Graduate Dirac's derivation of the action/Lagrangian for a free particle

[Kostik]

TL;DR

The action for a free particle is $I=-m\int{ds}$, hence the Lagrangian is $L=-m(ds/dt)=-m/\gamma=-m\sqrt{1-v^2}$. Dirac infers this by checking that it gives the correct momentum $p^k = \gamma mv^k$ -- "in the case of special relativity". Why?

The action for a free particle is $$I=-m\int{ds} = \int \left(-m\frac{ds}{dt}\right) dt \quad\quad\quad(*)$$ hence the Lagrangian is $$L=-m\frac{ds}{dt}=-\frac{m}{\gamma}=-m\sqrt{1-v^2} .$$ Dirac ("General Theory of Relativity", p. 52) infers this by checking that it gives the correct spatial components of the 4-momentum: $$p^k = \frac{\partial L}{\partial \dot{x}^k} = \gamma m \frac{d x^k}{dt} .$$ But Dirac prefaces this by saying "We see the need [in the action] for the coefficient $-m$ by taking the case of special relativity, for which the Lagrangian would be the time derivative of $(*)$."

Why does he say "by taking the case of special relativity"? Isn't it always true that $$\frac{\partial}{\partial t}\int{L}\,dt = L ?$$

 

[Kostik]

I can see that the assumption $$ds^2=\eta_{\mu\nu}dx^\mu dx^\nu = dt^2 - dx^2 - dy^2 - dz^2$$ assumes flat spacetime (special relativity). Therefore, I think Dirac is correct that his confirmation works in the case of special relativity. But I still don't see why he says "by taking the case of special relativity, for which the Lagrangian would be the time derivative of $(∗)$."

 

[Kostik]

I can see that the assumption $$ds^2=\eta_{\mu\nu}dx^\mu dx^\nu = dt^2 - dx^2 - dy^2 - dz^2$$ assumes flat spacetime (special relativity). Therefore, I think Dirac is correct that his confirmation works in the case of special relativity. But I still don't see why he says "by taking the case of special relativity, for which the Lagrangian would be the time derivative of $(∗)$."

As noted in the original post, Dirac's "General Theory of Relativity", p. 52.

 

[jbergman]

As noted in the original post, Dirac's "General Theory of Relativity", p. 52.

I don't think the comment is probably that deep. As you say that is generally the case that the action is of the form of $A = \int L dt$. However, I don't think it's universal. Two cases that come to mind are maybe he was thinking of the field case where one has Lagrangian Densities like the Einstein Hilbert Action. Or alternatively in some cases one can choose an affine reparametrization of time.