# Does light really travel more slowly near a massive body?

It is a routine problem for beginners in general relativity to calculate the coordinate velocity of light for the Schwarzschild metric. Starting from the metric:

$$ds^2 = -\left(1-\frac{r_s}{r}\right)c^2dt^2 + \frac{dr^2}{1-\frac{r_s}{r}} + d\Omega^2$$

We use the fact that light travels on a null geodesic so $ds^2 = 0$. This immediately gives us for a radial light ray:

$$\frac{dr}{dt} = \pm c \left(1 - \frac{r_s}{r}\right) \tag{1}$$

But this is a coordinate speed, a three-vector, not a covariant object and so it has no absolute meaning. Is there an easy way to see that this is just the speed for a particular observer and that other observers will measure a different speed?

• I would expect that correct way to discuss (any) speed as measured by observer is tetrad formalism. – A.V.S. Apr 14 '18 at 8:49
• @A.V.S. please feel free to add another answer. The answer I've posted is what I think is the simplest way to explain the issue to beginners in GR but a more sophisticated answer would be very welcome. – John Rennie Apr 14 '18 at 8:57
• I wish I could upvote this Q/A twice! It fills an extremely important gap on this site, and I'm sure it'll be very useful for many. – knzhou Apr 15 '18 at 13:05
• What is a 3-vector? I guess it is the components of a 4-vector expressed in a local orthonormal frame, but dropping the "time" component. MTW section 2.8 seem to use it this way. – Colin MacLaurin Apr 18 '18 at 0:14

The simple way to show that the speed derived from the Schwarzschild coordinates has no absolute meaning is to derive an expression for the speed measured by a different observer and show that they disagree. In particular we will choose a shell observer i.e. an observer hovering at fixed $r$, $\theta$ and $\phi$ (presumably using some form of rocket motor). Once again we will consider a radial light ray.

We will use $t’$ and $r’$ for the time and radial coordinates in the shell frame, and $R$ for the radial distance of the shell observer measured in the Schwarzschild coordinates.

In the rest frame of the shell observer we consider the infinitesimal proper time $dt’$. Comparing this with the Schwarzschild metric we find at the position of the shell observer:

$$dt’^2 = (1 - r_s/R)dt^2$$

giving us:

$$\frac{dt’}{dt} = \sqrt{1 – r_s/R} \tag{2}$$

The sharp eyed among you will spot that this is just the well known expression for the gravitational time dilation at a distance $R$. A similar argument for the infinitesimal proper distance $dr’$ gives:

$$\frac{dr’}{dr} = \frac{1}{\sqrt{1 – r_s/R}} \tag{3}$$

which is just the corresponding equation for the radial dilation. Using equation (1) from the question and equations (2) and (3) we can now calculate the speed of light at the position of the shell observer using the chain rule:

\begin{align} \frac{dr’}{dt’} &= \frac{dr}{dt} \frac{dr’}{dr} \frac{dt}{dt’} \\ &= \pm c \left(1 - \frac{r_s}{R}\right) \frac{1}{\sqrt{1 – r_s/R}} \frac{1}{\sqrt{1 – r_s/R}} \\ &= \pm c \end{align}

And there is our first result. The shell observer measures the speed of light at their location to be $c$, and this is independent of $R$ so it is true for all shell observers.

I must emphasise that I have made no assumptions in this working. It is pure algebra and allows no room for subterfuge. The two observers really do find different results for the speed of light. Neither is right and neither is wrong - it just shows that the coordinate speed of light is observer dependent not an absolute quantity.

But we can do better than this. We can extend our analysis to find the speed of light in the shell coordinates at radial distances greater and less than the shell distance. The argument is essentially the same as above so I’ll just give the result:

$$\frac{dr’}{dt’} = \pm c \frac{1- r_s/r}{1 – r_s/R} \tag{4}$$

And this looks like (for $R = 2r_s$):

Like the Schwarzschild observer the shell observer sees the coordinate speed of light fall when the light is closer to the massive object than they are, but the shell observer sees the light move faster than $c$ when the light is farther from the object than they are. So the shell and Schwarzschild observer agree on the speed of the light nowhere (except at the event horizon if one exists) but they both agree that the speed of light at their location is $c$.

And this makes the point. Shell observers are not some theoretical fiction - you and I are shell observers by virtue of our constant distance from the centre of the Earth and equation (4) gives the speed of light that you and I would observe. The point is that the coordinate speed of light depends on the observer and has no absolute meaning.

• And the general proof of this would be – invoke equivalence principle to pass to a locally Minkowski (aka Galilean) frame, the existence of which means that the speed of light in all directions is $c$. – Prof. Legolasov Apr 14 '18 at 8:50
• @SolenodonParadoxus I wanted as explicit a demonstration as possible because this issue has been a source of contention on the site. However if you'd like to add another answer please go for it! – John Rennie Apr 14 '18 at 8:56
• Sure, and I agree 100%. Just wanted to point this other thing out mostly for OP. Don't think it is enough for a second answer. – Prof. Legolasov Apr 14 '18 at 8:57
• Dear John Rennie, you are saying "but the shell observer sees the light move faster than c when the light is farther from the object than they are". So you are saying it is possible to measure light travel faster than c? – Árpád Szendrei Apr 14 '18 at 9:05
• @ÁrpádSzendrei I'm committing the usual sin of using the terms see vaguely when I'm really referring to the assignment of the trajectory of the light to a set of points in the coordinate system. I'm doing this because the answer is aimed at beginners in GR and I didn't want to confuse them with overly technical language. If you're asking could we construct an experiment to measure the speed of light that would give a result greater than $c$ then I think it can be done but I would have to think about the details. Maybe you could ask that as a separate question. – John Rennie Apr 14 '18 at 9:22