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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:

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

i.e.

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

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.

Qmechanic
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Alex Zeffertt
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  • Hint: why do you think that $g_{t\theta} \neq 0$??? – MariusMatutiae May 17 '16 at 17:09
  • 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. – Alex Zeffertt May 17 '16 at 17:18
  • zero divided by a negative number is still zero, not infinity – fqq May 17 '16 at 17:20
  • Yes, zero would be the correct solution for the stationary point of $ds^2(\omega)$, but that may be a minimum not a maximum. – Alex Zeffertt May 17 '16 at 17:22
  • Hint 2: That's only because you are not minimizing correctly... – MariusMatutiae May 17 '16 at 17:22
  • Ah. If I minimize $ds$ instead of $ds^2$ and assume $g_{t\theta}=0$ I get $0 = \frac{\omega g_{\theta\theta}}{\sqrt {g_{tt}+g_{\theta\theta}\omega^2}}$ which is actually optimized if $\omega =\pm\sqrt{\frac{-g_{tt}}{g_{\theta\theta}}}$ Is that close? – Alex Zeffertt May 17 '16 at 17:35
  • Scrap that. What I just wrote was nonsense. The solution would still be zero. I can't believe it's just calculus I'm getting stuck on. (Though it was a long time ago I did it at uni.) – Alex Zeffertt May 17 '16 at 17:43
  • Possible duplicates: http://physics.stackexchange.com/q/211930/2451 and links therein. – Qmechanic May 17 '16 at 18:03
  • The only thing I can see is that if $g_{t\theta} = 0$ then we must have $\left|\omega \right | < \sqrt{\frac{-g_{tt}}{g_{\theta\theta}}}$ for $ds$ to be a valid time-like line segment. That suggests $\pm \sqrt{\frac{-g_{tt}}{g_{\theta\theta}}}$ as the solution. But that's not quite right since those values would give $ds = 0$ and the geodesic ought to optimise the proper time not minimise it. Still confused. PS: Not actually homework - I haven't done that for 20 years! – Alex Zeffertt May 17 '16 at 18:37
  • I think I now understand - I had the premise wrong. The geodesic is the path that optimises $\int_{t_0,\theta_0}^{t_1,\theta_1} ds$. So I was supposed to choose start and end values for my coordinates and then look for the correct path between them. However, what I did was to choose start and end values of $t$ only and ask what the change in $\theta$ would be - which isn't how it is supposed to work. – Alex Zeffertt May 17 '16 at 18:52

1 Answers1

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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).

Alex Zeffertt
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