# Speed in general relativity

APhysicist
Hi all,

I'm starting to learn a bit of general relativity now, but I'm a bit confused as to the measurement of speed.
Let's say we use the Schwarzschild metric.
Now, I can parametrize the wordline of a certain object by giving a
parametrisation (t(tau),r(tau),theta(tau),phi(tau)), where tau is the proper time of
that object. Now, I would like to know how to measure the speed of the object;
that is, the magnitude of its 3-velocity. I can find the 4-velocity, but can I deduce in a meaningful way the speed, i.e. as observed by an observer 'at rest' in the frame
(of course, 'at rest' is a relative concept - I'm referring to the observer at infinity, where the Schwarzschild metric is more or less flat)?

From special relativity, I recall that you could write the 4-velocity
as (gamma,gamma*V) where gamma is the usual gamma factor and V is the three-velocity.
I'm looking for something analogous - if that's meaningful.

Thanks,

APhysicist

## Answers and Replies

Staff Emeritus
Gold Member
From special relativity, I recall that you could write the 4-velocity
as (gamma,gamma*V) where gamma is the usual gamma factor and V is the three-velocity.
I'm looking for something analogous - if that's meaningful.

This is meaningful in general relativity if the two observers are coincident.

Altabeh
Hi all,

I'm starting to learn a bit of general relativity now, but I'm a bit confused as to the measurement of speed.
Let's say we use the Schwarzschild metric.
Now, I can parametrize the wordline of a certain object by giving a
parametrisation (t(tau),r(tau),theta(tau),phi(tau)), where tau is the proper time of
that object. Now, I would like to know how to measure the speed of the object;
that is, the magnitude of its 3-velocity. I can find the 4-velocity, but can I deduce in a meaningful way the speed, i.e. as observed by an observer 'at rest' in the frame
(of course, 'at rest' is a relative concept - I'm referring to the observer at infinity, where the Schwarzschild metric is more or less flat)?

Thanks,

APhysicist

In general relativity, all we know about the proper velocities $$u^{\alpha}=\frac{dx^{\alpha}}{d\tau}$$ is that they satisfy

$$g_{\alpha\beta}u^{\alpha}u^{\beta}=c^2.$$

For instance, if we have

$$ds^2=c^2dt^2-dx^2-dy^2-dz^2,$$

then

$$c^2(u^{0})^2-(u^{i})^2=c^2.$$

AB

Staff Emeritus
Gold Member
In general relativity, all we know about the proper velocities $$u^{\alpha}=dx^{\alpha}\ d\tau$$ is that they satisfy

$$g_{\apha\beta}u^{\alpha}u^{\beta}=c^2.$$

For instance, if we have

$$ds^2=c^2dt^2-dx^2-dy^2-dz^2,$$

then

$$c^2(u^{0})^2-(u^{i})^2=c^2.$$

AB

But if two observers are coincident at an event, then the proper velocity of one observer can be used to express the proper velocity of the other observer in such a way that gamma and 3-velocity naturally appear.

Altabeh
But if two observers are coincident at an event, then the proper velocity of one observer can be used to express the proper velocity of the other observer in such a way that gamma and 3-velocity naturally appear.

I talked about a general way of exposing 4-velocity in GR! I don't know how one can make the gamma and 3-velocity appear in GR through an assumption like yours, but I know if we take the particles to be moving on geodesics for which all three spacelike components remain constant by assuming that the observer is co-moving with particles, then the 4-velocity takes a form

$$u^{\alpha}=(c,0,0,0).$$

We can only have the gamma and 3-velocity appear if we are in a locally inertial frame, measuring the 4-velocity of a particle from there.

AB

Staff Emeritus
Gold Member
I talked about a general way of exposing 4-velocity in GR!

So am I!
I don't know how one can make the gamma and 3-velocity appear in GR through an assumption like yours

You should  . This is related to the way I solved the SR elastic collision problem.
We can only have the gamma and 3-velocity appear if we are in a locally inertial frame, measuring the 4-velocity of a particle from there.

In both my previous posts in this thread, I said the that two observers are coincident. Are you saying that if observer A is coincident with observer B, then A can't measure B's physical speed (relative to A)?

For two such observers in general relativity,

$$\gamma = \mathbf{g} \left( \mathbf{u}_A , \mathbf{u}_B \right).$$

Notice that this expression does not explicitly make a choice of coordinate, but physical speed is used in the definition of $\gamma$.

Gold Member
For the benefit of the original questioner, let me spell out more explicitly George's method.

Let v be the 4-velocity of an object to be measured, and u be the 4-velocity of any local observer "at rest" at the event where the measurement is required.

In the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event, $u^\alpha = (c, 0, 0, 0)$ and $v^\alpha = (\gamma c, \gamma v, 0, 0)$, where v is the magnitude of the 3-velocity that you want to measure (let's assume it's along the x-axis) and $\gamma = (1-v^2/c^2)^{-1/2}$. Therefore

$$g_{\alpha\beta}u^{\alpha}v^{\beta} = \gamma c^2$$​

But $g_{\alpha\beta}u^{\alpha}v^{\beta}$ is invariant, i.e. the same in all coordinate systems, so you can also calculate it in Schwarzschild coordinates, and equate the two values.

Altabeh
For the benefit of the original questioner, let me spell out more explicitly George's method.

Let v be the 4-velocity of an object to be measured, and u be the 4-velocity of any local observer "at rest" at the event where the measurement is required.

In the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event, $u^\alpha = (c, 0, 0, 0)$ and $v^\alpha = (\gamma c, \gamma v, 0, 0)$, where v is the magnitude of the 3-velocity that you want to measure (let's assume it's along the x-axis) and $\gamma = (1-v^2/c^2)^{-1/2}$. Therefore

$$g_{\alpha\beta}u^{\alpha}v^{\beta} = \gamma c^2$$​

But $g_{\alpha\beta}u^{\alpha}v^{\beta}$ is invariant, i.e. the same in all coordinate systems, so you can also calculate it in Schwarzschild coordinates, and equate the two values.

I think your explanation is clear and catchy (though I see something wrong), I want to do the calculations my way that sounds more comfortable at least for me. I take the EP for granted as an accurate principle locally and assume the free-falling observer is comoving with the observer, so $$u^{\alpha}=\frac{dx^{\alpha}} {d\tau}=(1,0,0,0).$$ The observer is measuring the velocity of a particle moving along x axis. Now we know that

$$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=c^2,$$

On the other hand the particle is experiencing the local Minkowski frame, thus

$$ds^2=c^2dt^2-dx^2-dy^2-dz^2=c^2dt^2(1-v^2/c^2),$$

where $$v$$ is the particle's 3-velocity (according to the observer) and by definition, $$v^0=dt/dt=1.$$ So for the particle we can write

$$\gamma^2=\frac{dt^2}{d\tau^2},$$

with $$\gamma$$ being the Lorentz factor. Introducing this result and the velocity $$u^{\nu}$$ into the first equation yields

$$g_{\mu\nu}v^{\mu}u^{\nu}=\gamma^{-1}c^2.$$

This is my method and yet is different than yours! Now let's set an example for this. I would like to take into account the linearized isotropic Schwarzschild metric,

$$ds^2=c^2(1-2m/r)dt^2-(1+2m/r)(dx^2+dy^2+dz^2).$$

For this metric, our formula gives

$$c^2(1-2m/r)v^{0}u^{0}=\gamma^{-1}c^2$$ or
$$(1-2m/r)v^{0}=\gamma^{-1}$$

and finally

$$\frac{v^2}{c^2}=\frac{2m}{r}.$$

(I don't this gives the right answer. Could someone check it?)

Remember that this is the same as the coordinate velocity because we are assuming the observer to be comoving with particle. I don't know if I've not made a mistake here and of course it would be great to tell me if there's any! I am definitely not familiar with George's method so my understanding of how actually he calculates things like this is so limited.

AB

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Gold Member
$$g_{\mu\nu}v^{\mu}u^{\nu}=\gamma^{-1}c^2.$$
I don't understand your logic for saying this. You just showed in the line above that $dt/d\tau = \gamma$ for the observed particle (v0), and earlier that $dt/d\tau = 1$ for the observer (u0) from which it follows that

$$g_{\mu\nu}v^{\mu}u^{\nu}=\gamma c^2$$​

Now, in Schwarzschild coordinates $(t, r, \theta, \phi)$, $g_{00} = c^2(1 - 2GM/rc^2)$ and the other components of the metric don't matter in this case. We have

$$u^{\alpha} = \left(\frac{1}{\sqrt{1 - 2GM/rc^2}}, 0, 0, 0\right)$$​

(The last three components must be zero and the first is chosen to ensure that $g_{\mu\nu}u^{\mu}u^{\nu} = c^2$.) And

$$v^{\alpha} = \left(\frac{dt}{d\tau}, \frac{dr}{d\tau}, \frac{d\theta}{d\tau}, \frac{d\phi}{d\tau}\right)$$​

And so

$$\frac{c^2}{\sqrt{1-v^2/c^2}}= g_{\mu\nu}v^{\mu}u^{\nu} = c^2\frac{dt}{d\tau}\sqrt{1 - 2GM/rc^2}$$​

if I haven't made a silly mistake in the algebra.

That tells you how to calculate v from $dt/d\tau$ and r for any particle.

I don't really follow what you were trying to do in the last part of your post. You seem to have made some assumption about the motion of the particle, whereas the original questioner did not.

Altabeh
I just took $$v^{\alpha}$$ to be the coordinate velocity just because the comoving observer would be found at rest relative to the particle, so

$$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}u^{\nu}=g_{\mu\nu}\gamma v^{\mu}u^{\nu}= c^2.$$

I think your equation sounds flawless! But it is clear that, for example, in the case of an orbiting particle, we must be given the value of $$\frac{d\phi}{d\tau}$$. But if the motion occurs in the equatorial plane $$\theta={\pi}/ 2$$ and the particle is hovering at $${ (r_0,\theta_0,\phi_0)},$$ then

$$v^2/c^2 = 2GM/c^2{r_0}=2m/r_0,$$

which is the same as the result given in my earlier post:

and finally

$$\frac{v^2}{c^2}=\frac{2m}{r}.$$

But as I said, I don't think this is true as it gives such a very high value for the speed $$v$$ which is about 11,250 m/s for a particle hovering at the surface of earth unless my algebra would have went wrong somewhere! Does this sound logical?

AB

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Gold Member
I just took $$v^{\alpha}$$ to be the coordinate velocity just because the comoving observer would be found at rest relative to the particle, so

$$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}u^{\nu}=g_{\mu\nu}\gamma v^{\mu}u^{\nu}= c^2.$$

I think your equation sounds flawless! But it is clear that, for example, in the case of an orbiting particle, we must be given the value of $$\frac{d\phi}{d\tau}$$. But if the motion occurs in the equatorial plane $$\theta={\pi}/ 2$$ and the particle is hovering at $${ (r_0,\theta_0,\phi_0)},$$ then

$$v^2/c^2 = 2GM/c^2{r_0}=2m/r_0,$$

which is the same as the result given in my earlier post:

But as I said, I don't think this is true as it gives such a very high value for the speed $$v$$ which is about 11,250 m/s for a particle hovering at the surface of earth unless my algebra would have went wrong somewhere! Does this sound logical?

AB
I understand your confusion now. I am using $u^{\alpha}$ and $v^{\alpha}$ to refer to 4-velocity which is defined to be $dx^{\alpha}/d\tau$, not $dx^{\alpha}/dx^0$. (The expression $dx^{\alpha}/dx^0$ does not transform correctly to be a tensor.)

Note that as

$$g_{\alpha \beta} \, \frac{dx^{\alpha}}{d\tau} \, \frac{dx^{\beta}}{d\tau} = c^2$$​

then the value of $d\phi/dt$ has an effect on the value of $dt/d\tau$, and therefore affects the value of v in the formula I gave before.

Staff Emeritus
Gold Member
But as I said, I don't think this is true as it gives such a very high value for the speed $$v$$ which is about 11,250 m/s for a particle hovering at the surface of earth unless my algebra would have went wrong somewhere! Does this sound logical?

According to Newton, what is the escape speed for a particle that starts on the Earth's surface?

Altabeh
According to Newton, what is the escape speed for a particle that starts on the Earth's surface?

A genius point, man! :tongue:

I just took v to be the coordinate velocity
is equivalent to what you, DrGreg, said about my confusion! Anyways, thank you guys!

AB

Staff Emeritus
Gold Member
A genius point, man! :tongue:AB

Far from it! I've worked with this expression many times in a variety of ways; for example, in terms of frames in

In this, $\mathbf{e}_0$ is the 4-velocity of the hovering observer, and $\mathbf{e}'_0$ is the the 4-velocity of an observer who falls freely from rest at infinity.

Staff Emeritus
Gold Member
yuiop
...
Now, I would like to know how to measure the speed of the object;
that is, the magnitude of its 3-velocity. I can find the 4-velocity, but can I deduce in a meaningful way the speed, i.e. as observed by an observer 'at rest' in the frame
(of course, 'at rest' is a relative concept - I'm referring to the observer at infinity, where the Schwarzschild metric is more or less flat)?

My guess is that the 3 velocity of a particle in Schwarzschild coordinates according to an observer at infinity is:

$$\sqrt{ g^2 u_x^2 + g u_y^2 +g u_z^2 }$$

where g = $(1-2m/r)$, r is the radial location of a stationary local observer that measures the local velocity of the particle to be

$$\sqrt{ u_x^2 + u_y^2 + u_z^2 }$$

and u_x is the radial component and u_y and u_z are the horizontal components of the particle's velocity.

I hope that is some help (and correct).

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Gold Member
kev, [STRIKE]it's not clear to me how you are defining ux, uy, uz. You need to spell it out a bit more explicitly. In what coordinate system and relative to whom?[/STRIKE] EDIT: I didn't read your post carefully enough, I'll now go away and think about it.

(The original questioner asked for something in terms of a parameterisation $(t(\tau), r(\tau), \theta(\tau), \phi(\tau))$.)

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yuiop
(The original questioner asked for something in terms of a parameterisation $(t(\tau), r(\tau), \theta(\tau), \phi(\tau))$.)

My original equation for the 3 velocity (V) of a particle in Schwarzschild coordinates according to an observer at infinity:

$$V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 }$$

where g = $(1-2m/r)$

could be re-expressed as:

$$\sqrt{ g^2 (dr/dt)^2 + g (r d\theta/dt)^2 + g (r sin(\theta) d\phi/dt)^2 }$$

if that makes things clearer, but to me it is a bit messy and detracts from the trivial point I was trying to make that the vertical velocity component is treated differently to the horizontal velocity components.

The OP did express an interest in the 3 velocity in Schwarschild coordinates which no one seems to have responded to so far and how the 3 velocity relates to the 4 velocity.

Locally, the four velocity W is the same as in SR, i.e.:

$$W = c = \gamma_v(c, u_x, u_y, u_z) = \gamma_v\sqrt{ c^2 - u_x^2 - u_y^2 - u_z^2 }$$

where $\gamma_v = 1/(1-v^2/c^2)^{1/2}$ and v is the local 3 velocity $(u_x, u_y, u_z) = (u_x^2 + u_y^2 +u_z^2)^{1/2}$.

The transformed four velocity W' in Schwarzschild coordinates to an observer at infinity is:

$$W' = c = \gamma_v'(c, u_x', u_y', u_z') = \gamma_v'\sqrt{ c^2 - u_x '^2 - u_y' ^2 - u_z '^2 } = \gamma_v'\sqrt{ c^2 - g^2 u_x^2 - g u_y^2 - g u_z^2 }$$

where $\gamma '_v = 1/(1-V^2/c^2)^{1/2}$ and V is the transformed 3 velocity defined at the top of this post.

Not so sure about the last transformation (but it does satisfy the requirement that W = W' =c) and sorry if the notation or formalism is not correct. I am just feeling my way, so feel free to correct as you see fit.

P.S. It turns out that the above equations are only true for the 3 and 4 velocity of a photon which affords a simplification because d(tau) can be set to zero. For the motion of a particle with mass and arbitrary velocity it gets a lot more complicated. I will have to come back to that when I have more time.

P.P.S. Actually, right now I am not sure any of the above is correct. Maybe someone can help sort out my mess :)

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Altabeh
Far from it! I've worked with this expression many times in a variety of ways; for example, in terms of frames in

In this, $\mathbf{e}_0$ is the 4-velocity of the hovering observer, and $\mathbf{e}'_0$ is the the 4-velocity of an observer who falls freely from rest at infinity.

I have a question here: How do you get the following relations?

\begin{align*}\mathbf{e}_{0}^{\prime} & =\left( 1-\frac{2M}{r}\right) ^{-1}\frac{\partial}{\partial t}-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\frac{\partial}{\partial r}\\\mathbf{e}_{1}^{\prime} & =-\left( \frac{2M}{r}\right) ^{\frac{1}{2}}\left( 1-\frac{2M}{r}\right) ^{-1}\frac{\partial}{\partial t}+\frac{\partial}{\partial r}\\\mathbf{e}_{2}^{\prime} & =\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\\\mathbf{e}_{3}^{\prime} & =\frac{1}{r}\frac{\partial}{\partial\phi}.\end{align*}

Far from it! I've worked with this expression many times in a variety of ways.

Deep down, I did mean that though the escape velocity is something so referred to in the everyday physics, but it didn't really come to my mind at the time I dealt with that problem straightly while it came to yours! We all work in the same branch of science; one is genius and the others are less so!

AB

Altabeh
The OP did express an interest in the 3 velocity in Schwarschild coordinates which no one seems to have responded to so far and how the 3 velocity relates to the 4 velocity.

Yeah, and we give a relation of the form

$$g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$$ (1)

where in the $$\gamma_v$$, the appearing v is the local 3-velocity

$$v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau).$$

Take the sides of the last equation in (1) to the power 2 and write $$(v^0)^2=V^2-v^2$$ where $$V^{\alpha}$$ denotes the 4-velocity of the particle. Hence

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and with $$u^2=(u^0)^2$$ we obtain finally

$$V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

If your method doesn't lead to this equation, then it doesn't work!

AB

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amyfrog
Hi. does Schwarzschild use electomagnetic radiation (induction) to describe light.

Hi,

I also have a question about this. What about the speed of light in general relativity? Already in special relativity accelerating observers won't measure the speed of light to be c, right? But how can you explicitly show this?

In general relativity I would expect that only freely falling observers (following geodesics) would locally measure the speed of light to be c. Again: how can I show this explicitly, and is my intuition right?

I'll think about this, but any input would be nice :)

Altabeh
In general relativity I would expect that only freely falling observers (following geodesics) would locally measure the speed of light to be c. Again: how can I show this explicitly, and is my intuition right?

Who says that the second postulate of Relativity does no longer work in GR? The speed of light is always c in any branch of science and in GR all accelerating or non-accelerating observers find light travelling at c.

amyfrog
"The special theory of relativity has crystallised out from the Maxwell-Lorentz theory of electromagnetic phenomena. Thus all facts of experience which support the electromagnetic theory also support the theory of relativity." (Einstein, section 16)

amyfrog
"According to the theory of relativity, action at a distance with the velocity of light always takes the place of instantaneous action at a distance or of action at a distance with an infinite velocity of transmission." (Einstein, chapter 15).

Who says that the second postulate of Relativity does no longer work in GR? The speed of light is always c in any branch of science and in GR all accelerating or non-accelerating observers find light travelling at c.

Well, to start with, in GR vectors are defined in tangent spaces and in general there is no unique way to compare different tangent spaces. So the question itself is already subtle.

I have here the GR book by Gron and Hervik, which write down an explicit example of a photon described in a uniformly accelerating reference frame, in which they also mention the fact that the speed of light is only c locally, but in general it will differ due to the horizon involved.

I'm not sure how you would justify what you're saying, also because the mentioned problem with comparing velocities in GR at different places in spacetime.

Staff Emeritus
Gold Member
Physical speed is measured locally, and always is found to be c. In general, spatial distance is not defined in general relativity, so speed is not defined. Coordinate speed can be anything.

yuiop
The speed of light is always c in any branch of science and in GR all accelerating or non-accelerating observers find light travelling at c.

I think it would be better to say: ..in GR all accelerating or non-accelerating observers allways measure the local speed of light (in a vacuum) to be c.

I think it would be better to say: ..in GR all accelerating or non-accelerating observers allways measure the local speed of light (in a vacuum) to be c.
But why the accelerating observers? Is it not that accelerating observers in Minkowski spacetime don't measure the speed of light to be c?

Frame Dragger
I think it would be better to say: ..in GR all accelerating or non-accelerating observers allways measure the local speed of light (in a vacuum) to be c.

'c' IS the constant for the speed of light in vacuum and there is no such thing as "local c", but rather "local lightspeed". c = 299,792,458 m/s.

Altabeh
But why the accelerating observers? Is it not that accelerating observers in Minkowski spacetime don't measure the speed of light to be c?

It is a constant number and a number would not change frame by frame! The accelerating observers do measure the speed of light to be c if we all agree on the second postulate of relativity! In GR since we don't rely on a particular frame, the constancy of the speed of light can be safely used within any acceleraing frame!

AB

yuiop
'c' IS the constant for the speed of light in vacuum and there is no such thing as "local c", but rather "local lightspeed". c = 299,792,458 m/s.

The speed of light in a vacuum (in flat space) is a constant. (True locally and non-locally.)

The speed of light in a vacuum (locally) is a constant. (True in flat and curved space.)

The speed of light in a vacuum is a constant. (False in curved space for a non local region.)

I never said "local c".

I said "local speed of light" which is not much different in meaning to your "local lightspeed".

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yuiop
But why the accelerating observers? Is it not that accelerating observers in Minkowski spacetime don't measure the speed of light to be c?

That is true over extended spatial regions, but for very local measurements (infinitessimal) the speed of light is 'c' even for accelerating observers in Minkowski spacetime or for an observer on the surface of a very massive gravitational body.

For example, let us say an accelerating observer measures the speed of light over an extended distance dx to be (c+de), then in the limit as dx goes towards zero, the speed of light (c+de) goes towards c.

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yuiop
... Hence

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and with $$u^2=(u^0)^2$$ we obtain finally

$$V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

If your method doesn't lead to this equation, then it doesn't work!

AB

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and $$u^2=(u^0)^2$$

then I obtain:

$$V^2={\frac{{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

where does the extra $-{v}^{4}{g_{00}}^{2}{u}^{2}$ quantity come from?

yuiop
...
where does the extra $-{v}^{4}{g_{00}}^{2}{u}^{2}$ quantity come from?

I must have been tired when I posted the above. I have double checked the algebra and found there is nothing wrong with Altabeh's original result. Oops.
$$\frac{c^2}{\sqrt{1-v^2/c^2}}= g_{\mu\nu}v^{\mu}u^{\nu} = c^2\frac{dt}{d\tau}\sqrt{1 - 2GM/rc^2}$$​

if I haven't made a silly mistake in the algebra.

That tells you how to calculate v from $dt/d\tau$ and r for any particle.

In other words,

$$d\tau = dt \sqrt{1 - 2GM/rc^2}{\sqrt{1-v^2/c^2}}$$​

where v is the 3 velocity according to a local observer that is stationary in the Schwarzschild coordinates at radius r.

This is a nice result because it answers a question asked in this recent post https://www.physicsforums.com/showthread.php?t=385822 and confirms a result obtained (by a much longer method) by pervect and myself in a much older thread here https://www.physicsforums.com/showthread.php?t=355378 where we concluded that total time dilation is the product of gravitational and velocity time dilation rather than the sum.

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