Another christoffel symbols from the metric question

[Ai52487963]

Another "christoffel symbols from the metric" question

Homework Statement

Find the Christoffel symbols from the metric:

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

Homework Equations

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

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?

 

[Dick]

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}

 

[Ai52487963]

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}

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.

 

[Dick]

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}