Affine Parameters and Euler-Lagrange Equations

Earlier today I was struggling to see why I couldn’t derive the general geodesic equation (*)

\frac{\textrm{d}}{\textrm{d}\lambda}(g_{ab}\dot{x}^b)-\frac{1}{2}g_{bc,a}\dot{x}^b\dot{x}^c=f(\lambda)g_{ab}\dot{x}^b

where f a general nonzero function. I had been trying to do this by varying the action

S=\int L \textrm{d}\lambda with Lagrangian L=\frac{\textrm{d}s}{\textrm{d}\lambda}=\sqrt{g_{ab}\dot{x}^a\dot{x}^b}.

The standard technique is use as a Lagrangian L'=L^2 instead, and claim that this produces the same results as we would have got with L. I’d always accepted this as gospel, but a simple calculation shows that we need an additional assumption for this to be true.

Indeed the Euler-Lagrange equations for L' give

\frac{\textrm{d}}{\textrm{d}\lambda}(2L\frac{\partial L}{\partial\dot{x}^a})=2L\frac{\partial L}{\partial x^a}

To regain the Euler-Lagrange equations for L we want to cancel our 2L factors. Clearly a sufficient condition for this is that L is constant. But by the definition of L this means that \frac{\textrm{d}s}{\textrm{d}\lambda} is constant, which precisely says that \lambda is an affine parameter.

Hence, by the use of this method we lose the generality needed to obtain equation (*).

Nevertheless it is easy to derive (*) from the affine geodesic equation (**)

\frac{\textrm{d}}{\textrm{d}\lambda}(g_{ab}\dot{x}^b)-\frac{1}{2}g_{bc,a}\dot{x}^b\dot{x}^c=0

Indeed let \mu be a general parameter and write \lambda = \lambda(\mu). Then rewriting the geodesic equation (**) in terms of \mu using the chain rule yields (*) with f(\mu)=\frac{\lambda ''}{\lambda '} (try it).

Alternatively you can reach a manifestly reparameterisation invariant version of (*) by directly varying our original action S. It’s not pretty though!

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