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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?

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How does Goldstein obtain an equation from the second expression, which has no equals sign? –  wj32 Nov 2 '12 at 10:19
    
@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.. :( –  sunrise Nov 2 '12 at 10:26
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1 Answer

up vote 1 down vote accepted

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,

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thank you! but why does $[\delta q \frac {\partial V}{\partial \dot q}]$ disappear? –  sunrise Nov 2 '12 at 12:27
    
@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' and the tutorials linked there too (see too action). –  Raymond Manzoni Nov 2 '12 at 12:59
    
You're right! :) Thanks a lot! –  sunrise Nov 2 '12 at 13:27
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@sunrise: You are welcome ! –  Raymond Manzoni Nov 2 '12 at 13:32
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