# In general relativity what is the relative velocity between a fixed observer and one free-falling from a finite radius?

This question is the last missing link for me to solve my other question here.

In general relativity, what is the relative velocity between an observer in free fall from a distance $$r_1$$ from a mass $$M$$ and an observer stationary at distance $$r$$ when the first one crosses the sphere of radius $$r$$?

This is definitively not the value of $$dr/dt$$ for the free-falling observer.

Indeed, I have found that, if $$r_S={2GM}/c^2$$ is the Schwarzschild radius of the black hole with mass $$M$$, (even if the object of mass M is not particularly compact, but one is completely outside it) for an observer falling from infinity with zero initial velocity (that would take some time... )

$$\frac{dr}{dt} = c\left(1 - \frac{r_S}{r}\right) \sqrt {\frac{r_S}{r}}$$

while the quantity I am looking for, the relative velocity with an observer stationary at radius $$r$$ is just

$$V = c\sqrt {\frac{r_S}{r}}$$

But my question is, what about an observer that falls from a distance $$r_1$$ with zero initial velocity?

By intuition I have convinced myself that the answer is $$V = c\sqrt {\frac{r_S}{r}}\sqrt {\frac{r_1-r}{r_1-r_S}}$$

which coincides with the previous value when $$r_1$$ increases to infinity and coincides with the Newtonian value

$$V = \sqrt {\frac{2GM}{ r}-\frac{2GM} {r_1}}$$

when $$r$$ and $$r_1$$ are both much larger than $$r_S$$. But this is not the unique expression that satisfies these two constraints. For instance, if the last factor was $$\sqrt {\frac{r_1-r}{r_1}}$$ the $$r_1\to \infty$$ and Newtonian limits would be unchanged. But this "feels" wrong.

Can anyone confirm that my intuitive answer is indeed the correct one ?

• There is no concept of relative velocity between two separated points in GR. In general, when this happens you need to go back and figure out why exactly you're looking for this concept, and this will tell you the question you should actually be asking. Commented Jan 4, 2022 at 13:23
• You're right, I'm sorry, I read your question too quickly. Commented Jan 4, 2022 at 18:24
• @Javier It's OK, I just deleted my comment. Thanks for confirming my guess. I did expect this relative velocity to be exactly $c$ at $r=r_S$ for finite jump as it is for jump from infinity. But I didn't know how to prove it. Commented Jan 5, 2022 at 1:26

## 1 Answer

Your intuitive guess is indeed correct. To find the velocity, we first use conservation of energy. The energy per unit mass of a particle is given by

$$e = -g_{tt} \dot{t} = \left(1 - \frac{r_s}{r}\right) \dot{t}.$$

When the particle is at rest at $$r_1$$, its four velocity $$u^\mu = (\dot{t}, \dot{r}, \dot{\theta}, \dot{\phi})$$ has only a $$t$$-component, and from the normalization $$g_{\mu\nu} u^\mu u^\nu = -1$$ we can find that

$$\dot{t}|_{r_1} = \frac{1}{\sqrt{1-r_s/r_1}}.$$

Conservation of energy then lets us find $$\dot{t}$$ at an arbitrary radius:

$$\dot{t} = \frac{\sqrt{1-r_s/r_1}}{1-r_s/r}.$$

Next, we need the four-velocity of the observer, which by the same argument has only a $$t$$-component given by

$$U_o^t = \frac{1}{\sqrt{1-r_s/r}}.$$

We're ready to put everything together: the four-velocity of the observer defines the $$t$$-axis of a local inertial frame, and we want the $$t$$-component of the particle's velocity in that frame. Since $$U_o^\mu$$ is normalized, this is just given by

$$u^t|_{\text{inertial}} = -g_{\mu\nu}U_o^\mu u^\nu|_\text{Schwarzschild},$$

where $$u^\nu|_\text{Schwarzschild} = (\dot{t}, \dot{r}, \dot{\theta}, \dot{\phi})$$. A lot of terms are zero, so we just have

$$u^t|_\text{inertial} = -g_{tt} U_o^t \dot{t} = \left(1-\frac{r_s}{r}\right) \frac{1}{\sqrt{1-r_s/r}} \frac{\sqrt{1-r_s/r_1}}{1-r_s/r}.$$

But in a local inertial frame we know that special relativity applies, so that the $$t$$-component of the four velocity is the Lorentz factor $$\gamma = (1-v^2)^{-1/2}$$. Solving for the velocity $$v$$ gives your expression as the answer.