# Redshift due to a static gravitational field and the conservation of energy [duplicate]

I am standing on the surface of some planet. Gravity is described via General Relativity with some static metric (e.g. the Schwarzschild metric, so static means no time dependence, but the metric may vary from place to place). I send a blue photon up to my friend, who is x meters above me in some tower (we are both at rest relative to each other). He measures the photon and finds out it is red. We both conclude that a gravitational redshift occured. However, where did the energy go? In GR there is no gravitational energy so the photon did not trade "light energy" with potential energy.

I found several threads about this, but often they viewed this topic from a cosmological point of view where the metric does depend on time and thus Noether does not work to argue for a conservation of energy. Arguments without cosmology used the explanation via potential energy (which is not a thing in GR, as far as I know). So, since the metric is still time independent the energy should be conserved according to Noether. What is going on?

Edit: On the Einstein thought-experiment in the linked question:

This does not explain why energy is not conserved from a mathematical or physical point of view. This could also be viewed as a reason why you can not turn photons into matter (and vice versa) without losing energy.

## marked as duplicate by ACuriousMind♦, Daniel Griscom, user36790, yuggib, Kyle KanosJan 15 '16 at 11:11

In a static spacetime, there is (by definition) a timelike Killing vector field $\xi^\mu$, which implies that geodesics with four-velocity $u^\nu$ have a conserved quantity $\epsilon = -g_{\mu\nu}\xi^\mu u^\nu$. For example, in Schwarzschild spacetime, this is $$\epsilon = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\text{,}$$ where $\lambda$ is an arbitrary affine parameter for the geodesic. This is the correct generalization of conservation of orbital energy in the general-relativistic context.

For a massive particle in Schwarzschild space, for which we can take $\lambda$ to be the proper time $\tau$, this leads to a direct analogue of Newtonian orbital energy corrected by an extra $r^{-3}$ term, up to the understanding that the Schwarzschild $r$ and $\tau$ have somewhat different meaning than they do in Newtonian theory: $$\text{const} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 + \frac{1}{2}\frac{l^2}{r^2} - \frac{GM}{r} - \frac{GMl^2}{c^2r^3}\text{,}$$ where $l = r^2(\sin^2\theta)(\mathrm{d}\phi/\mathrm{d}\tau)$ is the specific angular momentum, another conserved quantity for Schwarzschild spacetime. The first two terms would constitute the Newtonian kinetic energy per mass, decomposed into radial and angular components.

For a massless particle, this is a little different: $$\epsilon^2 = \left(\frac{\mathrm{d}r}{\mathrm{d}\lambda}\right)^2 + \frac{l^2}{r^2}\left(1-\frac{2M}{r}\right)\text{.}$$

In GR there is no gravitational energy so the photon did not trade "light energy" with potential energy.

But in static spacetimes, we can always define a conserved total orbital energy for the geodesics.

I just read the derivation of your formulas in Carrolls GR book. You get conserved quantities but not really an exact energy term. You get something like energy per unit mass. To turn this into energy seems critical since photon mass is zero (and we look at photons in this example). Mathematically we have a conserved quantity but I don't see how we determine that this is really the energy and not something else.

A conserved quantity generated by time-translation invariance is how conservation of energy works in modern physics; that's the moral of Noether's theorem and the geometrical meaning of having a timelike Killing vector field.

Having it specific energy (i.e., per-mass) is exactly how gravitational potential works even in Newtonian theory, and even for a photon it can be interpreted as relative to energy at infinity, or adapted to a scale set anywhere along the orbit, for that matter. Operationally, it is exactly the specific energy measured by a family of static observers (comoving with the Killing vector field), since an inner product with a four-velocities gives the relatively Lorentz factor, or equivalently, the time-component of one in a local inertial frame comoving with the other.

• So the redshifted photon gains orbital energy? Also, the expression for the massless particle is for the energy squared? Is dr/d(lambda) c? – Thomas Elliot Jan 16 '16 at 7:36
• @Thomas Elliot : no the redshifted photon doesn't gain energy. Nor does a massive particle when it falls down - gravitational potential energy is converted into kinetic energy, that's all. When you descend to a lower elevation this kinetic energy is typically dissipated, whereupon you comprise less total energy that at the higher elevation (check out the mass deficit). Then when you measure the selfsame photon, it appears to have gained energy. The descending photon appears to be blueshifted. But it hasn't gained energy. You've lost it. – John Duffield Jan 16 '16 at 18:40
• Your post reads like "energy is conserved" but your comment reads like "energy is not conserved" (in my head). I also don't know how to interpret "the descending photon appears to be blueshifted." it ether is or is not blueshifted - or not? Could you perhaps elaborate on this things a little more in your post (not in this small comment section). – Thomas Elliot Jan 17 '16 at 15:45
• @ThomasElliot I think you're mixing me up with someone else. The total energy is conserved. The square is correct; actually, if you go through the steps to derive the massive particle case, the constant on the left is $\frac{1}{2}(\epsilon^2-1)c^2$. ... Yes, descending photons are blueshifted; I have no idea what anything 'appearing blueshifted but not really blueshifted' would even mean, and I can't clarify that because it's not my statement. Otherwise, the answers in linked question are good: there is energy in the system due to particle position, and it can be exchanged with kinetic energy. – Stan Liou Jan 17 '16 at 20:22
• I just read the derivation of your formulas in Carrolls GR book. You get conserved quantities but not really an exact energy term. You get something like energy per unit mass. To turn this into energy seems critical since photon mass is zero (and we look at photons in this example). Mathematically we have a conserved quantity but I don't see how we determine that this is really the energy and not something else. – Thomas Elliot Feb 2 '16 at 14:16

I send a blue photon up to my friend, who is x meters above me in some tower (we are both at rest relative to each other). He measures the photon and finds out it is red. We both conclude that a gravitational redshift occurred. However, where did the energy go?

It didn't go anywhere. The ascending photon didn't lose any energy. There is no magical mysterious mechanism by which an E=hf photon get's "tired" and loses energy. Your friend measured the photon to have a lower frequency than you because his clocks are running faster than yours. Because of gravitational time dilation. It's important to note that gravitational energy isn't about a photon losing energy when it ascends. It's about the photon being emitted at a lower frequency. See Einstein saying that here:

"An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated".

In GR there is no gravitational energy so the photon did not trade "light energy" with potential energy.

Correct. A photon is "all kinetic energy". There's no conversion of kinetic energy into potential energy. Moreover the ascending photon doesn't slow down. Instead it speeds up. See Einstein talking about the speed of light varying with position here:

"As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable".

I found several threads about this, but often they viewed this topic from a cosmological point of view where the metric does depend on time and thus Noether does not work to argue for a conservation of energy. Arguments without cosmology used the explanation via potential energy (which is not a thing in GR, as far as I know). So, since the metric is still time independent the energy should be conserved according to Noether. What is going on?

Energy is conserved. If you send a 511keV photon down into a black hole, the black hole mass increases by 511keV/c². The descending photon didn't gain any energy. The ascending photon doesn't lose it. Anybody who tells you otherwise is wrong. Alfred Centauri's update in the alleged duplicate is correct.

This could also be viewed as a reason why you can not turn photons into matter (and vice versa) without losing energy.

When you convert photons into matter and vice versa via pair production and annihilation, energy is conserved. I know of no situation in which energy is not conserved. If anybody else does, I'd be pleased to hear about it.