0
$\begingroup$

The following argument is used to determine the unknown factors (e.g., $A(r)$ and $B(r)$) in the Schwarzschild metric. $$ \lim_{r \to ∞}A(r) = \lim_{r \to ∞}B(r) = 1 \space\space\space\space\space\space \Rightarrow \space\space\space\space\space\space A(r)=1/B(r). $$ [For example, Weinberg, Steven. Gravitation and cosmology. p. 186.]

Because of this step, it seems to me that the geodesic equation is valid only for a motion of a free-falling particle, which starts with zero velocity at infinity in the Schwarzschild space.

For example, a planet bound to a star appears to be an object that does not meet the above condition of "zero velocity at infinity".

Is this my understanding correct?

If so, how is this problem solved in known gravity theories?

$\endgroup$
  • $\begingroup$ The geodesic equation is just the statement that the four acceleration is zero, and it's always valid in any metric. Are you thinking of a different equation maybe? One of the equations for the Schwarzschild metric describing an infalling observer? $\endgroup$ – John Rennie Oct 29 '18 at 5:21
  • $\begingroup$ From the comment given by @JohnRennie, I found that I was confused about something and that the above question did not adequately describe my difficulty. So, I want to close this question. Can I just delete this post? $\endgroup$ – SOQEH Oct 29 '18 at 7:16
  • $\begingroup$ @SOQEH if you still have a question it would be better to edit this question to explain exactly what you want to know. If your question is resolved then yes you can delete this. $\endgroup$ – John Rennie Oct 29 '18 at 7:27
1
$\begingroup$

The geodesic equation in general relativity works to find the trajectory of any free point particle with any initial conditions (as long as the metric is $C^{1,1}$, anyway). While the existence and uniqueness of such geodesics is guaranteed, their solvability is another matter, which is why Schwarzschild geodesics are typically very specific examples.

You may remember that even in classical mechanics, the solution of an orbiting body cannot be found in closed form, that is, $r(t)$ has no closed form solution in general (this is why it's usually solved as $r(\theta)$ instead). The situation is not made any better in GR, and there is no generic solution to the geodesic equation for Schwarzschild. But there are certainly orbital solutions of the form $r(\theta)$ for orbital mechanics, as well as a few solvable orbits as long as you make some assumptions on the initial conditions.

$\endgroup$
  • $\begingroup$ Your answer is helpful to me. However, the gist of the question is changed, and so I will postpone accepting. I hope you understand this. $\endgroup$ – SOQEH Oct 29 '18 at 12:33
0
$\begingroup$

The geodesic equation and its initial conditions are not used here. Firstly,

$$ A(r)B(r)~=~\text{constant} \tag{8.2.4}$$

follows from the vanishing of the Ricci-curvature. Secondly,

$$\lim_{r \to \infty}A(r) ~=~ \lim_{r \to \infty}B(r) ~=~ 1\tag{8.2.5}$$

follow from assuming the metric approaches the Minkowski metric as $r \to \infty$. Together they imply

$$ A(r)B(r)~=~1. \tag{8.2.6}$$

See also this related Phys.SE post.

$\endgroup$
  • $\begingroup$ As you said, the geodesic equations are not used to determine the A and B factors. However, when we try to use the geodesic equations to determine a motion of an object, the metric tensor obtained by the above process will play a role. I wondered if it would be appropriate to use the Schwarzschild metric, which was obtained under the condition of $r \to ∞$, for any motion. Anyway, in the answer by @Slereah, I found that this process has similar aspects to defining and using a gravitational potential in the classical mechanics, and my question about this part was resolved. $\endgroup$ – SOQEH Oct 30 '18 at 3:35
  • $\begingroup$ Thank you for your kind advice and, in particular, your link on the Birkhoff's Theorem. $\endgroup$ – SOQEH Oct 30 '18 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.