What is the relationship between speed and coordinates 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

 

[George Jones]

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

 

[George Jones]

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

 

[George Jones]

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.

 

[DrGreg]

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

 

[DrGreg]

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

 

[DrGreg]

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.

 

[George Jones]

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!

I just took v to be the coordinate velocity

is equivalent to what you, DrGreg, said about my confusion! Anyways, thank you guys!

AB

 

[George Jones]

A genius point, man! AB

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

https://www.physicsforums.com/showthread.php?p=848684#post848684.

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.

 

[George Jones]

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

https://www.physicsforums.com/showthread.php?p=848684#post848684.

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.

And, more informally, at the bottom of

https://www.physicsforums.com/showthread.php?p=621802#post621802.

 

[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).

 

[DrGreg]

kev, [STRIKE]it's not clear to me how you are defining u_x, u_y, u_z. 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)).)

 

[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 Schwarzschild 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 :)

 

[Altabeh]

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

https://www.physicsforums.com/showthread.php?p=848684#post848684.

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 Schwarzschild 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}^{<br /> 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

 

[amyfrog]

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

 

[haushofer]

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 traveling 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).

 

[haushofer]

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 traveling 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.

 

[George Jones]

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 traveling 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.

 

[haushofer]

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".

 

[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.

 

[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}^{<br /> 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

If I start with:

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}^{<br /> 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.

 

[DrGreg]

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}^{<br /> 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

In a private message, kev asked me to comment on this post. But I can't work out what

(v^0)^2=V^2-v^2

is supposed to mean.

I'm not sure where this is going anyway. It seems to me that the original question was answered by my post #9 (& comments in #11).

 

[yuiop]

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}^{<br /> 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!

Hi Altabeh,

I would like to understand your notation better as I would like to know if my method outlined in #18 agrees with your method and because I think I might learn something from you. It might help people that refer to this thread in the future too. You never know, the OP might even take a look at this thread.

Could you state the exact location and state of motion of the observer for each velocity measurement u^0, v^0, u^i and v^i and if you are using proper time in each case.

Could you also confirm if all these velocities are 3 velocities (i.e. everything other than V)?

 

[Altabeh]

In a private message, kev asked me to comment on this post. But I can't work out what (v^0)^2=V^2-v^2 is supposed to mean.

Here v=(v^1,v^2,v^3) is the proper 3-velocity and V represents the magnitude of the 4-velocity v^{\alpha}.

I'm not sure where this is going anyway. It seems to me that the original question was answered by my post #9 (& comments in #11).

This doesn't have anything to do with the original question. I actually answered by this to a question brought up by kev in post #18.

Hi Altabeh,

Hi

I would like to understand your notation better as I would like to know if my method outlined in #18 agrees with your method and because I think I might learn something from you. It might help people that refer to this thread in the future too. You never know, the OP might even take a look at this thread.

In the setup I use to denote the 4-velocities of observer, u^{\alpha}, and particle, v^{\alpha}, I assume the observer is instantaneously at rest at the event the proper 4-velocity v^{\alpha} is observed. Thus we have

u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)

at the event of observation. From now on it is easy to calculate the result already obtained by DrGreg, i.e. d\tau = dt \sqrt{1 - 2GM/rc^2}{\sqrt{1-v^2/c^2}}.

But about your method I think you should first elaborate the steps you take to for example get the magnitude of the proper 3-velocity, V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } . And then explain what do you mean by 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 }? Here I see some misleading arguments like equality of a 4-vector with a scalar value! After doing so, I'd be able to get the idea of your calculations!

Could you also confirm if all these velocities are 3 velocities (i.e. everything other than V)?

The only 3-velocity, which is basically a coordinate velocity, is the velocity we use in the Lorentz factor \gamma_v=\frac{1}{\sqrt{1-v^2/c^2}}. Other than that, all velocities used in calcualtions are proper 4-velocities!

AB

 

[yuiop]

Hi Altabeh,

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.

Here v=(v^1,v^2,v^3) is the proper 3-velocity and V represents the magnitude of the 4-velocity v^{\alpha}.

OK, (v^0)^2 is a 4 velocity so it has the value c^2 and V^{2} is also a 4 velocity so it also has the value c^2 by definition so:

(v^0)^2=V^2-v^2 means c^2 = c^2-v^2 which is only true when v=0. Is that what you intended?

In the setup I use to denote the 4-velocities of observer, u^{\alpha}, and particle, v^{\alpha}, I assume the observer is instantaneously at rest at the event the proper 4-velocity v^{\alpha} is observed. Thus we have

u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)

at the event of observation.

I take "observer is instantaneously at rest at the event" to mean that the measurement of the particle's 4 velocity (v^{0})^2 is made by an observer with 4 velocity (v^{0})^2 who is at rest in the Schwarzschild coordinates. (i.e this observer is an accelerating hovering observer). Can I assume that (v^{i})^2 is the 4 velocity of the particle according to an observer at infinity and and (u^{i})^2 is the 4 velocity of the hovering observer at r according to the observer at infinity?

But about your method I think you should first elaborate the steps you take to for example get the magnitude of the proper 3-velocity, V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } .

\sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } . was meant to be the magnitude of the coordinate 3 velocity according to an observer at infinity in Schwarzschild coordinates. It is not intended to be a proper 3 velocity.

It is obtained by starting from the local coordinate 3 velocity of the particle

\sqrt{( u_x^2 + u_y^2 + u_z^2 )} = \sqrt{ (dr &#039;/dt &#039;)^2 + (dy &#039;/dt &#039;)^2 + {dz &#039;/dt &#039;)^2 }
as measured by an observer that is stationary at r with respect to the Schwarzschild coordinates. In other words this local observer has dr = d(\Omega) =0. The primed coordinates are meant to indicate local measurements made using local rulers and clocks. Note that the local geometry can be described by Minkowski space and this local static observer does not need to concern himself with the radius or angular measurements.

Now the Schwarzschild metric in units of G=c=1 is:

dS = (1-2m/r) dt^2 - dr^2/(1-2m/r) - r^2d\theta^2 - r^2 sin^2(\theta) d\phi^2 }

can be expressed in a slightly simpler form as:

dS = (1-2m/r) dt^2 - dx^2/(1-2m/r) - dy^2 - dz^2 }

if we align the equator and the principle meridian with the location of the particle for convenience, a bit like we often align the boost direction in a Lorentx transformation in Minkowski space with the x coordinate.

Now we know when transforming from the coordinates of the particle (dt,dx,dy,dz) as measured by a observer at infinity to the coordinates of the same particle (dt',dx',dy',dz') as measured by a stationary observer at r in Schwarzschild coordinates that:

dt &#039; = dt\sqrt{1-2m/r}
dx &#039; = dx/\sqrt{1-2m/r}
dy &#039; = dy
dz &#039; = dz

using the above relations, it is straightforward to work out that for a local coordinate 3 velocity of

\sqrt{ (dx &#039;/dt &#039;)^2 + (dy &#039;/dt &#039;)^2 + {dz &#039;/dt &#039;)^2 }

that the coordinate 3 velocity according to an observer at infinity :

\sqrt{ (dx /dt)^2 + (dy /dt)^2 + (dz /dt)^2 }

is equal to:

\sqrt{ (dx &#039;/dt &#039;)^2(1-2m/r)^2 + (dy &#039;/dt &#039;)^2(1-2m/r) + (dz &#039;/dt &#039;)^2(1-2m/r) }

which is basically the expression I gave earlier for the coordinate 3 velocity for an observer at infinity:

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

and because it is a coordinate 3 velocity that does not involve proper times, the components are completely independent of each other unlike the components of a 4 velocity.

And then explain what do you mean by 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 }? Here I see some misleading arguments like equality of a 4-vector with a scalar value! After doing so, I'd be able to get the idea of your calculations!

OK, I think I should have used c^2 rather than c for the 4 velocity so I will restate it as:

W^2 = c^2 = (\gamma_v)^2(c, u_x, u_y, u_z) = (\gamma_v)^2( c^2 - u_x^2 - u_y^2 - u_z^2 )

Now \gamma_v is 1/\sqrt{1-v^2/c^2} (where v is the local coordinate 3 velocity) and \gamma_v = dt &#039;/d\tau so:

W^2 = c^2 = ( c^2dt &#039;^2/d\tau^2 - u_x &#039;^2 dt &#039;^2/d\tau^2 - u_y &#039;^2 dt &#039;^2/d\tau^2 - u_z &#039;^2 dt &#039;^2/d\tau ^2 ) =

( c^2dt &#039;^2/d\tau^2 - d_x &#039;^2/d\tau^2 - d_y &#039;^2 /d\tau ^2 - d_z &#039;^2 /d\tau^2 )

which is the more familiar way of expressing a 4 velocity.

The only 3-velocity, which is basically a coordinate velocity, is the velocity we use in the Lorentz factor \gamma_v=\frac{1}{\sqrt{1-v^2/c^2}}. Other than that, all velocities used in calculations are proper 4-velocities!

OK, I thought you were talking about 3 velocities because you defined v^{i} as

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

which only has 3 components.

Hopefully we are getting nearer to understanding each other's notations and calculations and most of the confusion is probably due to me because I am not familiar with the more formal notation style.

[P.S.] In #20 you also said

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

which seems to imply g_{ii}u^iv^i = 0.

Have I read that right?

 

[Altabeh]

Hi Altabeh,
OK, (v^0)^2 is a 4 velocity so it has the value c^2 and V^{2} is also a 4 velocity so it also has the value c^2 by definition so:

(v^0)^2=V^2-v^2 means c^2 = c^2-v^2 which is only true when v=0. Is that what you intended?

Actually v=v^i:=(v^1,v^2,v^3) and V is the magnitude of the four-velocity v^{\alpha} and is by definition equal to \sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=c.

I take "observer is instantaneously at rest at the event" to mean that the measurement of the particle's 4 velocity (v^{0})^2 is made by an observer with 4 velocity (v^{0})^2

Your notations are not correct. The particle's 4-velocity is v^{\alpha} not (v^{0})^2 and the observer's 4-velocity is u^{\alpha} not (v^{0})^2.

(who is at rest in the Schwarzschild coordinates. i.e this observer is an accelerating hovering observer).

A hovering observer is not at rest unless it is said that the observer is hovering at rest which I think is what you meant.

Can I assume that (v^{\alpha}) is the 4 velocity of the particle according to an observer at infinity and and (u^{\alpha}) is the 4 velocity of the hovering observer at r according to the observer at infinity?

I edited the quote above (note that the Latin symbols mostly run over the values 1,2,3 and the Greek ones take the values 0,1,..,3.) You can assume that the (u^{\alpha}) is the 4 velocity of the radial hovering observer at infinity and consequently (v^{\alpha}) is the 4 velocity of the particle according to him.

It is obtained by starting from the local coordinate 3 velocity of the particle

\sqrt{( u_x^2 + u_y^2 + u_z^2 )} = \sqrt{ (dr &#039;/dt &#039;)^2 + (dy &#039;/dt &#039;)^2 + {dz &#039;/dt &#039;)^2 }

Now you can suppose that \sqrt{( u_x^2 + u_y^2 + u_z^2 )} is measured by the observer at infnitiy.

Note that the local geometry can be described by Minkowski space and this local static observer does not need to concern himself with the radius or angular measurements.

This is even possible for the observer who is at rest at infinity and for a Schwarzschild metric the outside observer being at rest at an infnitely enough distance from the gravitating body spacetime can be taken to be Minkowski. So I don't know how you are going to define a transformation between a local stationary observer at r and an observer at rest at infinity. Let's first clear this up and then go on!

AB

 

[yuiop]

Actually v=v^i:=(v^1,v^2,v^3) and V is the magnitude of the four-velocity v^{\alpha} and is by definition equal to \sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=c.

When I asked you to define your velocity symbols u^0, v^0, u^i and v^i in #37, you replied that other than v used in the Lorentz factor, "all velocities used in calculations are proper 4-velocities", but here you seem to be making it clear that v^i is a 3 velocity and from #38 you said "Here v=(v^1,v^2,v^3) is the proper 3-velocity" I think we can establish that by v^i you mean the proper 3 velocity or what is sometimes called the spatial component of the 4 velocity. Since we are talking about a 3 velocity it is not invariant and we need to make clear who makes measures v^i, but I think from the context of your earlier posts it is the measurement made by a stationary observer at radial coordinate r.

At least we are agreed that V^2 has a magnitude of c^2 and so we can say:

(u^o)^2 = c^2 -v^2

Now if we assume by v^2 you mean ( v^1(\tau), v^2(\tau), v^3(\tau) ) then the quantity on the right of the above equation is not invariant and nor is it invariant if we assume you mean (v^1(t), v^2(t), v^3(t) ).

You still have not explicitly stated what you mean by the symbols u^0, v^0, u^i.

Your notations are not correct. The particle's 4-velocity is v^{\alpha} not (v^{0})^2 and the observer's 4-velocity is u^{\alpha} not (v^{0})^2.

The particle's 4-velocity is v^{\alpha} (as measured by who?) not (v^{0})^2 and the observer's 4-velocity is u^{\alpha} (as measured by who?) not (u^{0})^2.. Thank you for telling me what (v^{0})^2 and (u^{0})^2. do not mean, but that still does not tell me exactly what you do mean by the symbols u^0, v^0, u^i.

I edited the quote above (note that the Latin symbols mostly run over the values 1,2,3 and the Greek ones take the values 0,1,..,3.) You can assume that the (u^{\alpha}) is the 4 velocity of the radial hovering observer at infinity and consequently (v^{\alpha}) is the 4 velocity of the particle according to him.

In #37 you said

In the setup I use to denote the 4-velocities of observer, u^{\alpha}, and particle, v^{\alpha}, I assume the observer is instantaneously at rest at the event the proper 4-velocity v^{\alpha} is observed. Thus we have

u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)

at the event of observation.

which appears to imply by the observer "at rest at the event", that u^{\alpha}, and particle, v^{\alpha} are measured by a local observer at r rather than at infinty as you state in comment above. Anyway, this still does not clarify what you mean by your symbols u^0, v^0, u^i.

I have plenty more to say and I will put that in my next post, but for now I do not want anything to distract you from clearly defining what you mean by your symbols u^0, v^0, u^i and v^i that you used in the your equation g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2 back in #18.

 

[yuiop]

This is even possible for the observer who is at rest at infinity and for a Schwarzschild metric the outside observer being at rest at an infnitely enough distance from the gravitating body spacetime can be taken to be Minkowski. So I don't know how you are going to define a transformation between a local stationary observer at r and an observer at rest at infinity. Let's first clear this up and then go on!

In units of G=M=c=1, if we had a non rotating gravitational body of radius r=4M and I told you the vertical velocity of a particle on the surface of this body is zero and the horizontal velocity is 0.9c according to the measurements in coordinate time of a local stationary observer on the surface of that body, then I am sure you could tell me what the coordinate 3 velocity of the particle is according to an observer at infinity, even though I have not specified the local horizontal velocity in terms of r and d\theta or d\phi.

We are agreed that any observer can treat the local geometry as Minkowski spacetime but if an observer at infinity is making measurements of distant events at r then that is non-local and he can not use Minkowski coordinates and in this case he hase to use Schwarzschild coordinates.

If it helps please assume that when I used ux, uy and uz to describe velocity components I mean:

ux = dr/dt
uy = r d\theta/dt
uz = r sin(\theta) d\phi/dt

and by dx, dy, dz I mean:

dx = dr
dy = r d\theta
dz = r sin(\theta) d\phi

By making those substitutions my equations are perfectly conventional and my use of shorthand notation of ux, uy, uz for velocity components is not conceptionally different to your notation of v^1, v^2, v^3 for the velocity components.

 

[Altabeh]

When I asked you to define your velocity symbols u^0, v^0, u^i and v^i in #37, you replied that other than v used in the Lorentz factor, "all velocities used in calculations are proper 4-velocities", but here you seem to be making it clear that v^i is a 3 velocity and from #38 you said "Here v=(v^1,v^2,v^3) is the proper 3-velocity" I think we can establish that by v^i you mean the proper 3 velocity or what is sometimes called the spatial component of the 4 velocity. Since we are talking about a 3 velocity it is not invariant and we need to make clear who makes measures v^i, but I think from the context of your earlier posts it is the measurement made by a stationary observer at radial coordinate r.

You seem to be understanding me badly. I have not yet entered the Schwarzschild metric and my notations are so clear. I told you that the quantities with only Latin indices all refer to the 3-quantities whereas quantities with Greek indices include one more component, the null time component so when encountering them you should bear in mind that they are all 4-quantities. For example, v^i is a 3-velocity while v^{\alpha} is a 4-velocity (which is measured according to a comoving observer).

At least we are agreed that V^2 has a magnitude of c^2 and so we can say:

(u^o)^2 = c^2 -v^2

This is not true unless I accept that u^0 is just a typo and it must have been v^0.

Now if we assume by v^2 you mean ( v^1(\tau), v^2(\tau), v^3(\tau) ) then the quantity on the right of the above equation is not invariant and nor is it invariant if we assume you mean (v^1(t), v^2(t), v^3(t) ).

By v^2 I mean (v^1)^2+(v^2)^2+(v^3)^2.

You still have not explicitly stated what you mean by the symbols u^0, v^0, u^i.

I ignore my own observer and rather go right into the setup you are looking for. (Please use this notation to keep being in agreement with the previous notations used in this thread!) The coordinate 4-velocity u^{\alpha}=(u^0,u^i)=(1,0,0,0) belongs to the local stationay observer at r i.e. observer A. And the 4-velocity v^{\alpha}=(v^0,v^i)=(1,v^i) is for a particle moving in the Schwartzschild spacetime measured by the observer A. The 4-velocity v&#039;^{\alpha}=(v&#039;^0,v&#039;^i)=(1,v&#039;^i) is measured by an observer at infinity, i.e. observer B.

The particle's 4-velocity is v^{\alpha} (as measured by who?) not (v^{0})^2 and the observer's 4-velocity is u^{\alpha} (as measured by who?) not (u^{0})^2..

Do you think we need to clarify who is measuring the coordinate velocity of an observer!?

Now take the linearized isotropic Schwartzschild metric:

ds^2 = (1-2m/r) dt&#039;^2 - (1+2m/r) (dx&#039;^2+ dy&#039;^2 +dz&#039;^2 ).

So that the measurements of observer A are related to those of B by

dt= \sqrt{1-2m/r}dt&#039;,
dx= \sqrt{1+2m/r}dx&#039;,
dy= \sqrt{1+2m/r}dy&#039;,
dz= \sqrt{1+2m/r}dz&#039;.

One can check the validity of these formulae by knowing that 2m&lt;r and thus \sqrt{1-2m/r}&lt;1, \sqrt{1+2m/r}&gt;1. meaning that the time dilation and length contraction occur correctly.

By looking at the invariant proper velocity formula,

V=\sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=\sqrt{(1-2m/r)-(1+2m/r)(v&#039;^i)^2}=\sqrt{1-(v&#039;^i)^2-(2m/r)(1+(v&#039;^i)^2)}.

and neglecting the terms being of the order 2 in 2m/r we get

V=\sqrt{\frac{(1-4m/r)-(1+4m/r)(v^i)^2}{1-2m/r}}.

Assuming that the 3-velocity v^i is small, this approximately is

V=\sqrt{\frac{1-4m/r}{1-2m/r}}+O((v^i)^2),

implying that the circumferences under which observer B measures the particle's velocity has really tiny effects on the measurements due to being at an infinitely large distance from the particle. Also you can see that if m=0, one would get the formula

(v^0)^2=V^2+(v^i)^2.

AB

 

[yuiop]

Hi Altabeh,
Thanks for clearing some things up, but I now think the main reason your post in #20 (quoted below) has been confusing the heck out of me is that you made a mistake in the middle of your derivation.

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}^{<br /> 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

When Dr Greg asked:

...
I can't work out what (v^0)^2=V^2-v^2 is supposed to mean.

you replied:

Here v=(v^1,v^2,v^3) is the proper 3-velocity and V represents the magnitude of the 4-velocity v^{\alpha}.

So we establish that by v^2 in your equation (v^0)^2=V^2-v^2 you mean v^2 = (v^i)^2 = (dx^1/d\tau)^2 + (dx^2/d\tau)^2 + (dx^3/d\tau)^2.
Taking that into account, what you should have said is:

(v^0)^2= V^2 + v^2

where the (+) sign rather than the (-) sign that you used in your orginal equation is the key correction.

This is the proof that you should have used a (+) sign:

c^2 = (v^0)^2 - (v^i)^2

(v^i)^2 = [ (v^0)^2 - c^2 ]

Substitute the above equation into the equation below:

(v^0)^2 = V^2 + (v^i)^2

and you obtain:

(v^0)^2 = V^2 + [ (v^0)^2 - c^2 ]

Now we have agreed V^2 = c^2 so the equation above is now correct.

Later on in this equation:

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

you use v^2 to mean proper 3 velocity (v^i)^2 on the left of the equation and also use v^2 to mean coordinate 3 velocity on the right of the equation, so you are using the same symbol to mean two different things in the same equation which leads to errors.

This is not true unless I accept that u^0 is just a typo and it must have been v^0.

Yes I made a typo. I meant to quote your equation (v^0)^2=V^2-v^2.

By v^2 I mean (v^1)^2+(v^2)^2+(v^3)^2.

I was trying to clarify whether the v^2 you used in the (v^0)^2=V^2-v^2 was in terms of coordinate time or proper time and this answer does not make it make it clear but from the context of other posts I am now convinced you mean v^2 in terms of proper time in that particular equation.

Now take the linearized isotropic Schwartzschild metric:

ds^2 = (1-2m/r) dt&#039;^2 - (1+2m/r) (dx&#039;^2+ dy&#039;^2 +dz&#039;^2 ).

Please not the isotropic metric! I know Eddington came up with these sorts of metrics but they really are a horror and they mean the local speed of light is neither constant nor isotropic or the local observer has to use different length rulers for vertical and horizontal measurements. In this forum it often stressed that the local speed of light is always c and introducing metrics where that is not true is going to confuse things. I think this is worthy of further discussion but we should probably start a new thread for that to avoid going too far off topic.

 

[Altabeh]

Hi Altabeh,
... is that you made a mistake in the middle of your derivation.

Yes, the sign was incorrect and thank you for noticing it. I also corrected this in my recent post.

you use v^2 to mean proper 3 velocity (v^i)^2 on the left of the equation and also use v^2 to mean coordinate 3 velocity on the right of the equation, so you are using the same symbol to mean two different things in the same equation which leads to errors.

I believe that my notation is flawed in this case and I'd be glad if you could open a new thread and write up in a clear way all these things we talked about in like 10 or more posts!

AB

 

[DrGreg]

I believe that my notation is flawed in this case and I'd be glad if you could open a new thread and write up in a clear way all these things we talked about in like 10 or more posts!

I was getting very confused, so I think it's an excellent idea to start again, but we don't need a new thread if we're still on the same subject.

To avoid confusion I propose to adopt the convention of using upper-case for 4-vectors. We should also take care not to use the same letter with different meanings and whenever we refer to a 3-vector, or components of a 4-vector, we make it very clear which coordinate system we are talking about. I hope also everyone is happy to work with the convention that G = 1 = c.

To set the ball rolling and establish some notation I'll rewrite my previous solution using the method George Jones first mentioned. Recall the original question was

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)?

Let V be the 4-velocity of an object to be measured, and U be the 4-velocity of any hovering 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, \textbf{U} = (1, 0, 0, 0) and \textbf{V} = (\gamma, \gamma w, 0, 0), where w is the magnitude of the 3-velocity that you want to measure (let's assume it's along the x-axis) and \gamma = (1-w^2)^{-1/2}. Therefore

g(\textbf{U}, \textbf{V}) = g_{\alpha\beta}U^{\alpha}V^{\beta} = \gamma​

But g(U,V) is invariant, i.e. the same in all coordinate systems, so you can also calculate it in Schwarzschild coordinates, and equate the two values.

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

\textbf{U} = \left(\frac{1}{\sqrt{1 - 2M/r}}, 0, 0, 0\right)​

(The last three components must be zero and the first is chosen to ensure that g(U,U) = 1.) And

\textbf{V} = \left(\frac{dt}{d\tau}, \frac{dr}{d\tau}, \frac{d\theta}{d\tau}, \frac{d\phi}{d\tau}\right)​

And so

\frac{1}{\sqrt{1-w^2}}= g_{\mu\nu}V^{\mu}U^{\nu} = \frac{dt}{d\tau}\sqrt{1 - 2M/r}​

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

If the question had specified that the path was parameterised by Schwarzschild-coordinate-time instead of proper time, the final step would be to note that

\textbf{V} = \frac{dt}{d\tau} \left(1, \frac{dr}{dt}, \frac{d\theta}{dt}, \frac{d\phi}{dt}\right)​

and as g(V,V) = 1 this gives dt/d\tau in terms of r, \theta, dr/dt, d\theta/dt, d\phi/dt.

Now kev, if you want to discuss an alternative method we can do so.

 

[Altabeh]

This is a decent notation and it sets the stage for kev's further insights on the subject to be given systematically.

AB

 

[yuiop]

Hi Altabeh and Dr Greg,

I accept the challenge but I am I am a little short on time right now. I would also like to agree some further conventions first.

Can we assume that when talking about 4 velocities or components of 4 velocities that the velocity components are in terms of proper time and when talking about 3 velocities we should explicitly state whether we are talking about proper velocity (the 3 spatial components of a 4 velocity) or coordinate velocity or agree a notation to make the difference clear.

There is also the issue of primed variables. Should primed velocity components indicate the measurements of a local observer at r in Schwarzschild coordinates or the measurements of an observer at infinity or should we just reley on the context of the surrounding text to make the difference clear?

 

[DrGreg]

Can we assume that when talking about 4 velocities or components of 4 velocities that the velocity components are in terms of proper time

4-velocities are always differentiated with respect to proper time, so yes.

...and when talking about 3 velocities we should explicitly state whether we are talking about proper velocity (the 3 spatial components of a 4 velocity) or coordinate velocity or agree a notation to make the difference clear.

There is also the issue of primed variables. Should primed velocity components indicate the measurements of a local observer at r in Schwarzschild coordinates or the measurements of an observer at infinity or should we just reley on the context of the surrounding text to make the difference clear?

My suggestion is to always spell out explicitly what you mean, to avoid the confusions that previously arose in this thread.

 

[yuiop]

NOTE: After comments by Dr Greg and Altabeh it has become clear to me that I my conclusions in this post are not correct and I will be posting corrections in a later post.

This is a decent notation and it sets the stage for kev's further insights on the subject to be given systematically.

In this thread I am not offering insights, but just trying to come to a better understanding of 3 and 4 velocity in the context of the Schwarzschild geometry. If my learning curve helps someone else in the process then all well and good.

... always spell out explicitly what you mean, to avoid the confusions that previously arose in this thread.

In the interests of being explicit I would like to adopt the primed convention suggested by Altabeh in an earlier post of indicatiing the measurements made by an observer at infinity in Schwarzschild coordinates by primed symbols and local measurements by a an observer at r (or measurements by an observer in Minkowski spacetime) by unprimed symbols. Using this notation the classic Schwarzschild metric would be written as:

d\tau^2 = (1-2M/r) {dt &#039;}^2 - \frac{1}{(1-2M/r)}{dr &#039;}^2 - r^2{d\theta &#039;}^2 - r^2 sin^2(\theta) {d\phi &#039;}^2 }

Using this convention, the 4 velocity of a stationary observer at r, according to the observer at r is:

\textbf{U} = \left(1, 0, 0, 0\right)​

and the magnitude of the same 4 velocity is:

||\textbf{U}|| = 1​

The 4 velocity of the stationary observer at r, according to the observer at infinity is:

\textbf{U} &#039; = \left(\frac{1}{\sqrt{1 - 2M/r}}, 0, 0, 0\right)​

The magnitude of U' according to the observer at infinity is:

||\textbf{U} &#039;|| = \frac{1}{\sqrt{1 - 2M/r}}​

The significant observation I make here is that while the magnitude of a 4 velocity is always 1 in Minkowski coordinates, the same is not true in Schwarzschild coordinates. Agree or dissagree?

Now let us say the 3 velocity of a vertically falling test particle is w, then the 4 velocity according to a local stationary observer at r is

\textbf{V} = (\gamma, \gamma w, 0, 0)​

where \gamma = (1-w^2)^{-1/2}, then the 4 velocity of the same test particle according to the observer at infinity is:

\textbf{V} &#039; = \left(\frac{\gamma}{(1-2M/r)}, \frac{\gamma w}{(1-2M/r)}, 0, 0\right)​

and the magnitude of the 4 velocity of the falling particle, according to the observer at infinity is:

||\textbf{V} &#039;|| = \frac{1}{\sqrt{1 - 2M/r}}​

which is at least in agreement with the magnitude given earlier for the 4 velocity of the stationary observer at r, according to the observer at infinity.

This means that the equations I gave for 4 velocity in Schwarzschild coordinates in #18 of this thread are wrong because I was assuming the mgnitude of a 4 velocity is always c. (I think the equations I gave for coordinate 3 velocity transformations in Schwarzschild geometry in the same post are still good.)

...

\frac{1}{\sqrt{1-w^2}}= g_{\mu\nu}V^{\mu}U^{\nu} = \frac{dt}{d\tau}\sqrt{1 - 2M/r}​

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

Using primed notation for explicitness again, I can write the equation given above by Dr Greg as:

\frac{1}{\sqrt{1-w^2}}= \frac{dt &#039;}{d\tau}\sqrt{1 - 2M/r}​

making clear that w is a local 3 velocity according to a stationary observer at r and (dt'/dtau) is the time dilation according to an observer at infinity.

The OP asked:

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.

If my reasoning above is correct, then the answer is yes there is something anaogous in Schewarzschild coordinates, if we understand gamma to mean 1/(1-2M/r)*1/(1-v^2) and not just 1/(1-v^2) as in the Special Relativity context. I have changed v to lower case in the above quote, in line with our agreed notation for 3 velocity.)