PDA

View Full Version : fast moving batteries


Joseph14
Feb9-11, 12:52 AM
Lets say I'm in a train going the 1/2 the speed of light. I have a light source that releases signal photons.

I aim the light source in the direction i'm going and release 1 photon. Now my energy supply for the lamp has decreased by hf. From a person on the platform the frequency is shifted so that the energy in the photon is >hf.

Now when the train returns to platform how can the different energy losses be reconciled, so that both people agree on the energy remaining in the power supply.

Mentz114
Feb9-11, 01:22 AM
If the platform observer measures frequency f' and they know the speed of the train they can work out the frequency f in the rest frame of the emitter using the relativistic doppler formula. See, for instance.

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

DaleSpam
Feb9-11, 06:34 AM
Lets say I'm in a train going the 1/2 the speed of light. I have a light source that releases signal photons.

I aim the light source in the direction i'm going and release 1 photon. Now my energy supply for the lamp has decreased by hf. From a person on the platform the frequency is shifted so that the energy in the photon is >hf.

Now when the train returns to platform how can the different energy losses be reconciled, so that both people agree on the energy remaining in the power supply.The photon also has momentum. From the recoil, by conservation of momentum, the speed of the train is slightly reduced. The extra energy which goes into the photon comes out of the KE of the train.

Joseph14
Feb9-11, 04:59 PM
The photon also has momentum. From the recoil, by conservation of momentum, the speed of the train is slightly reduced. The extra energy which goes into the photon comes out of the KE of the train.

Thanks.