# 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.