I saw somewhere on here that you can't exactly talk about length contraction in General Relativity, you can however talk about time dilation. So I was wondering if I made a light clock, with a bouncing photon between two mirrors, would I be able to tell the amount of length contraction in a curved spacetime.

So if I had two such clocks, and I sent one around a massive object and back and the amount of time recorded on that clock is off what should be my conclusion. Because I know a photon will travel in a straightest path possible in a curved spacetime and I have seen comparisons made between the trajectory of a photon and a geodesic.

If the clock comes back and it's off can I say that there was a distortion in distance between the two mirrors of the clock. Or should I think that the speed of light itself was somehow distorted.

I know that if I was travelling with the clock the speed of light inside the clock would not change, but then nothing else travels faster than the speed of light, I wouldn't be able to tell if the light slowed down or the space shrunk or grew.

Can someone help me with this, thanks. ( Do not be afraid to use math. )


1 Answer 1


General relativity doesn't have a fundamental notion of distance, but the same is true of special relativity. The big problem being that, unlike inertial observers in special relativity, there's no "canonical" foliation of the spacetime into spacelike hypersurfaces.

In other words, for some inertial observer $\gamma$ in special relativity, there is a canonical foliation of Minkowski space $M$ into the spacelike hypersurfaces $\Sigma_t$, with a time function $t$, which is defined by the inertial frame in which $\gamma$ is at rest.

This doesn't entirely fix the foliation, but if we add the Einstein synchronization condition, which is that for a light clock emitting at point $p_1$, reflecting at $q$, and received at $p_2$, then we have a time function such that

$$t(q) = \frac{1}{2}(t(p_1) + t(p_2))$$

which leads to the Synge formula for distance, where the distance (at time $t(q)$) of the observer to the point $q$ is

$$d(q) = c \sqrt{(t - t_1)(t_2 - t)}$$

There is no such canonical foliation in general relativity. If our spacetime is nice enough, we can define a time function which assigns a time value to every point, and then compute the distance by simply considering the usual distance function on a spacelike hypersurface, with the usual formula for distances in Riemannian manifolds :

$$d(p,q) = \min_{\gamma}(\int_{p}^{q} \sqrt{g(\dot{\gamma}(\lambda), \dot{\gamma}(\lambda))}d\lambda)$$

ie the smallest proper length to that point. But of course, that distance is somewhat arbitrary, as we can choose another time function to give us a different distance. Note also that this distance is dynamic : even if we consider two objects at rest, they may still drift apart, because the metric tensor may change with time (this happens for the expansion of the universe, for instance).

Light clocks can be used to define such canonical coordinates, called radar coordinates. After all, these correspond to physical measurements, and these are coordinate invariants. But beware that you can't actually make an entire map of a spacetime with radar coordinates in general, as they may not work out for very large distances. On the other hand, there is always some neighbourhood where those coordinates exist, and on a more practical level, they tend to be valid at pretty large distances if the gravity is low enough.

You can then define a synchronization, a time function, and a distance as in the case of special relativity, with which you can check for length contraction. As general relativity has to reduce to special relativity locally, length contraction does indeed happen, and you could if you wish check the length contraction occuring between two "inertial" (unaccelerated) observers. The distance will indeed change depending on the speed of the observer, but also depend on the curvature of the spacetime.

Locally, the geometry of times and distances of general relativity is almost entirely identical to special relativity. If we have three points $p, q, r$ connected by three geodesics, their distances are related by

\begin{equation} d^2(p,q) = d^2(r,p) + d^2(r,q) - 2 d(r,p) d(r,q) \cos(\theta) + \phi \end{equation}

where the distance $d$ can be positive, negative or null, depending on how those points are organized, and $\theta$ is the angle between $rp$ and $rq$. This is exactly the same formula as the elementary trigonometry formula for Minkowski space, except for the last term $\phi$, which depends on the spacetime curvature. It's a rather complicated term but it can be roughly considered as the integral of a series of products of the Riemann tensor.

So, for $\phi \approx 0$, you will get your usual length contraction, implying a very small distortion due to the curvature. For a larger $\phi$, the difference of length between two points will not be solely due to length contraction as it is understood in special relativity : the change of the metric tensor will also play a role, making that distance shorter or longer (usually even shorter).

If you'd like more derivations regarding distances in general relativity, there is quite a lot in Synge's book "Relativity, the general theory".

  • $\begingroup$ Thanks very much for your comprehensive reply Slereah. One part I'm confused about is the part of the foliation of a curved manifold. Since all point on a Riemann manifold can be mapped into a flat chart into R^d and these charts overlap other points I should be able to add all of these charts and recreate the manifold which in turn should allow me to take measurements of entire swatches of the manifold. Say if I had a spacetime and I map into R^4 charts. If the manifold has a normal metric like Schwarzschild outside of the event horizon that does not change with time. $\endgroup$
    – sqljunkey
    Sep 20, 2020 at 20:17

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