# Getting geodesic from variational principle

• LCSphysicist

#### LCSphysicist

Homework Statement
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Relevant Equations
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The metric is $$ds^2 = \frac{dr^2 + r^2 d\theta ^2}{r^2-a^2} - \frac{r^2 dr^2}{(r^2-a^2)^2}$$

I need to prove the geodesic is: $$a^2 (\frac{dr}{d \theta})^2 + a^2 r^2 = K r^4$$

My method was to variate the action ##\int\frac{(\frac{dr}{d\theta})^2 + r^2 }{r^2-a^2} - \frac{r^2 (\frac{dr}{d\theta})^2}{(r^2-a^2)^2} d \theta##

But the equation i am getting is $$a^2 (\frac{dr}{d \theta})^2 + r^2 = K (r^2-a^2)^2$$
Which can not be reduced to the answer.

I am a little confused, i could simpy calculate the Christoffel symbol, but i think this variation method easier, yet i am not sure how to use it.

So basically, my guess on why i have got the wrong answer is that the "lagrangean" i am variating is wrong. So what lagrangean indeed give us the right answer? Only ##g_{ab} \frac{dx^a}{ds} \frac{dx^b}{ds}= 1 ## and ##\sqrt{g_{ab} \frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda}}##? Or the lagrangean i used above is right, and i have done some algebric error?

Your approach failed because in general the Euler-Lagrange equations produced from $L_1(r, dr/d\theta)$ are not the same as those obtained from $L_2(r, dr/d\theta) = \sqrt{L_1}$: $$\begin{split} \frac{d}{d\theta} \left( \frac{\partial L_2}{\partial r'}\right) - \frac{\partial L_2}{\partial r} &= \frac{d}{d\theta} \left( \frac{1}{2L_2} \frac{\partial L_1}{\partial r'} \right) - \frac{1}{2L_2}\frac{\partial L_1}{\partial r} \\ &= \frac{1}{2L_2} \left[\frac{d}{d\theta} \left( \frac{\partial L_1}{\partial r'}\right) - \frac{\partial L_1}{\partial r}\right] + \frac{\partial L_1}{\partial r'} \frac{d}{d\theta}\left( \frac{1}{2L_2} \right) \\ &\neq \frac{1}{2L_2} \left[\frac{d}{d\theta} \left( \frac{\partial L_1}{\partial r'}\right) - \frac{\partial L_1}{\partial r}\right] \end{split}$$ If you take $s$ as the independent variable then you can work with $f(r) r'^2 + g(r) \theta'^2$ directly because by definition its total derivative with respect to $s$ is zero and you end up with the same Euler-Lagrange equation as you obtain from $\sqrt{f(r) r'^2 + g(r) \theta'^2}$.
$$L=\dot r^2\Big(\frac{1}{r^2-a^2}-\frac{r^2}{(r^2-a^2)^2}\Big)+\frac{r^2}{r^2-a^2}\dot\theta^2.$$
$$\frac{\partial L}{\partial \dot \theta}=const$$
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