# Clarification on a Goldstein formula steps (classical mechanics)

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?

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

The $\frac{1}{2}$ is due to the differentiation rule
$$\frac{\partial }{\partial \dot q_j}({\bf v}_i \cdot {\bf v}_i ) ~=~2{\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial \dot q_j},$$
$$\frac{\partial }{\partial q_j}({\bf v}_i \cdot {\bf v}_i ) ~=~2{\bf v}_i \cdot \frac{\partial {\bf v}_i}{\partial q_j}.$$