# How to show that conformal spaces share the same null geodesics?

Two metrics $$ds^2 = g_{\mu\nu} dx^\mu dx^\nu$$ and $$d\tilde s^2 = \tilde g_{\mu\nu} dx^\mu dx^\nu$$ in the same coordinate system are said to be conformally related if ̃

$$\tilde g_{\mu\nu} (x) = \psi(x)g_{\mu\nu}(x)$$

for some positive function $$\psi(x)$$ of the coordinates.

I am trying to prove that if the curve $$x^\mu(\lambda)$$ is a null geodesic of the metric $$g_{\mu\nu}$$ , it is also a null geodesic of the metric ̃$$\tilde g_{\mu\nu}$$ . (By this it's meant the same physical curve in the spacetime manifold M , with an affine curve parameter which is a suitable function of $$\psi$$, in general not coinciding with $$\lambda$$.) That is, conformal spaces share their null geodesics.

The following fact can be used:

whenever a curve $$x^\mu(\omega)$$ satisfies:

$$\frac{d^2x^\mu}{d\omega^2}+\Gamma^\mu_{\nu\lambda}\frac{dx^\nu}{d\omega}\frac{dx^\lambda}{d\omega}=\phi \frac{dx^\mu}{d\omega}$$

for some function $$\phi$$ on $$M$$ , it also describes a geodesic (obviously, in a non-affine parametrization) in the sense that there is a curve parameter $$s = s(\omega)$$, depending on the function $$φ$$, such that $$x^\mu(\omega(s))$$ viewed as function of $$s$$ satisfies the usual geodesic equation.

I have not been able to get to anything meaningful, how should I do this?

• It indeed remains a null-curve, but why would you expect it to remain a geodesic? Commented Oct 17, 2023 at 7:35
• @MoisheKohan I am sorry I don't share your intuition to answer that, the aim is to do an explicit computation and show that Commented Oct 17, 2023 at 7:48
• @darkside I stand with Moishe. Two conformal metrics will share the same set of null curves, but not the same set of (parametrized) geodesics. Commented Oct 17, 2023 at 7:58
• Then why don't you do an explicit computation starting with the flat Lorentzian metric $ds^2=dxdy$ on ${\mathbb R}^2$, where null-geodesics are vertical and horizontal straight lines. Then pick a generic function $\psi(x,y)$ and compute Christoffel symbols and geodesic equation for the conformal metric $\psi(x,y)dxdy$, as explained in the answers here. Commented Oct 17, 2023 at 8:00
• @Didier This is what the exercise says I am not inventing it, are you saying the exercise is wrong? Commented Oct 17, 2023 at 9:48

Let us write $$\psi = e^{2f}$$, so that $$\tilde{g} = e^{2f}g$$. Let $$\gamma$$ be a null-geodesic for $$g$$. This means that $$g(\gamma',\gamma') = 0$$ and $$\nabla_{\gamma'}\gamma' = 0$$.
By the formula derived in this answer, we have \begin{align} \tilde{\nabla}_{\gamma'}\gamma' &= \nabla_{\gamma'}\gamma' + df(\gamma') \gamma' + df(\gamma') \gamma' + g(\gamma',\gamma') \mathrm{grad}(f) \\ &= 0 + 2 df(\gamma') \gamma' +0 \\ &= 2df(\gamma') \gamma'. \end{align} Therefore, $$\tilde{\nabla}_{\gamma'}{\gamma'} = 0$$ if and only if $$\gamma'$$ is everywhere in the kernel of $$df$$. Hence, for a generic conformal metric $$\tilde{g} = e^{2f}g$$, $$\gamma$$ is not a geodesic for $$\tilde{g}$$, even though it is a null curve.
However, we can reparametrize $$\gamma$$ in order to produce a null-geodesic for $$\tilde{g}$$. Here is the trick: let $$\varphi$$ be any reparametrization of the interval of definition of $$\gamma$$ and consider $$\tilde{\gamma} =\gamma\circ \varphi$$. Then $$\tilde{\gamma}' = \varphi'\cdot(\gamma'\circ \varphi)$$. It is already a null-curve. Let us derive an equation on $$\varphi$$ in order for $$\tilde{\gamma}$$ to be a geodesic.
By the linearity of the Levi-Civita connection in the lower entry, it is sufficient that $$\varphi$$ is such that $$\tilde{\nabla}_{\gamma'\circ \varphi} (\varphi'\cdot (\gamma'\circ \varphi)) = 0.$$ Note that \begin{align} \tilde{\nabla}_{\gamma'\circ \varphi} (\varphi'\cdot (\gamma'\circ \varphi)) &= \varphi''\cdot (\gamma'\circ \varphi) + \varphi' \tilde{\nabla}_{\gamma\circ\varphi}(\gamma \circ \varphi)\\ &= \varphi'' \cdot (\gamma'\circ \varphi) + \varphi' \cdot 2df(\gamma'\circ\varphi) (\gamma'\circ \varphi) \\ &= (\varphi'' + 2df(\gamma'\circ\varphi)\varphi')(\gamma'\circ \varphi), \end{align} so that it is sufficient to solve the differential equation $$\varphi'' + 2df(\gamma'\circ \varphi) \varphi'= 0$$, which can be done easily in coordinates.
• Why do I need to do this $\psi = e^{2f}$? It should be positive yes, but I don't see why is this an obliged step Commented Oct 17, 2023 at 17:03
• @darkside No one forces you to do so. Sometimes, the formulas are nicer by using $\psi=e^{2f}$ instead of (Kozsul formula), sometimes they are nicer when using $\psi = u^{\frac{4}{n-2}}$ (Yamabe problem)... It really depends on the context. Here, you can translate the formula by replacing $f$ with $\frac{1}{2}\ln \psi$, but you will end up with something not so nice Commented Oct 17, 2023 at 17:07