- #1

Ai52487963

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**Another "christoffel symbols from the metric" question**

## Homework Statement

Find the Christoffel symbols from the metric:

[tex]ds^2 = -A(r)dt^2 + B(r)dr^2[/tex]

## Homework Equations

[tex]\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a} [/tex]

## The Attempt at a Solution

Pretty straight forward, but I'm not seeing where one of the cross-terms comes from. I used the euler-lagrange eqns to rearrange for the equation of motion so I could just read the christoffel symbols off:

the t-component of the E-L eqns is zero, so moving on to the r...

[tex]\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r} [/tex]

[tex]\Rightarrow \frac{d}{dt} \left( 2B(r)\dot{r} \right) = -A'(r)\dot{t}^2 + B'(r)\dot{r}^2[/tex]

[tex]\Rightarrow 2B'(r)\dot{r} + 2B(r)\ddot{r} = -A'(r)\dot{t}^2 + B'(r)\dot{r}^2[/tex]

[tex]\Rightarrow 2B(r)\ddot{r} + 2B'(r)\dot{r} - B'(r)\dot{r}^2 + A'(r)\dot{t}^2 = 0[/tex]

[tex]\Rightarrow \ddot{r} + \frac{B'(r)}{B(r)}\dot{r} - \frac{B'(r)}{2B(r)}\dot{r}^2 + \frac{A'(r)}{2B(r)}\dot{t}^2 = 0[/tex]

But I'm not seeing how the answer given tells us that

[tex] \Gamma^r_{tt} = \frac{A'(r)}{2B(r)} [/tex]

[tex] \Gamma^r_{tr} = \frac{A'(r)}{2A(r)} [/tex]

[tex] \Gamma^r_{rr} = \frac{B'(r)}{2B(r)} [/tex]

From my EOM's it looks like I have the r_rr and r_tt symbols, but I'm not seeing where this r_tr symbol is supposed to come from. I have an extraneous rdot floating around that doesn't have a corresponding tdot , but even if it did, there has to be an A' in the numerator of that factor, so what gives? Where'd I go wrong?