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  1. Reading about the Schwarzschild metric in general relativity I see that sometimes $$L=g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$$ and sometimes $$L=\sqrt{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}.$$ Which is the right way?

  2. Also how is the energy $E$ defined as $$E=-\frac{\partial{L}}{\partial{\dot{t}}}=\left(1-\frac{2M}{r}\right)\dot{t}~?$$ Because $E$ here doesn't have units of energy. Am I missing something here?

MrDi
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  • First subquestion (v3) is essentially a duplicate of http://physics.stackexchange.com/q/149082/2451 and links therein. – Qmechanic Jun 17 '15 at 14:33
  • Let us continue this discussion in chat. – Kyle Kanos Jun 17 '15 at 15:21
  • $E$ is considered to be the energy per unit mass. Also note that this isn't the true Lagrangian, but the Lagrangian of a point particle moving in a fixed background geometry. Properly, you should express this as $S = mc^{2}\int\sqrt{g_{ab}{\dot x}^{a}{\dot x}^{b}}d\tau$, but most people aren't interested in tracking all of this, and so use $S = \frac{1}{2}\int d\tau, g_{ab}{\dot x}^{a}{\dot x}^{b}$, which has the same equations of motion, and is much easier to take the variation of.. – Zo the Relativist Jun 17 '15 at 15:24
  • Observe that the second choice gives a zero Hamiltonian – Phoenix87 Jun 17 '15 at 23:00

1 Answers1

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The correct way is to define the reparametrization-invariant action

$$ S[X] = \int d\tau \sqrt{g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau} }. $$

Note that the choice of $\tau$ is arbitrary. The system has a large group of gauge symmetries - those are reparametrizations of the worldline (different choices of $\tau$).

One way to deal with this is to gauge-fix the system. For example, we can choose to set $\tau$ to be a proper time along the geodesic (induced by the metric):

$$ d\tau = \sqrt{g_{\mu \nu} (X(\tau)) dX^{\mu} dX^{\nu}}. $$

But this immediately implies that the square root in the action is constrained to be equal to $1$. This is why it is convinient to drop the square root ($\sqrt{1} = 1$, right?) and write

$$ L = g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau}. $$

But this only works if $\tau$ is the proper time.

Also, you should multiply the Lagrangian by the overall factor of $m$. This will restore the units of energy.

Prof. Legolasov
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