1
$\begingroup$

Goldstein in "Classical Mechanics" (1ed) obtains

$$-\int \sum_j \left(\frac {\partial V}{\partial q_j} \delta q_j + \frac{\partial V}{\partial \dot q_j} \delta \dot q_j\right) dt=-\delta \int V dt$$

from

$$- \int \sum_j \delta q_j \left( \frac {\partial V}{\partial q_j}- \frac {d}{dt} \frac {\partial V}{\partial \dot q_j}\right) dt.$$

Could you explain me a little about the steps?

  • 0
    How does Goldstein obtain an equation from the second expression, which has no equals sign?2012-11-02
  • 0
    @wj32 Goldstein says that he has "reversed" integration by parts of the left-hand of first expression and so he has obtained the right-hand of first expression. And he also says that he started from the second expression. I don't know anything else.. :(2012-11-02

1 Answers 1

1

This is a classical trick used in calculus of variation i.e. integrate by parts using $\,\frac d{dt} \delta q=\delta \dot q$ :

$$\int \delta q \frac d{dt} \left(\frac{\partial V}{\partial \dot q}\right)\, dt=\left[ \delta q \frac{\partial V}{\partial \dot q}\right]-\int \frac{\partial V}{\partial \dot q}\frac d{dt}\delta q \, dt=-\int \frac{\partial V}{\partial \dot q} \delta \dot q \, dt$$

(using too the hypothesis made for large values).

I wrote only the term at the right (the left one $\ \delta q \frac{\partial V}{\partial q}\ $ is unchanged). Hoping this clarified,

  • 0
    thank you! but why does $[\delta q \frac {\partial V}{\partial \dot q}]$ disappear?2012-11-02
  • 0
    @sunrise: because at both (fixed!) endpoints $\delta q=0\ $ I think (I don't have Goldstein's book here...). You may see Wikipedia's article concerning ['calculus of variation'](http://en.wikipedia.org/wiki/Calculus_of_variations#Euler.E2.80.93Lagrange_equation) and the tutorials linked there too (see too [action](http://en.wikipedia.org/wiki/Action_%28physics%29)).2012-11-02
  • 0
    You're right! :) Thanks a lot!2012-11-02
  • 1
    @sunrise: You are welcome !2012-11-02