Bianchi-identity and null-geodesics

Can you show that the null-condition \(\mathrm{d}s^2=0\) follows from the Bianchi-identity in the case of photons? What about gravitational waves - would the argument apply to those as well?

If one uses the Maxwell-equation \(\partial_\mu F^{\mu\nu} = 0\) in vacuum or the Bianchi-identity \(\partial_\mu\tilde{F}^{\mu\nu}=0\) (which are equivalent, due to duality) and the definition for the Faraday-tensor \(F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu\) one obtains a wave equation \(\Box A^\mu =0\) in Lorenz-gauge \(\partial_\mu A^\mu=0\). Solving the equation with \(A^\mu\propto \exp(\pm\mathrm{i}k_\mu x^\mu)\) yields the null-condition \(k_\mu k^\mu\) for the wave vector \(k^\mu\), which in turn is defined as the tangent \(k^\mu = \mathrm{d}x^\mu/\mathrm{d}\lambda\) to the trajectory \(x^\mu(\lambda)\). Therefore, \(\mathrm{d}s^2 = k_\mu k^\mu \mathrm{d}\lambda^2\), which needs to come out as zero.

Weak perturbations \(h_{\mu\nu}\) of the metric follow in the traceless and transverse gauge exactly the same wave equation, \(\Box h_{\mu\nu}=0\), so that the argument applies in exactly the same way to gravitational waves.