The motion of a point particle in curved spacetime can be obtained by extremising
$$S = \int L d\lambda= \int \left( \frac{m}{2}g_{\mu\nu}(x) \dot{x}^\mu \dot{x}^\nu \right) d\lambda,\tag{1}$$ where $\dot{x}^\mu =dx^\mu/d\lambda$ and $\lambda$ is an (affine?) parameter for the trajectory. Using the e.o.m one can show that $L$ itself is a constant of the motion (I think corresponding to the translational symmetry in $\lambda$):
$$\frac{1}{2} g_{\alpha \beta,\mu}\dot{x}^\alpha \dot{x}^\beta -\frac{d}{d\lambda}(g_{\mu\alpha}\dot{x}^\alpha) = 0.\tag{2}$$ contracting with $\dot{x}^\mu $ and rearranging one has: $$\implies \frac{d}{d\lambda}(\dot{x}^\mu \dot{x}_\mu )=0.\tag{3}$$ Thus, the tangent vector norm is constant along the (geodesic) curve. We then say $$\dot{x}^\mu \dot{x}_\mu = \mp 1,0\tag{4}$$ depending on whether it is timelike, spacelike or null.
It is my understanding that in the presence of an EM field $A_\mu$ the Lagrangian becomes:
$$L = \int \left( \frac{m}{2}g_{\mu\nu}(x) \dot{x}^\mu \dot{x}^\nu + q A_\mu\dot{x}^\mu \right) d\lambda,\tag{5}$$
The trajectory should now deviate from the geodesic curve due to a force. Finding the equations of motion
$$\frac{m}{2} g_{\alpha \beta,\mu}\dot{x}^\alpha \dot{x}^\beta +q A_{\alpha,\mu} \dot{x}^\alpha -\frac{d}{d\lambda}(mg_{\mu\alpha}\dot{x}^\alpha + qA_\mu) = 0,\tag{6}$$
contracting again with $\dot{x}^\mu$ the terms with $A$ cancel and I again find that the norm of the tangent vector is conserved as before $$\dot{x}^\mu \dot{x}_\mu = c.\tag{7}$$
How do I interpret $c$ in this case? Can I still use it to distinguish massless from massive worldlines?
