EoM in Schwarzschild geometry: geodesic v Hamilton formalism

[Paul Colby]

It becomes the same equation, given that for an affine parameter A=constA=constA=\text{const}.

Ah, the missing fact, thanks.

 

[stevendaryl]

Possibly I'm confused, but something seems wrong here.

Let's take the flat space case when $g_{\mu\nu} = \eta_{\mu\nu}$ = diag(-1,1,1,1), so we have a -+++ metric signature.

We know that the relativistic Lagrangian for a free particle in the above flat space-time should be $L = -m\,c^2 \sqrt{1-\beta^2}$, where $\beta = v/c$. If we use geometric units so that c=1, we can write this as:

$L(t,\textbf{x},\frac{d \textbf{x}}{dt}) = -m \, \sqrt{ -g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}}$

It's one of those lucky miracles that you get the same equations of motion using:

L = -m \sqrt{g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}}

or

L =\frac{m}{2} g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}

(Actually, the first gives more solutions than the second, but every solution to the first can be converted into a solution of the second by reparametrizing s)