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Homework Help: Another christoffel symbols from the metric question

  1. Mar 29, 2012 #1
    Another "christoffel symbols from the metric" question

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

    Find the Christoffel symbols from the metric:

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

    2. Relevant equations

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

    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...

    [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?
  2. jcsd
  3. Mar 29, 2012 #2


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    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 [itex]\eta[/itex]:

    [itex]\frac{d}{d \eta} \left( \frac{\partial L}{\partial \dot{x^a}} \right) = \frac{\partial L}{\partial x^a}[/itex]
    Last edited: Mar 29, 2012
  4. Mar 29, 2012 #3
    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:

    [tex]L = -A\dot{t}^2 + B\dot{r}^2[/tex]

    applying partial wrt rdot:

    [tex] \frac{\partial L}{\partial \dot{r}} = 2B\dot{r}[/tex]

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

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

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

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

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

    gives us the Christoffel symbols:

    [tex]\Gamma^r_{tr} = \Gamma^r_{rt} = \frac{B'}{2B}[/tex]
    [tex]\Gamma^r_{tt} = \frac{A'}{2B}[/tex]
    [tex]\Gamma^r_{rr} = \frac{B'}{2B}[/tex]

    I'm thinking the mixed term has to be a typo, I can't see what I did wrong otherwise.
  5. Mar 30, 2012 #4


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    Re: Another "christoffel symbols from the metric" question

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