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Goldstein Mechanics example motion of one particle in polar coordinates

  1. Oct 4, 2014 #1
    I have a course next semester on Classical Mechanics (mostly Lagrangian problems), for a second time. I'm ok for the theoretical preparation, but I'm trying to work ahead on problems and exercises, which was badly explained and without much of any resources. So, one of the sources to exercise on my own is Goldstein's book, and am just working through the examples of the first chapter.
    However, for the second example I stumble across a derivation I'm confused how the author got to that one.

    Example: motion of one particle in polar coordinates, page 27 (3rd ed), for the theta equation.

    For the derivative of $$(mr² \dot \Theta)$$ he finds: $$mr² \ddot \Theta + 2 mr \dot r \dot \Theta$$.
    While I expect it to be: $$mr² \ddot \Theta + 2 mr \dot \Theta$$
    Where does Goldstein get the $$\dot r$$ from?

    Edited: rewrote the question in LaTex notation
    Last edited: Oct 4, 2014
  2. jcsd
  3. Oct 4, 2014 #2
    The second term is taking into account that the radius can vary in time. Use the chain rule.

    You can also see that your second term doesn't have the right units, and you need a second time derivative somewhere.
  4. Oct 4, 2014 #3
    Thank you: yes you are right, I don't have the same units in my second term as the first one. I'll make a note of it on the chain rule. I assume that would be the following one:

    ##\sum_{j} \frac{\delta^{2}L}{\delta q_{j}\delta \dot q_{i}} \dot q_{j}+\sum_{j} \frac{\delta^{2}L}{\delta \dot q_{j} \delta \dot q_{i}} \ddot q_{j}+\frac{\delta^{2}L}{\delta \dot q_{i} \delta t}##

    Edited: added \dot LaTex notation
    Last edited: Oct 4, 2014
  5. Oct 4, 2014 #4
    writing it out in a chain rule that would make for:

    ##\frac{d}{dt}(mr² \dot \Theta)= \frac{\delta}{\delta \dot \Theta}(mr² \dot \Theta)\frac{d \dot \Theta}{dt}+\frac{\delta}{\delta r}(mr² \dot \Theta)\frac{dr}{dt}=mr²\ddot \Theta+2mr \dot \Theta \dot r##

    Is that correct?

    Edited: added \dot LaTex notation
    Last edited: Oct 4, 2014
  6. Oct 4, 2014 #5
    That looks right to me.

    I should note that in LaTex you can write time derivatives using the \dot, e.g., $$\dot r.$$ For higher time derivatives just put as many 'd's as there are time derivatives, e.g., \dddot r is : $$\dddot r.$$
    Last edited: Oct 4, 2014
  7. Oct 4, 2014 #6
    Thanks! That'll help with future notations.
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