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Homework Help: Finding the geodesic equation from a given line element

  1. Jun 11, 2016 #1
    1. The problem statement, all variables and given/known data
    We've got a line element [tex]ds^2 = f(x) du^2 + dx^2[/tex] From that we should find the geodesic equation

    2. Relevant equations
    Line Element:
    [tex]ds^2 = dq^j g_{jk} dq^k[/tex]
    Geodesic Equation:
    [tex]\ddot{q}^j = -\Gamma_{km}^j \dot{q}^k \dot{q}^m [/tex]
    Christoffel Symbol:
    [tex]\Gamma_{km}^j = \frac{g^{jl}}{2} \left( \frac{\partial g_{lk}}{\partial q^m} + \frac{\partial g_{lm}}{\partial q^k} - \frac{\partial g_{km}}{\partial q^l}\right)[/tex]
    Lagrangian in this case (no potential, just free particle)
    : [tex]L = T = \frac{m}{2}\dot{q}^j g_{jk} \dot{q}^k[/tex]
    Coordinates: u and x

    3. The attempt at a solution

    I am stuck because I am not getting the same equations of motions for my coordinate u from the Lagrangian (with Euler-Lagrange) and the Geodesic equation, and I think, it is because I got the wrong inverse metric tensor.
    My metric tensor looks like this: \begin{equation} \begin{pmatrix}f(x) & 0 \\ 0 & 1\end{pmatrix}\end{equation} .
    The inverse: \begin{equation}\begin{pmatrix}1/f(x) & 0 \\ 0 & 1\end{pmatrix}\end{equation}

    I got 8 possible Christoffel-Symbols, but only 2 will be non-zero:
    \begin{equation}\Gamma_{ux}^{u} =\Gamma_{xu}^{u} = \frac{1}{f(x)} \frac{1}{2} \frac{\partial f(x)}{\partial x}\end{equation}
    \begin{equation}\Gamma_{uu}^{x} = -\frac{1}{2} \frac{\partial f(x)}{\partial x}\end{equation}

    If I plug those into the geodesic equation, I get 2 equations of motion. If I use the Euler-Lagrange-Equation to derive equations of motion from the lagrangian, I get the same equation of motion for x, but not for u.
    Geodesic u:
    \begin{equation}\ddot{u} + \frac{1}{2}\frac{1}{f(x)}\frac{\partial f(x)}{\partial x} \dot{u}\dot{x} = 0\end{equation}
    Euler-Lagrange for u:
    \begin{equation}\ddot{u}\dot{x}\frac{\partial f(x)}{\partial x} = 0 \end{equation}
    I think it has to do with the inverse, cause for x I get the same equation for both ways:
    \begin{equation}\ddot{x} - \dot{u}^2\frac{1}{2}\frac{\partial f(x)}{\partial x} = 0\end{equation}

    So my question is, is the metric tensor and it's inverse correct? If yes, where am I wrong?
    Thank you ^^!
  2. jcsd
  3. Jun 11, 2016 #2


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    Staff Emeritus
    Science Advisor

    First of all, you are off by a factor of 2 in your geodesic equation. The geodesic equation is:

    [itex]\ddot{x^\mu} + \Gamma^\mu_{\nu \lambda} \dot{x^\nu} \dot{x^\lambda} = 0[/itex]

    So for [itex]u[/itex], we have:

    [itex]\ddot{u} + \Gamma^u_{\nu \lambda} \dot{x^\nu} \dot{x^\lambda} = 0[/itex]

    The second term means sum over all possible values of [itex]\nu[/itex] and [itex]\lambda[/itex]. There are two nonzero values:
    [itex]\nu = u, \lambda = x[/itex] and [itex]\nu = x, \lambda = u[/itex]. So you have:

    [itex]\ddot{u} + \Gamma^u_{u x} \dot{u} \dot{x} + \Gamma^u_{xu} \dot{x} \dot{u} = 0[/itex]

    The second and third terms are equal, so you have:
    [itex]\ddot{u} + 2 \Gamma^u_{u x} \dot{u} \dot{x} = 0[/itex]

    The second problem is with your Euler-Lagrange equation for [itex]u[/itex].

    You have a Lagrangian: [itex]L = \frac{m}{2} (\dot{u}^2 f + \dot{x}^2)[/itex]

    Taking it in three steps:

    [itex]p_u = \frac{\partial L}{\partial \dot{u}}[/itex]
    [itex]F_u = \frac{\partial L}{\partial u}[/itex]
    [itex]\frac{d p_u}{dt} = F_u[/itex]

    What do you get for [itex]p_u[/itex] and [itex]F_u[/itex]?
  4. Jun 11, 2016 #3
    \begin{equation}p_u = \frac{\partial L}{\partial \dot{u}} = m\dot{u}f\end{equation}
    \begin{equation}\frac{\partial L}{\partial u} = 0\end{equation}
    \begin{equation}\frac{dp_u}{dt} = \frac{d}{dt}(m\dot{u}f) = m\ddot{u}f + m\dot{u}\frac{\partial f}{\partial x}\dot{x} = 0\end{equation}

    Dividing by m and f will get me to the same geodesic (corrected) equation.
    Edit: Ahh I messed up, I forgot the product rule

    Thank you very very much!
    Last edited: Jun 11, 2016
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