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How do bicycle spokes work? As you know, it is quite obvious that bicycle spokes attach the hub in the center to the rim. What else do they do? If you compare the wheels today with the ones from ancient times, there are more spokes now on motor bikes and bicycles than of a wheel of a chariot. Why is that? What effect does it have on the vehicle if there is a higher number of spokes on the wheels?
| If there is weight on the axle the rim gets pushed down into the ground and tries to deform by flattening on the bottom and bulging right besides the ground. Properly tensioned spokes will counteract this bulging and lessen the deformation allowing for an easier and smoother ride.
This means that the rim does not have to be super resistant to deformation which allows it to be much lighter compared to a column-type rim.
| {
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Why are two independent sources incoherent? Coherent sources are produced from a single parent source. But, why are two independent sources always incoherent? Two sources can produce light of the same frequency. Then, I guess the problem is with phase. Are two independent sources always out of phase? why so?
| We usually generate light by black body radiation, that is by heating something until it glows. There are many sources for black body radiation, but the dominant one is usually random thermal motion causing random transient electric dipoles within the black body. The changes in these dipoles generates eletromagnetic radiation, and because the dipole changes are random so is the EM radiation they generate.
If you take two points in the black body that are close compared to the wavelength of the lattice vibrations then their motion, and hence the EM generated, will be closely correlated. However as you increase the separation between the two points the correlation will decrease and at macroscopic distance will be essentially zero.
| {
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Do neutrinos travel faster than light in air? I read in wiki that the speed of light is 88km/s slower in air than it is in a vacuum.
Do neutrinos travel faster than light in air?
| The answer is yes. Neutrinos will travel faster than light in a medium with a refractive index ($n$) greater than one (which is the case of air). Indeed the speed of light in that medium will be $v_{\text{medium}}=c/n$ where $c=2.998\times10^8$ m/s and $n>1$.
Then, because neutrinos interacts only very weakly (only through the weak nuclear force) with the medium, neutrinos will barely be slowed compared to how much light is slowed and thus will go faster than light. Remember that neutrinos are almost massless and thus already travel to nearly the speed of light.
--- New Edit ---
Indeed, the neutrino speed will depend on it's energy (as pointed out in comments). But I think that in most process in which neutrinos are produced (take for instance a beta-decay), the energy of a neutrino is enough to consider it as going to nearly the vacuum speed of light. So strictly speaking, the answer is that it depends on the neutrino energy and what type of medium you are in.
| {
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If a material is built to handle tension, would removing the tension damage it? If an object is designed to cope with large forces such as tension, would removing these forces risk damaging the object?
For example: The neck of a guitar is built to handle the tension of steel strings (~800 Newtons),
if you removed/reduced the tension (removed the strings) for a long period of time would this risk damaging the guitar neck?
| A piston-cylinder pressure vessel is an example of an object that commonly fails in this way. Carbide has a high compressive strength but lower strength in tension. Pressure vessels may be pre-loaded externally, with a confinement ring and interference fit, so that they can withstand higher internal pressure. If the carbide creeps during a pressurization experiment (say at elevated temperature) - residual strains may cause the vessel to fail in tension upon unloading. When I used such pressure vessels we stored them in a metal box to protect persons in the lab from the possibility of injury from an unexpected failure of an empty pressure vessel. Failure has occurred like this on occasion.
| {
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Alternative Critical Dimensions in String Theory Is it possible to write down a Lagrangian for a string theory with a critical dimension different than the familiar 10 or 26? How could one find a string theory Lagrangian for a given dimension? Could you prove that no string theory exists for a given critical dimension?
| You write a lagrangian for a string propogating in $(d,1)$ dimensional spacetime and notice that it is equivalently a 2d conformal field theory on the worldsheet of the string. Consistency conditions (vanishing of the conformal anomaly) then imply that necessary dimension of spacetime be 26 for bosonic string theory and 10 for superstring theory.
I wouldn't go as far as to say that string theory cannot exist in any other dimension. The conservative statement to make would be that so far, we only know how to make it work sensibly in the specific cases mentioned above and from what we understand today, seems like it won't work in other dimensions.
| {
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During reflection does the emitted photon have same properties? When light (photon) is reflected the the original photon is absorbed by an electron and then emitted again. Does this "new" photon have the same wavelength, frequency etc. as the original?
| Your title and question has some in-consistency, for example, photon can also have spin, and the field itself can also carry orbital angular momentum et.al. But from fresnel equation:
r=(n-1+ik)/(n+1+ik)
where n and k are the real part and imaginary part for the materials, you can see the reflection spectra already contain the dispersion of the materials, and not only the amplitude, the reflection phase is also changed, from those information one can detect the properties of the materials: what human eyes see most is those reflected light, right? The photon reflected can also carry information for the spin and orbital angular momentum of the materials et.al.
| {
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Angular speed of the planets Do all the planets in our solar system have the same angular speed? Physics teacher says yes, my research is not crystal clear. I want to make sure I have the right information for future reference.
| No, planets in the solar system do not have the same angular speed.
Firstly, strictly speaking, the angular speeds of the planets change with time because they travel along ellipses, not circles. But the eccentricities of many of the planets are close to $0$ (see e.g. Why are the orbits of planets in the Solar System nearly circular?), so to a reasonable approximation, we can approximate them as traveling along circles to get an idea of their orbital angular velocities.
Kepler's Third Law tells us that the square of the orbital period of a planet is proportional to the cube of its orbit's semi-major axis;
\begin{align}
T^2\propto a^3
\end{align}
Since the semi-major axes of the orbits of the different planets are different, so are their orbital periods, and consequently their angular velocities are different as well. In particular, the Third Law shows that for each planet,
\begin{align}
T_\mathrm{planet} = \left(\frac{a_{\mathrm{planet}}}{a_\mathrm{Earth}}\right)^{3/2}T_\mathrm{Earth}
\end{align}
This allows us to determine how much longer and/or shorter the orbital periods of certain planets are, and how much smaller and/or larger their corresponding angular velocities are.
For example,
\begin{align}
a_\mathrm{Earth} &\approx 15.0\times 10^{10}\,\mathrm{m} \\
a_\mathrm{Mars} &\approx 22.8\times 10^{10}\,\mathrm{m}
\end{align}
so
\begin{align}
T_\mathrm{Mars} \approx \left(\frac{22.8\times 10^{10}\,\mathrm{m}}{15.0\times 10^{10}\,\mathrm{m}}\right)^{3/2}(1\,\mathrm{yr}) \approx 1.87\,\mathrm{yr}
\end{align}
and consequently, since the angular velocity $\omega$ and orbital period $T$ are related by $T=2\pi/\omega$, the angular velocity of Earth is roughly 1.87 times the angular velocity of Mars.
You can find a table of more info on this stuff at the end the of this hyperphysics page.
| {
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Chiral Spin Liquid(CSL), Chern number, and the ground state degeneracy(GSD) Consider a 2D gapped CSL with a nonzero Chern number $m$, then is the GSD of the system on a torus directly related to the Chern number $m$?
For example, see this article, in the last paragraph on page 7, the authors give the 4-fold GSD from the Chern number $m=\pm2$ for a CSL. I can not understand the explanation, can anybody present an intuitive illustration or a simple mathematical proof ? I will be very appreciated, thank you very much.
| Based on the paper, the answer is $|m|^2$. They suggest in their p.8, Eq.36, the effective theory is a Chern-Simons theory
$$
\frac{1}{4\pi}\int K_{IJ} a_I \wedge d a_J
$$
with the $ K_{IJ}$ bilinear K matrix as
$$K_{IJ}={\begin{pmatrix}m & 0\\ 0 & -m\end{pmatrix}}$$.
The up $m$ labels one sector and the lower $m$ labels the other sector. The degeneracy(GSD) is computed by a generalizing level-$k$ U(1) Chern-Simons theory(GSD=$k$) to a bilinear K matrix U(1)$^n$ Chern-Simons theory. GSD=$|\det(K)|=|m|^2$. This GSD result for GSD=$|\det(K)|$ is a well-known fact.
| {
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Gravitational collapse and free fall time (spherical, pressure-free)
A very large number of small particles forms a spherical cloud. Initially they are at rest, have uniform mass density per unit volume $\rho_0$, and occupy a region of radius $r_0$. The cloud collapses due to gravitation; the particles do not
interact with each other in any other way.
How much time passes until the cloud collapses fully?
(This was originally from a multiple-choice exam - I solved the problem via dimensional analysis on the options then. I'm wondering how it might be solved directly now).
The answer is $$t = \sqrt{\frac{3\pi}{32G\rho_0}}. $$
| There is a well-known solution for this problem which is as follows : Gravitationally the outer layer of the cloud is influenced by the rest just as the rest were compressed into a point mass. Therefore we have Keplerian motion: the fall of any part of the outer layer consists in a halfperiod of an ultra-elliptical orbit ; one focus is at the center of the cloud (by Kepler’s 1st law) and the other one is at $r_0$,
The period of the orbit is determined by the longer semiaxis of the ellipse (by Kepler’s 3rd law). The longer semiaxis is $\frac{r_0}{2}$. and we are interested in half a period. Thus, the answer is equal to the halfperiod of a circular orbit of radius $\frac{r_0}{2}$.
$$(\frac{2\pi}{2T})^2~\frac{r_0}{2}=\frac{Gm}{(\frac{r_0}{2})^2}$$
Hence $$T=\pi\sqrt\frac{r_0^3}{8Gm}$$
Or,$$T_{collapse}=\sqrt\frac{3\pi}{32G{{\rho}_0}}$$,independent of $r_0$.
| {
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How is dark energy calculated This should be a very simple question. What would be the proper way to calculate Dark Energy in Joules at any point in history and that is consistent with the Standard Model? I'm thinking that knowing the mass-energy of matter (after estimating the Mass of the Universe):
$$E_m = mc^2\tag{1}$$
and knowing,
$$ \rho = \frac{\rho_{m}} {\Omega_{m}}\tag{2}$$
$$\rho_{\lambda} = \rho \Omega_{\lambda}\tag{3}$$
of the universe $M$, I could get the total Energy of the Universe combining (1) and (2):
$$E_{tot} = \frac { mc^2}{ \Omega_m} $$
And then multiply by the $\Omega_{\lambda}$ factor in order to get the final dark energy.
$$DE = mc^2\cdot\frac {\Omega_{\lambda}}{ \Omega_m} $$
Of course, $\Omega_{\lambda}$ and $\Omega_m$ change overtime, so I should be able to find what they are at the given time and figure out what $DE$ is at that time. Does anyone see a problem with this calculations?
| It's simpler, since the amount of dark energy remains the same while de Universe evolves. Then, you can calculate it today: $\rho_\Lambda(t)=\rho_\Lambda(today)$.
Today $\Omega_\Lambda=\rho_\Lambda/\rho_c$, where $\rho_c$ is the critical energy density today, and $\Omega_\Lambda$ is measured to have a value $\Omega_\Lambda\approx0.685$. The critical density today is given by
$$\rho_c=\frac{3 H_0^2}{8\pi G}\approx4800\,\mathrm{eV\cdot cm^{-3}}$$ so that $$\rho_\Lambda=\Omega_\Lambda\rho_c\approx3300\,\mathrm{eV\cdot cm^{-3}}.$$
Remember that the $\rho$'s are not energies, but energy densities, i.e. energy over volume. You can's treat with total masses or total volumes. Is the Universe finite? You can only use energy densities, and of course, calculating dark energy in joules is impossible unless you define a volume $V$, such as the volume of your house or something like that.
| {
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White light diffraction I have a hard time understanding why light waves of different wavelengths diffract in a different manner. According to Huygens' principle, every point on the wavefront is a source of a secondary wave. So if we have a white light going through, say, a single slit (light rays parallel to each other and perpendicular to slit's plane), all what's supposed to happen is a plain diffraction, just like of any other wave. That is, the wave will progress spherically, but it will still be a white light. Why instead we get a splitting of different wavelengths? In other words, how does light color affect diffraction geometrically?
| I do not think Huyghens principle can be applied to white light, only to simple harmonic waves. Waves of light with different color have different wavelength, which will affect the radius of sphere drawn in the Huyghens construction. Around obstacles, waves with different wavelengths will move differently.
| {
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Does a moving object curve space-time as its velocity increases? We always hear how gravity bends space-time; why shouldn't velocity?
Consider a spaceship traveling through space at a reasonable fraction of the speed of light. If this spaceship, according to special relativity, gains mass as a factor of y as it approaches c, then its gravitational field should increase in strength as well. Hence, space-time should warp.
Note: Changes in space-time, gravity and mass should only be measureable by an outside observer with a different velocity. Those inside of the ship moving with it would not be able to measure the change in these properties.
| Equivalence requires acceleration to curve space the same way gravitating mass does. Hermann Weyl, Zur Gravitationstheorie, Annalen der Physik, 54, 117, (1917) argued that kinetic energy should curve space just as gravity and electromagnetic fields do by entering into the stress energy tensor.
The concept of inertial mass increasing at high speed has been used a lot in science, but is often replaced by non linear terms in the energy momentum equation which has better verification. An argument is made that particle accelerators running at high kinetic energy would not be able to levitate a beam of particles if they had relativistic gravitating mass unless the electric charge was also increasing relativisticly, which is not observed. To conserve the equivalence principle many scientists no longer use the concept of relativistic inertial mass.
In summary an object that is changing speed or direction is expected to generate a gravity wave, usually very small. It is proposed but not proven that an object of high kinetic energy could contribute to space curvature.
| {
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Measuring background radiation We tried to measure background radiation using a geiger counter for a experiment at school. The meter showed $0.12$-$0.21$ microSv/h during the day averaging at about $0.14$ mcSv/h.
As we tried to see ways how to shield incoming radiotion nothing seemed to work. Taking a cue from nuclear power plant we put the meter into a watertigt glasscontainer and sank it in the gym pool so that the container had a good three feet of water all around. The measurement did not budge at all. We tried all settings fast/slow/auto. The geiger meter should be okay as it had recently been calibrated.
Can someone explain.
| Thans for all answers and links.
If cosmic radiation is about 0.03 mcSv/h we should still see a drop.
There is no power plant in the area.
The pool was an indoor pool and it is in the basement which has a solid concrete ceiling.
How would I set this experiment to show that the counter goes down preferably to zero.
| {
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How do electrons get the energy to jump from one orbital to the next when in stationary orbits the electron does not radiate energy My Question is how do electrons get energy/lose energy to jump up or down an orbital as in a stationary orbit they do not gain energy and their energy is finite?
| Caution: simplified but hopefully not over-simplified answer approaches!
If you take a hydrogen atom, that is the system of one electron and one proton, and you solve the Schrodinger equation for it then you get a number of eigenfunctions. These are the well known hydrogenic orbitals; 1s, 2s, 2p, etc. These solutions are time independant so an excited hydrogen atom like $2p^{1}$ will remain stable forever. I would guess this is what you mean in your question.
But in emission and absorption the system is not time independant because now we have a three component system of an electron, a proton and an electromagnetic wave (or photon if you prefer), and of course the EM wave has to be time dependant because it travels at $c$. The Schrodinger equation for this system is not the same as the Schrodinger equation for an isolated hydrogen atom, and the solutions are not the hydrogenic orbitals.
This is probably more intuitive for absorption rather than emission. In absorption the EM wave approaches the atom and perturbs it. This mixes up the hydrogenic orbitals so the atom is in a combination of different states, and as a result there is a non-zero probability that it will absorb the EM wave and settle into an excited state. But the equations of (non-relativistic) quantum mechanics are time symmetric, so if there is a finite probability for the atom to absorb a photon and change to an excited state, there is also a non-zero probability for an atom in an excited state to emit a photon and change to a lower energy state.
| {
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Centre of instantaneous rotation problem Is there a point of Centre of Instantaneous Rotation (CIR) for every type of motion or only for cases of rolling?
| The fact you are stating is quite general in fact and even extends in a related form to 3 dimensions also.
It is known as Chasles's rotation theorem:
Any general displacement of a rigid body can be represented by a translation plus a rotation.
In the case of motion of a body in a plane,the axis intersects the given plane in a point which we can call the instantaneous centre of rotation.Even in the case if doesn't intersect,we say the centre of rotation is at infinity.
So,yes any motion of a body in a plane has an instantaneous axis of rotation.
| {
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Electrons on Stern–Gerlach experiment My questions about spin and negative charge of electrons. Stern-Gerlach experiment is very famous in order to find spin of electron. this video created by paris-sud university really well-explained on this experiment except one thing.watch the video first.
http://en.wikipedia.org/wiki/File:Quantum_spin_and_the_Stern-Gerlach_experiment.ogv
The thing I confuse second step of experiment they use electrons they goes two different ways. Aren't electrons have negative charge only? How they can go two different ways?
| Neat video.
That experimental setup is not measuring the charge of the particles. If it were, you would be correct; you'd get the same result each time for all of the identical particles. Instead, that setup measures some other property of the particles. Physicists gave that property the name spin. Spin is not charge, and charge is not spin.
Spin is a property that is intrinsic to particles. By that, I mean just like all electrons all have some given mass and charge, they all also have a given spin value. What's different about spin (well, one of the differences) is that when you measure which way it's oriented (like in the video), your possible results are always "discreet". That's a fancy way of saying the spin can only be measured to be one of several possible values; a measurements of spin never results an in-between or "continuous" result like the classical magnets showed in your video.
| {
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Can a liquid insulator be electrically charged by touching a charged conductor?
Can a liquid insulator be electrically charged by touching a charged
conductor?
I understand that solid insulator will only be charge on the surface where it is touch, but the case is different from liquid which it circulates, so I imagine that slowly liquid insulator will be fully charge due to circulation.
| Whether charge is transferred between an insulator and a conductor depends on the field strength. If it is high enough charge will move over due to corona discharge. That even works when the two objects are not touching because ions can move through the air.
So yes, if the field strength is sufficient you can charge a liquid insulator but because it is liquid, the charge will immediately flow to the surface and so you won't manage to charge the entire volume.
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Why does a flat clover-like shape fall slower when it is rotating? The plastic flat clover-like shape pictured below falls noticeably slower when it is rotating fast. I wonder why. Its three edges are flat, so I don't think that the rotation makes it act like a propeller. An ideal explanation would include an analogy that a young child can understand (e.g. rain bouncing on a car's wind-shield).
Here is a photo of the object.
| There is an amount of lift that each blade gets, and it depends on the angle at which the air hits the blade.
When you just drop the toy, the air hits each blade directly at 90 degrees.
In that case, all you get is drag.
In airplane terms, the blade acts like a stalled wing.
It's more like a parachute than a wing, and it's not really big enough to make a very good parachute.
However, when it's turning, the air is coming up, and the wing is moving forward fast enough so that the air hits the wing at a shallow angle.
When this happens, the blade actually has lift.
In airplane terms, it's like gliding, at sufficient speed that the angle of the air against the wing has the wing acting like a wing, not like a parachute.
In that case, it will fall more slowly.
It's exactly the same in an airplane.
If the speed of your glide is above stall speed, you come down slowly.
If the speed is below stall speed, you come down quickly and out of control.
Every airplane has a stall warning horn, and every pilot is trained what to do.
When that goes off, you shove the nose down to reduce that angle, and pick up speed.
Even if you don't want to, because you're close to the ground, it's your only hope.
Fortunately, most airplanes, if they're properly loaded, are very resistant to stalling, because if they get too slow, they automatically drop the nose.
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Why is $U(\Lambda)^{-1} = U(\Lambda^{-1})$ for a unitary representation? This is from the beginning of Srednicki's QFT textbook, where he writes (approximately):
In QM we associate a unitary operator $U(\Lambda)$ to each proper orthochronous Lorentz transformation $\Lambda$. These operators must obey the composition rule
$$U(\Lambda'\Lambda) = U(\Lambda')U(\Lambda).$$
So far OK.
But where does he get the following from?
$$U(\Lambda)^{-1} = U(\Lambda^{-1})$$
| Indeed, it is also necessary to assume that $U(I)= I$ where the former $I$ is the identity matrix in the group and the latter is the identity operator in the Hilbert space.
So, taking advantage of your first identity you have: $$U(\Lambda^{-1})U(\Lambda) =
U(\Lambda)U(\Lambda^{-1}) = U(I)=I$$ and it implies $U(\Lambda^{-1})= U(\Lambda)^{-1}$ because of uniqueness of the inverse operator.
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Faster than light galaxies/clusters? A few years ago in an astronomy course, we calculated some (transverse?) velocity of a moving object and got super luminal results. The answer was apparent and not physical velocity of the object. Hence no problem. But at the moment, I don't recall the solution to this apparent issue. Anyone?
| This question already has two good answers, but I needed an excuse to learn tikz, so here is my answer.
This "transverse" speed can be faster than the speed of light $c$ if the object is coming towards you sufficiently fast. To see how this works, look at the diagram below.
Here the object is moving at a speed $\beta c$, but it is not moving perpendicular to your line of sight. It is moving at an angle $\theta$ to the line of sight.
How would the speed of this object be estimated? A naive way would be to divide the observed transverse displacement by the observed time duration.
Let's see what these two quantities are when the actual time duration is $ T_{\mathrm{actual}}$. In this time the object will move an actual distance $v T_{\mathrm{actual}}$. Its apparent transverse displacement will be $d=v T_{\mathrm{actual}} \sin \theta$.
What will the observed time duration be? It won't be $ T_{\mathrm{actual}}$ because the light rays from the final position get a head start relative to the light rays from the initial position. The length of this head start is $v T_{\mathrm{actual}} \cos \theta$. Since light moves at $c$, this translates into a time speed-up of $v T_{\mathrm{actual}} \cos \theta /c$. That is how much faster than it "should" the final light gets there. Thus the observed time difference is shorter than the actual time difference by this amount. We get that the observed time difference is $T_{\mathrm{observed}} = T_{\mathrm{actual}} - v T_{\mathrm{actual}} \cos \theta /c = T_{\mathrm{actual}}(1 - v \cos \theta /c)$.
The observed speed is then $\frac{v T_{\mathrm{actual}} \sin \theta}{T_{\mathrm{actual}}(1 - v \cos \theta /c)} =\frac{v \sin \theta}{1 - \frac{v}{c} \cos \theta}$. Plugging in $v=\beta c$, we get that the observed speed is $c \frac{\beta \sin \theta}{1-\beta \cos \theta}$. To get an observed speed greater than $c$, we need $\frac{\beta \sin \theta}{1-\beta \cos \theta}>1$. Picking $\theta = 45^\circ$, this becomes $\frac{\beta } {\sqrt{2}-\beta }>1$. This inequality is achieved for $\beta > \frac{1}{\sqrt{2}}$, so superluminal speeds can be "observed".
| {
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If one is travelling at a significant fraction of $c$, will the length of the trip be shortened? Assuming two stars are 1 light year apart and a traveler is travelling at 0.75 of $c$, from the point of view of the traveler what would be the observed time en route? Also, if a vehicle is constantly accelerating, will it reach 0.75 of $c$ within a reasonable amount of time? What would the Lorenz transforms look like for these situations?
Be gentle, I'm not a physicist. I'm writing book, and I'm trying to devise a scenario whereby a vehicle with a practically unlimited amount of energy can travel between stars. I don't care about the actual travel time only that experienced by the travelers.
| The time measured on board the ship during such a trip would be 0.88 years (as calculated by Alex A). If you factor in constant acceleration during the whole trip (starting from zero and passing by the target at maximum speed), the time on board would be 1.9 years and you would be passing by the target at 0.75 c (coincidently), if you accelerate by 1 G. Other acceleration values will result in different times.
You can play with the parameters in various relativistic online calculators:
This one allows for two-way trip calculation
This one involves acceleration during the trip
| {
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Proof of minimum distance between object and image of a convex lens image being 4*focal length I can prove that the minimum distance between an object and its image, through a convex lens is 4*focal length, if I assume that the distance between the object and the lens is the same as the distance between the image and the lens, i.e. if:
$$1/u + 1/v = 1/f$$
that $u=v$, but is there a way of showing that the distance $u+v$ is minimised at $u=v$, because I can't work one out.
| Although the other answers are good, I will add my slightly different one. Rearranging the image equation gives;
$$
u+v=\frac{1}{f}uv
$$
On the left hand side you have the perimeter of a rectangle (well half of the perimeter, but it doesn't matter for the proof), and on the right hand side you have the area of the rectangle. It is a reasonably easy proof in geometry to show that a rectangle of fixed perimeter has minimum area when it is a square, i.e. when $u=v$. There are a couple of proofs of this fact here at math.SE.
Hence, the minimum of $u+v$ occurs when $u=v$. QED :)
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Energy of an inductor I know that for an inductor having self inductance $L$ energy stored in its steady state when a current $I$ has been established is given by $U = \frac{LI^2}{2}$.
But after this current has been established, if we suddenly cut the wires attaching the inductor to the potential source or short the circuit, what happens to the energy ?
It must not be stored anymore as $\frac{LI^2}{2}$ as there can be no $I$, could not have decayed as heat because we cut off the wires and did not have any circuit which may have allowed for reverse flow of current.
I have one thought that it might have gone as EM radiation but I am not sure.
| Satwik's answer is correct but I want to add a practical example. When we switch off some electric device(say an DC motor) there is a spark is produced ,this is where the energy is getting lost. This spark is so high that can be seen even with naked eyes at the switch.
PS: Do this experiment in vaccum then there be a part of energy dissipated as heat and the remaining will store in setting up an $\vec E$ in the wire and will cause the electrons to accumulate at the surface.
The capacitive energy will be $\dfrac{1}{2}CV^2$ ( $C$ is the stray capacitance of wire) and the remaining as heat energy produced by the movement of electrons from the body of conductor to the surface at a high current in a short time. Energy will not emit as EM waves in vaccum.
EM waves only generates due to corona discharge if there is some medium.
| {
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Why am I not burned by a strong wind? So I was thinking... If heat I feel is just lots of particles going wild and transferring their energy to other bodies, why am I not burned by the wind?
When I thought about it more I figured out that wind usually carries some humidity, and since particles of liquid are moving same speed as the wind, they are basically static relative to each other, so no energy is transferred between them (wind and water particles). And if that water sticks to my skin and wind blows, it'll evaporate thus taking energy from my skin and make me feel cold.
Thing is, I don't think that's really the case but even if it is, if I somehow dry out the wind, will it burn me if it's strong enough? And winds can reach some pretty high velocities (though I must admit I'm not sure if they are comparable to movement of atoms in warm bodies etc...).
So. Bottom line. Can I be burned by wind in some perfect scenario?
| The other answers address your question quite well. Just as a reminder of the ability to be burned by a strong enough wind, the image below shows the Chelyabinsk meteor during entry into Earth's atmosphere last year over Russia. :)
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Weinberg dimension 5 operator How to prove that the $\Delta L=2,$ dimension=5 Weinberg operator $LLHH$ is the unique operator which violates lepton number by two units, without derivative couplings, etc.??
| This $LLHH$ is not a unique operator. It just the operator with a lowest dimension using only Standard Model particles and giving Majorana neutrino masses. The way to understand this is as follows. Majorana neutrino mass must be of the form
\begin{equation}
\nu _{ L }^{\,\, T } m C \nu _L
\end{equation}
This term breaks $ SU(2) _L $ and $ U(1) _Y $ and so cannot be written down in a $ SU(2) _L \times U(1) _R $ invariant theory. However, It may come from a spontaneously broken term, just as the other Standard Model fermion mass terms do. In order to find what sort of form the $ SU(2) _L \times U(1) _Y $ invariant term we have, we consider the charges of the low energy term. It has a $ T _3 $ value of $ 1 $ and $ U(1) _Y $ value of $ - 2 $ (this may change slightly depending on your conventions). Since the low energy term has $ T _3 = 1 $, it can only arise from a term that transforms like a vector under $ SU(2) _L $ transformations. The lowest order such term is (you can easily check that this transforms as a vector under isospin transformations)
\begin{equation}
L ^T i \sigma _2 {\vec \sigma}C L
\end{equation}
To make an $ SU(2) _L $ invariant out of this term that gives a mass we must take the scalar product with another isospin triplet that gets a VEV. Since we don't have an isospin triplet Higgs in the SM we need to use a product of Higgses. Thus the lowest order term that gives neutrino Majorana masses is given by,
\begin{equation}
m\left( L ^T i \sigma _2 {\vec \sigma}C L \right) \cdot \left( \phi ^T i \sigma _2 {\vec \sigma} C \phi \right)
\end{equation}
This term is also invariant under $U(1)_Y $. Note, that it is helpful to go ahead and write down many different terms to convince yourself this operator is truly unique at this order.
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Euler-Lagrange equations and friction forces We can derive Lagrange equations supposing that the virtual work of a system is zero.
$$\delta W=\sum_i (\mathbf{F}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=\sum_i (\mathbf{F}^{(a)}_i+\mathbf{f}_i-\dot {\mathbf{p}_i})\delta \mathbf{r}_i=0$$
Where $\mathbf{f}_i$ are the constrainded forces and are supposed to do no work, which it's true in most cases. Quoting Goldstein:
[The principle of virtual work] is no longer true if sliding friction forces are present [in the tally of constraint forces], ...
So I understand that we should exclude friction forces of our treatmeant. After some manipulations we arrive to:
$$\frac{d}{dt}\frac {\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}=Q_i$$
Further in the book, the Rayleigh dissipation function is introduced to include friction forces. So given that $Q_i=-\frac {\partial \mathcal{F}}{\partial \dot q_i}$ and $L=T-U$, we get:
$$\frac{d}{dt}\frac {\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}+\frac {\partial \mathcal{F}}{\partial \dot q_i}=0$$
Question: Isn't this an inconsistency of our proof, how do we know the equation holds? Or is it just an educated guess which turns out to be true?
| The main point is that Goldstein is not saying we must exclude friction forces in our treatment, but we must place them in the tally of applied forces (that we keep track of in D'Alembert's principle) and not in the other bin of the remaining forces, see this and this Phys.SE posts.
Of course, there does not exist a generalized potential $U$ for the friction forces ${\bf F}=-k {\bf v}$, only the Rayleigh dissipation function, see this Phys.SE post and this mathoverflow post.
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Why does a star collapse under its own gravity when the gravity at its centre is zero? The gravity at the centre of a star is zero as in the case of any uniform solid sphere with some mass. When a massive star dies, why does it give rise to a black hole at it's centre?
I know how to derive the field equations for gravity inside a star assuming the star as a uniform solid sphere of mass M and radius R. I need to know how to find the expression for the total pressure due to gravity at the centre.
| Since every particle attracts all other particles, there is a net force directed towards the center of the star (or any object), for any particle not at the center. Therefore, the particles will move towards the center (collapse), unless some opposing force prevents it. In the case of a star, the kinetic energy of the particles creates the opposing force, until the energy "runs out" and the collapse fallows.
| {
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Non-Locality of Space - QFT (Srednicki's book) I was going through Mark Srednicki's book on QFT. It says in the relativistic limit the Schrodinger equation becomes something like :
$$ i\hbar\frac{\partial}{\partial t} \psi(\vec x,t) = \sqrt{-\hbar^2c^2\nabla^2+m^2c^4}\psi(\vec x,t) $$
Now he says that if I expand the square root (say binomially) it will have infinite no. of spatial derivatives acting on $\psi(x,t)$; this implies that
equation is not local in space.
What exactly does it mean to say the equation is not local in space?
| If you consider a standard differential operator $B$ working on functions defined in $\mathbb R^n$, like $\partial/\partial x_i$ or a polynomial of partial derivatives, and pick out a sufficiently smooth function $f$ vanishing in a neighbourhood $\Omega$, you see that also $Bf$ vanishes therein. This is the relevant notion of locality for operators.
In the RHS of the equation you wrote down an operator shows up which does not fulfil locality in the sense I said.
That equation is, in fact, the equation satisfied by the positive energy solutions of Klein-Gordon equation.
The operator in the RHS cannot be defined by formal Taylor expansion (it works only formally), but one has to use spectral theory. In the considered case it is equivalent to translate that equation in Fourier transform.
Non locality arises here due to a known property of the operator $A:= \sqrt{-\Delta + aI}$ and, more generally, for $(\Delta + aI)^\nu$ with $\nu \not \in \mathbb Z$. This property is called anti locality (I.E. Segal, R.W. Goodman, J. Math. Mech. 14 (1965) 629) and is related to the famous Reeh and Schlieder property in QFT.
Anti locality means that If both $f$ and $Af$ vanish in a bounded region $\Omega \subset \mathbb R^3$ then $f$ is everywhere zero.
If $f$ has support included in a bounded open set $\Omega$, then, remarkably
and very differently from what happens for standard differential operators, $Af$ does not identically vanish outside $\Omega$ otherwise $f$ would be the everywhere zero function.
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Which electron gets which energy level? Electrons sit in different energy levels of an atom, the farther the higher energy is. Every electrons have the same structure, they can gain energy from environment, electrons which gained energy could jump to a higher energy level and will finally fall back again.
I'm wondering why some electrons have the "right" to "store" that high energy since every electron is the same. Why do those electrons can have more energy and sit in higher energy level than other electrons?
| I'm surprised that no one has mentioned that there is really no such thing as "this electron" or "that electron" in an atom. Those are useful approximations that help us visualize energy levels; but the actual quantum-mechanical theory of, for example, a carbon atom with six electrons, is based on a single electron wave function in 18-dimensional phase space.
Or look at a Helium atom if you prefer, with only two electrons. You cannot solve for the wave function of the first excited state by saying "one electron is in the s-state and the other one is in the p-state." You have to write a function in six-dimensional wave space, and it has to be symmetrical in both electrons...so that if you switch "them" around, it's exactly the same function except for a 180 degree phase change.
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Propagator of Maxwell-Chern-Simons theory I need to compute the "topologically massive photon" propagator.
I've started with:
$$
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{\mu}{4}\epsilon^{\mu\nu\lambda}A_\mu\partial_\nu A_\lambda
$$
$$
=A_\mu\underbrace{[\frac{1}{2}g^{\mu\lambda}\partial^2+\frac{\mu}{2}\epsilon^{\mu\nu\lambda}\partial_\nu]}_{(\Delta^{-1})^{\mu\lambda}}A_{\lambda}
$$
So how can I invert The under braced part which will yield the topologically massive photon propagator?
| You might try it by the standard way of calculating Green's function (which is what the propagator is) of linear differential operators. For this purpose you have to derive the equations of motion from the given Lagrangian, which will be some operator acting on the gauge field. The crucial observation is now that the Green's function can be calculated by solving the equation
$$L G(x,y)=\delta(x-y),$$
which schematically shows that the differential operator's action on the Green's function gives a delta function. This equation can be solved for example by transforming it to momentum-space.
Note that I have not tried it for your specific example, so there might me caveats to my approach I am not aware of.
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Admixtures of longitudinal and timelike photons! In the quantization of electromagnetic field the physical states $|\psi\rangle$ are found to obey the following relation:
$[a^{(0)}(k)-a^{(3)}(k)]|\psi\rangle=0$
It is explained as the physical states are admixtures of longitudinal and timelike photons. What do longitudinal and timelike photons physically mean? Why the polarizations, $\epsilon^{(0)}$ and $\epsilon^{(3)}$, timelike and longitudinal photons, are called unphysical?
| The total field consists of the "near" field like the Coulomb one and more generally (and loosely) a retarded Coulomb field, which are always "attached" to the charge, and the photon (radiated) field with different polarization orientations. The near field is always present, its "photons" are not created and annihilated. The corresponding "photons", when introduced for "symmetry", should not modify the existing near field. That is why the creation/annihilation of them is restricted in some way.
The "near field" exists between two charges. It means their "virtual photons" are absorbed in the Feynman diagrams. The real photons are emitted or scattered, not completely absorbed.
| {
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Frequency of rotating coil Given a coil initially in the x-y plane, rotating at angular frequency $ \omega $ about the x-axis in a magnetic field in the z-direction. This uniform time varying magnetic field is given by $B_z (t)=B(0)cos(\omega t) $ I am required to show that there is a voltage of frequency $2\omega $ across the loop. Clearly when t=0 the flux is at a maximum, but I dont understand how to relate to the frequency?
If the frequency is just the inverse of the period then $f=\omega / 2\pi $ ?
Clearly I am not understanding something. How does the voltage affect the frequency?
| User31782 gave the right answer, but it's quite hard to read because of formatting. Let me repeat the argument for you:
The coil rotates at $\omega$, and the field is also changing at $\omega$.
At any moment in time, the area of the coil normal to the direction of the field is
$$A = A_0 \cos(\omega t)$$
and the field is
$$B = B_0 \cos(\omega t)$$
And so the instantaneous flux, which is the dot product of field and area, is
$$\Phi = A_0 B_0 \cos^2\omega t$$
The voltage is proportional to the time derivative of flux:
$$V \propto 2 \sin \omega t \cos \omega t = sin(2\omega t)$$
by trig identity.
To persuade yourself why this is so: when the coil has turned 180 degrees, the field is once again pointing in the same direction as the normal to the coil - so while the flux goes to zero when the coil is at 90 degrees to the XY plane, it's positive whenever the coil is in the XY plane.
| {
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If Earth was the size of an orange, what consistency would it be? I understand that the Earth is composed of a thin crust "floating" over a semi-solid layer of mantle.
I was wondering how it would feel at a human scale, say the size of an orange in your hand. Could you somewhat squeeze it or would it rather be totally solid?
| At first you'd burn your hand, then it would feel like a normal rock.
An orange sized Earth would cool very rapidly.
If an object gets twice as big, its volume increases by $2^3$, but its surface increases only by $2^2$. You can only lose heat at the surface but you 'hold' all your heat in your interior. Simply said, the bigger something is, the harder it is to lose heat. That's also why small mammals eat so much - mice eat their own body weight in 6 days! - they lose heat to their environment very easily.
If you consider the alternative - increasing your hand to the scale of the Earth - it won't end well for you.
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Why the speed of light is represented by $c$? In almost every textbook, I've found that the speed of light is $c \approx 3 \times 10^8\: \mathrm{m/s}$. I wonder why it's just $c$ ?
| It's c for constant or celeritas, which means speed in Latin. Everyone uses it because it's convention. You could use $\xi$ or $\zeta$ or $\gamma$ or any other symbol you wanted, but then you'd have to explain what it meant, and people would have to go through the trouble to remember this every time they read your papers. Better to go with convention and save everyone the headache
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In what way are the Mathematical universe hypothesis and A New Kind of Science connected The Mathematical universe hypothesis, mainly by Max Tegmark and A new Kind of Science, mainly by Stephen Wolfram both claim (as least as I understand it) that at its innermost core reality is mathematics.
Can this statement be made more precise, i.e. what is the exact relationship between these two hypotheses?
| I think that Wolfram is arguing that the study of cellular automata and perhaps similar computational systems could serve as an organizational principle, providing a coherent framework to look at different problem (just like the more familiar frameworks provided by physics and chemistry). This explains the title of his new book, A new kind of Science (i.e. the study of the above-mentioned structures).
On the other hand, Tegmark argues that our Universe is one big mathematical structure. This may be difficult to wrap your head around, but it would mean that we are just mathematical structures that are complex enough to be self-aware and do everything we do. I assume this would not have any observational consequences (as we cannot proof that something cannot be described by mathematics, exactly because we need mathematics to prove anything) and is therefore purely speculative.
As you can see, Wolfram is calling for a new framework to conceptualize and study problems, while Tegmark is positing a theory of the Universe. In my opinion, these are two completely different things. Disclaimer: I have not read the book by Wolfram, nor was I previously familiar with Tegmark's proposal.
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Does there exist a single plate capacitor(conductor)? Does there exist a single plate capacitor(conductor)? if yes
How will you define the capacitance and potential(difference)
of such conductor?
| Yes. Capacitance is very well defined for a sphere in vacuum, and can be extended to other media and shapes. The following is a simple abd correct definition of capacitance. Adding other conductors disturb the capacitance of a single conducting body, giving rise to mutual capacitance.
The following is exerpted and edited from The Great Soviet Encyclopedia (1979). Search that term for more complete answer.
Capacitance
The capacitance C of an isolated conductor is equal to the ratio of the charge on the conductor to the conductor’s potential.
Capacitance is determined by the size and shape of a conductor and by the electrical properties of the surrounding medium, that is, by the medium’s dielectric constant. Capacitance of a conducting sphere in vacuo is equal to the radius of the sphere. Capacitance is measured in centimeters in the cgs system and in farads (1 farad = 9 × 10^11 cm) in the International System of Units (SI).
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What's the dimensionality of a solid angle? I haven't seen this explained clearly anywhere. Solid angles are described usually as a fraction of the surface area of a unit sphere, similar to how angles are the fraction of the circumference of a unit circle. However, I don't know how solid angles are actually quantified.
Are solid angles just a single number, the describes this fraction of the area? It's confusing to me since often times, I've seen integrals that integrate over a sphere using solid angles, which seems to imply that solid angles are multi-dimensional quantities (e.g. when integrating using spherical coordinates, the solid angle would have to consist of the azimuthal and polar angles covered by the differential solid angle).
Following from this, how would you write down a solid angle that covers the entire surface of a unit sphere?
| The solid angle is defined as the area on the unit sphere subtended by the angle divided by one unit area. It's a ratio so it's a single dimensionless number.
I see why you think it should be a 2D quantity, because the surface of a sphere, and any patch on it, is a 2D manifold and you need two quantities (traditionally $\theta$ and $\phi$) to map it. When you calculate an area on the sphere you are basically calculating a definite integral over $\theta$ and $\phi$, and the result is of course just a single number. You do lose information in the process - for example you just know the total area not the shape of the patch on the sphere.
The solid angle that covers the whole sphere is of course $4\pi$/1 or $4\pi$.
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Where is the energy lost in a spring? Thinking about springs, and their extensions, I recently came to a confusion which I hope this wonderful community can help me solve.
The question is this. When the block is initially attached to the spring, the spring has some extension $x_0$. Now the spring gets extended to some extension $x=\frac{mg}k$ by an external force maintaining equilibrium at all the points such that $KE=0$ at the bottom.
As my reference is the line shown in the figure, the initial potential energy $U$ is 0 due to both gravity and spring potential energy($x=0$).
Now as the block comes down, the spring potential energy is:
$U_(spring)=\frac12kx^2$. Final extension is $\frac{mg}k$. So spring potential energy is $\frac{m^2g^2}{2k}$
But the decrease in gravitational potential energy is $mgx$ which equals $\frac{m^2g^2}k$.
This means that potential energy has decreased. Intitially, $U_{net}=0$ but finally $U_{net}=-\frac{m^2g^2}{2k}$.
Where if any, did this energy get compensated(to ensure COE is still true)?
| You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds.
It needs to be moving because, if it were not, then there must be some force (your hand?) holding it in place. This would count as an external force, and when such a force acts, CoE does not hold.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relativity of simultaneity and cause-effect Relativity says that two events simultaneous in one reference frame might not be simultaneous in another reference frame. Can we extend this idea a little and say that the order of two events might be be reversed in going from one reference frame to another? And if so, what if they have a cause-effect relation ship in the first frame. Isn't reversing the two in the second frame a contradiction?
| No, ordering of timelike-separated events is always preserved in special relativity.
First, let's agree on some definitions. Say event $O$ occurs at the origin of our 1D coordinate system. Consider another event $A$ in spacetime, at time $t$ and position $x$ in some pre-fixed coordinates. There possibilities for how $x$ and $t$ relate are as follows:
*
*$\lvert t \rvert < \lvert x \rvert$: $A$ is spacelike separated from $O$;
*$\lvert t \rvert = \lvert x \rvert$: $A$ is lightlike (or null) separated from $O$:
*
*$t = \lvert x \rvert$: $A$ is in $O$'s future;
*$t = -\lvert x \rvert$: $A$ is in $O$'s past;
*$\lvert t \rvert > \lvert x \rvert$: $A$ is timelike separated from $O$:
*
*$t > \lvert x \rvert$: $A$ is in $O$'s future;
*$t < -\lvert x \rvert$: $A$ is in $O$'s past.
Now apply a boost of velocity $v$, where $\lvert v \rvert < c \equiv 1$. In this new frame, we have
\begin{align}
t' & = \frac{t-vx}{\sqrt{1-v^2}}, \\
x' & = \frac{x-vt}{\sqrt{1-v^2}}.
\end{align}
As you can check, $(x')^2 - (t')^2 = x^2 - t^2$ independent of $v$. This quantity is the invariant interval, and the fact that it doesn't change between reference frames means the relation between $O$ and $A$ cannot shift amongst the different classifications given above.
The important lesson here is that the future is the future, the past is the past, and everything else is everything else (note I'm not calling this the "present," which doesn't mean anything in relativity), and these definitions are frame-independent.
| {
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Capacitors: why is the energy not stored in a magnetic field? When a capacitor is charging, the rate of change $dE/dt$ of the electric field between the plates is non-zero, and from the Maxwell-Ampère equation this causes a circulating magnetic field.
Now, since a magnetic field exists, why is the energy of a capacitor only stored in the electric field?
Usually the formula for the energy stored goes as $ W = \pi d A \times \frac{1}{2}\epsilon_0 E^2$, where the first term is the volume and latter is the electric field energy density.
In Poynting's theorem, the electro-magnetic field energy density is $ \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2 $, i.e. there is also the magnetic field B present.
In a capacitor B is non-zero, so why do we not include it in the calculation of the energy stored?
In other words, why is the energy stored in a capacitor just $[ (volume) \times \frac{1}{2}\epsilon_0 E^2 ] $ and not $[ (volume) \times (\frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2) ] $ ?
| For a constant potential on the capacitor, there is no B-field and that is the case usually considered for this calculation. When charging a capacitor, the currents will generate a B-field and there is stored energy in that field (same as for an inductor). But once the charging stops, the B-field will "collapse" and cause currents to flow in the wires, dissipating that energy. Real capacitors will have some inductance and so will the wires feeding the capacitor and yes, you might need to include the effects if they are large enough (and they often do get included when analyzing circuits with real, not ideal, components).
| {
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Can zinc nitride be used as light emitting material? I cannot manage to find any journal papers about the applicability of zinc nitride as active layer of an light emitting diode (LED). But certain papers got mention that zinc nitride with a direct bandgap can be fabricated with potential applications in optoelectronics.
Can anyone give me some opinions on this subject?
| I have recently published a paper on Zn3N2 nanocrystals (http://pubs.rsc.org/en/content/articlelanding/2014/tc/c4tc00403e#!divAbstract). The material appears to have a direct band gap around 1eV and makes nice nanophosphors so if you can work out how to p and n dope it I am sure you could make an LED.
| {
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$(\frac{1}{2},\frac{1}{2})$ representation of $SU(2)\otimes SU(2)$ The representation $(\frac{1}{2},\frac{1}{2})$ of the Lorentz group correspond to a four- vector or a spin-one object. Right? Does it imply that any four-vector is identical to a spin-one object or any scalar is identical to a spin-0 object? This can't be correct, right? Because although $A^\mu$ is a four vector and a spin-one object at the same time (which is photon), there is no concept of spin associated with $p^\mu$ or $J^\mu$. I'm confused by terminologies of representation.
Edit- How can I show that $A^\mu$ represent a spin-1 object?
| The problem here is with the identification of the $(A,B)$ values of a representation with spin. $A$ and $B$ do not correspond to spin (they are not even Hermitian!), they just happen to obey $SU(2)$ Lie algebras, and as such they add up in the same way that spins do. When we say that $A_\mu,J_\mu,p_\mu,...$ are all in the $(\frac{1}{2},\frac{1}{2}) $ representation of the Lorentz group we mean that they transform as a four-vector, that's all. People may get lazy and say they are spin 1 objects, but what they really mean is $(A,B)$ spin 1 objects.
| {
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Parity violating Dirac particle We normally write down the Dirac Lagrangian as
\begin{equation}
{\cal L} _D = \bar{\psi} ( i \partial _\mu \gamma ^\mu - m ) \psi
\end{equation}
but are the Lagrangian's,
\begin{equation}
\bar{\psi} ( i \partial _\mu \gamma ^\mu \gamma ^5 - m ) \psi , \quad \bar{\psi} ( i \partial _\mu \gamma ^\mu - m \gamma ^5 ) \psi , \quad {\cal L} _D = \bar{\psi} ( i \partial _\mu \gamma ^\mu \gamma ^5 - m \gamma ^5 ) \psi
\end{equation}
all equally as good but just don't agree with Nature? Furthermore, how would you check that the propagator is or isn't Parity invariant?
| All the alternatives to the Dirac Lagrangian are actually forbidden by the requirement of requiring the hamiltonian to be well behaved (bounded from below and unbounded from above) and hermiticity of the action. To see this most simply we write the Lagrangian in terms of the fundamental left and right handed fields, $ \psi \equiv \left( \begin{array}{c}
\psi _L \\
\psi _R ^c
\end{array} \right) $. For the modified kinetic term,
\begin{align}
i\bar{\psi} \partial _\mu \gamma ^\mu \gamma ^5 \psi & = i \psi _L ^\dagger \partial _\mu \bar{\sigma} ^\mu \psi _L - i \psi _R ^{c, \dagger } \partial _\mu \bar{\sigma} ^\mu \psi _R ^{ c}
\end{align}
We see that the left handed kinetic term is well behaved but the right handed kinetic term has the wrong sign. Having a negative in front of the kinetic term results in an unphysical spectrum (see for example, this question).
Now we consider the proposed modification of the mass term. The complex conjugate of this term is:
\begin{equation}
\left( m \bar{\psi} \gamma ^5 \psi \right) ^\dagger = m ^\ast \psi ^\dagger \gamma ^5 \gamma ^0 \psi = - m ^\ast \bar{\psi} \gamma ^5 \psi
\end{equation}
Therefore, in order for this term to be hermitian we must have purely imaginary $ m $ (one could imagine having real $ m $ and just adding an hermitian conjugate but then the term vanishes identically). To see the meaning of this mass term we write it in terms of two component fields:
\begin{equation}
{\cal L} _{mass} = i \left| m \right| \left( \psi _L ^\dagger \psi _R ^c - \psi _R ^{ c \, \dagger } \psi _L \right)
\end{equation}
We can consider a field redefinition, $ \psi _R ^c \rightarrow - i \psi _R ^c $,
\begin{equation}
{\cal L} _{ mass} \rightarrow \left| m \right| \left( \psi _L ^\dagger \psi _R ^c + \psi _R ^{ c \, \dagger } \psi _L \right) = \left| m \right| \bar{\psi} \psi
\end{equation}
Thus the parity-violating mass term is actually equivalent to the canonical one after a redefinition of the fields.
| {
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Eigenvalue spectrum of $L_x+iL_y$ Is it possible to find out the generic eigenvalue spectrum of the non-Hermitian operator $L_x+iL_y$, without using any representation?
| I am assuming that $(L_x,L_y,L_z)$ satisfy the usual angular momentum algebra. Then, one knows that all states can be labelled by the eigenvalues of $(L^2,L_z)$. Pick such a state and label it as $|j,m\rangle$ where the $L^2$ eigenvalue is $j(j+1)$ and the $L_z$ eigenvalue is $m$. Note that we have not made any assumption about the allowed values of $(j,m)$. The action of $L_\pm$ produces states $|j,m\pm1\rangle$. This can be continued to get an infinite set of states. Several possibilities occur:
*
*The process of raising using $L_+$ terminates at some state -- call it the highest weight state (the norm of $L_+$ on the highest weight state is zero) .
*The process of lowering using $L_-$ terminates at some state -- call it the lowest weight(the norm of $L_-$ on the lowest weight state is zero).
Recall that the groups $SU(2)$ and $SL(2,\mathbb{R})$ arise from the same Lie algebra (over $\mathbb{C}$). So the above construction holds for both of them. Unitarity (each state in the irrep has positive norm) decides on the allowed values of $(j,m)$. For $SU(2)$, all unitary irreps have an highest and lowest weight state. However, for $SL(2,\mathbb{R})$, the unitary irreps are different. In particular, they are infinite dimensional. For instance, one could have a highest weight state but no lowest weight state. You can look up unitary irreps of $SL(2,\mathbb{R})$ to see all the cases. See here for example: https://en.wikipedia.org/wiki/Representation_theory_of_SL2%28R%29
| {
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Range Of An Interaction Why is the Compton wavelength $\lambda_c=\frac{\hbar}{mc}$ used as a sensible measure for the range of an interaction, where $m$ is the mass of the corresponding mediator?
| For massive force carriers, one finds (in natural units) an exponential dependence $\mathrm{e}^{-mr}$ that prevents long-range forces with massive mediators. Restoring SI units, one sees that the compton wavelength is the length at which the damping is exactly $\mathrm{e}^{-1}$.
| {
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Magnetic field due to electron in Hydrogen? We can calculate the current density $\mathbf{j}$ of the electron in Hydrogen, and it is given by:
$$
j_\phi=-e\frac{\hbar m}{\mu r\sin\theta}\left|\psi_{nlm}\left(r,\theta,\phi\right)\right|^2
$$
(derivation found here on page 6)
How can I calculate the magnetic field produced by this current density?
I could use the Biot-Savart law,
$$d\textbf{B} = \frac{\mu_0}{4\pi} \frac{1}{r^2} \int Id\textbf{s}$$
where the integration should be (at least classicaly) the along the current loop, and $I$ = $\int \textbf{J} \cdot d\textbf{S}$. What should I use as $d\textbf{S}$ and $d\textbf{s}$ for an electron in Hydrogen?
| Actually I think it is explained fairly well in these lecture notes that you linked. Since the current j only has a φ component, you have to integrate over circles like the ones shown in the figure. These circles have a fixed radius and θ coordinate. Only φ varies.
The radius of that circle is $r \sin\theta$ and therefore $ds=r \sin\theta d\phi$.
You should now be able to calculate $dB$. To calculate the total magnetic field then, if you wish to do so, you will have to integrate dB over all $r$ and $\theta$.
Hope this helps!
| {
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Number of Parameters of Lorentz Group We embed the rotation group, $SO(3)$ into the Lorentz group, $O(1,3)$ : $SO(3) \hookrightarrow O(1,3)$ and then determine the six generators of Lorentz group: $J_x, J_y, J_z, K_x, K_y, K_z$ from the rotation and boost matrices.
From the number of the generators we realize that $O(1,3)$ is a six parameter matrix Lie group.
But are there any other way to know the number of parameters of the Lorentz group in the first place?
| You've got two very good answers from Hunter and NowIGetToLearnWhatAHeadIs. However, it's probably useful to know that this beast $O(1,3)$ is isomorphic or locally isomorphic (i.e. has the same Lie algebra) to a surprising number of other interesting groups, which each give you a slightly different way to think about it. First note that its identity connected component $SO^+(1,3)$ of orthochronous, proper Lorentz transformations (those that keep the orientation of space and time the same, also called the "restricted" Lorentz group) of course determines the Lie algebra.
*
*$SO^+(1,3)\cong {\rm Aut}(\hat{\mathbb{C}}) \cong PSL(2,\mathbb{C})$ is isomorphic to the Möbius group of all Möbius transformations, in turn isomorphic to the group of all conformal transformations of the unit sphere. So it is defined by $z\mapsto \frac{a\,z+b}{c\,z+d}$ with $a,\,b,\,c,\,d\in\mathbb{C}$ and $a\,d-b\,c=1$. So there are three independent complex parameters, i.e. six independent real parameters;
*The double cover of $PSL(2,\mathbb{C})$, namely $SL(2,\mathbb{C})$ (still locally isomorphic to $SO^+(1,3)$) is the group of all $2\times 2$ matrices of the form:
$$\exp\left(\frac{1}{2}\left[\left(\eta^1 + i\theta \gamma^1\right) \sigma_1 + \left(\eta^2 + i\theta \gamma^2\right) \sigma_2 + \left(\eta^3 + i\theta \gamma^3\right) \sigma_3\right]\right)$$
where $\sigma_j$ are the Pauli spin matrices, $\theta$ is the angle of rotation, $\gamma^1,\,\gamma^2,\,\gamma^3$ are the direction cosines of the rotation axis and $\eta^1,\,\eta^2,\,\eta^3$ the components of the rapidities of the Lorentz transformation. So it's just like the general matrix $\exp\left(\frac{\theta}{2}\left(\gamma^1 \sigma_1 + \gamma^2 \sigma_2 + \gamma^3 \sigma_3\right)\right)$ in $SU(2)$ but with three complex parameters, rather than three real ones ($\theta \gamma^j$) for $SU(2)$. So again we see six real parameters.
| {
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why galaxies look like discs rather than spheres? Black holes have much gravitation to hold all stars and nebulas
but why they are aligned in disc type shapes rather than spheres
because gravitation is everywhere around black holes. Even on upper and lower sides.
Even if rotation is the cause why cant they move above black holes
in smaller circles.
| TL;DR They can orbit in any direction but they usually stick to the disk.
This is an excellent question.
It happens to be for the same reason that the earth and all of the planets all orbit the sun in the same direction on the same plane, I.E. the solar system is a disk as well.
First imagine a large ball of gas in space that is roughly spherical. It is all spinning around an axis at uniform angular velocity. We let gravity act on this spinning sphere of gas. Everything is attracted to the centre, but the equator is spinning and so opposes this attraction (it keeps missing the centre) but the poles are moving quite slowly so fall inward.
Letting this evolve and a small amount of damping with collisions this will end up as an accretion disk, of gas and dust, orbiting a dense core, the sun in the case of the solar system and the super-massive black hole (and dark matter and light matter) in the case of the galaxy. This dust then forms planets in the case of the solar system and solar systems in the case of galaxies.
Even if rotation is the cause why cant they move above black holes in smaller circles.
If we look closely at the centre of the galaxy at the super-massive black hole we do see many stars orbiting it in many different directions and in different planes, this sphere around the super-massive black hole does exist and is called the galactic halo.
| {
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Molecules and electrons energy types What are the types of energy that an atom or a molecule could have?
For example they have kinetic energy, could they also have other types?
| The Lagrangian $\mathcal L$ and Hamiltonian $\mathcal H$ are mathematical objects that can be used to describe the behaviour of dynamical systems.
In classical systems the Lagrangian is the Kenetic energy minus the potential energy, whereas the Hamiltonian is the the Kenetic energy plus the potential energy.
Most systems can be fully described by the ammount of energy they have from being in a given position / state (potential energy) and the ammount of energy it has by travelling at a given speed (its kenetic energy).
The potential energy is often expanded for gasses and chemical reactions into different types of potentials such as each of the chemical potential.
There also exists internal energy $E=mc^2$ which may be overlooked as usually it does not contribute to the dynamics of the system since it usually does not change, and dissappears when taking the derivative.
In the Standard model there exists an equation describing everything that we know about particle physics. Here is an image of the Lagrangian, each term being added together is an energy term, the energy that a particle can have from each of the different fields.
| {
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The shape of speaker cones This is related to another question I just asked, but they are different enough I thought it deserved its own spot. Speaker elements seem to always be shaped like a cone with a portion of a sphere at the center. What is the physics reasoning behind this shape being so standard? Is it a precise shape that all speaker elements share (e.g. parabaloid + sphere), or just kind of a general shape that "works well"? Does it work just as well for any light solid, or is it actually tuned to the speed at which sound propogates through the speaker element?
| I believe it's to do with the fact that the speaker's function is to propagate pressure waves through the medium (air).
So, it's mainly a mechanical concern: you want something to push air, and you do not wish to expend much energy. So it has to be light and rigid, which the cone manages to fulfil due to its shape. A plane sheet, for instance would undergo buckling. [1]
The size of the cone depends on the wavelength you wish the speaker to transmit. Deep cone for low frequencies, shallow ones for higher frequencies, and curvilinear to cover a range [2].
As for the spherical portion, it's just a dust cap, and I have never come across literature claiming a 'significant' effect due to its shape.
Note: alls of this applies to dynamic speakers. Others, such as electrostatic speakers use different principles.
[1] http://www.vias.org/crowhurstba/crowhurst_basic_audio_vol1_044.html
[2] http://www.ewp.rpi.edu/hartford/~ernesto/SPR/Miller-FinalReport.pdf
| {
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If an object moved 5 meters in a second, how can its velocity be 10 m/s? On physicsclassroom.com, an assignment gives data for an object experiencing acceleration:
This is the corresponding chart for the object's velocity:
At $0$ seconds the position of the object is $0$ meters. And the velocity is $0$ m/s which makes sense. After $1$ second, the object has traveled $5$ meters. The chart says its velocity is $10$ m/s at this time.
How is this possible if the object only travelled $5$ meters within that second? Shouldn't the velocity be $5$ m/s and not $10$ m/s?
| One must distinguish between instantaneous velocity and average velocity.
| {
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Vertex operator - state mapping in Polchinski's book In Polchinski's textbook String Theory, section 2.8, the author argues that the unit operator $1$ corresponds to the vacuum state, and $\partial X^\mu$ is holomorphic inside couture $Q$ in figure 2.6(b), so operators $\alpha_m^\mu$ with $m\ge0$ vanishes.
I am a bit confused about why $\partial X^\mu$ has no pole inside the contour. Before this section $\partial X^\mu$ always has the singularity part ($1/z^m$). Therefore would it be possible for you to give a more mathematical argument what condition requires $\partial X^\mu$ having no poles in this case?
| The main point is that the operator-state correspondence maps all the annihilation operators to zero, so that an operator-valued Laurent series in $z$ and $\bar{z}$ maps to a ket-state-valued power series in $z$ and $\bar{z}$.
| {
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Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.
| The key is... the closer the mass to the axis of rotation, the easier it is to add angular velocity to the body.
For instance a figure skater rotates faster when she puts her limbs closer to her body.
Let's see how it works in a more intuitive fashion:
For instance, in the figure bellow, trying to lift up table (A) would be easier than table (B).
In both cases the mass of each individual box is the same, but in (A) you have a better lever because of the distance from the border where the force is being applied, to each box.
Therefore, table (B) would be harder to lift up, even when R (length of the table) and M (total mass of the four boxes) are the same.
Now let's see how it works in the case with the spheres:
*
*Let's make the sphere a disk, and then divide it in pieces.
*Make the center of mass of the disc fixed, and move all the pieces to
one side.
*Now we have a similar scenerario to the one with the tables.
Both spheres, the solid and hollow one, rotate around their center of mass in the same way that the table rotates around the legs at the opposite side to where the force is being applied.
To make sense of step 2, where the mass of the all pieces is collapsed, think on a Merry-Go-Round where all the kids move to one side keeping their distance to the axis of rotation fixed.
| {
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Maximum Weight Sustained by wire / Breaking limit of wire I know that breaking stress depends on material of wire, not on the diameter or the length of wire. and also that Breaking load depends on the area of cross section of wire. But still i have a confusion in this question.
a steel wire can sustain at the most 100 kg weight without breaking.if the wire is cut into equal parts , each parts can sustain at the most a weight of a) 400 kg b)100 kg c) 50 kg d) 200kg
Here the area of cross section remains the same so the breaking load also remains the same. And The lenght is halved but breaking stress does not depend on lenght of wire so it also remains the same. Does it mean that Both the halved wires can still sustain 100kg? Please correct me if i am wrong.
| You are correct: both halves of the wire will still support the same weight. Look at it this way: the tension in each bit of the wire is the same regardless of whether the weight is hanging directly from it or suspended by another bit of wire in the middle. To put it in terms of stress, the force remains the same and the area remains the same, so the applied stress is also constant.
| {
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Eigenfunction associated with the $\hat{x}$ operator Consider the following operator $\hat{x}=i\hbar \frac{\partial}{\partial p}$.
I am trying to show that the eigenfunctions of $\hat{x}$ are not square-normalizable. I am interested in doing so since theoretically, we notice that the eigenfunction of the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$ is not square-normalizable. How do we set up the equation for the $\hat{x}$ operator?
| The eigenvalue equation
$$\tag{1} \hat{x}\psi(x)~=~x_0\psi(x)$$
in the standard Schrödinger position representation
$$\tag{2} \hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x},$$
reads
$$\tag{3} (x-x_0)\psi(x)~=~0,$$
which has general solution
$$\tag{4} \psi(x) ~\propto~ \delta(x-x_0). $$
| {
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What is decay associated spectra? What is decay associated spectra?
Suppose we measure the fluorescence intensity over different wavelengths and over time, we get:
$$I(\lambda,t) = \sum_i^n \alpha_i(\lambda) \exp(\frac{-t}{\tau_i}).$$
The assumption is that there are n component,species, in the $I(\lambda,t)$. If we fit the right hand to the experimentally obtained $I(\lambda,t)$, and get $\alpha_i$ and $\tau_i$, then people call $\alpha_i$ the decay associated spectra.
Now, if we integrate over time we get the steady state emission spectra. The thing that I cannot understand is the decay associated spectra. What does it mean? If it is the steady state spectra of species i, then why does it become negative sometimes? People, say that when it becomes negative, it indicates energy transfer between species. Could someone please elaborate this concept more?
| Originally the term DAS (rather than LAS) was meant to also include multiexponential decay functions (e.g. the rise then fall of an acceptor species) -- but it was used so often for diagonal (heterogeneous mixture) work that the term SAS was subsequently proposed to distinguish the off-diagonal dynamics spectra from those of ordinary decay lifetimes. SAS were then obtained from global target analysis.
| {
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"timestamp": "2023-03-29T00:00:00",
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Explain the microscopic nature of Electric current? Explain the microscopic nature of Electric current?i.e
What is is average current and Instantaneous current? A microscopic view what really happens?
| Electrical current can be carried by conduction electrons, or by 'holes'.
For ordinary matter, there is roughly one electron per two daltons of matter, which is to say, $\frac 12 N_Ae$. This roughly works out at $480*10^6$ coulombs per kilogram. Many of these electrons are bound in the inner orbitals, but there are still plenty of conduction electrons per kilogram, that a 10-ampere current is not going to move in a day.
The bulk of electric current is then carried by random jumps of a very tiny amount of conduction electrons, which slush through the metal, or a similar process of an absence of electrons (holes), bubbling the other way through the metal, under the effect of $E$ along the conductor. The exact nature of the carrier is found by the Hall effect.
The field, on the other hand, travels very fast. But it is a field, not a flow of electrons. This mainly happens, from each segment of the wire, and is replenished by similar currents started in the previous segment of wire.
The energy is carried outside the conductor, in the poynting vector. The purpose of the conductor is to shape these fields.
| {
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Eigenenergies and eigenkets given the Hamiltonian For a two level system the Hamiltonian is:
$$
H=a(|1\rangle \langle1|-|2\rangle\langle2|+|1\rangle\langle2|+|2\rangle\langle1|)
$$
where $a$ is a number with the dimension of an energy.
I need to find the energy eigenvalues and the corresponding eigenkets (as a combination of $|1\rangle$ and $|2\rangle$).
I used:
$$
H |ψ\rangle=E|ψ\rangle
$$
And using the fact that : $|a\rangle = \sum_i c_i |a_i\rangle$
I wrote $|ψ\rangle$ as a combination of the two system kets $|ψ\rangle=c_1|1\rangle + c_2|2\rangle$ ($c_1$,$c_2$ are complex numbers).
so
$$\begin{aligned}
H|ψ\rangle =& a(|1\rangle\langle1|-|2\rangle\langle2|+|1\rangle\langle2|+|2\rangle\langle1|)\cdot(c_1|1\rangle+c_2|2\rangle)
\\ =& a(c1|1\rangle-c_2|2\rangle+c_2|1\rangle c_1|2\rangle)=a ((c1+c2)|1\rangle+(c1-c2)|2\rangle)
\\ =& E|ψ\rangle.
\end{aligned}$$
How do I continue?
| Finding eigenvalues of matrices is a straightforward process, so to solve this problem we'll begin by writing the Hamiltonian in a matrix form in the basis of $|1\rangle$ and $|2\rangle$.
To find the matrix form of any linear transformation in linear algebra, we can apply the transformation to the basis vectors. In our case, we find $H|1\rangle$ and $H|2\rangle$:
$$\begin{aligned}
H|1\rangle &= a(|1\rangle \langle1|-|2\rangle\langle2|+|1\rangle\langle2|+|2\rangle\langle1|)|1\rangle \\
&= a(|1\rangle \langle1|1\rangle -|2\rangle\langle2|1\rangle + |1\rangle\langle2|1\rangle+|2\rangle\langle1|1\rangle)
\end{aligned}$$
Since $|1\rangle$ and $|2\rangle$ are orthonormal basis vectors, we know that the inner product of two different vectors is 0 and the same vectors is 1. We can use this fact to greatly simplify the above:
$$a(|1\rangle \langle1|1\rangle -|2\rangle\langle2|1\rangle + |1\rangle\langle2|1\rangle+|2\rangle\langle1|1\rangle) = a(|1\rangle + |2\rangle)$$
A similar process reveals that
$$H|2\rangle = a(|1\rangle - |2\rangle) $$
Now, we can write the Hamiltonian as a matrix in the basis provided:
$$H = a \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}$$
Find the eigenvectors of this matrix to determine the eigenkets.
(By the way, you may notice some parallels between the given Hamiltonian expression and calculated Hamiltonian matrix. Think about how this could be used to expedite the process of finding the Hamiltonian matrix for problems in this format.)
| {
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Is the uncertainty principle axiomatic or derived? To take an example,
Feynman Lectures Vol 3 13-1
Let's think of an electron which ban be in either one of two positions [...] There are two possible states of definite energy for the electron. Each state can be described by the amplitude for the electron to be in each of the two basic positions. In either of the definite-energy states, the magnitudes of these two amplitudes are constant in time, and the phases vary in time with the same frequency. On the other hand, if we start the electron in one position, it will later have moved to the other, and still later will swing back to the first position.
(Feynman derives this behaviour from first principles in chapter 8-6 by solving equations for the state of an ammonia molecule, but that's a nice summary).
This sounds like the uncertainty principle - that you can know energy or position but not both. Is it possible to derive the uncertainty principle from such an analysis? Or is the uncertainty principle somehow axiomatic in the way this behaviour is derived?
| There are several mathematical proofs for the Uncertainty Principle, although it is also based out of intuition. A good, fundamental proof of the mathematical sort is found here: http://www.tjhsst.edu/~2011akessler/notes/hup.pdf.
| {
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How do greenhouse gases trap heat? I am looking for a molecular-level understanding of the greenhouse effect.
What is it about the carbon-dioxide molecule (and methane, and water, etc) that is different from other gasses (particularly, N2 and O2) such that it works in the atmosphere to trap heat?
Is it, say, the distance between nuclei in the molecules relative to the wavelengths of infrared light? Dipolarity of the molecule? A combination of various factors?
| To absorb infrared light, a stretching or bending vibration of the molecule must change the molecule's dipole moment. In $N_2$ and $O_2$ there is no dipole moment regardless of how you stretch the bond. On the other hand, O=C=O can change dipole moment by the C moving toward one O and away from the other O, or by bending with the C becoming a vertex of an obtuse angle. Water and methane molecules can also change dipole moment.
| {
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Why does foam dull knives? I have recently taken up prop making and just started my first foam-built costume from a video game. These kinds of costume armour builds are often built out of the various foam floor mats you can buy in say Home Depot or BJ's or Five Below for around $1 for 4 sq. foot tile.
The one thing that puzzles me is why foam dulls a sharp steel blade so easily and so quickly.
I have tried x-acto knives, those snap-off knives, even surgical scalpels. Regardless, after cutting several linear feet (maybe in the neighborhood of 5-10?), the knife blade begins to dull rapidly and no longer produces nice cuts, but begins to tear the foam instead. You can always sharpen your blade on a stone, but in short order it will again dull and tear.
I generally do all my cutting on one of the green, self-healing cutting mats. This could explain the tip dulling, but it is actually the whole length of the blade exposed to the foam that gets dull.
It happens across types, thicknesses and densities of foam. The floor mats are about 0.5
" and medium density. The craft foam from Michaels is 6mm, high density and rigidity. The roll I have is 0.25" low density, low rigidity
It absolutely baffles me how soft squishy foam just destroys the edge of a metal knife.
I am guessing it has something to do with the molecular organization of the foam or something, but would love to know if anyone has a good explanation for why and how this happens.
| Having previously done academic research on various aspects of knives, I agree with @MSAlters - the actual polymer you're cutting is quite tough. The practical solution is to do what professional knife-users, like butchers, carpet layers, whatever, do: run the blade through a hand-held sharpener after every few strokes. I now do this when indulging hobbies, whether cooking or carpentry. Works!
| {
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Which angle should it be? in the formula
$$dB = \frac{\mu_0l ~|dl \times r|}{4 \pi r^3} $$
and the image
where dl is in y-z plane and dB is in x-y plane. the ring conductor is in y-z plane carrying current I in direction as mentioned
EDIT: also point p can move in the circular ring
EDIT 2:To clear the confusion...The dl vector is having (L alphabet) and current is I (i alphabet).
I want to know that is the angle between dl and r is 'Theta' ? how?
| the angle between $dl$ and $r$ is $\pi/2$, which is not $\theta$.
$\theta$ is the angle between $r$ and y-z plane. if you know what is $x$ and $r$, then $\sin\theta=x/r$, where $x$ is the distance from the origin to the point of intersection of $r$ and x-axis, and $r$ is the distance from p point to the same place
UPDATE.
$\theta$ is important because your $dB$ is at this angle to x-axis. so when you add up all $dB$ resulting from all points p, only the $\cos\theta dB$ parts will contribute, because the part of $dB$ which is in y-z plan will cancel each other. for each point p, there's is an opposite point on the ring, they cancel each other's $dB$ on y-z plane, but not on x-axis. that's why resulting $B$ will be on x-axis.
| {
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Divergent path integral What does it mean to have a divergent path integral in a QFT?
More specifically, if
$$\int e^{i S[\phi]/\hbar} D\phi (t)=\infty $$
What does this mean for the QFT of the field $\phi $?
The field $\phi$ has action
$$S[\phi]=\int\left(\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)\right)\mbox{d}vol$$
where we use Minkowski signature $(+,-,-,-)$.
| If the path integral itself diverges, it means that the v.e.v. diverges. That by itself is bad, because then any arbitrary $n$-point function vanishes. Recall that to compute correlation functions, we append a $J(x)\phi(x)$ to the action and calculate
$$
\frac{\delta^n}{\delta J(x_1)\ldots \delta J(x_n)} \int e^{i S[\phi]/\hbar + J(x)\phi(x)}\mathcal{D}\phi = \langle\phi(x_1)\ldots \phi(x_n)\rangle
$$
which is normalized by the v.e.v.. Thus, you wouldn't be able to calculate anything sensible.
e.g. a v.e.v. might diverge when upon Wick-rotating to Euclidean time, the action might be unbounded from below (as @Alex points out)- that would typically happen when the potential is bad.
| {
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Why is a wine glass shaped the way it is? Why is a wine glass shaped the way it is? And why are there different shapes for different wines? Is this a tradition, or is there any scientific reason behind it?
| A little bit of physics - but mostly just fashion.
Red wines (supposedly) need to oxidize to release the flavo(u)r so red wine glasses have wider necks to allow more air. White wines don't and so the glasses have narrower necks.
If there really was a significant difference then you would drink white wines from a sealed pouch through a straw and shake red wines like a cocktail. So it's probably now more a question of fashion and pretend insider knowledge.
| {
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Does a mirror help a near-sighted persion see at a distance clearer? A near-sighted person without eye-glasses can not clearly see things at distance.
If he takes a photo of the things at distance, he can see the things from the photo much clearer, because he can place the photo much closer to his eyes.
If he turns his back at the things at distance, and holds a mirror close to his eyes in a position so that the mirror reflects the things at distance behind him, will he see the things much clearer than if he looked at the things at distance directly?
| Not necessarily. It depends on where the image is formed from the mirror. Depending on the radius of curvature (assuming spherical curvature) the image ould form anywhere, but the person would want it to form on their retina. You can calculate this using the mirror equation
$\frac{1}{d_{0}}+\frac{1}{d_{i}}=\frac{1}{f}$
Where $d_{0}$ is the distance (from mirror) to the object, $d_{i}$ is the location of the image (w.r.t to mirror), and $f$ is focal length.
So it does not matter how close the mirror is to the persons face, it only matters where the image is formed.
| {
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Coupling constant is turned off adiabatically? To me, adiabatic processes are idealisation. What do people mean with statements such as: "turning off the coupling constant (in QED say) adiabatically"?
| In general it means varying (or turning off an interaction in your particular question) a parameter on a time scale that is much larger than the smallest energy separation of your Hamiltonian. More explicitly : Suppose you have a Hamiltonian $H$ with energy levels $E_n$ and suppose that $\left|E_a-E_b\right|$ ($a\neq b$) is your smallest level splitting (assume no degeneracy for simplicity). Then suppose you turn on some knob to include a magnetic field in your system : $H_{\text{total}}=H+H_B(t)$, where $H_B(t)$ is the time dependent part of your Hamiltonian that couples to the magnetic field $B(t)$. The adiabatic limit is when the time scale, $\tau$ over which you turn on $H_B$ satisfies $\tau\gg\hbar/\left|E_a-E_b\right|$.
Of course this is all discussed in more details for instance on the wikipedia page.
| {
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Why isn't the electrical field between two parallel conducting plates quadrupled? For the sake of clarity and brevity, here is the description from my textbook (Halliday & Resnick 9th):
This description states that the electric field is doubled. I am seriously confused. When the two plates are brought together, shouldn't the field quadruple? The surface density doubles on each inner surface, doubling their generated electric field. Additionally, these two fields should obey superposition rules of vectors, doubling the net-field, for a quadrupled total?
| Since the plate is a conductor the surface charge is spread across both sides of plate (see fig 23.16 a or b). Hence this gives that the electric field in between the two plates is actually
$$E=\frac{\sigma}{2\epsilon_{0}}$$
This comes from Gauss' Law $\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enclosed}}{\epsilon_{0}}$
Now, the electric field for two capacitors of $\sigma$ surface charge has an electric field inside of
$$E_{1}=\frac{\sigma}{2\epsilon_{0}}-\left(\frac{-\sigma}{2\epsilon_{0}}\right)=\frac{\sigma}{\epsilon_{0}}$$
So now if we double the surface charge we get
$$E_{2}=\frac{2\sigma}{2\epsilon_{0}}-\left(\frac{-2\sigma}{2\epsilon_{0}}\right)=\frac{2\sigma}{\epsilon_{0}}=2E_{1}$$
| {
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Focused beam of light I'm trying to understand what happens to photons when a beam of light is focused down to its waist. In the image attached, do photons take the path 1 or 2. That is, do the photons cross or just get 'deflected'.
|
do photons take the path 1 or 2. That is, do the photons cross or just get 'deflected'.
They take path 1 and travel in straight lines.
However due to the finite effective size of the photon and Heisenberg Uncertainty Principle it is not possible for all of the lights straight lines to pass through the centre point.
See above image (ignore the circle and dot) for a series of rays passing through as close to the centre point as possible before diverging again.
| {
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How does a half-life work? Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?
| Imagine a sample of 1000 atoms with a half-life of 1 hour.
That means every hour, the sample is reduced to 50% of its size.
After one hour, you are left with 500 atoms.
How much time for that new sample (500 atoms) to be reduced to 50% (250 atoms) ?
In your interpretation :
For the new sample to be reduced to 50%, it needs to lose 250 atoms.
Since it lost 500 atoms in 1hour, it should take 30 minutes to lose 250 atoms. And that's where you're wrong. It still needs 1 hour for half of the atoms to decay.
You assume that the number of atoms decaying by time (500/hour) is constant, but it isn't.
What is constant is the probability for each atom decaying in an hour : 50%
(In this exemple, that means we can expect 250 atoms to have decayed after 1 hour, and it gets way more precise with "real" atoms number on a longer period)
| {
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How to differentiate between ferri-magnetic and para-magnetic substances? Both are feebly attracted by a magnetic field. I know the difference between these substances(on why these get attracted to magnetic field) but as both get feebly attracted how to find out whether a given substance is ferri-magnetic or para-magnetic?
Is it that all ferri-magnetic materials are compounds of different metals like magnetite and we can differentiate in this way?
| Refer NCERT class 12 book 1 chemistry...Solids chapter!
It's that ferri magnetic have some moment..but the net moment of all particles cancels out!
In paramagnetic...it has unpaired electrons..so there is a moment!
| {
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In calculating resistivity of skin do I use the body's surface area?
Most of the resistance of the human body comes from the skin, as the interior of the body contains aqueous solutions that are good electrical conductors. For dry skin, the resistance between a person’s hands is measured at typically $500\: \mathrm{k \Omega}$. The skin varies in thickness, but on the average it is about $2.0\: \mathrm{mm}$ thick. We can model the body between the hands as a cylinder $1.6\: \mathrm{m}$ long and $14\: \mathrm{cm}$ in diameter with the skin wrapped
around it.
(a) What is the resistivity of the skin?
So for this, I used the equation $R = (\rho*L)/A$. For Area, should I use the entire surface area of the body a.k.a cylinder?
| In general, when formulae call for "area" they are talking about the surface area on which the force or whatever is occurring. Note that this is not always the entire surface area, it might just be the curved part of a cylinder for example. This is one such example, because we actually aren't interested in the end effects at the bases of the cylinder.
| {
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Naive visualization of space-time curvature With only a limited knowledge of general relativity, I usually explain space-time curvature (to myself and others) thus:
"If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again as it approaches the ground.
"In reality, the ball in moving in a straight line at constant velocity, but the space-time curvature created by the Earth's gravitation makes it appear as if the ball is moving in a curved line at varying velocity. Thus the curvature of space-time is very much visible."
Is this an accurate description, or is it complete nonsense?
| Yes, that's a fair description of what happens though of course from the ball's perspective it isn't moving - the rest of the universe is moving around it.
However statements like this, while true, give little feel for what's going on. Actually it's extraordinarily difficult to get an intuitive feel for the way spacetime curvature works (or at least I find it so!). The notorious rubber sheet analogy gives a fair description of the effect of spatial curvature, but neglects the curvature in the time coordinate and the time curvature is usually dominant since $dt$ gets multiplied by $c$ in the metric.
The motion of the ball is described by the geodesic equation, but a quick glance at the article I've linked will be enough to persuade you this is not an approach for the non-nerd. I have never seen an intuitive description of how the geodesic equation predicts the motion of a thrown ball.
| {
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Renormalizing IR and UV divergences In lectures on effective field theory the professor wanted to find the correction to the four point vertex in massless $\phi^4$ theory by calculating the diagram,
$\hspace{6cm}$
We consider the zero external momentum limit and denote $p$ as the momentum in the loop. Then we get,
\begin{align}
\int \frac{ d ^d p }{ (2\pi)^4}\frac{1}{p ^4 } & = \frac{ - i }{ 16 \pi ^2 } ( 4\pi ) \Gamma ( \epsilon ) \mu ^\epsilon \\
& = - \frac{ i }{ 16 \pi ^2 } \left( \frac{1}{ \epsilon _{ UV}} - \gamma + \log 4\pi - \log \mu ^2 \right) \\
& = \frac{ i }{ 16 \pi ^2 } \left( \frac{1}{ \epsilon _{ UV}} - \frac{1}{ \epsilon _{ IR}} \right)
\end{align}
where we introduced $\mu$ as an IR cut-off and then take $\log \mu ^2 $ as a $\frac{1}{\epsilon_{IR}}$.
This is fine, however the professor then goes on to say that this diagram is zero since the two divergences cancel. Why would this be the case? The two divergences arise for completely different reasons. The UV divergence is due to a UV cutoff (possibly from new high energy particles arising at some high up scale) and the second is a consequence of studying a massless theory.
For more context the lecture notes are available here under Effective Field Theory (Eq. 4.17)
| I think you misunderstood what the professor wanted to say. To understand this, let us evaluate the integral more thoroughly (your expressions contain some mistakes). If we use the dimensional regularization prescription $d\rightarrow d-2\epsilon$ and an additional mass scale $\mu$, we get for the integral in question the following result:
$$\int \frac{d^{d-2\epsilon}p}{(2\pi)^{d-2\epsilon}}\frac{1}{(p^2+\mu^2)^2}=\frac{\Gamma(2-d/2+\epsilon)}{(4\pi)^{d/2-\epsilon}}\mu^{-2(2-d/2+\epsilon)}.$$
For $d=4$ we get
$$\int\frac{d^{4-2\epsilon}p}{(2\pi)^{4-2\epsilon}}\frac{1}{(p^2+\mu^2)^2}=\frac{\Gamma(\epsilon)}{16\pi^2}\left(\frac{\mu^2}{4\pi}\right)^{-\epsilon}.$$
Expanding this at $\epsilon\rightarrow 0,$ we arrive at
$$\int\frac{d^{4}p}{(2\pi)^{4}}\frac{1}{(p^2+\mu^2)^2}\approx\frac{1}{16\pi^2}\left[\frac{1}{\epsilon}-\gamma+\log(4\pi)-\log(\mu^2)\right].$$
In the massless limit, i.e. $\mu\rightarrow 0$, the logarithm diverges. So what can we say about the nature about this divergence?
As can be concluded from powercounting, a positive $\epsilon$ corresponds to curing UV divergences, while a negative one cures IR divergences. First, let us assume that that we deal with UV divergences and identify $\epsilon=\epsilon_{UV}.$ What can we say about the remaining divergent term? We can observe that the whole integral has to vanish (which is proven earlier in the lecture), and this happens only when the divergent term is equal to minus the $1/\epsilon$ term, i.e.
$$\frac{1}{\epsilon_{UV}}=\gamma-\log(4\pi)+\log(\mu^2).$$
Next, let us assume at we are dealing with divergences from the infrared, and identify $\epsilon=\epsilon_{IR}.$ We now have to observe that evaluating the integral gives us just the same result, but with $\epsilon_{UV}$ and $\epsilon_{IR}$ exchanged. The condition for vanishing of the integral is now
$$\frac{1}{\epsilon_{IR}}=\gamma-\log(4\pi)+\log(\mu^2).$$
But the right hand side is just the same as in the condition for the UV! This means we actually get
$$\epsilon_{UV}=\epsilon_{IR}.$$
As the lecturer has pointed out, this can be interpreted as dimensional regularization "taming" both the UV and the IR simultaneously.
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electron levels in a high voltage conductor what is the electron energy level in a 300,000 volt power line? This voltage is way above the ionisation potential but electrons are not emitted from the wire.
| (As far as I know, 300kV is the exclusive domain of HVDC so I will not mention AC here.)
Ionization potential is the energy difference between a free electron at rest outside an atom, and a particular electron energy state inside the atom. If you zap a bound electron with enough energy, it can assume a new, unbound wavefunction and travel away.
But, tunneling effects aside, that wavefunction is still localized to the same general place, which is inside the insulator (or air) surrounding the wire. If the electric potential at that point in space is little different from inside the metal, ionization is not energetically favorable.
Consider the instant the power line is energized. Suppose it immediately jumps to –150kV (relative to its surroundings, in equilibrium with the earth). Electrons will be torn from its surface, and migrate outward. Upon encountering air molecules, they form negative ions which migrate outward as a wind. The wind encounters physical resistance, which sets up a strong electric field. Over a little while the system reaches a dynamic equilibrium, because the air must support an electric field to keep the electrons inside the wire. (Or, for the positive line, to keep electrons out.)
If the wire is insulated, the insulator will contain some of the electric field as well, and provide an unfavorable environment for free electrons (raising the ionization potential).
In fact, this effect is called corona discharge and it is a major source of loss for HVDC transmission.
| {
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Schrödinger's Equation and its complex conjugate I would like to know why there is a minus sign on the right-hand side of the Schrödinger's complex conjugate equation, whereas in the Schrödinger's equation there isn't. I know it is a simple question, but I don't know where this comes from.
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t}
$$
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t}
$$
| Just because $\psi$ is a solution to the Schrodinger equation doesn't mean its complex conjugate is. After all, the Schrodinger equation is a pretty strong restriction, and not any random function is a valid wave function. So while it may be tempting to just substitute $\psi^*$ for $\psi$ in the Schrodinger equation, that's not something you're allowed to do.
But something you can do is take any arbitrary equation $A=B$ and take the complex conjugate of each side to conclude $A^*=B^*$. If you do that to the Schrodinger equation, you end up replacing all the $\psi$s with $\psi^*$s, leaving all the real bits alone (most of the constants and $V$)... and replacing $i$ with $-i$.
| {
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Bogoliubov transformation with a slight twist Given a Hamiltonian of the form
$$H=\sum_k \begin{pmatrix}a_k^\dagger & b_k^\dagger \end{pmatrix}
\begin{pmatrix}\omega_0 & \Omega f_k \\ \Omega f_k^* & \omega_0\end{pmatrix} \begin{pmatrix}a_k \\ b_k\end{pmatrix}, $$
where $a_k$ and $b_k$ are bosonic annihilation operators, $\omega_0$ and $\Omega$ are real constants and $f_k$ is a complex constant.
How does one diagonalise this with a Bogoliubov transformation? I've seen an excellent answer to a similar Phys.SE question here, but I'm not quite sure how it translates to this example. Any hints or pointers much appreciated.
| Hamiltonian is already diagonalized by momentum. You need to define new Bose-operators
$c_k = u_k a_k + v_k b_k \\
d_k = w_k a_k+x_k b_k $
This is general form, with some complex constants $u_k, v_k, w_k, x_k$ for each $k$ independently. There are also $c^+_k$ and $d^+_k$, conjugated with previous one. Now you need $c_k$ and $d_k$ correspond to some quasi-particles, so
$[c_k, c_k^+] = 1 \\
[d_k, d_k^+] = 1 \\
\text{(all other commute to zero)} $
This equation give you some constraint on constants $u_k, v_k, w_k, x_k$. But to find them definitivly, you must substitute them to hamiltonian. After that, you must obtain
$ H = \sum_k C_1 c^+_k c_k + C_2 d^+_k d_k + C_3 c^+_k d_k + C_4 d^+_k d_k. $
Constants $C_1, C_2, C_3, C_4$ derived from $\omega_0, \Omega, f_k$ and $u_k, v_k, w_k, x_k$. You must then solve $C_3 = 0, C4 = 0$ equations to obtain $u_k, v_k, w_k, x_k$. Then you'll get
$ H = \sum_k C_1 c^+_k c_k + C_2 d^+_k d_k, $
with found $C_1, C_2$. That completes diagonalization.
| {
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Field from non-conducting plate?
For a non-conducting sheet, the electric field is given by:
$$E = \frac{\sigma}{2\epsilon_0}$$
where $\sigma$ is the surface charge density.
This equation holds well for a finite nonconducting sheet as long as we are dealing with points close to the sheet and not too near its edges.
Why does the equation hold better with points closer to the sheet? I understand why the approximation worsens near the edges (because symmetry fails and causes fringe effects) but why is the approximation better near the sheet?
| Consider a square sheet with edges located at $(a,0)$, $(-a,0)$, $(0,a)$ and $(0,-a)$. Suppose, we wish to find the electric field at a point $(0,0,z)$. By symmetry, this electric field will point solely in the $z$-direction. To find the electric field, consider a small element on the sheet located at $(x,y)$ of area $dx dy$. The charge of this element is $\sigma dx dy$. The magnitude of the electric field at $(0,0,z)$ due to this element is then (treating the element as a point charge)
$$
dE = \frac{1}{4\pi \epsilon_0} \frac{\sigma dx dy}{x^2 + y^2 + z^2 }
$$
The $z$-component of this electric field is
$$
dE_z = \frac{1}{4\pi \epsilon_0} \frac{\sigma z dx dy}{\left( x^2 + y^2 + z^2 \right)^{3/2}}
$$
Integrating this over the sheet, we find the total electric field at $(0,0,z)$ as
$$
E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{a^2}{z \sqrt{ 2a^2 + z^2 } } \right]
$$
Let us now take the limit of small $z$. However, $z$ is a dimensionfull quantity, and you can't discuss the largeness or smallness of dimensionfull quantities, only dimensionless numbers. The only dimensionless number that I can construct using $z$ is $\frac{z}{a}$. So, when I say, $z$ is small, I really mean $\frac{z}{a}$ is small.
In this limit, we find
$$
E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{1}{(z/a)\sqrt{ 2 + (z/a)^2 } } \right] = \frac{\sigma}{2\epsilon_0} + {\cal O}(z/a)
$$
Thus, when we are sitting close to the sheet, the field takes the form you described above.
But, here's the important thing. We didn't really care if $z$ itself is small (that sentence doesn't even make sense). What we really care about is if $z/a$ is small. Now, there are two ways to make this small -
*
*Make $z$ small compared to $a$, i.e. move in very close to the sheet.
*Make $a$ large compared to $z$, i.e. make the sheet very very large.
Both the statements above are completely equivalent.
| {
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Superconducting wire in a Magnetic Field? A superconducting wire($SC$) is moved rapidly in a magnetic field( $1$ $Tesla$), what would happen to the wire? Are there any forces induced of attraction or repulsion?
In a typical conductor, we know that if it is moved around a magnetic field $-V$ is induced within the wire based on Faraday's law, however, with the condition of the $SC$ what could happen if $R = 0$ $ohms$?
Will Faraday's law still be applied to that wire with no resistance?
moving a $SC$ in a magnetic field will not induced $EMF$?
| I suppose that the moving wire is a closed circuit and that the magnetic flux enclosed is time-dependent. The Faraday’s law is of course always applicable. The current will not be infinite. Yes R=0 but, what about the self inductance L? It is never zero, in such a manner that the total E field will be null. If you make the calculations the electric field in the superconducting wire has two contributions. The first one related to the variation of the external magnetic flux and the second associated to the time-dependent current by means of the self inductance L.
| {
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Killing vector contractions along isometric curves Imagine $\xi_{\nu}$ is a Killing vector field on a manifold. Does $\xi_{\nu}\xi^{\nu}$ remain constant along any isometric curve defined by the Killing vector field?
My guess is that yes since as you move along an isometric curve every point "around" you looks pretty much as it did a differential step ahead, but I am don't have a full grasp on this and would thank some clarification.
| Let $\lambda$ be an affine parameter of the integral curves of $\xi^{\nu}$ then you question translates as
$$
\frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = \xi^\mu \nabla_\mu(\xi_{\nu}\xi^{\nu}) = (\xi^\mu \nabla_\mu\xi_{\nu})\xi^{\nu} + \xi_{\nu}(\xi^\mu \nabla_\mu\xi^{\nu})
$$
if the connection is Levi-Civita (i.e metric compatible)
$$
\frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = 2\xi^{\nu}\xi^\mu \nabla_\mu\xi_{\nu}
$$
but now since $\xi^\nu$ is a Killing vector and it satisfies Killing's equation
$$
\nabla_{(\nu}\xi_{\mu)}=\nabla_\mu\xi_{\nu} + \nabla_\nu\xi_{\mu} = 0
$$
and since $\xi^{\nu}\xi^\mu = \xi^{(\nu}\xi^{\mu)}$ is symmetric we have that
$$
\frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = 2\xi^{\nu}\xi^\mu \nabla_\mu\xi_{\nu} = 2\xi^{(\nu}\xi^{\mu)} \nabla_{(\mu}\xi_{\nu)}=0
$$
| {
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Measurements for thermal diffusivity of graphene? We have known for a long time that graphene has in-plane thermal conductivity ranging between 2000 and 4000 $W m^{-1} K^{-1}$. But in order to model heat transport on a sheet of graphene, we need more than the conductivity: we also need specific heat in order to obtain the thermal diffusivity that is used in the equation.
I couldn't find any measurement results online for this quantity. I've only seen some crude estimates based on phonon transport, but even so, no specific figures.
| According to this article, the molar heat capacities of graphite and graphene should be identical above roughly 100 Kelvin. The heat capacity of graphene is dominated by phonon contributions above roughly 1 Kelvin (below that, free electron heat capacity becomes a significant contributor).
The molar heat capacity of graphite at room temperature is $8.53\text{ J mol}^{-1}\text{K}^{-1}$.
| {
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How does one exert greater force on the ground by jumping? When one jumps, how does he/she manage to exert greater force on their ground than their weight?
Also, what is normal force and the reaction force (are they the same thing?) and by newton's third law, shouldn't the reaction(weight) when we are standing on the ground that the ground exerts on us send us flying above the ground- why doesn't the law apply here?
Finally, when we drop a hard stone on the ground why doesn't it bounce? Plus, why is the force exerted by the stone on the ground greater than its weight?
| When you jump, your muscles are using their chemical energy to contract or extend appropriately to create an additional force on the ground; by Newton's third law, the ground exerts additional force on you, overcoming your weight and pushing you off the ground.
In this case, the normal force is indeed the reaction force, as it is equal and opposite to (the perpendicular component of) the force you exert on the ground. There is indeed a normal force even when you don't jump, but this is balanced by the force of gravity pulling you towards the Earth. When you jump, these forces are still present, but the force of your jump increases the normal force such that it is greater than your weight (the force of gravity) and there is a net upward force, resulting in a jump.
The reason any object bounces is because it deforms very slightly when it hits the ground, but its natural elasticity sort of "pops out" the deformation (think of a basketball), pushing on the ground in a similar manner to a jump. Stones, however, are terribly rigid and inelastic, so by the time the ground applies enough force to deform it, it tends to simply chip or shatter rather than deform elastically.
I don't know why you say the force of the rock on the ground is greater than its weight, so I'm not sure how to answer that bit.
| {
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When I stretch a rubber band, it breaks. When I hold the broken ends together, why doesn't it join again? The question is simple. When we join the two broken surfaces, what is it that keeps the surfaces from connecting with each other, while earlier they were attached to each other? Also, would the two sides join again if I hold them together for an infinite amount of time?
| Your question is perfectly valid despite most people may think it's odd as reason looks obvious.
There's a property of system called entropy which must not decrease for any process (do some research on web; You may have problem in understanding it as you're in high school). Only those processes in the universe happen in which entropy of system either stays same or increases. For all real-world processes, entropy increases (stays same thing is for ideal reversible processes).
When you touch something hot, heat is transferred from that object to your body. You can argue about that: "Why not opposite?" Your argument would be valid as energy conservation law wouldn't be violated even if heat transfers from your body to that hot object. You can do such argument on almost all real life processes.
In case of your broken rubber band, it won't be attached again to become as before even if you try to undo all energy changes you did at molecular level (holding pieces together is also such try).
The key to all is entropy change. So, the answer to your question: The rubber band won't be attached again automatically because entropy of system would be decreased by that.
And, don't underestimate entropy thing. It is discussed in high-level theories like string theory, QFT. Entropy change is sometimes called "arrow of time".
| {
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Magnetic field inside and outside cylinder with varying current density I am reading through Introduction to Electrodynamics by David J. Griffiths and came across the following problem:
A steady current $I$ flows down a long cylindrical wire of radius $a$. Find the magnetic field, both inside and outside the wire if the current is distributed in such a way that $J$ is proportional to $s$, the distance from the axis.
If $J$ is proportional to the distance from the axis $r$, then we have:
$$\vec{J}(\vec{r})=kr\,\boldsymbol{\hat{z}}$$
We also have:
$$\iint_{\Sigma} \vec{J}(\vec{r})\cdot\:\mathrm{d}\vec{A}=I$$
We therefore have:
$$\int_{0}^{2\pi}\int_{0}^{a}kr^{2}\:\mathrm{d}r\:\mathrm{d}\theta=\frac{2\pi k a^{3}}{3}=I $$
And therefore we have:
$$\vec{J}(\vec{r})=\frac{3Ir}{2\pi a^{3}}\,\boldsymbol{\hat{z}}$$
Using Ampére's law we have:
$$2\pi r B = \mu_{0}I \implies \vec{B}(\vec{r})=\frac{\mu_{0} I}{2\pi r}\,\boldsymbol{\hat{\theta}}$$
If we take a Ampérian loop inside the cylinder, we have:
\begin{align}
2\pi r B &= \mu_{0} \iint_{\Sigma}\vec{J}(\vec{r})\cdot\:\mathrm{d}\vec{A} \\
&= \mu_{0}\int_{0}^{2\pi}\int_{0}^{r}\frac{3 I r'^{2}}{2\pi a^{3}}\:\mathrm{d}r'\:\mathrm{d}\theta \\
&= \frac{\mu_{0} I r^{3}}{a^{3}}
\end{align}
And therefore we have:
$$\vec{B}(\vec{r})=\begin{cases}\frac{\mu_{0} I r^{2}}{2\pi a^{3}} \,\boldsymbol{\hat{\theta}} & r < a \\
\frac{\mu_{0} I}{2\pi r}\,\boldsymbol{\hat{\theta}} & r \geq a
\end{cases}$$
Is this the correct magnitude and direction of the magnetic field?
| Those answers are correct. A first check is to see if the units match. You can always check direction by the right hand rule. The part for outside the wire is the same as if the current were uniform, because the enclosed current is all that matters when you have enough symmetry for Ampére's Law. For the part inside the wire, check to see if the function makes sense: for a uniformly distributed current, the magnetic field grows linearly with the distance from the axis, so it makes sense that for this current it would grow like the square of the distance from the axis.
| {
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Action and reaction pair problem If I sit on a chair, the first action and reaction pair force is my weight and the force acting on the earth by me. And the 2nd pair is the force acting on the chair by me and the opposite force.
According to newton third law, the action and reaction force must be equalI. 'm confused because how come the force acting on the chair by me is equal to normal force?
F=gmm/r2
| Where does the 'pairing' in your first pair come into place? I.e. what is the force counteracting the earth's gravity pull on you (in your theory)?
The relevant force pair in your example is the attractive force between your body and the earth (gravitational pull) and the repulsive force between your body and the chair's surface (its lack of compressability).
You can either see the chair as part of the earth in this scenario OR you can use a force chain in which the repulsive force the earth has on the bottom of the chair transfers via the chair to your body.
It's getting more complicated by the fact that thereby the force the earth has on you is mostly translated into the deformation of your body.
The application of Newton's laws is very much about abstraction and simplification or in other words macroscopic effects, that are in fact the result of a LOT of microscopic effects. (Electromagnetic repulsion vs. attraction on atomic level vs. gravity making up the bulk but not all of the forces ar work here.)
| {
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Creating electricity from mains water pressure. Could someone cleverer than me help me out?
I had a crazy thought going through my head the other day and I can't lay my mind to rest until I get an answer.
Q. How much energy could be produced by using mains water pressure to turn a generator? And would it be feasible to install a system to feed whatever is produced back to the grid? Assuming that the system would be installed in a building where a constant water supply is needed so the generator would be turning continuously, and a rough water pressure of around 3-4 bar.
Thanks in advance for any help
| We pay 3 dollars per cubic meter for water where I live. At 400 kPa (60 psi) that's 400 kJoules per cubic meter maximum theoretical power. But that's only about a tenth of a kilowatt-hour, which costs about a penny at 10 cents/kW-hr. So it's not that good a proposition.
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How jet engine works in vacuum Its not about availability of oxygen for combustion of gases. I want to know since there is no air or friction providing things in vaccum, how the force applied by a jet engine get can make thrust to move the vechile
| Think about the recoil of a gun. The gun shoots some mass (the projectile) with very high momentum into negative direction. Because of the balance of momentum the gun gets a momentum into positive direction.
The jet engine does the same with highly accelerated gas which has also mass and therefore also a high momentum at high speed.
| {
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How is it possible for light to be a wave and a particle? I have always been interested in Physics, and lots of people say that light is a particle and a wave.
How is it possible? How can a photon (a light particle) be a wave as well, when its a particle?
I have tried looking this up, but I do not really understand the answers, as they are quite complicated to me. If you could answer this, please could you explain it in quite a simple way!
| First, we need to ask ourselves what exactly do we mean when we say that something is a wave or a particle. Something is a wave when it oscillates through a medium. Something is a particle when it has a definite size and position at a given time in space.
Now, when photons interact with anything (say to excite an electron in an atom), they behave as if they are a particle. Because that electron has a definite position in space, to excite it, the quanta of light must hit it at that exact point.
But, when light travels through space, it does so in a wave, ie. there is an oscillation of its electric and magnetic fields. The photons are no longer like solid particles travelling through space.
Note: This explanation is from a macro perspective of what it means to be "solid" or "hit something". Also, wave-particle duality in matter is ignored.
| {
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In the Dirac equation, do $\alpha$ and $p$ commute? The Dirac Hamiltonian is given as $H = \vec \alpha·\vec pc + \beta mc^2$ , Do the alpha and beta operators commute with the momentum operator? If yes then how?
| Yes, they do. All three of the $\alpha$ matrices, as well as the beta matrix, are operators on the 'internal' degree of freedom of the electron (that is, its spin and the electron-positron distinction), while the momentum operator operates on its spatial degree of freedom. These are independent degrees of freedom, and operators on their different sectors of Hilbert space always commute.
| {
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Without seeing the lightning, can you tell how far away it struck by how the thunder sounds? Is there any way to tell how far away a lightning strike is by how its thunder sounds? I thought one way might be by using the fact that higher frequencies travel faster than lower frequencies. Would you have to correct for the fact that thunder may not take a straight path? (If so, this would affect the distance calculation based on the time between lightning and thunder as well.)
| I presume you did not see the flash, so cannot use that as a timing mark. From Wikipedia "The dependence on frequency and pressure are normally insignificant in practical applications. In dry air, the speed of sound increases by about 0.1 m/s as the frequency rises from 10 Hz to 100 Hz. For audible frequencies above 100 Hz it is relatively constant. " You might try to use the fact that high frequencies are attenuated more than low frequencies, but that seems very dependent on the terrain.
| {
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How does particles gain electrical charges and repel each others? (electrostatic stabilization) When I study electrostatic stabilization, I understand that the particles have same charge and thus repel others, this is how colloid is stabilize. But how does particles gain electrical charges and repel each other in the first place? I think that there are weak electrostatic force on the particle and when electrolytes were added, it gain more charges. Thank you.
| There are two mechanisms. Which one is more important depends on the colloid.
A large number of colloidal particles have ionisable groups on their surfaces. These are usually salts of carboxylic acids. The vast majority of organic colloids (e.g. milk) are in this class, as are colloids prepared from acidic monomers like the acrylates. In water the surface groups dissociate in exactly the same way that sodium acetate dissociates in water. The cations form hydrated ions in the water and the surface is left covered in negatively charged carboxylate ions.
The second class is where the colloidal particles have no surface groups that can dissociate, but they can adsorb ions from solution. The obvious example is if there is an anionic (or less commonly cationic) surfactant in the solution, but the particles may adsorb much smaller ions like sulphates. Colloids made from polyvinyl acetate/alcohol are probably the most common examples of solid/liquid colloids in this class, and emulsions are the most obvious liquid/liquid colloids.
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Neutron-Antineutron creation/annihilation data What is the most precise data for neutron-antineutron production by one photon (hitting a target in the laboratory system)?
and/or
What is the most precise data for neutron-antineutron
annihilation to two photons?
Is this data available online?
| There is no process $ \gamma \to n $ at all, nor $ \gamma \to n \bar{n}$ with an on-shell photon. The first violated multiple quantum number conservation rules and the second conservation of four-momentum.
The two-photon process
$$ \gamma + \gamma \to n + \bar{n} \,,$$
has allowed quantum numbers but will be exceedingly rare.
It is worth noting that photons do no "annihilate" in the usual sense, and two photon processes imply multiple interaction vertexes.
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What is sound in terms of acoustic sources? Sound is nothing more than small amplitude, unsteady pressure perturbations that propagate as a longitudinal wave from a region in space which created it (called the source region) into a quiescent (still) region where it is observed by hearing.
It is at this point where I get slightly confused in that we may distinguish between two types of pressure perturbations: hydrodynamic pressure perturbations and acoustic pressure perturbations. The acoustic perturbations are what we term sound, as they are characterized by their ability to propagate into the hearing region. The hydrodynamic pressure perturbations could be a consequence of fluid flow simply changing in the source region, and this may not necessarily propagate and become what we term sound.
Perhaps evanescent waves can serve as some insight to what I am trying to explain: with reference to this link the pressure fluctations created by the subsonically pulled wavy plate decay exponentially in the upward direction and so do not constitute a sound.
Is my interpretation so far correct? If so are there any rigorous methods for determining if pressure fluctations will propagate (thus becoming sound), or is sound simply that which we can hear outside a source region? Are there other cases besides the linked example for which pressure fluctations are created, but do not propagate as sound?
| Hydrodynamic perturbations = change in pressure due to a flow velocity (particles don't return to equilibrium positions).
Acoustic perturbations = change in pressure due to the fact the particles undergo an elastic restoring force (for a compressible fluid) which causes perturbations to travel at the speed of sound.
Any change in the pressure/velocity field will experience acoustic perturbations. The hydrodynamic fluctuating velocity therefore causes acoustic perturbations. Lighthill's analogy explains the equivalent source term in the wave equation, which is a weakly radiating quadrupole source for turbulence, depended on the magnitude of the fluctuating hydrodynamic velocity. The acoustic perturbations are of much smaller magnitude than the hydrodynamic perturbations.
If you define sound as the change in pressure at a receiver, then sound = hydrodynamic + acoustic perturbations. The hydrodynamic part only exists within the source region.
| {
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Expectation Value of a Dynamical Variable In quantum mechanics, we generally take about "expectation values of dynamical variables". However, by the postulates of quantum mechanics, every dynamical variable in quantum theory is represented by its corresponding operator.
Is it therefore, incorrect to talk about "expectation value of an operator" (rather than "expectation value a dynamical variable")? Is is just semantics or is something more going on here?
In other words is it incorrect to write:
$$<\widehat{A} > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}$$
instead of
$$<A > =\int \psi \widehat{A}\psi^{*}\mathbf{ d^{3}r}?$$
| Remember, operators are nothing but maps. Expectation value of an operator is pretty much defined (I guess) in general operator theory. It just turns out that in QM (Hermitian) operators correspond to dynamical variables. In general you can also calculate expectation values of operators like $L_+$ and $a^{\dagger}$ etc., which don't have any dynamical variables associated !!
On a more general note, if you want to get a representation for the operator (given some basis, which could be the eigenset of an operator), the expectation values would be the diagonal elements of the representation.
It is related to Linear algebra, This might throw some light on Operator theory. Linear operators (bounded of course) are maps defined in LVS that take one vector to another vector in the same LVS.
$$ \hat O : V \rightarrow V$$
$$ \hat O \left|\psi\right> = \left|\phi\right> $$
Then you have the inner product defined on the LVS, that takes two vectors to one complex number :
$$ \left<\psi|\phi\right> : V^* \times V \rightarrow \mathbb{C} $$
Using these two, one can define the expectation value of an operator to be,
$$ <\hat O> = \left<\psi\right|\; \big(\hat O\left|\psi\right>\big) = \left<\psi|\phi\right> \in \mathbb{C} $$
| {
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How to show the invariant nature of some value by the group theory representations? Let's have Dirac spinor $\Psi (x)$. It transforms as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$ representation of the Lorentz group:
$$
\Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot {a}}\end{pmatrix}, \quad \Psi {'} = \hat {S}\Psi .
$$
Let's have spinor $\bar {\Psi} (x)$, which transforms also as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$, but as cospinor:
$$
\bar {\Psi} = \begin{pmatrix} \kappa^{a} & \psi_{\dot {a}}\end{pmatrix}, \quad \bar {\Psi}{'} = \bar {\Psi} \hat {S}^{-1}.
$$
How to show formally that
$$
\bar {\Psi}\Psi = inv?
$$
I mean that if $\Psi \bar {\Psi}$ refers to the direct product (correct it please, if I have done the mistake)
$$
\left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right]\otimes \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right],
$$
what group operation corresponds to $\bar {\Psi} \Psi$?
This question is strongly connected with this one.
| short answer if $ \hat {S}^{-1} S = \mathbb{I}$
I can give you a general example of $\psi^\dagger\psi$ not being invariant.
because for Dirac spinor $\psi$ whe have the following transformation rules
$$\psi(x) \rightarrow S[\Lambda] \psi(\Lambda^{-1}x)=S[\Lambda] \psi(x^\prime) \\
\psi^\dagger(x) \rightarrow \psi^\dagger(\Lambda^{-1}x) S[\Lambda]^\dagger $$
So $\psi^\dagger\psi \rightarrow \psi^\dagger(\Lambda^{-1} x)S[\Lambda]^\dagger S[\Lambda] \psi(\Lambda^{-1} x) $ is invarieant if and only if $S[\Lambda]^\dagger S[\Lambda] = \mathbb{I}$
however for the case where $S[\Lambda]$ are formed by the Clifford algebra it can be shown this is not they case. I do not have the capability to show you that they dirac adjoint does satisfy this condition.
| {
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What's this about kinetic energy increasing with the fifth power of length? I don't quite understand this quote from Stephen J. Gould's Ever since Darwin, where he talks about the compensating physical characteristics of organisms for their size.
Other essential features of organisms change even more rapidly
with increasing size than the ratio of surface to volume. Kinetic
energy, in some situations, increases as length raised to the fifth
power. If a child half your height falls down, its head will hit
with not half, but only 1/32 the energy of yours in a similar fall. In return,
we are protected from the physical force of its tantrums, for
the child can strike with, not half, but only 1/32 of the energy
we can muster.
Mass increases with the third power of length, and even if we allow for a shorter fall (even though that is not clear from the wording) for the child (who is half as tall), I can only come up with the fourth power.
| Model the child as a rod of mass $m$ and length $l$ standing on the ground vertically, with center of mass at height $l/2$, with feet glued to the ground but the rest of the body able to rotate.
When upright, the potential energy is $mgl/2$. When lying on the ground, the potential is 0. So when falling, the child hits the ground with energy $mgl/2$. Equating this with $\frac{1}{2}I\omega^2$, with $I=\frac{m l^2}{3}$ (the moment of inertia of a rigid rod rotating about its end) one gets
$$\omega=\sqrt{\frac{3g}{l}}.$$
The head hits the ground with velocity $l\omega$. Letting the mass of the head be $m_h$, the kinetic energy of the head hitting the ground is
$$T=\frac{1}{2}m_h(l\omega)^2=\frac{3}{2} g l m_h.$$
Since $l\rightarrow \lambda l$ and $m_h\rightarrow\lambda^3m_h$ under a scale transformation $\lambda$, we have
$$T\rightarrow\lambda^4T.$$
So unless I'm overlooking something, I can't quite see where Gould's extra factor of $\lambda$ comes from. Assuming this is right and Gould is mistaken, I suspect he got the $\lambda^5$ figure by looking at $E=\frac{1}{2}I\omega^2$ and noting that $I\rightarrow\lambda^5I$, but forgot to look in detail as to how $\omega$ transforms.
NOTE: If it makes it easier to swallow, this analysis can also be carried out by assuming the child is a cylinder of radius $R$, length $L$ and constant density $\rho$ which is free to rotate about its base. The inertia tensor is
$$\mathbf{I}=\left(
\begin{array}{ccc}
\frac{1}{12} L \pi R^2 \left(4 L^2+3 R^2\right) \rho & 0 & 0 \\
0 & \frac{1}{12} L \pi R^2 \left(4 L^2+3 R^2\right) \rho & 0 \\
0 & 0 & \frac{1}{2} L \pi R^4 \rho \\
\end{array}
\right)$$
and repeating the previous analysis using $I=\mathbf{I}_{xx}$ and $m=\pi L \rho R^2$ gives a head energy which scales as
$$T=m_h(L\omega)^2=\frac{6 g L^3 m_h}{4 L^2+3 R^2}\rightarrow\lambda^4T$$
under scale transformation.
| {
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Why is the Gibbs Free Energy $F-HM$? With magnetism, the Gibbs Free Energy is $F-HM$, where $F$ is the Helmholtz Free Energy, $H$ is the auxiliary magnetic field, and $M$ is magnetization.
Why is this? Normally, in thermodynamics, we Legendre Transform the various free energies into each other to maximize the global entropy. In these cases, we subtract $TS$ when we are imagining a system exchanging heat with a thermal reservoir (i.e. heat bath at constant temperature $T$), add $PV$ when we exchange volume $V$ with a constant pressure reservoir at pressure $P$, and subtract $\mu N$ when we exchange particles with a chemical reservoir at constant chemical potential $\mu$.
In every other case, we exchange heat, volume, and particles with the reservoir. How do we justify writing $G=F-HM$. Though it is true that $H$ is maintained constant, we don't exchange magnetization with a "magnetic reservoir".
| Without magnetism, you have $F(T, V, N)$ and $G(T, P, N)$. So you did a Legendre transform so that your potential depends on $P$ instead of $V$. It also means that you move from an extensive to an intensive quantity.
The free energy $F$ comes via $U = TS$ from the internal energy, which depends purely on extensive quantities. By going from $U$ to $F$, you exchange $T$ and $S$. $U$ will depend on $M$, which is also extensive. If you want $G$ to depend mostly intensive quantities, you will have to transform $M$ as well. That is why you get an additional $- HM$.
In my lecture, I was just told that we could transform the potentials into each other so that they depend on the variables that we are interested in. $U$ is fine for no interaction, $F$ for energy exchange and $\Omega$ for energy and particle exchange, and they are related to the micro canonical, canonical and grand canonical treatment. You choose the parameters such that the entropy is maximized, but the Legendre transform will not do that, since it leaves the value of the function unaltered.
| {
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"timestamp": "2023-03-29T00:00:00",
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How exactly is the Poisson bracket of the modes of a classical string defined? In the theory of a classical bosonic string, we have expressions like:
$$ \{\alpha^\mu_m,\alpha^\nu_n \} = - i m \delta_{m,-n} \eta^{\mu \nu} $$
were $\alpha^\mu_n$ are the Fourier modes of the string. How is this Poisson bracket defined? The definition from analytical mechanics involves partial derivatives with respect to the generalized coordinates and momenta. Since the modes $ \alpha^\mu_n $ are constants, their partial derivatives are zero, so there must be something I'm missing here.
| Concerning OP's last sentence (v1), the Fourier modes $\alpha^{\mu}_{m}$ are (some of) the fundamental variables of the string. Phrased equivalently, the Poisson bracket reads
$$ \{F(\alpha),G(\alpha)\}~=~ \sum_{m\in\mathbb{Z}}
\frac{\partial F(\alpha)}{\partial \alpha^{\mu}_{m}} (-im \eta^{\mu\nu})
\frac{\partial G(\alpha)}{\partial \alpha^{\nu}_{-m}}.$$
The above Poisson bracket can be derived via the usual Legendre transformation from Lagrangian to Hamiltonian formulation, and by imposing CCR. This is done in many textbooks on string theory.
| {
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Deriving $F = ma$ - Newton's Second Law of Motion Context:
In my textbook it is given: 'momentum' short for 'linear momentum':
Mass = $m$, momentum is $p=mv$. In time $\Delta t$, momentum changes by $\Delta p$, the rate of change of momentum is:
$$\frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$
My Doubts:
*
*Isn't a $\Delta$ sign missing beside the $t$ in the second fraction, and thus it should be $\frac{\Delta(mv)}{\Delta t}$
*How did they derive the third fraction from the second. I tried a lot but can't seem to get that.
My Work:
I have looked at this question - How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?, but it's a totally different equation.
My Final Question:
Can someone please clear my doubts about this equation and help me understand how does:
$$\frac{\Delta(mv)}{t} = m \frac{\Delta v}{\Delta t}$$
Thanks a lot !
| You are right, there is a $\Delta$ missing in front of the $t$.
$\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps.
The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like that one does appear in the complete analysis of systems of variable mass (like a rocket with its propellent being exhausted from the rear). It is also sometimes called Newton' second law in incorrect analyses of the rocket problem.
| {
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"question_score": "3",
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What will happen to water at $0^\circ$ Celsius kept in large evacuated chamber Suppose some water is kept at $0^\circ$, in a glass (ideal container, if necessary). Now it is placed in a large evacuated chamber. What will happen?
I think some water will vaporize, some will freeze and rest will remain liquid. But then I think how can water exist in liquid phase in such a high volume. Then I think I am confused, so I must ask it here.
| When one says the boiling point is x degrees at such a pressure, what it means is that it holds a partial vapour pressure up to this. Heat is fed in to make water into steam, even at room temperature, but it's so slight you don't notice it in the speciic heat.
When you have a large volume, a large amount of water is turned into steam, until the correct vapour pressure is met. For tiny volumes, the pressure can get high quickly, and water can be heated beyond 100 centigrade, right up to 373 c.
For a small amount of water at 32 Fahr, it will all vapourise if the pressure is less than the boiling pressure at 0c.
| {
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Stern Gerlach with spin in opposite directions So for the Stern-Gerlach apparatus, we assume that we either have a particle spin up or spin down. We also have the varying field, $\partial B/\partial z$. This initial configuration results in the particle wither going to plus $\hbar$ or minus $\hbar$.
Suppose instead of having spin up/down in the z direction, I sent it through with an initial spin aligned in the x direction (same exact configuration)? The Hamiltonian is given (for a linear B) as
$$H=\frac{1}{2m}(p_x^2+p_z^2)-\mu \sigma_z(B_0+B'z)$$
So my equations of motion for the z direction would just give me $p_0t/m+z_0$ and $\dot p_x=0$. Do I need to account for the spin x now instead, or will the particle go undeflected?
| $\newcommand{\ket}[1]{\left| #1 \right>}$$\newcommand{\bk}[2]{\left< #1 | #2 \right>}$Notice that the eigenvectors of the operator $S_z$ spans the whole space, which means that you can write any state as a superposition, (if you prefer as a linear combination) of these states. The situation is akin to the basis vectors of usual 3d Euclidean space. You can choose tree basis vectors there, usually the following is chosen:
$$\vec x = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\quad \vec y = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad \vec z = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
Then you can write any vector $\vec v$ as a linear combination of these vectors that is
$$\vec v = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \implies \vec v = a \vec x + b \vec y + c \vec z$$
Notice that the choice of basis vectors is not unique. I could have chosen the following basis vectors if I wanted to:
$$\vec x' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\quad \vec y' = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad\vec z' = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
for which the vector $\vec v$ can be written as
$$\vec v = a\vec x' + (b-a) \vec y' + c\vec z'$$
The situation with spin 1/2 particle is almost the same. You can choose your basis vectors to be the eigenvectors of $S_z$ operator in which case you write
$$\ket{z+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \ket{z-} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$
$$\implies \ket{x+}= \frac{1}{\sqrt 2} \Big( \ket{z+}+ \ket{z-} \Big)\quad \ket{x-}= \frac{1}{\sqrt 2} \Big( -\ket{z+} + \ket{z-} \Big)$$
Notice that $\ket{x\pm}$ and $\ket{z\pm}$ are the eigenvectors of $S_x$ and $S_z$ operators respectively, as we wanted.
If you send a particle with spin $+x$ in a SG machine aligned in the $z$ direction, then half the time you get $+z$ and half the time you get $-z$. You can immediately see that by taking the inner product of the state $\ket{x+}$ with the states $\ket{ z \pm}$.
$$\left| \bk{z+}{x+} \right|^2 = \left( \frac{1}{\sqrt 2} \right)^2 = \frac{1}{2} \qquad{\text{and similarly}} \quad \left| \bk{z-}{x+} \right|^2 = \frac{1}{2}$$
You can also choose your basis vectors to be the eigenvectors of $S_x$ operator but everything gets messy for this problem and you don't, in general, want to do that.
| {
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