Timelike Geodesics: Solving Reissner Nordstrom Line Element

[alex_b93]

Homework Statement

Using the Reissner Nordstrom line element, which I've given in the relevant equations section, I'm looking to show that the time like Geodesics obey the equation again show below.

Homework Equations

Line Element[/B]
$ds^2= - U(r)c^2dt^2 +\frac{dr^2}{U(r)} +r^2(d\theta^2 + sin^2(\theta)d\phi^2)$
$U(r)=1-\frac{r_s}{r}+\frac{G^2Q^2}{r^2}$

Equation to Obey
$\frac{1}{2} (\frac{dr}{d\tau})^2 +V(r) = \varepsilon$

The Attempt at a Solution

I've presumed as we are looking for a $dr'$ the Euler Lagrange equation we would be interested would be[/B]
$\frac{d}{d\tau}(\frac{\partial L^2}{\partial r'}) - \frac{\partial L^2}{\partial r}$

If I work this through my answer doesn't really resemble the equation I'm looking for, I get the $(\frac{dr}{d\tau})^2$, but I can't get the $\frac{1}{2}$ factor, plus I have other terms in the denominator.
I also have a lot of other terms but they could possibly be grouped into $V(r)$.

I was hoping somebody could confirm whether the method I'm attempting is correct, as then I'll know if I'm incorrectly calculating it or it is something else.

Many thanks.

 

[stevendaryl]

You're on the right track, but why don't you first write down what you think L is. Your equations don't define L.

 

[stevendaryl]

Here are a couple of other hints:

First, \tau and s are the same thing, so \frac{ds}{d\tau} = 1. So if you take the expression for s, this gives you one "constant of the motion".

Second, if you have a Lagrangian of the form L(r, \frac{dr}{d\tau}, t, \frac{dt}{d\tau}, \theta, \frac{d\theta}{d\tau}, \phi \frac{d\phi}{d\tau}), and L doesn't mention \tau, then the following quantity is conserved (has the same value for all \tau):

H = (\sum_j P_j U^j) - L

where U^j = \frac{d x^j}{d\tau} and P_j = \dfrac{\partial L}{\partial U^j}.

So H gives you a second constant of the motion. So H = E, for some constant E