# Finding the geodesic equation from a given line element

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1. Jun 11, 2016

### Christoffelsymbol100

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

2. Relevant equations
Line Element:
$$ds^2 = dq^j g_{jk} dq^k$$
Geodesic Equation:
$$\ddot{q}^j = -\Gamma_{km}^j \dot{q}^k \dot{q}^m$$
Christoffel Symbol:
$$\Gamma_{km}^j = \frac{g^{jl}}{2} \left( \frac{\partial g_{lk}}{\partial q^m} + \frac{\partial g_{lm}}{\partial q^k} - \frac{\partial g_{km}}{\partial q^l}\right)$$
Lagrangian in this case (no potential, just free particle)
: $$L = T = \frac{m}{2}\dot{q}^j g_{jk} \dot{q}^k$$
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: $$\begin{pmatrix}f(x) & 0 \\ 0 & 1\end{pmatrix}$$ .
The inverse: $$\begin{pmatrix}1/f(x) & 0 \\ 0 & 1\end{pmatrix}$$

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

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:
$$\ddot{u} + \frac{1}{2}\frac{1}{f(x)}\frac{\partial f(x)}{\partial x} \dot{u}\dot{x} = 0$$
Euler-Lagrange for u:
$$\ddot{u}\dot{x}\frac{\partial f(x)}{\partial x} = 0$$
I think it has to do with the inverse, cause for x I get the same equation for both ways:
$$\ddot{x} - \dot{u}^2\frac{1}{2}\frac{\partial f(x)}{\partial x} = 0$$

So my question is, is the metric tensor and it's inverse correct? If yes, where am I wrong?
Thank you ^^!

2. Jun 11, 2016

### stevendaryl

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

$\ddot{x^\mu} + \Gamma^\mu_{\nu \lambda} \dot{x^\nu} \dot{x^\lambda} = 0$

So for $u$, we have:

$\ddot{u} + \Gamma^u_{\nu \lambda} \dot{x^\nu} \dot{x^\lambda} = 0$

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

$\ddot{u} + \Gamma^u_{u x} \dot{u} \dot{x} + \Gamma^u_{xu} \dot{x} \dot{u} = 0$

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

The second problem is with your Euler-Lagrange equation for $u$.

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

Taking it in three steps:

$p_u = \frac{\partial L}{\partial \dot{u}}$
$F_u = \frac{\partial L}{\partial u}$
$\frac{d p_u}{dt} = F_u$

What do you get for $p_u$ and $F_u$?

3. Jun 11, 2016

### Christoffelsymbol100

$$p_u = \frac{\partial L}{\partial \dot{u}} = m\dot{u}f$$
$$\frac{\partial L}{\partial u} = 0$$
$$\frac{dp_u}{dt} = \frac{d}{dt}(m\dot{u}f) = m\ddot{u}f + m\dot{u}\frac{\partial f}{\partial x}\dot{x} = 0$$

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