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?


2 Answers 2


I think you are here "rediscovering" the fact that worldlines in GR always have the same size of 4-velocity, i.e. $-c^2$ for timelike worldlines, as long as we take 4-velocity to be the tangent vector using proper time as affine parameter, and zero for null lines. So yes, the constant you called $c$ (an unfortunate choice of letter here by the way!) does distinguish between massless and massive worldlines.

  • $\begingroup$ This must only hold for certain types of 'forces' though? If instead of the extra EM term $q A_\mu \dot{x}^\mu$ I had added a 'potential-like' term $V(x)$ then the tangent vector would not have been constant. On the other hand, if we define (usually called proper time/ distance) $d \tau^2 = g_{\mu\nu}dx^\mu dx^\nu$, then it would seem to me that by definition any path, as long as it was never null, would have tangent vector $dx^\mu/d\tau$ with constant unit norm everywhere. $\endgroup$
    – Rudyard
    Aug 3, 2019 at 17:48
  • $\begingroup$ Perhaps we should start with a reparametrisation invariant $L$ really (The ones I wrote down are not). Because otherwise there's no guarantee that $\tau$ can always be identified with the parameter $\lambda$ in my question... $\endgroup$
    – Rudyard
    Aug 3, 2019 at 18:02
  • $\begingroup$ @Rudyard Yes any non-null path parameterised by proper time must have unit norm tangent. So I guess this amounts to noting that there's no guarantee EL eqns give correct path unless the Lagrangian is constructed the right way. $\endgroup$ Aug 3, 2019 at 19:17
  • $\begingroup$ Agreed. How do we know that the increment g_ab dx^a dx^b along a non-geodesic worldline of a massless particle is always zero? I feel it must be true but I wonder what is the easiest/clearest way of seeing that. $\endgroup$
    – Rudyard
    Aug 4, 2019 at 11:41
  1. OP's eq. (6) is the relativistic Newton's 2nd law for a massive point charge in the presence of the Lorentz force. There is no reason why the 4-speed (7) should be constant.

  2. OP's Lagrangians only work for a massive point particle. To describe a massless point particle, it is necessary to use a Lagrange multiplier, cf. e.g. this & this Phys.SE posts.


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