Is ##p^k = \partial L / \partial \dot{x}^k## true for all ##L##'s?

[Kostik]

TL;DR

Is the relation Is $p^k = \partial L / \partial \dot{x}^k$ true for all Lagrangians?

Using the Lagrangian $$L=T-U=\frac{1}{2}mv^2-U$$ we clearly have $$ \frac{\partial L}{\partial \dot{x}^k} = m\dot{x}^k = p^k $$ i.e., the $k$'th component of momentum. How does one show that the relation $$p^k = \frac{\partial L}{\partial \dot{x}^k} $$ holds for all Lagrangians?

 

[anuttarasammyak]

“The generalized momentum "canonically conjugate to" the coordinate qi is defined by

“ https://en.m.wikipedia.org/wiki/Generalized_coordinates

 

[Kostik]

Thanks. I did not phrase the question very well. I have made a more detailed post of the question here:

Graduate Thread 'The Lagrangian of a free particle $L=-m \, ds/dt$'

Jul 18, 2025

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...

• Kostik
• Replies: 5
• Forum: Special and General Relativity