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- 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##."
\, .$$ To confirm that ##-m## is the correct coefficient, he assumes flat spacetime (special relativity) and calculates $$\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##."
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