Energy loss of a photon moving against gravity

[nonequilibrium]

General relativity doesn't have global conservation laws that make sense in all spacetimes. It does have local conservation laws.

Hm, I'm just reminded of the fact that in Newtonian Mechanics non-inertial reference frames don't obey the principle of conservation of energy. So in relativistic mechanics we lose even more? In relativistic mechanics the old adagium "In an inertial frame of reference the total amount of energy is constant" fails? Might there be something else accounting for the loss of energy? If not, are there also situations in which we gain energy instead of lose it (like with redshift)?

EDIT: strange thought experiment: let a photon near a galaxy, it will experience redshift. Now turn back time, you see a "blueshift". Yet if you'd shoot a photon away from a galaxy, you'd experience redshift, right? Huh?

 

[bcrowell]

Hm, I'm just reminded of the fact that in Newtonian Mechanics non-inertial reference frames don't obey the principle of conservation of energy. So in relativistic mechanics we lose even more?

You actually gain something, because you still have a local conservation law, even if you choose a noninertial frame. All you lose is the ability to make global conservation laws: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.5

In relativistic mechanics the old adagium "In an inertial frame of reference the total amount of energy is constant" fails? Might there be something else accounting for the loss of energy? If not, are there also situations in which we gain energy instead of lose it (like with redshift)?

It's not that there's a loss or gain of energy, it's that it's not even possible to define the amount of energy in any useful way.

 

[Naty1]

Ben..so glad you posted

http://www.lightandmatter.com/html_b...tml#Section4.5

to remind me I had promised myself I would read the whole thing...that will be my next "project" for myself...

can you put that link in your signature so when you make posts it appears at the bottom...or did the lords of the forum removed all signatures?? I see mine no longer appears...

 

[Naty1]

Post # 12: TriKri...since this is your thread...

I'm confused ... does the speed of light vary at other locations than the local or not? Jonathan says it does, but Naty1 says it doesn't ... or have I misunderstood anything?

I hope I did not say that. Read my post #8 and Jonathan's post #12 ...I interpret them to say the same thing...distant observers may observe different speeds of light...Only inertial observers in flat spacetime (no gravity) see both local and distant light as "c"...

 

[Austin0]

This is incorrect.

Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by anyone observer. When we talk about redshift increasing with potential, what we actually mean is that a series of different local observers at different potentials would observe a photon passing their own location to have different redshift relative to a local reference photon. However, from the point of view of any single observer, the photon has constant frequency and energy when moving freely in a static gravitational field.

In a static gravitational field, using an isotropic coordinate system, the momentum of a free-falling particle increases downwards with time, even if it is a photon, but the energy is constant. This is related to the effect that in such coordinates the speed of light at a location other than the observer's own varies a bit depending on the gravitational potential.

Hi 1) I have searched for a definition of an isometric coordinate system without result.
Intuitively I can picture what this would mean ;essentially a Cartesian system but am not sure if that's what your talking about in this context.
2) In this circumstance how do you differentiate between momentum and energy wrt a photon?
I can see why the energy would apparently increase (blueshift) relative to local electron frequencies but wouldn't this also be equivalent to an increase in momentum??
Also; viewed from a locale of higher potential wouldn't the speed of light decrease at lower levels?
If this is the case, then how would this be related to an increase in momentum?
Thanks

 

[Jonathan Scott]

Hi 1) I have searched for a definition of an isometric coordinate system without result.
Intuitively I can picture what this would mean ;essentially a Cartesian system but am not sure if that's what your talking about in this context.

Your searching may be more successful if you look for isotropic coordinates.

It does not necessarily mean flat (Minkowski) space. "Isotropic" means for example that if you observe a light speed signal expanding from a point (creating a sphere expanding at c according to a local observer) then in the coordinate system the light also expands at the same rate in all directions, still creating a sphere, but at a coordinate speed which is not necessarily exactly equal to c.

In more general coordinate systems, the speed of light is not necessarily the same in different directions, so you can't even talk about the coordinate speed of light unless you also specify a direction.

2) In this circumstance how do you differentiate between momentum and energy wrt a photon?
I can see why the energy would apparently increase (blueshift) relative to local electron frequencies but wouldn't this also be equivalent to an increase in momentum??
Also; viewed from a locale of higher potential wouldn't the speed of light decrease at lower levels?
If this is the case, then how would this be related to an increase in momentum?
Thanks

The magnitude of the momentum of an object is Ev/c² where v is the speed of the object and c is the speed of light. For a photon the speed is equal to c, so the magnitude of the momentum is E/c. Relative to an isotropic coordinate system, c decreases with potential, so the momentum increases if the photon falls down to a lower potential, or decreases if the photon rises to a higher potential. If the photon moves sideways along a line of constant potential, it gets deflected a bit downwards by the field.

The interesting thing is that regardless of the direction in which the photon is travelling, the coordinate rate of change of momentum with time at a given point is always the same downwards pointing vector, equal to exactly twice the Newtonian force that would be experienced on an object of the same total energy at that point in the field. In general, for an object traveling at speed v, it is simply (1+v²/c²) times the Newtonian force, at least in fields where the weak approximation holds (that is, nowhere near a neutron star or worse).