Are photon energies conserved in general relativity?

As I understand it, both Maxwell's wave equation and the null geodesics of general relativity are scale invariant.

Thus an electromagnetic wave can be shifted along a null geodesic without changing the laws of physics that it obeys.

Does this imply that the energy/momentum of photons is conserved as they travel through spacetime relative to the co-ordinate system of some observer A?

The phenomenon of gravitational redshift could be due to the difference between an observer's local energy, as demonstrated by the frequency of his clocks, and the constant energy of the photon, both described relative to observer A's co-ordinates.

• redshift means less frequency and then , less energy – user46925 May 28 '15 at 13:19
• See addition above. – John Eastmond May 28 '15 at 13:23
• I'm not sure to understand. Do you mean a variation of the Planck constant ? it's not enough for all the observations, including time dilatation – user46925 May 28 '15 at 13:28
• If Observer B is near a black hole then his clocks are running slow according to a distant observer A. If B's clocks run slow then their energies are reduced according to observer A. – John Eastmond May 28 '15 at 13:37
• Invariance of the equations under a transformation does not imply invariance of the solutions - only that they still obey the equation after the transformation. – ACuriousMind May 28 '15 at 13:58

In a curved spacetime, the wave four-vector $k^a$ for a solution to Maxwell's equation (in the appropriate geometric-optics limit) satisfies the geodesic equation: $$(k^a \nabla_a )k^b = 0.$$ Effectively, this means that "the derivative of $k^a$ in the direction $k^a$ is zero". (Note that the first term is just a directional derivative in the direction of $k^a$. What's more, it is always possible to set down a set of coordinates in a "tube" surrounding the worldline of this geodesic, called Fermi normal coordinates, such that all the Christoffel symbols vanish at points along the geodesic. (See here for a reference to how this is done for null geodesics.) In this case, if $\lambda$ is our coordinate along the null geodesic, then the geodesic equation as expressed in these coordinates is just $$\frac{d k^\mu}{d\lambda} = 0.$$ In other words, the components of the wave vector are constant in these coordinates. So in this sense, the energy & momentum of the photon is conserved in the "photon frame" (whatever that means.)

But of course, we're not photons, and so we observe gravitational redshift. In a lot of cases, though, we have a stationary spacetime, in which (roughly speaking) there's a time coordinate $t$ that the metric doesn't depend on. Formally speaking, we want there to be a vector field $t^a$, pointing in the direction of increasing $t$, such that $\nabla_a t_b + \nabla_b t_a = 0$. In this case, $t^a$ is called a Killing vector field, and it's not too hard to show in such an instance that $$k_a t^a = \text{const.}$$ In this sense, a quantity related to the energy of the photon is conserved as the photon travels. For example, if we're just dealing with flat spacetime, then $t^\mu = (1,0,0,0)$ and we have $k^t = \text{const.}$, i.e., the energy of photons is constant for observers whose four-velocity points in the $t$-direction. More generally, in a static spacetime, we'll have something like $k_t = k^t g_{tt} = \text{const.}$ instead. Similarly, if we have a spacelike Killing vector field (for example, $x^a = (0,1,0,0)$ in flat spacetime), then we can define a conservation law for particles that travel along geodesics which reduces to conservation of momentum in the case of flat spacetime.

Finally, it's important to note that the actual energies & momenta that an observer will measure are related to the observer's four-velocity $u^a$ and their set of spatial basis vectors. For example, we have $\omega = u^a k_a$. Since $u^a u^b g_{ab} = -1$ by definition, this means that in general an observer whose four-velocity is in the $t$-direction will have $u^\mu = ((-g_{tt})^{-1/2},0,0,0)$, and such an observer will then see the photon as having $$\omega = k_t u^t = \frac{\text{const.}}{\sqrt{-g_{tt}}}.$$ In this interpretation, the gravitational redshift is in fact solely due to the clocks of various observers "running slow" due to the $tt$-component of the metric being different. However, it's important to note that the conserved quantity $k_t$ is not necessarily an energy as seen by any particular observer (even the observer who emitted it in the first place.)

A good reference for this business with Killing vectors and conserved quantities is Hartle's Gravity: an Introduction to Einstein's General Relativity. Killing vectors are explained in Section 8.2.

• What happens if we have the FRW metric of an expanding Universe? In that case there is only a conformal time-like Killing vector field. Is there an interpretation of the cosmological redshift, similar to the one you describe above, where there is a conserved quantity associated with the photon and the redshift is due to changes in the observer's clock? – John Eastmond May 29 '15 at 16:40
• Yes and no. For a null geodesic in a spacetime with a conformal Killing vector $\tau^a$, you can work out that $k^a \nabla_a (k^b \tau_b) = \frac{1}{2} k^a k^b \mathcal{L}_\tau g_{ab} = \frac{\alpha}{2} k^a k^b g_{ab} = 0$ if $k^a$ is null. So in such a case, you have a conserved quantity $k_a \tau^a = k_\tau$ associated with the photon, and you can say that the redshift is due to observers' clocks running slower with respect to conformal time as the Universe expands. This is perhaps of limited utility, though, since I'm not sure how one could measure conformal time for comparison. – Michael Seifert May 29 '15 at 17:25
• Did you mean that the redshift is due to observers' clocks running faster with respect to conformal time as the Universe expands? Maybe conformal time $\tau$ is our physical time because there is a conformal Killing vector associated with it leading to a conserved quantity. Cosmological time $t$ doesn't have any type of Killing vector. – John Eastmond May 30 '15 at 22:50
• The relationship between conformal time and observer time is $d\tau/dt = a(t)$, so you can't actually say whether conformal time or coordinate time is running faster without knowing the actual value of the scale factor. Moreover, in flat FRW models (like the one we seem to live in), the actual value of $a(t)$ is arbitrary up to an overall rescaling (since the metric is invariant under the transformation $\vec{r} \to \lambda \vec{r}, a \to \lambda^{-2} a$. So there's enough freedom in how you define conformal time that you can't really say which time is "running faster". – Michael Seifert May 31 '15 at 13:03

Gravitational red/blue-shifting conserves energy. It merely swaps potential energy for kinetic energy. A photon outside a gravity well has zero gravitational potential energy. As it falls into the well, it gains momentum and loses potential energy, similar to anything else falling into a gravity well. At least, that's a simplistic way of looking at what's happening.

On local scales that you would expect to see gravitational red/blue-shifting, energy is a conserved quantity. However, general relativity does not require energy to be conserved globally. It only requires that the stress-energy-momentum tensor is conserved. In a universe like ours, that means that energy and momentum need not be conserved independently on the largest scales. In fact, the expansion of space is widely held to be an example of the non-conservation of energy at large scales. The total matter energy of the universe is (more or less) conserved, the total radiation energy decreases linearly with the increase in size along any one spatial dimension, but the total amount of dark energy increases proportional to the volume of the universe. This means the total amount of energy seems to end up with a net increase, thus showing it as non-constant.

I seem to have gotten a bit off track. Remember how I said the total radiation energy decreases? That right there is the answer to your question. A phenomenon called Cosmological Redshift happens because as the universe expands, is also expands the wavelength of light travelling through it. This causes a net decrease in the energy of a photon over time and was described through general relativity. So the answer to your question is no, general relativity does not conserve photon energies.