Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
How can two charged black holes merge despite electrostatic repulsion? I have read this question:
Collision of charged black holes
And it made me curious.
I understand that the charged black holes do have negative EM charge, and they repel.
This EM interaction and repulsion between electron fields around nuclei, cause matter to have spatial extent. This EM repulsion is why atoms can't get closer to each other then a certain distance. And why atoms in molecules can't get closer to each other then a certain distance.
This is the reason why matter is 99% space.
Of course, the Heisenberg Uncertainty principle has an effect on this too.
Question:
*
*How can two charged (negative EM charge) black holes merge? How can gravity overcome the EM repulsion and the Heisenberg uncertainty principle?
*Do these electron fields pass through each other (merge too) when the holes merge? Does the Heisenberg uncertainty principle and the Pauli exclusion principle for fermions not apply anymore?
| Two charged black holes are very much like two charged droplets: they can merge, as long as the charge is not strong enough to repel them. The result is a bigger hole with the sum of the charges. It is not like the EM repulsion goes to infinity as they approach each other: they have a spatial extent, and the surface electric fields will be roughly following Coulomb's law (with some corrections due to curved spacetime).
Approximately, two black holes of mass $M$ and charge $Q$ each will repel each other if $GM^2/r^2 < kQ^2/r^2$, or $Q/M > \sqrt{G/k}\approx 10^{-10}$ C/kg. For a solar mass black hole that is about $10^{20}$ C. This produces a field at the horizon that is way past the Schwinger limit where the electromagnetic field becomes nonlinear, so it is likely that long before this the whole system destabilizes in a shower of electrons and positrons.
The Heisenberg uncertainty for macroscopic black holes is so small that it is negligible. Whether black holes can be treated as fermions looks uncertain.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Concise way to express commutators of two-qubit quantum gates I would like to calculate the commutator of quantum operations
*
*$C^{(ij)}$ (by that I mean CNOT with $i$ being the control qubit and $j$ being the target qubit; please correct me if there's a better and more common notation) single qubit operations, $\sigma_{x,y,z}^{(i)}$
and
*
*single qubit operations, $\sigma_{x,y,z}^{(i)}$ and $\sigma_{x,y,z}^{(j)}$. So far, I have not come up with anything better than doing this in a brute force way, by multiplying $4\times4$ matrices and then decomposing them. I'm wondering if anyone could check my results or, even better, suggest a more appropriate way to approach the problem.
Conventions:
$$
C^{(12)}=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&0&1\\
0&0&1&0
\\\end{pmatrix}
\quad,\quad
\sigma^{(1)}_{x,y,z}=\sigma_{x,y,z}\otimes\hat{1}
\quad,\quad
\sigma^{(2)}_{x,y,z}=\hat{1}\otimes\sigma_{x,y,z}
\quad.
$$
Results so far:
$$
\begin{alignedat}{99}
&[C^{(12)},\sigma^1_x] = - i \sigma_y\otimes(\hat{1}-\sigma_x)
\quad&&,\quad
[C^{(12)},\sigma^1_y] = - i \sigma_y\otimes(\hat{1}-\sigma_x)
\quad&&,\quad
[C^{(12)},\sigma^{(1)}_z] = 0
\quad&&,\\
&[C^{(12)},\sigma^{(2)}_x] = 0
\quad&&,\quad
[C^{(12)},\sigma^{(2)}_y] = -i(1-\sigma_z)\otimes\sigma_z
\quad&&,\quad
[C^{(12)},\sigma^{(3)}_z] = i(1-\sigma_z)\otimes\sigma_y
\quad&&.
\end{alignedat}
$$
(too ugly to be true...)
It would be great to know if there's a more appealing and universal expression for the commutator between $\sigma^{i,j}_{x,y,z}$ and their controlled versions.
| Remember that $$\operatorname{CNOT}=\lvert0\rangle\!\langle0\rvert\otimes1+
\lvert1\rangle\!\langle1\rvert\otimes X.$$
Because $[1,\sigma]=0$ for any Pauli matrix $\sigma$, we immediately have
$$[\operatorname{CNOT},1\otimes\sigma]=\lvert1\rangle\!\langle1\rvert\otimes [X,\sigma],$$
which is consistent with your results noting that $\lvert1\rangle\!\langle1\rvert=(1-Z)/2$.
Furthermore, because
$\lvert0\rangle\!\langle0\rvert=(1+Z)/2$ and $\lvert1\rangle\!\langle1\rvert=(1-Z)/2,$
we see that
$$[\operatorname{CNOT},\sigma\otimes1]=\frac{1}{2}[Z,\sigma]\otimes(1-X).$$
The two expressions can be made to look even more similar by writing them as
\begin{align}
[\operatorname{CNOT},1\otimes\sigma]&=Z^-\otimes [X,\sigma],\\
[\operatorname{CNOT},\sigma\otimes1]&=[Z,\sigma]\otimes X^-,
\end{align}
where $\sigma^\pm\equiv(1\pm\sigma)/2$ are the projectors onto the $\pm$ eigenvectors of $\sigma$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Are two charges attracting each other constantly accelerating? Using Coulomb's Law: $\frac{k_eq_1q_2}{r^2}$, say we have $2$ charges of $1\ \rm C$ each, separated by a distance of $1\ \rm m$. The force would be $8,987,551,787.3681767\ \rm N$, considering $k_e = 8,987,551,787.3681767\ \rm N m^2 C^{-2}$. So if we were to now set the distance between the $2$ coulombs at $1\ \rm mm$, the force would then be $8,987,551,787,368,176.7\ \rm N$. So my question is, if we were to release the $2$ charges from a distance of $1\ \rm m$, would the acceleration be constant, and more importantly, would that mean the force is also constantly increasing? What would happen when they collide? Does this have to do with what Coulomb's Constant actually represents?
| Yes, there is a net force on the charges, so they will accelerate until they collide. The acceleration will obviously not be constant as they move toward each other, the distance is decreasing, which means that the electric force acting on them is increasing. Acceleration is directly proportional to the net force. What happens after the collision depends on what the objects carrying the charge are made of. If they are conductors, both will be neutralized, but if they are insulators, they will simply stick together.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does work depend on distance? So the formula for work is$$
\left[\text{work}\right] ~=~ \left[\text{force}\right] \, \times \, \left[\text{distance}\right]
\,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
| You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 11,
"answer_id": 2
} |
How to resolve impulse on a free-floating body into translation and rotation I have free-floating in 2D a long thin homogeneous rectangular body with the center of gravity at its geometric center. It seems to me intuitively that if an impulse is applied to the middle of a long edge, parallel to the edge, there will be more translation and less rotation that if a similar impulse is applied parallel to the middle of a short edge. Is this correct? And more importantly, how do I quantize this? In other words, how do I apportion translation and rotation from an impulse?
| The general motion of a planar rigid body is a rotation about a point. If you consider an impulse $\hat{j}$ that passes on a line a distance $a$ from the center of mass, then the center of rotation is going to be a distance $b$ on the other side of the center of mass as seen below
The relationship between the two distances is
$$ b = \frac{I}{m a} \tag{1}$$
or if you use the radius of gyration $r$, and use $I = m r^2$ then
$$ b = \frac{r^2}{a} \tag{2}$$
For example for a rectangle of length $\ell$ and height $h$ the MMOI is $ I = \tfrac{m}{12} ( \ell^2 + h^2 )$, or $r = \tfrac{\sqrt{3}}{6} \sqrt{ \ell^2+h^2}$
The great thing about the above relationship is that it is purely geometrical. It also has the property that the less $a$ is the larger $b$ is, and vice versa.
There are three special cases to (2)
*
*$a=0$, the impulse goes through the center of mass, and the body translates since $b=\infty$.
*$b=0$, the body is rotating about the center of mass when $\hat{j}=0$ and $a=\infty$. This situation corresponds to a pure impulsive torque on the body.
*$a=r$ the impulse is tangential to the circle of gyration, then $b=a$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
States created by translation operator Quantum Mechanics Volume One page 188 by Claude Cohen Tannoudji.
In $q$ and $p$ state vectov formalism.
$QS(\lambda) |q\rangle=(q+\lambda)S(\lambda)|q\rangle$, where $S(\lambda)=e^{-i\lambda P/\hbar}$.
Thus when only consider $p$ and $q$, one may effectively think $S(\lambda)|q\rangle=|\lambda+q\rangle$.
However, what if there is a third set of eigenvectors say $l$, does $S(\lambda)|p\rangle$ still holds? Meaning if there is a third set of eigenvectors say $l$, can one still regard there is no difference between $S(\lambda)|q\rangle$ and $|\lambda +q\rangle$?
(consider $[S(\lambda),L]$)
| The properity closely related to the fact that $|q>$ in heliber space is considered as a complete set of basis(Use of 'complete' as in 'complete set of states' or 'complete basis'), that is $S(\lambda)|q>$ as a states with eigenvalue of $\lambda +q$, there is "nothing more to say" about the vector $S(\lambda)|q>$ as it has been fully expressed in the $|q>$ space.
However, this is only valied in the space vector formalism, and once the language translated into wave foramilism or expression of finite dimesion of matrix, thre results may not hold.
More mathematical proofs are still welcome.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modes in optical fibers I am trying to understand the modes in step-index optical fibers and I saw that they say the electric field distribution in the core and cladding is as bellow. my question is that which component of electric field is this?
| As The Photon already pointed out, "TE" means "Transverse Electrical", which is just telling you that there is no component along the direction of propagation - $z$, in this case. See here for the distinction.
Now, if you want to find the functional form of your electric field, you have to solve the wave equation.
Since your problem is cylindrically symmetric around $z$, the direction of propagation, your solution will exhibit the same symmetry.
These solutions have a name, the Laguerre-Gassian modes, which look like this:
$$ E(r, \phi, z) \propto e^{-\frac{r^2}{w^2(z)}}\cdot e^{-il\phi}\cdot e^{-ikz}. $$
You can see that the $\phi$ dependence is in a pure phase factor.
What you actually see on a camera is the intenstiy $I \propto E^{\dagger}E$ so $$ I \propto e^{-2\frac{r^2}{w^2(z)}}.$$
So only the radial part remains, and you can drop the $\phi $ depdendence from $E(r, \phi, z) \propto e^{-\frac{r^2}{w^2(z)}}$. Which is exactly what is plotted in your drawings. The $y$ axis is $r$.
(I have just written the fundamental mode, which is a gaussian, like your $TE_0$, on the Wiki link you can see that there is an $l$ dependent spatial part, which is what gives you the different shape of the other $TE$ modes: for details look at the Mathematica script below:
)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/428903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Which field Passes Through the Polarization Film? Why Isn’t the Perpendicular Field Stopped? When one EM field is aligned so that it can pass through a polarizing lens the other field (E or B) is 90 degrees out. Is only one of the EM fields affected by a polarizing lens or film? How is it that one field is stopped yet the other seems unaffected?
| One way of making a polarizing film is having a lot of small narrow conducting 'rods' inside it. The rods are much longer than they are wide, and are mostly oriented in the same direction. The spacing between the rods is on the order of the wavelength of light. Imagine, for lack of a better analogy, the bars on a prison cell.
When light enters the polarizing film, the mode with electric field oriented along the direction of the rods excites the electrons inside the rods and vibrates them, in much the same way as happens in an antenna. Thus the light interferes with itself and cancels out.
The mode with electric field oriented perpendicular to the direction of the rods is not able to excite the electrons inside very well, as they are restricted to only moving in a very small length (the width of the rods). Thus that mode is unaffected and passes through.
This is a very very simplified picture and there's lots of other ways to polarize light without the use of conducting material at all. But it's useful to get a general idea of what happens.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/429005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Mixed canonical transformation Wikipedia and most authors denote four types of canonical transformations: $F_1(\mathbf{q},\mathbf{Q})$
, $F_2(\mathbf{q},\mathbf{P})$, $F_3(\mathbf{p},\mathbf{Q})$ and $F_4(\mathbf{p},\mathbf{P})$. However, is it possible (or necessary, perhaps) to define a transformation that mixes both "old" and "new" canonical variables?
For instance, the following transformation is canonical:
$$
\begin{matrix}
Q_1=q_1, & P_1=p_1-2p_2,\\
Q_2=p_2, & P_2=-2q_1-q_2.
\end{matrix}
$$
I have found that a possible generating function for this transformation would be
$$
G(q_1,q_2,P_1,Q_2)=P_1q_1+Q_2(2q_1+q_2),
$$
which doesn't fit any of the four traditional types.
P.S.: I found $G$ by (i) making a canonical tranformation $\tilde{Q}_2=-P_2$, $\tilde{P}_2=Q_2$, (ii) finding a $F_2$-type function and (iii) reverting back to $Q_2,P_2$.
| Classical Mechanics (Third Edition), H. Goldstein, Chapter 9, Page 374:
[...]
Finally, note that a suitable generating function doesn't have to conform to one of the four basic types for all degrees of freedom of the system. It is possible. for some canonical transformations necessary, to use a generating function that is a mixture of the four types. To take a simple example, it may be desirable for a particular canonical transformation with two degrees of freedom to be defined by a generating function of the form $$F'(q_1,p_2,P_1,Q_2,t)$$[...]
I think that essentially answers your question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/429197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple DC motor with one bar magnet https://www.instructables.com/id/The-Simple-DC-Motor/
I'm struggling to get my head around how this motor works.
Let's say for the sake of argument that the field lines are running from the LHS of the bar magnet round to the right and that current flows through the coil in a CW direction.
In the bottom right of the coil this would give us a force out of the page. but as the current travels round to the bottom left side of the coil it now has a vertical upward component and the force would be in to the page.
What am I missing?
| You are missing the fact that this simple motor has a commutator which does not allow current to flow during half a revolution of the coil so the direction of the force/torque is never reversed it just goes from being in one direction to zero to being in the same direction as before to zero etc.
Hold the loop vertically by placing your thumb through the center of the rotor Place one
of the straight sections of wire on a flat surface. Using a razor blade, strip ONLY the
TOP surface of the wire. Be sure not to strip the sides or the bottom, just the top. Strip
the wire from the coil all the way to the end of the straight section
is the first instruction which enables the current to the coil to be switched on and off during one revolution.
The other important feature of this motor is that it uses a magnadur magnet which has magnet poles top and bottom as shown in the diagram below.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/429409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can the Earth's negative charge make negatively charged objects levitate? Since the earth's surface has a negative charge, could it repel a large highly negatively charged body? An example is an object on stilts carrying a negative charge that is heavier than air, maybe multiple tons in weight.
| The strength of the electric field near the surface of the Earth due to its negative charge is reported to be around $100 \frac V m$.
This field should easily lift an electron, since the force acting on an electron, $eE=1.6\times10^{-19}C\times 100 \frac V m=1.6\times 10^{-17}N $ is orders of magnitude greater than the weight of an electron $m_e g=9.1\times10^{-31}kG\times 9.8 \frac m {sec^2}=8.9\times10^{-31}N$.
This does not scale favorably, however. As the of an object increases, its mass grows as a cube of its linear dimension, while its charge density as a square.
As a result, for a large object to be lifted by the Earth's electric field, it would have to be charged to an unsustainable voltage level.
As an example, we can consider a conductive sphere with a diameter of $1m$ and assume that it weighs $1kG$ or about $10N$. The charge required to develop $10N$ force in the $100 \frac V m$ electric field, would be $Q=\frac F E=\frac {10N} {100 \frac V m}=0.1C$.
The capacitance of such sphere is $4\pi \epsilon_0R \approx 55pF$, which means that, if it is going to be charged to $0.1C$, its potential will be $V=\frac Q C=\frac {0.1C} {55\times 10^{-12}F}=1.8\times 10^9V=1.8GV$, which is obviously impractical.
Lifting objects "multiple tons in weight" would be even less realistic.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/429508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What happens at the molecular level during evaporation? During boiling the liquid gets its energy from the heat source so it can break the force between molecules and turn into gas while evaporation happens at the top of the liquid at any temperature. Where does it get it's energy, I've read that when molecules bump they exhange kinetic energy and if one gets enough energy they turn into the gas state if it is correct why does it only occur how is it affected by humidity and wouldn't the air molecule bump into the liquid molecules so they go downwards not upwards. Also how would humidity affect evaporation if they have enough energy to escape why would saturated conditions prevent them from escaping. I've read a lot of answers but some of them deny other answers. It is been frustrating me that I can't even under stand simple concepts.,
| Evaporation is not simple to understand, even wikipedia omits a major factor in evaporation called entropy which is powerful force. The above explanations focus on the kinetic aspect of things but entropy is just as important. Entropy basically says that atoms and molecules will mix together and once this happens you can not reverse it unless you apply a lot of energy to repurify. In other words the energy of the system has deceased while entropy has increased. Entropy is taught in 2nd year university typically in a thermodynamics course, it's a tough subject. An example is putting salt of ice(say -5c), the ice melts, forming a solution colder than -5C.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/429640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can velocity be a tensor? I have just begun studying general relativity and have a question.
I know that if a tensor is zero in one coordinate system, it will be zero in all coordinate systems.
So how can the $\mu$'th component of the velocity $dx^\mu/d\tau$ be a tensor?
There must be something I have misunderstood.
| In your question, you ask "how can the $\mu$-th component of $\frac{d x^{\mu}}{d \tau}$ be a tensor?"
To put it simply, it is not a tensor. The thing that is actually the tensor is the four-velocity $v$. The numbers $\frac{d v^{\mu}}{d \tau}$ are the components of this tensor in some particular coordinate system $x^{\mu}$. This could be Cartesian, spherical, etc.
You are correct that if you have an equation that sets the components of a tensor equal to zero, i.e.
$$
A^{\mu\nu} = 0, \text{ for a rank (2,0) tensor, or } B^\mu = 0 \text{ for a rank (1,0) tensor,}
$$
then these components are zero in every coordinate system. However, for the four-velocity there is no such true equation. You can't write $dx^{\mu}/d\tau = 0$, because even if the object is at rest (i.e. it has zero 3-velocity $\vec{v}$ ), the four velocity will be
$$
v^\mu = \begin{cases}c & \mu = 0 \\ 0 & \mu = 1 \\ 0 & \mu = 2 \\ 0 & \mu = 3 \end{cases}
$$
(This is not typical notation, usually people say would $v^\mu = (c,0,0,0)$, but for clarity I write it as a statement for each value of $\mu$). This is because the four velocity generally looks like this in Cartesian coordinates:
$$
v^\mu = \begin{cases} \gamma c & \mu = 0 \\ \gamma v_x & \mu = 1 \\ \gamma v_y & \mu = 2 \\ \gamma v_z & \mu = 3 \end{cases},
$$
with $\gamma = \left(\sqrt{1-|\vec{v}|^2/c^2}\right)^{-1}$ the Lorentz factor, and $v_x, v_y, v_z$ the usual components of the 3-velocity in Cartesian coordinates.
Thus the four-velocity can never be entirely zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Equation for probability amplitude of a free particle given a mean position, a mean velocity, and the mass of the free particle? The uncertainty principle can be expressed using the equation $\sigma_x\sigma_p\geq\frac{h}{4\pi}$ with $\sigma_x$ being the uncertainty in position, $\sigma_p$ being the uncertainty in momentum, and $h$ being the plank constant. The uncertainty in velocity would be given by the equation $\sigma_v=\frac{\sigma_p}{m}$ with $\sigma_v$ being the uncertainty in velocity and $m$ being the mass. So the uncertainty principle could also be expressed using the equation $\frac{\sigma_x\sigma_v}{m}\geq\frac{h}{4\pi}$.
Assuming that both the uncertainty in position and uncertainty in momentum are both at a minimum what is the equation for the probability amplitude of a free particle at a chosen position and momentum, given the mean position, and mean momentum of that particle?
| A usual definition of the wavefunction of a free particle is simply $\psi(x) = \exp(-ikx)$. This of course corresponds to a particle with definite momentum that is infinitely 'smeared out' in space ($\delta p = 0$, $\delta x = \infty$). I assume you're asking about the wavefunction of a free particle with the constraint that it's localized in space. A unique localized particle wavefunction does not exist because it depends on how you write down your Hamiltonian. You can get for instance a 'wave packet'-like wavefunction, defined as (in 1D):
$$
\Psi(x,t) = \left({a \over a + i\hbar t/m}\right)^{3/2} e^{- {x^2\over 2(a + i\hbar t/m)} }.
$$
Which gives a Gaussian wave packet, with width $a$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lorentz Velocity Transform With Tensor Notation So I'm attempting to prove the Lorentz Velocity tranform:
$${v_x}' =\frac{v_x-u}{1-v_xu/c^2} $$
using tensor notation. In this case obviously $\beta = u/c$ and $\gamma=(1-\beta^2)^{-1/2}$. The velocity transform tensor can be represented as
$$\Lambda = \begin{pmatrix}
\gamma & \beta \gamma & 0 & 0 \\
\beta \gamma & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} $$
assuming the the boosted frame $F'$ is moving in the x-direction from frame $F$. I'm also using the following two facts:
$${\partial_v}' = {\Lambda^u}_v \partial_u \hspace{20mm} {x^v}'={(\Lambda^{-1})^v}_u x^u$$
where
$$\partial_u = \left(\frac{1}{c}\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right)$$.
My proof begins as follows:
$${v_x}' = c{\partial_0}'{x^1}' = c({\Lambda^{u}}_0 \partial_u)({(\Lambda^{-1})^{1}}_{\sigma}x^{\sigma}) $$
I use $\sigma$ to keep the summation seperate. Now I expand:
$${v_x}' = c({\Lambda^{0}}_0 \partial_0+{\Lambda^{1}}_0 \partial_1)({(\Lambda^{-1})^{1}}_{0}x^{0}+{(\Lambda^{-1})^{1}}_{1}x^{1}) $$
Subbing in the appropriate elements from the tensor yields:
$${v_x}' = c(\gamma \frac{1}{c}\frac{\partial}{\partial t}+\beta \gamma \frac{\partial}{\partial x})(-\beta \gamma c t+\gamma x) $$
$$=c(-\beta \gamma^2 \frac{\partial t}{\partial t}+\frac{\gamma ^2}{c}\frac{\partial x}{\partial t} - \beta^2\gamma^2c\frac{\partial t}{\partial x}+\beta \gamma^2 \frac{\partial x}{\partial x})$$
at this point I make the (perhaps incorrect) assumption that $\partial t/\partial x=1/v_x$. Canceling out the obvious terms leaves me with
$${v_x}'=\gamma^2v_x -\frac{\beta ^2 \gamma^2 c^2}{v_x} $$
which I know to be incorrect.
| Einstein velocity addition equation in Minkowski space
\begin{align*}
&\text{I) We want to show that The velocity addition equation is:} \\
&v_g=\frac{v_1+v_2}{1+\frac{v_1\,v_2}{c^2}}\\
&\text{ We take the coordinate transformation between $(t'\,,x')$ and $(t\,,x)$ (c=1)}\\\\
&\begin{bmatrix}
t' \\
x' \\
\end{bmatrix}=\underbrace{\gamma(v)
\begin{bmatrix}
1 & -v \\
-v\ & 1\\
\end{bmatrix}}_{L(v)}
\begin{bmatrix}
t \\
x \\
\end{bmatrix}\\
&\text{$L(v)$ is the Lorentz transformation matrix.} \\\\
&\text{II) we take additional coordinate transformation between $(t"\,,x")$ and $(t'\,,x')$ we get }
\end{align*}
\begin{align*}
\begin{bmatrix}
t" \\
x" \\
\end{bmatrix}=L(v_2)
\begin{bmatrix}
t' \\
x' \\
\end{bmatrix}=
L(v_2)\,L(v_1)\,
\begin{bmatrix}
t \\
x \\
\end{bmatrix}
&=\underbrace{\gamma(v_1)\,\gamma(v_2)\begin{bmatrix}
1 & -v_1 \\
-v_1\ & 1\\
\end{bmatrix}\begin{bmatrix}
1 & -v_2 \\
-v_2\ & 1\\
\end{bmatrix}}_{L(v_1\,,v_2)} \begin{bmatrix}
t \\
x \\
\end{bmatrix}\\
&=\underbrace{\gamma(v_1)\,\gamma(v_2)\,(1+v_1\,v_2)}_{\gamma(v_1\,,v_2)}
\begin{bmatrix}
1 & -\frac{v_1+v_2}{1+v_1+v_2} \\
-\frac{v_1+v_2}{1+v_1+v_2} & 1 \\
\end{bmatrix}
\end{align*}
\begin{align*}
&\text{with:}\\\\
&\gamma(v_1\,,v_2)=\frac{1}{\sqrt{1-v_1^2}}\,\frac{1}{\sqrt{1-v_2^2}}
(1+v_1\,v_2)=\frac{1}{\sqrt{1-\left(\frac{v_1+v_2}{1+v_1\,v_2}\right)^2}}\\\
&\Rightarrow\\
&\text{The Lorentz transformation matrix between $(t"\,,x")$ and $(t,x)$ is:}\\\\
&L(v_1,v_2)=\frac{1}{\sqrt{1-v_g^2}}
\begin{bmatrix}
1 & -v_g \\
-v_g & 1 \\
\end{bmatrix}\quad, \text{with:}\\\\
&\boxed{v_g=\frac{v_1+v_2}{1+\frac{v_1\,v_2}{c^2}}}\quad \text{Einstein velocity addition equation in Minkowski space}
\end{align*}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If we push a syringe with gas in it In space , would its temperature increase permanently , Violating Boyles Law? On Earth
If we push a gas syringe very fast Inwards , work is done on the gas by
W = PV
which increases the Kinetic energy of molecules in it and since Temperature is Average Kinetic Energy thus Temperature increases However due to neighbouring atoms(around syringe) the kinetic energy is conducted outwards until Thermal Equilibrium is reached and this is how Boyles Law is maintained
But In space there is no air , so the Temperature would Increase Permanently
Hence Violating Boyles law ,
So my question is If my conclusion is correct, do gas Laws only work where there is AIR ?
| This will happen only if the container in which gas is present conducts 0 heat. This means that their is no exchange of heat from inside to outside. This will also happen on earth if we use that type of container and it is not the violation of boyle's law
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/430895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Non-acceleration and 0 force If a mass is moving at the rate of 30/ft. per minute for 5 minutes on a straight line and it strikes a second stationary mass and effects a change of position to this second mass, then we know from F=ma that the force is 0 since the acceleration is 0. Then why do we say because it is 0 that there is no force when in fact the first mass changed the position of the second.
| It appears you are mixing the force at one given time with the acceleration at a different given time.
Don't feel bad; it is a common beginner mistake to think that there is "one force" and "one acceleration" in a given problem. Part of learning physics is learning how to treat values from different times, as well as values from different objects.
During the 5-minute interval, the first mass stays at the same velocity. This means its acceleration is zero. By $F_{net}=ma$, the net force on this object during that time interval is zero.
During this same time interval, the second mass stays at rest. Its acceleration is zero, and thus the net force on it is zero.
The collision occurs at a different time. Neither force nor acceleration care about what happened during the 5-minute interval, they only depend on what is happening now. The first object is now slowing down -- a form of acceleration -- so $a \ne 0$. This means the net force on the first object $F_{net}=ma\ne 0$. This net force comes from the second mass pushing on the first mass.
By the Third Law, during the collision the first mass pushes back on the second mass. This creates a net force on the second mass, which accelerates the second mass during the collision.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Once a black hole is formed, is there anything other than Hawking radiation which shortens its life? Hawking radiation is supposed to very slowly evaporate a black hole (terms and conditions apply :] ).
Apart from Hawking radiation, is there any mechanism or effect that can make a black hole cease to exist? Or once they are formed are they expected to exist in this form "forever"?
| There is speculation that a black hole which somehow manages (or is made) to rotate fast enough could turn into a naked singularity. The missing event horizon of such a speculative anomal anomaly would imply that "stuff" could escape from it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Why doesn't using more appliances at home decrease the electricity bill? I know that at home the electric circuits are parallel, and this explains why if one appliance (eg bulb) fails, everything else continues to work, but if more devices are added in parallel to each other, their combined resistance should decrease, and thus the total power supplied by them should increase, as $\mathrm{Power} = V^2/ R$. It doesn't look like that's the case, what am I getting wrong?
| The power will increase .See the formula given carefully $V^2/R$. For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think in the way if overall resistance decreases then current increases and hence $P=VI$ increases
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Symmetries in quantum mechanics I have been studying symmetries in quantum mechanics and I have come across two types:
*
*Given some transformation on a Hilbert space of states $\mathcal{H}$, an operator $U: \mathcal{H} \rightarrow \mathcal{H}$ is a symmetry operator if
$$ U^\dagger H U = H$$
where $H$ is the Hamiltonian of the system.
*A symmetry transformation is either a unitary or anti-unitary operator on $\mathcal{H}$ (Wigner's Theorem).
Are these two definitions of symmetry equivalent?
| In the first definition you have not made $U$ preserve the $\mid \langle \psi \mid \phi \rangle \mid$. As an example, suppose $H=0$ (a trivial system), then you would get all operators $U$ as symmetries even if they aren't even unitary or anti-unitary. Those shouldn't have been called symmetries. You want both conditions not either.
The implications go the other way
Symmetry => Unitary or Anti-unitary
Symmetry => Preserves Hamiltonian under $U^\dagger H U$
You need both conditions to say something might be called a symmetry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
What is the difference between reflection and emission? What is the difference between a body that simply reflects the light that falls upon it and a body that absorbs and emits it (like a black body in thermal equilibrium)?
How can one experimentally distinguish these two scenarios by simple measurements?
(I think that reflected and transmitted light do not change the temparature of the body where as emitted and absorbed light do. Is this correct and is it possible to experimentally tell the difference between emission and reflection using this information?)
And what are the mechanisms behind reflection and emission?
( I think emission is related to the electron transissions. But how does a photon decides whether to be reflected or transmitted if it is not absorbed ? )
| In both cases, light falling on an object is a vibrating electromagnetic field. This makes electrons in the object vibrate and may make them more energetic.
A mirror is a smooth object with free electrons. Light makes them vibrate. Vibrating electrons are accelerated. Accelerated electrons radiate. For a mirror, the radiation is the same as the incoming radiation, but in a different direction. If the surface isn't smooth, it is a lot of little mirrors in many directions.
In some objects light gives the electrons energy, and the electrons pass the energy on to atoms around them. This makes the atoms vibrate more, making them hotter. The energy gets passed around randomly. In some cases, energetic electrons emit radiation. In this case, the emitted radiation is not so directly tied to the incoming radiation. The outgoing radiation depends mostly on the temperature of the object.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Instantaneous velocity So here’s a question I’ve been thinking of for a while. Suppose we say, “an object is having an instantaneous velocity along a particular direction ( say 10 m/s along the $x$-direction)” . Is it fair to conclude that it is traveling in a straight line along the $x$-axis? Well my opinion on this is,
For instance, a projectile ( on earth ) , the instantaneous velocity ( which is constant through out the journey ) is always in the $x$-direction while the body is executing a parabola in the $x$-$y$ plane? Please acknowledge me if I’m wrong.
| In a parabola, the instantaneous velocity isn't constant or only acting in one direction while travelling through it's path. It is always tangent to the parabola.
An object which has an instantaneous velocity in one direction does not have to keep travelling in that direction as long as there are other forces acting on it (for example, at the top of a parabola, the instantaneous velocity is purely horizontal, but then as soon as you go away from the top, there is a downwards component and it's not longer just in the x direction.
Instantaneous velocity is just what it's name suggests; the velocity of something at the exact instant you are analyzing it. With only information about the instantaneous velocity, that's not enough to say the velocity at another point in time, as forces can change the velocity of the object.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/431908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Are Minkowski and Schwarzschild spacetimes diffeomorphic? Another mathematical question, arising from GR. Some days ago I wrote, in an answer to 1, that they are. Then @magma commented they are not. He promised a proof, but none appeared. After magma's comment I have some doubts about my intuition and I would like to see a proof, in one sense or the other, or at least a reliable reference.
| They aren't. That is because Kruskal-Schwartzschild is diffeomorphic to $S^2 \times R^2$, while Minkowski is $R^4$. This is roughly similar to the difference between a 2D cylinder $S^1\times R$ and $R^2$.
The formal reason is that pointed out by G. Golfetti in a comment above: the second homotopy groups are different, as is expected by the presence of the factor $S^2$. In M. space you can shrink two-spheres continuously into points, in K-S space you cannot: when you encounter the bifurcation surface (the 2D event horizon) you cannot go further and you pass from the right wedge to the left wedge. The standard global Kruskal coordinates map K-S manifold to $S^2\times R^2$. There is no way to make vanishing the radial coordinate. It would vanish on the singularity, but it does not belong to the manifold. At the center of the manifold it takes the Schwarzshild radius value which is strictly positive. There is no center of those spheres. Similar coordinates can be defined on Minkowski spacetime but they are not global since they do not cover the set $r=0$ which is part of the manifold.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is Chirped Pulse Amplification, and why is it important enough to warrant a Nobel Prize? The 2018 Nobel Prize in Physics was awarded recently, with half going to Arthur Ashkin for his work on optical tweezers and half going to Gérard Mourou and Donna Strickland for developing a technique called "Chirped Pulse Amplification".
In general, optical tweezers are relatively well known, but Chirped Pulse Amplification is less well understood on a broader physics or optics context. While normally the Wikipedia page is a reasonable place to turn to, in this case it's pretty technical and flat, and not particularly informative. So:
*
*What is Chirped Pulse Amplification? What is the core of the method that really makes it tick?
*What pre-existing problems did its introduction solve?
*What technologies does it enable, and what research fields have become possible because of it?
| As an addendum to @EmilioPisanty’s excellent review, I’d just like to mention one more application of CPA lasers, which may be overlooked from a theorist’s perspective:
Ultrafast Spectroscopy
Sometimes lower-order nonlinear processes like second-harmonic generation are enough; you just need them done efficiently for practical purposes.
Lasers based on chirped-pulse amplifiers, which have pulses with energy ~5 mJ and duration ~30 fs are the backbone of all but the simplest ultrafast pump-probe spectroscopic techniques. This is because such pulses allow for efficient low-order nonlinearities such as sum-frequency generation, optical parametric amplification, etc., which have greatly expanded the parameter space of these types of experiments. People are clever, and when given more than enough laser power to work with, they come up with increasingly more complex tricks to ferret out the truth.
Without launching into a review of thousands of ultrafast spectroscopic research results, let it suffice to say that humanity knows a lot more about how the world operates on picosecond and sub-picosecond time scales (in fields as disparate as condensed matter physics and quantum biology) thanks to CPA-based lasers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "299",
"answer_count": 2,
"answer_id": 1
} |
What happens on the atomic level that allows us to see objects that have the same order of magnitude as visible light's wavelength? By virtue, human eyes can only see EM waves in the visible light region. From my understanding of 'why we see things', it is because light reflects off an object, and the lights' diffraction patterns that enter our eyes determine our ability to resolve it.
My textbook then says "the wavelength of visible light is too long for light to interact effectively with individual atoms and molecules."
However, what exactly does "interact effectively" mean? Does it mean light doesn't reflect from individual atoms or molecules? If so why would the wavelength of a wave affect its ability to reflect off a small thing e.g. an atom?
|
the wavelength of visible light is too long for light to interact effectively with individual atoms and molecules
That's some extremely clumsy wording on the part of your text. The wavelength of visible light is indeed much, much longer than the characteristic sizes of atoms and molecules (half a micron to half an angstrom) but that doesn't mean that they cannot interact.
It does mean that as far as the light is concerned both atoms and molecules are point sources, so the light won't be able to interact with the specific shape of the molecule or image it.
However, if the frequency of the light is correctly tuned to the frequencies at which the molecule can respond (sharp levels in gas phase, broader bands in condensed matter), then the light will induce charge oscillations in the molecule and those will feed back to the EM field as scattered or reflected light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Time evolution of eigenstates superposition If a system is in a state $\psi$ which is superposition of, let's say two, energy eigenfunction, namely $\psi_1$ and $\psi_2$, so that $$\psi(t)=\psi_1e^{-i\omega_1t}+\psi_2e^{-i\omega_2t}$$ (I am omitting normalization constants for simplicity) then $$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2+\psi_1^* \psi_2 e^{i(\omega_1-\omega_2)t}+\psi_2^* \psi_1 e^{i(-\omega_1+\omega_2)t}$$
right?
But isn't it true that since $\psi_1$ and $\psi_2$ are orthogonal eigenfunctions then $$\psi_1^* \psi_2$$ and $$\psi_2^* \psi_1$$ must be zero and so one would have:
$$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2$$
How does it work? thanks
| Your notation is ambiguous so it's not quite clear what you're actually asking. If by $\psi_1$ you mean the position basis-wavefunction $\psi_1(x)$, then as the other answers point out, only the integrals of the cross-terms with respect to $x$ vanish, not the pointwise products. If by $\psi_1$ you mean the basis-independent ket $|\psi_1\rangle$, and by $\psi_1^*$ the corresponding bra $\langle \psi_1|$, then your last equation is correct as written (with no need for any integration), and simply says that the magnitude of $|\psi\rangle$ is preserved over time, so if it's correctly normalized at one time then it remains correctly normalized at all other times.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Speed of EM waves differ from GWs? I understand that light travels at speed c in vacuum, when measured locally. This speed has an exact value, not an approximation, because it is defined as 299 792 458 m / s.
It is an exact value, because the meter is defined as the distance that light travels in a 1/299 792 458th of a second.
Now as per SR, all particles that have no rest mass must travel at speed c in vacuum, when measured locally.
This goes for photons (that build up the EM waves), and the (hypothetical) gravitons (that build up GWs). So both photons and gravitons have to travel at speed c. Both EM waves and GWs have to travel at this speed in vacuum when measure locally.
So theoretically there can be no difference between the speed of EM waves and GWs (assuming they propagate both in vacuum).
Then I found this:
http://iopscience.iop.org/article/10.3847/2041-8213/aa920c/pdf
And it talks about an experimental evidence about non-zero difference between the speed of EM waves and GWs. How is that possible? I thought the speed of light is an exact value, and it has to go for all massless particles.
Question:
*
*How can there be experimental evidence of non-zero difference between the speed of EM waves and GWs?
| Because GR is just one of the many models of gravity. From the observation of gravity's long range, classical GR is based upon masslessness of graviton. GR is the unique theory of massless spin 2 particles.
But you can give up masslessness and study the ensuing consequences of such a modified theory of gravity. Such models have massive gravitons. This is called Massive gravity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Confusion with regards to uncertainty calculations Let’s say we have a scenario of a ball being released from the top of the building. This can be modeled simply with the kinematics equation $S=ut +\frac{1}{2}at^2$, which reduced to $S=\frac{1}{2}at^2$. We are given $\Delta t, t, \Delta S, S$, are we are to find $a, \Delta a$.
Firstly, I have no problems calculating the absolute portion of the uncertainty.
Here is my problem: Differentiating $S=\frac{1}{2}at^2$ gives me $\frac{\Delta S}{S}=\frac{\Delta a}{a}+2\frac{\Delta t}{t}$. However, substituting these values gives me a wrong value of $\Delta a$.
The correct approach should have been to rearrange the equation to $a=\frac{2S}{t^2}$, and then solve $\frac{\Delta a}{a}=\frac{\Delta S}{S}+2\frac{\Delta t}{t}$. As can be seen, there appears a contradiction.
Further substitution of $S=82m,\Delta S=1m,t=4.1s, \Delta t=0.2s$ to solve for $a, \Delta a$ using the second equation and then putting this value back into the first gives me a contradiction.
I would like to know which one is correct and which should be used because both seem correct to me.
I have discovered that the addition/subtraction of uncertainties is as follows. Let’s say $(A\pm\Delta A)+(B\pm\Delta B)=(C\pm\Delta C)$.
Then $C_{max}=(A+\Delta A)+(B+\Delta B), C_{min}=(A-\Delta A)+(B-\Delta B)$. Referring back to the definition of uncertainty, $C+\Delta C$ is the average of the minimum and maximum of $C$, thus giving us $C=A+B$ and $\Delta C=\Delta A+\Delta B$.
Using this principle, I am however confused by what I get. $C_{max}=(A+\Delta A)(B+\Delta B), C_{min}=(A-\Delta A)(B-\Delta B)$. Expanding, I got $C=AB +\Delta A\Delta B$, which was contradictory to what I have learnt. I got $\Delta C=A\Delta B + B\Delta A$, which was correct though... This raises a new problem, as I am now unsure as to why the rule applies to multiplication.
| So I am pretty sure the difference between your two equations depends on what you actually measured and what you are calculating from those measurements.
Think of it this way: One of your two equations (before your addition to the question), tells you how to calculate an uncertainty from the uncertainties of your measurements. In other words, these equations are not relationships between any uncertainties you want. You use the measured quantities on the right to find the uncertainty of the value on the left. You never actually measured the value on the left. (Although if you did a different experiment to directly measure the value on the left, then you would want it to be consistent with said calculations).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/432831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why the rubber ball bounced higher than the glass ball? For a teaching degree we did the following experiment:
Drop a rubber and a glass ball of approx. same size from approx. same height onto laminate flooring.
As we expected, the rubber ball bounced higher than the glass ball.
Our explanation: The rubber ball is more efficient in converting the kinetic energy into deformation energy and back. So the rubber ball transforms less energy into heat.
But alas, attempted verification with a heat camera showed otherwise: While there was a big blotch of heat left behind on the table by the rubber ball, the mark by the glass ball was barely visible.
The lecturer was very vague and eluded our question for an explanation.
So my question is:
What is the correct explanation of this?
And why does the glass ball not produce at least an equal amount of heat?
Before posting here I googled this problem and to my dismay found the following blog entry which contradicts our experiment:
https://sciencenotes.org/why-a-glass-ball-bounces-higher-than-a-rubber-ball/
| There is a property of materials called the coefficient of restitution that is given by the relative velocity after a collision divided by the relative velocity before the collision. This coefficient is often denoted by e and comes from "Newton's Law of Restitution". This number is proportional to the height of an object as it bounces off the floor divided by the distance from which it was dropped. $$ e = \frac{y_1}{y_0}$$ $$where\ y1\ =\ maximum\ height\ of\ ball\ after\ 1st\ bounce\ off\ floor$$ $$ y_0 = height\ from\ which\ ball\ is\ initially\ dropped $$
A ball with a coefficient of restitution = 1 will return to the starting point when dropped. A hard rubber ball at room temperature has $e \approx
0.9$ while a glass ball has e between 0.65-0.7. Hence the rubber ball bounces higher. There are tables of values of e on the internet and in the Wikipedia reference given below.
Some useful references are here, here, and here.
I'm not sure why the paper you linked had the reverse answer. This reference has measurements for balls of different material and include a steel ball. Maybe the study done in the linked paper used a very soft rubber ball. All things being equal, a hard solid rubber ball ( think superball) will bounce much higher than a steel or glass ball.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Reverse engineering a time dilation problem Consider the following problem
A rocket with a clock moves at $0.8c$ relative to the earth. An observer A on the rocket measures a time interval of 6 seconds. With respect to an observer B on the earth, the time interval is $$\frac{6}{\sqrt{1-(0.8)^2}}=10$$ seconds.
I got confused if I reverse engineer the problem as follows.
The observer A knows that the observer B will measure 10 seconds. With respect to A, the observer B moves at $0.8c$. The observer A calculates
$$\frac{10}{\sqrt{1-(0.8)^2}}=\frac{50}{3}$$ seconds which is not the same as the original one (6 seconds).
Question
What is wrong in my understanding?
| $\let\D=\Delta \let\g=\gamma$
Your question (and your misunderstanding) is recurring over and over. Your first application of time dilation formula is right, the second is wrong. Why?
Simply because some simple points must be kept, referring to reference frames and events.
You have two frames, say A (the rocket) and B (the Earth). Then you think of a clock at rest in frame A measuring an interval of 6 seconds. Time interval between what? Between two events, happening in frame A: starting ($E_1$) and stopping ($E_2$) of the clock.
B observes the same events, but is reckoning their times with two clocks, standing in frame B. Two clocks are needed, since events $E_1$ and $E_2$ do not occur in the same place of B. In this situation it correct to write
$$\D t_{\rm B} = \g\,\D t_{\rm A} \tag 1$$
as you did.
Your second calculation instead cannot refer to the same events. It would apply if you were talking of a clock on earth, measuring an interval of 10 s and observed from the rocket. Now events are no longer $E_1$ and $E_2$ but two other, say $F_1$ and $F_2$: start and stop of clock on earth. These events happen in the same place of frame B, but if you want to observe them from A and measure their interval, you need two clocks standing in frame A. Then again you may use time dilation, in reverse:
$$\D t'_{\rm A} = \g\,\D t'_{\rm B}.\tag 2$$
I've used $\D t'$ and not $\D t$, because the former intervals, eq.
(1), referred to events $E_1$ and $E_2$. Intervals in eq. (2)
refer to events $F_1$ and $F_2$ instead.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How is momentum conserved in this example? Suppose a sticky substance is thrown at wall. The initial momentum of the wall and substance system is only due to velocity of the substance but the final momentum is 0. Why is momentum not conserved?
| You should also consider what the wall is attached to. And obviously it is the Earth. If we assume the Earth's velocity is zero after the substance is thrown, since there is the force that slow down the substance at the moment of impact, there is also the reaction force on Earth with the same magnitude and opposite direction. So Earth will gain velocity and final momentum of combined Earth and substance system will be equal to the intial momentum of the substance.
And also we can look at the situation in a bit different way. When we stand on the floor and throw the substance, there appears a friction force between our feet and the floor and it acts on us in the throw direction. So the friction force on Earth will be opposite to the throw direction and Earth will pick up speed towards the substance, too. And at any moment, Earth plus substance system will have zero momentum. The substance and the Earth will move towards each other and after the impact their speed will be zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Why do we use the RMS but not the fourth root mean quad? Why do we use the power of $2$? What is the relation between this and having the same heat energy in both AC and DC?
|
what is the relation between this and having the same heat energy in
both AC and DC?
First, the context here is the voltage across and current through a resistor. The power delivered to a circuit element is the product of the voltage across and current through. So, for a resistor (only), the instantaneous power is
$$p_R(t) = v_R(t) \cdot I_R(t) = \frac{V^2_R(t)}{R} = I^2_R(t)\cdot R$$
So you can already see why we're interested in the square of a voltage across (or current through) and not the cube or higher powers.
For a periodic voltage across with period $T$, we can ask what the average (mean) power is over the period and find
$$\langle p_R(t)\rangle = \frac{1}{T}\int_0^T\mathrm{d}t\,\frac{v^2_R(t)}{R}$$
Note that the average power (over a period) is proportional to the mean of the square of the voltage across.
Now we can ask what constant voltage across $V_R$ would give the same power as the average power of the periodic, time varying voltage.
The answer is, by inspection, the root of the mean of the square of the periodic, time varying voltage:
$$V_R = \sqrt{\frac{1}{T}\int_0^T\mathrm{d}t\,v^2_R(t)} = V_{rms}$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is the dark matter present only around the galaxies? Dark matter is believed to give galaxies its shape and prevent them from flying apart.
Is dark matter present only on outer edges of the galaxy or is it present throughout the galaxy?
| In order to explain the rotation curves of galaxies, the dark matter would have to be distributed throughout the galaxy and not only in the outer edges. More detail below:
The basic experimental results that lead to the discovery of dark matter is the rotation curve of galaxies. The rotation curve is basically a graph that plots speed of rotation (of material in the galaxy) at various distances from the center of the galaxy. We can theoretically calculate this rotation curve based on how much mass is encircled at each radius and using Newton's law of gravitation. When we went to do this calculation, we found that the theoretical results of the calculations didn't match the actual rotation curves of the galaxies. This led to the hypothesis that there is extra mass (dark matter) that isn't accounted for in the luminous matter of the galaxy. In order to explain the discrepancy of the calculated rotation curve to the real rotation curve, the dark matter would have to be pretty much distributed throughout the galaxy and not only at the outer edges.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/433941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question on convective heat transfer Why is the following formula used when we define heat transfer due to convection on a flat plate?:
$$ q=h(T_s-T_\infty)$$
Where $T_s$ is the temperature of surface at a particular location and $T_\infty$ is the temperature of fluid flowing with free stream velocity.
Why are we using $T_\infty$ here and not any other temperature of the fluid flowing over the plate?
I read in a book that this expression is derived from newtons law of cooling i.e $dT/dt=-k(T_s-T_\infty)$, where $T_\infty$ is the temperature of surroundings?
| The way I think about $t_\infty$ is that we could pick it by going to extremes: $t_\infty$ should either be the temperature of the fluid right above the plate or it should be the temperature of the fluid far away from the plate. I will argue that it is way more useful to choose the temperature of the fluid far away from the plate.
Assume the plate velocity = zero. If the fluid is not slipping, the velocity of the fluid just barely above the plate is the same as the velocity of the plate: zero. Therefore, theoretically since the fluid above the plate is not moving, the temperature of the fluid at that point will eventually reach the temperature of the plate, thus $t_\infty$ = $t_s$. If $t_\infty$ = $t_s$ then we would have no net heat transfer at all and the equation would be useless. So it is better to pick the temperature of the fluid far away from the plate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Conceptual question in thermodynamics about isothermal processes If a process is isothermal then Δ U is zero. So ΔQ is non zero. But isn't ΔQ=nCΔT. Implying that ΔQ is zero. Where am I wrong? (Considering ideal nature)
| The thing is that the relation $\Delta Q = nC\Delta T$ is true for some particular processes -- in an ideal gas, for instance, it is true when the system passes through a process at constant pressure (in which case $C = C_p$) or in constant volume (in which case $C = C_V$), but this is not true in general. In an adiabatic process, for example, you have $\Delta Q = 0$, but $\Delta T \neq0 $ -- and that does not contradict the relation above because neither pressure nor volume are constant in an adiabatic process in an ideal gas. The same holds for an isothermal process, which is the case you are interested in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why position and momentum operators are both continuous spectrum while angular momentum is discrete? We know that position $\hat{r}$ and momentum $\hat{p}$ are both continuous spectrum operators, i.e.
$$\hat{r}|r'\rangle=r'|r'\rangle, \quad \hat{p}|p'\rangle=p'|p'\rangle.$$
But the angular operator $\hat{L}=\hat{r}\times\hat{p}$ is not:
$$\hat{L}^2|l\rangle=\hbar^2 l(l+1)|l\rangle.$$
Could anyone give some explanation?
| One possible answer: boundary conditions.
If we think of the classical to quantum transition to be from point-like particles to waves which describe probability densities, then the position and momentum of a free particle does not have any explicit boundary conditions. That is, there is no restriction on the specific character of the probability density wave.
On the other hand, for a particle in orbit, with well-defined angular momentum, when replacing the point particle with a wave means the wave has to satisfy certain conditions for stability - namely, it must be continuous. That restricts the wavelengths to only be certain multiples of the radius - and viola, discreteness.
Of course, we aren't actually talking about analogous systems here - I said "free particle" for position and momentum, whereas the angular momentum quantum number only occurs for bound states, but the origin of both is boundary conditions. The free particle has none, and therefore there are no conditions set on the continuity of the waves. The bound particle must stay bound, which results in restrictions on what wavelengths of waves can actually be present.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Problem regarding finding work done This problem is given in my physics book. How much work will be done if anyone wants to stack up 12 bricks? Given that each brick is $10 cm$ high and each brick's mass is $2 kg$. ($g=9.8$)
Now in finding my answer I used the mass and height of all the bricks. But in my book they used the mass of all the bricks. But in using the height they used the height of only 11 bricks. So my result is $141.12 J$ and the book's result is $129.36 J$. Why is that?
| We are supposed to stack 12 bricks on one another. One brick is unmoved. The centre of mass of 11 bricks is 60 cm or 0.6m above initial level after stacking.
Now, the work done against the force of gravity is:
$W = mg \Delta h = 2(11)(9.8)(0.6) = 129.36J $
Another method is to calculate work done individually
$W = mg(\Delta h_1+\Delta h_2+\Delta h_3+\Delta h_4+\Delta h_5+\Delta h_6+\Delta h_7+\Delta h_8+\Delta h_9+\Delta h_{10}+\Delta h_{11})$
$\implies W = 2(9.8)(0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9+1+1.1)$
$\implies W = 129.36J$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does the Boltzmann equation deal with single-particle phase space density? Why does the Boltzmann equation deal with single-particle phase space density $\rho_{1}(\textbf{r}_1,\textbf{p}_1,t)$ rather than the N-particle phase space density $\rho(\{\textbf{r}_i,\textbf{p}_i,t\})$? How'll deal with Boltzmann equation for a system of $N$ interacting particles?
| That's the whole point. Rather than dealing with the complicated N-particle distribution we try to find an equation of motion for the 1-body distribution
$$
f_1(r_1,p_1,t)=\int d\Gamma_2\ldots d\Gamma_n\, f_N(r_1,p_1,r_2,p_2,\ldots).
$$
Boltzmann brilliantly guessed an equation for $f_1$, and standard textbooks try to at least outline a proof that starts from the Liouville (classical) or von-Neumann (quantum) equation, but obviously there are some subtleties (how do you get the non-reversible Boltzmann equation from reversible microscopic dynamics), and much ink has been spilled discussing these subtleties.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pressure on side walls of container My textbook mentions the standard derivation of finding pressure at a depth of a liquid by considering a cylindrical portion of liquid an then using the equilibrium of forces and hence pressure comes out to be $$hdg$$.What I want to ask is that the above method is correct but then it is mentioned that pressure on a point of side wall at a depth of $$h$$ would also be $$hdg$$ without giving any explanation. If I try to use the above stated derivation then there is problem that I can't use the weight since gravity acts downward and also I don't know the force exerted by fluid on side walls. So how can I say that pressure at depth is same as pressure on side walls?
| Pressure at a depth in a liquid is equal to the pressure exerted by the liquid on a surfce at that depth in the liquid.Because pressure at a point is the same in all directions and if the liquid particles exert a pressure on each other then they exert the same pressure on any other surface immersed at that depth.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Orientation of momentum of a virtual photon Picture a wire with current directed out of the page. The electrons in the wire emit virtual photons in all directions which mediate the produced magnetic field. What is the orientation of the momenta of the emitted virtual photons, or are they randomly oriented?
| I disagree with the other answer. If the electrons are moving with a constant velocity (i.e. not accelerating), then there are no real photons emitted.
In the rest frame of the moving electrons, the electrons will set up an electric field (in this case a field that will point inwards and drop like a log of the distance from the line of charge). The Fourier transform of this field will give the direction of the momenta of the virtual photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/434983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why naturalness is an issue in particle physics? It is not in fluid dynamics I have often heard that naturalness is a strong argument in favor of Supersymmetry. They say that there must be energy scales not too far away the EW-energy so that the quantum corrections of the heavy particles (Higgs, top quark, etc) do not diverge.
The argument goes like this: the Plank Energy scale is so huge that if nothing else exists in between the EW scale and the Plank scale, renormalization would pull all "natural numbers" from the EW to the Plank scale.
However, there are some interesting problems in physics where two extremely large numbers almost exactly cancel: in a 2D turbulent flow of an imcompressible fluid, the integrated vorticity is almost zero, even though positive and negative vorticity numbers are huge.
| I think that this is a misguided example. In isotropic turbulence the mean vorticity is zero (by rotational invariance), but the mean square vorticity is not. This means that there is indeed a symmetry that ensures that the mean vorticity is much smaller than the root mean square vorticity. This is precisely what naturalness says: If a number is small, there should be a symmetry reason.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Confusion while deriving kinetic-molecular theory of gases
How many of the molecules will collide within a given area A of the
wall normal to the x-direction in some fixed time interval Δt ? On average half the molecules have a positive x-component of velocity. Therefore half the molecules contained inside a cylindrical volume of cross sectional area A and length
vΔt will strike the wall within a given area A during the time
interval Δt.
So, the number of collision with A during $\Delta{t}$ is
$\frac{1}{2}\frac{NAv_x\Delta{t}}{V}$
So my question is how can we assume that half of the molecules are moving one way and half the other? I know that we are focusing on one co-ordinate i.e. x-coordinate, thus, we are not focusing in other directions like up and down etc.Nevertheless,I wondered, if I were the physicist coming up with this derivation, why would I divide the number of collisions by 2 saying that half of the particles go one way and half the other way. How can it be just half not one third any other probability.
Thanks for the time and help.
| A simple way of thinking about this is to think about the motion of particles through an imaginary plane somewhere inside the fluid. The number of particles crossing the plane from one side to the other must be the same as the number of particles crossing in the other direction. This must be the case because first of all the gas is homogeneous and it has the same properties (e.g. average kinetic energy, particle velocity, etc.) everywhere. If the particles moved from one side to the other more than in the other direction, you'd have a build-up of particles on one side, violating the second law of thermodynamics.
For particles near a wall, they don't 'know' they are near the wall until they hit it. So the same rule must apply to particles near a wall.
Viewing it from the perspective of the 2nd law might be the easiest way of thinking about it, but you can also view it purely from the basic assumptions of an ideal gas: a bunch of non-interacting particles. If you had just a single particle in the box, the same law would apply. On average, over long enough time, the particle would find itself moving toward a wall the same number of times it would find itself moving away from it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is extracting energy via the Penrose Process only possible with Kerr black holes? I have been learning about extracting energy via the Penrose Process. Everything I read mentions leveraging the ergosphere of a rotating black hole.
But the ergosphere and the mechanisms of extracting energy sound a lot like frame dragging. Is it possible to extract energy from all rotating massive bodies, or must it be a rotating black hole?
| Energy extraction from Kerr black holes is possible not only through Penrose process, but from a variety of other processes. Penrose process is also not astrophysically feasible because it has been shown by Bardeen that for Penrose process to occur, the relative velocity of the split should be around $c/2$, i.e., the process must be relativistic. This is possible only if the disintegration process must convert most of the rest mass energy of the initial body into kinetic energy for any extraction of energy to become possible. Such conclusion might be avoided if one is willing to accept the existence of naked singularities or wormholes, where the $g_{tt}$ component of the metric can in principle become very large.
Alternative processes:
*
*Superradiance: Superradiance is nearly same as old as the Penrose process. This is a wave-analog of the Penrose process and is a fascinating area of research at present. Applications of this process is not only limited to General Relativity, but also to other areas of research like Quantum Field Theory, String Theory, Dark Matter etc.
*Magnetic Penrose Process: This is a modified version of the original Penrose process. The papers are papers I and paper II. Here the authors had incorporated the effect of magnetic field that significantly improves the energy extraction efficiency. This is an almost unexplored area but might prove useful in the study of relativistic jets.
*Blandford-Znajek process: This is a very successful and popular process for energy extraction from Kerr black holes. This is one of the best explanations for the way quasars are powered. This process exploits the magnetic field around a rotating black hole for the purpose of energy extraction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
What part of special relativity is factored in in relativistic redshift velocity equations? While doing research into redshift equations (doppler redshift and cosmological redshift), both types of redshift had two equations for finding recession velocity: a 'non-relativistic' equation and a 'relativistic' equation. I looked up SR, and found out about time dilation and length contraction. My question is which of these phenomena have been factored in in relativistic equations (maybe both, or possibly a third phenomena I'm unaware of?)
I'm doing this for a high school project, so please explain in very simplistic language. Also showing a bunch of complicated equations will seriously fly right over my head, so I'd greatly appreciate it if you could please use words only :)
| Speaking of Special Relativity, there are three basic relativistic effects: time dilation, length contraction, and relativity of simultaneity. Of those three the relativity of simultaneity is the most difficult for new students to understand, and it is the source of most of the errors that new students make in relativity.
These three effects are combined in the Lorentz transform. All of the relativistic formulas can be derived from the Lorentz transform. So all the relativistic formulas account for all three effects: time dilation, length contraction, and relativity of simultaneity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the meaning of $PV^\gamma=$ constant in an adiabatic process? In an adiabatic process, heat transfer doesn't occur and hence $\Delta U=-W$ (an increase in internal energy due to work done on the system or a decrease in internal energy due to work done on the surrounding).
My textbook then says (for monatomic gas) adiabatic process takes the form: $PV^{5/3} =$ constant
What does this equation means? From my understanding, "$PV^{5/3} =$ constant" pinpoints to a specific Pressure and Volume of the system and implies a specific Temperature of the system. This is pretty much equivalent to $PV=nRT$ where a specific Pressure and Volume implies a specific Temperature of the system.
But in an Adiabatic process the Temperature, Pressure and Volume can change (otherwise if nothing changes, what's the point of there being a process? and from $\Delta U=-W$ there should be something going on, the temperature should change which will cause a change in the volume of the system [but not a change in pressure since we're assuming the process is quasi-static and thus pressure is constant]) and thus what is the meaning of the equation? Why is this equation specific to adiabatic process?
| You are correct in saying that, in an adiabatic reversible process, the pressure, volume, and temperature are changing in tandem. However, the temperature is changing in such a way that, along the process path, the equation $PV^{\gamma}=const$ is always satisfied. And, in an adiabatic process, the parameter $\gamma$ is equal to the ratio of the specific heat at constant pressure to the specific heat at constant volume.
When you mentioned a quasi-static process path, I think you were thinking of a stair-step arrangement in which, over each step, the incremental change takes place at constant pressure. This is only an approximation to a perfectly reversible process in which the applied pressure is changing very gradually (with time). However, it is close enough to say that, if you determine the pressure and volume at the beginning and end of each incremental step, to an excellent approximation $P_iV_i^{\gamma}=P_fV_f^{\gamma}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do charges lose electrical potential energy when going through a resistor? I can understand that the charges need to do work against the resistance - which transfers energy to forms such as light, heat, etcetera - using the electrostatic force provided by the battery terminals. But how does this result in a loss of electrical potential energy? The charges would only lose electrical potential energy if they were closer to the end terminal - in this picture, the voltage drop is like a boulder going down a smooth hill, but in the correct picture, the boulder instantaneously falls off a cliff as soon as it reaches the resistor.
Help!
| The charges lose potential energy by moving from a higher potential energy to a lower one. In the mean time they are accelerated and scattered. The result is an equilibrium drift speed, while the excess kinetic energy is transformed into heat.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why are tall block stacks so hard to make? Consider a stack of wood chips: each 0.5cm thick and 2x2 cm in length and width. There are 200 of them all stacked on each other. For some reason they all instantly fall. Evwn though their centre of gravity is at the centre of the stack and they have the extra added help of friction the further down the chips you go (because the second to bottom chip is squashed against the first because of the heavy load). So why does it fall?
| First, let's assume the tower is a single solid rod. The tower will fall over when the center of mass extends past the base. For a longer tower, a smaller angular displacement is needed for this. This is because we want the horizontal displacement of the center of mass to be
$$x=\frac 12 H\sin\theta<\frac 12 w$$
where $H$ is the length of the tower, $w$ is the width of a block, and $\theta$ is the angle the tower makes with the vertical.
Therefore:
$$\sin\theta<\frac HL$$
The larger $H$ is, the smaller $\theta$ needs to be. This means that for a taller tower, smaller disturbances can cause it to fall.
Now our actual tower is not a single solid rod, but the same idea applies. Additionally, if the tower fails this at any section it will fall over, with taller portions falling onto ones below. This is why it seems like everything falls at once, as you have said. I'm not sure the friction between the blocks really matters, since there probably isn't a lot of shearing going on here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/435982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the relative speed of light really invariant, irrespective of the motion of the observer? If 3 observers are on a planet which 100 light years from a star, and the star goes supernova, if one observer moves towards the star and one moves in the opposite direction, each observer will see the explosion at a different time. The observer who stays on the planet, sees the explosion a hundred years after it occurs. The person moving towards the star, sees it in less than a hundred years, while the person moving in the opposite direction, will see it more than 100 years after it occured. Doesn't this prove that the speed of light, relative to a moving observer, is not invariant.
| You are reasoning in the frame of the observers and while the speed of light is the same in that frame, the distance to the supernova is different in the frame of the moving observer so they are not all "100 light years away" at the time (on the observer clock which is different for the 3 observers) the light is emitted by the supernova.
Let us assume that observer 1 is at the origin of the "rest" frame and the supernova is along the X axis at coordinate $x_1$.
At t = 0 (clock of observer 1), t' = 0 (clock of observer 2), observer 2 is coinciding with observer 1 but moving towards the supernova with constant velocity V.
The supernova explodes at t=0.
Using the Lorentz transformation, we can compute the explosion time in observer 2 clock:
$t'_1 = -\gamma V x_1 / c^2$
(while for observer 1 the explosion is simultaneous with the passage of observer 2, for observer 2 it happened before)
and its distance to observer 2 in its local reference system
$x'_1 = \gamma x_1 $
with:
$\gamma = 1/\sqrt(1-V^2/c^2)$
For small velocity $\gamma$ is close to 1, then:
$t'_1 = - V x_1 / c^2$
In the newtonian approximation the observer moving towards the source will see the event happen $x_1/c - x_1/(c+V) \approx V x_1/c^2$ before. Same as what we just derived in the approximation $\gamma = 1$. Here the travel time is the same because we neglected length contraction assuming $\gamma = 1$ and of course the speed of light is the same, but the explosion happened "before" for observer 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
What is the most useful to learn out of complex analysis and differential equations for undergraduate studies in physics? Next year I'm planning to start on my bachelor's in physics, however, I have already started taking some undergraduate courses in mathematics and next semester I will have to choose between complex analysis and differential equations - I will take the other course at some other time, but that will at least be a year away.
I know both subjects are important in physics, but my question is which subject will have the largest impact when I start on my bachelor's degree.
| Complex analysis is unlikely to be of any use at all in undergraduate physics classes. Differential equations are used constantly. All of physics is expressed in terms of differential equations, such as Maxwell's equations and the Schrodinger equation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Work done by a gas In the expression for work done by a gas,
$$W=\int P \,\mathrm{d}V,$$
aren't we supposed to use internal pressure?
Moreover work done by gas is the work done by the force exerted by the gas, but everywhere I find people using external pressure instead of internal pressure.
| At the interface with the surroundings, by Newton's third law, the force per unit area exerted the gas on its surroundings is equal to the pressure of the surroundings on the gas. But, in an irreversible expansion or compression process, the pressure of the gas may not be uniform within the cylinder. So the pressures match only at the interface. In addition, viscous stresses contribute to the force per unit area exerted by the gas at the interface (as well as throughout the cylinder), so the equation of state (e.g., ideal gas law) cannot be used to establish the gas pressure within the cylinder or at the interface. The ideal gas law applies only if the gas is at thermodynamic equilibrium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
Does measurement with unknown outcome transform a pure superposition state into a mixed state? $\def\+{\!\!\kern0.08333em}$
Say I have a spin-1/2 particle in a general, pure superposition state
$$
|\psi\rangle=\alpha|\+\uparrow\rangle+\beta|\+\downarrow\rangle,
$$
or equivalently
$$
\rho=(\alpha|\+\uparrow\rangle+\beta|\+\downarrow\rangle)(\alpha^*\langle\uparrow\+|+\beta^*\langle\downarrow\+|)=
\begin{bmatrix}
|\alpha|^2 & \alpha\beta^*\\
\alpha^*\beta & |\beta|^2
\end{bmatrix}.
$$
When I measure this state in the $\{|\+\uparrow\rangle,|\+\downarrow\rangle\}$ basis, I get $|\+\uparrow\rangle$ with probability $|\alpha|^2$ and $|\+\downarrow\rangle$ with probability $|\beta|^2$.
Now assume that this state was measured, but I do not know the outcome (e.g. my friend measured it but did not tell me the result, and will never do so). In this case, it seems appropriate, from my point of view, to describe the post measurement state as the mixed state $\rho=|\alpha|^2|\+\uparrow\rangle\langle\uparrow\+|+|\beta|^2|\+\downarrow\rangle\langle\downarrow\+|$. After all, I know that the measurement yielded either $|\+\uparrow\rangle\langle\uparrow\+|$ or $|\+\downarrow\rangle\langle\downarrow\+|$ with the given probabilities.
Is this reasoning correct? If no, why not?
| Yes, this reasoning is correct.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is the mass of Tachyons real or imaginary? I have always considered quantities like mass, charge, momentum etc. to be real quantities as them being imaginary doesn't make much sense to me.
But for tachyons to exist, they should have imaginary mass compatible with special relativity. Or their rest mass should be imaginary for their masses to be real. So is the mass of tachyons real or imaginary? And if it's imaginary, can there be an imaginary charge, momentum etc.?
| First let's clear up the possible confusion about rest mass and mass. All working physicists these days define mass to be the same thing as what used to be referred to as "rest mass" back in the 1950's. The modern convention has slowly filtered down into textbooks and is now a universal standard except in popularizations.
The definition of mass is $m^2=E^2-p^2$ (in units with $c=1$). Mass is not additive.
A tachyon has a spacelike world-line by definition. By symmetry, its energy-momentum is parallel to the tangent vector to its world-line. Therefore its energy-momentum is spacelike, and $m^2<0$.
And if it's imaginary, can there be imaginary charge, momentum etc?
We don't do general relativity on complex manifolds, we do it on real manifolds, so for compatibility with GR, the energy-momentum must have real components.
I'll leave it to others to discuss whether imaginary charge makes sense. This might depend on what you pick as your definition of charge, whether you're talking about classical or quantum physics, and what kind of distasteful outcomes you are or are not willing to accept.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are uncertainties higher than measured values realistic? Whenever I measure a positive quantity (e.g. a volume) there is some uncertainty related to the measurement. The uncertainty will usually be quite low, e.g. lower than 10%, depending on the equipment. However, I have recently seen uncertainties (due to extrapolation) larger than the measurements, which seems counter-intuitive since the quantity is positive.
So my questions are:
*
*Do uncertainties larger than the measurements make sense?
*Or would it be more sensible to "enforce" an uncertainty (cut-off) no higher than the measurement?
(The word "measurement" might be poorly chosen in this context if we are including extrapolation.)
| Something which has an exponential distribution (so positive) with expected value $\mu$ also has standard deviation $\mu$ - there are other distributions on positive values where the standard deviation can be many times the size of the expected value
By a vague recollection from the normal distribution, you might naively think there could be roughly a $95\%$ chance that a observation of this would be in the range $\mu \pm 2 \mu$. This is a wrong approach in general, but in this special case you would turn out to be correct; more precisely this is the probability of an observation below $3\mu$, which is $1-e^{-3} \approx 0.9502$
The mistake, if any, is thinking that the interval of uncertainty should always be symmetric about the central value
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/436747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Perfect Vacuum Beneath an Object Recently I became interested in how suction cups work by creating a region of space with lower pressure than their surroundings, thus sealing the suction cup to the contact surface. Extrapolating this idea further, does this mean that a cube resting on a perfectly flat surface would be unable to be lifted since by doing so, you would be creating a region with zero pressure? Also, keep in mind that I'm speaking in purely theoretical terms and I understand that no object is perfectly flat and whatnot.
| Even though real objects cannot be perfectly flat, they could be flat enough to stick together pretty well.
Common examples are two pieces of glass or gauge blocks wrung together (video). Besides vacuum action, other forces, like molecular attraction (adhesion) may play a significant role.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can any body be uniform in the universe? If I take any body in the shape of a rod and stretch that, after it reaches breaking stress it breaks at one point.
Even though we apply the same the stress on each and every part of the rod it broke at one point. If it's uniform it should break at all points because breaking stress is same for all the parts of body as it's uniform.
| Uniform bodies are idealizations like frictionless surfaces or no air resistance. They make the work easier. In reality there will be slight deviations in material properties (such as density, tensile strength, etc.) along various parts of the body. In your example these deviations become more and more important and extreme as the body is stretched (which the stretching method will also have some asymmetries as well) until we get a single breaking point.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Faxen's Law Proof Been working on this for a couple weeks and have had no dice. Does anyone know of a nice way to prove Faxen's law (force on a sphere in a fluid flow)? Particularly using the reciprocal theorem? Here is the link to what Faxen's law is: https://en.wikipedia.org/wiki/Fax%C3%A9n%27s_law
In particular the first law:
$$\mathbf{F} = 6 \pi \mu a \left[ \left( 1 + \frac{a^2}{6} \nabla^2\right) \mathbf{u}' - (\mathbf{U} - \mathbf{u}^\infty) \right]$$
So far everything I have researched seems very complicated involving an extremely long employment of stokeslets and stresslets. Any guidance would be great.
| The most straightforward development I've seen of the Faxén laws, at least for the first one you mention, is in Kim and Karrila's book which precisely uses the Lorentz reciprocal theorem.
Although I won't reproduce it exactly, I'll mention the key conceptual aspects and you can take it from there:
*
*Consider the general form of the Lorentz reciprocal theorem: $$\oint_S \vec{v}_1\cdot\left(\bar{\bar{\sigma}}_2\cdot\vec{n}\right)dS\,-\int_V \vec{v}_1\cdot\left(\nabla \cdot\bar{\bar{\sigma}}_2\right)dV = \oint_S \vec{v}_2\cdot\left(\bar{\bar{\sigma}}_1\cdot\vec{n}\right)dS\,-\int_V \vec{v}_2\cdot\left(\nabla \cdot\bar{\bar{\sigma}}_1\right)dV$$ where $\vec{v}_i$ represents the fluid velocities and $\bar{\bar{\sigma}}_i$ represents the fluid stress tensor for the $i$th solution of a given Stokes flow problem.
*Take your $\vec{v}_1$ and $\bar{\bar{\sigma}}_1$ to represent the solution for a spherical particle translating with constant velocity $\vec{U}$ in a quiescent infinite fluid. Note $\nabla\cdot\bar{\bar{\sigma}}_1 = \vec{0}$ everywhere in the fluid.
*Take your $\vec{v}_2$ and $\bar{\bar{\sigma}}_2$ to represent the solution for a stationary spherical particle embedded in a fluid with a Stokeslet present at some position $\vec{y}$ not inside the spherical particle. Note $\vec{v}_2 = \vec{0}$ everywhere on $S$.
*Simplify the expression above by using the singularity solutions for a sphere, in particular for $\vec{v}_1$, in order to pull out the first Faxén law for a sphere.
The derivation of the other Faxén laws follow from similar constructions when one considers the singularity solutions for the object under consideration (in your case, a sphere).
Hope this helps!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/437737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Color and the absorption of light in quantum mechanics In an answer to a another question, the poster states without sources the following:
From a quantum mechanical perspective, all light scattering is a form
of absorption and re-emission of light energy. Photons don't bounce
off a surface.
If this is true, what is the evidence for it, or is it a theoretical postulate under quantum mechanics?
The reason I ask this is that the statement would seem to be contradicted by the phenomena of specular reflection. Since specular reflections always faithfully reproduce the spectrum of the incident light, that would suggest that no light is be absorbed and that, indeed, photons are "bouncing" off of the surface.
| The fundamental theory that described the interaction between light and matter is quantum electrodynamics (QED). It has been exhaustively tested and found to be in excellent agreement with experimental observation.
According to QED the fundamntal interaction is given by a vertex with three legs: two for the charged fermion (electron) and one for the photon. It means that the photon is either absorbed or emitted - not reflected.
But here is the catch: a single vertex cannot describe a physical process, because by itself it cannot satisfy momentum conservation. Therefore, a physical process will always involve an even number of vertices, the simplest being a case with two vertices, connected by a (virtual) particle that is "off-shell" (it does not obey the dispersion relations). So, the combination of two vertices can represent a "reflection." Moreover, because the combination of the two vertices must be computed together, the coherence of the photon can be maintian during the whole process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can a particle in circular motion about a fixed point accelerate, if the point doesn't too? When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But I was taught in school this is not possible because of the string constraint:
The accelerations of the ends of a string are the same if the string is not slack.
Where am I wrong?
| To distill (previous answers are correct, but perhaps unnecessarily long if I understand your point of confusion):
For a rigid system moving without rotation, all points in the system move with the same acceleration.
In any other case (i.e. if there is any change in orientation of the system) this is no longer true.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
How does Newton's corpuscular theory explain the speeding up of corpuscles when entering a denser medium? I can't find an explanation for this anywhere. Intuition would imply that the corpuscle would slow down. I mean a person running at a constant speed enters a crowd of people or a forest. The presence of obstacles would cause a reduction in velocity. How did Newton think it would speed up the particles ?
| In order to explain the refraction of light in going from air into a medium Newton assumed that the medium applied an attractive force on the corpuscles which was perpendicular to the surface.
This force accelerated the corpuscles as they entered the medium in a direction which was perpendicular to the surface of the medium whereas the component of velocity parallel to the medium was unchanged.
So for Newton his intuition was that if there was this attractive force then the corpuscles would speed on entering a medium and from that assumption Snell's law could be derived with the refractive index being the ratio of the speed in the medium relative to the speed in the air.
Newton had no explanation for the speeding up other than if he made that assumption he could use his principles of mechanics to derive Snell's law.
For him the good news was that using his mechanics with some assumptions Newton could also derive the laws of reflection.
Your intuition is perhaps biased by a whole body of knowledge regarding the atomic theory of matter and how the wave theory of light correctly predicts the slowing down of light as it enters a medium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
BB84 protocol with cloning I am taking a Quantum Information course and I stumbled onto this problem, which I am unsure about. The situation assumes BB84 protocol, but with Eve doing a cloning operation (let's assume that she's doing it in states $|+>$ and $|->$ which are orthogonal) instead of the measurement. She keeps the cloned qubit. Would it mean that Bob and Alice won't know if the qubit was intercepted?
Also, if Bob and Alice are working in the same basis, would it actually mean that Eve intercepted the qubit without them knowing and has perfect knowledge of the qubit's state?
| Cloning an arbitrary quantum state is forbidden. See here for a proof https://en.wikipedia.org/wiki/No-cloning_theorem
It is not clear what you mean by "assume she does it (cloning) in the $\vert +\rangle, \vert -\rangle$ basis" but she is unable to measure (as that is detectable) and unable to clone as that is forbidden by quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can someone explain why there is water in this vacuum chamber? I work at a water treatment plant that uses a vacuum system to pull the water into the filters and backwash the filters. In the middle of the vacuum system there is the vacuum chamber. What I don't understand is why is there water in the bottom of the chamber? When the water rises, there is "high vacuum" and when the water is low there is "low vacuum." There are two vacuum pumps that turn on when the vacuum pressure drops to a certain level. So why exactly is there water inside this chamber?
| The vacuum process in the drawing is configured in a way that allows you to "pull" a low vacuum by opening the appropriate valve, and it allows you to "pull" a high vacuum by opening a different valve, and you can get these two different vacuum settings from the same vacuum pumps. A high vacuum setting will pull water up the high vacuum seal pipe, but only to a certain level which depends on hydrostatics and ambient atmospheric pressure. When the valving is switched to the low vacuum setting, the vacuum pump suction pressure doesn't change, but the path for air flow forces any gas entering this system to bubble down through the tube that is in the high vacuum seal pipe, which means that the entering gas has to have enough pressure to overcome the static head of liquid in that pipe. This guarantees that the equipment on the low vacuum side of this system sees a somewhat higher pressure than the vacuum pump suction pressure.
You haven't included a drawing of any process equipment around the vacuum pumps, so I have to guess that the designers of this system wanted one vacuum system to produce low vacuum and another vacuum system to produce high vacuum. Rather than buy and install two different vacuum systems, the designers configured what is shown in the drawing, such that you can get two different vacuum settings from one vacuum system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Why are bend losses higher at higher wavelengths in fiber optic cables I have been reading about fiber optic cables for the past few months, specifically about bend losses. It is often mentioned that bend losses are higher at longer wavelength light and indeed this is the case, but I am curious as to why this is so. In my research, I have not been able to find any physical explanation for this phenomenon.
If anyone knows the answer it would be of great help.
| An intuitive explanation for why this occurs is as follows. If you think of the wavelength of the light as the limiting length scale for it to resolve changes in the guiding structure then as the wavelength increases resolution decreases. Since the resolution has decreased the bend in the guide appears to become more abrupt and causes a larger perturbation in the propagating mode resulting in more radiated power from the optical fiber.
Consider how the radius of curvature (R) would appear to change as the wavelength increases. To do this we can normalize (R) to the resolving ability of the light, the wavelength ($\lambda$). Looking at
$R/\lambda$ it is apparent that this normalized version of the radius of curvature becomes smaller at larger wavelengths.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/438913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does gauge invariance in electrodynamics mean that there are redundant degrees of freedom? It is possible to choose different gauges in electrodynamics. I am familiar with two of them: Coulomb gauge and Lorenz gauge. Let us stick to the Coulomb gauge. It sets $$\nabla\cdot\vec{A}=0.$$ The wisdom is that with this choice the physical electric and magnetic fields $\vec{E},\vec{B}$ do not change. But there is more to it. It is also important for me to understand why this gauge condition implies that there are superfluous degrees of freedom.
What are these superfluous and non-superfluous degrees of freedom? With which mathematical quantities should we identify them?
First of all, at each spacetime point, we have four numbers $$\phi(\vec{x},t),A_1(\vec{x},t),A_2(\vec{x},t),A_3(\vec{x},t).$$ I understand these four numbers as the four degrees of freedom. Now, Coulomb gauge means that the latter three can be related, without any loss of generality, through the differential equation $$\partial_1A_1+\partial_1A_2+\partial_3A_3=0.$$ Given this, how to understand the rest?
| Gauge invariance just happens to be the technical term people chose to indicate that redundant electromagnetic degrees if freedom. I guess that answers the question in the title.
In your example you as they say fixed the gauge by choosing $\vec \nabla \times \vec A=0$. That is all there is to it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Relativistic space rocks. Are they possible? I was thinking that the Universe is full of extreme events. Colliding galaxies, exploding stars, colliding planets (like in one of models of our Moon formation), colliding black holes...
Can it be possible, that such event would accelerate some rocks to let's say $0.1c$ ?
Do we observe such objects? If not, why?
| The most likely way for a rock to reach that speed a long distance from any stars would be as a result of a close encounter with a close binary system - or even better, with a pair of black holes in close orbit. Think of the rock as a baseball and one of the pair of black holes as a baseball bat. The rock could fall in from lightyears away, swoop around the "bat", and fly away with a net speed increase of twice the "bat's" orbital velocity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How do we know quantum entanglement works no matter the distance? It is said quantum entanglement works regardless of distance. 2 particles can be entangled and information is shared instantaneously, even if they are lightyears away from each other.
But how do we know this still works with such a vast distance between both particles?
I can image experiments in a lab, or even on opposites sides of the planet, but not with light years between them.
So how do we know?
| Personally I would like to say the statement is not exact or at least not clear so far. If generally we believe the structure of spacetime is closely related with entanglement, then we should be very careful to talk about 'distance/time/velocity' in a case when entanglement is involved.
My opinion is that when we are playing with two entangled particles, ER=EPR seems to claim that the 'normal' spacetime structure is broken (or at least perturbed) by the entanglement. Under this situation, it's not very strict to say 'the two particles are far away' since the tiny wormhole between them might just provide a shortcut so that their 'distance' is 0. In fact, the observed fact that entangled particles show an instantaneous correlation seems to claim that they ARE the same point.
Finally, I believe before we can complete the quantum gravity to integrate QM and GR, there should be no definite answer to this question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 10,
"answer_id": 7
} |
How many eyes are needed to see a four-dimensional world? If a four-dimensional world were to exist, how many eyes would a creature minimally need to see it (in three dimensions)?
Three? Four?
(Bonus question: how should these eyes be spatially configured?)
| One eye would be enough. (Try closing one eye and you can experience the result for a 3-D world).
But I guess you had in mind what is the value of $N$ such that $N$ eyes give geometric info not available with $N-1$ eyes.
First suppose that in the 4D world, an eye has a 3D volume into which it projects whatever light arrives along a given direction in the 4D space, and the brain has access to the result at each point in this 3D volume. Then 2 such eyes enable the brain to do the triangulation to get distances as well as directions, so 2 is a sufficient number of eyes to get a complete 4D picture.
Now suppose that the eyes are more primitive. Say they project incoming light onto a plane, and the brain scans this plane. (You can compare this with what our eyes would be like if they could determine left from right but not up from down.) Label the 4 directions $w,x,y,z$ and suppose there are some coloured dots to look at, all of different colours or brightness. One eye can determine the $w,x$ direction of a given dot. In principle a single further eye could determine the $y,z$ direction of the dot of the same colour, but such an eye would have to be a long way from the first one. If Instead we suppose the eyes are not far from one another, then I suppose the second eye would be oriented so as to determine the $w,y$ direction, if we take it that $z$ is the direction extending away from the location of the eyes, in the direction they are looking. Thus for any given dot the brain can now figure out $w,x,y$. It can then also get $z$ by using triangulation since it has the $w$ information twice. So once again two eyes are sufficient.
Having said that, I must admit that I wrote the above without drawing any diagrams and I am not absolutely confident whether I have missed something.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What would happen to 2 separate photons on an uninterrupted path So I am a middle schooler and from what i understand, photons can interact with surrounding particle via gravity. If this is true, would 2 photons on an uninterrupted and completely isolated path eventually gravitate towards each other and pass each other, then gravitate again in an oscillatory fashion, while the distance from an imaginary midpoint slowly decreases. If this would happen, wouldn’t we be incapable of telling this combination of two photons apart from a single photon other than its increased energy than a normal photon and increased gravitational pull.
| That is an insightful question. To answer the first part fully would require a LOT of math. Your basic thought is right: photons would attract each other gravitationally. However, the attraction would be very small; much too small to observe or measure using any instrument currently imagined. The second part of your question is
If this would happen, wouldn’t we be incapable of telling this combination of two photons apart from a single photon other than its increased energy than a normal photon and increased gravitational pull.
If there were a pair of gravitationally bound photons, a detector would presumably record a double event, having twice the energy of a single photon detection event. But it is not known if it's possible for two photons to be gravitationally bound. Some relevant discussion can be found here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is there a traditionally accepted threshold probability at which highly unlikely becomes impossible? Many events which by any practical definition are impossible have extremely low but nonzero probability of occurrence. For instance, the positions of oxygen molecules in a room are basically random and independent, but it is clearly impossible that they all end up in one half of the room at the same time. Is there a standard order of magnitude of probability where the transition from unlikely to impossible occurs? In terms of coin flips, I don’t think 50 consecutive heads would be considered impossible, but I guess that 500 consecutive would.
I was hoping for a general answer, application independent, but it seems the consensus is that there isn’t one.
|
Is there a traditionally accepted threshold probability at which highly unlikely becomes impossible?
No, there isn't. People will in practice set certain probability thresholds that are specific to certain applications. For example, suppose a drug company is doing an automated trial of 10,000 new substances in vitro as possible drugs for killing antibiotic-resistant bacteria. Then they probably will not follow up on one of these substances unless it shows an effect that would have a probability $\lesssim 10^{-4}$ of occurring by chance. Setting a less stringent threshold would essentially guarantee that they would be doing follow-ups on a whole bunch of substances that don't work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/439909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the $\,\phi=0\,$ gauge called? In electromagnetism textbooks, the gauges most often talked about are the Lorenz gauge and Coulomb gauge. Sometimes it's convenient to work in a gauge in which there is only the vector potential $\vec{A}$ but no scalar potential $\phi$. The following gauge transformation transforms a general pair of potentials $(\vec{A},\phi)$ into $(\vec{A}',0)$, such that
$$\vec{A}'=\vec{A}+\int_0^t\nabla\phi\,dt,\quad \phi'=0.$$
Then one could work with only the vector potential $\vec{A}'$ to produce both the electric field
$$\vec{E}=-\frac{\partial\vec{A}'}{\partial t}=-\frac{\partial\vec{A}}{\partial t}-\nabla\phi,$$
and the magnetic field
$$\vec{B}=\nabla\times\vec{A}'=\nabla\times\vec{A}.$$
The above procedure seems to work generally without assuming there being no electric charge (which would produce the retarded scalar potential in Lorenz gauge). Is there a name for this $\,\phi=0\,$ gauge?
| Your specified gauge is actually incomplete. In general, though, any gauge with $\phi = 0$ is a Weyl gauge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why isn't the emissive power of a black body 1? My text book has a question which says that the emissive power of a black body isn't one but the answer states that the absorptive power is 1, considering that $$e=a \tag{Kirchoff's law}$$
and a black body is defined as an object which has $$e=1$$ Then why isn't $a=1$
*Choose correct options
(a) Good absorbers of a particular wavelength are good emitters of same wavelength.This statement was given by Kirchoff.
(b) At low temperature of a body the rate of cooling is directly proportional to temperature of the body.This statement was given by Newton.
(c) Emissive power of a perfectly black body is 1
(d) Absorptive power of a perfectly black body is 1
The answer is given as (a,d)
Waves and Thermodyanamics by DC Pandey 15th edition
| For blackbody $a=e=1. a=1$ implies that the blackbody absorbs all radiation falling on it. For anything that is not a blackbody, $0≤e=a<1$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Energy density in magnetic field If an electric field $\mathbf{E}$ exists at a point in free space then the energy density in that point is $\frac{1}{2} \epsilon_0 |\mathbf{E}|^2$.
When a field with same magnitude $|\mathbf{E}|$ exists in a material with relative permettivity $k$, the energy density becomes $\frac{1}{2} \epsilon_0 k |\mathbf{E}|^2$, i.e. it increases $k$ times.
A similar analysis for energy density in magnetic field $\frac{1}{2 \mu_0} |\mathbf{B}|^2$ states that: if the field $\mathbf{B}$ exists in a material with relative permeability $k$ (consider a ferromagnetic material, for example) then $\mu = k \mu_0$ and thus the energy density decreases $k$ times (?). But this does not seem right because inductors with ferromagnetic cores store more energy.
| Inside a dielectric, an externally-applied electric field is reduced. Therefore, a higher energy density is required to generate a given electric field inside the dielectric. This means that for a given electric field inside a dielectric, the associated energy density will be larger than in the vacuum; intuitively, you have to "work harder" to produce an electric field in a dielectric, because a dielectric opposes the applied fields.
Inside a ferromagnet, an externally-applied magnetic field is amplified. Therefore, a lower energy density is required to generate a given magnetic field inside the ferromagnet. This means that for a given magnetic field inside a ferromagnet, the associated energy density will be smaller than in the vacuum; intuitively, you "don't have to work as hard" to produce a magnetic field in a ferromagnet, because a ferromagnet amplifies the applied fields.
Along the same lines, ferroelectric materials, which have dielectric constants less than 1, amplify fields like ferromagnets, so the energy density of the electric field inside a ferroelectric material will be reduced relative to the vacuum. Likewise, diamagnetic materials, which have relative permeabilities less than 1, reduce fields like dielectrics, so the magnetic field inside a diamagnetic material will be increased relative to the vacuum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to suggest a physical experiment in which measurement errors are dramatically increasing? (due to the accumulation of errors) Is it possible to suggest a physical experiment in which measurement errors are dramatically increasing? (due to the accumulation of errors)
Maybe some magnetic field can be used here? Or any other idea?
| Here is a simple experiment which will demonstrate this effect.
You can buy a device called a pedometer, which measures how far you have walked on a hike by detecting each time you take a step. The pedometer has a calibration procedure in which you tell it how many inches there are in one of your steps, and it then adds that number of inches to the distance total each time you take a step.
If you enter 33 inches for your stride length but it is really 30 inches, then it registers a two inch error for every step you take and those errors accumulate- so if you walked 3 miles in reality, the pedometer would tell you 3.3 miles instead.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What would happen if light was completely blocked from an object? Here's the scenario:
There's an object that has been cased with a device that shoots off an 'anti-light' wave (complete reverse of light waves, destructive interference). Say, you have it set up to automatically detect and fire upon a a light wave in that proximity so that no light wave reaches that object.
What would happen? Would the object be unseeable? Or would it be a black space.
I've been wondering what would happen since color is based off of light and black is when all light is absorbed and not reflected, but there is no light reaching that object.
|
There's an object that has been cased with a device that shoots off an 'anti-light' wave (complete reverse of light waves, destructive interference). Say, you have it set up to automatically detect and fire upon a a light wave in that proximity so that no light wave reaches that object.
You've just described a layer of a material which is perfectly absorptive, which surrounds your object. This is precisely how absorption works: the incoming electromagnetic wave excites charge oscillations in the object, and those charge oscillations emit radiation which interferes destructively with the incoming beam.
Since you've encased your object in a fully opaque case, you won't be able to see it from outside.
As to what that will "look like", it depends on whether you have an outer casing that's able to reflect light coming from outside. If you don't, then it will basically look like vantablack.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relation of Hydrogen orbitals to its spectral series? I'm looking for the link between the Rydberg formula for hydrogen spetcral series
$$\frac{1}{\lambda_{\mathrm{vac}}} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
and this image.
Is it right to say that the Balmer Series is created by all transitions from (x, _, _) to (2, _, _) where x>2 and _ are "don't cares"?
Are the three (2, _, _) states all of the same energy?
| The 'link' is, in short, the Schrödinger equation.
The orbitals plotted in the image ─ the wavefunctions $\psi_{n,l,m}(r,\theta,\phi)$ ─ are the hydrogenic solutions to the hydrogen Schrödinger equation,
$$
\left[
-\frac{\hbar^2}{2m}\nabla^2 - \frac{e^2}{4\pi\epsilon_0} \frac{1}{r}
\right]\psi_{n,l,m}(r,\theta,\phi) = E_{n,l,m} \:\psi_{n,l,m}(r,\theta,\phi),
$$
for which you require the energy to be
$$
E_{n,l,m} = E_{n} = -\frac{m e^{4}}{(4\pi\epsilon_0)^{2}\hbar ^{2}}{\frac {1}{2n^{2}}},
$$
independently of $l$ and $m$.
Thus,
Are the three (2, _, _) states all of the same energy?
Yes.
Is it right to say that the Balmer Series is created by all transitions from (x, _, _) to (2, _, _) where x>2 and _ are "don't cares"?
Yes, that is correct (though it's important to note that selection rules generally apply, and not all the transitions within that set actually contribute any significant signal).
For further details, see any introductory textbook on quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can we enhance evaporation rate with a vacuum pump? First, consider having water at 100C and 1atm and a heat source. If more heat added to water, we have more evaporation rate according to the following formula:
$$Q_{in}=m_{vapor}*h_{fg}$$
THEN
Consider having water at 100C and 1atm and a vacuum pump.
Can we change the water evaporation rate with the vacuum pump as we did change the evaporation rate with the heat source?
What are the formulas and calculations?
Thx
| I dont quite see the purpose of the vacuum pump if you stay at the same 1 atm pressure. With a simple fan though you will accelerate evaporation because the air is never saturated
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Use of generating function in canonical transformation In the theory of Canonical transformations, initially we use the fact that the new and the old system of $(q_i, p_i)$ with the Hamiltonian $H$ satisfy the modified Hamilton's principle. Now here, the use of the theory of generating functions makes sense. But after the development of Poisson bracket formalism of canonical transformations, what exactly is the use of generating function formalism? As such even, if it was tough to guess the transformation equation $$P_i=P_i(q_i,p_i,t) ; Q_i=Q_i(q_i,p_i,t)$$ as such, how would it be any easy to guess the generating function?
| In canonical perturbation theory, the generating function (of the $F_2$-type) is determined order by order from the perturbative term, i.e. if
$H(q,p,t)=H_0(q,p)+\epsilon H_1(q,p,t)$ then one can write a series expansion
$$
S= qP+ \epsilon S^{(1)}(q,P,t)+\epsilon^2 S^{(2)}(q,P,t)+\ldots
$$
and determine $S^{(1)}$ given $H_1(q,p,t)$ and $H_0(q,p)$. Note that $S^{(1)}$ usually involves some averaging process. From $S^{(1)}$ one can then determine the corrections to the unperturbed motion (to first order in this case) and so forth to higher order. It gets messy but it's doable.
Possibly the cleanest example is seen in the action-angle formalism. If $K(I)$ is the transformed Hamiltonian in the new action-angle coordinates $(I,\theta)$, we have
\begin{align}
K(I)&=K_0(I)+\epsilon K^{(1)}(I)+\ldots\, ,\\
H(\phi,J)&= H_0(J)+\epsilon H^{(1)}(\phi,J)
\end{align}
with $(\phi,J)$ the "old" action angle variables. Then
$$
S^{(1)}(\phi,I)=\displaystyle\int_0^{2\pi}
\frac{K^{(1)}(I)-H^{(1)}(I,\phi)}{\omega_0(I)}d\phi
$$
with $K^{(1)}(I)$ the average of the perturbation
$$
K^{(1)}(I)=\frac{1}{2\pi}\int_0^{2\pi} d\theta H^{(1)}(\theta,I)\, .
$$
expressed in terms of the new variables, with $J=I, \phi=\theta$ to leading order in $\epsilon$.
A good discussion of this can be found in the book by Jose and Saletan, Classical Dynamics. Also some examples with one position and one momentum can be found in the book by Percival and Richards Introduction to dynamics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Why can't the collision be elastic if after the collision the two objects move together with the same velocity? I don't quite understand why two bodies can't 'stick' together in elastic collision. I have found several sources that say that if two objects collide and end up moving at the same velocity their collision is perfectly inelastic. I don't get why :(
| By definition, elastic collision is one in which total mechanical energy is conserved. That means kinetic energy of the system before and after collision will be same.
Other way to look at it is using coefficient of restitution. If coefficient of restitution is 0, there is zero recoiling (bouncing back) and the collision is called perfectly inelastic. If the coefficient of restitution is anything greater than 0, it will push the balls (or objects) away from each other upon collision and both the objects can not have same velocity.
You might like to see this - https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3D8ko3qy9vgLQ&ved=2ahUKEwjNnvn059feAhUKUI8KHb59D9AQt9IBMA56BAgHEDY&usg=AOvVaw3TiCtJvtPRhMntkq0Cmryf
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Will a warm object in a vacuum cool more quickly than a warm object surrounded by an atmosphere containing CO2? I have gotten stuck on an endless thread about Global Warming, in which skeptics sometimes make the claim that an object surrounded by a "heat absorber" like CO2 will cool more quickly than one not surrounded by something that absorbs heat, because the "heat absorber" will absorb heat from the warm object, speeding its cooling. I think this is likely complete nonsense, but would like to see what people with greater expertise have to say. If it is false, a nice, simple explanation to fire back with and try to set these people straight would be much appreciated!
Could this be tested by lab experiment? Has it been? any links?
| Depends on the situation!
If you have a thermos flask setup, where your warm object is surrounded by some medium and then by air, then the carbon dioxide cools it faster. This is because there's better thermal contact - the carbon dioxide molecules collide with the warm object and carry away some energy, whilst if there's a vacuum then all cooling has to be done by radiation, which is slow.
If, on the other hand, you have a setup more like Earth's climate, where you have a warm object, a boundary layer, and a vacuum, and a constant inflow of energy (from the Sun), then you find that the warm object stays warm for much longer if there's carbon dioxide. This can be shown with a fairly simple treatment, like the idealised greenhouse model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why we can't take torque equation for the mid point of the ladder in the question **Source IIT JEE 2005 PHYSICS** Question (Source IIT JEE 2005 Physics)
Two identical ladders, each of mass M and length L are resting on the rough horizontal surface as shown in the figure. A block of mass m hangs from P. If the system is in equilibrium, find the magnitude and the direction of frictional force at A and B.
Attempt
For the equilibrium of the whole system:
$ΣFy=0$
So,$$N=\frac{(2M+m)g}{2}$$
And for rotational equilibrium
$$mg\frac{lcos(\theta)}{2} +N(\frac{lcos(\theta)}{2}-f\frac{lsin\theta}{2}=0$$ This what I have written for either rod's rotational equilibrium about the mid point of the rod since the rod is in equilibrium.
Main Difficulty
When I approached my teacher for the answer, mine was wrong and then my teacher tried to explain why it should not be correct for the rotational equilibrium equation for the rod, I was not able to understand. This step is my main difficulty or obstacle. Please help me in clearing this doubt.
Thank you for helping me.
| The left ladder presses the right ladder at point P. This is a force you need to include as a torque unless you select P as your axis of rotation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why gas molecules move with different speed at a given tempreture? As per my understanding we know that molecules of an ideal gas are identical in all aspects (size, shape, mass). Since collisions are elastic in nature, they don't lose their kinetic energy. That means that kinetic energy of each molecule doesn't change over time. Then how do the molecules move with different velocity regardless of possessing same mass and kinetic energy ?
| Here is the misunderstanding:
Since collisions are elastic in nature, they don't lose their kinetic energy
Only in the center of mass of two colliding particles the collisions have equal and opposite energy , not in the laboratory frame of the containing box. When one puts all the "identical molecules of an ideal gas" means the "molecules" not the energy momentum vector of each molecule in the laboratory frame of the box.When introduced in the box they will have an average kinetic energy according to the temperature, but there will be a distribution of possible energies and momenta. The elastic center of mass collisions of individual pairs will transform back to the lab with different energies due to the angles of scattering.
It gets worse, because of the spill over electric fields of molecules , the collisions quantum mechanically will allow for radiation, black body radiation, which will eventually lower the temperature to an equilibrium with the outside the box temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/441735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the difference between the voltage in the electrical circuit and electrostatics? In electrostatics it depends on the distance from the charges, should it also be in the circuit? But in practice, the voltage depends on the resistance, for example on a resistor. And the distance itself does not play any role. Why so?
| I'm not quite sure, what do you actually mean. Voltage is voltage, it does not "know" where it comes from. Also in case of induced electricity. Ohm's law (a relationship between voltage, resistance and current) remains in force in both cases too.
The main difference is that although a static voltage can be quite high, it typically involves relatively small charges because capacities in normal circumstances are very low. So if it has an opportunity to discharge - through a spark or by touching a grounded element - the voltage decreases very rapidly, even to zero, and the current consequently decreases very rapidly as well. On the other hand, in case of circuits an element is being constantly recharged from the other parts of the circuit and the voltage remains high. That's why although a woolen sweater can charge to thousands volts without doing any real harm to you (integrated circuits are a different story... they are very sensitive), while a mere 240V from the mains can kill you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Will the SI units need redefining ever again? Up until recently, there were obvious problems with the SI definitions of fundamental units, like bits rubbing off the kilogram prototypes (or mercury vapour absorption), and the water used for the kelvin definition being only found in Switzerland.
From wikipedia: "On 16 November 2018, the 26th General Conference on Weights and Measures (CGPM) voted unanimously in favour of revised definitions of the SI base units, which the International Committee for Weights and Measures (CIPM) had proposed earlier that year. The new definitions will come into force on 20 May 2019."
With these new definitions mostly defining the units in terms of fundamental constants, will there be a need for any further revision to the SI definitions? There are no more physical artefacts/prototypes to remove from the definitions, so unless we abandon SI units altogether, does anything else need to change?
| One possible redefinition that is foreseen, is that the second may be redefined. Currently, it is defined in terms of a radio frequency of the Cesium atom. However, there are many optical clocks (clocks based on some optical frequency of some atom) that can measure time intervals more accurately. Perhaps in the future, the definition of the second may be based on one of these optical clocks.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Focusing Sunlight with Flat Mirrors- solar barbecue Say somebody had 48 1sqft glass mirrors (almost 4.5 sq.m total), because these tiles cost about $1 each.
If mirrors are all tediously aligned to redirect sunlight onto a Weber BBQ, the air temp inside the BBQ quickly rises to 260F (400K), and the irradiated surface (interior caked with years of BBQ crud) smokes profusely.
The question for y'all is:
How complex of a model is needed to explain the current findings?
Could the system by upgraded/expanded to actually BBQ- air temps near 574F (also K)?
Should this hypothetical buffoon pointing mirrors at his BBQ: a)buy waaaay more mirrors, b)try to improve the efficiency of the receiver (the Weber)?
Should I include more data (ambient air temp, irradiated surface temperature)?
I'm hoping I can turn solar BBQ'ing into a 5 person game before summer rolls around.
| Considering the discussion to date- it's looking like I'll need both more mirrors and to improve the receiver.
It sounds like I'll need at 10-15 square meters of mirror to rival the "BTU rating" of a similar gas grill.
It sounds like the BBQ desperately needs to insulated on the backside. Also, I should probably scuff up the gloss-black paint. I'm having trouble interpreting the "gloss numbers" and "light reflected values" listed for paints... but the suns reflection off the BBQ is currently strong enough to hurt my eyes, so I bet there is substantial room for improvement here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inner product in QFT When we write inner product in QM for example $\langle\psi\rvert x \psi\rangle$it means (in position space) $\int\psi^*(x,t)x\psi(x,t)dx$. But when we write, in QFT, $\langle0\rvert0\rangle=1$ what does it mean? As $\rvert0\rangle$ stands for vacuum state and since there is no particle as of now (since no creation operator is operated on it yet), why are we choosing it to be one?
| In the standard treatment of quantum field theory, states are treated as abstract vectors in a large Hilbert space (specifically, a Fock space), and no "wavefunction" interpretation is given to them like it is in 1D quantum mechanics.
For applications to black hole physics (and other areas of quantum gravity), however, one often refers to the "wave functional" (the probability amplitude for the field to be in a specific configuration) $\Psi[\phi(\textbf{x})]$, for which a matrix element is written as a path integral
$$\langle\Psi|\mathcal{O}|\Phi\rangle=\int\mathcal{D}\phi(\textbf{x})\,\Psi[\phi(\textbf{x})]^*\mathcal{O}[\phi]\Phi[\phi(\textbf{x})],$$
which is calculated in perfect analogy to expectation values in 1D quantum mechanics.
When we write $\langle 0|0\rangle=1$, what we're saying is that the vacuum wave functional $\Psi_0[\phi(\textbf{x})]$ is normalized with respect to this (somewhat ill defined) inner product.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is $E=mc^2$ and not $E=m\frac{c^2}{2}$? Kinetic energy for a moving object is the integral of force with respect to distance, often given as:
$$E=m\frac{v^2}{2}.$$
This would imply that for mass moving at the speed of light, the kinetic energy would be:
$$E=m\frac{c^2}{2}.$$
This puts it off from the Einstein result by a factor of two. Why the discrepancy?
| $E=mc^2$ isn't supposed to be the body's kinetic energy. In that equation, $c$ is the speed of light, but in the formula for kinetic energy, $c$ (or preferably $v$ is the velocity of the body. Furthermore, there's no body which can be accurately described as having kinetic energy $E=\frac{1}{2}mc^2$, because that implies a massive body travelling at the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Killing vectors - Schwarzschild metric Given the Schwarzschild metric, $$\mathrm{d}s^{2}=-\left(1-\frac{R_s}{r}\right)\mathrm{d}t^{2}+\left(1-\frac{R_s}{r}\right)^{-1}\mathrm{d}r^{2}+r^{2}\mathrm{d}\theta^{2}+ r^2 \sin^{2}\theta\mathrm{d}\phi^{2},$$ I'm asked to show that $$K^{\mu}=\left(1,0,0,0\right),\; R^{\mu}=(0,0,0,1)$$ are Killing vectors, i.e. they satisfy the Killing equation $$\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0.$$
First of all, using the metric I lower the indices: $$K_\mu=\left(-\left(1-\frac{R_s}{r}\right),0,0,0\right),\; R_\mu=\left(0,0,0,r^2\sin^2\theta\right).$$ Then, for $R_\mu$ I tried to compute $$\nabla_\mu R_\nu\equiv\partial_\mu R_\nu - \Gamma^\lambda_{\mu\nu} R_\lambda=\partial_\mu R_\nu - \Gamma^\phi_{\mu\nu} R_\phi;$$ so $$\nabla_r R_\phi=r\sin^2\theta, \; \nabla_\theta R_\phi=r^2\sin\theta\cos\theta.$$ But now it's not clear for me what should I do next.
| Your equation for $R_\mu$ reads $$\nabla_\mu R_\nu + \nabla_\nu R_\mu =\left(\partial_\mu R_\nu -\Gamma^\lambda_{\mu\nu} R_\lambda\right) + \left(\partial_\nu R_\mu - \Gamma^\sigma_{\nu\mu}R_\sigma\right)=\partial_\mu R_\nu + \partial_\nu R_\mu -2\Gamma^\phi_{\mu\nu} R_\phi.$$
If you use $\mu=r$ and $\mu=\theta$ (the only non-vanishing terms), you will find that Killing equation for $R_\mu$ it's satisfied.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/442983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can we say that bosons attract each other? We know that bosons donot follow Pauli exclusion principle, thus they can occupy the same state. But is it equivalent to say that bosons attract each other?
| Using a statistical physics approach, you can indeed show that the properties of bosons and fermions result in an effective potential that is attractive for bosons and repulsive for fermions:
In a statistical mechanics framework, you can construct the partition functions for both classical, interacting particles in a potential $V$
$$Z_N=\frac{1}{N!} \bigg( \frac{V}{\lambda^3} \bigg)^N \int \frac{\mathrm{d}^{3N}q}{V^N}e^{ -\beta \sum_{i<j}V(\vec{q_i}-\vec{q_j}) } $$
If we now do the same for quantum gases, we get
$$Z_N \approx \frac{1}{N!} \bigg( \frac{V}{\lambda^3} \bigg)^N \int \frac{\mathrm{d}^{3N}q}{V^N}e^{ -\beta \sum_{i<j}V^S(\vec{q_i}-\vec{q_j}) } $$
Comparing these two equations, we see that the statistics of bosons
and fermions express themselves as an effective potential under a
configurational integral. This we call the statistical interaction
$$V^S(q)= -k_BT \log (1 \pm e^{-2 \pi (q/\lambda)^2}) $$
with the thermal de-Broglie wavelength $\lambda=\sqrt{\frac{2 \pi \hbar^2}{mk_BT}}$ giving the range of the interaction.
The effective potential, attractive for Bosons ("B") and repulsive for Fermions ("F") is shown below. Note, that this "potential" follows only from the statistical properties of the particles.
For a detailed derivation, I recommend Huang's Statistical Mechanics textbook, Figure 9.3 and onwards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Easiest Way to understanding coefficient of friction? So I am having issues understanding how to find the coefficient of friction, I should have understood this couple chapters ago but I couldn't. I am not understanding the difference between the static coefficient and the kinetic coefficient (I know static is not moving and kinetic is moving). Maybe you can help me understand it better than my teacher explained?
| The coefficient of friction is just the proportionality constant between the normal force and the friction force magnitudes between two surfaces: $f=\mu N$
Kinetic Friction
This is, in my opinion, the easier one to understand. If the two surfaces are moving relative to each other, then kinetic friction will be the force opposing this relative motion. The magnitude of the friction force is given by $f=\mu N$, where $\mu$ is our coefficient of friction, and $N$ is the magnitude of the normal force between the two surfaces.
Static Friction
This is a bit different than kinetic friction. It is at play when two surfaces are at rest relative to each other, but there is still some external force being applied that is attempting to cause the surfaces to slide past each other.
In this case, instead of saying $f=\mu N$, we have $f\leq\mu N$. What this means is that we are given the largest the static friction force can be before it fails and the surfaces start sliding past each other. Before this is reached, the static friction force is whatever it needs to be to stop the object from moving (imagine a string that will hold an object in place until you pull on it too hard).
A point to be made. This is an extremely simplified model of friction. It does not describe what happens on smaller scales; it is a macroscopic model. But it still does a good job at describing friction between many types of surfaces.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why can't we see images reflected on a piece of paper? Why can't you see a reflected image on a piece of paper? Say you put a pen in front of the paper, even when light rays are coming from other sources, hitting the pen, reflecting back, and hitting the paper, there is no reflection.
What's wrong with the following "ray diagram" and why such even don't happen and the image of the pen don't form on the paper (right side is a paper)?
When then can you see the image of a torch when you shine it on the paper?
When you put a convex lens in front of the pen, why you can now see the image of the pen on the paper?
| You will not see a real image of the pen forming on the paper, nor will you see a virtual image from the reflection of the pen from the paper.
The reason you will not see a real image forming on the paper is that the light scattering from the pen is diverging, and in order to see an image all rays from a single point on the pen must converge onto a single point on the paper. That's why placing a converging lens at the right place will form an image.
The reason you will not see a virtual image from reflection is that the reflection from the paper is diffusive and not specular, i.e. the incident rays are scattered randomly instead 'angle of incident equals to the angle of reflection'.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "72",
"answer_count": 9,
"answer_id": 5
} |
Derivation of 2nd law of Thermodynamics from ergodicity assumption In Wikipedia it is claimed that:
Assumption of the ergodic hypothesis allows proof that certain types of perpetual motion machines of the second kind are impossible.
Since perpetual motion machines are machines which would violate the 2nd law, shouldn't it then from this be possible to derive the 2nd law from the ergodic hypothesis?
| Ergodicity is a property used to allow the use of ensemble averages in place of time averages. As such, it is only indirectly related to the second law.
In order to provide a statement equivalent to 2nd law, statistical mechanics has to show that the relevant fundamental equation (entropy, Helmholtz free energy, grand potential,..) has got the correct convexity properties. For instance, entropy per particle must be a concave function of its natural variables (energy per particle , volume per particle). Assuming the ergodic hypothesis, statistical mechanics can prove existence and the right convexity of the entropy per particle, under a few general conditions on the interactions and under the condition of dealing with an infinite system (the so called thermodynamic limit). These last conditions are necessary to ensure that the statistical behavior of a mechanical system would reproduce the macroscopic thermodynamics, including the second law..
There is no observable (no function of the microstate) which can be used to define the entropy of a single configuration. Whatever approach can be used, it turns out that entropy depends on all the microstates. So, neither the reversibility of dynamics nor the initial condition is really an issue.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Mathematical question on Mathisson-Papapetrou-Dixon equations I am studying about Mathisson-Papapetrou-Dixon equations which govern the motion of a test particle around a central massive object in the pole-dipole approximation.
Given that $S_a=-\frac{1}{2}\epsilon_{abcd}V^bS^{cd}$ I want to prove that $S^{cd}=-\epsilon^{cdaj}S_aV_j$ where $\epsilon_{abcd}$ is the Levi-Civita symbol, $V^b$ is the four velocity of the particle and $S^{cd}$ is it's spin tensor.
Some useful relations I am about to use are $S^{cd}=-S^{dc}$, $V_aV^a=-1$ and $V_aS^{ka}=0$. The latter one is called spin supplementary condition and it is imposed in order to have a closed system of equations.
To prove the desirable result I contract $S_a$ with $\epsilon^{ajkl}$ so we have $\epsilon^{ajkl}S_a=-\frac{1}{2}\epsilon^{ajkl}\epsilon_{abcd}V^bS^{cd}$.
But $\epsilon^{ajkl}\epsilon_{abcd}=-\delta_b^l\delta_c^k\delta_d^j+\delta_b^k\delta_c^l\delta_d^j+\delta_b^l\delta_c^j\delta_d^k-\delta_b^j\delta_c^l\delta_d^k-\delta_b^k\delta_c^j\delta_d^l+\delta_b^j\delta_c^k\delta_d^l$.
That means that $\epsilon^{ajkl}S_a=-\frac{1}{2}(-V^lS^{kj}+V^kS^{lj}+V^lS^{jk}-V^jS^{lk}-V^kS^{jl}+V^jS^{kl})$ so $\epsilon^{ajkl}S_a=-(V^lS^{jk}+V^kS^{lj}+V^jS^{kl})$ which comes from the antisymmetry of the spin tensor.
Finally contracting with $V_j$ and using the fact that $V_aS^{ka}=0$ we get $\epsilon^{ajkl}V_jS_a=S^{kl}$ which means that $S^{kl}=\epsilon^{klaj}S_aV_j$ or even better $S^{cd}=\epsilon^{cdaj}S_aV_j$ which clearly differ with the relation I want to prove by one minus sign.
Can anybody help? Thanks in advance!
| So what you basically mean following your notation is that $\epsilon_{abcd}\epsilon^{aijk}=-$$\begin{vmatrix}
4 & \delta_b^a & \delta_c^a & \delta_d^a \\
\delta_a^i & \delta_b^i & \delta_c^i & \delta_d^i\\
\delta_a^j & \delta_b^j & \delta_c^j & \delta_d^j \\
\delta_a^k & \delta_b^k & \delta_c^k & \delta_d^k\\
\end{vmatrix}$$ $ . Is that right?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can I apply the standard Runge Kutta 4th order method to the Langevin Equation? If I have a Langevin Equation with an external force term (which may be time dependent), is it possible for me to apply the standard 4th order Runge Kutta algortihm to solve it numerically?
Edit:
I would like to mention the use case for the question asked! I am trying to simulate the motion of a Brownian Particle, which is Harmonically bounded but driven with a sinusoidal force (apart from the noise and the friction)!
| No, you cannot directly apply a deterministic method such as 4th order Runge-Kutta to the integration of stochastic differential equations, in general. This is only possible, in the way described by alephzero, if the stochastic Langevin force can be considered a very weak perturbation compared to the deterministic part of the dynamics. The reason for this is that the impulse generated by the stochastic force changes at each time step by a Wiener increment ${\rm d}W \sim {\rm d} t^{1/2}$, where ${\rm d} t$ is the time step. So, a first-order method to integrate a stochastic differential equation needs to include terms of second order in ${\rm d}W$, or more precisely in $\int_t^{t+{\rm d}t}{\rm d} W(t)$, which can be achieved with the Milstein algorithm. A fourth order method would need such terms at eighth order, which becomes extremely complicated (see Kloeden and Platen). If possible, integrating the equivalent Fokker-Planck equation is probably preferable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/443986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two People Pushing Off of Eachother, Newton's Third Law, and Unbalanced Force Different versions of this question have come up all over the internet. Usually it deals with tension in a rope or two people pushing on each other with the same force. I am trying to understand 2 people pushing each other with different forces.
Say two people of equal mass are standing on a frictionless surface, touching palm to palm, and they push off of each other at the same time with different amounts of force. Say the person on the left pushes with 100 N and the person on the right pushes with 70N. What happens to this system? It seems like there should be a net force of 30 N to the right, but at the same time, each person pushing should feel the same force pushing back as per Newton's 3rd Law.
If I push on you with 100 N then I also feel the 100 N from the force pair and we slide apart each having been accelerated by 100 N. If you push on me with 70 N then I feel the 70 N force and I exert 70 N back on you, we slide apart each having felt 70 N this time. But what happens when we both push with a 100 N and 70 N respectively?
Also, I have seen elsewhere on this site that if 2 people were to push on each other with the same force under conditions like those in the scenario above, they each feel the same force but fly apart at double the final velocity because the same force has been applied for twice the distance. Is this actually the case?
| The human body is terrible laboratory for Gedanken experiments, and this problem proves it. What exactly does pushing with 70N mean? Is it active? Is it passive? Nothing feels correct.
This is why Newton's Laws are quite often imagined with masses and springs, which is the only way to save this problem from comments explaining the answer is insufficient or feels wrong.
So: Imagine each person pushes not with a hand, but with a giant spring--a giant calibrated spring that measures force. If person A pushes with 100 N and person B does nothing, both their springs compress to read 100 N. (This is passive pushing for B). Now B fights back and actively adds 70 N to his spring: both springs compress to read 170 N.
Equal and opposite. Always.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to do dimensional analysis? $$mgh = \frac{mc^2}{\sqrt{1-(v/c)^2}}-mc^2.$$
In dimensional analysis do we just ignore the square root? Or do we solve what’s inside first then we do the square root? Do we say $(v/c)^2$ is 1 as dimensions cancel? Then say each term now has the same dimension, so this is correct? I’m so confused please help!
| Since:
$$ \beta = \frac v c $$
and
$$ \gamma = \frac 1 {\sqrt{1-(\frac v c)^2}} $$
are both dimensionless, that are not amenable to dimensional analysis.
In fact, their lack of dimension makes them great "scale factors" telling you how relativistic your system is. $\beta$ goes from $\epsilon$, or Newtonian mechanics, up to $1-\epsilon$, which is ultra relativistic. (Here, $\epsilon$ is the traditional "small number").
Meanwhile $\gamma \in [1, \infty)$ tells you how hyperbolic your geometry is.
If one were to apply dimensional analysis to particles of mass $m$ and speed $v$, you'd come up with:
$$ p \propto mv $$
and
$$ E \propto mv^2 \propto \frac{p^2} m$$
which are correct for Newtonian mechanics.
In relativity, at $v=0$, you get:
$$ E_0 \propto mc^2 $$
and at nonzero velocity:
$$ p \propto \gamma^n\beta^k mv $$
and kinetic energy:
$$ T \propto \gamma^{n'}\beta^{k'}pc $$
There's no dimensional analysis that will tell you $n$, $k$, $n'$, $k'$;
rather, you just have to say $k, k' = 0$ and then guess, "let's add them in quadrature because it's geometry" (ikr: a real stretch):
$$ T = E-E_0 = \sqrt{a (pc)^2 + b(mc^2)^2 } -bmc^2 $$
$$ T = bmc^2[-1 + \sqrt{1 + \frac{a(pc)^2}{b^2(mc^2)^2)} }]$$
and then take the limit for small $p$ and set it equal to Newton's value:
$$ T \rightarrow bmc^2[-1 + 1 + \frac 1 2 \frac{a(pc)^2}{b^2(mc^2)^2}] = \frac{a(pc^2)}{2bmc^2} = \frac a {2b} \frac{p^2} m \equiv \frac 1 2 \frac{p^2}{2m}$$
So in summary: dimensional analysis is too naive to predict special relativity. In any system where you have dimensionless combinations, it's likely that dimensional analysis will not save they day.
On the other hand, if you were guess the potential energy of a mass in a gravitational field, you would do:
$$ U \propto mgh $$
which is pretty good, or, say it's on a string of length $L$, what is the period:
$$ t = m^a g^b L^c $$
you'd find $a=0$, $b=-\frac 1 2$, and $L = \frac 1 2 $; also pretty good.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If there is no atmosphere on a planet that is rotating on an axis and a rocket is launched "Straight up" from its surface, If there is no atmosphere on a planet that is rotating on an axis and a rocket is launched "Straight up" from its surface, won't the rocket still have an angular/orbital velocity because the planet surface's rotation has imparted it's rotation to it?
Also, would that orbital/angular velocity increase the higher rocket goes straight up?
(Kind of like a person walking straight out from the center of a merry-go-round covers more distance in each rotation the further out they walk -- thus the faster they go on the outer arms of the merry-go-round.)
Then also would not the rocket be in Stationary orbit straight up above that same point on the planet?
Please help me to understand this not so "straight up" concept.
| It all depends on what "straight up is", and also the location of the launch complex. Let's put it on the equator on a prograde planet. A the rocket rises, the guidance will have to account for the Coriolis force pushing the rocket westward in order to keep it above the launch pad. That will require increasing the eastward speed so that the angular velocity remains fixed at one rev per (sidereal) day.
If the rocket reaches the height of the geosynchronous orbit (and has no vertical speed), it will remain orbiting directly above the pad.
Note that presence or lack of atmosphere has no conceptual bearing on this problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Time taken by a Magnifying Glass to heat Metal I would like to know how long would it take for a small or medium sized mag. glass to heat a black painted metal particularly aluminum since I can't do the maths.
| In your comment, you say you are trying to heat the following object...
a soda can tightly selaed and painted black at the bottom ... inside the can is water.
using a 70mm magnifying glass.
In this case, the magnifying glass will probably not raise the average temperature of the can at all! The magnifying glass focusses a 70mm circle of sunlight onto a smaller circle. This means that most of your can will be in shadow. In other words, the glass will make your can hotter at the focal point but cooler far from the focal point.
Most of the light focused by the glass would have hit the can anyway. On the other hand, the glass would be very useful in heating a very small object.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Are photons and electromagnetic waves the same thing? I have a little bit of confusion: From my understanding (which could be totally wrong), photons are the same as electromagnetic waves but are called photons because their frequency is in the realm of what we can see.
Seems too simplistic. Can anyone enlighten me?
|
are photons and electromagnetic waves the same thing?
Photons are quantum mechanical entities, a particle in the standard model of particle physics. Light is perfectly modeled as an electromagnetic wave with classical Maxwell equations , so they are not the same. They are related similar to the way bricks are related to a building.
Photons are spin one zero mass particles , described by a wavefunction from a quantized form of Maxwell's equations. The quantum mechanical superposition of the wavefunctions of zillions of photons builds up the total wavefunction of a light beam, and the connection with the values measure macroscpically and fitted with the classical Maxwell's equation comes through the probability distribution for the ensemble of photons. A quantum field theoretical analysis can be seen here.
Because of the same equation giving classical and quantum descriptions the frequency found in $E=hν$ of the photons is the same as the frequency $ν$ of the classical electromagnetic wave.
The visible is a very small part of the frequencies in the electromagnetic spectrum. All the light spectrum is described by the superposition of photons with a given frequency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/444917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the function of this complicated tensioning system? I saw this arrangement for tensioning overhead cables from my train window (schematic below). Why not just have one pulley wheel leading directly to the weights? What function do the additional pulleys serve? For that matter, what are the cables for? They're clearly not power lines.
| Railway overhead electrification is a complex business, I am far from an expert but this is my simplified understanding.
There will normally be at least two wires, a "contact" wire which runs horizontally to make contact with the train's pantograph and a "catenary" wire which supports the contact wire. An appropriate tension must be maintained in the wires, too tight and the stresses from trains would be unacceptable, too loose and the wires would droop.
Wires contract and expand with temperature, this implies two things.
*
*A tensioning system is needed that can maintain tension as the length of the wires changes.
*The distance between tensioning points is limited. Otherwise distortions to the geometry from expansion and contraction would become unacceptable.
So periodically along the length of the railway the wires will pull off to a tensioning device at the side (or sometimes overhead). There will be an overlap between sections of wire so there is always a wire to hold the pantograph in place.
In this case the tensioning device consists of three pulleys and a set of weights, the pulleys will reduce the weight needed to achieve the desired tension in the wires. Then there appears to be an insulator and then a bar joining the catenary and contact wires.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 1
} |
How do we know that light cannot travel faster than it does? We assume the speed of light in vacuum is its maximum speed but can we not assume that it could be faster, or slower?
| Light can be treated as an electromagnetic wave. Electric and magnetic fields have some properties: if electric field changes the magnetic field appears around, if magnetic field changes - electric field appears. There are accurate mathematical formulas which describe these laws (Maxwell's laws). There are some constants in these formulas which depend on the properties of surrounding substance. A consequence of these mathematical formulas is that there electric and magnetic field can exist without charges and propagate with some constant speed that depends on the constants in formulas.
If properties of surrounding substance are different, constants in Maxwell's laws are different, electromagnetic waves propagate with different speed.
Properties of vacuum are the same everywhere. (Why - it's a different question, I do not know). So, the speed of light in vacuum is constant everywhere.
In some substance light can propagate slower than speed of light in vacuum $c$. But not faster. It's a consequence of special relativity theory.
Special relativity theory theory states that time and space are not independent. If there are two events it's not always possible to tell which event happened earlier. It may happen that in one frame of reference the first event happened earlier, in some other frame of reference - the other one happened earlier. The formulas which translates time and coordinates of some event in one frame of reference to time and coordinates in another frame of reference contain some constant speed $c$. Would it be possible to send information faster than $c$ it would be also possible to transmit information between several stations A -> B -> C -> A so that A receives the information before it was sent. Time machines like this would bring miscellaneous paradoxes, so moving faster than that speed $c$ from special relativity formulas is not possible.
Why the speed of light in vacuum and the constant $c$ from the special relativity formulas are the same thing - it's yet another separate question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is torque defined as $r × F$ and not $F × r$? Is it merely due to popular convention or does it supply any special clarification regarding other physical quantities?
| Maybe we can make a quick comparison of both.
Let us define the torque $\boldsymbol M = \boldsymbol r \times \boldsymbol F$. Let us further consider $\boldsymbol r$ and $\boldsymbol F$ lying in a plane and $\boldsymbol M$ facing upwards. The absolute value of $\boldsymbol M$ becomes $M=r\cdot F\cdot \sin\theta$ with $r=|\boldsymbol r|$, $F=|\boldsymbol F|$, $M=|\boldsymbol M|$, and $\theta$ the angle between $\boldsymbol r$ and $\boldsymbol F$. Now we use $\boldsymbol M = \boldsymbol r \times \boldsymbol F = -\boldsymbol r \times \boldsymbol F$ and realise that $M$ remains unchanged.
Let us define now the torque as $\boldsymbol M = \boldsymbol F \times \boldsymbol r$. Again, $\boldsymbol M$ is facing upwards from the plane. And again, $M$ has the same value as before.
What do we learn from this? The order does not matter. It is just a convention. If you'd like to use another convention, feel free to do so, but make sure to be consistent!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/445322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.