The Lagrangian of a free particle ##L=-m \, ds/dt##

[Kostik]

TL;DR

How to show that $L=-m \, ds/dt$ for a free particle.

In Dirac's "General Theory of Relativity" (p. 52), he postulates that the action for a free particle of mass $m$ is $$I=-m \int ds$$ hence the Lagrangian is $$L=-m\frac{ds}{dt} = -m\frac{\sqrt{\eta_{\mu\nu}dx^\mu dx^\nu}}{dt}
\, .$$ To confirm that $-m$ is the correct coefficient, he assumes flat spacetime (special relativity) and calculates $$\frac{\partial L}{\partial \dot{x}^k} = -m\frac{\partial}{\partial \dot{x}^k}\left( \frac{ds}{dt}\right) = m\frac{ \dot{x}^k }{ ds/dt } = m \frac{dx^k}{ds}$$ which is the correct formula for relativistic 4-momentum $p^k$. ("As it ought to be", says Dirac.)

$\qquad$ But doesn't this assume that $$p^k = \frac{\partial L}{\partial \dot{x}^k} \quad ?$$ This is true for the non-relativistic Lagrangian for a free particle $$L=T-U = \frac{1}{2}mv^2 \qquad (*)$$ but is it true for a relativistic particle?

$\qquad$ (Landau-Lifshitz give what seems to be a more convincing confirmation. They also postulate that $L= \kappa \, ds/dt$, and assume flat spacetime (special relativity) where $ds^2 = \eta_{\mu\nu}dx^\mu dx^\nu$. Then $$L=\kappa \frac{ds}{dt}=\kappa \sqrt{1-v^2} \, .$$ Hence, for velocities $v \ll 1$, we have $$L = \kappa - \frac{1}{2}\kappa v^2 + O(v^4) \, .$$ The nonrelativistic Lagrangian is shown above in $(*)$, so to get the correct kinetic energy term, we must have $\kappa = -m$.)

$\qquad$ I'd like to understand Dirac's confirmation that $\kappa = -m$. How does he know in advance that $$p^k = \frac{\partial L}{\partial \dot{x}^k} $$ for a relativistic particle? It seems like he's using circular reasoning.

$\qquad$ I think what Dirac forgot to add was "Since $p^k = \partial L / \partial \dot{x}^k$ in the case $v \ll 1$, we see that $\kappa = -m$."

 

[dextercioby]

IIRC, Dirac uses Greek for 0-3, while Latin for 1-3, unlike L-L who use Greek for 1-3, and Latin for 0-3.

So if we try to switch from the Lagrangian density \mathcal{L} to the Hamiltonian formalism, we define the 4-momentum as

\begin{equation} p_\mu =: \frac{\partial \mathcal{L} \left(x,\dot{x}\right)}{\partial \dot{x}^\mu}\end{equation},

where the dot is the worldline parameter (most commonly chosen as the proper time).

Why would going Greek > Latin be circular reasoning?

 

[Kostik]

I think it's circular reasoning because $p^k = \partial L / \partial \dot{x}^k$ is derived for a non-relativistic free particle, where $L=mv^2/2$. But, otherwise, $p^k \equiv \partial L / \partial \dot{x}^k$ is the definition of the "conjugate momentum". We cannot simultaneously define $p^k \equiv \partial L / \partial \dot{x}^k$ and $p^k \equiv m\, dx^k/ds = mv^k$.

I think Dirac needs to reduce to the non-relativistic case (like L-L) in order to determine $\kappa$.

$\qquad$ I think what Dirac should have done is this: Consider the case $v \ll 1$, where $L = T = mv^2/2$. Then we confirm by calculation that $$\frac{\partial L}{\partial \dot{x}^k} = \frac{m}{2}\frac{\partial v^2}{\partial \dot{x}^k}=m\dot{x}^k \,. $$ Comparing this with $$\frac{\partial L}{\partial \dot{x}^k} = -\kappa \frac{dx^k}{ds} $$ and using $ds=dt$ when $v \ll 1$, we obtain $\kappa = -m$.

 

[anuttarasammyak]

TL;DR Summary: How to show that $L=-m \, ds/dt$ for a free particle.

But doesn't this assume that pk=∂L∂x˙k? This is true for the non-relativistic Lagrangian for a free particle

This is also true for SR Lagrangian. For SR Lagrangian pk has the factor \frac{1}{\sqrt{1-v^2/c^2}} as we see in experiments.　Analytical mechanics holds also in special relativity only by changing the Lagrangian functions.