# Null geodesic given metric

I (desperately) need help with the following:

What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$

I don't know how to transform a metric into a geodesic...! There is no need to start from the Lagrangian. I know that $$0=g_{ij}V^iV^j$$ where $V^i={dx^i\over d\lambda}$ where $\lambda$ is some parameter. But I don't know what that parameter is nor how to find the geodesic.

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A geodesic is just a special type of curve, and the parameter is, well, just a parameter parametrizing that curve, and so doesn't matter ultimately. Also, geodesics are only unique if you specify more about them, like a point on the curve and a direction on that curve. Make sure to understand what these objects are before diving into symbol manipulation. – Chris White Feb 22 '13 at 2:43

As you mentioned a null geodesic implies:

$$g_{\mu \nu} U^{\mu} U^{\nu}=\left (\frac{ds}{d \lambda} \right )^2=0$$

where $\lambda$ is some affine parameter. If you take $\lambda=t$, then this implies:

$$-x^2+\left ( \frac{dx}{dt} \right )^2=0$$

So $~dx/dt=\pm x$ is a null geodesic. (Take a second time derivative to get the actual geodesic.)

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Thanks, elfmotat. What do you mean by the "actual" geodesic? – hetherson Feb 22 '13 at 0:06
I mean the second-order differential equation, which is called the "geodesic equation." – jld Feb 22 '13 at 0:18

The easiest way to solve this is to pretend that it IS a Lagrangian:

$$I = \frac{1}{2}\int d\lambda \left(-x^{2}{\dot t}^{2} + {\dot x}^{2}\right)$$

Where both $t$ and $x$ are functions of $\lambda$, and ${\dot x} \equiv \frac{dx}{d\lambda}$.. Take the variation of the action, find the minimum, and then set your constants so that your geodesic is null.

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