2

I need to find the partial derivative of the action $S$ with respect to the generalized coordinate $q(t_f)$ and according to my textbook, it should equal the generalized momentum $p(t_f)$.

I have a solution but I'm not sure if it is valid.

Equations are:

$$S = \int_{t_i}^{t_f} L \, dt$$

and

$$q_i = 0 \, .$$

I took the partial derivative of both sides with respect to the generalized coordinate $q$ to get

$$\frac{\partial S}{\partial q}= \int_{t_i}^{t_f} \frac{\partial L}{\partial q} \, dt = \int_{t_i}^{t_f} \dot{p} \, dt = p(t_f) - p(t_i) \, .$$

Since $q_i = 0$ then $p(t_i) = 0$ and therefore $(\partial S / \partial q)(t_f) = p(tf)$.

Is this a possible solution?

DanielSank
  • 24,439
shinobi20
  • 73
  • 3
    Welcome to Physics Stack Exchange. Whenever you have equations to include in a post, please type them. Pictures of equations are strongly discouraged. This site supports mathjax. You can learn how to use it in the help center. – DanielSank Oct 04 '15 at 07:00
  • 3
    This question (v2) is explained in a Lemma of my Phys.SE answer here. – Qmechanic Oct 04 '15 at 07:22
  • This site supports math formatting, as I already noted. I fixed it for you and you should please click the "edit" button to see how it works. Please do this yourself in the future. – DanielSank Oct 04 '15 at 19:12

0 Answers0