Recent content by soupdejour

I am stuck with Gaussian units. If the field in vacuum is E = \frac{Q}{r^2} Is the field in a dielectric E = \frac{Q}{\epsilon_r r^2} , or is it unchanged?

Yes I want the boundary conditions satisfied.

I think I understand how superposition (of the electric field) applies for when I have two (or more) charges. BUT, If I have only one charge, but two regions of dielectric \epsilon_a and \epsilon_b, can I superimpose the solutions for the charge Q in just \epsilon_a with the solution of Q in...

Awesome, I understand, Thanks!

To make things straightforward, I just want the new frame to be moving along one dimension, say z. My guess is ct' = \gamma (ct - \beta z) r' = r \theta' = \theta z' = \gamma (z - \beta ct)

I know the Lorentz transformation in rectangular coordinates ct' = \gamma (ct - \beta x) x' = \gamma (x - \beta ct) y' = y z' = z I want to do this same transformation, but from cylindrical coordinates (r,\theta,z) to (r',\theta',z'). Any ideas?

Right. So for a given frequency of light, it isn't necessarily possible to produce N at rest. Only one frequency can produce N at rest. Thanks!

A closely related follow-up question: If I have a photon colliding with this pion, and I want to try to find the minimum energy of the pion to create some particle N. I know the mass of N and the wavelength of the photon. \gamma + \pi \rightarrow N For the minimum energy of pion to produce...

OK, I think I can visualize the wavelength of light changing in different frames. I was confused that the momentum was different in different frames even though the speed was the same (c). But it makes sense that a photon just has a special four vector with "length" zero. Thanks to both...

I have never done relativistic collisions when a photon is involved, and it's messing wtih me. For example, I have a photon colliding with a pion moving at a speed 3c/4. Is it possible to use a frame so that the pion is at rest? I think not, because the momentum of the photon is the same in...