0
$\begingroup$

Let $L$ be as smooth as needed a function of the arguments $(q_1,\dots,q_n,\dot q_1,\dots,\dot q_n,t)$, where the dot denotes the derivetive with respect to $t$. Let $\delta$ denote the variation of any function due to a small variation of its arguments.

Then $$\delta L=\sum_{r=1}^n\frac{\partial L}{\partial q_r}\delta q_r+\frac{\partial L}{\partial\dot q_r}\delta\dot q_r$$

Would anyone give the derivation of the equation $$\delta L=\delta\sum_{r=1}^n\frac{\partial L}{\partial\dot q_r}\dot q_r+\sum_{r=1}^n(\frac{\partial L}{\partial q_r}\delta q_r-\dot q_r\delta \frac{\partial L}{\partial\dot q_r})$$

$\endgroup$
1
$\begingroup$

Just apply the product rule to the first term:

$$ \delta \sum_r \frac{\partial L}{\partial \dot q_r} \dot q_r = \sum_r \delta \left(\frac{\partial L}{\partial \dot q_r} \dot q_r \right) = \sum_r \left( \frac{\partial L}{\partial \dot q_r} \delta\dot q_r + \dot q_r \delta \frac{\partial L}{\partial \dot q_r}\right).$$

The second term will cancel the last term on your desired equation, giving your original expression for $\delta L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.