# Curves under conformal trans.

1. Mar 2, 2014

### spookyfish

Hi. This problem is about General Relativity. I am not actually taking a course, but studying myself. I tried to solve Hobson 2.7:

1. The problem statement, all variables and given/known data
A conformal transformation is not a change of coordinates but an actual change in
the geometry of a manifold such that the metric tensor transforms as $\bar{g}_{ab}(x)=\Omega^2(x) g_{ab}(x)$, where $\Omega (x)$ is some non-vanishing scalar function of position. In a pseudo-Riemannian manifold, show that if $x^a(\lambda)$ is a null curve with respect to $g_{ab}$ (i.e. $ds^2=0$ along the curve), then it is also a null curve with respect to $\bar{g}_{ab}$. Is this true for timelike curves?

2. Relevant equations

for null curves:
$$ds^2=g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=0 \\ \therefore g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}=0$$

3. The attempt at a solution
after the conformal transformation, we have
$$ds^2=\Omega^2 g_{\mu \nu}dx^\mu dx^\nu =g_{\mu \nu}(\Omega dx^\mu) (\Omega dx^\nu)$$
and along the path $x^\mu(\lambda)$:

$$ds^2=g_{\mu \nu} \left[\frac{d}{d\lambda}(\Omega dx^\mu) \frac{d}{d\lambda}(\Omega dx^\nu) \right] (d\lambda)^2 \\=g_{\mu \nu}\left(\frac{d\Omega}{d\lambda}dx^\mu +\Omega \frac{dx^\mu}{d\lambda} \right) \left(\frac{d\Omega}{d\lambda}dx^\nu +\Omega \frac{dx^\nu}{d\lambda} \right) (d\lambda)^2 \\=g_{\mu \nu} \left\{ \left(\frac{d\Omega}{d\lambda}\right)^2 dx^\mu dx^\nu +\Omega \frac{d\Omega}{d\lambda} \left(dx^\mu \frac{dx^\nu}{d\lambda} + \frac{dx^\mu}{d\lambda} dx^\nu \right) +\Omega^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right\} (d\lambda)^2 \\= g_{\mu \nu} \left\{ \left(\frac{d\Omega}{d\lambda}\right)^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^4 + 2\Omega \frac{d\Omega}{d\lambda} \left(\frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right) (d\lambda)^3 + \Omega^2 \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2 \right\} \\= \left( g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} \right) \left( \left(\frac{d\Omega}{d\lambda}\right)^2(d\lambda)^2+ 2\Omega \frac{d\Omega}{d\lambda}d\lambda +\Omega^2 \right)(d\lambda)^2 = 0$$

now in the last line, the first term vanishes because of the null curve for the original $g_{\mu \nu}$, so the whole $ds^2$ after the transformation vanishes, which means it is a null curve.

Is this solution correct?

Last edited: Mar 2, 2014
2. Mar 2, 2014

### TSny

Hello, spookyfish.

It seems to me to be more straightforward. Before the transformation you have a curve for which any infinitesimal displacement $dx^\mu$ along the curve satisfies
$$ds^2= g_{\mu \nu}dx^\mu dx^\nu = 0$$
After the transformation you have a new metric $\overline{g}_{\mu \nu} = \Omega^2 g_{\mu \nu}$. But the spacetime coordinates of points in the manifold have not changed their values.

So, what is the value of $d\overline{s}^2= \overline{g}_{\mu \nu}dx^\mu dx^\nu$ for the same $dx^\mu$?

3. Mar 2, 2014

### spookyfish

The value would be
$$d\bar{s}^2 =\Omega^2 g_{\mu \nu} dx^\mu dx^\nu$$
but we need to check $d\bar{s}^2$ of a curve. Can I say that
$$d\bar{s}^2 =\Omega^2 g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=0$$
because $g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2=ds^2=0$?
Also, using this would not imply that timelike curves and spacelike curves remain the same, because $\Omega^2$ doesn't change the sign of $ds^2$?

4. Mar 2, 2014

### TSny

Yes, that all sounds correct.

I don't see the need to explicitly introduce the curve parameter $\lambda$, because $d\bar{s}^2 =\Omega^2 g_{\mu \nu} dx^\mu dx^\nu$ and $d\bar{s}^2 =\Omega^2 g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} (d\lambda)^2$ have the same value. But, you can if you want.