Say an object is moving towards me and is emitting visible light, the light i see will be shifted to the blue end of the spectrum and so its frequency will be greater.
Well since E = (planks constant) x f
Where does the extra energy come from than causes F to increase?
ps. This is isnt (as far as im aware) an exam question i was just wondering.
Awesome question, the reason I could think of is that energy is dependent on your frame of reference.
The energy/frequency of the photon is dependent on the relative speed between the observer and the source, in perhaps a similar way to how the kinetic energy of a car is dependent on how fast it's travelling relative to you, since there's no preferred frame of reference, the measurements of energy are equally valid.
The energy in E=hf, is the energy of the wave relative to itself. Therefore the value of f is dependent on the energy for that certain wave, so if the frequency of the wave (relative to itself) changes, then the energy will change by the same amount. However it shifts blue because the apparent change in the frequency of a wave is caused by relative motion between the source and the observer. In this case when the source of the waves is moving toward the observer, each successive wave 'start point' is emitted from a position closer to the observer than the previous wave. Therefore each wave takes slightly less time to reach the observer than the previous wave. Therefore the time between the arrival of successive waves at the observer is reduced, causing an increase in the frequency, hence causing a blue shift.
hmmm alot of that went straight over my head.
I think i've figured it out.
If something is traveling towards me and emitting light, then the light becomes blushifted by a certain amount.
However, if i was to be on the opposite side of the source and it was therefore moving away, the light would be redshifted by the same amount.
So i would measure a energy increase of +E on the blueshifted side and an energy decrease on the redshifted side of -E.