I have a question about applying the variational principle to obtain the geodesic equation for null geodesics. Specifically, I am unsure about the justification for the choice of the Lagrangian. I will briefly explain taught to derive the geodesic equation for timeline geodesics.
The metric for a given spacetime geometry is given by $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu},$$ which can we rewritten as: $$ds=\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda.$$ We can therefore compute the length of the spacetime interval by integrating over $\lambda$. Therefore:
$$S = \int_{\lambda_0}^{\lambda_1}\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda.$$
A geodesic is a curve that leads to the shortest possible spacetime interval, so we can obtain it by applying the Euler-Lagrange equation to $S$. However, the square root makes the calculation difficult. For time-like geodesics, there is a trick we can use to eliminate the square root, since we can choose the proper time as a special parameter, and then minimize a perturbation to the geodesic. For a time-like geodesic, the action is the proper time $\tau$.
$$\tau = \int_{\lambda_0}^{\lambda_1}\sqrt{-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda = \int_{\lambda_0}^{\lambda_1}\sqrt{-\mathcal{L}}d\lambda.$$
We now find the first-order deviation from the geodesic, $\delta S$ by Taylor expanding $$\sqrt{-\mathcal{L}} = \sqrt{-\mathcal{L}_0} -\frac{1}{2\sqrt{-\mathcal{L}_0}}\delta \mathcal{L}.$$ We can now extremize the perturbation:
$$\delta \tau = -\frac{1}{2}\int_{\lambda_0}^{\lambda_1}\frac{1}{\sqrt{-\mathcal{L}_0}}\delta \mathcal{L}d\lambda.$$
We can now chose the parameter to be $d\lambda = d\tau$. This choice will necessitate that the lagrangian $\mathcal{L}=-1$.
$$\delta \tau = \frac{1}{2}\int_{\tau_0}^{\tau_1}\delta \mathcal{L}d\tau$$ $$\delta \tau = \frac{1}{2}\delta \left[\int_{\tau_0}^{\tau_1} \mathcal{L}d\tau\right]$$
Therefore the Euler-Lagrange equations we would obtain from extremizing the original spacetime interval are the same as we would get if we extremized: $$\tau = \int_{\tau_0}^{\tau_1}\mathcal{L}d\tau$$
Therefore we can find the geodesic by extremizing $\mathcal{L}$ instead of $\sqrt{\mathcal{L}}$, which is much simpler.
However, this trick does not work for null geodesics, since there is no parameter that we can choose that will simplify the Lagrangian to $\mathcal{L}=-1$. Another problem that I ran into when I tried to do it, was that any null interval has the property that $$ds=\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}d\lambda = 0,$$ so when I got to this stage, I had a zero in the denominator:
$$\delta S = \frac{1}{2}\int_{\lambda_0}^{\lambda_1}\frac{1}{\sqrt{\mathcal{L}}}\delta \mathcal{L} d\lambda.$$
Yet from my lecture notes (and past exam papers), it seems that you can still find null geodesics by extremizing $\mathcal{L}$ instead of $\sqrt{\mathcal{L}}$. I absolutely do not see that can be justified.
