Thursday, 21 February 2019

classical mechanics - D'Alembert's Principle and the term containing the reversed effective force



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)?



Answer




Why is there a subtraction in Eq. (1.50)?




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.


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