Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $r$:


Then, using energy conservation:


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:


for the proper time for one circulation.

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

Where do I have wrong?

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

1 Answer 1

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.

share|improve this answer
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

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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