Null geodetic on different metric relation between affine parameter If i have a null geodetic on a $g_{\mu \nu}$ metric, it's also null in a generic $\hat{g}=\exp{2\omega(x)}g_{\mu \nu}$ for any $\exp{2\omega(x)}\geq 0$? 
What's the relation between affine parameter?
 A: Consider a null geodesic on $g$ with affine parameter $\tau$. We have the following equations
$$
\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} = 0 , \qquad g_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} = 0 . 
$$
For the metric ${\tilde g} = e^{2\omega} g$, we have
$$
{\tilde \Gamma}^\mu_{\nu\rho} = \Gamma^\mu_{\nu\rho} + \delta^\mu_\nu \partial_\rho \omega + \delta^\mu_\rho \partial_\nu \omega - g_{\nu\rho} \nabla^\mu \omega  .
$$
Thus, the geodesic equation on the conformal metric takes the form
\begin{align}
\frac{d^2x^\mu}{d\tau^2} + {\tilde \Gamma}^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} &=   \frac{d x^\mu}{d\tau} \left( 2  \partial_\rho \omega \frac{d x^\rho}{ d \tau} \right) + \left( \frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} \right)  -  \nabla^\mu \omega \left(  g_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau} \right) \\
&=  \frac{d x^\mu}{d\tau} \left(  2 \frac{d \omega}{d\tau} \right) .
\end{align}
Thus,
$$
\frac{d^2x^\mu}{d\tau^2} + \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tau} \frac{d x^\rho}{ d \tau}  =  \frac{d x^\mu}{d\tau} \left(  2 \frac{d \omega}{d\tau} \right). 
$$
Thus, we find that same curve $x^\mu(\tau)$ is a geodesic (also null - reader should verify!) in the new metric as well! However, $\tau$ is not an affine parameter on the new metric. Suppose that the affine parameter is ${\tilde \tau} \equiv {\tilde \tau}(\tau)$. Then, writing out the equation in terms of ${\tilde \tau}$
\begin{align}
\frac{d^2x^\mu}{d \tilde \tau^2} + \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d\tilde \tau} \frac{d x^\rho}{d\tilde \tau} &= \frac{d^2\tau}{d\tilde \tau^2}  \frac{dx^\mu}{d \tau} + \left( \frac{d\tau}{d\tilde \tau} \right)^2  \left( \frac{d^2x^\mu}{d\tau^2}  +  \tilde \Gamma^\mu_{\nu\rho} \frac{d x^\nu}{d \tau} \frac{d x^\rho}{d \tau} \right) \\
&= \left[ \frac{d^2\tau}{d\tilde \tau^2}  + 2 \frac{d\tau}{d\tilde \tau}  \frac{d \omega}{d\tilde  \tau} \right]  \frac{dx^\mu}{d \tau}.
\end{align}
Then, ${\tilde \tau}$ is an affine parameter iff.
$$
\frac{d^2\tau}{d\tilde \tau^2}  + 2 \frac{d\tau}{d\tilde \tau}  \frac{d \omega}{d\tilde  \tau}  = 0 \quad \implies \quad \frac{d^2 {\tilde \tau} }{ d \tau^2 }  - 2 \frac{ d {\tilde \tau } }{ d \tau } \frac{d \omega}{d\tau} = 0 . 
$$
This solves to
$$
{\tilde \tau}(\tau) = a + b \int_0^\tau e^{2 \omega(x(\tau')) } d\tau'.
$$
The constants $a$ and $b$ represent the freedom in the definition of the affine parameter of the form $\tau \to a \tau + b$. These can be fixed by picking an origin and a scale for the affine parameter along the geodesic. For instance, pick two points $x_0$ and $x_1$ on the geodesic and we choose our parameterization so that $x(\tau=0) = x_0$ and $x(\tau=1) = x_1$. Suppose that in the new affine parameterization, we wish to have the same choices, i.e. $x(\tilde \tau=0) = x_0$ and $x(\tilde \tau=1) = x_1$. In other words, we need ${\tilde \tau}(0)=0$ and ${\tilde \tau}(1) = 1$. It follows that $a=0$ and $b= \frac{1}{ \int_0^1 e^{2 \omega(x(\tau')) } d\tau'}$ so that
$$
{\tilde \tau}(\tau) =  \frac{\int_0^\tau e^{2 \omega(x(\tau')) } d\tau'
}{ \int_0^1 e^{2 \omega(x(\tau')) } d\tau'}.
$$
