# Is there a difference between what Schwarzschild vs Gullstrand-Painleve predict for c?

When we set the interval(ds) for a given coordinate(i.e. Schwarzschild coordinates) to 0 to calculate for the velocity of light in a gravitational field, we can arrive at an equation such as the one John Rennie finds.

The final result is:

$$\frac{\mathrm{d}r}{\mathrm{d}t} = v_c = \left(1-\frac{r_s}{r}\right)$$

(c=1)

But when I compare this with that of the Gullstrand-Painleve metric, we find there are two roots:

$$\frac{\mathrm{d}r}{\mathrm{d}t_r} = v_c = ± 1 - \sqrt{\frac{r_s}{r}}$$

The plus and minus 1 is dependent on the direction that the light is traveling.

If we suppose that the Schwarzschild metric should have one more solution/root, I get the following graph:

Since 2 is the event horizon, we can see light going in either direction slows down to a complete stop at the horizon.

While if I graph the Gullstrand-Painleve solution, I arrive at the following:

We can see that the direction light is traveling affects the velocity seen by the coordinate, which has been described as the river model. If the photon is traveling away from the singularity, it will sit at the event horizon until the singularity gains more mass. But if it is moving the opposite direction, it will "flow with the river" and continue at 2c through the event horizon.

So my question summed up is are these in agreement? Or is my second root for the Schwarzschild solution incorrect?

Having a tough time getting my head around what seems to be a discrepancy between the two. My main point of interest is right before the event horizon where we don't have maximal time dilation, yet.

Yes, your two results are in agreement.

It seems unintuitive that in one case the ingoing and outgoing velocities have the same magnitude while in the other case they don't, but this is simply because the time coordinate is different in the two cases.

It's important to realise that the Gullstrand-Painlevé velocity $dr/dt_r$ is not a velocity that any observer could measure, because the $r$ and $t_r$ coordinates are taken from the coordinates of two different observers. It is a purely abstract velocity and we have to be very careful about attaching a physical significance to it. All observers making a local measurement of the speed of light, i.e. a measurement of the speed of light at their location, will find the velocities are always $\pm c$. The difference in the ingoing and outgoing velocity is purely an artefact of the coordinates.

This is invariably glossed over in the proofs that light cannot escape from an event horizon. We tend to say oh look, $dr/dt_r=0$ so the light can't escape and leave it at that. I have to confess I did this as well in my treatment of the problem.

• Thank you for your answer, John. My perspective on this(for what it's worth) is that it shows what a distant observer far away would see of the velocity of light at each point if they could somehow watch the light beams going in or away from the bh(which is impossible, of course). I'm just initially surprised that the ingoing light of the sc metric also comes to a halt, instead of falling in faster. But you are right, it must do so or the local observer would measure light moving faster than c. Thus time dilation of gravity effects matter and light the same. – user2299067 Jan 18 '18 at 12:29
• Is it possible that the change in the speed of light is due to less density of space? For example, in air, if the density is lower then the speed of sound slows down, while if air is more dense, the speed of sound increases. Thus in flat space, the speed of light is c, but at the same time, matter undergoes the same effect as it's just another form of light(matter/antimatter;left-handed vs right-handed), and so measures no difference locally. But we can measure Shapiro delay of light traveling in and out of the gravitational field. – user2299067 Jan 23 '18 at 2:58
• @user2299067: I wrote a long and involved discussion of the issue here if you're interested. – John Rennie Jan 23 '18 at 6:09
• #John Rennie - Your simplification of the velocity of c throughout the metric got me thinking. QM teaches us that matter are composed of matter waves. Given the intrinsic 2 pi in the Planck constant, the energy seems to be moving in a circle. If it's a wave, then it ought to follow Snell's law of refraction at each point.Thus a gradient speed of c would seem to cause an acceleration toward the lower c-speed region. That acceleration just comes about by applying Snell's law of refraction to each part of that circular wave path. Could it be that this is the very underlying "cause" of gravity? – user2299067 Mar 5 at 6:13
• @user2299067 the $2\pi$ in $\hbar$ doesn't mean anything is travelling in a circle. The $2\pi$ appears because plane waves (moving in a straight line not a circle) are described by an equation like: $\sin(2\pi f t - 2\pi x/\lambda)$. – John Rennie Mar 5 at 6:46