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Nov
2
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Apr
27
comment Feynman diagram for annihilation
@Harold, You would need two photons to conserve 4-momentum while satisfying $E = pc$, but because it's a virtual photon (an internal line) you're allowed to violate $E = pc$.
Apr
25
comment Feynman diagram for annihilation
@Harold What physical process you're talking about is identified by the initial and final states. If in an experiment you have two electrons going in and two electrons going out, you can't look at it and say "the intermediate state is a single photon". That is just one diagram that contributes to the process. At tree-level you also have the diagram where a photon is exchanged (the t-channel), and there are higher order diagrams as well.
Apr
25
comment Why the name “optical phonon”?
@Seanny123 Yes, it's a very important fundamental concept in solid state physics. I'm not sure if I can do justice to it in a comment, perhaps you should post a separate question on that? (Or have a look at an introductory text on solid state.) The short version is that it's a cell of the reciprocal lattice (the lattice of k-vectors that arises from the Fourier transforms of the states in your original lattice). It is relevant both to the case where the primitive cell contains one atom or multiple atoms.
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
This paper discusses the quantum mechanical case: academic.csuohio.edu/deissler/PhysRevE_77_036609.pdf
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
As a simplified (classical) example, consider a loop of current-carrying wire dangling from a string (so it's free to rotate) in an external magnetic field. As the loop rotates to align with the field, it gains rotational kinetic energy, but at the same time there's an induced EMF due to the change in magnetic flux through the loop. This opposes the motion of the circulating charges in the loop, thus robbing them of some kinetic energy. I believe if you do the calculation you'll see the amount of energy removed is the same as the amount of energy added.
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
This answer doesn't seem quite right to me. When a magnetic field gives kinetic energy to an object, it induces an EMF that cancels out an equal amount of internal energy of the object. So the net work done by the magnetic field is truly zero.
Mar
20
comment Capacitor with dielectric between the plates
@guru Sorry, I believe you are incorrect. You get an attraction from inducing dipoles in the dielectric. The negative charges in the dielectric are pulled closer to the positive plate, so their attraction to it outweighs the repulsion felt by the more distant positive charges. Likewise, the positive charges are pulled closer to the negative plate, so their attraction towards it outweighs the repulsion felt by the more distant negative charges.
Mar
20
comment Capacitor with dielectric between the plates
@DumpsterDoofus That sounds right to me. Why does it seem strange?
Mar
20
comment Why friction is zero when wheel slip is zero?
Hmm, I don't think a rack and pinion requires kinetic friction either. Kinetic friction arises when one surface is sliding against another. Looking at the animation in this article, for example, I don't see anything I'd describe as sliding: en.wikipedia.org/wiki/Rack_and_pinion
Mar
20
comment Atwood machine with spring
Yeah, that line was a bit misleading.
Mar
20
comment Capacitor with dielectric between the plates
Sure. The charge of the plates causes the dielectric to become polarized: positive charges in the dielectric get pulled a little bit towards the negative plate, and negative charges in the dielectric get pulled a little bit towards the positive plate. These induced dipoles in the dielectric are attracted by the electric field of the plates.
Mar
20
answered Capacitor with dielectric between the plates
Mar
20
comment Atwood machine with spring
The last bob is $\frac{1}{2}m(\dot{x}-\dot{y})^{2}$ because if the big bob moves $\Delta x$, then the string moves $\Delta x$, but the spring can stretch by $\Delta y$ to make the last bob move less. Maybe it will be more clear if you imagine what would happen if $x$ and $y$ increased at the same rate: the right-most bob should be stationary then, right?
Mar
20
comment Why friction is zero when wheel slip is zero?
I'm sorry, I don't understand the comment. "No slip" for a tire means "rotating in the normal way", such that a point on the tire's surface follows the cycloid curve. You can still have static friction in this case, and this is needed to prevent slipping when accelerating (e.g. going around a turn, speeding up, slowing down, etc.) I suppose that would resist wind as well.
Mar
20
revised Atwood machine with spring
added 2 characters in body
Mar
20
comment Atwood machine with spring
Ah, I get it. Normally for these problems you write $L = T - V$ and use Euler-Lagrange, but this is just part 1. I've given you a gist of how to do it in my answer, while trying not to give away the whole thing.
Mar
20
answered Atwood machine with spring
Mar
20
comment Atwood machine with spring
Have you tried writing an expression for the potential energy (since you already have kinetic energy), and then using the Euler-Lagrange equation?
Mar
20
revised Why friction is zero when wheel slip is zero?
added 687 characters in body