Local speed of light and accelerated observers in general relativity

Question

Background:

By the equivalence principle, an observer (time-like geodesic + orthonormal frame (3 spacelike, 1 timelike) which serves as their measurement standard) in gravitational free fall sees physics which is locally indistinguishable to that they would see in an inertial frame in flat spacetime. As we well-know, measurements in general relativity can only be made locally due to the anholonomy of curved spacetime (ex. we can project the tangent vector of another particle flying by into the observer's physical space subspace of the tangent space at a point. But how do we make a sensible definition of the speed of a particle whose four-velocity is in another tangent space?).

Thus by Einstein's postulate of Special Relativity, the speed of light measured by any observer (time-like geodesic) is $c$ as their measurements are local, and so they might as well be in an inertial frame in Minkowski space. Mathematically: the curved spacetime's metric can always be made close to the Minkowski metric at a point, and so geodesics will locally behave in that famous congruence of inertial observers way - like Einstein and his books in free fall in an elevator.

A lemma that needs to be proven: I've read that measuring a passing worldline's speed (spatial projection of four-velocity) as $c$ literally means its tangent is null. (Why?) The spatial velocity the observer gets for a passing object is defined as $$v = \varepsilon^{\alpha} (v_{\delta,\delta(t_1)}) e_{\alpha}$$ where $e_{\alpha}$ is the observer's orthonormal basis for the "space" directions $\alpha = 1,2,3$ and $\epsilon^{\alpha}$ are the pointwise covector counterparts to these vectors. The tangent vector being acted on is that of the beam of light whose world line I've called $\delta$. How do we prove $v = c \implies g(v_{delta,\delta(t_1)},v_{delta,\delta(t_1)}) = 0$ where $t_1$ is the time of the observer's and light beam's worldlines crossing so the meausurment can be done.

By the geodesic postulate then, the beam continues on through spacetime after its encounter with you always with a null tangent (parallel transport preserves length of tangent). This is one way to see why light beams follow null geodesics in GR.

So here's my question: curved spacetime has geometry pointwise always sufficiently similar to Minkowski, but one needn't be in gravitational free fall. So an accelerated (wrt free fall) observer with $\nabla_X X \neq 0$ shouldn't necessarily measure a local speed of light equal to $c$.

Yet light beams are traveling on null geodesics one way or another, and so as long the observer's tangent is time-like, the measurement should come out to $c$ (once we've proven the above lemma). What is going on here?

Even forgetting curved spacetime, applying this logic to a Rindler observer makes this all the more clear: local measurements of the speed of light are always going to be $c$.

Answer 1

All timelike observers measure the local speed of light as $c$ - even accelerating observers! Two useful concepts here are local flatness and the construction of momentarily comoving reference frames (MCRFs).

Local flatness ensures at any point $P$ in the manifold, there exists coordinates such that the spacetime locally looks like that of special relativity: Minkowski space! At point $P$, you may actually find such coordinates with 6 degrees of freedom left over. These correspond to the 3 boosts and 3 rotations you may perform on your coordinate system.

Now there's obviously no inertial frame in which an accelerating observer is always at rest. However there is an inertial frame which momentarily has the same velocity of the observer. This is the MCRF. Of course, a moment later this frame is no longer comoving with the observer.

Now consider an event $P$ in an arbitrary curved spacetime where an accelerating observer passes right by a light beam. From local flatness we know we may find coordinates about $P$ where the spacetime looks Minkowskian. Moreover, we may use one of our six degrees of freedom to boost into the MCRF of the accelerating observer at this point. The MCRF is locally an inertial frame which momentarily describes the accelerating observer at $P$. So via the postulates of special relativity, we are guaranteed this accelerating observer measures the speed of light as $c$ at point $P$! Here we've had to use “local” both in the sense of space - the light beam passes right by the observer, and in time - our MCRF represents the observer only for a moment (but that's all we need).

Concerning the lemma, the four-velocity of a photon is not well-defined. Typically, the components of the four-velocity for some observer $\mathcal{O}$ with coordinates $(t, x, y, z)$ are

$$\vec{u}\rightarrow \left(\frac{dt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right)$$

where $\tau$ is the proper time of observer $\mathcal{O}$. The issue arises because there is no inertial frame where light is at rest, and hence proper time is ill-defined for a photon. You might get around this using what's known as an affine parameter $\lambda$. However I believe it's simpler to use the four-momentum which is well-defined for a photon.

Suppose in the MCRF of the accelerating observer described above with coordinates $(t, x, y, z)$ the light beam propagates in the $x$-direction. The energy, $E$, and momentum, $p$, of a photon are related by $p = E/c$ and in units where $c = 1$: $p = E$. Then the components of the four-momentum in the MCRF are

$$\vec{p}\rightarrow (E, p, 0, 0) = (E, E, 0, 0)$$

and as the MCRF is locally inertial, we may use the metric for Minkowski space to compute

$$\mathbf{g}(\vec{p},\vec{p}) = \eta_{\mu\nu}p^\mu p^\nu = -E^2 + E^2 = 0$$

which highlights the null trajectory of the photon in this frame.

Lastly, there are many popular systems where accelerating observers or observers in curved spacetimes will perceive light traveling slower or faster than $c$ at a distance (e.g. Rindler coordinates for Minkowski space, recessional velocities of distant galaxies in FRW cosmologies, etc). Ultimately, this comes down to the fact that it is impossible to cover these spacetimes with global coordinates that represent an inertial frame.