Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

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


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

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

share|cite|improve this question
Another question about the same equation in Goldstein: – Qmechanic Oct 28 '12 at 20:58
@Qmechanic thank you. I was sure that I have already read a similar question but I couldn't find it! :) – sunrise Oct 29 '12 at 11:32

1 Answer 1

up vote 1 down vote accepted

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

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


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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