Why does ∂L/∂q equal the time derivative of momentum?

[iScience]

hi, silly question but would someone please show me how $\frac{∂L}{∂q}$=$\dot{p}$?

L being the lagrangian, p being the momentum, and q being the general coordinate.

 

[iScience]

(i hope this doesn't qualify as a homework question this actually has nothing to do with my homework.. I'm just trying to derive the hamiltonian equation and this was just part of the steps)

 

[UltrafastPED]

please show me how $\frac{∂L}{∂q}$=$\dot{p}$?

$\frac{∂L}{∂q}$=$\dot{p}$ because the Lagrangian, L=T-U, depends on q only in the potential energy, U.

$-\frac{∂U}{∂q}$=$\dot{p}$ is Newton's 2nd law of motion expressed in terms of the potential.

 

[iScience]

thanks!

also, i had another question i hope i can just ask it in the same thread; if not let me know (moderators/admins) and i'll just make a new thread.



how do i know when there is translational symmetry? in other words what quantity has to be zero?

 

[UltrafastPED]

You want p=constant (p-dot = 0) for the momentum conjugate to that coordinate.

 

[iScience]

"momentum that is conjugate to that coordinate" meaning just the momentum corresponding to the particular coordinate at hand right? (just checking)

 

[vanhees71]

The canonical momentum of the generalized coordinate $q$ is by definition given by
$$p=\frac{\partial L}{\partial \dot{q}}.$$
It's important to keep in mind that also for a Cartesian coordinate it is the canonical and not necessarily the mechanical momentum. An interesting example for that both need not be the same is the motion of a particle in a magnetic field.