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Center of Force, Generalized Potential

  1. Sep 9, 2014 #1
    The problem states that a particle moves in a plane under the influence of the following central force:
    [tex]
    F = \frac{1}{r^2}\Big(1 - \frac{\dot{r}^2 - 2\ddot{r}r}{c^2}\Big)
    [/tex]
    and I am asked to find the generalized potential that results in such a force. Goldstein gives the following equation involving generalized forces obtained from a potential ##U(q_j , \dot{q}_j)##:
    [tex]
    Q_j = -\frac{\partial{U}}{\partial{q}_j} + \frac{d}{dt}\Big(\frac{\partial{U}}{\partial\dot{q}_j}\Big)
    [/tex]
    This question is apparently just solved by "guessing" the potential. The answer is:
    [tex]
    U(r, \dot{r}) = \frac{1}{r} + \frac{\dot{r}^2}{c^2r}
    [/tex]
    And this can be checked by taking the appropriate derivatives and plugging them into the general Lagrange equation above. However, I want to make sure this is the only way to arrive at it; just making a lucky guess seems pretty unsatisfying. Therefore, I tried the following approach:

    Suppose that the generalized potential is a sum of two potentials, namely:
    [tex]
    U(r, \dot{r}) = U_1(r, \dot{r}) + U_2(r, \dot{r})
    [/tex]
    Where we consider the following Lagrange equation:
    [tex] F = -\frac{\partial{U}}{\partial{r}} + \frac{d}{dt}\Big(\frac{\partial{U}}{\partial{\dot{r}}}\Big) [/tex]
    Which is just the general Lagrange equation specific to the problem. I then equate the following, on a hunch:
    [tex] -\frac{\partial{U_1}}{\partial{r}} = \frac{1}{r^2} [/tex]
    and
    [tex] \frac{d}{dt}\Big(\frac{\partial{U_2}}{\partial{\dot{r}}}\Big) = \frac{-\dot{r}^2 + 2\ddot{r}r}{c^2r^2} = \frac{-\dot{r}^2}{c^2r^2} + \frac{2\ddot{r}r}{c^2r^2} [/tex]
    The first equation is easily solved such that:
    [tex] U_1(r, \dot{r}) = \frac{1}{r} [/tex]
    But the second equation is where I may be using dubious methods. It stands to reason that:
    [tex] \frac{\partial{U_2}}{\partial{\dot{r}}} = \int\Big[\frac{-\dot{r}^2}{c^2r^2} + \frac{2\ddot{r}r}{c^2r^2}\Big]dt [/tex]
    I now take an uneasy step. Treat ##r## as a constant such that I pull them out, break up the integral, and write the time derivatives in their proper forms:
    [tex] \frac{\partial{U_2}}{\partial\dot{r}} = -\frac{1}{c^2r^2}\int\Big(\frac{dr}{dt}\Big)^2dt + \frac{2r}{c^2r^2}\int\frac{d^2r}{dt^2}dt = -\frac{1}{c^2r^2}\int\frac{dr}{dt}dr + \frac{2r}{c^2r^2}\int\frac{d^2r}{dt^2}dt [/tex]
    The second integral is easy to interpret; it's just ##\dot{r}##. But for the first, I use integration by parts and reintroduce the dot notation to get:
    [tex] \frac{\partial{U_2}}{\partial{\dot{r}}} = -\frac{1}{c^2r^2}\Big[\dot{r}r - r\int\frac{d^2r}{dt^2}dr\Big] + \frac{2\dot{r}r}{c^2r^2} = \frac{2\dot{r}}{c^2r} [/tex]
    where I collected terms of ##\frac{\dot{r}r}{c^2r^2}## and then simplified. The potential can now be solved as:
    [tex] U_2(r, \dot{r}) = \frac{2}{c^2r}\int\dot{r}d\dot{r} = \frac{\dot{r}^2}{c^2r} [/tex]
    By adding ##U_1## and ##U_2##, I have arrived at the correct answer! But this could be problematic. After all, ##r## is dependent upon ##t##, but I treated it as a constant during my integrations. Thus, my question is: is this a happy accident as a result of bad mathematics or does my method have some justification and I've just left out some details?

    Assistance would be greatly appreciated!
     
  2. jcsd
  3. Sep 15, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
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