Speed of light in the Schwarzschild metric or Gullstrand metric

In the Schwarzschild metric, the speed of light will be given by:

$$\frac{\mathrm{d}r}{\mathrm{d}t}= \Big(1-\frac{2M}{r}\Big)$$

and in the Gullstrand metric, it will be given by:

$$\frac{\mathrm{d}r}{\mathrm{d}t}= 1- \sqrt{\frac{2M}{r}}$$

As is evident they are clearly different from $$c$$ in suitable regions.

But SR tells us that the speed of light is an invariant and a constant of the universe.

So what is happening here? Why is it not an issue that we are getting that the speed of light is different from $$c$$ in different coordinates?

I will focus on the part of your question regarding the Schwarzschild metric.

I assume that what you have done to find that expression is to set $$\mathrm{d}s^2 = 0$$, considered radial motion only, and simplified the expression for the metric to get the ratio $$\left(\frac{\mathrm{d}r}{\mathrm{d}t}\right)^2 = \left(1 - \frac{2M}{r}\right)^2.$$

This does result (up to a $$\pm$$ sign) in what you have written, but it is not the speed of light!

Instead, this is the coordinate speed of the light ray: it moves $$1-2M/r$$ units of $$r$$ per each unit of $$t$$. This is not the speed of light since neither $$r$$ nor $$t$$ correspond directly to physical length or physical time.

Physical radial distance is measured by the coordinate $$\tilde{r}$$ defined by $$\mathrm{d}\tilde{r}^2 = \mathrm{d}r^2 / (1 - 2M /r)$$, and physical time is measured by the coordinate $$\tilde{t}$$ defined analogously. These coordinates create issues if you try to extend them to the whole spacetime, but locally they work well enough. The reason for this is that the line element definitionally describes physical distances: if you want to measure them, you always need to integrate it. In our case, we are interested in distances along a certain coordinate axis, so we can simply consider that element of the metric.

So, locally the metric will look like $$\mathrm{d}s^2 = - \mathrm{d}\tilde{t}^2 + \mathrm{d}\tilde{r}^2$$, and with the same manipulation you have done you find $$\frac{\mathrm{d} \tilde{r}}{\mathrm{d}\tilde{t}} = \pm 1$$ as expected.

• 2 small follow on issues- 1) what’s the issue in extending them to extend a bit more rather than being just defined locally? 2) the speed of light in accelerating frames is greater or lesser than c. Isn’t that an issue? Feb 17 at 19:36
• One issue (regarding 1) is the fact that since the coordinate multiplying the angular part is $r$ and not $\tilde{r}$ if you measure the (physical!) area of a sphere at coordinate radius $r$ you get $4 \pi r^2$, while the physical radius would be given by integrating $\tilde{r}$. Feb 17 at 19:40
• Plus, the coordinate $\tilde{r}$ has the issue of diverging at the horizon... Feb 17 at 19:42
• Regarding 2, the physical speed of light in a local inertial frame is always measured to be $c$. You can describe the trajectory of a light ray with many small inertial frames, so it makes sense to say that the physical speed is always $c$ locally (or equivalently, $\mathrm{d}s^2 = 0$ always) --- however, you often cannot find a global reference frame with wich to state that "total distance over time" is $c$. Feb 17 at 19:45

But SR tells us that the speed of light is an invariant and a constant of the universe.

So what is happening here? Why getting the speed of light different from c in different coordinates not an issue?

SR only tells us that the speed of light in an inertial frame is equal to c. Neither the Schwarzschild nor the Gullstrand coordinates are inertial everywhere. Even in flat spacetime it is easy to get speed of light violations in non-inertial coordinates, and in curved spacetime there are no global inertial frames so it is essentially unavoidable in global coordinates.

• So will the speed of light in an accelerating coordinate be greater or lesser than c. Isn’t that an issue? Feb 17 at 19:22
• It can be both. The easiest example is a rotating coordinate system. There the tangential speed will be greater than c in one direction and less than c in the opposite direction. No, it isn't an issue as long as you don't accidentally use inertial frame equations in non-inertial frames.
– Dale
Feb 17 at 19:27