Equation of null geodesics

Question

Consider a 2d metric given by:

$$g = dx^2 - x^2 dt^2$$

Then my notes mention that null geodesics are governed by the following equation:

$$\dot{x}^2 - x^2 \dot{t}^2$$

where the dot indicates differentiation with respect to some affine paramter $\lambda$. Can someone explain how is this equation derived? Why must null geodesics obey this equation?

Answer 1

Null geodesics follow $g= 0$, or equivalently

$$ 0 = ({\rm d}x)^2 - x^2({\rm d}t)^2 \tag{1} $$

Also, for some affine parameter $\lambda$ you have

$$ {\rm d}x = \frac{{\rm d}x}{{\rm d}\lambda}{\rm d}\lambda = \dot{x}{\rm d}\lambda \tag{2} $$

with a similar expression for ${\rm d}t$, if you replace that in Eq. (1) you get

$$ 0 = ({\rm d}\lambda)^2 \left(\dot x^2 - x^2\dot{t}^2 \right) \tag{3} $$

and from here you recover the conclusion of your notes

Answer 2

If $(M,g)$ is any even-dimensional pseudo-Riemannian manifold, then every mid-dimensional null submanifold $N$ is totally geodesic, simply because $g(\nabla_XY,Z)=0$ for all $X,Y,Z \in \mathfrak{X}(N)$ (proof: said expression is symmetric in $X,Y$, skew in $Y,Z$, hence vanishes).

So in a $2$-dim Lorentz manifold, null curves are necessarily reparametrized geodesics. This means that you can just look for solutions to $g_{\alpha(\lambda)}(\dot{\alpha}(\lambda),\dot{\alpha}(\lambda))=0$, which in this case gives the equation $$\dot{x}^2-x^2\dot{t}^2=0,$$where the dot stands for $\lambda$-derivative. Namely, write $\dot{\alpha} = \dot{x}\partial_x+\dot{t}\partial_t$ and use ${\rm d}x(\partial_x)=1$, etc.