# Proper time of circular motion under Schwarzschild metric

I'm trying to calculate the proper time of a massive particle circulating Schwarzschild black hole, using EL equation of the following Lagrangian:

$$L=-\frac{m}{2}\left(1-\frac{2M}{r}\right)\dot{t}^{2}+\frac{m}{2}\left(1-\frac{2M}{r}\right)^{-1}\dot{r}^{2}+\frac{m}{2}r^{2}\dot{\theta}^{2}+\frac{m}{2}r^{2}\sin^{2}\theta\dot{\varphi}^{2} .$$

At first I get from the Euler-Lagrange equation of $r$:

$$\pm\sqrt{\frac{M}{R^{3}}}\dot{t}=\dot{\phi}.$$

Then, using energy conservation:

$$\dot{t}\equiv\frac{E}{m\left(1-\frac{2M}{r}\right)}.$$

Now, I thought to integrate over $\tau$ (the proper time), as LHS does not depend on it, while at RHS it turns to integration over $\phi$ form $0$ to $2\pi$ .

Eventually, I end up with:

$$2\pi\left(\sqrt{\frac{R^{3}}{M}}\frac{m}{E}\left(1-\frac{2M}{R}\right)\right)$$

for the proper time for one circulation.

This result doesn't seem to make a lot of sense.

Where do I have wrong?

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Hi @Nillls Zhuaberg, click here if you want to merge your accounts. – Qmechanic Jun 10 '13 at 9:38

As you say the E-L equation gives $$\dot{\phi}^2 = \frac{M}{R^3}\dot{t}^2$$ where $\dot{a}$ denotes $\frac{d}{d\tau}a$ for some $a$. The metric tells us that $$d\tau^2 = \left(1-\frac{2M}{R}\right)dt^2 - R^2d\phi^2 \implies 1 = \left(1-\frac{2M}{R}\right)\dot{t}^2 - R^2\dot{\phi}^2$$ applying the first equation and rearranging this gives $$1 = \left(1-\frac{2M}{R}\right)\frac{R^3}{M}\dot{\phi}^2 - R^2\dot{\phi}^2 \implies \dot{\phi}^2 = \left[\frac{R^3}{M}-3R^2\right]^{-1}$$ integrating using separation of variables yields: $$\tau_{orbit} = 2\pi\sqrt{\frac{R^3}{M}}\sqrt{1-\frac{3M}{R}}$$

In short, there is nothing wrong with your solution, if you just plug in $\frac{E}{m} = \frac{\left(1-\frac{2M}{R}\right)}{\sqrt{1-\frac{3M}{R}}}$ (this comes from replacing $\dot{\phi}^2$ with $\frac{M}{R^3}\dot{t}^2$ in my second equation and comparing to your definition of $\frac{E}{m}$) into your result, you get my result.

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How can one recover the Newtonian approximation from that result? – user25600 Jun 10 '13 at 9:25
It should be rather obvious from my answer above. The weak field limit is $\frac{M}{R}\rightarrow0$. And look, that's exactly what gives us the Newtonian result $2\pi\sqrt{\frac{R^3}{M}}$ (in units where $G=1$). – Will Jun 10 '13 at 12:13
Will, it is indeed obvious from your answer, I had the Newtonian result wrong in my notebook. thanks for all. – user25600 Jun 10 '13 at 15:49

## protected by Qmechanic♦Jun 10 '13 at 15:57

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