Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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How is quantum teleportation possible when there's a continuum of possible states to send over? I'm trying to understand quantum teleportation and I was wondering if anyone could provide an intuition about it. I have seen the derivation but it still bugs me.
You start with an entangled pair of qbits in a Bell state: $$\frac{|00\rangle+|11\rangle}{\sqrt{2}}$$
You want to 'teleport' the qbit $\psi$ to the second entangled qbit. $\psi$ is in an arbitrary quantum state.
In the end, the quantum teleportation protocol requires you to perform only one of four deterministic operations on the resulting qbit and it will become $\psi$.
Here is the problem I'm having: This whole procedure puts the resulting qbit in one of 4 predetermined states even though there is a continuum of possibilities for the original state $\psi$.
Any guesses to what I'm missing?
| I do not know if this can be regarded as an answer. Anyway this is my picture to understand it.
If we check the computation procedure of the teleportation. Starting from ABC are initially in a product state:
(1) AB are entangled to $|00+11\rangle$ by a gate $U_{entangle}$.
(2) C is transformed to an unknown state $|a0+b1\rangle$.
(3) AC carries a scrambling operation $U_{scramble}$.
(4) AC are measured to get 4 possible results.
(5) B is transformed to recover C.
From the recent GM=GR idea, these operation changes the state of the 3 qubit and therefore constructs new geometry. After the measurement of AC, the geometry of the state of ABC is:
AC is cut off from the geometry due to the measurement. B is connected with the original state of C $|a0+b1\rangle$ by two 'gates', one is the gate to entangle AB, $U_{entangle}$, the other is the gate of $U_{scramble}$. $U_{entangle}$ generate a wormhole and its length is 0 (AB are completely entangled), $U_{scramble}$ is a wormhole with a length 1 (Without the information of AC, $U_{scramble}$ can not be inversed). With the measurement result, the 2 classical bits, this wormhole can be shrinked (by the post-operation on B, which is essentially to inverse $U_{scramble}$ using the two classical bits) to a 0 length and therefore the information $|a0+b1\rangle$ appears from the other end of the wormhole at the position of B.
So the information of C is there, teleportation just digs it out.
| {
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Could the big bang singularity have been set into motion by a drop in the higgs field? Could the big bang have been set into motion by a drop in the energy level of the Higgs field?
| For the Big Bang to be set in motion by the breaking of the electroweak symmetry this requires that there was some stable state of the universe that existed at energies of around the electroweak symmetry breaking energy. While I suppose this is possible no-one I know considers it likely. It seems far more likely that any such stable state would have an energy of around the Planck energy. We have no theory of quantum gravity to describe this state so it is impossible to make a definitive statement either way.
However it has long suggested that the Higgs is responsible for inflation. So while the Higgs field didn't start the expansion it produced the exponential expansion described by the theory of inflation.
If you use the term Big Bang to describe the moment that inflation ended then you could say the Higgs field was responsible for the Big Bang. I have seen the term used this way though personally I take it to mean the singularity at time zero.
The idea that the Higgs is responsible for inflation is usually referred to as Higgs inflation. A search will find plenty of papers on the subject though I get the impression the idea has faded in popularity recently.
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If the wire of the secondary circut is thicker than that of the primary in a transformer, what type of transformer is this and why? There is this question in my physics book, and two teachers (a private teacher of a friend of mine and the school teacher) say that it's a step down transformer, while two other teachers say that it's niether of them, since a transformer's type is only determined by the number of turns.
I dont really know which one is correct and why, so if someone could explain id appreciate it.
| In UPS the wires from transformer connected to the 12V battery are thicker(E72332 AWM 1015 12AWG 600V.) Normally it is a step down transformer to charge 12V battery. In case of power failure it acts as a step up transformer to power the system. The wires are thinner (LL34978 18AWG 600V). Any comment. Thanks for your help.
| {
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What is the relationship between gravity and inertia? What is the relationship between Gravity and Inertia? Einstein told us that gravity and inertia are identical. And from the fact that two different masses fall at the same rate, I believe we can say that gravity and inertia are equal (That is, the inertia of a dropped larger mass is exactly sufficient to slow it’s acceleration to the same level as a dropped smaller mass, regardless of them being dropped on the Earth or on the Moon). But is this where we are left hanging: that gravity and inertia are both identical and equal? Is gravity inertia? Or is inertia gravity? What is the next step beyond saying that gravity and inertia are both identical and equal?
| Gravity and inertia are not the same. Inertia is the “change” of the center of gravity. If gravity and inertia are the same, then there is no difference between a fast ball and a curve ball!
| {
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Why do cold chocolate bars crack randomly outside the crevasses when twisted? Why is it that an ordinary chocolate bar (e.g. 4 x 12 pieces) breaks easily where it is supposed to when the chocolate bar is at room temperature or higher but breaks "randomly" when the chocolate is cold? By breaking randomly I mean that it cracks outside the crevasses created at the factory.
Below is a piece of a chocolate bar which has cracked "randomly" outside the crevasses.
Below is a picture of a bar which breaks as it is supposed to at the crevasse (the bar is likely at or above room temp).
| Good question. Yes you would expect stress to concentrate, the material to yield at the minimum sections in the material. Our everyday experience with other materials tells us that. But chocolate seems to be different.
Although I don't know the answer, I'll take a stab at a hypothesis:
When you cool the bar below some temperature you create random crystalline dislocation boundaries that are not necessarily in the crevasses. So these actually become the 'weak links'. At higher temperatures the chocolate becomes semi-fluid and the dislocations disappear; the bar becomes monolithic.
And to test this hypothesis - sorry not sure how to go about designing an experiment. You'll have to ask a chemist or materials engineer to answer that question.
| {
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When wood absorbs water, expands, and breaks stones, how is conservation of energy working? There is a traditional stone cutting method consisting of making a series of small holes in the stone, then inserting wood into the holes, and then adding water to the wood. When the wood expands, it will break the stone.
How is energy being conserved, when the stone is being broken by the expansion of the wood?
| Dry wood consists of air-filled pores surrounded by cellulose fibers and lignin. As such it has extremely high internal surface area, all of which possesses a certain amount of surface energy, which has the ability to perform work. An easy way to get it to perform that work is by bringing the wood into contact with water. that surface energy then does work by drawing water up into the open-celled structure of the wood to wet out those surfaces and fill all the pores. Since the pores are very small, the radius of curvature of the air/water meniscii that are in contact with the cell surfaces in the wood are tiny, and the Young-Laplace pressure jump across them is correspondingly large. This causes the water to wedge the cell structure open as it wets out the wood, causing it to swell. the forces developed in this process are sufficient to break rocks.
The classic demonstration of this effect is one in which a cylindrical container of dry pinto beans is filled with water, fitted with a piston, and slid under the rear axle of a car to produce a snug fit. 12 hours later, after the beans have rehydrated themselves, the car's axle will have been lifted up by several centimeters by the expansion of the beans.
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Bogoliubov transformation for fermion (exercise in Piers Coleman) I am trying to solve the exercise 3.2 in Piers Coleman's Introduction to many body physics. It's about fermionic Bogoliubov transformation with only 2 fermion operators $a_{1}^{\dagger}$, $a_{2}^{\dagger}$, $a_{1}$, $a_{2}$.
The canonical transformation is
\begin{align}
& c_{1} = u a_{1} + va_{2}^{\dagger} \\
&c_{2}^{\dagger} = -va_{1} + ua_{2}^{\dagger}
\end{align}
The starting Hamiltonian is
\begin{equation}
H = \epsilon(a_{1}^{\dagger}a_{1} - a_{2}a_{2}^{\dagger}) + \Delta(a_{1}^{\dagger}a_{2}^{\dagger} + H.C.)
\end{equation}
I have already successfully transformed the above $H$ into
\begin{equation}
H = \sqrt{\epsilon^{2} + \Delta^{2}} (c_{1}^{\dagger}c_{1} + c_{2}^{\dagger}c_{2} - 1)
\end{equation}
where $u=cos\theta$ and $v=sin\theta$ with $\theta$ satisfies $tan2\theta = \frac{\Delta}{\epsilon}$.
I am now struggling with finding the ground state of the Hamiltonian in terms of (1) vacuum state of $a_{1}$ and $a_{2}$ (2) $a_{1}^{\dagger}$ and (3) $a_{2}^{\dagger}$.
I know that the ground state energy is $-\sqrt{\epsilon^{2} + \Delta^{2}}$ and the ground state should also satisfies
\begin{align}
& c_{1}\left| {G} \right\rangle = 0\\
& c_{2}\left| {G} \right\rangle = 0
\end{align}
I have browsed some website and it seems that it is possible that I assume the ground state as
\begin{equation}
\left| {G} \right\rangle = x \left| {0,0} \right\rangle + y \left| {1,1} \right\rangle
\end{equation}
where
\begin{align}
& a_{1} \left| {0,0} \right\rangle = 0\\
& a_{2} \left| {0,0} \right\rangle = 0\\
& a_{1}^{\dagger} a_{2}^{\dagger} \left| {0,0} \right\rangle = \left| {1,1} \right\rangle
\end{align}
But I found that I can not get a proper $\{x,y\}$ coefficients to make
\begin{equation}
\left| {G} \right\rangle = x \left| {0,0} \right\rangle + y \left| {1,1} \right\rangle
\end{equation}
well-defined.
I think this is a rather easy exercise but I just can not find the ground state haha. I will be extremely grateful for any suggestion! :)
Edit :
It turns out that I find that answer and it is rather easy, as follow.
We want to find a state $\left| {G} \right\rangle$ that satisfies
\begin{align}
& c_{1} \left| {G} \right\rangle = 0\\
& c_{2} \left| {G} \right\rangle = 0
\end{align}
The key properties are $\{ c_{1},c_{1}\} = 0$ & $\{ c_{2},c_{2}\} = 0$.
We now construct a state $\left| {\psi} \right\rangle = c_{1}c_{2} \left| {0} \right\rangle$
where $a_{1}\left| {0} \right\rangle = a_{2} \left| {0} \right\rangle = 0$, meaning that $\left| {0} \right\rangle$ is the vaccum w.r.t the original fermionic operators $a_{1}$ & $a_{2}$.
Let's see whether this trial $\left| {\psi} \right\rangle$ can satisfy what we require for the ground state.
First, we apply $c_{1}$ on $\left| {\psi} \right\rangle$ and we have $c_{1}c_{1}c_{2} \left| {0} \right\rangle = 0$ according to $\{ c_{1},c_{1}\} = 0$.
Similarly we apply $c_{2}$ on $\left| {\psi} \right\rangle$ and we again get $0$.
This proves that $\left| {\psi} \right\rangle = c_{1}c_{2} \left| {0} \right\rangle$ is the $\left| {G} \right\rangle$ we are searching for.
So, if we write $\left| {G} \right\rangle$ explicitly, it will be
\begin{align}
c_{1}c_{2} \left| {0} \right\rangle &= (u a_{1} + va_{2}^{\dagger})(-va_{1}^{\dagger} + ua_{2}) \left| {0} \right\rangle \\
& = ( -uva_{1}a_{1}^{\dagger} + u^{2} a_{1}a_{2} -v^{2} a_{2}^{\dagger}a_{1}^{\dagger}+uva_{2}^{\dagger}a_{2} ) \left| {0} \right\rangle\\
& = (-uv - v^{2}a_{2}^{\dagger}a_{1}^{\dagger} )\left| {0} \right\rangle \\
& \propto (u + va_{2}^{\dagger}a_{1}^{\dagger} )\left| {0} \right\rangle
\end{align}
By acting the Hamiltonian on the $\left| {G} \right\rangle$ we can easily confirm the ground state energy is exactly $-\sqrt{\epsilon^{2} + \Delta^{2}}$
I think the reason that I can not get the correct coefficient is because I view the $a_{2}^{\dagger}a_{1}^{\dagger} \left| {0} \right\rangle$ as $\left| {1,1} \right\rangle$. However, because of the anticommute properties for fermion, the arrangement of the operator, and the $+$ or $-$ sign become very important. So it will be safer to write $a_{2}^{\dagger}a_{1}^{\dagger} \left| {0} \right\rangle$ rather than $\left| {1,1} \right\rangle$.
| Take $\psi=x \psi_{00}+w \psi_{01}+z\psi_{10}+y\psi_{11}$ then apply $c_1, c_2$ set equal zero. Then you get $w=z=0$ and one of $x,y$, and remember you need normalization condition to determine the last one.
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Why does the kinetic energy of a photo-electron vary? Provided it is above the threshold frequency of the metal, when electromagnetic radiation is shone onto a metals surface photo-electrons are emitted. This occurs because 1 photon is absorbed by 1 electron giving it enough energy to be ejected.
We know that the energy of the incident photons are all equal from the equation E = hf. If this is so why does the kinetic energy of the emitted photons vary? Why is there a maximum kinetic energy, is it not the same amount every time?
| There are a couple of reasons for this. First and foremost, the electrons are ejected from the surface of the metal in random directions. When you measure things like the "stopping potential" you're only sensitive to motion in directions that would carry the photo-electron to the anode. Because you're only sensitive to motion in particular directions, you only see the part of the kinetic energy along that particular direction. It is this kinematic messiness that makes measuring the reverse potential needed to stop all current the preferred method of measuring the photoelectric effect - the last electrons stopped will be the ones where the largest possible fraction of the photon's energy went in to propelling the electron toward the anode.
Secondarily, the valence electrons in metals have energies in what are known as the "conduction band", which means the electron's energy can exist in a continuum. That fact, combined with the random messiness inherent in thermodynamics, means that no two electrons will have the same kinetic energy before the photons hit them. Thus, each electron will have a slightly different kinetic energy after it gets ejected from the metal.
Of approximately equal weight is the spread in frequencies for the incident light. See, even if your light is produced by a nice sharp atomic line, like in a low pressure mercury lamp, the mercury atoms in the gas will be undergoing thermal motion, leading to Doppler broadening of the line.
| {
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What can be the simplest way to find the thickness of a soap bubble? Is it possible to measure the thickness of a soap bubble without using any sophisticated instruments such that anyone can do it?
| I actually did this as an experiment. So you need a scale, a ruler, a paper towel, and a bubble wand. Blow a large bubble to catch it on the bubble wand. Measure it using the ruler. Now set the bubble aside and measure the paper towel rolled up in a ball. Now try to fling the bubble so that it floats downward on the paper towel, go ahead and measure it (the change in mass). The change in mass will be the volume of the bubble which is approximately the density of water (mass x 1ml/ 1 g). Anyways, using geometric formula for the area of the sphere one can close in on the thickness of a bubble, which turns out is around 1000 nanometers.
| {
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Kinetic energy when observer is moving and object is stationary I know kinetic energy is due to motion of an object. But what if, I, the observer of an object is moving and that object appears to be moving for me, then which one has kinetic energy. is it me or that object?I mean the object doesn't actually move but it appears to be moving because I am moving.
| Kinetic energy depends on the reference frame of an observer. Therefore, kinetic energy is not a property of an object only: If you are moving along with an object and you define yourself as reference, then the kinetic energy of the object is zero (in this special reference frame).
The English Wikipedia article on Kinetic Energy has a section "Frame of reference" where this is explained in detail.
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How to determine the wavefront In case of symmetry around a point we can consider a sphere on whose surface points the waves are equidistant. But how to determine the wavefront around a linear source or a cylinder or triangle or of these or of an complex figure. Should I get the symmetry around every every point and integrate that? How to do it? And is there any alternative process to find it? Please help!
| For an alternate process: According to Huygens Principle the wavefront at time $t$ would be the envelope (or tangent surface) of all the Huygens wavelets originating from each point on the surface of the 'complex figure'. The radius of the wavelets would be $ct$.
For an approximate graphical solution, you can draw a figure and work out the results for the 'wavefront' by drawing circles of radius ct centered at closely spaced points on the 'complex figure' and then draw their tangent surface.
Note that the complete 'wavefield' is more involved. The 'wavefront' is the furthest extent of the wave, which is governed by causality. The 'wavefield' includes what follows after the 'wavefront'.
Google Huygens Principle to get images and further explanation. The images illustrate use of Huygens Principle.
Re. your "Should I get the symmetry around every every point and integrate that? How to do it?": To solve for the wavefield of a geometrically complex source using math would involve numerically solving Kirchoff's or Poisson's formula--you can Google that too.
| {
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Is the Rindler vacuum invariant under Poincare symmetries? More generally, when we quantize fields in the Rindler space and obtain the Fock space of Rindler particles - does that carry a unitary representation of the Poincare symmetries? It should not, because Rindler spacetime is not invariant under global translations. At the same time the Killing vectors in Rindler space are well defined except at the horizon.
Is there some subtlety related to this?
| It depends on the requirements you impose on such a representation. So we assume that there is a strongly continuous unitary representation ${\cal P} \ni p \mapsto U_p$ of the orthochronous proper Poincaré group working on the Fock space constructed upon the Rindler vacuum $\Omega_R$ exploiting the standard static space quantization (the notion of time being the boost one) and we assume that $$U_p\Omega_R= \Omega_R\quad \forall p \in \cal P\:.$$
In principle this is possible. What it is not possible is that the action of that $U$ on Rindler fields $\phi(f)$ (constructed out the standard procedure with creation of annihilation operators referred to $\Omega_R$) smeared with smooth functions $f$ supported in $\cal W_R$, also implements $\cal P$ on the algebra of fields into the natural geometric way:
$$U_p\phi(f) U_p^* = \phi(f\circ p^{-1})\:.$$
This is impossible because $p(\cal W_R) \not \subset \cal W_R$ for some $p \in \cal P$. It is possible if restricting $\cal P$ to the three dimensional Lie subgroup generated by the boost along $x$ and by the spatial translations along $y$ and $z$.
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Fictitious forces and internal forces Say I have two blocks on top of each other and the bottom one is accelerated (relative to the ground) with a horizontal acceleration $a$. I would like to understand what the maximum acceleration $a_{max}$ can be such that the top body does not move relative to the bottom one.
If I work in the accelerated frame, I can assume the friction force $F_f = \mu mg$ can balance the fictitious translational force (magnitude $ma$) such that block stays still. This means:
$m a_{max} = m g \mu_{max}$ and thus $a_{max} = g \mu_{max}$.
Where $\mu$ is the coefficient of static friction .
How would I do a similar analysis from an inertial reference frame?
| In the ground frame:
The upper block experiences Friction in the direction of motion of the lower block. This is the only force which acts on the upper block.
so $m a_{max} = m g \mu$ where a is the acceleration of the upper block equal to acceleration of lower block.
PS- The forces on the lower block will be the external force that pulls it, and the friction opposite to the external force from the block above.
| {
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What is a "decade" as a unit of measure (ex. a decade of the EM spectrum)? Reading through papers and online sources about radio galaxies, I kept stumbling across a term--a "decade" of the electromagnetic spectrum. Radio galaxy emission encompasses "11 decades of the EM spectrum". Or this quote from NASA:
Astronomers have made observations of electromagnetic radiation from cosmic sources that cover a range of more than 21 decades in wavelength (or, equivalently in frequency or energy)!
Source.
What exactly does this term correspond to?
Note: I used the electromagnetism tag because of the context, but I am not sure if the unit can be used outside of the field. Feel free to edit away!
| From 10Hz to 100Hz is a decade (on a logarithmic axis this is $10^2$ to $10^3$).
| {
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Reparameterization and conformal transformation in SYK I have a simple question about SYK model.
For SYK in the IR limit, the Schwinger-Dyson equations have some so called rearamerization invariance
$$\psi_i(\tau)\rightarrow \psi_i(f(\tau))=f'(\tau)^{-\Delta}\psi_i(\tau)$$
I want to whether this transformation is related to conformal transformation,
since I often heard that SYK has IR emergent conformal symmetry. But from my knowledge, the conformal trnasformation is not only related to dimension but also to spin which is not appear here.
| The SYK model is a (0,1)-dim model. This means that because there is only a time dimension that there is no notion of spin. In one dimension there is also no notion of an angle so that every smooth transformation is conformal (Diff$(\mathbb{R})\cong$ Conf$(\mathbb{R})$). In the IR the solution to the Swinger-Dyson equations $G_c(\tau,\tau')$ spontaneously breaks the Diff$(\mathbb{R})$ symmetry down to a SL$(2,\mathbb{R})$ symmetry. This spontaneous symmetry breaking produces Goldstone modes and it can be shown that their dynamics are governed by an EFT described by a Schwarzian action.
I can recommend a thesis by Eric Marcus where these properties of the SYK are discussed in quite some detail.
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How do charged particles interact with each other? As we know, charged particles have polarity and like charges repel each other and unlike charges attract each other, and we have Coulomb's law to find that force. But how does it work? Does it work like gravity, like when two like particles are apart do they still repel? Is there any observational data relating to it?
| We have frameworks in physics where questions can be answered: the classical electrodynamics framework, the quantum framework, the special and general relativity framework etc. Frameworks differ in the variable's range of validity, but blend smoothly in the overlap region.
In current day physics particle is a name given to electrons, muons etc, in the particle data table of the standard model of particle physics.. This is the quantum mechanics framework.
Your question is in the classical electromagnetism framework, so these are classically defined particles carrying charge .
As we know charged particles have polarity
yes, if by polarity you mean + or - charges.
and like charges repel each other and unlike charge attract each other, and we have coulomb's law to find that force.
Correct.
But how does it work, does it work like gravity, like when two like particles are apart do they still repel,
Yes, the field of each particle overlapping generates a repulsive force, following Coulomb's law, theoretically no matter how far apart they are.
Note the term law. Laws are axioms in a physics theory, they tie measurements and observation to the mathematical model, in this case the 1/r^2 law.
is there any observational data relating to it.
A lot. Classical electrodynamics is a theory with Maxwell's equations that describes and predicts all possible situation macroscopically, and has not been falsified by measurements. This includes Coulomb's law.
Now let us go back to particles in the microscopic, quantum mechanical framework, like two electrons repelling each other :
The mathematical formulae represented by this feynman graph, for large distances will display Coulomb law behavior. At small distances it is complicated and needs the study of new mathematical tools as displayed by the graph above.
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Newton's 3rd law of motion, related to Earth's gravity When a stone falls from a certain height above the Earth's surface, it accelerates towards the center of Earth under the influence of Earth's gravity. According to Newton's 3rd law, the stone also exerts an equal force on the Earth, but towards itself. So the Earth accelerates towards the stone, even though it is very insignificant.
What type of force does the stone exert on the Earth?
I know it is a reaction force, but what type of force is it?
Is it also gravity? Or am I misunderstanding this?
| Consider the Earth and the falling stone as one system with no external forces acting on the system.
There are two internal forces which are the force on the Earth due to the gravitational attraction of the stone and the force on the stone due to the gravitational attraction of the Earth.
These two forces are a Newton third law action and reaction pair of the same type (gravitational) and must be equal in magnitude but opposite in direction.
Because there are no external forces the centre of mass of the Earth & stone system does not accelerate and so if the stone is accelerating towards the Earth the Earth must be accelerating towards the stone.
If the mass of the stone is $m$ then the force on the stone due to the Earth is $mg$ where $g$ is the acceleration of free fall of the stone towards the Earth.
If we assume $g \approx 10\, \rm m\,s^{-2}$ and $m \approx 0.1 \,\rm kg$ then the force on the stone due to the Earth is $10 \times 0.1 = 1 \, \rm N$.
This is the force on the Earth due to the stone and if the mass of the Earth is $6\times 10^{24} \approx 10^{25} \,\rm kg$ then the acceleration of the Earth is $\frac {1}{10^{25}} = 10 ^ {-25} \,\rm m \, s^{-2}$ which is such an extremely small value that it is normally neglected.
| {
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How can 2D flat holograms be considered holograms? When I think of a hologram I imagine this
where you shine a beam of laser light onto a holographic film to reproduce an object beam and see a virtual image of an object that can be viewed from many angles.
I don't understand how flat holograms, like that seen on money (shown below), can even be in the same category as the hologram in the image above. I am very very confused any clarification would be much appreciated.
| The term "hologram" is used in a lot of ways. There are "3D" holograms, "2D" holograms, "E-BEAM" holograms, all of which use diffraction to produce an image and/or color. The term has been stretched in recent years to include anything that gives the impression of a 3D image.
So, don't worry about it. The diffractive images on money are usually made by electron beam lithography, but could as well be made by interfering object and reference laser beams. It is okay to call them "holograms".
Originally, though, the term applied only to the kind of image Denis Gabor envisioned, which you yourself would happily call a hologram.
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Derivative of proper time w.r.t. time? On page 86 of Theoretical Minimum (part 3 on special relativity) by Susskind, he writes:
$$\frac{d\tau}{dt}=\sqrt{1-v^2}$$
Where $v$ is the velocity of a moving reference frame relative to the restframe (which is the frame of coordinates $t,x$). He derives this from the equation (I don't know where he is getting it from):
$$d\tau = \sqrt{dt^2-dx^2}$$
But the relation for the proper time should be
$$\tau=\sqrt{t^2-x^2}$$
And if I differentiate this w.r.t. $t$ I get:
$$\frac{d\tau}{dt}=\frac{t-xv}{\sqrt{t^2-x^2}}\neq\sqrt{1-v^2}$$
Why the discrepancy? What am I misunderstanding?
Edit: I now see that the inequality I wrote doesn't hold, because $v=x/t$, which is due to the fact that the path of the reference frame is linear.
IThough, I still don't get how to go from
$$\tau=\sqrt{t^2-x^2}$$
To
$$d\tau = \sqrt{dt^2-dx^2}$$
Also, surely the derived relation no longer holds when we are describing a particle that takes a nonlinear trajectory?
| I'm not sure $\tau = \sqrt{t^2 + x^2}$ is the relation you're looking for, since you would be looking for $\Delta\tau$ rather than $\tau$. In any case, where his relation comes from is the Minkowski metric, which in one space dimension can be expressed $ds^2 = -c^2dt^2 + dx^2$ where $ds$ is the proper time, where $d\tau = \frac{ds}{c}$, where the sign difference in your expression comes from convention.
A metric is just a way to relate coordinate distance to some proper length for a given geometry; i.e. in 2-D Cartesian space the metric would be $ds^2 = dx^2 + dy^2$. If we 1 space 1 time with 'Galilean' reference frames (non-special relativity), we'd have a metric $ds^2 = c^2dt^2 + dx^2$ (which you can recognize as the triangle inequality $a^2 + b^2 = c^2$); the difference with the Minkowski metric is the sign difference on the $dt$ term. The Minkowski metric describes geometry in flat spacetime, which is another way of saying it handles how length and time change in special relativity.
| {
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Position when potential energy and kinetic energy of a spring are equal
I've been given that there is a spring holding a $.25kg$ mass where $k=10N/m$ and is held at $40cm$ and then let go. I've found that the max velocity is $2.53m/s$ and that when the spring is at $20cm$ the velocity is $2.19m/s$
What I'm having a hard time wrapping my head around is finding the position when the kinetic energy and potential energy are equal.
I keep trying to arrange it like so but end up with two unknowns
$.5(10N/m)(\Delta x)^2 = .5(.25kg)v^2$
| You know that energy is always conserved so at the point where $U$ and $K$ are equal a.k.a $$U = K$$, you also know that $$U + K = \text{Total Energy}$$. Now you have two variables two equations and I'll leave the rest for you to solve :D Good Luck! Does this help?
| {
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Why do turkey bacon slices form bubbles in the same places? Often, turkey bacon forms large bubbles when cooking. It's simply a fact of how it cooks. More interestingly however, I was making turkey bacon this morning when I noticed that all of the slices formed bubbles in nearly exactly the same place along the slice, regardless of where on the pan the bacon was situated. What would account for this?
What structural details of the turkey would cause the slices to form bubbles in the same place?
| Maybe it's a trick of perspective, but it seems to me that it is an effect due to the fact that the bacon is bulging outwards in some parts (the pink-ish ones), and therefore the fat/water mixture that comes out because of the heat slides downwards, and then partly evaporates forming bubbles.
The bulging may be due to the fact that low fat (pink-ish) and high-fat (withe-ish) regions of the meat deform in different ways when their temperature is raised, but I have no idea about the physical mechanism behind this phenomenon...
| {
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Velocity of light in Galilean transformation What is the velocity of light in Galilean transformation? Is it infinity?
| Galilean transformations are a low speed approximation and therefore they are not a valid description of the physics when speeds approach the speed of light. Since light travels at the speed of light (obviously :-) that means it cannot be described by Galilean transformations. This is true as long as the speed of light is a universal constant, and it doesn't depend on the numerical value of that speed.
It is often said that special relativity approaches Newtonian mechanics if we take the speed of light to infinity, but all this means is that the range of speeds for which Galilean transformations can be used increases without limit as we increase $c$ towards $\infty$. We can't simply set $c = \infty$ as we can't do arithmetic with $\infty$.
| {
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Proof for uniqueness of transformation between relativistic frames My understanding of the Lorentz transformation is that to ensure that laws of physics remain frame-independent, a transformation was devised, which we call today by the name Lorentz Transformation.
But how do we know that there does not exist any other transformation which ensures that the laws remain frame independent, or that Lorentz Transformation is a special case for some other transformation?
| Assuming
*
*The Principle of Relativity (that the laws of physics are the same in all inertial frames),
*The isotropy and homogeneity of space,
*The transformations form a group, and
*Respect causality,
gives the Lorentz transformations with a free parameter C which specifies a maximum speed. If C is infinite you get the Galilean group, while for C finite you get the usual Lorentz transformations. (C is later identified physically with the speed of light in vacuum if you accept that the photon is massless).
So there are only two possibilities under those above four assumptions. The derivation has been rediscovered many times since Ignatowski did something similar it in 1910, see eg http://o.castera.free.fr/pdf/One_more_derivation.pdf
If you relax any one of the assumptions above you get generalisations which might be useful when searching for potential deviations from special relativity. See eg
https://arxiv.org/abs/1302.5989
| {
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Concept regarding Venturi Tube-Bernoulli application I was recently studying applications of Bernoulli Equation and came across the Venturi tube. This is diagram I have used to analyse the venturimeter. I understand how we obtain the first equation using bernoulli theorem which is
$$P_1 - P_2 =(1/2)ρ(v_2^2 - v_1^2) \tag{1}$$
and also the continuity equation
$$A_1 v_1 = A_2 v_2. \tag{2}$$
However, I am unable to process how to obtain the following third equation
$$ P_1 - P_2 = ρgh $$
where $h$ is difference in the heights of the liquid level in the two tubes and $ρ$ is density of fluid.
Lets say the atmospheric pressure at the top of each tube is $P$. Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.
$$P_1 +ρ(v_1^2)/2 = ρg(h_1) + P \tag{3}$$ and
$$P_2 + ρ(v_2^2)/2 = ρg(h_2) + P \tag{4}$$
Here $P_1$ and $P_2$ are the pressures at the points in the tube and constriction respectively and the points are at the SAME HORIZONTAL LEVEL.
Subtracting (3) and (4) and even using (1) does not yield me
$$P_1 - P_2 = ρgh $$
rather gives me $0 = ρg(h)$ which makes no sense whatsoever.
What am I missing here and how do I obtain the right result ?
|
Lets say the atmospheric pressure at the top of each tube is . Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.
1+ρ(21)/2=ρ(ℎ1)+ (3)
and
2+ρ(22)/2=ρ(ℎ2)+ (4)
3 and 4 are incorrect. Bernoulli theorem goes as:
1/2ρ v ^2 + ρgh + P = constant
Your equation 3 and 4 does not equate the LHS to constant but to Pressure which is incorrect.
Note that P + ρgh1 = P1 in equation 3 which gives us velocity as zero.
| {
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Paradox in special relativity involving capacitor In the laboratory reference frame (LRF), a horizontally moving (with constant speed) flat capacitor would have a different size of plates therefore resulting in different capacity $C'$, namely
$$
C' = \frac{1}{\gamma}C,
$$
where C is a capacity in its own reference frame. The energy of a capacitor is
$$
W' = \frac{q^2}{2C'} = \gamma W.
$$
Here I assumed the value of a charges remain constant in different inertial reference frames (otherwise we could distinguish one reference frame from another). So if the capacitor is closed on a resistor in his reference frame then for me in LRF would be seen like there was more heat produced on a resistor since $Q=W'$ no matter what current was. So, wouldn't it be the way I distinguish one inertial reference frame from another?
|
So if capasitor is closed on a resistor in his reference frame then for me in LRF would be seen like there was more heat produced on a resistor since $Q=W'$ no matter what current was. So, wouldn't it be the way I distinguish one intertial reference frame from another?
In the frame where the capacitor with resistor are at rest, all EM energy in the capacitor eventually transforms into increase in internal energy and manifests as increased temperature and increased mass of the system.
In the frame where the capacitor with resistor move, EM energy available is higher, but here not all EM energy needs to transform into increase of internal energy; some of it may transform into increased kinetic energy.
So different frames imply different division of available energy. This does not mean, as I see it, any indication that some reference frames are preferred in the sense of ether theories.
| {
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The photon propagator term in Peskin & Schroeder Eq. 6.38 In Peskin and Shcroeder, when calculating the one-loop vertex correction, the line above Eq. (6.38) reads
$$ \rightarrow \int \frac{d^4 k}{(2\pi)^4} \frac{-ig_{\nu\rho}}{(k-p)^2 + iϵ} \bar{u}(p') (-ie\gamma^\nu) \frac{i(\displaystyle{\not} k' + m)}{k'^2 - m^2 + i\epsilon} \gamma^\mu \frac{i(\displaystyle{\not} k + m)}{k^2 - m^2 + i\epsilon}(-e\gamma^\rho)u(p)$$
I am trying to figure out where does the $\gamma^\mu$ between the $k'$ and $k$ propagators come from. What I see is that this should be the propagator of the real photon, so the term should be $-ie\gamma^\mu$. Where did the $-ie$ go?
| I found the answer. The original expression for all vertex corrections is $-ie\Gamma^\mu$. The expression on the left should hence be proportional to $-ie\delta\Gamma^\mu$ and the $-ie$ term on the left is removed with the same term from the photon propagator on the right.
| {
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Space and spin symmetry in light baryons In Particle Physics by Martin and Shaw in Ch.6 they show how the baryon supermultiplets are built from the assumption that the space and spin wave functions are symmetrical, so eg. $uud$ has spins up,up,down so that under exchange of the $u$ quarks the wavefunction doesn't change sign as the spins are the same, but if a $u$ and $d$ is exchanged the spin changes so I assume the space wavefunction must introduce another minus from the flavour change to compensate the spin asymmetry and make the whole wavefunction symmetric. However, if say the $uds$ baryon has spin up,up,down then exchanging $u$ and $d$ particles leaves the spin wavefunction unchanged but if the $u$ and $d$ are exchanged it's not symmetrical due to the flavour part. So the $uud$ case requires an asymmetrical flavour wavefunction but $uds$ needs to be symmetrical under $u$ and $d$ exchange but antisymmetrical under exchange of the $s$ with the $u$ or $d$. So do you need to construct a spacial wavefunction for each spin state to make the whole thing symmetric? I feel like i'm missing something, it seems weird to have flavour change be asymmetric only sometimes, if someone could tell me where my thinkings going wrong or maybe justify the spacial wavefunction change I'd really appreciate it.
| Chapter 6 of Martin & Shaw does seem to say that the $aa$ quark pair in an $aab$ baryon must be in a spin-1 state, e.g. that the spin-up proton wave function is $u{\uparrow}\;u{\uparrow}\;d{\downarrow}$. You are right to be confused by this, since the actual proton flavour-spin wave function is
$$
\frac{1}{\sqrt{18}}(2\;u{\uparrow}~u{\uparrow}~d{\downarrow} - u{\uparrow}~u{\downarrow}~d{\uparrow} -u{\downarrow}~u{\uparrow}~d{\uparrow} \\ \quad+ 2\;u{\uparrow} ~ d{\downarrow} ~ u{\uparrow} - u{\downarrow} ~ d{\uparrow} ~ u{\uparrow}-u{\uparrow} ~ d{\uparrow} ~ u{\downarrow} \\ \quad\;\;+ 2\;d{\downarrow} ~ u{\uparrow} ~ u{\uparrow} -
d{\uparrow} ~u{\downarrow}~u{\uparrow} - d{\uparrow} ~ u{\uparrow} ~ u{\downarrow} ).
$$
See, for example, the answers to Proton spin/flavor wavefunction or page 222 of this handout by Prof. Mark Thomson.
| {
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Can all waves interfere with each other? What conditions must two waves have such that they interfere? Do they need to have the same frequency or amplitude? Should they pass through a given space at the same time? Should they have the same sources?
| For two waves to interfere (by which I mean that they produce a stable interference pattern on a screen) they must be coherent i.e., they must maintain a constant phase difference and have same frequency.
However, the explanation above is naive because the real situation is a bit more complicated due to the fact that no source emits perfectly monochromatic waves. I'll try to give a more detailed answer.
| {
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How does holding a glass prevent it from falling? When I hold a glass of water,
$\hspace{1.5cm}$,
I am applying a force horizontally, but its weight acts downwards. Should it not fall?
How do you describe the equilibrium?
| In addition to the answer of @The Photon, consider that for many glasses, the sides of the glass taper inward towards the bottom of the glass.
For such a glass, friction is not needed. Simply holding your fingers a fixed distance apart will exert a force on the wall of the glass with an upward component, stopping the glass from continuing downward.
Of course, in the absence of friction, you run the risk of squeezing the tapered glass too hard, and squirting it upward...
| {
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Projectile motion Need hint
stone is dropped from a cliff of height h at the same moment as another stone is thrown vertically upward from the bottom of the cliff with an initial velocity u. The stones are at the same horizontal level after a time t. Show the condition necessary for the stones to have equal speeds at this level.
Need hint.
| You could try thinking in terms of kinetic and potential energy. If they have the same speed at a given level that means that their potential energies are equal and their kinetic energies are equal too. Think of initial conditions that would satisfy this at some point.
| {
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Why is magnetic field zero for open circuit and electric field zero for short circuit? I read this statement in a book for a simple circuit (1 resistor connected to a voltage source). If we short out the resistor, the E field is said to be zero and if the circuit is broken (open circuit), the magnetic field is said to be zero. Why is this ? How can I prove this to myself ?
| The comment below is restricted to stationary or very low frequency circuits where any electric field is localized within a resistive wire via Ohm's law or between capacitor plates, while the magnetic field can be found only in a coil.
The short circuit case is easier to visualize. Assuming an ideal short such that its resistance is zero then the electric field parallel within the wire must also be zero for any finite current density. When you have a coil (or just a piece conducting wire) and a current passing through it you will have a magnetic field surrounding the coil, so for that to be zero the current must be zero and that means an open circuit for any finite voltage across the terminals.
For an arbitrary distributed parameter circuit it is conventional to say that an ideal conducting wall provides an "electric wall", ie., one so that the E field has no tangential component with the wall. This is the case of a capacitor where the static field is always orthogonal to the metal plates but the idea carries over to arbitrary frequencies.
By analogy it is conventional to call magnetic wall a surface to which the H field is orthogonal, ie., the tangential H component is zero. This can never happen over a metallic surface ("metallic short") at any frequency but can be made to happen by some symmetric excitation, for example. At any rate, it happens in empty space, ie., an "open circuit".
| {
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In pair production does gamma rays actually strike a nucleus? I think the quetion is vagueless.I simply want to ask it is often printed that there must be a heavy nucleus present for pair production.It is answered that its needed to make electron and positron existable after prodution.Major question in that gamma rays actually strike nucleus for pair production?If not then how is pair formed?
| No, the gamma ray doesn't strike the nucleus. You need something around to make it possible to conserve momentum and energy and to obey the relationship between them for every particle. In sum, there is no way to make two massive objects, like the electron and positron, move that will make the pair of them act like they have zero mass the way a photon does.
See, for a photon the momentum and energy obey the relationship
$$E_\gamma = p_\gamma c. \tag1$$
For an electron and positron, they obey
$$E_{\pm}^2 = (m_e c^2)^2 + (p_{\pm}c)^2. \tag2$$
Now, the conservation laws require that the total momentum and energy are unchanged, so
\begin{align}
E_\gamma &= E_+ + E_-,\ \mathrm{and} \tag3\\
\vec{p}_\gamma &= \vec{p}_+ + \vec{p}_- . \tag4
\end{align}
If you square (3) and $c$ times (4) and subtract them you get
$$E_\gamma^2-p_\gamma^2c^2 = E_+^2 + 2 E_+E_-+E_-^2 - c^2p_+^2 - 2 c^2\vec{p}_+\cdot\vec{p}_- - c^2p_-^2. \tag5$$
Applying (1) and (2) (the mass shell relations) to (5) gives
$$0 = (m_e c^2)^2+ \sqrt{\left[(m_ec^2)^2+(p_+c)^2\right]\left[(m_ec^2)^2+(p_-c)^2\right]} - c^2p_+p_-\cos\theta. \tag6$$
What you'll find is that no matter what real values of $p_+$, $p_-$, and $\theta$ you put in to (6) you cannot get the right hand side to equal the left.
In order to end up with something that has non-zero $(E_++E_-)^2-\left(\vec{p}_+c +\vec{p}_-c\right)^2$ you need $(E_1+E_2)^2-\left(\vec{p}_1c +\vec{p}_2c\right)^2$ to be non-zero to begin with. The main way to accomplish that is to have another photon moving in roughly the opposite direction before the collision.
In the case of the nucleus, the nucleus supplies a virtual photon, one that doesn't obey (1) and so cannot exist for very long. So, the gamma ray does not strike the nucleus, directly, but the nucleus does recoil under the force supplied by a virtual photon, and the virtual photon supplies the needed energy and momentum to make a virtual electron or positron real.
| {
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Lorentz transformation of polarisation vector I know that the polarisation vector is not lorentz invariant. But how exactly can I derive the lorentz boosted polarisation vector for a spin-1 particle (say photon)?
What about spin-2 particle's polarisation?
| The polarization properties of masselss particles with any spin (and hence in particular for spin 2) are discussed in detail in Weinberg's papers
S. Weinberg,
Feynman rules for any spin II. Massless particles,
Phys. Rev. 134 (1964), B882--B896.
S. Weinberg,
Feynman rules for any spin III,
Phys. Rev. 181 (1969), 1893--1899.
Everything is roughly analogous to the well-known spin 1 case.
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Question on Eigenstate thermalization hypothesis My understanding of a system that satisfies the eigenstate thermalization hypothesis (ETH) is the following. If we consider a local operator, $O_i$ we can relate its expectation value with respect to some eigenstate, $|\epsilon\rangle$ of the Hamiltonian, $H$ to a thermal expectation value via a thermal density matrix, $e^{-\beta H}$.
$\langle\epsilon| O_i | \epsilon \rangle = \frac{1}{Z} Tr(O_i e^{-\beta H})$.
The inverse temperature is presumably defined implicitly by the consistency condition
$\epsilon = \frac{1}{Z} Tr(H e^{-\beta H})$.
Let us now look at a system with a bounded spectrum, like the Hamiltonian of a spin chain with the lowest energy $\epsilon_{min}$ and highest energy, $\epsilon_{max}$. I will assume that there are no degeneracies for simplicity. I want to simply understand what temperature corresponds to different energies as defined by the above equation. That is $\beta(\epsilon)$. Zero temperature, $\beta \rightarrow \infty$ corresponds to $\epsilon = \epsilon_{min}$ as expected. Infinite temperature, $\beta = 0$, corresponds to the mean energy, $\bar{\epsilon} = \frac{\sum_i \epsilon_i}{\sum_i 1}$.
My question is how should I think about the other eigenstates above $\bar{\epsilon}$. This seems to need negative temperatures which is unphysical. ($\epsilon_{max}$ corresponds to $\beta \rightarrow - \infty$). Or maybe the way I am thinking about how to assign temperature itself if wrong?
| @DominicElse answer in the comment was great. Just a bit of add up:
Consider $DoS(\epsilon)$ as the density of state for your system.
If you chose an eigenstate in part of the spectrum where $dDoS(\epsilon)/d\epsilon > 0$, the temperature associated with that eigenstate is positive. Otherwise, in the region with a negative slope for $DoS(\epsilon)$, $T$ is negative.
In everyday life, we are not dealing with systems with an upper bound on its energy spectrum. So we assume that $DoS$ increases exponentially and then, we always correctly expect a positive temperature. But it is different in discrete systems.
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Does a gap closing mean an occurrence of a quantum phase transition? If we have observed a closing of the excitation gap in the energy spectrum of a certain model, can we safely conclude that a quantum phase transition occurs?
| No. For instance, you could have just touched a phase boundary and returned into the same phase. Or you could have a short-range correlated system (such as a Toric Code wavefunction) which can occur both as a ground state of a gapped and a gapless Hamiltonian (see e.g. https://arxiv.org/abs/1111.5817), and interpolate between those Hamiltonians -- this path would close a gap without the ground state changing at all.
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Is it possible to make a laser by sending sunlight through an optical apparatus? 1) Is it possible to create a laser from focused sun light by separating and using only one wavelength of light as a laser and using the proper mechanism to polarize it and make it coherent?
2) If so, would it be possible to use some type of a wavelength filter enabling it to focus different wavelengths for different applications?
| If you're asking about the possibility of making coherent light by filtering incoherent, broad-spectrum sunlight, the answer is "no". You could filter the light down to a very narrow wavelength band by throwing away all the other wavelengths, and end up with nearly monochromatic light. That leaves about 1/10000 of the original light power. Now, to obtain spatial coherence (which allows a laser to be focused to a small spot), you would need to put a tiny pinhole (a micron or two wide) at the focus of the filtered sunlight. Try focusing sunlight and you'll find that it is difficult to get a spot smaller than about a millimeter. So the pinhole throws away at least 99.999 % of the monochromatic light. Now you've got only 1/10,000,000,000 of the light you started with -- and it's still not as coherent as a laser.
On the other hand, if the objective is simply to get highly monochromatic light for experiments such as testing spectral response of photosynthesis or light sensors, all you would need is a prism or diffraction grating, a slit filter, and a lens.
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Gauge invariance lost using Maxwell's equations I read a paper on superconductivity, and the calculations use Newton's second law and the Maxwell-Faraday law to write the "free electron current" as
$$
\textbf{J}_e=-\frac{ne^2}{m_e}\textbf{A}
$$
where $\textbf{A}$ is the vector potential ($\textbf{B}=\textbf{rot A}$). After that, it said that the Coulomb gauge is chosen ($\text{div }\textbf{A}=0$), and it's verified since $\text{div }\textbf{J}_e\propto \partial_t \rho =0$. But it also says that we have lost gauge invariance. Why?
Moreover, I don't think I understand properly what implies gauge invariance. Can someone help me?
| You have lost gauge invariance because a specific gauge has been chosen, the Coulomb gauge $\nabla \cdot {\bf A} = 0$.
Gauge invariance of a theory implies some redundancy in how certain fields in the theory are defined, in that the physics is unchanged by different definitions of these fields. In classical electromagnetism, one has $\nabla \times {\bf A} = {\bf B}$. Notice that we can add the gradient of an arbitrary scalar function, $f$, to the vector potential, ${\bf A}' = {\bf A} + \nabla f$, without changing the magnetic field, since the curl of the gradient of a scalar function vanishes.
Via Maxwell's equation, $\nabla \times {\bf E} = -\partial {\bf B}/\partial t$, and the fact that ${\bf E} = -\nabla V - \partial {\bf A}/\partial t$, you can see a similar thing happen with the scalar potential, $\phi$: $\phi' = \phi - \partial f/\partial t$. This redundancy in the scalar potential is perhaps most intuitive, since only changes in the electric potential are ever physically measurable, and changes are unaffected by the addition of a constant.
Often when doing calculations it's possible to simplify a problem by fixing the gauge, that is, by choosing a specific function $f$. In the Coulumb gauge, $f$ is chosen so that $\nabla^2 f = 0$, giving $\nabla \cdot {\bf A} = 0$. Once we've selected a gauge, though, the theory is no longer gauge invariant. This makes it sometimes difficult to know whether the result obtained is specific to the gauge choice, or a result true of the gauge invariant theory.
| {
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Are materials which are bad at conducting heat always bad at conducting electricity also? When defining a material's conductivity, we usually consider its conductivity of heat and conductivity of electricity separately. However, I realize that materials like metal conduct both heat and electricity well. In contrast, materials like wood and glass conduct both heat and electricity poorly. Therefore can we conclude that if a material is bad at conducting one kind of "flow of energy", then it will also be bad at conducting another kind of "flow of energy"? Thanks a lot.
| Diamond is a good thermal conductor but a poor electrical conductor.
Diamond at Wikipedia
| {
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In Atwood machine, does the tension of the rope do work? It seems in all Atwood machine exercises I can find, no one ever takes into account the tension of the rope when solving with conservation of energy. Why is this? Shouldn't the tension be a non-conservative force contributing to the net work?
| Depends on what particular subsystem you're looking at. The tension does work on each of the blocks, but since it's an internal force, it does no work on the Atwood machine as a whole.
| {
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Is Yellow a monochromatic light? I have got a serious doubt.
I have read,
"yellow light from a low pressure sodium vapour is monochromatic"
How can it be monochromatic when yellow light is a combination of red and green primary colours?
| The key here is that there are a basically infinite number of different mixings of photons of different wavelengths that will be perceived as yellow by our eyes or a camera. This is because our eyes are not spectrometers, and use a relatively crude three-color system (red, green, and blue-sensitive cone cells) to identify colors. (Incidentally, this is why color TVs are feasible to make - they use essentially the same crude setup as our eyes, with red, green, and blue-emitting pixels.) But there is definitely a physical difference between 550nm monochromatic yellow light and the mix of red and green light that will appear to be the same color from your eyes. The monochromatic light will not disperse in a prism, while the mix of red and green will disperse into its red and green components.
| {
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Human body water storage effects on weather patterns There are 7.614 billion humans and counting. On average a human being consists of 60% water. With that water being locked away from natural circulation for an average of 80 years. Are there any adverse effects this could have on global weather patterns?
| Humans are not locking up water. You can use mass conservation and Kirchoff's first law (where you replace charge with water) to see why: if water was accumulating in human bodies, they would tend to get bigger. Over the span of a year a human consumes somewhere around half a ton of food and water, yet our weight usually doesn't go up that much per year.
Second, the total mass of all human water is on the order of ($7.2\times 10^9$ people)(62 kg)(65% water) =$2.9\times 10^{11}$ kg. This is a bit more than the water stored in the London storage reservoirs (0.2 km$^3$), nothing significant globally when you consider real lakes, rivers and seas.
(Atoms that get integrated in DNA in long-lived cells like neurons are taken out of circulation for a few decades, but the total mass is tiny.)
| {
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What is the quotient of two quantum operators? It's probably useful to explain the context, which led me to this question. We were asked the following question:
By writing ${L}^2 = \sum_{ijklm}\epsilon_{ijk}{x}_j{p}_k\epsilon_{ilm}{x}_l{p}_m$ show that: $$p^2 = \frac{L^2}{r^2}+\frac{1}{r^2}\left\{(\textbf{r}\cdot \textbf{p})^2-i\hbar(\textbf{r}\cdot \textbf{p})\right\}$$
I got up to this point: $${L}^2 = {\textbf{r}}^2{\textbf{p}}^2-\left(\textbf{r}\cdot\textbf{p}\right)^2+i\hbar\textbf{r}\cdot\textbf{p}$$
However now my question is, am I allowed to just divide by $\textbf{r}^2$ and if yes what is the interpretation of a division of two quantum operators? After all $L,\textbf{r},\textbf{p}$ are all quantum operators so I'm quite worried about just applying "normal" Algebra rules and callying it a day.
| Yeah, it's a subtle issue. You can't just divide two non-commuting operators - you need to specify whether you're left-multiplying or right-multiplying the numerator by the reciprocal of the denominator. I would avoid ever using "division" notation and only multiply operators and their reciprocals, for clarity. You can left-multiply your operator expression by $\left({\bf r}^2 \right)^{-1}$ to get one particular quantization of the final result that you're supposed to show.
Strictly speaking, in $d$ spatial dimensions the domain of the operator $r^{-2}$ is the subset of the Hilbert space $L^2 \left( \mathbb{R}^d \right)$ which $r^{-2}$ takes to $L^2 \left( \mathbb{R}^d \right)$, i.e. the set of square-integrable functions $\psi({\bf r})$ such that
$$\left \langle \psi \middle| \left(r^{-2} \right)^\dagger r^{-2} \middle| \psi \right \rangle = \int d^dx\ \frac{|\psi({\bf r})|^2}{r^4} = \int d\Omega \int r^{d-1} dr \frac{|\psi(r, \Omega)|^2}{r^4}$$
is finite. This is the set of functions $\psi({\bf r})$ that go to zero at the origin at least as fast as $r^p$ for some power $p > 4 - d$. In three spatial dimensions, this means that $\psi({\bf r})$ must go to zero faster than $r$ near the origin.
| {
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Could we construct an experiment to measure the speed of light greater than c? As I understood we were not able but then i read John Rennie's question and answer here:
Does light really travel more slowly near a massive body?
And got me a little bit curious.
He says:
We can extend our analysis to find the speed of light in the shell coordinates at radial distances greater and less than the shell distance. The argument is essentially the same as above so I’ll just give the result:
$$ \frac{dr’}{dt’} = \pm c \frac{1- r_s/r}{1 – r_s/R} \tag{4} $$
And this looks like (for $R = 2r_s$):
Like the Schwarzschild observer the shell observer sees the coordinate speed of light fall when the light is closer to the massive object than they are, but the shell observer sees the light move faster than $c$ when the light is farther from the object than they are.
Question:
*
*Is it really possible to construct an experiment where we would measure the
speed of light faster then c?
| You don't need to do a new experiment in order to see the coordinate velocity of light be different from $c$. A coordinate velocity can be any number you like, simply by your choice of coordinates. Pick any experiment that measures $c$, change coordinates, and express the speed of light in those new coordinates. For almost any choice of coordinates, the coordinate speed of light will be unequal to $c$.
The coordinates in which the speed of light equals $c$ are the local Minkowski coordinates of an inertial observer.
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Effective action: Cancellation of odd powers (Altland & Simons) Consider a system of interacting electrons. Using the path integral formalism, we introduce the Hubbard Stratonovich transformation to decouple the interaction in the density channel. Then, we integrate out the fermionic degrees of freedom and extremize the action. The new effective action involves a term of the form
$$
\ln\left(-\hat G^{-1}\right) + \ln\left(1 - i\hat G\hat\phi\right)\,,
$$
where $\hat G$ is the diagonal bare propagator for the electrons and $\hat\phi$ is the auxiliary field. The first term just gives the partition function for the non-interacting system. The trace of the second term can be expanded as
$$
\mathrm{tr} \sum_n \frac{\left(-i\hat G\hat\phi\right)^n}{n}\,.
$$
The first term is
$$
\mathrm{tr}\left(\hat G\hat\phi\right) = \phi_0\sum_n G_n\,,
$$
where $G_n$ is the diagonal element of the propagator matrix.
The second term is
$$
\sum_q \phi_q \phi_{-q}\left(\sum_p G_p G_{p+q}\right)\,.
$$
The stuff inside the parentheses is the RPA polarization bubble. So far so good.
The third term becomes
$$
\sum_{kp}\phi_k\phi_p\phi_{-k-p}\left(\sum_qG_qG_{k+q}G_{q-p}\right)\,.
$$
This is a triangular loop diagram. Intuitively, it seems that it should cancel. Odd powers just seem, well, odd. In fact, consulting Altland and Simons book (second edition, after Eq.(6.6)), it does say that the odd powers cancel by symmetry. However, I don't see it. Am I missing something obvious?
Thank you
| I think the reason is rooted in the interpretation of the fluctuation around mean-field.
Back to the Hubbard-Stratonovich transformation, if we substitute $\phi_q$ with $-\phi_q$, the physics should not change. From another perspective of view, we can say the positive flucuation around the mean-field $\hat \phi=0$ is equivalent to its negative counterpart.
From above we can say odd powers of $\phi$ (ignoring $\phi$'s indices) must vanish.
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What exactly is deterministic in Schrödinger's equation? I have read the following on Wikipedia but I can't understand it:
In quantum mechanics, the Schrödinger equation, which describes the continuous time evolution of a system's wave function, is deterministic. However, the relationship between a system's wave function and the observable properties of the system appears to be non-deterministic.
–"Deterministic system", Wikipedia [links omitted]
How can a system be deterministic and not deterministic at the same time?
Can anyone explain simply?
| Mathematically speaking, the Schrodinger equation is a type of linear PDE known as the wave equation. The wave equation, which you may be familiar with from other studies of PDEs, is deterministic, because it evolves in the sense that if you know the initial conditions at an initial time, you know exactly what the wave described by this equation will be at a later time. The state at a later time is determined by the initial conditions and time-evolves toward that state according to the wave equation: you freeze time at an instant and you 'know' what it will be.
In the case of the Schrodinger equation, the wave (as it was original conceived of) is a probability distribution (this is the origin of the term 'wave function'). The probability distribution evolves in a deterministic way according to a linear PDE (probability distributions can do this in general), but the results of measurements are obviously 'not deterministic', because you 'don't know what you are getting' when you dip your hand in the bag, although you do know the probabilities of what you can get.
There are more complicated ways of coming to the same conclusion, but I think is the simplest way (without getting into the foundations of QM).
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Are there any electric fields associated to the scalar fields in quantum field theory? We know that electric field due to stationary charges is given by gradient of phi.
In quantum field theory the scalar fields associated with bosons doesn't have any electric fields but while dealing with em fields we get vector E as conjugate momenta to A_u. why there are not any electric fields associated to such scalar fields?
| I don't quite understand your question, perhaps you can clarify. Also, the boson mediating the electromagnetic interaction is a spin-1 particle, not a scalar which would be spin-0.
First of all, a electric field and a field in quantum field theory are not the same thing. The former is a measurable quantity for instance, while the second is a second-quantized field of a classical field theory. In classical electrodynamics the electric field is given by
$$\vec{E} = - \nabla \phi - \frac{\partial}{\partial t} \vec{A}$$
If we put $\phi$ and $\vec{A}$ together in the 4-vector $A_\mu$ we find the electric fields in the field strength tensor $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$.
From the $F_{\mu\nu}$ we can construct a classical field theory which if second-quantized correctly (which does lead to some problems) gives quantum electrodynamics QED where we identify the gauge bosons with photons, the quanta of electric fields. Note that we only talk of bosons once we quantize and find a spin-1 particle in the spectrum.
So it is by construction that the electric field is in this way associated to the Lorentz vector $A_\mu$. There are certainly fields associated to other particles such as the scalar Higgs field.
| {
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What is the cosmological redshift drift effect? What is the redshift drift effect in cosmology?
What are the necessary cosmological conditions for there to be a measurable redshift drift effect?
| Redshift drift is the name given to the effect whereby as the universe expands, the redshift of an object will change with time. i.e. It will drift.
A galaxy at a fixed co-moving distance will have a redshift that changes with time and the first and second time derivatives of the redshift may prove to be valuable probes of cosmological models (e.g. Meliá 2016).
The size of the effect is of order 10 cm/s/yr, with the difference between different cosmological models being smaller than this. It is expected that the E-ELT telescope may be able to measure redshift drift over 5-year periods, using large ensembles of galaxies at similar redshifts.
An expression for the first order redshift drift is
$$\frac{dz}{dt_0}= (1+z)H_0 - H(z),$$
where $H_0$ is the current Hubble parameter at time $t_0$ and $H(z)$ is the value of the Hubble parameter at an epoch corresponding to a redshift $z$.
The condition that there be a redshift drift is therefore that
$$ H(z) \neq (1+z)H_0$$
which is certainly the case for arbitrary values of $z$ in $\Lambda$CDM cosmology$^*$ where
$$H(z) = (1+z) H_0\left[ \Omega_r (1+z)^2 + \Omega_m (1+z) + \Omega_k + \Omega_{\Lambda}(1+z)^{-2}\right]^{1/2}$$
and the condition for a redshift drift is that
$$\left[ \Omega_r (1+z)^2 + \Omega_m (1+z) + \Omega_k + \Omega_{\Lambda}(1+z)^{-2}\right] \neq 1$$
$^*$ Other cosmologies are possible...
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How the current of a wire pass so quickly if it has that less drift velocity of electrons.? The rate of flow of electrons is known as current. When we switch on a light bulb, then it lights immediately. If drift velocity of electrons is so small--at least on the supply on 220V (supply of my home)--then why does the light bulb light up so quickly?
| The conduction electrons that are already in the wire are moved by the electro-magnetic wave that can propagate once the circuit is closed.
With a 220V ac supply, actually the electrons are moving back and forwards, their net drift velocity is zero.
There are many other analogies to this. Sound moves in the air at 300m/s, but it is not an individual air molecule that travels from the source to your ear in that time, but a wave of pressure.
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Can there be general relativity without special relativity? Can General Relativity be correct if Special Relativity is incorrect?
| From a theoretical point of view, one may argue that it depends on how one defines the words GR & SR, and how abstract one wants to be. For starters one may generalize to other than 4 spacetime dimensions. Moreover, the local spacetime model could in principle be different from Minkowski spacetime. Instead of Einstein gravity, one may consider e.g. conformal gravity, SUGRA, Newton-Cartan gravity, etc.
| {
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Comparing work in thermodynamics with work done in mechanics Let us the consider a gas as our system enclosed in a cylinder with piston.
1st case(Expansion of gas):
Here force on the piston is exerted by the gas in upward direction and during expansion piston moves up. So, the work done here is positive(force and displacement in same direction). Also the relation W=PΔV (with their usual meanings) also satisfies the "positive" sense of work, since the volume increases during expansion.
2nd case(Compression of gas):
Here, the surrounding exerts the force on the piston and compresses the gas. Since, the direction of force by surrounding on the system and displacement of piston(both downwards) are in same direction, should not the work done by the surrounding on the system should be positive?
But, W=PΔV gives -ve work, since volume decreases during compression. Why does the mechanical concept of work and W=PΔV does not give same result?
(In Physics)We are usually told that work done by the system is positive and work done by the surrounding on the system is negative.
But
2nd picture shows exactly what I am confused with. During compression, it is the gas that does negative work not the surrounding does the work in the gas?
1st picture is screenshot of book University Physics.
2nd picture is from here.
| Normally both results are the same for work but it is defined in another way as you wrote: $dw = -pdV$ so the work you can get by integration.
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Rotational mechanics theory This is a statement given in my text book:
"The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axis need not be stationary."
I don't understand this properly. Please explain.
| If it does not interact with other objects, a rigid body can be reduced to its center of mass. Then you can describe the motion of this point as the motion around an axis (in case of straight travel it's an infinitely far away axis). If for some reason you need to know more about what the constituent masses of the rigid body are doing, you can describe the motion of the object as a dance around an axis that goes through the center of mass. This axis can behave rather weird, for example if you throw a brick, that bricks' center of mass will go in a parabola, but the brick will probably tumble, because it has three very different main axis, and rotation is only stable around the longest and the shortest. Spin around the middle axis will soon become spin around one of the others. --- I said the center of gravity will move in a parabola - this is only describable as a rotation if the axis of the rotation moves nearer to the object all the time. So both axis move, and we can seperately describe the rotations around those axis.
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How to understand $L_4$ and $L_5$ Lagrange points gravity balance? It's relative easy to understand gravity balance of Lagrange points $L_1$, $L_2$ and $L_3$. But I am having a hard time to understand how a body would be "kind of" balanced out on Lagrange points $L_4$ and $L_5$.
| a small body occupying L4 or L5 point around a planet completes the full circle both around the planet and the star while the planet itself completes the full circle around the star in the same time span. By the way, L4 and L5 are the only stable Lagrange points, if there is a small perturbation on a body in L4 or L5 point, it must be above some threshold for the body to leave the Lagrange point it was resting in.
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Group representations and active/passive transformations Suppose we are in Euclidean 3-space with coordinates $x$ and a scalar function $\phi(x)$ defined on it, and consider the group of rotations $SO(3)$ for simplicity. Take a rotation matrix $R \in SO(3)$; then the usual explanation for the difference between active and passive transformation is:
*
*Active: If we picture the function $\phi(x)$ as having a "bump" somewhere, an active transformation moves this bump around. We now have a new function $\phi'$ of the same coordinates $x$, given by $\phi'(x) = \phi(R^{-1}x)$.
*Passive: We leave the bump fixed and instead make $R$ rotate the coordinate axes to get a new set of coordinates $x'$ on the same physical space. In terms of the new coordinates, the field is expressed as a function $\phi'(x') = \phi(Rx')$ or $\phi'(x) = \phi(Rx)$ if you prefer, since the name of the coordinates doesn't make any difference.
This makes perfect sense but there's a problem: the passive transformation is not a representation of the group of rotations. If we are to think of transformations as an action of $SO(3)$ (or the Lorentz group or whatever your favorite group is) on the space of functions, we can only use active transformations, since only they are actually a group action. Does this imply that if we study fields as representations of some group we are restricting ourselves to active transformations only? Shouldn't the active and passive viewpoints be equivalent?
Edit: the passive transformation is not a group representation because if we define $(\rho_R \phi)(x) = \phi(Rx)$, we get $\rho_{R_1} \rho_{R_2} = \rho_{R_2 R_1}$ instead of the other way around.
| I think an "active" transformation is a left action, and a "passive" transformation is a right action.
Active transformation:
$$(R_1 \cdot (R_2 \cdot \phi)) (x) = (R_2 \cdot \phi) (R_1^{-1} x) = \phi(R_2^{-1} R_1^{-1} x) = ((R_1 R_2) \cdot \phi) (x)$$
Passive transformation:
$$(R_1 \cdot (R_2 \cdot \phi)) (x) = (R_2 \cdot \phi) (R_1 x) = \phi(R_2 R_1 x) = ((R_2 R_1) \cdot \phi) (x)$$
Usually physicists don't ever think about right actions, but mathematically there's no reason not to.
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Imagine a steel bar floating in space. Assuming the bar wouldn't break or bend, if I shoot a bullet at one end, would it rotate, fly away, or both? Is there a formula to calculate both the translational and rotational velocity? Does the bar always bend, and if so is there a formula on how it bends (maybe related to the velocity/force of the bullet)? What if the bar is very long, to the point where if the bar rotate it'd be at relativistic velocity?
What if I shoot at a string of cloth instead? Would the behavior of the bar/string be different if there's air drag?
Sorry for asking too many questions, but if you can answer any of them I'd be grateful.
|
Imagine a steel bar floating in space.
With you so far.
Assuming the bar wouldn't break or bend, if I shoot a bullet at one end
What do you mean "one end"? Be specific. Precisely describe the bar and where the bullet hits it.
would it rotate, fly away, or both?
That depends entirely on where the bullet hits it. If you hit it in the middle of the end, then it simply moves. If you hit it on the side of the end, it rotates too. Which is trivially calculated using basic trigonometry.
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How the velocity of the air is slowing down in different distances from a fan? Is it determined completely by the air pressure and the temperature? Do I need to take into account also the geometry of the "air cone" of the fan or is it determined by the speed of the air when it leaves the fan?
Practically I would like to solve questions e.g. which Cubic Feet per Minute (CFM) ensures a given air speed at a given distance.
| A general approach of finding a required CFM based on a desired air speed at a given distance from a fan comes down to knowing the shape of the cone, which is a function of the fan construction. Presumably, you can get this information from a manufacturer or from some other sources.
If you know the shape of the cone, you can calculate the cone area at the distance of interest and from there the CFM of the fan as a product of that area and the desired air speed.
This method is concisely described in this fan application guide from Greenheck.
| {
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Active and passive pulling - is there a physical difference? Let's suppose we have two people Alice and Bob in a vacuum. The are connected by a rope.
At first, Alice pulls Bob with force $F$ and Bob only clings to the rope. Afterwards, the experiment is repeated with Bob pulling with force $F$ and Alice only clinging to the rope.
The two situations seem different because in one case, Alice seems to be the active part and in the other case, Bob seems to be the active part. My question is: Is there a way to describe this difference in physical terms?
It can't be the force, because due to the third law $F_{Alice \rightarrow Bob}$ = - $F_{Bob \rightarrow Alice}$ in both cases. Also the work done by the active person seems to be equal to the work done by the passive person because both the forces and the displacements are equal in magnitude. Is this correct? Does it take exactly the same amount of energy to pull as it takes to being pulled?
| If my understanding of this experiment is correct, a person A that pulls the rope remains stationary and a person B, that clings to the rope, moves.
We should assume that person A holds on to something or relies on the static friction - otherwise, he or she will move as well.
Person B will accelerate and gain some kinetic energy all due to the work performed by person A, unless person B is helping somehow to move things along.
Ignoring the intricacies of muscle contractions, this scenario, from the forces and energy prospective, should not be any different than the scenario when person A is pulling a sled: person A is applying the force and spending energy, which goes toward increasing the kinetic energy of the sled and overcoming friction.
Of course, for person B to move, the tension force has to be greater than the effective static friction, while, for person A to remain stationary, the effective static friction has to be greater than the tension.
| {
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EM Plane wave, the changing electric field is in all directions right? I just want to confirm this, because this type of diagram seems pretty popular.
The electric field and magnetic field actually surround in all directions orthogonal to x axis, right? It is not just 2d pointing only in the y direction and z direction respectively.
| The wave in your figure is linearly polarized with E along the $\hat{y}$ axis and B along the $\hat{z}$ axis. That's built into the wave when it was created, and it stays that way. You won't find any E along $\hat{z}$ or B along $\hat{y}$ for that wave.
Some other wave might be different. If could be linearly polarized along some other axis, or it could be circularly polarized. A circularly polarized wave has E pointing in different directions at different times.
So no, in the wave you show, it doesn't surround the $\hat{x}$ axis. But other waves could do that.
| {
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How are real particles created? The textbooks about quantum field theory I have seen so far say
that all talk in popular science literature
about particles being created spontaneously out of
vacuum is wrong. Instead, according to QFT those virtual particles are
unobservable and are just
a mathematical picture of the perturbation expansion of the propagator.
What I have been wondering is, how did the real particles, which are observable, get created? How does QFT describe pair production, in particular starting with vacuum and ending with a real, on-shell particle-antiparticle pair?
Can anybody explain this to me and point me to some textbooks
or articles elaborating on this question (no popular science, please)?
| The difference between real and virtual particles in QFT has to do with whether the particles are represented by an internal line or whether they have at least one free end.
For real particles, their momenta and energy need to be considered in the overall conservation of energy and momentum. This is represented, ad-hoc so far as I understand, by the dirac delta in the propagator integral. For the internal lines, their momenta integrals are done over all of phase space.
So if you ask "How are real particles created in QFT?" As with everything else quantum, the answer to "how" is "it was possible and then it happened." If you mean "what is the difference between the mathematical treatment of virtual and real particles?" I would say, "there are extra constraints on the energy-momentum vector for real particles."
| {
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AP physics 1 rotation problem could someone help me with this problem?
the correct answers are a and d. one issue i have with it is that i just don't understand what the problem is asking. like what spool? what table? i tried making some sense of the question and the answers, but i can only see d moving the wheel clockwise. the answer explanation mentions,
The key is knowing where the “fulcrum,” or the pivot for rotation, is. Here, that’s the contact point between the surface and the wheel.
but why is that? isn't the pivot where the axle meets the wheel?
| The question is worded poorly and the diagram is not very helpful, the setup looks like this:
[Source]
Now imagine wrapping a rope around the axle like this:
[Source]
Now imagine those arrow represent pulling the free end of the rope. Hopefully your intuition can get you the rest of the way there.
Edit:
The second image shows how the rope pulls on the axle, but can be misleading, because in this problem the rotation axis is not the axle. At the instant that a force is applied, the pivot point will be the point where the wheels touch the ground. The "lever arm" is the line extending from the pivot point to the point at which the force is applied. If we draw this line for each of the cases then we see that the force will in general either be along the line (no torque), to the left of the line (counterclockwise torque) or to the right of the line (clockwise torque). (a) and (d) both have a clockwise torque, and thus we would expect them to begin rolling to the right, so long as the angle between the force vector and the lever arm is maintained. (b) has a counterclockwise torque while (c) has no torque, the force is directed along the lever arm.
The confusion of this question is if you mistake the second image I posted as being the setup. In that case, the rotation axis is the axle, and then the lever arm would be different and we would get a different answer. The key is to recognize that the pivot point/rotation axis is where the wheel touches the ground.
| {
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Electric flux through finite width/ infinitely long plane due to a point charge The question is as follows:
Consider a point charge Q placed at $(0,h,0)$ (Cartesian coordinates). Find the flux in an area formed by $y=0$, $z\leq0$, $x\geq l$ and $x\leq a$ $( l\leq x\leq a )$.
I tried this by considering the two extreme lines $x=l$ and $x=a$ and I constructed a cylinder (infinitely long) using the plane passing through $y=h$ and $x=l$. Then I tried taking some angles but I was unable to find any symmetry in the situation.
Can someone please help me solve this?
| If Q is a point charge in empty space then the flux through a surface $\:S\:$ is
\begin{equation}
\Phi=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}}
\tag{01}
\end{equation}
where $\:\Theta\:$ is the solid angle(*) by which the point Q "sees" the surface $\:S$.
Now, if $\:0=\ell \leq a \:$ and the strip has a finite height $\:b\:$ along the negative $\:z\:$ then according to the Proposition-Practical Rule in my answer here we have (Figure 01)
\begin{equation}
\tan\Theta\left(b\right)=\dfrac{s}{h\!\cdot\!d}=\dfrac{a\!\cdot\!b}{h\!\cdot\!\sqrt{a^{2}+b^{2}+h^{2}}}
\tag{02}
\end{equation}
so for an infinite strip (Figure 02)
\begin{equation}
\tan\Theta=\lim_{b\rightarrow \infty}\tan\Theta\left(b\right)=\lim_{b\rightarrow \infty}\dfrac{a\!\cdot\!b}{h\!\cdot\!\sqrt{a^{2}+b^{2}+h^{2}}}
\nonumber
\end{equation}
that is
\begin{equation}
\boxed{\:\:\color{blue}{\tan\Theta=\dfrac{a}{h}}\:\:} \qquad 0=\ell \leq a
\tag{03}
\end{equation}
If $\:0<\ell \leq a \:$ then we must subtract from the solid angle $\:\Theta_{1}\:$ of the infinite strip of width $\:a \:$ the solid angle $\:\Theta_{2}\:$ of the infinite strip of width $\:\ell \:$
\begin{equation}
\Theta=\Theta_{1}-\Theta_{2}
\tag{04}
\end{equation}
But
\begin{equation}
\tan\Theta_{1}=\dfrac{a}{h} \quad \text{and} \quad \tan\Theta_{2}=\dfrac{\ell}{h}
\tag{05}
\end{equation}
so
\begin{equation}
\tan\Theta=\tan\left(\Theta_{1}-\Theta_{2}\right)=\dfrac{\tan\Theta_{1}-\tan\Theta_{2}}{1+\tan\Theta_{1}\tan\Theta_{2}}=\dfrac{\dfrac{a}{h}-\dfrac{\ell}{h}}{1+\dfrac{a}{h}\dfrac{\ell}{h}}
\tag{06}
\end{equation}
that is (Figure 03)
\begin{equation}
\boxed{\:\:\color{blue}{\tan\Theta=\dfrac{h\left(a-\ell\right)}{h^{2}+a\ell}}\:\:} \qquad 0<\ell \leq a
\tag{07}
\end{equation}
If $\:\ell<0 \leq a \:$ then we must add to the solid angle $\:\Theta_{1}\:$ of the infinite strip of width $\:a \:$ the solid angle $\:\Theta_{2}\:$ of the infinite strip of width $\:\vert\ell\vert=-\ell \:$
\begin{equation}
\Theta=\Theta_{1}+\Theta_{2}
\tag{08}
\end{equation}
But
\begin{equation}
\tan\Theta_{1}=\dfrac{a}{h} \quad \text{and} \quad \tan\Theta_{2}=\dfrac{\vert\ell\vert}{h}=\dfrac{-\ell}{h}
\tag{09}
\end{equation}
so
\begin{equation}
\tan\Theta=\tan\left(\Theta_{1}+\Theta_{2}\right)=\dfrac{\tan\Theta_{1}+\tan\Theta_{2}}{1-\tan\Theta_{1}\tan\Theta_{2}}=\dfrac{\dfrac{a}{h}-\dfrac{\ell}{h}}{1-\dfrac{a}{h}\dfrac{-\ell}{h}}
\tag{10}
\end{equation}
that is
\begin{equation}
\boxed{\:\:\color{blue}{\tan\Theta=\dfrac{h\left(a-\ell\right)}{h^{2}+a\ell}}\:\:}\qquad \ell<0 \leq a
\tag{11}
\end{equation}
Combining (03),(07) and (11) we have in general
\begin{equation}
\boxed{\:\:\color{brown}{\tan\Theta=\dfrac{h\left(a-\ell\right)}{h^{2}+a\ell}}\:\:}\qquad \ell\leq a,\quad a\ge 0
\tag{12}
\end{equation}
(*)
Examples of solid angles $\:(\rho\pi, \rho=1/4,1/2,1,2,4)\:$ are shown in the the Figure Solid Angles under my answer here : Flux through side of a cube.
An other method to find a solution to this specific problem is based on the fact (could be proved easily) that in case of a solid angle $\:\Theta\:$ formed by three rays $\:\epsilon_{1},\epsilon_{2},\epsilon_{3}\:$ with $\:\epsilon_{3}\:$ normal to the plane of $\:\epsilon_{1},\epsilon_{2}\:$ (Figure 04) we have
\begin{equation}
\Theta=\phi \quad \Longrightarrow \quad \tan\Theta=\tan\phi
\tag{13}
\end{equation}
where $\:\phi\:$ the plane angle formed by the rays $\:\epsilon_{1},\epsilon_{2}$. These are the cases of Figures 02 and 03.
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Special Relativity Muon Decay When talking about how muons reach earth even though their half-life is very short the explanation of time dilation is given. From earth's frame of reference, the Muon's clock is "slowed down" so is has longer to live and keep racing at it's speed, I think I understand this.
However, will the length between the Muon and the earth when seen from earth's reference also contract? If so, why is this never included in the calculation of amount of muons reaching the earth?
Also, from the muon's perspective the earth contracts which I understand but nobody talks about the time dilation for the earth. (Although this I understand better because it doesn't really matter how "fast" time flows for the earth relative to the muon, the earth will still take the same amount of time to reach the muon wether time dilation is included or not)
Examples of places where they discuss these phenomena without including the length contraction (from earth's frame of reference):
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html
https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation
| You have to be consistent in each reference frame you consider.
1. In the earth frame the muon proper time runs slower, but the distance it covers is not length-contracted. The length contraction refers to the muon itself, in principle the muon dimension in the direction of motion is contracted, but that is irrelevant in this experiment.
2. In the muon frame the distance to the earth is contracted, that is why the muon life time is enough for it to reach the earth.
| {
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A gravitational field is path independent. Why does a rocket not fly in serpentine lines? in theory a gravitational field is path independent, a gravitational field is a gradient field and so conservative. why doesn't a rocket fly in serpentine lines to exit the gravitational field of the moon, as said the gravitational field of the moon is path independent and the moon has no atmosphere and so there is no aerodynamic drag?
https://en.wikipedia.org/wiki/Gradient_theorem
https://en.wikipedia.org/wiki/Conservative_vector_field#Path_independence
i know that the reason is that flying in serpentine lines would require more energy ( fuel ) however where is this surplus of energy for a serpentine line path reflected in the theories?
| What you say would be true if rockets were held up by, say, a very long string. Since real rockets don’t have such a string available they have to do something much less efficient: fight the gravitational force by pushing fuel downward. A rocket needs to exert energy just to stay in the same place, let alone move in a fancy pattern! This is why a space elevator would be so helpful; it’s essentially just a long string extending out to space.
| {
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Is this derivation for schwarzschild radius for a black hole of mass $M$ correct? Consider a body of mass $M$.
We know that light can’t escape a black hole.
Speed of light being the highest possible could be set as the escape velocity.(??)
Then
$$\text{Escape velocity}^2=(2GM/r)$$
Solving for $r$ we get
$$r=2GM/v^2$$
Since $v=c$;
$$r=2GM/c^2$$
My only problem with this derivation is that shouldn’t we be using relativistic mechanics instead of newtonian? If we do use Relativistic mechanics,is there any proof similar to this one?
| If you assume that a sufficiently compact object does have an event horizon, then the relationship has to be of the form
$$r=\alpha GM/c^2,$$
where $\alpha$ is a unitless constant. This is because there is no other way to combine the mass with the universal constants $G$ and $c$ in order to get units of distance. The fact that the Newtonian derivation gets $\alpha=2$ right is just luck.
| {
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Where do the particle and antiparticle wavefunctions originate from in the Klein Gordon equation? In my textbook (Sakurai) it is given that
$$\left(D_\mu D^\mu+m^2\right)\Psi(\mathbf{x},t)=0$$
where $D_\mu=\partial_\mu+ieA_\mu$ is the covariant derivative.
It states that since it is a second order differntial equation we must specify the wavefunction at its initial time as well as its first derivative. Alternatively we can reduce the second-order KG equation to two first order equations and intepret the result in terms of the sign of the electric charge.
We can use that $D_\mu D^\mu=D_t^2-\vec{D}^2$ to get the two new functions
$$\phi(\mathbf{x},t)=\frac{1}{2}\left(\Psi(\mathbf{x,t})+\frac{i}{m}D_t\Psi(\mathbf{x,t})\right)$$
$$\chi(\mathbf{x},t)=\frac{1}{2}\left(\Psi(\mathbf{x,t})-\frac{i}{m}D_t\Psi(\mathbf{x,t})\right)$$
but where do these two functions arise from? How do you get them from this information?
| The two equations you wrote at the bottom of are just definitions of $\phi$ and $\chi$. They have no physical interpretation at this point in Sakurai's argument. The physical interpretation comes later when he rewrites the Klein-Gordon 'probability' density in terms of them (see 8.1.20)
There is a standard way to turn a higher order differential equation into a system of first order equations. For instance if we give $D_t \Psi(x,t)$ a new name, $\Pi(x,t)$,
$$D_t \Psi \equiv -i m\, \Pi$$
then the Klein Gordon equation becomes a first order system in two functions $\Psi,\Pi$
$$D_t \Pi = \frac{i}{m}(\vec{D}^2 - m^2)\Psi$$
Substitute the definition of $\Pi$ to get the original Klein Gordon equation.
Now instead of the degrees of freedom $\Psi,\Pi$ we are free to define two independent linear combinations instead:
$$\phi\equiv\frac{1}{2}(\Psi+\Pi)\quad\chi\equiv\frac{1}{2}(\Psi-\Pi)$$
Then adding and subtracting the first two equations we get first order equations for $\phi$ and $\chi$ instead (see 8.1.16, or try it yourself).
| {
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Where does the work come from if tidal forces are stretching elastic objects? An elastic object e.g. a rubber band will be stretched
*
*in accelerated expanding FRW-spacetime
*during radial free fall in Schwarzschild spacetime
by tidal forces due to Ricci- and Weyl-curvature resp., until equilibrium with its internal cohesive forces is achieved.
The work done to stretch the rubberband is represented by the elastic potential energy stored in it.
Where does this work done come from in these two cases?
| In the case of a rubberband falling into a Schwarzschild blackhole, the work comes from the increasing differential in binding energy between the two ends of the string. This is completely analaguous to the Newtonian case. Binding energy is well defined due to the existence of a time translation symmetry (i.e. timelike Killing field.)
The situation in FRW is slightly different. The rubberband in an FRW metric will only expand if
*
*The expansion of the of the FRW spacetime is accelerating
or
*The intitial conditions of the rubberband are such that there is relative kinetic energy between the two ends of the rubberband.
If neither is the case the size of the rubberband will just stay the same, and no work is done.
In case 1, the work comes from the vacuum energy and effectively slows down the acceleration. Instead of a rubber band it is easier to think about an FRW metric filled with an elastic medium. The effect is that you get an FRW solution with a slightly different equation of state (the pressure is a bit larger).
In case 2, the work simply comes from the relative kinetic energy of the opposite ends. This slows down the expansion of the rubber band.
| {
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Can an electromagnetically induced dipole be explained with photon interactions? An incident electromagnetic wave will cause a dipole moment in the medium it passes through, displacing positive and negative charges in accordance with the EM field. How much of this interaction (if any) can be explained via the particle nature of light, i.e. photons?
| One of the main reasons we declare that light (and matter) carry both wave and particle properties is because in some situations, we observed that light behaved as a wave, and in others, we observed that it behaved as a particle. The resolution of this was to declare that it must be something that is both, and hence the wave-particle duality was formulated.
What you are asking is "How can we describe a wavelike property of light using a particle description?". If we could, we would have simply said that light is a particle and done away with the wave description of light.
| {
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What is cosmic ray albedo neutron decay (CRAND)? I need a small description of what is cosmic ray albedo neutron decay (CRAND), where it comes from and what effects they have?
| I've never heard the specific acronym CRAND.
However, cosmic rays which reach Earth's ground level lose energy by, among other processes, spallation on the heavy nuclei in the soil and rock. Some of these spallation neutrons will leak back out of the surface of the ground, having partially thermalized$^1$.
These neutrons basically form a very tenuous part of Earth's atmosphere, with a scale height somewhat higher than the nitrogen/oxygen/ozone parts of the atmosphere, because the neturons' temperatures are comparable to the other gas species, but the neutrons are less massive.
I think an astronomer might consider these "albedo neutrons from cosmic rays" (your CRAN).
A thermal neutron at the ground, with average speed $v\sim 2\rm\,km/s$, can certainly reach the ionosphere/exosphere within its 1000 second half-life. The fraction that decay above the atmosphere will contribute to the population of electrons and protons trapped there.
My dumb Google search also finds solar proton albedo neutron decay, which sounds like the same effect but triggered by the solar wind rather than by cosmic rays.
That would be more important on a planet without an atmosphere or magnetosphere to modulate the solar wind protons.
My slightly less dumb search finds this 2017 Nature paper
that says I mostly guessed right, but that my involvement of the ground rather than the upper atmosphere was some terrestrial chauvinism on my part.
$^1$ One of the early papers describing these neutrons has a great comparison between the detectable neutron flux in St. Petersburg in a boat in the harbor (where the spallation neutrons from the seabed are mostly captured in the seawater), over bare ground (poorly thermalized), and over a meter of snow (well-thermalized). All I'm finding today are balloon flights:
Flerov et al., JETP 9, 511 (1959); Kopff, Phys. Rev. 59, 949 (1941).
| {
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Reference Request: Basis-independent formulation of tensor networks I could not find any references for a basis-independent formulation of tensor networks: All papers I have found use pretty much (explicitly or implicitly) the canonical computational basis by defining tensors as lists of complex numbers.
While this is certainly ok from the perspective of "I want to compute stuff", it is not when trying to look at the bigger picture, i.e. realizing tensor networks as certain string diagrams in the category of finite dimensional Hilbert spaces.
Does anyone know of a resource that addresses this?
| https://arxiv.org/abs/1210.7710 develops a framework of MPS modulo their gauge degree of freedom (which includes basis choice) as a fibre bundle.
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How can electrons spin if they have no volume? My understanding of electrons is that they have no volume, eg they are point particles. If this is true, how can a point spin?
| If you ignore the location/velocity of an electron, each electron carries one bit (or, more precisely, q-bit) of information: whether it is spin up or spin down with respect to a chosen axis. This information is encoded in a vector with two components and complex components. That is, the "spin state" of the electron can be written as
$$(\alpha, \beta) \in \mathbb{C}^2$$
where $\alpha$ and $\beta$ are complex numbers, and $|\alpha|^2 + |\beta|^2 = 1$. There is nothing "spinning," nothing is "dynamical." If you rotate the electron (or even just rotate your reference frame) the spin state of the electron will have to "rotate" accordingly as well. (How do you rotate an electron? You can manipulate it with a magnetic field, or you can just give up and mathematically pretend you are rotating it.)
Any appropriate "rotation" of a vector in $\mathbb{C}^2$ will be an $SU(2)$ matrix. The 3 dimensional rotation group, $SO(3)$, must somehow be mapped into $SU(2)$. (The map can only be accomplished up to a physically unmeasurable sign ambiguity, but whatever. This is called the "spin 1/2" (projective) representation of $SO(3)$.)
If that went over your head, consider this: the spin state is, in an extremely crude sense, pointing in some direction, and this little arrow, pointing in space, must be rotated somehow. In classical mechanics, aside from velocity, the only arrow you can really associate with an object is its angular momentum, which will will only be non-zero if the object is "spinning." Unlike a classical object, which can be made to stop spinning, a spin 1/2 particle like an electron, always has this little arrow pointing somewhere that we call spin.
| {
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Is there anything like a "polarizer rejection spectrum"? In general, if you shine some EM radiation through an arbitrary polarizer which works for visible light, it's not guaranteed to polarize your EM radiation. E.g. you can take polarizing sunglasses and shine unpolarized X rays or radio waves through it, and this radiation most likely go through almost unaffected.
But I've never heard of any term which would refer to polarizer's rejection spectrum – i.e. how much of the "wrongly" polarized radiation of given wavelength it blocks. Is there any such term? What is it?
| The quantity I was interested in is called polarization extinction ratio. See the usage of this term e.g. in the description of this wire grid polarizer, like in the following graph taken from that page's Graphs section:
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From "Matrix" form to "Component" (tensor) form Given
$\omega=-\eta\omega^T\eta^{-1}=-\eta\omega^T\eta$,
where $\eta$ is the usual Minkowski metric.
Is the following logic correct?:
$$
{\omega^{~\mu}}_{\nu}= -{\eta_{\varepsilon\nu}}{\left(\omega^T\right)_{\sigma}}^{~\varepsilon}~\eta^{~\mu\sigma}= -{\left(\omega^T\right)^{\mu}}_{~\nu}=-{\omega^{~\nu}}_{\mu}
$$
and so
$$
{\omega^{~\mu}}_{\nu}+{\omega^{~\nu}}_{\mu}=0 .
$$
I'm not sure I'm translating the equation into component form correctly, the manipulation of indices I'm (in principle) happier with.
Thanks!
| No, as you can see your indices don't match up. You must have an equation of the form $A^\mu{}_\nu = B^\mu{}_\nu$.
The correct way to do it is
$$
\omega^\mu{}_\nu = - \eta^{\mu\rho} ( \omega^T)_\rho{}^\sigma \eta_{\sigma\nu}
$$
Note the way the indices are always adjacent since this is matrix multiplication. Next, we use
$$
( \omega^T)_\rho{}^\sigma = \omega^\sigma{}_\rho
$$
Then,
$$
\omega^\mu{}_\nu = - \eta^{\mu\rho} \omega^\sigma{}_\rho \eta_{\sigma\nu} = - \omega_\nu{}^\mu
$$
Thus, the correct equation is
$$
\omega^\mu{}_\nu + \omega_\nu{}^\mu = 0 .
$$
| {
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Bose-Einstein Condensation in lower dimensions Bose-Einstein condensation occurs at 3 dimensions. However, it is not possible to happen at 1 or 2 dimensions; in fact I am able to prove this myself. What is the explanation for this?
| For simplicity, I will consider non-interacting gases. The idea here is that there is a bound on the density of excited states in 3 and higher dimensions, whereas no such bound exists in 1 and 2 dimensions. This bound enforces a condensation of the gas particles into the ground state for 3 and higher dimensions but not in lower.
The details are a bit cumbersome to write down from scratch, so I will refer to some standard details and sketch out the non-trivial part. Check out eqns (7.31) and (7.32) of Mehran Kardar, Statistical physics of particles, where the average occupation number of a non relativistic gas is derived. Write down the same in $d$ dimensions. If you now convert the integral for the average occupation number into a dimensionless form, you will arrive at the generalization of eqn (7.34) for $d-$ dimensions given by
$$ n = \dfrac{g}{\lambda^d} f^1_{d/2}(z)$$
where
$$ f^1_{m}(z) = \dfrac{1}{(m-1)!} \int_{0}^{\infty} \dfrac{dx x^{m-1}}{z^{-1}e^x - 1}$$
This integral is finite for all values of $z$ if $d \geq 3$, and hence has a maximum value , whereas the same is not true for lower dimensions. In $d \geq 3$, we therefore have a bound on the density of excited states at $z = 1$, given by
$$ n _x = \dfrac{g}{\lambda^d} f^1_{d/2}(z) \leq n* = \dfrac{g}{\lambda^d} \zeta_{d/2}$$
where $\zeta_{d/2}$ is the max value of the number density of excited states at $z = 1$.
Since the integral is not finite and therefore no such maximum value exists for $d = 1,2$, hence there is no BEC in lower dimensions for this system.
| {
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Why is the relativistic adiabatic index 4/3? I was told that in the relativistic limit the adiabatic index approaches 4/3 for a monoatomic gas instead of 5/3 in the non-relativistic case. I was told this occurs due to a reduction in degree of freedom but this may be incomplete and does not quite explain the new expression since $ \gamma = (n + 2)/n$ where $n$ is the # of degrees of freedom. Thus I am wondering both quantitatively and qualitatively, why does the adiabatic index decrease, and to 4/3 specifically, in the relativistic regime for a monoatomic gas?
| I don't know what a reduction in degrees of freedom is in this case. But statistical physics has well known methods to compute properties of ideal gases. Ultra-relativistic particles dispersion relation has the form
$$
\varepsilon(\vec{p}) = c|\vec{p}|
$$
Correspondent partition function of a monatomic classical gas is
$$
Z = \frac{V^N}{N!(2\pi\hbar)^{3N}} \left(\int e^{-c|\vec{p}|/\theta} d\vec{p}\right)^N = \frac{V^N}{N!(2\pi\hbar)^{3N}} \left(\frac{8\pi\theta^3}{c^3}\right)^N
$$
Using standard thermodynamic relations we obtain equations of state (all specific values are computed per one particle)
$$
c_v = 3, \quad p = \frac{\theta}{v}
$$
Further
$$
c_p = c_v - \theta \left(\frac{\partial p}{\partial \theta}\right)^2_v\Bigg/\left( \frac{\partial p}{\partial v}\right)_\theta
$$
gives
$$
c_p = c_v + 1 = 4
$$
Due to ideal gas's thermal equation of state $p = \theta/v$ and $c_v = const$ the adiabatic process equation is
$$
pv^\gamma = const,
$$
where $\gamma = c_p/c_v = 4/3$.
The talk about degrees of freedom is relevant when the energy of a molecule is equal to a sum of quadratic terms. For an ultra-relativistic atoms this is not a case.
| {
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Reason for body attached to a string being in free fall? This is a question I found in a book:
A string is wrapped around a uniform cylinder as shown in diagram. When cylinder is released string unwraps without any slipping and the cylinder comes down.
I assumed that an equation $T - mg = ma$ could be formulated and the tension does negative work. However the answer is that the tension does zero work. This I understand is because the cylinder is in free fall, and the equation will be $T - mg = mg$ and therefore $T = 0$.
Is my assumption correct? If it is, why is this so? It doesn't make sense to me. Should the tension exert some upward force and the downward acceleration be at least a bit less than $g$?
| Your equation $T-mg=ma$ seems right to me. The reason the tension does not do any work is not because $T=0$, but rather because the point where $T$ acts does not move.
This is simply because of the "no slipping" condition : the point of contact of the cylinder has speed 0. Hence, the work produced by $T$ is simply $W =\vec{T}\cdot d\vec{x}$ where $d\vec{x}$ is the instantaneous displacement of the contact point. However, at any given instant, $\frac{d\vec{x}}{dt} = 0$ because of the no slip condition, hence $d\vec{x}=\vec{0}$, and so $W=0$.
This is a bit counterintuitive at first, because obviously the contact point does move along the string. However, at any given time the speed of the contact point is 0. In other words, the cylinder is not moving with respect to the string.
| {
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If the 3-Body force problem hasn't been solved, how do rocket scientists plan orbits of spacecraft? What methods would they use to predict what would happen in a situation when a probe is being acted upon by the gravity of two stars, say?
| The three body problem isn’t “solved” in the sense that there is no known closed form solution that works for any general initial conditions.
However, when you have two massive bodies and one that is considerably lighter, you can estimate the trajectory with almost any degree of accuracy. Furthermore, numerical techniques will allow you to do that with multiple bodies and without any restrictions on their mass.
The analysis roughly comes down to this:
*
*at a given instant the position and velocity of all objects is known (the initial conditions)
*from the positions we know the gravitational forces on each object (magnitude and direction)
*given these forces and the mass of each object, we can compute the acceleration at this instant
*from the position, velocity and instantaneous acceleration, we can compute the velocity and position a very short time later
*repeat for small time steps to compute the complete orbit (obviously, if you are firing the rocket you also need to take account of the thrust and the changing mass)
There is an entire field of scientific computing dedicated to doing this right. And the method briefly made an appearance in the movie “Hidden figures” - at that time these things were still under development, and of course computers were much slower, and had less memory, than today’s machines.
I have written several answers on this site that used such an approach. Probably the most relevant is this one
| {
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Statistical mechanics and thermal averages in $\mu-$space and $\Gamma-$space What is the relation between the thermal averages in $\mu-$space and $\Gamma-$space of a system having $f$ degrees of freedom in statistical mechanics? For a system with $N$ particles (and having $n=3N$ degrees of freedom) the thermal average is given by $$
\left\langle O \right\rangle_{\Gamma}
=\frac{\int O\left(q,p\right) \rho \left(q,p,t\right) \, \mathrm{d}^{n}q \, \mathrm{d}^{n}p}{\int {\rho\left(q,p,t\right) \, \mathrm{d}^{n}q \, \mathrm{d}^{n}p}}
$$while the thermal average in $\mu$-space is given by $$
\left\langle O \right\rangle_{\mu}
=\frac{\int O\left(q,p\right)f\left(q,p,t\right) \, \mathrm{d}^{3}q \, \mathrm{d}^{3}p}{\int f\left(q,p,t\right) \, \mathrm{d}^{3}q \, \mathrm{d}^{3}p}
$$where $f\left(q,p,t\right)$ represents the single-particle phase space density (used in deriving the Boltzmann equation).
What is the relation, if any, between these two averages? Sometimes one discusses statistical properties of a system using $\mu-$space and sometimes using $\Gamma-$space which confuses me.
| The precise relation is given by
$f(q,p,t)=\int \rho(q^N,p^N,t)\delta(q-Q)\delta(p-P)dq^Ndp^N$,
where $Q$ is the center of mass of $q^N=(q_1,\ldots,q_N)$, and $P$ is the total momentum, the sum of $p^N=(p_1,\ldots,p_N)$. With this identification, both formulas give the same expectation value when applied to a 1-particle operator.
| {
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Link between integrability and soliton solutions I have been doing some research on the properties and dynamics of solitons (in particular, solitons in superfluids) and several works and papers mention the link between solitonic solutions and integrability of the non-linear differential equation describing the physical system. However, I found that the explanation of what exactly it entails for a system to "be integrable" and what this has to do with solitonic solutions is often either quite vague or explained in very technical and mathematical terms, with little physical content. Therefore, my question is twofold:
a) Is it possible to explain in physical terms what it means for a differential equation and its underlying system to be integrable, e.g. the 1D Gross-Pitaevskii equation (describing a 1D Bose-Einstein condensate) is said to be integrable.
b) What is the link between integrability of a non-linear differential equation and solitonic solutions of this equation? I have come across non-integrable differential equations which also permit solitary wave solutions, but maybe these are not "true solitons" in the strictest sense of the word?
| There is plenty of integrable partial differential systems which are dispersionless a.k.a. hydrodynamic-type (i.e., can be written as first-order homogeneous quasilinear systems; apparently in the case of more than three independent variables the dispersionless systems form an overwhelming majority of all known integrable systems, see e.g. this article and references therein) and do not necessarily have multisoliton solutions but have some analogs thereof, multiphase solutions, cf.e.g. here.
| {
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Why is $ \frac{\vec{r}}{r^3} = \frac{1}{r^2} $? I know it's surely a beginner's question but I don't see why you can write
\begin{align}
\frac{\vec{r}}{r^3} = \frac{1}{r^2}\cdot \frac{|\vec{r}|}{r}
\end{align}
Could someone explain it please? It would help understand quite a few things ...
| $$\frac{\vec{r}}{r^3} = \frac{1}{r^2}\frac{\vec{r}}{r}=\frac{1}{r^2} \vec{u_r}$$
where $\vec{u_r}$ is an unitary vector with the direction of $\vec{r}$
| {
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How do you expand $\langle x'-\Delta x'\rvert \alpha\rangle$? In my textbook (Sakurai) the following identity is often used:
$$
\left< x'-\Delta x' \, \middle| \, \alpha\right>~=~\left< x' \, \middle| \, \alpha \right> - \Delta x'\frac{\partial}{\partial x'} \left< x' \, \middle| \, \alpha \right>
\,.$$
How can this identity be derived?
| you have to derive the taylors expansion. I wont give the full proof but I will give how you proceed.
$$f(x)-f(x_0)=\int^{x}_{x_{0}}f'dx$$
and
$$f'(x)-f'(x_0)=\int^{x}_{x_{0}} f''dx$$
then if you substitue latter to former you would have
$$f(x)-f(x_0)=\int^{x}_{x_{0}}f'(x_0)dx+\int^{x}_{x_{0}}\int f''(x)dx$$
and
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\int^{x}_{x_{0}}\int f''(x)dx$$
by induction you can find more terms but in your question the first two is enough.
then you have to treat your inner product as $f(x)$ that will give you the answer.
| {
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What atomic forces are acting to resist me pushing an air filled bottle underwater? Yes air is less dense than water but how does the bottle know to rise or indeed to move? It's not electromagnetism I think. Does this have a relation to gravitational forces? Is it to do with the number of protons in the molecules of the different substances?
| Archimedes figured this one out around 250 BC. If you push an empty bottle underwater you have to raise the surface of the water by the same volume. This is the counterforce that you feel.
| {
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Modifying the Bohr Model for Muonic Hydrogen I am trying to find the energy between the $n=2\leftrightarrow3$ transition for Muonic Hydrogen. My approach was to modify the Bohr model for standard hydrogen but taking the mass of the Muon $m_\mu\approx 207m_e$ instead of the mass of the electron $m_e$ and then substituting in the effective mass $m=\frac{m_\mu m_p}{m_\mu+m_p}\approx186.03m_e$ (I took $m_p\approx1836m_e$).
Now taking $R=\frac{m_ee^4}{8\varepsilon_o^2h^3c}\approx10.97\times 10^6$ it gives $\Delta E\approx340.12\times 10^6\text{ J}$
Is this correct? Because it's quite a large value (assuming I got the units right)
| $E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$) this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.
| {
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Why is a vacuum balloon impossible? According to Can Impossible Vacuum balloon be possible with this idea?, a vacuum balloon is impossible. Why is that?
| A simple vacuum balloon that consists just of a spherical shell of uniform thickness is not possible at all in air at 1 atm. I only know how to prove it can't be done with a material with uniform properties such as an amorphous solid but I believe it can be shown that it can't be done with any material. Suppose you have a vacuum balloon that's a spherical shell of an amorphous material. The problem is not that it doesn't have the strength to support the pressure difference. The problem is that the vacuum balloon would be in an unstable equilibrium because the material doesn't have a high enough shear modulus to stabilize it shape. The YouTube video Railroad tank car vacuum implosion shows that a solid steel tank car can implode when it's nowhere near thin enough to float in air. For a spherical shell of a given thickness to support 1 atm on the outside and 0 atm on the inside, the sheer modulus it must have varies as the inverse cube of the thickness. I once read that water is 800 times denser than air. If the shell is thin enough to float in air, then in order for it to support 1 atm on the outside and 0 atm on the inside, it must have a shear modulus of about 1,000,000,000 atm or more. I also read that diamond has the highest shear modulus of any material at 535 GPa which is far below the required shear modulus.
| {
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Why can a wave be expressed with a sine function? I see many expressions which express waves with the sine function like $y=\sin(kx-\omega t)$.
Waves really look similar to the shapes of a sine or cosine function, but does this guarantee that expressions that show wave-like movement are sine or cosine functions or is this just an approximation?
| Suppose $f(x)$ is a period-$2\pi$ function. Why would it be expressible in terms of only the $e^{ikx}=\cos kx+i\sin kx$ with $k\in\mathbb{Z}$? Well, define the Fourier transform $\tilde{f}(k):=\int_\mathbb{R}f(x)e^{ikx} dx$ so $$e^{2\pi ik}\tilde{f}(k)=\int_\mathbb{R}f(x)e^{ik(x+2\pi)} dx=\int_\mathbb{R}f(y-2\pi)e^{iky} dy=\int_\mathbb{R}f(y)e^{iky} dy=\tilde{f}(k).$$Thus $\tilde{f}(k)=0$ whenever $e^{2\pi ik}\ne 1$, i.e. $k\notin\mathbb{Z}$.
| {
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What is the evidence for a supermassive black hole at the center of Milky Way? Black holes cannot be seen because they do not emit visible light or any electromagnetic radiation. Then how do astronomers infer their existence? I think it's now almost established in the scientific community that black holes do exist and certainly, there is a supermassive black hole at the centre of our galaxy. What is the evidence for this?
| Sagittarius A* (the black hole at the center of our galaxy) has some of the best observational evidence for black hole I have ever seen. Here, check out the animations from UCLA made from our observations. This is from data taken over a span of 20 years. You can see the bright spots (stars) orbiting around a patch of nothingness. The ones that get really close whip around at some insane speed but slow down quickly as they move away. Obviously, whatever is at the center has got a respectable amount of mass to it. But notice also that the stars always seem to move around something that is at the dead center (and their orbits are ellipses, which shows that we aren't just moving the camera to keep it at the center). Consider that, the mass of those stars must be insanely small next to the mass of the central body, otherwise it would be flung out into space the other direction when one star gets really close.
So, here you can see massive stars orbiting something that gives off no light and must be orders of magnitude more massive than any of the stars around it. Well that seems to fit the profile of a black hole. Plus the mass that we calculate it must have is high enough that anything that massive and that compact would have to collapse into a black hole.
P.S. If you didn't check out the videos, do. They're great; I love them.
| {
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Is zitterbewegung physical or not? It appears that zitterbewegung, a frequency associated with the total energy of a particle or system, is widely considered to be an unphysical quantity (e.g., Kobakhidze et.al.), @Lubos Motl, McMillan). However, a few physicists including Hestenes, Recami et.al.,consider it to be a fundamental, physical quantity.
It appears that, for example, the photon emitted in transitions between two states in an atom has precisely the frequency of the "beat frequency" (difference frequency) between the zitterbewegung frequencies of the two states.
Why is zitterbewegung considered unphysical by most physicists, and has there been a solid refutation of Hestenes' position that zitterbewegung is physical?
| I think the answer to your question might be that no one yet knows whether zitterbewegung is physical or not because no one really knows what the physical underpinnings of quantum physics are yet.
For a discussion of what zitterbewegung might mean physically, I would study walking droplets / hydrodynamic quantum analogs. Here a particle (droplet) always has an intrinsic oscillation, as it bounces along the bath propelled by the Faraday waves from its impacts.
Disclaimer: I am not familiar with zitterbewegung in other contexts than HQA, so I don't know exactly how the original idea of zitterbewegung relates to this phenomenon in HQA.
| {
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What are the physical conditions for general ohms law? A problem presents Ohm's law as $$\vec{J}=\sigma \vec{E}$$ where $\sigma$ is the conductivity given by a scalar. The problem asks what physical conditions must be satisfied for the equation to hold.
I understand that the material must be isotropic for the conductivity to be represented by a scalar instead of a tensor. What other physical conditions must apply?
| A trivial condition is that the temperature should be constant with current. An important one is that the mean free path of the charge carriers should be much shorter than the distance between the contacts.
| {
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why does rolling without slipping imply no work done by the frictional force? Suppose we have a cylinder (mass $m$, radius $R$, and moment of inertia $I$) that rolls without slipping straight down an inclined plane which is at an angle $\alpha$ from the horizontal. The quickest way to solve for the motion is to use conservation of energy, assuming that there is no work down by the frictional force.
My question is: why does rolling without slipping imply no work done by the frictional force? One explanation you will sometimes read is that this is because the point of contact of the cylinder is instantaneously at rest relative to the ramp surface, so when we compute the work via $W = F\ dx$ we have $dx = 0$ and hence $dW = 0$.
However, consider a related problem. Suppose the cylinder is held fixed in place (say by a rod drilled down its middle) but is free to rotate around its center. Now suppose we take the ramp up against the cylinder and move the ramp so as to get the cylinder spinning. In this case the frictional force obviously did work on the cylinder, yet once again the contact point of the cylinder is at rest relative to the ramp. Question: what is the difference between the two cases?
| Friction without slipping can do work. Consider two blocks stacked on top of one another. If you push the bottom one, it will do work on the top one, and the two blocks will move together. The key here is that both blocks are moving, as in your second example.
Note that "work done" is a quantity that depends on the reference frame your are working in. Your argument says that there is no work done on the cylinder in the reference frame where $dx=0$. That is only the same as the lab frame if the ramp is not moving.
In your first example, the ramp isn't moving, and so the friction of an object rolling on it without slipping does no work.
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Help with D. Tong example on Noether in QFT In this lectures, example 1.3.2 on page 14 concludes that the Noether current is
But how can the current be a two index object when it is defined in eq. (1.38), which is
as a one index object? If I apply the formula I obtain something of the form $j^\mu$. Can someone make the calculations explicitly?
| It's OK to have multiple indices on the Noether current; you just need to know how to lose all but one of them to form the most general conserved current that results.
The Killing vectors satisfy $\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0$. (Replace all $\nabla$s in what follows with $\partial$s if you only care about flat spacetime.) Since $T^{\mu\nu}=T^{\nu\mu}=0$, $\nabla_\mu\xi_\nu T^{\mu\nu}=0$. The result $\nabla_\mu T^{\mu\nu}=0$ then implies $\nabla_\mu j^\mu=0,\,j^\mu:=\xi_\nu T^{\mu\nu}=\xi^\nu T^\mu_\nu$. Thus each $2$-index current gives a vector space of $1$-index conserved currents, whose dimension is equal to the number of linearly independent Killing vectors. This is why an $n$-dimensional is called maximally symmetric if its number of linearly independent Killing vectors is maximal (i.e. $\frac{1}{2}n(n+1)$).
| {
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How to draw Penrose diagram for a particle of mass $m$ in between two Schwarzchild black holes? I am reading Hobson's GR book. I am curious about Penrose diagrams so I asked.
| You can't draw a Penrose diagram for a spacetime that lacks the necessary symmetry and isn't conformally flat. Normally our first step is that we take a slice or projection to reduce the number of dimensions to 2. If the spacetime lacks symmetry, then this step leaves you with something that doesn't give you complete information about the whole spacetime. The next step is to do a conformal transformation in order to compactify the space. In 2 dimensions, all spacetimes are conformally flat.
For a general spacetime, you can't do the reduction to 2 dimensions without losing the picture of the whole spacetime, and if you can't reduce it to 2 dimensions, then you can't compactify it, because it isn't conformally flat.
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Does a black body have a higher temperature when moving away from us at high speed? When a black body moves away from us at high speed the average velocity of its constituents will be higher. This means the body's temperature has increased. The whole spectrum of the wavelengths though will have been increased when arriving on earth, suggesting the body's temperature has décreased. Now my question is which of the two is the case?
| If "high speed" means at some fraction of the speed of light, then the blackbody will appear to be cooler.
First off, as lesnik notes, the idea of measuring the temperature of a BB is basically a comparison of the average KE compared to itself. What you get is a curve, and that curve's shape will be independent of the overall gross velocity of the object. So in that respect, the measured temperature would not change.
However, at some point redshift/blueshift comes into play. In your case you have set it up so it's moving away, so it's redshift. So instead of seeing something that is, say, red-hot, the redshift will cause it to appear infrared-hot, which is, by definition, cooler.
In fact, it is precisely this effect that we use to measure the velocity of astronomical objects. In practice, stars are not perfect blackbodies (but pretty close) and we can often pick out specific spectrum lines. We assume (and have lots of evidence to conclude) that those spectral lines have the same frequency in distant stars as they do here on Earth. Thus, by measuring this shift, you can calculate the radial velocity.
| {
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"url": "https://physics.stackexchange.com/questions/407448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Redshift - How can $z$ be greater than 1? I'm having trouble understanding the equation for redshift:
$z = Δλ/λ ≈ Δf/f ≈ v/c$.
If $z = v/c$ and $c =$ speed of light,
how can $z>1$ (as nothing can exceed the speed of light)?
| If we denote z as a Doppler redshift (due to velocity of an object moving through space away from us) and z’ as cosmological redshift (due to the expansion of space), then conceptually, ONLY z’ should be applied to spatial expansion throughout the range of z’ and general relativity (GR) solved for distance as a function of z’. GR says nothing about velocities attached to spatial expansion, and the special relativity (SR) speed limit applies only to Doppler redshifts and objects moving through space. GR puts no limits on the increase in distance from galaxy A (suppose our Milky Way) to unbounded galaxy B in time t as the universe expands. Only due to experimental limitations in the past was z instead of z’applied when z’< 1. But now, it is possible to distinguish outcomes between using SR and z versus GR and z’ even when z’= 0.01. If GR is applied to z’ < 1, the SR speed limit doesn’t apply. Just imagine a bound galaxy cluster getting further and further apart from all other such clusters like dots on a balloon that is increasing in size. SR forbids any object or information from passing another object faster than c, but no galaxy that gets further and further away from all unbounded ones due to expanding space can ever pass another galaxy. The SR speed limit simply doesn’t apply to spatial expansion.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the equation of motion for a spring-damper system the same whether oriented upward or downward? So every spring-damper system I've found online has the equation of motion:
$$mx''+cx'+kx=0$$
I can understand how this is derived when downwards is positive, but what about when upwards is positive? Wouldn't it be
$$mx''-cx'-kx=0$$
| Let $\hat d$ and $\hat u$ be unit vectors in the down and up direction.
Looking at your right (correct) derivation you have made down $\hat d$ as positive and the displacement of the mass from the equilibrium position is $x \hat d$ leading to the velocity being $\dot x \hat d$ and the acceleration being $\ddot x \hat d$.
So your equation of motion is $mg \hat d - ks \hat d - kx \hat d - c \dot x \hat d = m \ddot x \hat d \Rightarrow m\ddot x \hat d + c \dot x \hat d + k \hat d = 0$
To change from having down as positive to having up $\hat u$ as positive all you have to do is note that $\hat d = - \hat u$ and substitute for $\hat d$ into the equation which was derived was having down as positive.
You will find that you get the same equation of motion.
So what was wrong with your left hand derivation?
You made the displacement of the mass $-x \hat u$ and the velocity of the mass $- \dot x \hat u$ but what you failed to do was to make the acceleration $-\ddot x \hat u$
| {
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"timestamp": "2023-03-29T00:00:00",
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Potential due to line charge Is it possible to calculate the electric potential at a point due to an infinite line charge? Because potential is defined with respect to infinity.
| Since this an infinite line - not an infinite sphere - there are plenty of points in space infinitely removed from it, which you can use as your zero reference points.
So, once you know how the field of the infinite charged line looks like (you can check here), you can calculate the electric potential due to this field at any point in space.
It is worth noting, that the electric field of an infinite line will be diverging, so, unlike the field of an infinite plane, it will be approaching zero at infinity and, therefore its potential at a random point in space won't be infinitely high.
| {
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"url": "https://physics.stackexchange.com/questions/407797",
"timestamp": "2023-03-29T00:00:00",
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How does emitting light while moving not break the law of conservation of energy? Say you have 2 identical LED monochromatic flashlights. The battery lasts exactly 1 hour. We are able to calculate the energy emitted as photons from the LED by Planck's Law: $E = h\nu$.
Now if we were to move one of our flashlights though a vacuum at half the speed light, we'd be able to calculate the change in in frequency $\Delta \nu$ using the Doppler Effect for an observed frequency: $\nu = (1 + \frac{\frac{1}{2}c}{c})\nu_0 = 1.5\nu_0$, meaning that each photon from our flying flashlight is blue shifted so that it carries 1.5 times the energy of the stationary flashlight.
Accounting for time dilation, a stationary observer would only observe the flying battery/LED emitting light for $1 hour * \sqrt{1-\frac{0.5c^2}{1c^2}} ≈ 0.866 hour$, but I feel this is a moot point because photons should be emitted from the flying flashlight at the same rate as the stationary flashlight, from an observer moving/standing with either flashlight. That is to say both flashlights will emit the same number of photons before the battery dies, it's just that the stationary flashlight will emit them over 1 hour and the flying flashlight will emit them over a period of 0.866 hours to a stationary observer.
Accounting for the energy for moving the flashlight, Newton's first law of motion states that an object in motion stays in motion and I feel that the energy required to accelerate the flashlight to $\frac{1}{2}c$, does not need to be accounted for here, as that energy is conserved until the flashlight needs to be slowed back down to stationary.
This says to me that the total energy emitted from the flying flashlight would increase just because it's moving? If the battery and LED in the flashlight were 100% efficient (in theory) while stationary, how could it emitting 1.5x as much energy while moving from the perspective of a stationary observer?
|
I feel that the energy required to accelerate the flashlight to $\frac 12 c$, does not need to be counted for here, as that energy is conserved until the flashlight needs to be slowed back down to stationary.
Actually, it does need to be counted. The energy(mass) of the flashlight is reduced after it emits the photons. So the energy you can recover from slowing down the flashlight to stationary is less that was put in to accelerate it. I think you'll find that it equals the gain in the energy to the beam of light.
| {
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"url": "https://physics.stackexchange.com/questions/408202",
"timestamp": "2023-03-29T00:00:00",
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Surface tension molecular theory How does the surface of water behave like a stretched membrane? What's the cause for that?
I know that the surface molecules experience a net force downward so there is dense layer of water molecules in the surface but how does this leads to minimizing the surface area? Please help me clear my doubts.
| Every one of the water molecules in the bulk liquid exerts a small amount of attractive force upon its nearest neighbors. The total energy of the ensemble of water molecules is minimized when the molecules are pulled close to one another by this force.
When a portion of liquid water is in contact with air, the molecules right at the surface have only half as many nearest neighbors with which to share that force of attraction, and as a result they attract one another more strongly at the surface than they do in the bulk. To minimize their energy, they want to minimize the distance between them, and this makes the surface of the water act like a stretched membrane which strives to minimize its area.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Data - Monte Carlo Correction techninques in Particle physics It is said that the data does not always match with the Monte Carlo simulations in particle physics.(I guess even in the Higgs to gamma gamma channel, the peak in real data was at about 127GeV and thus it was corrected) .Thus,I wished to know what are the different ways in which the differences in Data and MC simulations are corrected, if possible, for the case of jets like bjets.
| You have to understand that a Monte Carlo simulation is a way of integrating the predictions of a model.
Theoretical models are often calculable, and can be inserted with "simple" generators. i.e. every event is generated with the probability given by the quantum mechanical calculation.
This is not enough to allow comparison of data with theory. The data itself follows theoretical models, that describe particle interactions through the detector, with errors coming from statistical probability functions and estimates of these functions. All these errors need to be generated with the appropriate statistical probability width. The summation of events is the total integrals giving crossections for generating events in the experiment.
Thus a Monte Carlo program gives one event with the mathematically correctly combined probability of being observed in the detector. A simulation of one real event's probability.
When the Monte Carlo data do not fit the data, then two things are possible:
1) there is a programming error in the numerous generators or the code, which has to be found
2) there is a new discovery and everybody rejoices.
For example, back in the days when quantum chromodynamics was not even well formulated as a theory deep inelastic scattering on protons was thought to be an elementary interaction, and was modeled in monte carlos accordingly, with the appropriate theoretical scattering amplitude in the generator. The data showed deviation in the high momentum transfer region, not reproduced in the monte carlo, which led the way to verifying the complex quark internal structure of the proton.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In QM why can wavefunctions be written as $\psi (x,t)= Ae^{i(kx-\omega t+\phi)}$? In my textbook, a wave traveling in the positive $x$-direction can be described using $$\psi(x,t)=Ae^{i(kx-\omega t+\phi)}. $$
I understand that the equation for a classical wave can be extrapolated by solving the differential equation associated with simple harmonic motion, giving us $\psi(x,t)=A\cos(kx-\omega t +\phi)$. I know Euler's identity is $e^{ix}=\cos(x)+i\sin(x)$. It would seem that we could simply equate $\psi(x,t)=A\cos(kx-\omega t +\phi)$ with $\Re (Ae^{i(kx-\omega t+\phi)})$ by omitting the $i\sin(kx-\omega t +\phi)$ term. My book seems to equate $\psi(x,t)=A\cos(kx-\omega t +\phi)$ to the entire complex version of the wavefunction $\psi (x,t) = Ae^{i(kx-\omega t+\phi)}$ with the stipulation that you can simply extract the real portion of the function when needed. It seems like these two functions should not be able to be equated for reasons described above. Is this a typo or am I not understanding this correctly?
Also, why in QM do we need to include the imaginary terms associated with the $i\sin(kx-\omega t +\phi)$ to get an accurate picture of the wave? Can the above relation only hold because it is derived from a wave associated with simple harmonic motion with classical not quantum waves?
| As stated in the answer by @my2cts, $\psi(x,t)=Ae^{i(kx-\omega t+\phi)}$ is the solution of the free particle Schrödinger equation, but $\psi(x,t)=A\cos(kx-\omega t+\phi)$ is not, even if it is a solution to wave equation $\frac{\partial^{2}\psi(x,t)}{\partial x^{2}}-\frac{1}{v^{2}}\frac{\partial^{2}\psi(x,t)}{\partial t^{2}}=0$ because Schrödinger equation has the form $\frac{\partial^{2}\psi(x,t)}{\partial x^{2}}\propto\frac{\partial\psi(x,t)}{\partial t}$. Therefore, the solution to Schrödinger equation is a linear combination of sines and cosines (as you can show), which is represented as the imaginary exponential function by Euler's identity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Meaning of crystal orientation If I have a crystal, say (001)-oriented ABO3, what does the 001 imply, from an experimental perspective? I understand that the 001 is referring to a plane as described by Miller indices, but what is it about this plane that is relevant? Is it that this plane is the same plane as the surface of the sample?
| When someone says (hkl)-oriented anything, they almost always mean the surface normal of the sample is parallel to (hkl). So Si-(111) would mean silicon oriented so the surface normal is along the (111) direction.
If you have a layered material, then having a (001) orientation means the c-axis is perpendicular to the surface of the sample. This is an extremely common orientation for layered crystals for example.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Sound, Ultrasonic waves: Distance travelled and spreading angle I am working on a project related to measuring distance using sonic/ultrasonic waves(i.e. sending a pulse and calculating the time between the echo and then distance). I was reading a lot about the optimum frequency that I can choose for getting a sufficient maximum range lets say 100meter measurement. There are few things that i want to ask:-
*
*How much distance can these waves travel so as to generate enough echo to be detected.
*what are the factors affecting the distance traveled by these waves.
*what is the spreading angle of these waves and is it possible to generate a wave of narrow angle?
I am working on getting the water level height from top of the surface, where a narrow beam would be sent and the reflected echo would be used to find the distance divided by two to get the actual height.
Note that the diameter of inlet is atmax 10 inches and hence a narrow beam is needed. for the start however the spreading is not an issue for me since it is just for the test purpose.
thank you in advance.
| 1)"How much distance can these waves travel so as to generate enough echo to be detected." It depends on your receiving transducer and electronics noise levels, among other things.
For attenuation in air see:
https://physics.stackexchange.com/a/52246/45664
2)"what are the factors affecting the distance traveled by these waves."
Attenuation will be the primary factor.
3) "what is the spreading angle of these waves and is it possible to generate a wave of narrow angle?" The unfocused beam width in radians is approximately given by the wavelength divided by the aperture width. (more true for continuous rather than pulsed transmission) So you can make a narrower beam by increasing the frequency or increasing the aperture size (also by using some focusing). It is possible to have a narrow beam.
See: http://www.ndtcalc.com/utbeamspread.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How strict are the boundaries that divide dimensions? Is a single-layer sheet of graphene 2D or 3D? I would like to know if there is any theory that describes a set of rules that define the boundaries of dimensions.
For example, does a single layer sheet made of graphene considered a two or a three dimensional object?
Or does any object consisted of at least one atom considered by default a 3D object?
How about the elementary particles, do they qualify as 2D objects?
Since a 3D object can be projected to and be observed in a higher dimension, do we have any example of 2D objects projected to the third dimension?
| As anna already answered, rigorously, any object is either zero- or three-dimensional.
That said, it can be very useful to approximate a 3D object by a 2D or 1D one. Graphene in particular can often be well approximated as a 2D sheet.
Being an approximation, there's no strict rule about when it can be used: as long as it provides a description close enough (for your purposes) to reality, it's a good model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/409172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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