1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Another christoffel symbols from the metric question
Loading...