For our Classical Mechanics class, I'm reading Chapter 1 of Goldstein, et al. Now I come across Eq. (1.50). To put it in context:
$$\begin{align*}
\sum_i{\dot{\mathbf{p}_i} \cdot \delta\mathbf{r}_i}&=\sum_i{m_i\ddot{\mathbf{r}}_i \cdot \delta{\mathbf{r}_i}}\\
&=\sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j
\end{align*}$$Consider now the relation Eq. (1.50):
$$\begin{align*}
\sum_{i,j}{m_i\ddot{\mathbf{r}}_i} \cdot \frac{\partial\mathbf{r}_i}{\partial q_j}&=
\sum_i{\left[ \frac{d}{dt} \left( m_i\dot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \right) –
m_i\dot{\mathbf{r}}_i \frac{d}{dt} \left( \frac{\partial \mathbf{r}_i}{\partial q_j} \right) \right]}
\end{align*}$$
I'm at a loss for how he resolved it that way. He goes on to explain that we can interchange the differentiation with respect to $t$ and $q_j$. My question is: Why is there a subtraction in Eq. (1.50)?
Best Answer
Goldstein is using the Leibniz rule for differentiation of a product
$$ \frac{d (fg)}{dt}~=~\frac{d f}{dt}g + f\frac{d g}{dt} $$
with
$$f=m_i\dot{\mathbf{r}}_i $$
and
$$g=\frac{\partial \mathbf{r}_i}{\partial q_j}. $$
The minus is caused by moving a term to the other side of the equation.