# Reconciling Energy Losses on Fast Moving Batteries

• Joseph14
In summary, when a person on a platform sends a photon towards a train going 1/2 the speed of light, the photon has a higher energy than the photon emitted by the train. The train's kinetic energy is reduced when the photon returns, and the extra energy goes into the photon's momentum.

#### Joseph14

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.

Joseph14 said:
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.

DaleSpam said:
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.