First ever post - please be kind.

I'm trying to understand how General Relativity becomes equivalent to Newton's laws of motion, plus Newton's law of gravitational attraction in the limiting case of low speeds and low mass. But I've come unstuck even in the simplest case!

I am imagining a satellite in a perfectly circular orbit around a planet. It should be possible to choose co-ordinates $t,\theta,\phi,r$ such that $r,\phi$ and the metric values $g_{\theta\theta},g_{tt},g_{t\theta}$ are constant along the orbit.

Now, letting $dt$ be fixed, according to what I have understood GR says that the satellite moves along a geodesic, i.e. a path for which the line element $ds$ is optimised, where:

$$ds^2 = c^2 d\tau^2 = dt^2 g_{tt} + d\theta^2 g_{\theta\theta} + 2 dt d\theta g_{t\theta}$$

I thought that it should be possible to solve this and work out the angular velocity $\omega$ of the satellite in terms of $g$. So I rewrote the above as

$$ds^2 = dt^2 (g_{tt}+\omega^2 g_{\theta\theta} + 2\omega g_{t\theta})$$

and then I found the maximum by differentiating w.r.t. $\omega$ and setting to 0:

$$0 = 2 \omega g_{\theta\theta} + 2 g_{t\theta}$$


$$\omega = -\frac{g_{t\theta}}{g_{\theta\theta}}$$

But I must have gone wrong somewhere in my reasoning as there are two solutions for $\omega$ in Newton's theory (you can orbit clockwise or anti-clockwise) but the above gives only one.

  • $\begingroup$ Hint: why do you think that $g_{t\theta} \neq 0$??? $\endgroup$ – MariusMatutiae May 17 '16 at 17:09
  • $\begingroup$ If $g_{t\theta} =0$ wouldn't you get either $\omega = 0$ or $\omega = \infty$ (depending on whether $g_{\theta\theta}$ is positive or negative)? Neither of which match the Newtonian solution. $\endgroup$ – Alex Zeffertt May 17 '16 at 17:18
  • $\begingroup$ zero divided by a negative number is still zero, not infinity $\endgroup$ – fqq May 17 '16 at 17:20
  • $\begingroup$ Yes, zero would be the correct solution for the stationary point of $ds^2(\omega)$, but that may be a minimum not a maximum. $\endgroup$ – Alex Zeffertt May 17 '16 at 17:22
  • $\begingroup$ Hint 2: That's only because you are not minimizing correctly... $\endgroup$ – MariusMatutiae May 17 '16 at 17:22

Not sure about the etiquette of this but I think I can now answer my own question. Please post if there is a better answer.

The problem is that I misunderstood the meaning of the statement "the geodesic is the path the optimises the proper time". What this means is that given two endpoints, say $x_0^\alpha$ and $x_1^\alpha$ a geodesic is a path $x^\alpha(\lambda)$ with that satisfies

$x^\alpha(\lambda_0) = x_0^\alpha$

$x^\alpha(\lambda_1) = x_1^\alpha$

for some $\lambda_0, \lambda_1$ and optimises

$\int_{\lambda_0}^{\lambda_1}ds = \int_{\lambda_0}^{\lambda_1}\sqrt{g_{\alpha\beta}dx^\alpha dx^\beta}$

However, what I was doing was varying one co-ordinate of the endpoints, rather than fixing the endpoints and varying the path.

I guess the difficulty is in changing mindset from a Newtonian one (you know where you are and your velocity - now work out where you will be) to an Lagrangian one (?) (you know where you start and end - now find out the path taken).

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