- #1

- 466

- 5

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter iScience
- Start date

- #1

- 466

- 5

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

- #2

- 466

- 5

- #3

UltrafastPED

Science Advisor

Gold Member

- 1,912

- 216

please show me how [itex]\frac{∂L}{∂q}[/itex]=[itex]\dot{p}[/itex]?

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

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

- #4

- 466

- 5

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?

- #5

UltrafastPED

Science Advisor

Gold Member

- 1,912

- 216

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

- #6

- 466

- 5

- #7

- 17,546

- 8,539

[tex]p=\frac{\partial L}{\partial \dot{q}}.[/tex]

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.

Share: