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Neutrino flavor and mass eigenstates Neutrions are produced and detected as flavor eigenstates $\nu_{\alpha}$ with $\alpha=e, \mu, \tau$. These states have no fixed mass, but are the combinations of three mass eigenstates $\nu_{k}$ with $k=1, 2, 3$, with mass $m_1$, $m_2$ and $m_3$, respectively. My questions are:
a) do neutrinos travel from source to the detector as flavor eigenstates or mass eigenstates?
b) is it possible to know which mass eigenstate the neutrino is in?
| (a) They start as a flavor eigenstate, which is a super position of mass eigenstates. The mass eigenstates have different time evolution, hence the state is, in general, a mixed state in either basis.
(b) No. As an analogy, consider polarized photons and Faraday rotation--it may start out + polarized, rotate to a mixture of + & - (with coefficients a & b) and then at your + detector you see it $a^2$ fraction of the time, and $b^2$ you don't. In either case, you can't say which state a particular photon was in.
(b') Can we detect a $\nu_e$ and know it's mass? Can it have the mass of a $\nu_{\tau}$? The $\nu_e$ doesn't have "a" mass, it has 3:
$|\nu_e\rangle=0.82|\nu_1\rangle+0.54|\nu_2\rangle-0.15|\nu_3\rangle$
while a tau-neutrino:
$|\nu_{\tau}\rangle=0.44|\nu_1\rangle-0.45|\nu_2\rangle-0.77|\nu_3\rangle$
So, "yes", if we measure it's mass, then it will have a mass that a tau neutrino mass measurement could yield.
In theory: it's not a sensible question to ask, since flavor eigenstates aren't mass eigenstates.
In practice: we do not know the masses of the mass eigenstates, and their differences are much less than an eV--so how are you going the measure that?
| {
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Feynman Diagram with antineutrinos and neutrinos I just recently learned how to draw feynman diagrams by looking at an equation such as one for $\beta-$ Decay:
$$n = p + e^- + \bar{v}_e$$
I was wondering in $\beta-$ why an electron-antineutrino was produced and not just an electron-neutrino? Similarly, why does $\beta+$ produce an electron-neutrino and not an electron-antineutrino.
$\beta+$ Decay:
$$p = n + e^+ + v_e$$
Feynman Diagram for $\beta-$ Decay:
Feynman Diagram for $\beta+$ Decay:
|
I was wondering in $\beta^{-}$ why an electron-antineutrino was produced and not just an electron-neutrino?
Because lepton number has to be conserved. In the LHS the lepton number is zero so on the RHS the lepton number must also be zero. All leptons have assigned a value of $+1$ and antileptons have $-1$. The conservation of lepton number is satisfied by the fact that $\beta^{-}$ decay produces an antineutrino instead of a neutrino.
Similarly, why does $\beta^{+}$ produce an electron-neutrino and not an electron-antineutrino.
$\beta^{+}$ produces a positron and a neutrino. The explanation is the same as the one above. Positron has a lepton number equal to $-1$ (because its an antilepton) and neutrino has $+1$, so that the total lepton number on the RHS is equal to the total lepton number on the LHS.
The second Feynman diagram is wrong. $W^{+}$ should decay into a positron-neutrino pair:
$$W^{+}\rightarrow e^{+}\nu_{e}$$
| {
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Wrong sign in Conformal Casimir The quadratic conformal Casimir in $d$-dimensional Euclidean space is given by
\begin{equation}
C = \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 -\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right)
\end{equation}
as given for example in the beginning of lecture 6 here http://pirsa.org/C14038.
Since there is an isomorphism between the conformal group and $SO(d+1,1)$ it should be possible to get this result by simply expanding $\frac{1}{2} M^{ab}M_{ab}$ with the identifications (DiFrancesco Eq. (4.20))
\begin{equation}
\begin{split}
M_{-1,0} &= D \\
M_{-1,\mu} &= \frac{1}{2} \left( P_\mu -K_\mu \right) \\
M_{0,\mu}\ &= \frac{1}{2} \left( P_\mu +K_\mu \right) \\
M_{\mu \nu}\ &= L_{\mu \nu}
\end{split}
\end{equation}
and $\eta_{ab}= \mathrm{diag}(-1,1,...1)$. However absolutely every time I attempt to do this calculation I get
\begin{equation}
C = \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 +\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right).
\end{equation}
There are many different sign conventions out there but I don't think that's the problem because my wrong Casimir really does not commute with the elements of the algebra.
I know it's not the most exciting calculation to do but I would eternally grateful to whoever can point out where the flaw lies.
| To do the computation, considering
$$\frac{1}{2}M^{ab}M_{ab}=\frac{1}{2}(M^{\mu\nu}M_{\mu\nu}+M^{\mu0}M_{\mu0}+M^{\mu,-1}M_{\mu,-1}+M^{0\nu}M_{0\nu}+M^{0,-1}M_{0,-1}+M^{-1,\nu}M_{-1,\nu}+M^{-1,0}M_{-1,0})=\frac{1}{2}(L^{\mu\nu}L_{\mu\nu}-\frac{1}{2}(P+K)^2+\frac{1}{2}(P-K)^2-2D^2)= \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 -\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right)
$$
where it's crucial that the $-1$ signature in the metric is in the $0$ th direction.
| {
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Is it feasible/possible to use refraction for x-ray spectrum analysis? In x-ray spectroscopy Bragg reflection off of a crystal is used for spectral analysis. In x-ray diffraction the same principle is used for monochromatizing the x-ray beam from an x-ray tube. For visible light, however, an optical prism is used to decompose the spectrum. I understand that Bragg scattering would not be feasible for visible light as the wavelength range of visible light (390-700nm vs. 0.01-10nm for x-ray) is much larger than atomic spacing in most crystals (a few angstroms), and the Bragg angle has to be very shallow(?).
Is it feasible to use a prism instead of Bragg reflection in x-ray spectral analysis (i.e., is there a material with low x-ray attenuation and suitable refraction index)? Is there any table of refraction indices for different materials at various wavelengths (alternatively, a table of relative permittivity and relative permeability)?
| There is http://henke.lbl.gov/optical_constants/
But no, prisms are not suitable for x-rays: dispersion is small, there will always be absorption. And there are anomalous effects near absorption edges.
| {
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What is going on inside this disc I recently saw a video of a copper disc repelling paper pins which I suppose are made up of steel. By watching the video carefully in slow motion, I feel there is something inside this copper disc (As the pins are getting rotated and thrown out from the disc's surface).
My questions are :
*
*What is going on here?
*How is this working?
*What could be inside this disc?
*As far as I know, there shouldn't be a ferromagnetic substance inside it, as paper pins are getting repelled.
I have uploaded the video on YouTube.
| Some comments suggested that the effect is caused by diamagnetism, other comments contained doubts, as diamagnetism is weak. Let me note, first, that copper is indeed diamagnetic (although it is not mentioned in the video that the disk was made of copper, unless I missed something) and that effects of diamagnetism can be quite impressive (https://www.bing.com/videos/search?q=copper+diamagnetic&&view=detail&mid=67B0211024968A13130367B0211024968A131303&&FORM=VDRVRV ). So the video in question could be made with diamagnetic copper and magnetized pins.
| {
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Does salt affect the boiling time of water? If I have 1 cup of water on the stove and another cup of water with a teaspoon of salt.
would the salt change the boiling time of the water?
| Since you are interested in time two factors have to be considered:
1) The increase in the boiling point temperature for the salt solution. This requires that more energy (aka heat) be transferred to the solution than the pure water to reach boiling. If this were the only consideration then indeed it would take more time to reach boiling point for the salt solution.
2) The change in conductivity and viscosity of the solution over pure water. To change the temperature of the fluids, heat is transferred from the pan (or beaker?) to the fluid and the presence of ions from the dissolved salt actually promote heat transfer within the fluid. This affect would tend to shorten the time. But adding salt also increases fluid viscosity. This means that convective cells within the fluid which also help distribute heat would be slower in the salt solution, so increasing time to boiling point.
All other factors being the same, it's difficult if not impossible to predict analytically using first principles which would take more time to boil. You can only accurately predict that the boiling point will be raised for the salt solution.
So the best way to make this determination is by experiment.
| {
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Force on plate of parallel plate capacitor with dielectric If we have a parallel plate capacitor whose charge is +Q and the polarization charge as Qp as shown in the figure..
then while finding the force acting on the left plate of the capacitor for instance, shouldn't the force due to the polarized charge -Qp and +Qp together be zero and therefore the only force acting be due to the right plate of the capacitor and hence the total force acting on the left plate of the capacitor be independent of the dielectric constant of the medium?
| The nett force on the Dielectric due to the involved Electrostatic interactions is zero. There is, however, a nett force on the dielectric on the plates because the forces from both the face of the dielectric are not equal in magnitude and don’t cancel out.
I have assumed the numerator of both F(-q) and F(+Q) to be equal in magnitude.
As you can see, the forces are unequal because the two surfaces of the dielectric are at two different distances away from the Plate of the capacitor. Therefore, there is a nett force due to the dielectric on the capacitor plates.
| {
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Rotation in Higher Dimensions In a world of three spatial dimensions plus time, every atom rotates around a line, the axis of rotation.
In a world of $N$ spatial dimensions where $N$ is greater than 3, must every atom rotate, and if so does it rotate around a line, a plane, or a subspace of smaller number of dimensions?
| *
*One may show that a general rotation $R\in SO(N)$ in $N\geq 2$ spatial dimensions can be composed
$$R ~=~ R_1\circ \ldots\circ R_{k} $$
of at most $k=[\frac{N}{2}] $ pairwise commuting rotations $$R_1,\ldots, R_{k}~\in~ SO(N)$$ that each leaves a co-dimension-2 subspace invariant (although not necessarily the same subspace).
*More explicitly, given a rotation $R\in SO(N)$ there exists an orthonormal basis $(e_1, \ldots, e_N)$ [which may depend on $R$] such that the rotation $R$ is represented by a block-diagonal matrix of the form
$$ \begin{pmatrix} \cos\theta_1 & \sin\theta_1 & \cr
-\sin\theta_1 & \cos\theta_1 & \cr
&& \cos\theta_2 & \sin\theta_2 & \cr
&&-\sin\theta_2 & \cos\theta_2 & \cr
&&&& \ddots \cr
&&&&&\cos\theta_k & \sin\theta_k & \cr
&&&&&-\sin\theta_k & \cos\theta_k & \cr
&&&&&&&1\cr
&&&&&&&&1\cr
&&&&&&&&& \ddots \cr
&&&&&&&&&&1\end{pmatrix}. $$
*The rotation $R$ itself is only guaranteed to leave invariant a dimension-1 subspace (=a line through the origin) if the space dimension $N$ is odd.
| {
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Increasing distance between Earth and Moon I have a problem where a planet's rate of rotation is decreasing due to tidal effects of the moon. I know that the angular momentum of the system will be conserved. So, in order to conserve that the moon will recede away from the planet.
$$ L = mvr = m\omega r^2.$$
I'm not sure how to convert/translate this loss in angular momentum of the planet to the rate of recession of the moon.
| Whether the Earth slows down or speeds up isn't something you can just decide from basic logic. It is a mathematical question whose answer depends on the relative phase relationship of the driving force (the moon) and he driven object (the tide). If the frequency of the driving force is less than the natural frequency of the tides, the tidal bulge will lag and the effect will be to speed up the earth and pull the moon down into a lower orbit. (The lower orbit is actually faster, but the angular momentum goes down because the radius changes more than the velocity.) If the frequency of the driving force is greater than the natural frequency of the tides, the bulge will lead the moon and the effect will be to slow the earth down and drive the moon into a higher orbit. (The higher orbit will actually be slower, but the anuglar momentum goes up because of the radius.)
It is an experimental fact that the driving frequency of the moon (one rotation every 24 hours) is greater than the natural frequency of the tidal bulge (circe the earth in 2 or 3 days, or a wave velocity of several hundred mph). So the earth slows down and the moon gets farther away.
I analyze this in a series of blog posts starting here: How the Tides Slow Down The Moon
| {
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Special relativity and spinning tires I apologize for the subsequent headache. There is a person who claims quite adamantly that Einstein is wrong, using the following reasoning:
A clock on a train slows down the faster the train moves. Mechanical clocks are valid clocks too. The wheels of the train are mechanical clocks. Do they slow down?
The train moves at $0.5c$. The diameter of its wheels is $1$ meter for simplicity. The top of the wheel moves at $0.5c$ relative to the train, the lower point at $-0.5c$. The sides move vertically. From an observer on the ground the lower point of the wheel is stationary, just about to move upwards. The upper part has velocity
$$\frac{0.5c + 0.5c}{1+0.25c^2/c^2} = \frac{c}{1.25} = 0.8c.$$
With that said, we can see that the top of the wheel rotates at 80% of the angular velocity required to keep the train moving at $0.5c$ - the wheel is deformed.
What this shows further is that the wheels as a whole don't slow down. They have to keep spinning. So time is not running any slower in this sense. It's only deformed.
Also, this shows that this thing that we call time is not a real thing - because the real thing would be a lower rate of events, as in a lower angular velocity of the wheel. Instead, it's just a mathematical variable we can play around with.
I know he is wrong but I'm not an expert in this and I would enjoy an answer.
| One important feature of special relativity is that special relativity does not know bodies of absolute rigidity, the universe is composed of particles which are held together by finite forces.
Example: Einstein's length contraction considers rigid objects for purposes of simplification only, for a better understanding of the phenomenon. But there are no rods of absolute rigidity.
By consequence, in your example the wheel will deform much earlier before reaching 0,8 c. In turn, the rules of special relativity will fully apply, but you must consider each individual particle: Particles which are describing circles will be subject to higher time dilation and length contraction then particles which are located on the axis.
| {
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Why does a DC voltmeter show a zero reading when an AC voltage is applied? One of my books says that if an AC voltage is applied across a DC voltmeter, its reading will be zero.
I think that since average value of AC voltage(in a complete cycle) is zero, DC voltmeter measures it as zero.
But I couldn't find a reliable explanation on the internet. Is my logic correct ?
Thank you !
| Think of it his way.
Assume you have a really, really fast DC voltmeter (and really fast eyes as well). In that case you will see the voltmeter go up and down, positive and negative, as the input AC voltage fluctuates.
In reality, a DC voltmeter is not that fast, so it will start to move up a very tiny bit, and then it is being forced to move down again. The end result is that the voltmeter will simply sit at zero.
If the AC voltage is not symmetrical around zero, then the DC voltmeter will still try to follow it, but instead of sitting at zero, it will show the average of the up and down swings of the input. For a symmetrical AC input, the average is zero.
| {
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Do higher frequency/energy levels in the EM spectrum mean higher temperatures? I am trying to find concrete evidence that for example, light in the optical spectrum would be hotter than infrared light because it has a higher frequency, and that is directly proportional to energy. Is energy directly proportional to temperature?
If we are to split up the optical spectrum into its components, blue light has a higher frequency than red light, and blue light is hotter than red light. Does this work for the whole spectrum?
| Molecules of matter at given temperature have satisfy below relation,
$\Delta E=NkT\tag1$
where $N$ is number of particles, in this case number of molecules and $T$ is temperature.
From planck's law, energy difference of higher frequency level is given by,
$\Delta E=Nhf\tag2$
where $N$ is number of particles, here it is number of photons and $f$ is frequency of photon.
From (1) & (2),
$f=\dfrac{k}{h}T=bT\tag3$
where $k$ is boltzmann's constant, $h$ is planck's constant and $b$ is equivalent to wein's constant and called as wein's constant in frequency in terms of two constants.
From (3), frequency is directly related to temperature this is wein's law if ratio of $f$ and $T$ is equal to $b$. Then that frequency, $f_m$ is frequency at which intensity of radiation is maximum and it is different for different temperatures.
This is theoretical calculation, if one find it's not true then it's anomaly of theory.
| {
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Mercury's precession I read in an article about Mercury's precession that Newton's law of gravitation predicts such precession of planets ;but fails to caluclate the precession of Mercury.But most of popular science books or other articles on the internet suggest that Newton predicts identical ellipses whereas the real orbital shape is like a rose petal. To conclude, Newton's also thought that the orbits should be like rose petal because of the perihelion advancement due to precession .Right ?
Edit 1
the article that i read
| Newton's law of gravitation predicts a perfect ellipse for the orbit of a planet orbiting a star, only in the idealized case that there a just those two bodies (the planet and the star). However, in reality there will be many bodies orbiting the star, each with its own gravitational field that slightly perturbs the orbits of the other bodies, causing the orbits to deviate from perfect ellipses and slightly precess.
| {
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Clarification about what makes a system isolated I am a grade 12 physics student and I just need some clarification about what makes a system isolated. I've read the definitions online, but they still don't make a lot of sense. For example:
When people jump on a rotating merry-go-round, the angular momentum decreases. However, according to my physics teacher, when people jump off from a spinning merry-go-round, conservation of momentum does not hold because it is not an isolated system. Why is one case isolated and the other isn't?
Also, if an object suspended to the ceiling by a cord gets hit by an object (ex. a bullet), why is the moment of impact the only time the system is isolated, and not when the object+bullet moves towards the ceiling?
| It all depends on the system under consideration (that's fancy text for the system you're thinking of). In the first part, you could argue that when people jump off, the momentum of the entire Earth (which is what you landed on) is conserved because it's an isolated system. Your physics teacher is assuming that you aren't treating the entire Earth as part of the problem. This is actually rather sensible, since the mass of the Earth is $10^{24}$kg, while people are only ~$100$ kg. The momentum imparted to the Earth is way negligible then!
I don't understand the second part very well. Naively, the object + bullet should be isolated, simply because there's nothing else in the problem.
| {
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Physical meaning of gauge choice in electromagnetism In electromagnetism, it is often referred to gauges of the electromagnetic field, such as the radiation or Coulomb gauge. As far as I know, the definition of a gauge helps us to redefine the problem in terms of a vector potential and a scalar potential that, since we have some freedom in choosing them, can be chosen in cleverest way it is possible for the given problem.
Here comes my question: is the choice of the gauge a mere mathematical simplification of the given problem? Does this choice have a physical meaning?
My troubles are actually in understanding the physical meaning of this choice of the gauge and what will change if I choose a different gauge.
| In classical physics, and also quantum gauge field theory with an abelian gauge group (like QED), the choice of gauge has no physical significance whatsoever. It's basically just like choosing where to place the origin of your coordinate system. In nonabelian gauge quantum field theory the situation is a bit more subtle, because large gauge transformations take you between physically distinct states. But this is a rather technical detail, and for the most part you can safely think of gauge choices as completely physically irrelevant.
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Quantum State Representation with Commuting Operators Let $[A,B]=0$. Then, we can find a set of eigenvectors $\{|a_n,b_n\rangle\}$ common to both $A$ and $B$. According to this, and my own understanding, it makes sense to write an arbitrary quantum state as
$$\tag{1}|\Psi\rangle=\sum_n \sum_i c_n^i |a_n,b_n,i\rangle,$$
where the sum over $n$ goes over all the eigenvectors, and the sum over $i$ allows for degeneracy to exist.
To me, it seems like we're saying $|a_n,b_n\rangle$ is a single eigenvector common to $A$ and $B$, that could have very well be written as $|w_n\rangle$. This also makes sense.
Yet, Cohen's quantum mechanics text writes
$$\tag{2}|\Psi\rangle=\sum_n \sum_p \sum_i c_{n,p,i}\ |a_n,b_p,i\rangle.$$ This has greatly confused me as it seems like we are dealing with two different sets of eigenvectors, one for $A$ and one for $B$. This representation (at least to me) says for each $n$, we are going over all $p$ eigenvectors and account for their degeneracy. Whereas the representation in Eq. (1) says to simply go over the eigenvectors $|a_n,b_n\rangle$ and account for their degeneracy.
Any help in trying to understand where I'm going wrong is appreciated.
| In the first equation, n is the index of the pairs ${a_n,b_n}$ of simultaneous eigenvalues, and $i\geqslant 1$. In the Cohen's book equation, he has two indexes, n for $A$ eigenvalues and p for $B$ eigenvalues.
If we have a system with four states $|a>$, $|b>$, $|c>$, $|d>$, that have the property that:
$A|a>=|a>$, $A|b>=|b>$, $A|c>=3|c>$, $A|d>=3|d>$
$B|a>=0$, $B|b>=2|b>$, $B|c>=4|c>$, $B|d>=4|d>$
In first notation you mention, we will have $n=1,2,3$, and the states will be called $|{a_1,b_1,1}>=|a>$, $|{a_2,b_2,1}>=|b>$, $|{a_3,b_3,1}>=|c>$, $|{a_3,b_3,2}>=|d>$, with $a_1=1, a_2=1, a_3=3, b_1=0, b_2=2, b_3=4$.
In the second notation, we will have $n=1,2$, $p=1,2,3$, and the states will be called $|{a_1,b_1,1}>=|a>$, $|{a_1,b_2,1}>=|b>$, $|{a_2,b_3,1}>=|c>$, $|{a_2,b_3,1}>=|d>$.
I think that that's the idea...
Was this useful?
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Will current flow if there's no return path? Here is the problem I was trying to solve:
Find the potential difference between the points A and D
I used Kirchhoff's voltage law for the left loop and right loop and found out the current through the left loop to be $\frac{10}{2+3}$ A (2A) and for the right loop $\frac{20}{4+6}$ A (2A), both flowing clockwise. But this does not take into account the current between B and C (The connecting wire)? By book says current will not flow through BC and they proceeded to find the potential difference by adding\subtracting the potential drops along the way while taking current through that wire as 0.
One explanation was that it's because there's no return path for the current. But even during Earthing, there's no return path, yet charges flow for a short while.
My question:
Why does current not flow through the BC path? If there exists a potential difference between B and C of 4V, charges should flow, right? Shouldn't all the current eventually pass only theough the loop at a lower potential?
Edit: What about a case like this?
Will current flow now?
| If there were a constant current between B and C, then the left and the right part of the circuit would charge indefinitely with the charges of opposite signs, the potential difference between the left and right parts would increase and eventually the current between B and C will stop.
| {
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Trajectory of a rolling ball with uneven weight distribution A perfect ball is rolling on a plane. Without further forces, it would roll in a straight line, and that's it.
What, however, if the ball's weight distribution is uneven? For example, the ball might have a higher-density smaller ball placed within it, but being slightly off-center, so the center of gravity is not the geometric centroid of the ball. Note that I assume the ball to be solid, i.e., no moving parts within the ball.
How would this ball roll then? Which trajectories would be realizable by such a ball?
Sadly, I don't know enough mechanics to even write down the differential equations and simulate it on a computer...
| Well, as the center of mass is not coinciding with the normal force(vertically upwards), there will be a net torque about the point during the motion of the ball. When the heavier ball is towards the front, it will support rolling and increase the rotational speed of the ball. When it will go to back, it will give the opposite effect. As net energy is conserved, the ball will slip(considering no friction) if it started with pure rolling. You can write the equations as follow, just consider the motion of the center of the heavier ball. Let it's distance from the center of the main ball be r. Now, x will be equal to $r\cos \theta $,where theta is the angle from the horizontal axis passing through the center of the main ball. This will give you the torque as (M-m)gx where m is the mass of the equivalent part of the other side. This will give you $\tau=I\dot{w}$. Now, solve for the w where w is $d\theta/dt$. As I am not considering friction, the velocity in the horizontal direction will remain same. You should work out the case with the friction yourself.
| {
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How do electron microscopes not get obstructed by atoms in the air? How do different electron microscopes avoid just scanning the atoms between the probe and the surface of the object that is actually being scanned?
| Electron microscopes, Transmission Electron Microscope (TEM) and Scanning Electron Microscope (SEM), are both operated under high ($10^{-6}-10^{-8}$ Torr) or ultra high vacuum ($<10^{-9}$ Torr) conditions. There are two reasons for this:
*
*to keep the specimen clean
*to avoid scattering of electrons by residual gas
So the answer to the question
How do different electron microscopes avoid just scanning the atoms between the probe and the surface of the object
is that there is essentially no atoms between the probe (electron gun) and the surface of the object.
| {
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Transition amplitude of n-vacuua in QCD The vacuum state of QCD is a superposition of different ground states with non-tirvial topological charge $\propto n\in \mathbb{Z}$. We lable this vacuum configurations with $|{n}\rangle$. The transition amplitude is given by:
$$ \langle n|m\rangle = \int \mathcal{DA}_{n-m}\ \ \text{exp} (-iS) $$
where $\mathcal{A}_{n-m}$ denots the gauge field configuration to the topological charge $n-m$.
Maybe I missed something but I don't understand why this transition amplitude depends only on the difference $n-m$? Shouldn't one calculate the the vacuum-to-vacuum transition amplitude by the generating functional (including all configurations)?
| The path integral starts from a configuration $A^{(m)}$ of a topological charge $m$ and ends in a configuration $A^{(n)}$ of topological charge $n$.
The reason that it depends only on the difference is that both the QCD Lagrangian and the path integral measure are invariant under large gauge transformation, i.e., gauge transformation with nonvanishing winding number:
$$Q = \frac{1}{24 \pi^2} \int \mathrm{Tr}\left((g^{-1}dg)^3\right)$$
If we change the integration variables by performing a gauge transformation
$$ g\cdot A = g^{-1}Ag + g^{-1} dg$$
The Lagrangian will not change, while the boundary configurations will become of topological charges $m+Q$ and $n+Q$, respectively.
Since a transformation of the integration variables does not change the result. The final value should depend only on the difference $m-n$.
| {
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Frames of reference and inhomogeneity and anisotropy of space I was reading mechanics by Landau and Lifshitz where I encountered this statement, "If we were to choose an arbitrary frame of reference, space would be inhomogeneous and anisotropic." I tried to think of random frames but couldn't come up with something which could help me understand the above statement. Can someone explain the gist of the statement made above, please?
| Consider a noninertial frame (one accelerating in an arbitrary direction for instance). In this frame the expansion of space would appear different in different directions and would change with time.
| {
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Boltzmann's equation and Liouville's theorem in curved spacetime I have two related questions:
How Boltzmann equation can be written in a covariant (using differential forms, connections, and etc.) way in a classical (not quantum) but curved system?
How does the Liouville's theorem change in a curved background?
Any answers, suggestions, clarifications, or references are welcome.
| Liouville's Theorem is pretty easy because it's the same in curved spacetime as in flat spacetime. MTW gives "Liouville's theorem in curved spacetime" as
The volume $\mathscr{V}$ occupied by a given swarm of $N$ particles is independent of location along the world line of the swarm.
See section 22.6 for the discussion. Basically, the only things that we might consider changed are how we measure volume and how the world line evolves.
The Boltzmann equation is more complicated. For example, if we just have a relativistic gas of particles with the same mass (and we choose units where that mass is 1 unit), and the gas at some point has momentum $p^\alpha$, then Boltzmann's equation for the distribution function $f$ looks like
\begin{equation}
p^\alpha \frac{\partial f}{\partial x^\alpha} - \Gamma^\gamma_{\alpha \beta} p^\alpha p^\beta \frac{\partial f}{\partial p^\gamma} = Q(f, f),
\end{equation}
where $Q$ is a complicated collision operator and $\Gamma$ are Christoffel's symbols. An original reference for this is Bichteler's paper, but a more modern discussion can be found here, for example.
| {
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How does an electromagnetic wave move? Somewhere I found the explanation that the EM fields create and destroy each other during the oscillation (I suppose by Faraday's law) and this makes the wave "move".
I can't imagine this because unitary vectors in E,B and k directions are a right hand ordered set of vectors and by the fact of E and B are in phase.
So, this is my question, what is the reason because the wave moves?
| An electromagnetic wave is a change in the electromagnetic field of an object. The change in its electromagnetic field can only propagate at the speed of light. If the sun disappeared, it would take 499 seconds for the change in the gravitational field to propagate to earth and earth to go off at a trajectory; at the same time, earth would go dark due to the change in the electromagnetic field also reaching it at the same time. The further away from the source of the electromagnetic field, the weaker the field gets, so eventually the electromagnetic wave becomes too weak to detect. When an electromagnetic field is not changing, it does not induce a current in an antenna because the antenna is a capacitor, and current only flows when the voltage (charge of the capacitor) is changing, which is why voltage leads current by 90° – it is an AC circuit.
The change in the electromagnetic field strength clearly propagates as a wave outwards at the speed of light when you think about it. The field itself is made of photons.
| {
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Extreme life - energy source for living tens of kilometers underground? Living cells were found up to at least 12 miles underground (article), and in other extreme places (BBC survey article), for which beside the problem of just surviving in such extreme conditions, a basic physics thermodynamical question is: what energy source it is based on?
And in such extreme temperatures there is needed a lot of energy just to fight 2nd law of thermodynamics - actively protect cell's structures against thermalization.
Such energy source needs to be relatively stable for past billions of years - what seems to exclude chemical energy sources (?)
One stable energy source in such high temperatures are thermal IR photons, and thermophotovoltaics is generally able to harvest energy from them. However, cell living in such extreme conditions would rather have the same temperature, hence 2nd law seem to forbid harvesting energy from such IR photons?
| Chemical.
As the Wikipedia entry on Lithoautotroph puts it (restricting ourselves to the deep underground forms):
derives energy from reduced compounds of mineral origin
which they do through inorganic oxidation (see, e.g., Lessons from the Genome of a Lithoautotroph: Making Biomass from Almost Nothing) or other reactions, such as the reaction of formate (HCOO-) and water, to form bicarbonate and hydrogen (Extremophile microbes survive only on energy from formate oxidation).
| {
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Why is the normal force $(M+m)g$? I am trying to understand the solution to this problem. The problem asks to find F such that m stays fixed relative to M. In the solution, it is mentioned that the normal force for block M is (M+m)g, I don't understand that. I thought it is supposed to be only Mg.
The solution states - The normal force on the first block, M is Mg + u_2*F_bb = (M+m)g.
Normal Def from wiki - is that component of the contact force that is perpendicular to the surface that an object contacts.
Since block M is in contact with 2 surfaces, is that why they are adding the the Mg+ u_2*F_bb?
I think I am just confused about the definition of Normal Force and it's application in this problem.
| The normal force is the force the table (or surface) must exert on the block $M$ in order to keep it stationary in the vertical $y$ direction. This means that the normal force must be equal and opposite to the net downward force that is being applied on M. The net downward force being applied on $M$ in this question is composed of two parts. One part is the weight of $M$ which is given by $F_w=Mg$, the other is the force that the little block $m$ exerts on $M$ in the downward direction. By Newton's third law, this force must equal the force that $M$ exerts on $m$ in the upward direction. This force, as was argued earlier in the text, must have magnitude $mg$ in order for little $m$ to remain stationary in the vertical direction. Thus, the total force that is pulling down on $M$ is $F=F_w+mg=(M+m)g$. The normal force must equal this in order to counteract it so that $M$ does not move in the vertical direction.
| {
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How do you solve the Schrödinger equation with a position space delta function potential in momentum space? I am solving the Schrodinger equation in position space with an attractive delta function potential energy,
$$
-\frac{h^2}{2m} \frac{d^2}{dx^2} \psi(x)-\lambda \delta(x) \psi(x)=E \psi(x),
$$
for a bound state. I need to solve for E, using fourier transform. I took the fourier transform of the whole equation and ended up with,
$$
-\frac{h^2}{2m} (ik)^2\phi(k)-\lambda \delta(k) \phi(k)=E\phi(k).
$$
After using,
$$
\delta(k) =1 \text{ or } 2\pi
$$
I don't get the required energy for this attractive potential, I would be glad if anyone could help? As I remember the energy expression should be
$$
E=-\frac{m\lambda^2}{2h^2}.
$$
| Start from the position space SE with potential $-\lambda \delta(x)$. Define the Fourier transform as:
$$\tilde \psi(k)=\frac{1}{\sqrt{2\pi\hbar}} \int dx \,\,e^{-ikx/\hbar}\psi(x)$$
The Fourier transform of the product is given by:
$$\mathcal{F}[\delta(x)\psi(x)](k)=\frac{1}{\sqrt{2\pi\hbar}} \int dx \,\,e^{-ikx/\hbar}\delta(x)\psi(x)=\frac{\psi(0)}{\sqrt{2\pi\hbar}}$$
The momentum space SE therefore gives:
$$\tilde \psi(k)= \frac{\lambda\psi(0)}{\sqrt{2\pi\hbar}}\frac{1}{(k^2/2m-E)}$$
Inverting the transform:
$$\psi(x)= \frac{\lambda\psi(0)}{2\pi\hbar}\int dk \,\frac{e^{ikx/\hbar}}{k^2/2m-E}=\frac{m\lambda\psi(0)}{\hbar\sqrt{-2mE}}\exp{\left(-\frac{\sqrt{-2mE}}{\hbar}|x|\right)}$$
where the result of the integral in the second step is standard and can be obtained from tables of Fourier transforms. Setting $x=0$, we obtain $E$ in terms of $\lambda$:
$$E=-\frac{m\lambda^2}{2\hbar^2}$$
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Nuclear Fusion: Why is spherical magnetic confinement not used instead of tokamaks in nuclear fusion? In nuclear fusion, the goal is to create and sustain (usually with magnetic fields) a high-temperature and high-pressure environment enough to output more energy than put in.
Tokamaks (donut shape) have been the topology of choice for many years. However, it is very difficult to keep the plasma confined within the walls because of its high surface area (especially in the inner rings).
Why hasn't anyone used spherical magnetic confinement instead (to mimic a star's topology due to gravity)? - Apart from General Fusion
E.g. injecting Hydrogen into a magnetically confined spherical space and letting out the fused energy once a critical stage has been reached?
| Background: confining a plasma requires controlling all of an enormous spectrum of possible instabilities. The tokamak does a good job of stirring the flows so that no individual instability can grow so much that the plasma rushes out and contacts the wall (and thus is quenched). Regarding the particular question of a spherical tokamak: the answer just edited above points out that having a field everywhere tangent to a spherical surface is forbidden by the hairy ball theorem.
Magnetic confinement takes advantage of the fact that a charged particle's path is bent in a (strong) magnetic field; particle trajectories tend to be tight spirals about magnetic field lines owing to the V cross B force (Lorentz force): so particles cannot travel across the field lines unless they are scattered out of their spiraling motion. effectively allowing them to jump to a different line of force. For a spherical tokamak we would require an arrangement of field lines which was everywhere strong and tangent to the surface of the sphere. Such an arrangement would contradict the Hairy Ball theorem (mentioned elsewhere in this topic).
| {
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Why do fundamental particles have a specific size? If Quantum Field Theory is accurate, all particles are actually just excitations of the field in which the particle interacts.
Therefore, wouldn't it be possible to have particles of any conceivable size, provided the energy, couldn't you have a photon the size of a building? Or one unimaginably smaller than the accepted size of a photon?
Am I missing something, or is this one of those unanswered questions that linger in physics?
If there are hard limits on particle size, why those sizes, what makes them meaningful?
| Does a single photon have a size?
It depends on what you mean by size. When you look at a basketball and you think of its size, you are looking at the entire space that the entity exists in. But you could cut up pieces of that basketball, you would see that each piece of the basketball takes up space on its own. A photon sort of like a pixel of light. It can't be cut-up into smaller units so it can't simultaneously take up space in the same way a basketball can take up space.
But quantum mechanically it can sort of exist in multiple spaces at the same time. A photon can exist in a superposition of a number of different locations at the same time! And in fact, a single-photon pulse is a pulse of probability of finding one of these pixels of light. You could refer to the size of that pulse of probability as the photon's "size," but then you would learn that one, can in fact shrink that size to be as small as possible!
So if you think of the size of a single photon as the size of the probability pulse of that single photon, then such size has no limits. A photon can be arbitrarily large or small!
| {
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Quantum states in position and momentum phase space While studying introduction to statistical mechanics ,I came across a new idea phase space where we use both position and momentum coordinates to denote a system .In my book the author calculates the number of quantum states within the energy range between E to E+dE.But after a while he writes that the number of quantum states in both momentum and energy states are equal but isn't energy E=p^2/2m and so the results are not equal as there will be a constant multiplied. Why the number of quantum states under these circumstances be equal is unclear to me...
Any help will be appreciated
| The number of states with an energy between $E$ and $E+dE$ is: $$dN_E=D(E)dE$$ where $D(E)$ is the density of energy states. The number of available states in a system is one, and this number is the same no matter what you consider, whether it's energy or momentum. And it's from here that you impose that the number of energy states is equal to the number of momentum states. $$dN_E=D(E)dE=dN_p=D(p)dp$$ where $D(p)$ and $dN_p$ are respectively the density and number of momentum states. So since $dN_E=dN_p=dN$ you would have $$D(p)=D(E)\frac{dE}{dp}$$ and since $E=\frac{p^2}{2m}$ so $\frac{dE}{dp}=\frac{p}{m}$ you finally get $$D(p)=D(E)\frac{p}{m}$$ So yes, it's the density of energy and momentum states that is different and is multiplied by a constant as you said, but the number of states $dN$ is the same for momentum and energy.
Note: the $g(E)$ and $g(p)$ that you intend are just the integral of my $dN$ over the range of energy/momentum considered.
| {
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In semiconductors when electrons jump from VB to CB, do they leave behind their parent atom's nuclei? In semiconductors/conductors when electrons jump from VB to CB, do they leave behind their parent atom's nuclei?
If yes, when this happens in Si (electron jump from VB to CB) why don't they ionize Si to Si +?
If no, why do they ionize doped As atom when the unbonded free electron from VB goes to CB?
| When an electron with its negative elementary charge moves from the VB to the CB it leaves behind a positive elementary charge, a hole. The positive charge of the hole is, of course, related to the positive Si+ ions in the crystal lattice. But the hole is not localized at a specific Si atom. Its wave function is spread out over the crystal. Thus it can be found near any Si atom of the lattice just like an electron in the CB.
The case is different for an As atom occupying a lattice point. If its electron moves to the conduction band, it leaves behind a positive As ion fixed at that location.
| {
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Inverse Square relationship using paint problem confusion I want to ask a question about the inverse square relationship using an aerosol paint spray mentioned in my book.
I am reading the book Advanced Physics by Steve Adams, and it mentions this in the book.
Imagine you are holding an aerosol paint spray at $50$cm from a wall. By squirting it for one second, you make a circle of radius $10$cm.
Now, I am aware that I can find the area of the circle as follows:
$$A_{10} = \pi r^2 = \pi \times 10^2 = 100 \pi$$
The book next talks about increasing the distance from a wall:
Now imagine you move along the wall and stand twice as far from the wall - 100cm. You squirt for the same length of time.
Because he is standing twice as far away from the wall, the radius of the patch is doubled to 20cm.
This therefore means that the area of the circle is now:
$$ A_{20} = \pi r^2 = \pi \times20^2 = 400 \pi$$
Now, I have studied the inverse-law relationship ($I \propto \frac{1}{r^2}$ previously in regards to a light source, and I wanted to understand this concept more easily using this paint example.
However, I cannot understand why standing twice as far away from the wall, the radius of the patch is doubled to 20cm.
Can someone explain why this is the case please?
| It is square inverse intensity even if you can adjust the spread of spray like most real life paint sprayers to spray wider or narrower, even if the nuzzle has been damaged and sprays a wiggly circle like an amoeba with one or two spots even being sprayed outside of the main circle or holes left inside.
Let's assume that that this amoeba is measored horizontally 20 cm for a distance of 50 cm of sprayer from the wall. As long as we disregard the fact that spray shooting out of paint gun will go down gently in a parabolic path, we can assume the amoeba sprayed by this paint gun will be 40 cm wide at 100 cm distance.
Because the area of similar shapes is proportional to square of one of their sides, length, or width, the second amoeba will be four times bigger and hence will receive 1/4 paint per square cm.
As to the proof of why the area of similar shapes is proportional to square of their sides or length, let's assume we use tiles small enough to cover the suface of the first amoeba within acceptable precision and we call these unit tiles, the second amoeba will be covered by the same number and configuration of tiles except they are now each four times bigger than the original. Hence it's surface 4 times bigger.
| {
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Why does gravity need a graviton? Einstein theorized that gravity is a phenomena manifested by the curvature of spacetime, in effect it IS the curvature of spacetime. If this is so, why do we need a graviton to convey the force of gravity? If I have mis-understood Einstein then I would appreciate a little help in grasping the relationship between warped space and gravity.
| the (mostly) non-mathematical answer is that any time we have a field in physics, there will be defined for that field an associated quantum, which can be considered an excitation of that field. when that field is responsible for transmitting forces between objects, that process can be modeled as the exchange of those quanta between those objects. For the electromagnetic field, the associated field quantum is the photon. for the gravitational field, the associated field quantum is the graviton.
the mathematics of the process by which a certain field is quantized defines the characteristics of the quantum of that field. Even before we go out and try to catch one of those quanta as it propagates through the universe, we know in advance what its properties are.
in the case of the electromagnetic field, the quantum of the field must be massless and possess a spin number of one. for the gravitational field, it too must be massless and have a spin number of two.
in studying how gravity works, we can visualize the process of gravitational forces acting between two massive objects as occurring because mass bends spacetime and thereby alters the trajectories of those objects, or equivalently by visualizing it involving instead the exchange of gravitons between those objects.
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Solving the Lie algebra of generators: path from algebra to matrix representation Given the Lie algebra, what is the systematic way to construct the matrix representation of the generators of the desired dimension? I ask this question here because it is the physicists for whom representation of groups is more important than mathematicians.
Let us, for example, take $SU(2)$ for concreteness. Starting from the generic parametrization of a $3\times 3$ unitary matrix $U$ with $\det U=1$, and using the formula of generators $$J^i=-i\Big(\frac{\partial U}{\partial \theta_i}\Big)_{\{\theta_i=0\}}$$ one can find the $3\times 3$ matrix representation of the generators.
However, I'm looking for something else.
*
*Given the Lie algebra $[J^i, J^j]=i\epsilon^{ijk}J^k$, is there a way that one can explicitly construct (not by guess or trial) the $3\times 3$ representations of $\{J^i\}$?
*Will the same procedure apply to solve other Lie algebras appearing in physics such as that of $SO(3,1)$ (or $SL(2,\mathbb{C})$)?
| You got outstanding answers, but they are uncharacteristically abstract for physicists; they assume your students are comfortable with basic Lie algebra theory, as taught to mathematicians, by example, in their first week of such courses. Unfortunately, physicists, who are especially used to being taught by example even more, are not getting such in their first week of remedial group theory courses, as a rule.
So here is a "brass tacks" almost trivial example of a dimension 3 Lie algebra, albeit not familiar to most physicists, Winternitz's "lucky guess" algebra, (chosen to forestall "monkey-see-monkey-do", as virtually ritual in, e.g., rotations):
$$
[ X,Y]=Y, \qquad [ X,Z]=Z+Y, \qquad[Y,Z]=0. \qquad
$$
Its adjoint representation is the linear map of the 3-vectors of generators
$$ xX+yY+zZ= \begin{pmatrix}
x \\
y \\
z
\end{pmatrix},
$$
to another 3-vector in that space, so, effectively, a collection of 3×3 matrices.
By the definition $\operatorname{ad}(J) ~K\equiv [J,K]$, you simply compute the action
$ \operatorname{ad}(X) Y= Y $ , $ \operatorname{ad}(X) Z=Z+ Y $, and null acting on X, naturally. So it is just represented by the matrix
$$ X_a= \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{pmatrix}.
$$
Likewise, $ \operatorname{ad}(Y) X= -Y $, and null action on the rest, so
$$ Y_a= \begin{pmatrix}
0 & 0 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},
$$
and also $ \operatorname{ad}(Z) X= -Z-Y $, null on others, so
$$ Z_a = \begin{pmatrix}
0 & 0 & 0 \\
-1 & 0 & 0 \\
-1 & 0 & 0
\end{pmatrix}.
$$
By the shrewd definition of the adjoint action, the Jacobi identity, these three matrices are guaranteed to satisfy the posited Lie algebra.
The algebra has the distinction of not having been identified in physics, yet.
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Why does Griffiths's book say that there can be no surface current since this would require an infinite electric field for an incident wave? In sec. 9.4.2 Griffiths shows the well known boundary conditions for E and B fields, one of them is this:
$$\frac{1}{\mu_{1}}\textbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\textbf{B}_{2}^{\parallel}=\textbf{K}_{f}\times\hat{\textbf{n}}$$
Where $\textbf{K}_{f}$ is the free surface current. Griffiths says in this section:
"... For Ohmic conductors ($\textbf{J}_{f}=\sigma\textbf{E}$) there can be no surface current, since this would require an infinite electric field at the boundary."
I just can't understand it yet. Why is this true?
I have another question:
The boundary for the normal component of E is
$$\epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}=\sigma_{f}$$
Where $\sigma_{f}$ is the free surface charge.
When we treat with incident EM waves on a conductor, is it necessary to consider $\textbf{K}_{f}$ and $\sigma_{f}$ different to zero?
I am asking this because in this section of the book Griffiths made $\textbf{K}_{f}=0$ and $\sigma_{f}$ vanishes naturally because he only studies normal incidence, but my question goes to the most general case in which the normal component of E is nonzero.
| While Griffiths typically writes surface currents exclusively as $\mathbf{K}_f$, one can also write them as volume currents as $\mathbf{J}_f = \delta(s) \mathbf{K}_f$, where $s = 0$ corresponds to the surface being considered. For an ohmic material, this means an electric field $\mathbf{E} = \frac{1}{\sigma} \delta(s) \mathbf{K}_f$, which blows up at the surface because the volume current blows up at the surface due to the presence of a Dirac delta at the surface.
Notice Griffiths is not forbidding you from having surface currents in general. That is certainly possible. He's just forbidding you from trying to put a surface current on an ohmic material. If there is a surface current, it can't be ohmic.
As for the second question, I see no reason for why that would be necessary, but it is convenient to assume the material is overall neutral. If there were surface charges, you could simply sum their effects later, since Electromagnetism is a linear theory.
| {
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Unitary Transformation of Eigenstates Suppose I have two operators, $A$ and $B$, with eigenstates $A \lvert a \rangle = a \lvert a \rangle$ and $B \lvert b \rangle = b \lvert b \rangle$, where $a$ and $b$ are all unique. Furthermore, suppose that $A$ and $B$ are related by a unitary transformation $$A = U B U^{-1}.$$ This is equivalent to saying that the eigenstates are related as $$\lvert a \rangle = U \lvert b \rangle.$$
Then it seems I can prove the following: since $$A \lvert a \rangle = a \lvert a \rangle,$$ I also have $$A U \lvert b \rangle = U B U^\dagger U \lvert b \rangle$$ by inserting the identity, so that $$A U \lvert b \rangle = U B \lvert b \rangle = b U \lvert b \rangle = b \lvert a \rangle.$$ Thus, $a = b$.
Doesn't this imply then that the eigenvalues for corresponding eigenstates of $A$ and $B$ are equal, and therefore-- by the assumption that they are unique-- that the unitary transformation doesn't actually do anything?
| I’m not sure what you mean by “$a$ and $b$ are unique” but clearly if $A=UBU^\dagger$ and $U$ is unitary, $A$ and $B$ have the same eigenvalues but it doesn’t mean $U$ doesn’t do anything.
For instance, the Pauli matrices $\sigma_{x,y,z}$ all have the same eigenvalues, are related by a unitary transformation $U$, but are certainly different. The transformation $U$ is a change of basis, so if $B$ is initially diagonal, say
$$
B=\sigma_z=\left(\begin{array}{cc} 1&0 \\ 0&-1\end{array}\right)
$$
and $U=\left(\begin{array}{cc} \cos\theta/2&-\sin\theta/2\\ \sin\theta/2 &\cos\theta/2\end{array}\right)$ then
$$
U\sigma_zU^{\dagger}=\cos\theta\sigma_z+\sin\theta \sigma_x
$$
still have eigenvalues $\pm 1$ but obviously $U$ has done something.
Of course the eigenstates are no longer $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\end{array}\right)$.
| {
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Experiment on friction coefficient Here you can see the results of the experiment about a friction coefficient:
The mean of the friction coefficient becomes 0.262 but when I do a linear regression in the form of y=mx the slope is 0.31. Shouldn't it be the same? I used $F_N$ as x values and $F_D$ (friction force) as y values.
regression: https://www.desmos.com/calculator/njj4utvsdk
| You have a $F_D$ measurement issue because from the numbers it appears the linear regression gives you a negative y-offset. This means you are missing some force that is not measured.
Go with the linear regression slope, removing this error as any constant value in the measurements is taken out. This would be the 0.386 value according to Excel.
I suggest looking for ways to remove any other sources of error, like stickiness or friction in the areas that you are not interested in. Also, add the trailing zeros to $F_D$ because it is misleading they way presented here. A value of 0.3 can be really anything between 0.25 and 0.35, but if you specify the value of 0.3000, the range of possible values is 0.2995 to 0.3005 which is much more well defined than the values posted.
| {
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Validity of Boltzmanns Equation and $H$-function theorem? A while ago I came across a resource (which I have forgotten) on the validity of Boltzmann's equation. It talked about the fact that the Boltzmann's equation is valid at the extrema of the $H$-function. In the discussion there was a graph that looked similar to the following
but better drawn (clearly). With the dots indicating some of the locations where the Boltzmann equation holds.
This may not be the exact theory, but it is along the right lines. Does anyone know if it has a name and what the theory actually states? (Even better if you can name the actual source I was looking at.)
| Very late answer, but it may be Kerson Huang "Statistical Mechanics" fig. 4.7 pag. 89 (in second edition).
The dots are to evidence the local maxima in time, related to states of molecular chaos of the gas. The theory states that a system with a distribution function $f$ that satisfies Boltzmann's transport equation tends to diminish its $H(f)$ with time (ideally going towards Maxwell-Bolzmann distribution in which $H$ is a minimum and $\text{d}H/\text{d}t=0$): at the same time, the theory behind is intrinsecally statistical so you should expect some fluctuations of your quantity, that in fact are present. That's mainly all I know of the subject, hope this helps someone.
| {
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Interval Preserving in Minkowski Space The squared line element in any spacetime is given as $$ds^{2}=g_{ab}dx^{a}dx^{b}.$$ The use of tensors helps us to infer that the line element in some other frame would be $$ds'^{2}=g'_{ab}dx'^{a}dx'^{b}$$ where simply $dx'^{a}=\frac{\partial x'^{a}}{\partial x^{b}} dx'^{b}$.
My question is, in special relativity, there is further a condition on the line element that it should $$c^{2}(s-t)^{2}-(x_{1}-y_{1})^{2}-(x_{2}-y_{2})^{2}-(x_{3}-y_{3})^{2}=c^{2}(s'-t')^{2}-(x'_{1}-y'_{1})^{2}-(x'_{2}-y'_{2})^{2}-(x'_{3}-y'_{3})^{2}$$ which gives us the Lorentz transformations. How can we prove this condition using the postulates of special relativity?
Also where and how do we employ the condition that the frames we are transforming to are inertial?
| The reason you've been unable to find a derivation of the Lorentz transformation (relating, say, Bob's frame to Alice's) from the usual two postulates of special relativity is that the Lorentz transformation does not follow from those postulates. You're going to need some additional assumption.
You can, for example, add some assumption about homogenenity/isotropy (though it's not enough, of course, to just mumble the words "homogeneity" and "isotropy"; you need some careful statement of exactly what you're assuming), or you can start by assuming that Bob's measurement of Alice's velocity is constant over time. Or you can just assume that the transformation must be linear.
There might then be room for someone to argue that your new assumption is really just a consequence of the old assumptions (in particular that the laws of nature "look the same" to both Bob and Alice). Since the old assumption is pretty vague to begin with (never specifying exactly what counts as a law of nature), it's possible to defend this argument, but it's at least a stretch.
| {
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$i\varepsilon$ in momentum space propagator; is it actually needed? In (say) phi-4 theory the momentum space propagator is given by:
$$\frac{i}{p^2-m^2+i\varepsilon}$$
where I am using the signature $(+---)$. Now momentum space we can do momentum space integrals using the Schwinger Paramterization etc in which we do not take any contour integrals etc and as far as I could tell
$$\frac{i}{p^2-m^2-i\varepsilon}$$
would give exactly the same results.
My question is therefore when is it valid to take $\varepsilon \rightarrow 0$ in momentum space and why would we keep it?
| Yes, it is necessary. See, the real space propagator isn't actually analytic, and the poles in the momentum space propagator will make of your integrals diverge unless you move them off of the integration axis (equivalently, deform the contour around the poles) and take a limit that moves them back onto the integration axis. Because the real space propagator isn't completely analytic, the way you move the poles off of the integration axis in momentum space will affect which type of propagator you're using.
To see how this happens, take the Fourier transform version of the real space propagator
\begin{align}G(x,y) &= \int \frac{\operatorname{d}^4 p \operatorname{d}^4p'}{(2\pi)^4} \left(\frac{\delta^4(p-p')}{p^\mu p_\mu-m^2}\right) \mathrm{e}^{-ip^\mu (x_\mu-y_\mu)}\\
&= \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \int_{-\infty}^\infty \operatorname{d}p^0 \left(\frac{1}{p^0p^0 - p^ip^i-m^2}\right) \mathrm{e}^{-ip^\mu (x_\mu-y_\mu)}.
\end{align}
Notice how the $p^0$ integration has two order $1$ poles in it at $p^0 = \pm \sqrt{p^ip^i + m^2}$. This makes the $p^0$ integral, as written, ill defined because it basically involves subtracting two infinite quantities. Thus, the integral only becomes meaningful if we have some prescription for how the limits approach the poles (e.g. Cauchy principle value) or we move the poles off of the integration axis.
With the Feynman $i\epsilon$ prescription, one pole moves up and the other down, leading the residue theorem to give
$$G(x,y) = \ldots + \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \left(\frac{2\cos\left(\left[x^0-y^0\right]\sqrt{\mathbf{p}\cdot\mathbf{p} + m^2}\right)}{\sqrt{ \mathbf{p}\cdot\mathbf{p} + m^2}}\right) \mathrm{e}^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}.$$
If you use an $i\epsilon$ prescription that moves both poles in the same direction, e.g. $p^0 \rightarrow p^0+i\epsilon$, then you get something like
$$G(x,y) = \ldots + \int \frac{\operatorname{d}^3 p }{(2\pi)^4} \left(\frac{2i\sin\left(\left[x^0-y^0\right]\sqrt{\mathbf{p}\cdot\mathbf{p} + m^2}\right)}{\sqrt{ \mathbf{p}\cdot\mathbf{p} + m^2}}\right) \mathrm{e}^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \Theta\left(x^0-y^0\right).$$
Note that I've not been very careful, so there may be some sign errors or incorrect leading numerical factors. I've also hidden the expected light cone delta functions inside of the "$\ldots$" because I do not yet have the background in distribution theory needed to explain how they come from the integral in question.
Bottom line - while the $i\epsilon$ prescription doesn't make a difference to momentum space integrals, it defines which version of the propagator you're using in real space so it should be kept around until that transition in case you, or subsequent users of your work, want to make that transition at some point.
See also the Wikipedia propagator article.
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Does radiation cause a change in temperature? If yes, then is there a limit to the temperature decrease? If no, then can the body which radiates heat attain an absolute zero temperature?
| Yes, radiation can cause a change in temperature. It's a form of heat transfer, after all. Radiative heat transfer can cause an object to warm up or cool off.
The Earth's temperature hasn't changed all that much over the last several million years. (Global warming and ice ages represent smallish temperature changes.) This means the Earth is more or less in balance, with just about as much energy coming in from the Sun as is going out via thermal radiation, and all of that heat transfer (both directions) is radiative.
There is a lower limit for large objects, and that's the 2.725 kelvins cosmic microwave background radiation. A large object in empty space will eventually come into equilibrium with that radiation, and that equilibrium point will be 2.725 kelvins (currently). Small objects (sub-millimeter) have a harder time interacting with microwaves, so they can cool below the CMBR temperature.
| {
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Would "gravity" and the "law of gravity" have a meaning in a universe without matter? I was discussing the fact that if there was no matter in the universe, just vacuum and radiation, can we say that anything called gravity wouldn't exist?
In that universe, the Friedman equations would still be useful, but is it related with gravity? It just describes the expansion and the geometry of the universe, but it is developed from general relativity and has the $G$ constant in it. So, is gravity a valid thing in that universe?
| The Friedman Equations would still exist. In other words, $G_{\mu\nu}=8\pi G T_{\mu\nu}$ would still apply, but the assumption of vacuum means that $T_{\mu\nu}$ would be 0 (we can immediately see that the gravitational constant G falls out due to multiplication by zero). If I claim gravity is the curvature of spacetime, the $G_{\mu\nu}$ on the left hand side would still be around. (Also I'm assuming that the cosmological constant $\Lambda$ remains on the left hand side of the EFE since it originates in the gravitational action.)
Also, you said "just vacuum and radiation". Radiation is a source of gravity with equation of state $p=\frac{1}{3}\rho$. A universe with radiation in it is not truly a vacuum (e.g. $T_{\mu\nu}\neq 0$ in this case).
| {
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Physical significance of the zeroth component of 4-velocity and 4-force Is there any physical significance of the zeroth component of the four velocity vector and four force vector? I understand that the space part of u$^\mu$ is related to ordinary velocity and space part of F$^\mu$ is the usual force. But are there any physical quantity related to the zeroth component of u$^\mu$ and F$^\mu$?
The zeroth component of four momenta, p$^\mu$ is energy. So, similarly are there any physical significance to u$^0$ and F$^0$component?
| The zeroth component of the 4-velocity $u^a=(\gamma ,\gamma \vec v/c)c$ is essentially the time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$ (multiplied by $c$ for dimensional purposes). Using the rapidity $\theta$ (the Minkowski angle between two timelike vectors), that zeroth component is essentially $\cosh\theta$.
In practice, [in geometric units] the 4-velocity is a unit-timelike vector.
If drawn on a spacetime diagram, the tip would represent "one tick" of that object's clock. Thus, the time-component of the unit 4-velocity would be the apparent duration of that object's tick, namely the time-dilation factor multiplied by "one tick".
The zeroth component of the 4-force is the relativistic power.
| {
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Density Functional Theory for Quantum Field Theory vs fixed-particle-number Quantum Mechanics Introductions to Density Functional Theory (DFT) usually discuss the Hohenberg–Kohn theorems which prove that there exist universal functionals of density that can be used to determine ground state properties of a system. This has extensions for degenerate ground states, or incorporating spin, or magnetic interactions, or even time dependent systems and excited states.
But all these seem to arrive at the proof by discussing a system in terms of a fixed number of particles, and the wavefunction in terms of the particle positions. This means DFT has its basis in non-relativistic fixed-particle-number quantum mechanics.
Is it possible for density functional theory to also be applied to a quantum field theory, such as QED or QCD?
| I suspect this isn't quite what you're looking for, but it's too long to share in a comment:
Relativistic corrections can be added with augmentation methods. Usually though, relativistic corrections arise in the core of (typically heavy) atoms and may therefore be incorporated with small adjustments to the pseudopotentials in canonical DFT. This is discussed in Martin's Electronic Structure book.
As for a non-constant number of particles (electrons usually), there do exist methods for non-canonical ensemble sampling. See, for example, Mermin's approach to finite-temperature DFT.
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How to measure a static electric field? I looked up google but didn't find any design for measuring electric field that doesn't vary with time.
My own idea is to use two parallel plates (like a capacitor but without the dielectric). In an electric field E a potential difference V = Ed (d is separation between the plates) will develop, which can be measured using a voltmeter. Will this work?
| An old method to measure an electric field does, ideed, use two thin metal plates of area $A$ held on insulating handles. These metal plates are put in contact and the combined plates are inserted into the electric field so that the surfaces are normal to the field lines. Then a (positive and negative charge) of $Q=A\sigma$ will be induced on the upper and lower plate where $\sigma=\epsilon E$ is the induced charge per unit area on the upper and lower plate, $\epsilon$ is the total permittivity of the surrounding medium. One can then separate the two plates and measure the total charge $Q$ on one of them with a Faraday cup electrometer. The electric field strength is the given by $$E=\frac{Q}{\epsilon A}$$
Measuring the static voltage on a capacitor with a voltmeter is, in general, not a good idea because you will get a transient discharge current through the voltmeter.
| {
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Intrinsic Concurrent Pitch/Yaw/Roll Rotation Between Two Rotation Matrices Given an object with a rotation matrix, how do you calculate the pitch, yaw, and roll velocities that needs to be applied over time for the object to reach a goal rotation matrix given that:
*
*The x-Axis is left, the y-Axis is up, and the z-axis is forward
*Pitch is rotation about the x-Axis, yaw the y-Axis, and roll the z-Axis
*The rotation is intrinsic
*The required velocity is reached instantly
*The pitch, yaw, and roll velocities all need to be applied at the same time (not one after another)
*We do not need to worry about center of gravity or translation, just direction
| You can find the rotation axis $\vec{z}$ and angle $\theta$ between the two orientations and use this information to apply $\vec{\omega} = \frac{\theta}{\Delta t} \vec{z}$.
Given two 3×3 rotation matrices $\mathrm{R}_1$ and $\mathrm{R}_2$ the relative rotation matrix is
$$ \mathrm{R} = \mathrm{R}_1^\top \mathrm{R}_2 $$
Then you can use any number of available methods to convert into Euler angles and/or rotation axis-angle.
Finally you apply either the three Euler angles in sequence, or the single axis-angle rotation in one step.
| {
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What does Mobius group/transformations have to do with special relativity? The group of Mobius transformations, denoted by ${\rm Mob}(2,\mathbb{C})$, is isomorphic to ${\rm SL}(2,\mathbb{C}))/\mathbb{Z}_2$ which in turn is isomorphic to the Lorentz group ${\rm SO}^+(3,1)$.
This connection, to me, seems very intriguing. After all, Mobius transformation is the most general, one-to-one, conformal map of the Riemann sphere to itself, given by $$w=f(z)=\frac{az+b}{cz+d}\tag{1}$$ where $a,b,c,d$ are arbitrary complex constants satisfying $(ad-bc)=1$. Apparently, (1) has nothing to do with spacetime transformations.
But the aforementioned isomorphism makes me curious whether there is any deep physical consequence(s) related to this isomorphism.
| You'll find an intuitive way into your question via the Wiki article on Möbius.
"In physics, the identity component of the Lorentz group acts on the celestial sphere in the same way that the Möbius group acts on the Riemann sphere. In fact, these two groups are isomorphic. An observer who accelerates to relativistic velocities will see the pattern of constellations as seen near the Earth continuously transform according to infinitesimal Möbius transformations. This observation is often taken as the starting point of twistor theory."
See also the late section on Applications, which discusses the isomorphism of the Möbius group with the Lorentz group, SO+(1,3) and SL(2,C). Also the classification table near the end, which I find suggestive.
Hope this helps.
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Is this "Permanent magnet gun" real or fake? There are several videos on youtube describing linear accelerator built solely from permanent magnets, put like this:
I find it hard to believe that this can work because if it did, it would be exploitable to gain energy. Where would that energy come from then?
I think it's faked and that the magnet should find equilibrium in the middle of the contraption. Am I right or wrong?
I'm willing to test it if I don't get conclusive answer, but I thought I'd as first before wasting money on bunch of magnets.
| In the video I see that the author needs quite some force to load the gun. So here he puts the projectile in a high potential state, i.e. he provides the energy during the loading process. When released this energy accelerates the projectile. Hence, no violation of energy conservation.
Edit
Here a small representation with python and ten dipole magnets:
Top: field lines in top view. Bottom: x-Field (arb.u.) on projectile path.
If you came from the left you have to somewhat overcome the negative bump. You see that the last max is almost as high as the first min is deep. That's why lubricant is important, as metnioned by @xcoderx. Actually, the thing looks better if you make it shorter. On longer ones the max approaches a saturation value and you'll have more friction loss.
Extra
Just looked at the video again because the author inserts the magnet as [NS] and [SN]. The behavior corresponds to the lower graph. In one case one has to overcome the first bump by pushing, but the total distance is not very much as the magnet is later attracted by the second bump. In the second case, the reversed magnet is attracted by the first bump, the author then pushes it over the second bump, which the naturally results in a larger distance.
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Strouhal number motivation I am looking for a nice way to motivate the Strouhal number definition. Let me illustrate what I mean on the Reynolds number. (As ususal, $\mathbf{u}$, $p$, $\rho$, $\nu$ denote the flow velocity, pressure, density and kinematic viscosity respectively.)
Sure, there are multiple good ways to show the importance and properties of the Reynolds number. I particularly like the one based on the momentum equation (the Navier-Stokes equation) scaling. The equation reads:
$$
\frac{\partial \mathbf{u}}{\partial t} + \left( \mathbf{u} \cdot \nabla \right)\mathbf{u} = -\frac{1}{\rho}\nabla p + \nu \nabla^2 \mathbf{u}
$$
then by introducing $X_i = x_i/L$, $U_i = u_i/V$, $P = p/(\rho V^2)$, $\tau = \nu t/L^2$ we obtain:
$$
\frac{\partial U_i}{\partial \tau} + \text{Re}\left( U_k \frac{\partial U_i}{\partial X_k} + \frac{\partial P}{\partial X_i} \right) = \frac{\partial U_k}{\partial X_k \partial X_k}
$$
i.e. the Reynolds number $\text{Re} = \frac{UL}{V}$ appears naturally as the only control parameter of the scaled system.
And now, is there a way to obtain the Strouhal number by a similar procedure?
Notes and notions:
*
*The particular beauty of the aforementioned procedure is that "it is autonomous". I presume that for the Strouhal number there should be an assumption such as "let the flow instability be described by a time-harmonic function".
*It must be based on Euler equations rather then Navier-Stokes equations.
*Would recasting the momentum equation in Crocco's form be of any help?
| In your definition of the dimensionless time you have assumed that the characteristic scale for time is $\frac{L^2}{\nu}$. If you instead assume that the characteristic scale is the inverse of the vortex-shedding frequency $f^{-1}$ and redo the analysis you will retrieve the Strouhal number. You will need to rescale the characteristic scale for the pressure accordingly.
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What does covariance/non-covariance mean in QFT? I'm studying QFT using the book of Mandl and Shaw. In the first chapter they start by quantising the electromagnetic field, but in a "non-covariant" way. What do they mean by that?
They have a chapter about the covariant theory of photons (chapter 5). They say using the Coulomb gauge (as in chapter 1) results in a transverse an longitudinal split of the polarisation vectors, which is frame dependent and thus non-covariant. (?) But then later they use the same vectors again and say "this is the transverse part and that the longitudinal part" but now it suddenly is covariant. What do they mean by this?
| Covariance means Lorentz invariance in explicit form. For example, you may work with a specific coordinate system, and derive expressions in terms of these coordinates, but Lorentz invariance will no longer be obvious.
On the other hand, when all your formulae have is dot products ($p \cdot q$), derivatives ($\partial_\mu V^\mu$), etc - one can immediately tell that they are Lorentz-invariant. It is said that the formulae are manifestly Lorentz-invariant or just covariant in this case.
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Approximate Killing vector field in general relativity In this paper the authors consider an approximate Killing field $\chi$. It vanishes on a given 2 surface and its first order part is given.They say that if it obeys the Killing equation $\chi_{a;b}+\chi_{b;a} = 0 $ then its second order part vanishes.
Do you understand why?
| Zhen Lin shows here that
if a vector field $\chi$ obeys the Killing' equation $\chi^i_{;j} +\chi^j_{;i} = 0 $ then $\chi^a_{;bc} = R^a_{;bcd} \chi^d $everywhere.
As $\chi$ vanishes at p, we have at this point $\chi^a_{;bc} = 0. $
The second order part of the taylor serie being quadratic in the covariant derivatives, it equals zero.
Raf Guedens who is a co-author of the initial paper gives many details in (https://arxiv.org/abs/1201.0542)
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Why do Newton's laws have to be used only when working with a particle? I have a small understanding of physics but I am not studying the subject.
Whilst trying to model a plane landing in Differential equations (an A-level maths module), we were told that you have to assume that the plane is a particle to be able to apple newtons laws to it, is this the case? If so, why?
| Because any bigger system is made up of small particles and if want to apply Newton's equations to the whole system then we have to apply them on every individual constituent particle, that would be very tedious and lengthy. That's why we use them only for particles.
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Difference between vorticity and circulation The definition of vorticity is $\boldsymbol{\omega} = \nabla \times \mathbf{v}$, where $\mathbf{v}$ is the velocity vector field.
Now, if I look at a rotating flow in cylindrical coordinates I find that:
$$\nabla \times \mathbf{v} = \frac{1}{r}\frac{\partial (r v_{\theta})}{\partial r},$$
in case of a free vortex I also know that $v_\theta \propto 1/r$ and therefore the derivative in the above equation vanishes for a vortex centered at $r=0$. In other words, the vorticity is zero everywhere, $\boldsymbol{\omega} = \mathbf{0}$.
I can also look at the situation globally, and instead of the localised curl I take a line integral of the speed along a circular path distance $r$ from the centre. In this case I find that:
$$C = \int_{\text{circular path}}{\mathbf{v}\cdot d\mathbf{l} = 2\pi ru_{\theta}},$$
a finite constant. But from the Stoke's theorem I know that:
$$\int_{\text{enclosed area}}{\left(\nabla \times \mathbf{v}\right)\cdot d \mathbf{A}} = \int_{\text{enclosing curve}}{\mathbf{v}\cdot d\mathbf{l}},$$
but if the circulation is a finite non-zero constant, the curl must be also non-zero somewhere within the enclosed area! Thus the vorticity is non-zero somewhere in the velocity vector field!
These two findings seemingly contradict each other, where am I making a mistake?
| There is a singularity at the origin: a delta function in the vorticity field. The vorticity is zero (irrotational flow) everywhere but at the origin, where it is infinite. The circulation around any path not enclosing the origin is zero. The circulation around any path enclosing the origin is a constant (non-zero).
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Gravity in vector We know that gravity is a force. But what is it's direction? Can it be expressed by vector and how can we do that? This question can also be asked for Coulomb's Law.
| If we use the centre of the earth as origin, we have
$$\mathbf{F}=-\frac{GM_{\oplus}m}{r^3}\mathbf{r} \tag{$r>R_{\oplus}$}$$
where $\mathbf{r}=(x,y,z)$ and $\displaystyle \left| \frac{\mathbf{r}}{r^3} \right|=\frac{1}{r^2}$.
At the surface of the earth,
$$g=\frac{GM_{\oplus}}{R_{\oplus}^2} \approx 9.8 \text{ m s}^{-2}$$
where $M_{\oplus}$ and $R_{\oplus}$ is the mass and radius of the earth respectively.
We assume the earth and the test mass have spherical symmetry in their densities.
The Columb's law version is
$$\mathbf{F}=\frac{Q_1 Q_2}{4\pi \epsilon_0 r^3}\mathbf{r}$$
assuming point charges or negligible electrostatic induction.
See another answer with electrostatic induction here.
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Energy density in string wave The total energy density in a harmonic wave on a stretched string is given by
$$\frac{1}{2}p A^2 \omega^2 sin^2(kx-\omega t).$$
We can see that this energy oscillates between a maximum and a minimum. So the energy is maximum at 0 displacement when the string is stretched and at its maximum speed (both KE and PE density are maximum at the same time) and minimum when the displacement is maximum as it is unstretched and doesnt have any velocity.
This makes sense but I am having trouble merging this with SHM oscillations. In SHM the KE and PE are not in phase. And if we consider each particle of the wave acting as a shm oscillator then would the PE not be maximum at the maximum displacement?
| PE and KE that we are talking here are of a small part of string. PE and KE are maximum when the element passes through its mean position as velocity is maximum and string part is most stretched. At crest (or trough) velocity is zero and string part is not stretched so both PE and KE are zero of that string part.
Now to relate it with SHM, let us take SHM of spring-block system. In the string, the string part is like a block and rest of the string is like the spring. PE that we talk in SHM is of the spring and not stored in the block, whereas in the wave on string we are talking of PE stored in the string part i.e. block. That is why the two PE we are talking are different. The PE stored in the rest of the string is behaving as PE stored in the spring. Hope it helps!
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Does the annihilation of antihydrogen with heavier matter resulting in conversion of heavier elements back to hydrogen? If an antihydrogen atom annihilate with a heavier atom of matter, will the remaining nucleus of the heavier atom be disassembled into individual protons and neutrons?
If so, is this considered to be a process which convert(regenerate) hydrogen from heavier elements?
Is it allowed by the laws of thermodynamics?
| An antihydrogen beam is a very recent achievement in particle physics
The ASACUSA experiment at CERN has succeeded for the first time in producing a beam of antihydrogen atoms. In a paper published today in Nature Communications (link is external), the ASACUSA collaboration reports the unambiguous detection of 80 antihydrogen atoms 2.7 metres downstream of their production, where the perturbing influence of the magnetic fields used initially to produce the antiatoms is small.
Such a beam hitting nuclei , the individual antihydrogen will just annihilate with one proton or neutron of a nucleus and a lot of energy will be released. There will be fission of the original nucleus into nuclear fragments because of the very high energy released (~1800 MeV ) in the annihilation with respect to the binding energies of nuclei ( order of ten MeV).
There will be protons flying around, which will finally trap an electron and become hydrogen, but it will be a hugely inefficient way of generating hydrogen.
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Abrikosov's Vortex Lattice (Beta Parameter) In order to find the correct vortex lattice configuration (i.e. ground state) in Ginzburg-Landau theory (or the Abelian Higgs Model), it is standard practice to minimize the beta parameter:
$\beta=\frac{\langle |\phi|^{4}\rangle}{\langle |\phi|^{2}\rangle}$.
What is the difference between minimizing $\beta$ and minimizing $\langle |\phi|^{4}\rangle$? Why are they not equivalent?
| The answer can be found in Abrikosov's paper: Sov. Phys. JETP 5 1174 (1957). The free energy for the vortex lattice is
$F=H_{0}^{2}-\frac{(H_{c}+H_{0})(H_{c}-H_{0})}{\beta(2\chi^{2}-1)}$, where $H_{0}$ is the external magnetic field, $H_{c}\equiv\chi$ is the critical field above which the vortex lattice is unstable and the ground state is the normal vacuum, $\beta_{A}$ is the Abrikosov parameter that determines the lattice structure and $\chi$ is the only free parameter in the theory. The minimum value $\beta$ can take is $1.1596$ for a hexagonal lattice. This assertion is valid for $\chi>\frac{1}{\sqrt{2}}$. Therefore, minimizing $\beta$ gives the correct ground state, it is not equivalent to minimizing $\langle |\phi|^{4}\rangle$.
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Work and mechanical energy I have come across the following lines in "Introduction to Mechanics" by Kleppner and Kolenkow.
A peculiar property of energy is that the value of mechanical energy $E$ is arbitrary; only changes in $E$ have physical significance. This comes about because the equation $$U_b - U_a = -\int_{a}^{b} \vec F \cdot \vec {dr}$$
defines only the difference in potential energy between $a$ and $b$ and not the potential energy itself. We could add an arbitrary constant to $U_b$ and the same constant to $U_a$ and still satisfy the defining equation. However, since $E = K +U$, adding an arbitrary constant to $U$ increases $E$ by the same amount.
What does he mean by "$E$ is arbitrary; only changes in $E$ have physical significance."? What is this physical significance he is talking about?
The concept of potential energy is still unknown to me. The author has introduced this form of energy by mathematically restating the equation of work done by a conservative force. Potential energy will be discussed a little later.
| In this case, the "physical significance" is referred to the evolution of the system: it will be defined only by the differences between certain values of $E$, then $E$ is arbitrary since whatever constant you add to $E$ (i.e. you change $E=K+U$ with $E'=K+U+c$) it will not be relevant since the constant will cancel when you evaluate the difference between two value of $E'$.
Think about a problem involving the gravitational potential energy $U=mgh$ of a body of mass $m$ suspended on an height $h$: whatever problem will you think about, the solution will ever depend on the difference between potentials evaluated on different height, and it will never depend on a single potential evaluated in a point. Then, if we redefine $U$ as $U'=mgh+c$ (where $c$ is an arbitrary constant) the physical situation of your problem will never change, since evaluating the difference between $U'$ in two different point, the constant $c$ will cancel and the result will be identical to the one you can obtain with the original potential.
In this sense, "no physical significance" means that the constant $c$ is not relevant at all in the evolution of your system. Being $E=K+U$, the reasoning done on $U$ can be extended on $E$, since whatever constant $c$ you add to $E$, it can be considered, for example, as a constant added to $U$.
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dependence of fundamental frequency of vibration of a stretched string on the medium in which it is kept suppose a stretched wire's fundamental frequency in air is 280 Hz. What would be it's fundamental frequency in water ?
(all other conditions of the string remain same)
I looked into the laws of vibrations of stretched strings, but all of them give information on characteristics of string, but nothing about the surrounding medium.
Please help.
The answer to the question is 243.2 Hz, but I am unable to calculate it myself. I read all the texts of fundamental modes and harmonics but found no way forward.
| the surrounding medium has a characteristic acoustic impedance which can be calculated. if that characteristic impedance is close to that of the vibrating string, then two things will occur: first, the string will be strongly damped and second, the mass of the surrounding medium will begin to couple to the mass of the string and the string will act as if its vibrating mass is increased relative to its tension. both of these effects will reduce the natural frequency of the string.
Per Sammy Gerbil's suggestion, I will enlarge upon my answer in this edit:
When a resonant system is coupled to a load that extracts power from it, the width of its resonant response peak is broadened and the location of that frequency peak shifts down to a slightly lower frequency. Immersion in water will extract power from the vibrating string and dissipate it by a variety of mechanisms and therefore its resonant frequency will certainly be reduced upon immersion. The first thing I would try is plugging the string's characteristics into the resonance equation and add progressive amounts of damping, to see how strong the damping effect is and whether or not it can account for the frequency shift.
I'm going off-line now to search for resources and will edit again if I find clues to share.
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Harnessing permanent magnetism? Putting aside any energy generating schemes that would break the laws of thermodynamics, is it possible or is there a motor which generates power using a permanent magnet? So that the energy wouldn’t be coming from nothing but from the atoms in the magnet being misaligned.
| In my opinion the answer is: energy is conserved, if kinetic energy is extacted using a permanent magnet, it will be at the expense of demagnitization of the magnet. Here is a link with some estimates of the amount of energy stored in a permanent magnet.
There is energy stored in a permanent magnet which slowly becomes demagnetized if it is used for moving masses.
A magnetic circuit-based approach to deriving stored energy provides an intuitive understanding of stored energy in permanent magnets. The resulting energy expression is also consistent with all granularities of analysis, from magnetic circuits to 3D finite elements calculations.
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Exercise on bosonic vacuum
Consider bosonic canonical transformation, generated by operator $S = e^{\lambda (a^{\dagger})^2}$. Show, that
\begin{equation}
b \equiv SaS^{-1} = a - 2\lambda a^{\dagger}.
\end{equation}
Calculate the norm of transformed vacuum $S|0>$ and show that the norm is finite only if $\lambda < 1/2$.
since $a^\dagger$ commutes with itself $S^{-1} = e^{-\lambda (a^{\dagger})^2}$ and the first question is easily answered. Let's show that $S|0>$ is indeed a vacuum:
$$
b=SaS^{-1} \Rightarrow bS = Sa \Rightarrow bS|0> = Sa|0>.
$$
Since $a|0>=0$ we've proved that S|0> is a vacuum state.
But after that I can't see where I'm wrong. Let's call $S|0> = |0_b>$:
$$
<0_b|0_b> = <0|S^{\dagger} S |0> = <0| e^{\lambda a^2} e^{\lambda (a^\dagger)^2} |0>
$$
I am assuming $\lambda$ is a real parameter. Let's rewrite the insides of the average over vacuum:
$$
e^{\lambda a^2} e^{\lambda (a^\dagger)^2} = e^{\lambda (a^\dagger)^2} e^{\lambda a^2} e^{\frac{\lambda^2}{2} [a^2,{a^\dagger}^2]}
$$
Since $<0|e^{\lambda (a^\dagger)^2} = <0|$ we can drop it. If my calculations are accurate (I hope so), the commutator is:
$$
[a^2,{a^\dagger}^2] = 2(2a^{\dagger} a + 1)
$$
and
$$
<0_b|0_b> = <0| e^{\lambda a^2} e^{\lambda^2 (2a^\dagger a + 1) }|0> = e^{\lambda^2}
$$
Which is true for every $\lambda$, but this should have been an example of transformation that don't save the "finiteness" of vacuum norm, so I must have mistaken somewhere.
| From $(a^\dagger)^n|0\rangle=\sqrt{n!}|n\rangle$ we have $e^{\lambda (a^\dagger)^2}|0\rangle =\sum_{n=0}^\infty\frac{\lambda^n\sqrt{(2n)!}}{n!}|2n\rangle$, which has norm $\sum_{n=0}^\infty\frac{\lambda^{2n}(2n)!}{n!^2}$. The Stirling approximation gives $\frac{(2n)!}{n!^2}\approx\frac{4^n}{\sqrt{n\pi}}$ for large $n$, so the convergence condition is $4\lambda^{2}< 1$. (The case $\lambda =\frac{1}{2}$ doesn't work because $\sum_n n^{-1/2}$ diverges.) In fact, the sum is $(1-4\lambda^2)^{-1/2}$.
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Gravitational field strength Can I use $g=GM/r^2$ to calculate the gravitational field strength proton or electron or any other particles? If not then why? If yes then what would be that really mean?
| You could use Newton's classic equation, but, as illustrated in an answer here, its effects would be almost negligible.
Aside from that, we don't know if Newtonian gravity even applies to particles on that scale. To answer that question would require a theory of quantum gravity which, to date, has not yet been developed.
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Original 1925 paper by Einstein on Bose-Einstein Condensation? Does anyone know if it is possible to retrieve the original 1925 paper by Einstein on Bose-Einstein Condensation? Possibly a translation into english, but german would be fine if no translation is available. I have managed to find a translation of Quantentheorie des einatomigen idealen Gases (Quantum Theory of a Monoatomic Ideal Gas) from 1924 where it is referenced as Sitzungsberichte der Preussischen Akademie der Wissenschaften, Berlin, Physikalisch-mathematische Klasse, 1925, p. 3–14, but couldn't find it anywhere.
| For those looking for the original german A Scanned version is provided by the university of Münster
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What prevents two objects from falling toward each other faster than the speed of light? I was thinking about what happens when two objects fall toward each other in space. The farther apart they are when they begin falling, the faster they will be traveling when they hit each other due to gravity's acceleration. So, if they are far enough apart, what prevents them from falling towards each other faster then the speed of light?
Can someone explain where in my thought process I am going wrong?
| No, this is not the case at all. And, in fact, I can give you a concrete answer for a specific case (and from this you can extrapolate almost everything there is to it):
If you place a space ship infinitely far away from planet Earth (in an otherwise empty universe) and wait infinitely long, then the space ship will crash into Earth at ~ 11 km/s. Not quite infinitely fast or very fast at all. It's the escape velocity of Earth on the surface of Earth.
As a remark: You are right that the impact velocity keeps increasing the further away you place the space ship at the "start" - but it's not increasing unboundedly - merely asymptotically to Earth's escape velocity.
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Applying Kinematics to find retarding force in a medium Question
If an object free falls let's say off a cliff that is 3 meters high, clearly it increases velocity and if at the bottom of the three meter there was a bucket full of jelly which created a retarding force in which the object stops 1 meters in and if the object is 5kg, why wouldn't the following work?
a) Find Velocity right before hitting jelly
b)Then find out deceleration value inside of jelly
c) do Acceleration*5kg=Force
Because if I find velocity right before it falls into the jelly- $V_{before}$, then use that in $V_f^2=V_{before}^2+(2*a*\Delta{x})$, wouldn't that find the deceleration in the jelly?
What my friend suggested-
So I asked this to my friend and he said that I should set the $mgh$ on top of cliff compared work and solve for the force. Why is that supposed to work but not my way?
| Both methods will work. It's not uncommon that there are more than one different approaches to a problem and you can choose whatever method seems most convenient.
For example consider the initial drop from rest at height $h$ down to the surface of the jelly. The PE change is $-mgh$ and since the total energy is conserved that means the kinetic energy must have increased by $+mgh$, so we get:
$$ \tfrac{1}{2}mv^2 = mgh $$
or:
$$ v^2 = 2gh $$
which is exactly the same as your kinematic formula (with $v_i = 0$). So both methods give the same velocity.
In the case of the passage through the jelly you could once again use either method, but given you know $v_i$, $v_f = 0$ and $s = 1$m I would simply use your kinematic formula. That seems the quickest and simplest approach to me.
As a footnote: you're assuming the deceleration in the jelly is constant and for a real jelly that wouldn't be the case. You'd probably get something like quadratic drag i.e. $a \propto v^2$.
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What determines the direction of current in a superconductor Type I superconductors have no electric field nor magnetic field inside of them, when they are in the superconducting state. This means no voltage difference across any two points or regions inside of them. Yet they carry a current. This means the Cooper pairs (or the electrons responsible for the current) are moving in a particular direction.
My question is, what determine this particular direction, if there's no voltage involved?
Edit: If we have to apply a voltage initially, to get the Cooper pairs moving in a particular direction and then we remove that voltage and the current will still persist, then it would mean that the superconductor has a sort of "memory" in that it's possible to retrieve where (and how strong?) the voltage was applied? Does it also mean that the superconductor behaves the same way with and without the applied voltage? If so, that would be very strange and I'd like some clarifications.
| There is no electric field in a superconductor, but there can be a voltage across it. Recall that:
$$ \vec E = -\nabla\phi-\frac{\partial\vec A}{\partial t} $$
so the voltage need not be zero for the electric field to be zero.
Any superconducting loop has some inductance, so this voltage is required to get a current going. Since it has no resistance, a voltage is not required to keep the current going: in fact, keeping a voltage applied will continuously increase the current, as:
$$ V=L\frac{dI}{dt}$$
(Eventually, if the voltage is not removed, the superconductor will reach a critical current. At this point, it becomes normally conductive. This can be bad, since an awful lot of heat is released all at once.)
Once the current is going, there's nothing to resist it, and so it can keep flowing even if the voltage is removed.
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Why does doping a Sodium Iodide scintillator with Thallium result in a higher ratio of Compton interactions to total interactions? We are looking at designing a Compton camera for 662KeV photons and have been told that "the fraction of Compton interactions to total interactions in the photopeak is usually higher than fraction in an un-doped Sodium Iodide crystal".
I expected the converse to be true - Thallium has a high Z and photoelectric absorption has a higher Z dependence than Compton interactions. I have been doing some reading and it appears that the Thallium acts as an activator in the Sodium iodide crystal; the benefits of the crystal being transparent to the scintillation radiation is obvious but I cannot see how this affects the interaction probabilities of Compton to photoelectric interactions.
Any help would be very much appreciated!
| The doping of the NaI crystal with thallium improves the scintillation efficiency by improving the light emission due to the improved recombination by light emission of electrons and holes at the dopant site. Thallium in small concentrations in the NaI crystal is a so-called scintillation activator The effect of thallium doping is not related to any direct interaction of thallium with the gamma-rays. Here you'll find an article which describes the effect of thallium doping in NaI scintillators for gamma-ray detection.
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Moment of Inertia of an Equilateral Triangular Plate I was reading about moment of inertia on Wikipedia and thought it was weird that it had common values for shapes like tetrehedron and cuboids but not triangular prisms or triangular plates, so I tried working it out myself. I will post my attempt below, but for some reason I cannot find any source online that confirms or denies my solution. Please let me know if you find anything wrong with it. Thanks.
Q: What is the moment of inertia of an equilateral triangular plate of uniform density $\rho$, mass $M$, side length $L$, rotating about an axis perpendicular to the triangle's plane and passing through its center?
*
*First I modeled an equilateral triangle using three lines with its center of geometry at the origin as follows: $x=\frac{1}{\sqrt{3}}y-\frac{1}{3}L \\
x=\frac{1}{3}L-\frac{1}{\sqrt{3}}y \\
y=-\frac{\sqrt{3}}{6}L$
I used the fact that the circumradius of an equilateral triangle is $\frac{\sqrt3}{3}L$ and that its height is $\frac{\sqrt{3}}{2}L$ .
*Next, using the definition of moment of inertia ($I$) and with the help of Wolfram Alpha, I obtained the following result:
$$I=\int r^2 dm=\rho \int r^2 dA\\
=\rho \int_{-\frac{\sqrt{3}}{6}L}^{\frac{\sqrt{3}}{3}L} \int_{\frac{1}{\sqrt{3}}y-\frac{1}{3}L}^{\frac{1}{3}L-\frac{1}{\sqrt{3}}y} x^2+y^2 dxdy\\
=\frac{\rho}{16 \sqrt{3}}L^4=(\frac{4M}{\sqrt{3} L^2})(\frac{L^4}{16\sqrt{3}})\\
=\frac{1}{12}ML^2$$
| I can confirm your result. I can also suggest you a neater way to derive it inspired by David Morin - Introduction to Classical Mechanics, check it out in a library if you have access.
The main idea is to use the symmetry of the equilateral triangle and split it into 4 smaller equilateral triangles like this
Now analyse how the moment of inertia changes when we rescale its mass and sidelength,
i.e. if $ i = \alpha ml^2$ and $L = 2l, M = 4m$, then $I = \alpha (4m) (2l)^2 = 16i$,
where $m$ is the mass of a small triangle, $l$ is the sidelength of a small triangle and the capital $M, L$ correspond to the larger triangle. But the moment of inertia of the big triangle can be also split into $4$ moments of inertia. Be aware that we need to use the parallel axis theorem for the $3$ triangles which enclose the central triangle. Hence, $$I = 16i = 4i + 3m\left(\frac{l\sqrt{3}}{3}\right)^2,$$
which reduces to $i = \frac{1}{12} m l^2.$
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Is there evidence that a = dv/dt and a = F/m are always equivalent? If the rate of change in velocity in a particle (of mass m) caused due to a force F is dv/dt, then
F = m dv/dt
It may be argued that this is how we define force. But my question is:
Can there be any kind of force, which is so strange that no matter how we write the formula for the force, we will find that F = m dv/dt fails in at least some cases?
| This formular holds only true for time-constant masses. The original formular is $$\vec{F} = \dot{\vec{p}}$$
which, taking the derivative, leads with to
$$\vec{F} = m\dot{v}+\dot{m}v$$
So $$F=m\dot{v}$$ is only true if the second term is zero, so if $\dot{m}$ is zero.
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Conservative force definition Classical Mechanics, by John Taylor defines a conservative force $F$ as a force that satisfies:
*
*$F$ depends only on the particle's position and no other variables.
*Work done by $F$ is the same for all paths taken between two points
I'm wondering if this definition is redundant. Doesn't (1) imply (2) and vice versa?
If not, what is an example of a force that satisfies (1) but not (2) and an example of a force that satisfies (2) but not (1)?
| The comment of @probably_someone shows clearly the necessity of (1). It eliminates a possible force dependence on time, velocity or on any other parameters.
(2) does not follow from (1): Consider the force on one pole of a long thin bar magnet which is next to a current carrying wire. The work done moving it in a circle around the wire is different to the work done in a loop which doesn't go around the wire. The same would be the location dependent force on an object moved in a water whirl.
(1) doesn't follow from (2): When a charged particle moves in a magnetic field no work is done on the particle on going on any path from A to B. The force experienced by the particle is dependent on the velocity not only the position (inhomogeneous B).
| {
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Why do we use superposition instead of tensor product in interferometer? In the description of a neutron interferometer here, it says:
In an interferometer the incident beam is split into two (or more) separate beams. The beams travel along different paths where they are exposed to different potentials (which results in different phases). At some point the beams are brought together again and allowed to interfere. The resulting beam is the superposition of the separated beams:
$$ \psi = \psi_I + \psi_{II} $$
I am interested in why the total wavefunction is not written as $\psi=\psi_I\otimes \psi_{II}$? Because when they are separated, they should be considered in two physical systems and we should use tensor product to describe them, even they are later combined together, right?
| In a nutshell: these are not two different systems, but the probability amplitudes of the two different states of the same system.
I do agree that simple discussions of two interfering waves (electromagnetic or particle waves) are in practice only complicating the matter, as opposed to thinking of a single wave in a multiply connected geometry. They however make the math manageable.
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Energy in simple harmonic motion ─ where is the kinetic energy stored, and where is the potential energy? When a mass connected to a spring is in simple harmonic motion and somewhere between the mean and extreme positions the mass is cut from spring. Then instantaneously after cutting the mass will only have its kinetic energy right? (Or it will have total energy kinetic+potential?)
Or I mean to say that in a system in simple harmonic motion, the kinetic energy is stored in the mass while the potential energy is always stored in spring. Am I correct?
| While the mass is not cut from the spring, there is transfer of energy. This is the transfer of potential energy into kinetic energy of the mass.
When you cut the spring, the block will proceed to move with the kinetic energy it had before. The potential energy in the spring will not disappear or somehow suddenly transfer to the mass but it will remain in the spring. Ideally the spring will keep oscillating which maintains that same energy. In practice the energy will slowly dissipate until it stops oscillating due to friction and other forces. When the spring is cut the potential energy doesn't go anywhere, it remains in the spring system (from which the mass escaped). There is no way for energy to transfer from the spring to the mass because the mass is no longer in contact.
| {
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Why do charged particles deflect one way but not the other in a magnetic field? I am well aware that a charged particle moving in a magnetic field will experience a force perpendicular to that magnetic field. But why is it that positive and negative particles experience a force in opposite directions?
What exactly determines the direction that a given charge will experience a force? I.e. why does a negative particle experience a force in one direction and not the other?
| I've been told I can't just link to another site so I will try to paraphrase the article I linked to.
The magnetic force is perpendicular to the velocity of the particle it is acting on. That causes the direction of the particle to change and travel in a circular motion.
https://cnx.org/contents/bZRPyVNP@2/Motion-of-a-Charged-Particle-i
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How to reconcile infinite cross section of resonances with cross section formula from quantum mechanics? If we consider $s$-wave scattering for two scalar fields $\phi$ and $\chi$ with an interaction $\frac{g}{2}\phi^2\chi$, then the Lorentz-invariant scattering amplitude to second order is:
$\mathcal{M}_{fi} = \frac{-ig^2}{s - m_\chi^2}$, where $s$ is the Mandelstam-s ($E_{cm}^2$).
The cross section, which goes as $\left| \mathcal{M}_{fi}^2 \right
|$, is clearly singular at the $\chi$ resonance. But I don't know how to reconcile that with the scattering formulat given in Sakurai 7.6.17:
The cross section by $\left| f(\theta)\right|^2$, where $f(\theta)$ is given by the partial wave decomposition:
$f(\theta) = \frac{1}{k}\sum_l (2l+1)e^{i\delta_l}\sin\delta_l P_l(\cos\theta)$
For $s$-channel scattering, only $l=0$ contributes, and I do not see how we can have a singularity in $f(\theta)$ anywhere, as I would expect for resonance scattering.
| The issue is that the $\mathcal{M}_{fi}$ in the question is only calculated to second order. When you include loop corrections and "dress" the propagator, it takes on a different structure and the width of the peak becomes finite.
| {
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Thermal energy generated by collision observed from two different frames of reference An isolated system is composed of two bodies $A$ and $B$, with masses $m_A$ and $m_B$, $m_A \ne m_B$, which are in route of collision. The relative velocity between them is $v$. The collision is inelastic and I want to calculate how much thermal energy is generated at the collision.
I understand that the thermal energy $\Delta T$ generated equals the variation of the kinetic energy, so that the total energy is conserved. I expected $\Delta T$ to be independent of the reference frame.
However, if I consider a reference frame $S_A$ fixed in body $A$, I get the following change in kinetic energy of the system:
$$ \Delta T_A = \left( \frac{m_A 0^2}{2} + \frac{m_B v^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_B v^2}{2}. $$
And if I consider a reference frame $S_B$ fixed in body $B$, I get a different change in the kinetic energy of the system:
$$ \Delta T_B = \left( \frac{m_A v^2}{2} + \frac{m_B 0^2}{2} \right) - \left( \frac{m_A 0^2}{2} + \frac{m_B 0^2}{2} \right) = \frac{m_A v^2}{2}. $$
Could you please point out the flaw in this reasoning?
| The problem is that you assume that the reference frames $S_A$ and $S_B$ stay the inertial frames before and after the collision even though their velocities change abruptly at the collison. It would be better to use the center of mass reference frame which does not change with the collision.
| {
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Understanding the introduction of a symbol in Einstein's paper In Einstein's paper "On the Electrodynamics of Moving bodies" (1905); first he introduces two sets of coordinates, for two inertial frame moving with relative velocity of $v$: ($x$, $y$, $z$, $t$) and ($\xi$, $\eta$, $\zeta$, $\tau$).
Then he introduces one more coordinate, $x'$, as $x' = x - vt$.
A possible explanation is: Because of Galilean transformation $x = \xi + vt$, so that $x - vt$ is constant wrt to time in the moving frame k for a particle stationary in k. But is it fine to assume the Galilean transformations in order to derive that Galilean transformations must be replaced by Lorentz transformations?
| It seems to be a Galilean transformation between two coordinate systems with relative velocity v in x direction.
The Galilean transformations form a part of the symmetry group of Newtonian mechanics.
But is it fine to assume the Galilean transformations in order to derive that Galilean transformations must be replaced by Lorentz transformations?
Because of, for example, the Michelson-Morley-Experiment, Einstein, like a few other Physicists of that time, came to the conclusion, that the Newtonian concept of absolute space and time isn't true and therefore that the Galilean Transformations had to be replaced by some other ones.
If you consider non-relativistic speeds, it is fine to assume the Galilean Transformations, since they proved to be the right thing when dealing with velocities $v<<c$.
But they don't work with the first postulate of Special Relativity, namely that the speed of light in the vacuum is the same for all oberservers (as it was indicated by the Michelson-Morley-Experiment). Einstein derived the Lorentz Transformations based on his two postulates of Special Relativity and the fact that they had to
*
*reduce to the Galilean Transformations for low speeds
*be linear because the Galilean Transformations are.
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Electrons diffusion in gas with present of electric field For research purposes, I am looking for a way to calculate how far the electrons "spread" perpendicularly to the electric field in a chamber of gas.
For example, if a beam of alpha particles ionize the gas in a chamber and leave a track of electron-postive ion pairs, is there a way to calculate how far the electrons move away from the point of ionization (perpendicular to the electric field). Note: also present in the gas chamber is an electric field, so the electrons and positive ions would move in opposite direction. I am reading W. R. Leo's Techniques for Nuclear and Particles Physics Experiments for hints on how to calculate this, but I haven't found any mention of this yet. The closest thing I found was about electron diffusion with the absent of an electric field. Please point me to articles, book chapters, equations, or links regarding information of this topic. Thanks in advance.
| I have found the answer in the W. R. Leo's Techniques for Nuclear and Particles Physics Experiments. The equations that I needed are Eq. 6.18
$D=\frac1 3 v \lambda $ and Eq. 6.46 $\sigma=\sqrt{2Dx/\mu E}$.
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Does a gas of neutrons obey the ideal gas equation much better than hydrogen gas Consider a gas of neutrons. Does it obey the ideal gas equation much better than hydrogen gas at the same temperature and pressure?
| It will not be easy to establish a pure neutron gas in thermal equilibrium to test the ideal gas law.
(1) Free neutrons have a life time of only 14.7 minutes. They decay (predominantly) into a proton, an electron, and an electron antineutrino. So you will always have positive protons and electrons and thus hydrogen around.
(2) It is not easy to contain neutrons in a vessel because due to the missing electric charge they will easily escape through any walls.
The neutron-neutron scattering cross section is probably very low.
Added later:
The above does not yet answer the question, as @JohnRennie has rightly pointed out. To come closer to an answer, I add the following reasoning based on the van der Waals equation which has two constants $a$ and $b$ which describe the deviation from the ideal gas law.
The answer depends on the question how the "molecular volume" and the "intermolecular force" of thermal neutrons compares to those of molecular hydrogen. These two effects are represented by the constants $a$ and $b$ in the van der Waals equation $$(p+a/V_m^2)(V_m-b)=R T$$ and they describe the deviation from the ideal gas law. Here $p$ is the pressure, $V_m$ the molar volume, $R$ the ideal gas constant, $T$ the absolute temperature. The constant $a$ is a measure of the inter-molecular force, the constant $b$ the actual volume of a mole of the gas molecules.
I would consider it a very rough estimate to assume that the constant $b$ describing the molecular volume of the neutrons should be much smaller (many orders of magnitude) than the corresponding molecular volume of hydrogen. Also, the forces between hydrogen molecules expressed by the constant $b$ are probably stronger than between neutrons. But the latter is only a wild guess. Thus a neutron gas is probably closer to an ideal gas than the hydrogen gas.
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Is there a SI unit for space-time? Space and time are routinely combined into space-time nowadays, which implies that the SI meter and second should be combined into a single SI unit such as [meter-second]. So far, I haven't come across such a SI unit.
|
So far, I haven't come across such a SI unit.
And you won't find one. The problem is that meters and seconds are inconsistent with one another, much as are the customary units of mass and force. With mass in pounds mass, force in pounds force, and acceleration in feet per second squared, one has to use the ungainly $F=\frac{196133}{6096}ma$ rather than the much more handy and consistent $F=ma$ used in the metric system.
In a consistent, relativistically correct system of units, the speed of light is identically one, and it is unitless. The speed of light in the metric system is 299792458 m/s. Think of that 299792458 as equivalent to the ungainly factor of 196133/6096 that one runs into in American customary units.
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Gravitational potential at the center of a uniform sphere Feynman in second Messenger Lecture said that potential at the center of ball with small radius $a$ is equal to average potential on surface of ball minus $G$ times mass inside the ball divided by $2a$. You can see it on the picture. I don't understand that. I had an idea that it is from Taylor series of potential at the centre, but I can't figure it out. Can you help me?
| Feynman's answer does not refer to a sphere filled with uniform mass density. It is a completely general statement about the potential at the center of a (small enough) sphere given an arbitrary mass distribution. It is a different way of stating Newton's law of gravitation. It can be derived from the differential formulation of Newton's law (ΔΦ = 4πG ρ) with the help of the full-space Green's function G(r,r')=-1/(4 π |r-r'|).
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Trouble in understanding AM modulation Amplitude modulation is in fact, superimposing the low frequency transmission signal into a high frequency carrier signal, right?
So, if the transmission signal can be represented as $c(t)=A_c \sin (\omega_ct )$ and the carrier wave can be represented as $c(t)=A_m \sin (\omega_mt)$, then upon superimposing and simplifying the equation using trigonometric identities we will have
$$
c_m(t)=A_c \sin (\omega_ct )+\frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t-\frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t,
$$
but this equation has the carrier wave and two sinusoidal waves with frequencies $\omega_c-\omega_m$ and $\omega_c+\omega_m$. So where is the signal which is transmitted? What we've got is a corrupted signal with uneven frequencies i.e. $\omega_c-\omega_m$ and $\omega_c+\omega_m$
Correct me if I'm wrong at any point.
| The low frequency transmission signal means slowly changing $A_c(t)$, not fixed $A_c$ with fixed $\omega$ which does not carry any information.
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Are the electromagnetic waves transverse? The em waves are said to be the oscillations o electric and magnetic field perpendicular to each other and to the direction of propagation of wave and hence transverse.
However consider a charged particle oscillating along x axis with no motion along y and z axis.
Let it be at O. Consider a point P where we are considering the electric field due to charge.
When the particle moves from O to A, the electric field at P increases. And when it goes from O to B, the electric field at P decreases by an equal amount. Since the charged particle is in SHM, the electric field at P will vary sinusoidally. However the varying electric field is also in direction of X axis. And the wave also propagates in this direction. So, it comes out to be longitudinal. However since magnetic field variation will be in perpendicular direction to electric field, it will also be perpendicular to direction of propagation of wave.
So the wave should be partially longitudinal and partially transverse.
| You are right in your observation of the electric and magnetic fields at a point P. This is, however, a consideration of the so-called near-field of an oscillating charge. The near field doesn't constitute a freely propagating electromagnetic field. To get the freely propagating (far field) electromagnetic field, you have to consider distances much larger than the wavelength corresponding to the oscillation ferquency. Then you will see that the propagating field are transverse EM waves.
Note added later: Irrespective of the distance, the near field (electric and magnetic) decays as $1/r^2$ and the far field decays as $1/r$, corresponding to propagating electromagnetic fields. Thus far enough from the source, the far field is dominating.
If you look at the electric field derived from the Lienard-Wiechert potential (See Wikipedia, https://en.wikipedia.org/wiki/Liénard–Wiechert_potential) of a point charge moving (accelerating) in x-direction, there is only a near-field component in x-direction at any distance and no far-field component. In particular, there is no transverse electric or magnetic field.
| {
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Why do the following Network Transformations give different answers? I did a Star to Delta Network Transformation and a Delta to Star Network Transformation on different parts of the original circuit as shown in the image below.
It gave me two new circuits. On solving those circuits, I get different answers for the equivalent resistance. Why is this so?
| Because you've made a mistake in your application of the formalism.
Both of the transformations you have performed are guaranteed, when done correctly, to produce completely equivalent circuits. The only thing that can break here is the 'done correctly' part ─ so double- and triple-check all your algebra.
Ultimately, if you want a higher standard which you can use to hold both of those answers to, you need to do a full Kirchhoff's-Laws analysis of the circuit, which will (when done correctly) give an unambiguous answer for the total resistance of the circuit.
As to where you've made a mistake, that's not within our purview; please read these guidelines as well as these ones for more information.
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Definition of stress-energy tensor
The image from the wiki article on the stress energy tensor gives $T_{00}$ as $1/c^2$ times the energy density. I believe this is incorrect and that the $1/c^2$ factor should be dropped. Am I missing something?
| You are correct and Wikipedia is wrong. Energy density is $Nm/m^3 = N/m^2$ in SI units. Pressure is also $N/m^2$. Dividing $T^{00}$ by $c^2$ makes no sense.
| {
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Are all waves either transverse or longitudinal? So I recently searched up "em wave transverse proof", and I understood it pretty well enough I think.
After that, I just started to wonder if all waves are either transverse/longitudinal. If there are waves that are neither one of them, how do we put that in mathematical notation?
| I was wondering if there is a difference between a transverse and longitudinal wave...
Imagine a rubber rod, it flexible to the sides so you can bend it and oscillate it like a string if you do it fast enough. That would be its transverse wave behavior.
Now the same rubber rod can be compressed or decompressed and that would be its longitudinal wave behavior.
Now... what is the difference? in the longitudinal the atoms are contracting and expanding toward each other. And in the transverse movement it is doing the same but at the same time meaning when it bends some part of the radius is been compressed and the outer side is been expanded, but in the end, it is the same.
It could be said that the transverse wave is a complex longitudinal wave.
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Is the drag force on a thrown object higher in hot or cold air? Increased temperature lowers viscosity of gases like air but also decreases density.
So then drag force would be lower if an object is thrown in higher temperatures but what about viscosity? I thought viscosity was drag.
| Depends on how fast the object is traveling through the fluid a.k.a. the value of the Reynold's number.
If the flow around the object is turbulent (high Reynold's number) then density is the key fluid property. If Reynold's number is small then viscosity becomes relevant.
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Von Karman mixing length $l=k \frac{du/dy}{d^2u/dy^2}$
In a fully developed turbulent flow of an in-compressible fluid inside a pipe of radius $R$, velocity at the center is $U_m$. If we define $U^*=\sqrt{\tau_0/\rho}$, where $\tau_0$ is the wall shear stress and $\rho$ is the density, then find the velocity distribution as a function of $y=R-r$ distance from the wall.
Consider the $l=k \frac{du/dy}{d^2u/dy^2}$ as Von Karman mixing length.
Now, if we write $\tau \approx \tau_0=-\overline{\rho u' v'}= \rho l^2 (du/dy)^2$, then we get $$(U^*)^2=k^2 \left(\frac{du/dy}{d^2u/dy^2}\right)^2 (du/dy)^2$$ and $$U^*=k\frac{(du/dy)^2}{d^2u/dy^2}$$ Now let $p=u'$ to get $p'/p^2=k/U^*$. Integrating twice gives $$-1/p=\frac{k}{U^*} y+C_1$$and $$u=-\frac{U^*}{k} \ln \left( \frac{k}{U^*} y +C_1\right)+C_2. $$
Now, one of the conditions for finding $C_1$ and $C_2$ is $u(y=R)=U_m$. What will be the other condition? This is the problem I encountered solving a similar problem:
In a pipe with diameter $0.8 \ m$ water is flowing (turbulent) and velocity at $y=0.2 \ m$ is $2 \ m/s$. If the relation $u/U^*= C_1 \ln(y/R)+ C_2$ is true, then find $C_1$, $C_2$, and wall shear $\tau_0$ (notation is as same as above).
Should we relate this to viscous sub-layer somehow?
| try to write $$ y = k(R-r)(1-\frac{R-r}{R})$$ now find when y is maximum then you graph this function :$$u^+ = \frac{1}{\kappa} \ln\, ((R-r)y_m) + C^+$$ you sould get something like this 1) REICHARDT 2) PRESENT 3) UNIVERSAL
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Why is the internal energy of a real gas a function of pressure and temperature only? While studying thermodynamics, I read that the internal energy of an ideal gas is a function of temperature only. On searching the internet, i found an article which stated that the internal energy of a real gas is a function of temperature and pressure only. I could not find a proper reason for this.
So my question is: why is the internal energy of an ideal gas a function of temperature only and that of a real gas a function of temperature and pressure only?
Is this property of ideal gases and real gases derivable through any equation?
| In real gas Pressure and volume change when temperature changes so work and heat exchange with surroundings then internal energy changes during this process
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What is the shortest distance between electron and positron before they are annihilated? I just want to know how does an electron felt the presence of a positron before they are converted into energy? Also how does the electron tell if it is positron or proton if this makes any difference?
| As a greasy handed experimenter I have a simple answer to this. Look through all the high-energy, exclusive data you have available on $e^+ + e^- \to 2\gamma$ (or possibly $e^+ + e^- \to \mu^+ + \mu^-$), pick the event with the largest squared four-momentum transfer $Q^2 = -t$ (or invariant mass $s$), and use that to characterize a length scale
$$ L = \frac{hc}{\sqrt{-t}} \quad\text{or}\quad \frac{hc}{\sqrt{s}}\;,$$
which you declare to be the best experimental answer to date.
I assume the winning event is to be found somewhere in the LEP II datasets, but I have no idea what the answer might be.
However, there is no theoretical upper limit on those experiment observables and therefore no theoretical lower limit on the distance (short of the Planck scale). Finding such a lower limit would mean either a discovery that the charged leptons are not fundamental or that you've hit the scale of a more fundamental underlying theory (string theory, super-symmetry, etc.)
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Mass-Energy equivalence in case of minimal coupling The energy-momentum relation of a free particle is (in SI Units):
$$
m^2c^4 =- c^2 \vec{p}^2 + E^2
$$
Minimal coupling is a way to fix a gauge freedom for the choice of canonical momentum (which I can in special relativity give as $ p_{\mu} = \left( \begin{matrix} \frac{E}{c} \\ p_x \\ p_y \\ p_z \end{matrix} \right)$.
One way to write the minimal coupling would be by writing the canonical momentum now as:
$$ p_{\mu} = \left( \begin{matrix} \frac{E - e \Phi}{c} \\ p_x + A_x \\ p_y + A_x \\ p_z + A_z \end{matrix} \right)$$
(Here I emply kind of a relativistic hamiltonian formalism, where the movement of a particle in space time, parameterised by $s$, is given by the hamiltonian $\lambda (m^2 + p^{\mu} p_{\mu})$, with $\lambda$ being an abitrary positive function of s, that describes how fast the path in space time is gone through).
Does this mean that I can generalize the energy-momentum relation to:
$$
m^2c^4 = - c^2 (\vec{p} + \vec{A})^2 + (E-e\Phi)^2~?
$$
If I solve this for $E$ for $\vec{A} = 0$, it resolves to:
$$
E = e \Phi + \sqrt{m^2 c^4 + c^2 p^2}
$$
which to me seems plausible, and that's why I'm asking.
Edit: the question is not only wether the modified relation holds, but also wether the reasoning behind is right.
| Yes, indeed it is the expression of the energy of a relativistic massive charged particle in the presence of an electric potential. Consistently, if the expression is expanded in the limit $cp\ll mc^2$, it reduces to the known non-relativistic formula
$$
E = \sqrt{m^2c^4+c^2p^2} + e\Phi \simeq mc^2+ \frac{p^2}{2m}+e\Phi
$$
with the addition of the rest mass term.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Blocking WiFi with Faraday cage As part of a project I'm trying to prevent WiFi transmission of frequency 2.4 GHz from reaching a Raspberry Pi via a Faraday cage.
Would a 20 micron aluminum foil do the job?
| Yes.
The quantity that measures how far an electromagnetic wave of frequency $f$ can pass through a conductor is called the skin depth:
$$ d = \sqrt{{\rho \over \pi f \mu}} $$
Where $\rho$ is the resistivity and $\mu$ the magnetic permeability of the conductor. This equation is true at frequencies lower than $1/\rho\epsilon$ where $\epsilon$ is the permitivity.
If you don't feel like looking up those values for aluminum, use an online calculator. I get a skin depth of 1.647 microns. The skin depth will dictate how quickly the amplitude of the EM wave decreases, and with amplitude related to intensity, we can relate how many dB drop 20 microns of aluminum would generate:
$$\text{dB} = -\frac{10}{\ln{10}}\frac{20 \mu\text{m}}{1.647 \mu\text{m}} = -52 \text{dB}$$
Which is down in the noise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Is potential energy a type of energy at all? Is potential energy, whether it be that of a charge in an electric field or a mass in a gravitational field or anything like that, actually an energy that the particle itself contains, like kinetic energy? Or is it just a measure of its ability to do work?
Is it the case that instead of integrating conservative forces over distances to find work done, we use the fact that the work done by a conservative force doesn't depend on path and hence we just use the notion of 'potential energy' and it's variation with distance, and just take the difference between the potential energy at two points to easily find the work done? And hence, is potential energy nothing but a tool to calculate work done by conservative forces?
| Well, as the name suggests, its the potential of the body to store energy inside it or some other way (whatever you like), potential energy is actually potential to do work. But having a constant potential energy means nothing.
as F = - Del (V)
This means , its something , not nothing.
Just think of Gravitational potential energy (stored in a body), it will come down.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is the singlet state for two spin 1/2 particles anti-symmetric? For two spin 1/2 particles I understand that the triplet states ($S = 1$) are:
$\newcommand\ket[1]{\left|{#1}\right>}
\newcommand\up\uparrow
\newcommand\dn\downarrow
\newcommand\lf\leftarrow
\newcommand\rt\rightarrow
$
\begin{align}
\ket{1,1} &= \ket{\up\up}
\\
\ket{1,0} &= \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2}
\\
\ket{1,-1} &= \ket{\dn\dn}
\end{align}
And that the singlet state ($S = 0$) is:
$$
\ket{0,0} = \frac{\ket{\up\dn} - \ket{\dn\up}}{\sqrt2}
$$
What I'm not too sure about is why the singlet state cannot be $\ket{0,0}=(\ket{↑↓} + \ket{↓↑})/\sqrt2$ while one of the triplet states can then be $(\ket{↑↓} - \ket{↓↑})/\sqrt2$. I know they must be orthogonal, but why are they defined the way they are?
| Let's temporarily forget that the two $m=0$ states exist, and consider just the two completely aligned triplet states,
$\newcommand\ket[1]{\left|{#1}\right>}
\newcommand\up\uparrow
\newcommand\dn\downarrow
\newcommand\lf\leftarrow
\newcommand\rt\rightarrow
$
$\ket{\up\up}$ and $\ket{\dn\dn}$.
There's not any physical difference between these: you can "transform" your state from one to the other by changing your coordinate system, or by standing on your head. So any physical observable between them must also be the same.
Either of the single-particle states are eigenstates of the spin operator on the $z$-axis,
$$\sigma_z = \frac\hbar 2\left(\begin{array}{cc}1&\\&-1\end{array}\right),$$
and "standing on your head," or reversing the $z$-axis, is just the same as disagreeing about the sign of this operator.
But let's suppose that, on your way to reversing the $z$-axis, you get interrupted midway. Now I have a system which I think has two spins along the $z$-axis, but you are lying on your side and think that my spins are aligned along the $x$-axis. The $x$-axis spin operator is usually
$$\sigma_x = \frac\hbar 2\left(\begin{array}{cc}&1\\1\end{array}\right).$$
Where I see my single-particle spins are the eigenstates of $\sigma_z$,
$$\ket\up = {1\choose0} \quad\text{and}\quad \ket\dn = {0\choose1},$$ you see those single-particle states as eigenstates of $\sigma_x$,
\begin{align}
\ket\rt &= \frac1{\sqrt2}{1\choose1}
= \frac{\ket\up + \ket\dn}{\sqrt2}
\\
\ket\lf &= \frac1{\sqrt2}{1\choose-1}
\end{align}
If you are a $z$-axis chauvinist and insist on analyzing my carefully prepared $\ket{\rt\rt}$ state in your $\up\dn$ basis, you'll find this mess:
\begin{align}
\ket{\rt\rt} = \ket\rt \otimes \ket\rt
&= \frac{\ket\up + \ket\dn}{\sqrt2} \otimes \frac{\ket\up + \ket\dn}{\sqrt2}
\\
&= \frac{\ket{\up\up}}2 + \frac{\ket{\up\dn} + \ket{\dn\up}}2 + \frac{\ket{\dn\dn}}2
\end{align}
This state, which has a clearly defined $m=1$ in my coordinate system, does not have a well-defined $m$ in your coordinate system: by turning your head and disagreeing about which way is up, you've introduced both $\ket{\up\up}$ and $\ket{\dn\dn}$ into your model. You've also introduced the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$.
And this is where the symmetry argument comes in. The triplet and singlet states are distinguishable because they have different energies.
If you propose that the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$ is the singlet state, then you and I will predict different energies for the system based only on how we have chosen to tilt our heads. Any model that says the energy of a system should depend on how I tilt my head when I look at it is wrong. So the $m=0$ projection of the triplet state must be symmetric, in order to have the same symmetry under exchange as the $m=\pm1$ projections.
| {
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"timestamp": "2023-03-29T00:00:00",
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Adding energies vs. adding momenta Suppose we have a (time and space) translationally invariant system with the Fock space for a Hilbert space. Temporal and spatial translation invariance implies that energy and momentum are good quantum numbers. In the free theory, the momenta and energies of one-particle states simply add to get the total momentum and total energy of the system. Why is it that, for an interacting theory, we can (always?) add the momenta of the one-particle states to obtain the total momentum given that we can't do the same for the energy?
| In quantum mechanics, the linear momentum is represented by a one-body operator: the momentum of the $i$-particle only depends on the variables of this particle, $\hat{\bf p}_i = -i\hbar \nabla_i.$ So, the total momentum of the system is simply defined by the sum of the momentum of each particle. Of course, if the particles are immersed in some field (like an electromagnetic one), it would be necessary to sum the field momentum to the momenta of the particles.
On the other hand, the interaction energy terms correspond to two-body operators, for example, the repulsion between two electrons corresponds to $\hat{V}({\bf \hat{r}}_1, {\bf \hat{r}}_2) = \frac{e^2}{|{\bf \hat{r}}_1-{\bf \hat{r}}_2|},$ and it involves, naturally, the variables of both particles. For these reason, the total energy is not additive, as the interaction terms "belong" to the two-particles, and not to each particle separately. It doesn't matter if the particles form bound states.
Other energies are additive (one-body operators), for example, the kinetic energy, the potential energy in an external field.
| {
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"source": "stackexchange",
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What is the definition of the charge conjugation? I seem to have troubles finding definitions of the charge conjugation operator that are independant of the theory considered.
Weinberg defined it as the operator mapping particle types to antiparticles :
$$\operatorname C \Psi^{\pm}_{p_1 \sigma_1 n_1;p_2 \sigma_2 n_2; ...} = \xi_{n_1} \xi_{n_2} ... \Psi^{\pm}_{p_1 \sigma_1 n_1^c;p_2 \sigma_2 n_2^c; ...}$$
He does not really seem to specify what he means by "antiparticles" around there, but I'm guessing this is the one-particle state that is conjugate to this one. This assumes that it is possible to decompose everything into one-particle states.
Wightman seems to go with $C \gamma^\mu C^{-1} = \bar \gamma^\mu$, which isn't terribly satisfying and also only works for spinor fields.
I've seen thrown around that the $C$ conjugation corresponds roughly to the notion of complex conjugation on the wavefunction but never really expanded upon.
Is there a generic definition of charge conjugation that does not depend on how the theory is constructed? The CPT theorem in AQFT indeed seems to not have any of those extraneous constructions, but the action of the different symmetries is a bit hidden as
$$(\Psi_0, \phi(x_1) ... \phi(x_n) \Psi_0) = (\Psi_0, \phi(-x_n) ... \phi(-x_1) \Psi_0)$$
Is the action of $C$ symmetry $\Psi' = C \Psi$ just a state such that for any operator $A$,
$$(\Psi, A \Psi) = (\Psi', A^\dagger \Psi')$$
or something to that effect? From some parts seems like it may just be $C \phi C^{-1} = \phi^*$.
| All of your fields naturally lie in some representation of the group of all symmetries (these include gauge symmetries, global gauge transformations and global Lorentz transformations). Charge conjugation is simply passing to the conjugate representation of that group.
E.g. complex scalars are 1d irreps of $U(1)$, and the conjugate object is $\phi^{*}$. The same logic also works for spinors, gauge fields, etc.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What experimental bounds do we have on big $G$? I know that there has been a large amount of controversy surrounding the exact value of the gravitational constant $G$, but I know that there is not a substantial difference in the measured value. So I was wondering what experimental bounds we have on it so far?
| According to NISTconstants, as of 2017,
$$G = 6.674 08(31) \times 10^{-11} \space {\rm m}^3 {\rm kg}^{-1} {\rm s}^{-2} $$
which means the range is 6.67377 to 6.67439
It's not the easiest measurement to make, large forces from electrical, magnetic, and other sources have to be considered.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Potentials and position uncertainty In the Schrodinger equation, we have some potential $V(x)$. But generally, there is some uncertainty in the position with solutions to Schrodinger's equation. Classically, we would say that a particle at position $x$ is associated with the potential at that point -- is there a quantum analog when the position wavefunction isn't localized? What's to say if a particle is in a potential or not if it doesn't have a well-defined position?
| The schrodinger equation tells us that a particle interacts with a potential if its wavefunction has a non-zero amplitude in the region enclosed by the potential. That is to say, the wavefunction's time evolution is independent of the potential energy's value in regions where the wavefunction is zero.
However, the wavefunction may have non-zero amplitude in regions inside and outside the potential simultaneously. Therefore, we can not always say without performing a measurement whether or not the particle is inside or outside the potential.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why change in resistivity is proportional to the original resistivity? When there is a temperature change $\Delta T$, the change of resistivity is
(1) proportional to $\Delta T$
(2) proportional to the original resistivity $\rho_0$
Hence we can define the temperature coefficient of resistivity $\alpha$ so that
$$\Delta \rho = \rho_0 \alpha\Delta T$$
I searched on the internet about (2) but it is usually simply stated as a fact or "experiments show that", without explaining why.
Length expansion has similar property but I can understand why intuitively. For the same temperature change, doubling the length will double the change in length as well, because every part of the length expands.
But I don't understand why the change in resistivity should be proportional to the original resistivity.
| It's just a matter of definition. If we expand $R(T)$ in a Taylor series about $T_0$, the first two terms are
$$
\rho(T) \approx \rho(T_0) + \rho'(T_0) (T - T_0).
$$
The coefficient $\alpha$ is then defined so that
$$
\alpha = \frac{\rho'(T_0)}{\rho(T_0)},
$$
which yields $\Delta \rho \approx \rho_0 \alpha \Delta T$ (where $\rho_0 \equiv \rho(T_0)$.)
We could equally well define $\tilde{\alpha}$ to be $\rho'(T_0)$, and then we would have $\Delta \rho \approx \tilde{\alpha} \Delta T$. But the physics would be unaffected by this change of definition. The important part is that the change in resistance is roughly proportional to the change in temperature; how we decide to define the constant of proportionality in this relationship is just a matter of convention.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does the electric field strength for a dipole go as $1/r^3$? I've been given the following graphic to help wrap my head around this.
If the potential can be shown to represent a $1/r^2$ relation, then I'm more than happy to accept that the electric field is hence a $1/r^3$ relation, but I need to accept the first part first:
Since $V = \frac{kQ}{r}$, this basically implies that
$$r = \frac{bc}{c-b}$$
given the geometry of this graphic, yet I simply do not see it.
| The important physical interpretation that you need to keep in mind is that the charges are opposite, so the $1/r$ pieces of their potentials cancel out. We'll see this happen explicitly in the math in the correct derivation.
But it also explains why your logic isn't quite enough to understand what's going on: You really do need those two opposite charges to cancel out the $1/r$ pieces. So you shouldn't start from just $V = kQ/r$ and then find the approximation for $r$ by plugging in the correct answer that you don't yet understand. That's not a helpful way of going about it. Instead, you should do the full derivation yourself.
First of all, I hope you can see from the geometry that
\begin{equation}
r \approx b + \frac{LM}{2} \approx c - \frac{LM}{2}.
\end{equation}
The reason is that as you take $P$ very far away, those lines $c$, $b$, and $r$ all become basically parallel, and they squeeze down onto each other. So $c$, $b$, and $r$ just represent different points along (roughly) the same line.
Now, accepting the equation above, you can immediately see that $LM \approx c-b$. You can also see that
\begin{align}
r^2
&\approx \left(b + \frac{LM}{2}\right)\left(c -\frac{LM}{2} \right) \\
&\approx bc + (c-b)\frac{LM}{2} - \frac{LM^2}{4} \\
&\approx bc + LM\frac{LM}{2} - \frac{LM^2}{4} \\
&\approx bc + \frac{LM^2}{4}
\end{align}
But when $r \gg a$ (and because $a \geq LM$), we know that $LM^2/4$ must have a small effect on the result, so we can just ignore it: $r^2 \approx bc$. That explains the two approximations shown in your figure. But I argue that's not enough to actually understand the total potential.
So onto the real derivation. You'll agree that (without any approximation) the full expression for the potential is
\begin{align}
V
&=
\frac{kQ} {b} - \frac{kQ} {c}.
\end{align}
Note that crucial minus sign! Next, we can start inserting our approximations:
\begin{align}
V
&\approx
\frac{kQ} {r-LM/2} - \frac{kQ} {r+LM/2} \\
&\approx
\frac{kQ} {r-\frac{a\cos\theta}{2}} - \frac{kQ} {r+\frac{a\cos\theta}{2}} \\
&\approx
\frac{kQ} {r} \frac{1}{1-\frac{a\cos\theta}{2r}} - \frac{kQ} {r}
\frac{1}{1+\frac{a\cos\theta}{2r}}
\end{align}
In that last line, I haven't done anything fancy; I've simply pulled out the denominator. But now, that lets us use the approximation $\frac{1}{1 \pm x} \approx 1 \mp x$ for small $x$. In this case $x$ is $a\cos\theta / 2r$, and since $r \gg a$ this is small, so we can use that approximation:
\begin{align}
V
&\approx
\frac{kQ} {r} \left(1+\frac{a\cos\theta}{2r}\right) - \frac{kQ} {r} \left(1-\frac{a \cos\theta}{2r}\right) \\
&\approx
\frac{kQ} {r} \frac{a\cos\theta}{r} \\
&\approx
\frac{kQa\cos\theta} {r^2}.
\end{align}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/390938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the 4-spin vector of a photon? The photon, being a vector boson has 2 spin states, $\pm 1$.
In relativity, we can determine the four-spin vector $s^{\mu}$ of a particle (see e.g. Costa et al. 2017).
What would $s^{\mu}$ be for a photon? How does it relate to the 2 spin states?
Thanks
| There is no such thing as a rest frame for a photon, so there is no such thing as a 4-spin in the sense of the Wikipedia article you link.
In fact, there is no such thing as spin for a photon, cf. also this answer of mine for a lengthy elaboration on how the properties of masslessness, being a gauge boson and having no true "spin" are all basically equivalent. Massless particles are characterized by helicity, not spin, the "2 spin states" should more precisely be called "2 helicity states".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why do different materials reflect different light? So, as far as I understand, white light contains photons of all energy levels. These hit a material, say iron. The photons that are below the energy level to move electrons just pass through. The others deflect the electrons to another orbit and when the electrons go back to their original orbit, they emit a photon.
Is that the case?
If so, I can understand why a material will emit a certain color back. These are the photons emitted back. But, what happens to the photons that just passed through? Why don't I see the complement color from the other side of the iron? Is it because they too get absorbed and turn to infrared level photons? If so, how?
| The photons of visible light do not pass through. The energy that is not reemitted is absorbed as heat. That’s why black surfaces get hotter than white services.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/391335",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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