What is the relationship between speed and coordinates in general relativity?

[DrGreg]

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 '}^2 - \frac{1}{(1-2M/r)}{dr '}^2 - r^2{d\theta '}^2 - r^2 sin^2(\theta) {d\phi '}^2 }

To be consistent, you need to prime all the symbols:
d\tau^2 = (1-2m/r') {dt '}^2 - \frac{1}{(1-2M/r')}{dr '}^2 - r'^2{d\theta '}^2 - r'^2 sin^2(\theta') {d\phi '}^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​

Well, you haven't explicitly spelled out exactly which coordinates this observer is using.

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

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

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

||\textbf{U} '|| = \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?

First of all, 4-vectors are concepts that exist independently of coordinate systems. So really U and U' are the same 4-vector, but you are expressing the vector's components relative to two different coordinate systems. Loosely speaking, the "physicist's definition" of a 4-vector is something whose components transform under the rule

U'^{\alpha} = \frac{\partial x'^{\alpha}}{\partial x^{\beta}} U^{\beta}​

and this is true for 4-velocities.

Second, the "length" of a 4-vector is calculated via the metric

||\textbf{U}|| = \sqrt{|g_{\alpha\beta}U^{\alpha}U^{\beta}|}​

so, in your primed coordinates you have to use the Schwartzschild metric components and you still get an answer of 1.

The magnitude of a 4-velocity is always 1, no matter what coordinates you use. This follows from the two equations

||\textbf{U}||^2 = \left|g_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^<br /> ^{\beta}}{d\tau}\right|
d\tau^2 = |g_{\alpha\beta}dx^{\alpha}dx^{\beta}|​

Now let us say the 3 velocity of a vertically falling test particle is and 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)​

You will need to explain why you think that is true (which will involve clarifying what coordinates you are using). I won't comment any further on what you said after that.

 

[Altabeh]

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

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

You can't use the locally Minkowski metric where you're applying the Schwartzschild metric to the components of the 4-velocity U&#039; at the same time. The magnitude of U&#039; is

||\textbf{U} &#039;|| = \sqrt{g_{\mu\nu}U^{\mu}U{\nu}}=\sqrt{(1 - 2M/r)\frac{1}{1 - 2M/r}}=1.​

Agree or dissagree?

Disagreed for the reason given above.

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)​

The second component here is not correct as the factor \frac{1}{(1-2M/r)} has been written inversely. If this was corrected, then its magnitude would be

||\textbf{V} &#039;|| =||\textbf{V}||.​

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.)

I don't know what you mean by this paragraph but the above assumption you made with an observer at infinity instead of an instantaneously at rest observer only works when the 4-velocity V&#039; is defined to be of the same general form as in SR. The important observation here is that the 4-velocity of the observer at infinity must be U&#039;. And this is nothing new but transferring the stationary observer to an infinitely large distance from where it is!

AB

 

[dx]

What exactly is the problem here? Are you trying to express the 4-velocity of an infalling observer in a basis derived from the Schwarzschild coordinates?

An object falling radially inward in a Schwarzschild spacetime with a speed v (as measured by a local inertial frame) will have a 4-velocity

u = \gamma (1 + m/r) \partial_t - v\gamma(1 - m/r)\partial_r

(ignoring quadratic terms in m/r and setting G = c = 1)

 

[yuiop]

What exactly is the problem here? Are you trying to express the 4-velocity of an infalling observer in a basis derived from the Schwarzschild coordinates?

An object falling radially inward in a Schwarzschild spacetime with a speed v (as measured by a local inertial frame) will have a 4-velocity

u = \gamma (1 + m/r) \partial_t - v\gamma(1 - m/r)\partial_r

(ignoring quadratic terms in m/r and setting G = c = 1)

The trouble with this equation is that it does not seem to work, but maybe I am just reading it wrong. To clear things up, what does your equation predict for u, when v = 0.8 and r = 2 Rs where Rs = 2GM/c^2 ?

I calculate u = dr/dtau = 0.94281 using a different method. What do you get?

(I am assuming (1 +m/r) is an approximation of 1/(1-2m/r). Is that right?)

 

[yuiop]

I have been putting off replying to this thread, because I started reading up on matrix and tensor notation, which got me hung up on a couple of things and probably got me even more confused. Anyway, Dr Greg and Altabeh have put me straight on some things and I would like to thank them for that.

To be consistent, you need to prime all the symbols:
d\tau^2 = (1-2m/r&#039;) {dt &#039;}^2 - \frac{1}{(1-2M/r&#039;)}{dr &#039;}^2 - r&#039;^2{d\theta &#039;}^2 - r&#039;^2 sin^2(\theta&#039;) {d\phi &#039;}^2 }

Hmmm, I have always assumed that m = m', r = r' and \theta = \theta &#039;. Is that not the case? Anyway, the local observer can always make measurements in terms of locally Minkowsky coordinates and not concern himself with the values of m, r, \theta, m', r', or \theta &#039;, no?

Second, the "length" of a 4-vector is calculated via the metric

||\textbf{U}|| = \sqrt{|g_{\alpha\beta}U^{\alpha}U^{\beta}|}​

OK, I think I understand that now. g_{\alpha\beta} is the metric of the coordinate system in matrix form (usually called the metric tensor) and is required if you want to find the norm of a four vector.

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)​

Taking onboard what Dr Greg and Altabeh have told me, the above statements need to be corrected and I should have said something like:

Let the coordinate 3 velocity of a vertically falling test particle in locally Minkoskian coordinate terms be w, according to a local stationary observer at at distance r from the gravitational source. The 4 velocity according to the same observer is

\textbf{V} = (\gamma, \gamma w, 0, 0) = \gamma(1,w,0,0)​

where \gamma = (1-w^2)^{-1/2}.
The 4 velocity of the same test particle in Schwarzschild coordinates (i.e according to the observer at infinity) is then:

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

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

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

Hopefully, that is all a bit more correct.

 

[Altabeh]

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

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

Hopefully, that is all a bit more correct.

More correct but not correct at all! In Schwartzschild metric, finding the coordinate velocity is a little bit cumbersome and needs some nerves actually! This can be done for a radially free falling particle in the following way:

The metric is assumed to be in the form

ds^2=(1+f(r))dt^2-(1+f(r))^{-1}dr^2-r^2(d\theta^2+\sin^2{\theta}d\phi^2),

where I use the convention c=1, for the sake of simplicity and f being the relativistic correction to the Minkowski metric in a weak gravitational field.

First off, I have to write the time and radial geodesic equations:

\ddot{t}+\frac{f&#039;}{1+f}{\dot{t}\dot{r}}=0,
\ddot{r}+\frac{f&#039;}{2}(1+f)\dot{t}^2-\frac{f&#039;}{2(1+f)}-r(1+f)\dot{\theta}^2-r(1+f)\sin^2\theta}\dot{\phi}^2=0,

where overdots are derivatives wrt the affine parameter s and the primes stand for differentiation wrt the coordinate r. By integrating the first equation one gets

\dot{t}=\frac{k}{1+f},

for some integration constant k. Introducing this into the second equation and noticing that the motion is radial, then

\ddot{r}=-\frac{f&#039;}{2(1+f)}(k^2-\dot{r}^2)

which if integrated, it would yield the equation for the coordinate radial velocity in Schwarzschild coordinates:

v_r=\frac{dr}{dt}=(1+f)}(1-\frac{a}{k^2}(1+f))^{1/2}. (#)

Here a is some other integration constant. The constant \frac{a}{k^2} can be written in terms of the initial velocity (v_r)_{(initial)} where

(v_r)_{(initial)}=(1+f_i)}(1-\frac{a}{k^2}(1+f_i))^{1/2}.

It is apparent from (#) that if f=-1, then we have v_r=0 which means that the free falling particle has a zero coordinate velocity when it hits the singularity.

AB

 

[yuiop]

The metric is assumed to be in the form

ds^2=(1+f(r))dt^2-(1+f(r))^{-1}dr^2-r^2(d\theta^2+\sin^2{\theta}d\phi^2),

where I use the convention c=1, for the sake of simplicity and f being the relativistic correction to the Minkowski metric in a weak gravitational field.

OK, we are considering radial motion only here, so we can cut out the last term and write the above equation in this form:

1 = (1+f)\dot{t}^2}-(1+f)^{-1}\dot{r}^2

where I have written f as abbreviation for f(r) as you appear to have done in the remainder of your post. Is that correct?
GR is complicated enough without introducing unnecessary complications!

First off, I have to write the time and radial geodesic equations:

\ddot{t}+\frac{f&#039;}{1+f}{\dot{t}\dot{r}}=0,

OK, slow down a bit here. I get that when you differentiate 1 wrt some variable you get zero, but my calculus was never very good, so can you break down how you get the geodesic equations?

\ddot{r}+\frac{f&#039;}{2}(1+f)\dot{t}^2-\frac{f&#039;}{2(1+f)}-r(1+f)\dot{\theta}^2-r(1+f)\sin^2\theta}\dot{\phi}^2=0,

where overdots are derivatives wrt the affine parameter s and the primes stand for differentiation wrt the coordinate r. By integrating the first equation one gets

\dot{t}=\frac{k}{1+f},

for some integration constant k. Introducing this into the second equation and noticing that the motion is radial, then

\ddot{r}=-\frac{f&#039;}{2(1+f)}(k^2-\dot{r}^2)

My algebra is better and something is not right about the above. After the substitution the end result should be:

\ddot{r}=-\frac{f&#039;}{2(1+f)}(k^2-1)

or the initial equation (ignoring non radial spatial terms) should have been:

\ddot{r}+\frac{f&#039;}{2}(1+f)\dot{t}^2-\frac{f&#039;}{2(1+f)}\dot{r}^2 =0

 

[Altabeh]

OK, we are considering radial motion only here, so we can cut out the last term and write the above equation in this form:

1 = (1+f)\dot{t}^2}-(1+f)^{-1}\dot{r}^2

where I have written f as abbreviation for f(r) as you appear to have done in the remainder of your post. Is that correct?

Yes!

OK, slow down a bit here. I get that when you differentiate 1 wrt some variable you get zero, but my calculus was never very good, so can you break down how you get the geodesic equations?

I thought you're familiar with the derivation. You should put the right-hand side of the above equation in the Euler-Lagrange equations for the time coordinate, i.e.

\frac{d}{ds}(\frac{\partial F(x^\alpha,\dot{x}^\alpha,s)}{\partial \dot{t}})=\frac{\partial F(x^\alpha,\dot{x}^\alpha,s)}{\partial {t}}.

Then you'll get that equation of mine.

or the initial equation (ignoring non radial spatial terms) should have been:

Yeah, just forgot to write \dot{r}^2 in the second term.

AB

 

[yuiop]

Here a is some other integration constant. The constant \frac{a}{k^2} can be written in terms of the initial velocity (v_r)_{(initial)} where

(v_r)_{(initial)}=(1+f_i)}(1-\frac{a}{k^2}(1+f_i))^{1/2}.

It is apparent from (#) that if f=-1, then we have v_r=0 which means that the free falling particle has a zero coordinate velocity when it hits the singularity.

AB

I am not sure what you mean by "the singularity". Normally this means the singularity at the center of a Schwarzschild black hole, but you seem to be talking about the coordinate singularity at the event horizon, because f = -2GM/(rc^2) and f=-1 at the event horizon. Are you are claiming that a particle can remain stationary at the event horizon?

I am also not sure why you are attaching any significance to initial velocity, because this can be arbitrarily set to any value you like between 0 and c. The useful quantity is the difference between final and initial velocities.

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

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

Hopefully, that is all a bit more correct.

More correct but not correct at all!

I do not see why you think my equation is not correct. Please elaborate.

I think we are talking at cross purposes here and that is why our equations differ. I am talking about arbitrary 3 velocities being transformed from locally Minkowskian coordinates to Schwarzschild coordinates while you seem to be talking about terminal velocities of free falling particles.

 

[Altabeh]

I am not sure what you mean by "the singularity". Normally this means the singularity at the center of a Schwarzschild black hole, but you seem to be talking about the coordinate singularity at the event horizon, because f = -2GM/(rc^2) and f=-1 at the event horizon. Are you are claiming that a particle can remain stationary at the event horizon?

Here by singularity I mean the coordinate singularity or the hypersurface f=-1. This never means that the particle remains stationary at f=-1 but that its coordinate acceleration goes to infinity while the coordinate velocity gets zero at the time that the particle crosses the surface f=-1.

I am also not sure why you are attaching any significance to initial velocity, because this can be arbitrarily set to any value you like between 0 and c. The useful quantity is the difference between final and initial velocities.

I don't even know what this is supposed to imply! Did I insist on a certain value of initial velocity and said that for example v_i=v_0 knocks spots off v_i=v_1!?

I do not see why you think my equation is not correct. Please elaborate.

I think we are talking at cross purposes here and that is why our equations differ. I am talking about arbitrary 3 velocities being transformed from locally Minkowskian coordinates to Schwarzschild coordinates while you seem to be talking about terminal velocities of free falling particles.

See, for example, excercise 9.13 at page 225 of "General relativity" by Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby to find what your equation lacks in the radial coordinate.

AB