \pi r^3 \rho$$
So
$$\frac{t_0}{t_f}=\sqrt{1-\frac{8G \pi r^2 \rho}{3c^2}}$$
So the greater the density, the greater the time dilation.
Has any relation been observed or postulated to exist between the energy-density (or the surrounding space) of an object and time dilation?
Yes. In every single case, in fact - not just the spherically symmetric static cases, as I gave above.
1 Around a static, spherically symmetrical object.
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"timestamp": "2023-03-29T00:00:00",
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What types of fusion reactions happened in population III stars? I have read that, in smaller stars, such as our Sun, the fusion reaction that takes place is a proton-proton chain, or PP chain for short.
From what I have learned, in larger stars, a different process takes place, known as the CNO cycle, in which carbon, nitrogen and oxygen are utilized as a catalyst.
The first stars were many times more massive than the Sun, but pretty much the only elements they had to make use of were hydrogen and a little bit of helium. So, what types of fusion reactions went on in the first stars?
| I think you already know the answer...
Pop III stars, by definition, are born from primordial gas that is basically Hydrogen, Helium with trace amounts of deuterium, tritium, lithium and beryllium; they initially contain almost no C, N, or O. Therefore the primary fusion in massive Pop III stars has to be (well, initially the deuterium is burned but this is not energetically important) the pp chain, which produces more helium from hydrogen. The lower temperature dependence of the pp chain means that the core can get hotter and denser than would be usual in more metal-rich massive stars.
It is not until He burning is initiated that C, N, and O can form, but this actually takes place before the red giant tip is reached in Pop. III stars.
Your recent questions all have a theme. You might find this presentation I came across interesting.
EDIT: Further to this: Current thinking is that CNO burning will take place on the main sequence for Pop. III stars more massive than $20 M_{\odot}$. This is because they cannot be supported by the pp chain alone; they contract and heat up sufficiently to start the triple alpha He burning phase that produce carbon; once the carbon concentration builds up to about $10^{-10}$ of hydrogen then the CNO cycle takes over (Ekstrom et al. 2008; Yoon et al. 2012).
| {
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Normalization of wave function meaning...? I just have one question.
I'm doing a problem where I'm told to normalize a wave function, which is split up into two regions, namely where $r \leq r_0$ and $r > r_0$. My question is, why am I doing this? I'm not using any of the math I get out of it later on. The only thing I do after it is applying the continuous condition. So what I was thinking was, that by normalizing my wave function, I also showed that my wave function is in fact continuous. So is that the case, or am I just mistaken all together?
| Like gonenc pointed out your assumption that normalizing your wave function does not imply continuity. And yes you'll probably won't need the normalization factor in your further calculations.
The reason for you doing this could be consistency with the Interpretation of the wave function squared as a probability amplitude:
$$
P = \int|\psi(r,t)|^2 dr
$$
| {
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White, is it a colour or absence of colours? Our chemistry sir and we had an argument today at the lab, he says that white actually is not a colour, it is the abscence of colour, but we say that it is a colour and we gave the following point to substanciate our point that white is a colour:
When we see an object in red colour, it actually reflects red colour and absorbs all the other colours, in this point of view, a white object reflects all colours which fall on it, so it is a colour.
We do not know who is correct, I am posting this question in hope that I will get the correct answer.
| To add to Steeven's answer:
Any wavelength of input light will be perceived as lighter or darker in color depending on the intensity of the light. At high intensities, we only 'perceive' whiteness. I'm not sure whether the brain just ignores the overloaded cones (color-sensitive retinal elements) in favor of the rods (non-color elements), or whether a saturated cone-signal is interpreted directly as "white."
Further, it is the case that very low-intensity signals don't even stimulate the cones, so everything is interpreted as white/grey via the rods' signals.
| {
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Solving for the density operator in the quantum Brownian motion master equation I want to solve for the density operator in the quantum Brownian motion master equation,
\begin{align}
\begin{aligned}
\frac{d\rho_S(t)}{dt}=&-\left(\frac{i}{\hbar}\right)\Big[H_S+\frac{1}{2}M\widetilde{\Omega}^2X^2,\rho_S(t)\Big]-i\gamma[X,[P,\rho_S(t)]_+]\\
&-D[X,[X,\rho_S(t)]]-f[X,[P,\rho_S(t)]]\,,
\end{aligned}
\end{align}
where
\begin{align}
\begin{aligned}
H_S&=\frac{1}{2M}P^2+\frac{1}{2}M\Omega^2X^2\,,\\
\widetilde{\Omega}^2&=\frac{2}{M\hbar}\int_0^{\infty}\eta(\tau)\cos(\Omega\tau)d\tau\,,\\
\gamma&=\frac{1}{M\Omega\hbar^2}\int_0^{\infty}\eta(\tau)\sin(\Omega\tau)d\tau\,,\\
D&=\frac{1}{\hbar^2}\int_0^{\infty}\nu(\tau)\cos(\Omega\tau)d\tau\,,\\
f&=-\frac{1}{M\Omega\hbar^2}\int_0^{\infty}\nu(\tau)\sin(\Omega\tau)d\tau\,.
\end{aligned}
\end{align}
The only operators are $X$ and $P$ - everything else is a constant. If I want to write this master equation in the momentum basis, I know that I need to sandwich both sides of the equation with $\langle{p}|$ and $|{p}\rangle$. If I just focus upon $\Big[H_S,\rho_S(t)\Big]$, ignoring constants, I have
\begin{align}
\langle{p}|\Big[P^2+X^2,\rho_S(t)\Big]|p\rangle&=\Big[p^2+(i\hbar)^2\frac{d^2}{dp^2},\rho_S(p,t)\Big]\,,\\
&=\Big[(i\hbar)^2\frac{d^2}{dp^2},\rho_S(p,t)\Big]\,.
\end{align}
But now I'm not sure what to do. This final commutator gives me a derivative that's acting on nothing. How do I interpret this when solving for the density operator?
Since I am going to eventually solve for the density operator in the momentum basis, is there an easier way to do so than sandwiching both sides of the master equation with the bra and ket of the momentum?
| How come you took an expectation value of an operator and still got an operator instead of a c-number?
In order to get a c-number equation, you will get it for the matrix elements of the density matrix, so in position space this would be $$\left\langle x | \rho | x'\right\rangle \equiv \rho(x,x')$$
I'll give you an example for the last operator and you can do the rest (I set $\hbar=1$).
$$\left\langle p | \left[ X^2 + P^2,\rho\right] | p' \right\rangle = \left\langle p | \left[ X^2 ,\rho\right] | p' \right\rangle + \left(p^2 - p'^2\right)\left\langle p |\rho|p'\right\rangle$$
$$=\left(-\frac{\partial^2}{\partial p^2}\left\langle p \right| \right) \rho \left| p' \right\rangle + \left\langle p\right|\rho\left(-\frac{\partial^2}{\partial p'^2}\left| p' \right\rangle \right) + \left(p^2 - p'^2\right)\left\langle p |\rho|p'\right\rangle$$
$$=\mathscr{F}_{pp'}\left[ \left(x'^2 - x^2\right)\rho(x,x') + \frac{\partial ^2}{\partial x^2}\rho(x,x')-\frac{\partial ^2}{\partial x'^2}\rho(x,x')\right]$$
Where $\mathscr{F}$ stands for the double fourier transform.
By the way, you taking expectation value between the same state, you are only finding the differential equation for the diagonal elements. Using different $p$ and $p'$ gives you the general matrix element.
| {
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Estimating divergence of set of vectors I have a set of points where directions and intensities of a flow are given (in 3D). Is it possible to estimate the divergence of the flow defined by those vectors? I only need a rough estimate and I can assume continuity and smoothness.
I expected this to be a common question, but I wasn't able to find any literature, which makes me suspect I used uncommon phrases to describe my problem...
| The definition of divergence is
$$\textrm{div}\,\vec{F} = \lim_{V \to p}\iint_{S(V)} \frac{\vec{F}\cdot\vec{n}}{|V|}dS, \qquad [1]$$
where $\vec{F}$ is the vector field, $V$ is the volume surrounding the point $p$ where the divergence is calculated, $\vec{n}$ is a unit-length normal vector of the surface, $S(V)$, of the volume, and $|V|$ is the total volume.
To approximate this with a sampling of the vector field, instead of shrinking the volume to a point, $$\lim_{V \to p},$$ choose a small volume that surrounds the point in question and make it large enough to encompass many samples of the vector field (divergence is zero for a constant field).
$$\textrm{div}\,\vec{F} \approx \iint_{S(V)} \frac{\vec{F}\cdot\vec{n}}{|V|}dS.$$
If you are working in Cartesian coordinates, you could make your life easier and choose a cube aligned with your axes so $\vec{n}$ is constant on each side. Take the average of the vector field across the surface, interpolating as needed. If your cube as side lengths $x$, then your calculation becomes:
$$\sum_{i=1}^6 \frac{\vec{F_i}\cdot\hat{e_i}x^2}{x^3} = \frac{1}{x}\sum_{i=1}^6 \vec{F_i}\cdot\hat{e_i}$$
where $\vec{F_i}$ is the average vector on each surface and $\hat{e_i}$ is the unit normal vector on each cube face, $[\pm 1, 0, 0]$ for the faces perpendicular to the x-axis.
[1] http://en.wikipedia.org/wiki/Divergence#Definition_of_divergence
| {
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