Question

Atoms at rest emit photons isotropically. For an observer watching a beam of atoms moving at speed 0.9 c relative to the lab, does the emission appear isotropic to an observer in the lab? For photons emitted at angles 0,30,60,90,120,150,180 in the reference frame of the atoms, what angles of emission does the observer see? Describe qualitatively and draw the distribution of emission. Will atoms appear brighter as they approach or recede from the observer?

Answer 1

The Lorentz transformation for velocity is given by

where u is the velocity of the photon as observed by lab frame of reference, u' is the velocity of photon as observed by atom frame of reference, v is the velocity of the atom and c is the velocity of the light in free space.

Now, for the lab frame of reference, the velocity of photons released in the same direction of the velocity of the atom will be

Now, for photons that are released on the oppostite direction than that of the velocity of that of the atom, the observed velocity is given by

So, we have observed that the x component of the velocity changes from c to -c, as we change the angle wrt the to the atom.

This proves that even if the atom is having a relativistic velocity, the photons are observed to have been emitted at an observed velocity of c, in all direction, no matter what.

The x component of the velocity will be where theta is the angle of velocity of the photon makes with the x axis. and similarly, y component will be , but as we have seen, the magnitute of the velocity will be always c., and that it will be isotropic, that is equal in all directions.

inorder for atom to appear brighter, there should be more no. of photons per second, entering in a specific direction, than other direction. but since the emission will be isotropic, we wont be observing any such phenomenon. but what we can observe is the doppler shift in the frequency of the light emitted.