2
$\begingroup$

At page 20 of Classical Mechanics' Goldstein (Third edition), there are these two steps given between eqs. (1.51) and (1.52):

$$\sum_i m_i \ddot {\bf r}_i \cdot \frac{\partial {\bf r_i}}{ \partial q_j}= \sum_i [\frac {d}{dt}(m_i {\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial \dot q_j})-m_i {\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial q_j}]$$

and

$$\sum_j \{ \frac{d}{dt}[ \frac{\partial}{\partial \dot q_j}(\sum_i \frac{1}{2}m_i v^2_i)] - \frac{\partial}{\partial q_j}(\sum_i \frac{1}{2}m_i v^2_i)-Q_j \}\delta q_j .$$

Why does "$ \frac {1}{2}$" appear in the second formula?

$\endgroup$
2
  • $\begingroup$ Another question about the same equation in Goldstein: physics.stackexchange.com/q/12361/2451 $\endgroup$
    – Qmechanic
    Commented Oct 28, 2012 at 20:58
  • $\begingroup$ @Qmechanic thank you. I was sure that I have already read a similar question but I couldn't find it! :) $\endgroup$
    – sunrise
    Commented Oct 29, 2012 at 11:32

1 Answer 1

2
$\begingroup$

The $\frac{1}{2}$ is due to the differentiation rule

$$\frac{\partial }{\partial \dot q_j}({\bf v}_i \cdot {\bf v}_i ) ~=~2{\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial \dot q_j},$$

and

$$\frac{\partial }{\partial q_j}({\bf v}_i \cdot {\bf v}_i ) ~=~2{\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial q_j}.$$

$\endgroup$
1
  • $\begingroup$ Correction to the answer (v1): The word rule should be rules. $\endgroup$
    – Qmechanic
    Commented Apr 26, 2022 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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