# Homework Help: Another christoffel symbols from the metric question

1. Mar 29, 2012

### Ai52487963

Another "christoffel symbols from the metric" question

1. The problem statement, all variables and given/known data

Find the Christoffel symbols from the metric:

$$ds^2 = -A(r)dt^2 + B(r)dr^2$$

2. Relevant equations

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}$$

3. 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...

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r}$$

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

$$\Rightarrow 2B'(r)\dot{r} + 2B(r)\ddot{r} = -A'(r)\dot{t}^2 + B'(r)\dot{r}^2$$

$$\Rightarrow 2B(r)\ddot{r} + 2B'(r)\dot{r} - B'(r)\dot{r}^2 + A'(r)\dot{t}^2 = 0$$

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

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

$$\Gamma^r_{tt} = \frac{A'(r)}{2B(r)}$$
$$\Gamma^r_{tr} = \frac{A'(r)}{2A(r)}$$
$$\Gamma^r_{rr} = \frac{B'(r)}{2B(r)}$$

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?

2. Mar 29, 2012

### Dick

Re: Another "christoffel symbols from the metric" question

t is a coordinate. Just like r. Don't presume that the "dot" means derivative with respect to t. Write your Euler-Lagrange equation with "dot" meaning derivative with respect to some arbitrary parameter $\eta$:

$\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}$

Last edited: Mar 29, 2012
3. Mar 29, 2012

### Ai52487963

Re: Another "christoffel symbols from the metric" question

I agree and it makes sense that I should have a tdot * rdot cross-term, but I'm not seeing where this phantom A' is coming from in that cross-term.

If the partial of L with respect to rdot is really a partial, then it shouldn't act on the -A(r) term in the first part of the lagrangian:

$$L = -A\dot{t}^2 + B\dot{r}^2$$

applying partial wrt rdot:

$$\frac{\partial L}{\partial \dot{r}} = 2B\dot{r}$$

and lacking any A's in the numerator or denominator of the soon-to-be cross term

$$\frac{d}{dt} (2B \dot{r} ) = -A'\dot{t}^2 + B'\dot{r}^2$$

$$\Rightarrow 2B' \dot{t}\dot{r} + 2B\ddot{r}= -A'\dot{t}^2 + B'\dot{r}^2$$

$$\Rightarrow 2B \ddot{r} + 2B'\dot{t}\dot{r} + A'\dot{t}^2 - B'\dot{r}^2 = 0$$

$$\Rightarrow \ddot{r} + \frac{B'}{2B}\dot{t}\dot{r} + \frac{A'}{2B} \dot{t}^2 - \frac{B'}{2B}\dot{r}^2 = 0$$

gives us the Christoffel symbols:

$$\Gamma^r_{tr} = \Gamma^r_{rt} = \frac{B'}{2B}$$
$$\Gamma^r_{tt} = \frac{A'}{2B}$$
$$\Gamma^r_{rr} = \frac{B'}{2B}$$

I'm thinking the mixed term has to be a typo, I can't see what I did wrong otherwise.

4. Mar 30, 2012

### Dick

Re: Another "christoffel symbols from the metric" question

You are making some mistakes there. E.g. $\frac{d}{d \eta} (2B \dot{r} ) = 2B(r) \ddot{r} + 2 B'(r) {\dot r}^2$ There is no single $\dot r$ term. And the mixed term is coming from the other EL equation:
$$\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{t}} \right) = \frac{\partial L}{\partial t}$$