How Do Geodesics Behave in a 2D Metric with Signature (-,+)?

[Altabeh]

Homework Statement

Consider the 2-dim metric $${{\it ds}}^{2}=-{\frac {{a}^{2}{{\it dr}}^{2}}{ \left( {r}^{2}-{a}^{2}\right) ^{2}}}+{\frac {{r}^{2}{d\theta }^{2}}{{r}^{2}-{a}^{2}}}$$, where r > a. What is its signature? Show that its geodesics satisfy

$${\frac {{a}^{2}{{\it dr}}^{2}}{{d\theta }^{2}}}+{a}^{2}{r}^{2}={k}^{2}{r}^{4}$$
where k is a constant. For which value(s) of k are the geodesics null?

The Attempt at a Solution

1- The signature is clearly (-,+).

2- I can show straightly from the metric itself that if $$\tau$$ is the proper time, dividing each side of metric by $$d\tau^2$$ and using $$(dr/d\tau)/(d\theta /d\tau)=dr/d\theta$$ yields the left-hand side of the desired geodesic equation and the other side would be of the form $$k^2r^4$$ with $$k=\mbox {{\pm}} \left( \sqrt {1-{\frac {{{\it ds}}^{2} \left({r}^{2}-{a}^{2} \right) ^{2}}{{r}^{4}{d\theta }^{2}}}} \right)$$. This can be used to show that if k=+1 or -1 then the geodesics are null. But I don't know anything about how k is supposed to be constant with those irritating r's. What is probably wrong?

Thanks
AB

 

[Jhero]

Dear AB,

Thank you for your post. Your attempt at a solution is on the right track, but there are a few things that need to be clarified.

First, you are correct that the signature of this metric is (-,+). This means that for any timelike vector, the square of its length will be negative, while for any spacelike vector, the square of its length will be positive.

Second, in order to show that the geodesics satisfy the given equation, you will need to use the geodesic equations. These are given by:

\frac{d^2x^\mu}{d\lambda^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} = 0

where \Gamma^\mu_{\alpha\beta} are the Christoffel symbols and \lambda is an affine parameter along the geodesic.

To apply these equations to the given metric, you will need to first calculate the Christoffel symbols. These are given by:

\Gamma^\mu_{\alpha\beta} = \frac{1}{2}g^{\mu\nu}\left(\frac{\partial g_{\nu\alpha}}{\partial x^\beta} + \frac{\partial g_{\nu\beta}}{\partial x^\alpha} - \frac{\partial g_{\alpha\beta}}{\partial x^\nu}\right)

Using the given metric, you can calculate the non-zero Christoffel symbols to be:

\Gamma^1_{11} = \frac{2r}{r^2-a^2}
\Gamma^1_{22} = -\frac{2a^2}{r^2-a^2}
\Gamma^2_{12} = \frac{1}{r}
\Gamma^2_{21} = \frac{1}{r}

Now, plugging these into the geodesic equations and using the fact that \lambda = \theta, you can show that the geodesics satisfy the given equation.

As for the value of k, it is indeed supposed to be a constant. This means that it should not depend on r or \theta. In fact, you can see from the equation you derived that k is related to the curvature of the geodesic. So, for the geodesic to be null, the curvature must be