Im studying Mechanics form Goldstein. I cross this equation in "Derivation of Lagranges equation from d'Alembert's Principle",section 1.4. I have two questions from this derivation.

  1. The expression given is
    $$\sum_{i=1}^n \sum_{k=1}^n m_i \ddot r_i \frac{\partial r_i}{\partial q_k} \delta q_k.\tag{1.50}$$ This can we written as $$\sum_{i=1}^n \sum_{k=1}^n \left[ \frac{d}{dt} \left(m_i \dot{r_i}\frac{\partial r_i}{\partial q_k}\right)-m_i \dot{r_i} \frac{d}{dt}\left(\frac{\partial r_i}{\partial q_k}\right) \right] \delta q_k.\tag{1.50}$$
  1. How that expression can be written like this. I don't understand why we split into two parts and make subtraction. Is that any formula?
  1. How $$\frac{d}{dt} (m_i \dot{r_i}) \frac{\partial v_i}{\partial \dot{q_k}}=\frac{d}{dt} \frac{\partial}{\partial \dot{q_k}} \left[ \frac{1}{2} m_iv_i^2\right] ?$$

Any help?


2 Answers 2

  1. This is just restating product rule of differentiation:

$$\frac{d}{dt} \left(m_i \mathbf{\dot{r}_i} \cdot \frac{\partial \mathbf r_i}{\partial q_k}\right) = \color{blue}{m_i \mathbf{\ddot{r}_i} \cdot \frac{\partial \mathbf{r_i}}{\partial q_k}} + m_i \mathbf{\dot{r_i}} \cdot \frac{d}{dt}\left(\frac{\mathbf{\partial r_i}}{\partial q_k}\right). $$

What we are interested is in evaluating the term highlighted in blue. You will see that this manipulation allows one to express the principle in terms of kinetic energy $T$.

  1. One can write $$\mathbf{v_i} \equiv \frac{d \mathbf{r_i}}{dt} = \sum_k\frac{\partial \mathbf{r_i}}{\partial q_k}\dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t}.$$

Now do you see $\displaystyle\frac{\partial \mathbf{v_i}}{\partial \dot{q}_k} = \frac{\partial \mathbf{r_i}}{\partial q_k}$? It should be easy to realise the answer to second now. Make use of the fact that differentiating the kinetic energy $T = \frac{1}{2} mv^2$ wrt to $\dot{q_k}$ is nothing but using chain rule to obtain $\displaystyle m \mathbf{v_i}\cdot\frac{\partial{\mathbf{v_i}}}{\partial q_k}$.


The answer to the first question is just the product rule, as stated by Abhay Hedge. However, regarding your second question, I cannot find this formula, $$ \frac{d}{dt} (m_i \dot{\mathbf{r_i}}) \cdot \frac{\partial \mathbf{v_i}}{\partial \dot{q_k}}=\frac{d}{dt} \frac{\partial}{\partial \dot{q_k}} \left[ \frac{1}{2} m_iv_i^2\right], $$ anywhere in the textbook.

Did you mean $$ \frac{d}{dt} (m_i \dot{\mathbf{r_i}} \cdot \frac{\partial \mathbf{v_i}}{\partial \dot{q_j}}) =\frac{d}{dt} \frac{\partial}{\partial \dot{q_j}} \left[ \frac{1}{2} m_iv_i^2\right]? $$
If that's the case, the answer is pretty simple, just apply the chain rule: $$ \frac{d}{dt} \frac{\partial}{\partial \dot{q_j}} \left[ \frac{1}{2} m_iv_i^2\right] = \frac{d}{dt}(\frac{1}{2} 2 m_i\mathbf{v_i} \cdot \frac{\partial \mathbf{v_i}}{\partial q_j}) = \frac{d}{dt}(m_i\mathbf{v_i} \cdot \frac{\partial \mathbf{v_i}}{\partial q_j}) $$

  • $\begingroup$ Yes, I couldn't find the second equation OP asked. However, unless I'm mistaken, I wrote the same hint to use chain rule for second part as well. $\endgroup$
    – exp ikx
    Aug 15, 2021 at 7:28

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