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Kinetic Energy in Spherical Coordinates

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Derive the expression for kinetic energy of a classical particle in spherical coordinates.


    2. Relevant equations
    I believe the answer I am supposed to reach is:
    [tex]T=\frac{1}{2} m (\dot{r}^2 + r^2\dot{\theta^2} + r^2\dot{\phi ^2}sin^2\theta)[/tex]

    3. The attempt at a solution
    [tex]T=\frac{1}{2}mv^2[/tex]
    [tex]T=\frac{1}{2}m(\dot{x^2} + \dot{y^2} + \dot{z^2})[/tex]
    Knowing that:
    [tex] x=rsin\theta cos\phi [/tex]
    [tex] y=rsin\theta sin\phi [/tex]
    [tex] z=rcos\theta[/tex]

    After plugging these in and working it out, I came up with:
    [tex] T= \frac{1}{2}mr^2 [/tex]

    My question now is, how do I get from the answer that I currently have to the solution I am trying to get to (that I listed above in part 2)?
     
  2. jcsd
  3. Mar 7, 2013 #2

    TSny

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    Did you remember to take the time derivatives?
     
  4. Mar 7, 2013 #3
    I'm not quite sure what you mean by that?
     
  5. Mar 7, 2013 #4

    TSny

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    ##\dot x = dx/dt##. Did you find an expression for ##\dot x##? Note that your answer does not have the correct units [dimensions] for energy.
     
  6. Mar 7, 2013 #5
    I think I might understand now.

    Since:
    [tex]\dot{x}=\frac{dx}{dt} \ \ \ \dot{y}=\frac{dy}{dt} \ \ \ \dot{z}=\frac{dz}{dt}[/tex]

    Then:
    [tex]\dot{x}=\frac{d(rsin \theta cos \phi)}{dt} \ \ \ \dot{y}=\frac{d(r sin\theta sin\phi)}{dt} \ \ \ \dot{z}=\frac{d(rcos \theta)}{dt}[/tex].

    And since [itex]dr=dr \hat{r} + r d\theta \hat{\theta} + rsin \theta d\phi \hat{\phi} [/itex], this would then make [itex] \frac{dr}{dt} = \dot{r} + r \dot{\theta} + rsin\theta \dot{\phi} [/itex]

    Now, as I did before with squaring x, y, and z to get 1 (in my original post), the answer then becomes the solution of [itex] (\frac{dr}{dt})^2 [/itex].

    Which, when put back into my original answer is: [itex] T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 sin\theta^2 \dot{\phi}^2)[/itex].

    Is that more along the correct lines of thinking?
     
  7. Mar 8, 2013 #6

    vela

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    Did you work these out? You have ##x=x(r,\theta,\phi)##, so
    $$\dot{x} = \frac{\partial x}{\partial r} \dot{r} + \frac{\partial x}{\partial \theta} \dot{\theta} + \frac{\partial x}{\partial \phi} \dot{\phi}.$$ Just calculate the partial derivatives and grind it out.

    You're being too sloppy with your notation, so what you wrote doesn't make sense. You have dr on both the lefthand side and righthand side of the first equation, but they don't represent the same quantity. You should have written
    $$d\vec{r} = dr\,\hat{r} + r\,d\theta\,\hat{\theta} + r\sin\theta\,d\phi\,\hat{\phi}.$$ In the next equation, the unit vectors just mysteriously disappeared.
     
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