Energy loss of a photon moving against gravity

by TriKri
Tags: energy, gravity, photon, redshift, work
P: 1,162
 Quote by Jonathan Scott This is incorrect. Neither photons nor even massive objects change in frequency or energy when moving in a static gravitational field as observed by any one 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 thats 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
PF Gold
P: 1,164
 Quote by Austin0 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 thats 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.

 Quote by Austin0 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 travelling 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).

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