Photon frequency loss over time?

[ratfink]

So, would the photographic plate be exposed on both galaxies or only one?
This is what energy conservation is about (and forums for that matter too!)

 

[SpaceTiger]

This is incorrect. Energy conservation in GR is in dispute. Some say that energy is not conserved others say it is due to 'curvature'. I hope that this is not a 'warning' because you wish to avoid answering valid questions.

The only dispute is in appropriate definitions of energy in GR. See here:

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html"

Unless I'm to interpret your previous post as advocating a particular pseudotensor definition of energy, then it belongs in IR.

 

[ratfink]

This is not peer reviewed and who are they?

 

[ratfink]

OK,
Let me spell this out.
B/W paper is not sensitive to red light. That is why we have red safelights in photographic darkrooms.
Light from a distant galaxy is redshifted.
I say that there must be a point where the paper is not exposed anymore.
You and Garth say "Oh the pricinciple of conservation of energy does not apply".
I say B**-t either the paper is exposed or it is not. It is all to do with the photoelectric effect.
Nothing to do with reference frame - we all see the same result.
So I will ask again. If the photon has enough energy to excite and expose a photographic plate on galaxy X, will it once it has been redshifted, be able to expose it on galaxy y?

 

[Garth]

No need to be rude.

Energy is not generally conserved in GR because of the time dilation caused by space-time curvature. The energy of a system can only be consistently defined at null infinity where curvature effects and gravitational waves become insignificant.

The reason this is so is energy is a frame dependent concept, moving frames, or frames at different gravitational potential, measure energy differently from one another. We do not all see the same result - that is the whole point of relativity, certain measurements are relative.

In GR it is energy-momentum, or particle rest mass, that is conserved.

Your photographic plate question is irrelevant. The exposed plates X & Y both record the photon, but at a later cosmological time, after more cosmic expansion, the plate Y records a redder image. That is what Hubble red shift is all about.

Garth

 

[SpaceTiger]

That's enough, the OP's questions have long since been answered. I'm locking this.