Schwarzschild Metric Geodesic Eq: Qs & Answers

[exmarine]

TL;DR Summary

A term from a geodesic does not seem to match the metric

I have no idea if this is an “A” level question, but I will put that down.

From the Schwarzschild metric, and with the help of the Maxima program, one of the geodesic equations is:

(I will have to attach a pdf for the equations...)

I believe this integrates to the following, with :



(1) Is that correct so far?

(2) But when I “solve” for with the metric, I don’t get the same thing. Shouldn’t I get something “similar” or recognizable when all the other differentials are zero?Obviously g is not equal to the square root of g, so I must be missing something. Any help is appreciated.

 

[PeterDonis]

@exmarine, please post your equations directly in the thread using the PF LaTeX feature [1]. We have to be able to quote your equations in replies; we can't do that if they're in a separate attachment.

[1] https://www.physicsforums.com/help/latexhelp/

 

[exmarine]

Ok. Will take me a while.

 

[strangerep]

@exmarine : you didn't say what textbook(s) you're using, so I'll suggest Carroll, ch5.

Solving the equations of motion, even the Newtonian case, is quite nontrivial. You need to figure out the conserved quantities (constants of the motion) and work with those to simplify the DEs.

 

[PeterDonis]

when I “solve” for with the metric, I don’t get the same thing. Shouldn’t I get something “similar” or recognizable when all the other differentials are zero?

No. By setting all the other differentials to zero, you are restricting yourself to a curve along which only $t$ changes; but in general such a curve will not be a geodesic, and it certainly isn't in the case of Schwarzschild spacetime. So such a curve is not described by the geodesic equation and the expression you get for $dt / ds$ for such a curve will not match anything you get by solving the geodesic equation.

 

[Ibix]

Ok. Will take me a while.

Use the tex() command to get LaTeX you can cut and paste. So if the equation you are concerned about is the result of what you input at %i16 (your 16th line of input), just add tex(%o16); (which asks for the 16th line of output expressed as tex) to the end of your program and copy and paste the output, including the double dollar signs at each end. Or if you've defined a variable (e.g. F:GMm/r^2) somewhere then tex(F) will also work.

 

[vanhees71]

Motion in the Schwarzschild metric (the GR equivalent of the Kepler problem in non-relativistic Newtonian theory) is most easily analyzed using the "square form" of the Lagrange formalism where the world-line parameter is automatically affine and thus for a massive test particle can be chosen to be its proper time:
$$L=-\frac{m}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Then you can use the symmetries of the metric as written in the usual Schwarzschild coordinates to solve the problem in a very similar way as the Kepler problem in non-relatistic mechanics.

 

[exmarine]

From the Schwarzschild metric, and with the help of the Maxima program, one of the geodesic equations is:
$$0=\frac{d^2 t}{ds^2}+\frac{r_s}{r(r-r_s}\frac{dr}{ds}\frac{dt}{ds}$$
With $g=1-\frac{r_s}{r}$ I believe this integrates to:
$$\frac{dt}{ds}=\frac{k}{g}$$
(1) Is that correct so far?
(2) But when I "solve" for $(dt/ds)$ with the metric, I don’t get the same thing. Shouldn’t I get something "similar" or recognizable when all the other differentials are zero?
$$1=g(dt/ds)^2-0-0-0$$
$$\frac{dt}{ds}=\frac{1}{\sqrt{g}}$$
Obviously g is not equal to the square root of g, so I must be missing something. Any help is appreciated.
PS. I was not working with a specific textbook. I have about a dozen by now from MTW to Lieber, but I don't think I have Carrol. Will look for it. Thanks.
PPS. Looks like this worked. Now I need to read the latest responses. Thanks!

 

[Orodruin]

(2) But when I "solve" for $(dt/ds)$ with the metric, I don’t get the same thing. Shouldn’t I get something "similar" or recognizable when all the other differentials are zero?
$$1=g(dt/ds)^2-0-0-0$$

As was already pointed out in #5, the curve where all coordinates but $t$ are constant is not a geodesic in the Schwarzschild spacetime. It should therefore come as no surprise that it does not satisfy the geodesic equations.

 

[Ibix]

For any normalised four vector, such as the four velocity, $1=g_{\mu\nu}U^\mu U^\nu$ (give or take a minus sign). In the case of the four velocity, the components of $U^\mu$ are the change in the $\mu$th coordinate with respect to $s$ - i.e., $U^t=dt/ds$. So you've calculated the $t$ component of your four vector but not yet the $r,\theta,\phi$ components.

In your second expression, though, you've explicitly set those components to zero and asked what $t$ component results in a normalised four vector. Perfectly fine question to ask, but as others have noted, such a path (someone not orbiting or falling, just hovering) is not a geodesic. So you do not expect the results to satisfy the geodesic equation.

Put another way, I can write down $dy/dx$ for $y=x^2$ and for $y=\sin(x)$. Do you expect those two things to be equal just because they're both $dy/dx$?

 

[exmarine]

Ok I have traced my misunderstanding (apparently) back to Kevin Brown’s “Reflections on Relativity”, around page 409. (I think that book is now online at Math Pages?)

After inserting the squared derivatives of the time and angular coordinates into the metric, he says the following:

“We arrive at this same equation if we insert the squared derivatives of the coordinates into equation 3 (equation 3 is the second derivative of the radial coordinate), because one of the geodesic equations is always redundant to the line element.”

The italics are mine of course. Do I misunderstand that statement? Or is it incorrect?

Thanks again for the responses.

 

[Orodruin]

Do I misunderstand that statement?

You misunderstand the statement. The point being made is that $g_{ab} \dot x^a \dot x^b$ is a constant of motion, which makes one of the geodesic equations redundant. It does not mean that all but one of the coordinate derivatives is zero.

I do not have the particular book so it is difficult to understand exactly what is being said unless you write it out or provide a link.

 

[pervect]

From the Schwarzschild metric, and with the help of the Maxima program, one of the geodesic equations is:
$$0=\frac{d^2 t}{ds^2}+\frac{r_s}{r(r-r_s}\frac{dr}{ds}\frac{dt}{ds}$$
With $g=1-\frac{r_s}{r}$ I believe this integrates to:
$$\frac{dt}{ds}=\frac{k}{g}$$
(1) Is that correct so far?

As others have said, it is correct to say that for a particle following a geodesic in the Schwarzschild space-time that

$$\left( 1-\frac{r_s}{r} \right) \frac{dt}{ds}= k$$

where k is some constant of motion for the particle following the geodesic.

The expression that didn't agree with this is not correct for a particle following a geodesic.

 

[exmarine]

I should paste links to Brown’s online book in case anyone else is interested.

https://www.mathpages.com/rr/s6-02/6-02.htm

I see that he has revised this section. He removed the sentence I quoted and added an appendix discussing this very issue. (He also renumbered his equations, so it might be hard to exactly sync with my previous quotes.)

https://www.mathpages.com/rr/appendix/appendix.htm

I have not studied this appendix yet, but it appears more complicated than I misinterpreted his deleted sentence to mean. Here is a brief copy and paste. “As noted in Section 6.2, we apparently made use of only two of the three geodesic equations, excluding equation (8), when deriving the relativistic orbital precession. Recall that we integrated equations (7) and (9), and then substituted the results into the line element, which then gave the equation of motion represented by (10). To show that this is consistent with the remaining geodesic equation (8), we can differentiate equation (10) with respect to t and divide through by 2(dr/dt) to give​

​

​

​

This is the same equation we would get if we inserted the squared derivatives of the coordinates into equation (8). Thus we have consistency with all of the geodesic equations.​

​

In general, if a path satisfies all but one of the geodesic equations, the metric line element ensures that it satisfies the remaining equation as well. To show this, we can write the general metric in the form…”​

Thanks for all the responses.

PS. It looks like his equations won't paste here either, so follow the link if you want those.

 

[New Simplicio]

Focusing on the Fundamentals: The Schwazchild Metric doesn't contract the transverse direction

 

[Ibix]

Focusing on the Fundamentals: The Schwazchild Metric doesn't contract the transverse direction

Um... what? And it's Schwarzschild (schwarz is black in German, schild is shield or sign).