# Geodesic in 2D space

1. May 13, 2013

### vidi

1. The problem statement, all variables and given/known data
I am having trouble understanding how the following statement (taken from some old notes) is true:

>For a 2 dimensional space such that $$ds^2=\frac{1}{u^2}(-du^2+dv^2)$$
the timelike geodesics are given by $$u^2=v^2+av+b$$ where $a,b$ are constants.

2. Relevant equations
Euler-Lagrange, Normalisation condition

3. The attempt at a solution

When I see "geodesics" I jump to the Euler-Lagrange equations. They give me
$$\frac{d}{d\lambda}(-2\frac{\dot u}{u^2})=(-\dot u^2+\dot v^2)(-\frac{2}{u^3})\\ \implies \frac{\ddot u}{u^2}-2\frac{\dot u^2}{u^3}=\frac{1}{u^3}(-\dot u^2+\dot v^2)\\ \implies u\ddot u-\dot u^2-\dot v^2=0$$
and
$$\frac{d}{d\lambda}(2\frac{\dot v}{u^2})=0\\ \implies \dot v=cu^2$$
where $c$ is some constant.

Timelike implies $$\dot x^a\dot x_a=-1$$ where I have adopted the (-+++) signature.

I can't for the life of me see how the statement results from these. Would someone mind explaining? Thanks.

2. May 15, 2013

### Thaakisfox

Do not expand the Euler-Lagrange equations. But do a trick like this:

The Lagrangian is:

$$L = \frac{1}{u}\sqrt{\left(\frac{dv}{du}\right)^2-1}= \frac{1}{u}\sqrt{v'^2-1}$$

Now you see this doesnt depend on $v$. The Euler Lagrange equations then give:

$$0=\frac{\partial L}{\partial v}=\frac{d}{du}\frac{\partial L}{\partial v'} \Longrightarrow \frac{\partial L}{\partial v'}=C$$

Now calculate $$\frac{\partial L}{\partial v'}$$ from the Lagrangian and put it equal to the constant C. This will lead to a differential equation which gives the solution you were looking for.

Last edited: May 15, 2013
3. May 15, 2013

### vidi

Thanks, Thaakisfox, I've got it now!