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A Question on Goldstein and D'Alembert's Principle

  1. Sep 23, 2015 #1
    Hey all,

    I am reading Goldstein and I am at a point where I can't follow along. He has started with D'Alembert's Principle and he is showing that Lagrange's equation can be derived from it. He states the chain rule for partial differentiation:
    [tex]\frac{d\textbf{r}_i}{dt}=\sum_k \frac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k+\frac{\partial \mathbf{r}_i}{\partial t}[/tex]

    Then he states, by the equation above, that:
    [tex]\frac{d}{dt}\frac{d\mathbf{r}_i}{dq_j}=\sum_k \frac{\partial^2 \textbf{r}_i}{\partial q_j \partial q_k}\dot{q}_k+\frac{\partial^2 \mathbf{r}_i}{\partial q_j\partial t}[/tex]

    He further states from the first equation that:
    [tex]\frac{\partial \mathbf{v}_i}{\partial \dot{q}_j}=\frac{\partial \mathbf{r}_i}{\partial q_j}[/tex]

    I have tried to connect the dots but I cannot succeed. Any insight is greatly appreciated. Thanks!
  2. jcsd
  3. Sep 23, 2015 #2
    Think the first relation as a definition for an operator

    [tex]\frac{d}{dt}=\sum_k \frac{\partial}{\partial q_k}\dot{q}_k+\frac{\partial }{\partial t}[/tex]

    The second equation follows immdiately from applying ##\frac{d}{dt}## to ##\frac{d\mathbf{r}_i}{dq_j}## and the last one from applying ##\frac{\partial}{\partial \dot q_j}## to the first equation.
  4. Sep 23, 2015 #3

    So is the definition simply the chain rule of a function that depends on q_1, q_2,...q_N, and t? If the function had no explicit dependence on t, even though the generalized coordinates did, would you simply drop the partial with respect to t?
  5. Sep 24, 2015 #4
    Yes, certainly.

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