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Force and Potential Energy Coordinates

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Evaluate the force corresponding to the potential energy function ##V (r) = \frac{cz}{r^3}##, where ##c## is a constant. Write your answer in vector notation, and also in spherical polars, and verify that it satisfies ##∇∧F = 0##.

    2. Relevant equations
    ##F(x)=-\frac{dU}{dx}##

    Spherical Coordinates:
    ##r=\sqrt{x^2+y^2+z^2}##

    3. The attempt at a solution
    So I'm confused a bit here. I think the vector notation will only have a ##z## component. So differentiate to get ##F## and then as ##r## is a polar coordinate I just change ##r^3## to ##\sqrt{x^2+y^2+z^2}^3##. This doesn't look right though. I have a list of relations from cartesian to spherical coordinates but I don't understand how to answer the question with the ##\frac{1}{r^3}##. I know how to do the last bit which is the curl. Thanks in advance for any help in the right direction.
     
  2. jcsd
  3. Oct 7, 2015 #2

    ehild

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    Your relevant equation is not true for three dimension. So what is the relation between the force and the potential energy function?
     
  4. Oct 7, 2015 #3
    Ok is it:

    ##F(x,y,z)=-\Big(\frac{dU}{dx}+\frac{dU}{dy}+\frac{dU}{dz}\Big)##?
     
  5. Oct 7, 2015 #4

    ehild

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    No. The force is a vector. You wrote a scalar.
     
  6. Oct 7, 2015 #5
    ##F(x,y,z)=-\Big(\frac{dU}{dx}\hat{i}+\frac{dU}{dy}\hat{j}+\frac{dU}{dz}\hat{k}\Big)##?
     
  7. Oct 7, 2015 #6

    ehild

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    Yes.
    Now you can determine the force from the potential function.
     
  8. Oct 7, 2015 #7
    So ##F=\frac{c}{r^3}\hat{k}##?
     
  9. Oct 7, 2015 #8

    ehild

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    How did you get it?
    You have to differentiate the potential energy function with respect to all coordinates, x, y, z.
     
  10. Oct 7, 2015 #9
    Well there is no x or y component so its only the differentiating with respect to z that matters no?
     
  11. Oct 7, 2015 #10

    ehild

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    The potential energy is a scalar, it does not have "components". But it depends on all the three coordinates through r ##r=\sqrt{x^2+y^2+z^2}##
    What are the partial derivatives?
     
  12. Oct 7, 2015 #11
    So is ##r^3 = x^3+y^3+z^3##?
     
  13. Oct 7, 2015 #12

    ehild

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    NO!!!!!!!!!!!!!!!!!!!!!
    ##r^3=(x^2+y^2+z^2)^{\frac{3}{2}}##
     
  14. Oct 7, 2015 #13
    I know that's the obvious answer but I thought it was to do with what the power was. So I have to differentiate:

    ##\frac{cz}{(x^2+y^2+z^2)^{\frac{3}{2}}}##

    With respect to x,y and z?
     
  15. Oct 7, 2015 #14

    ehild

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    Yes. You need all partial derivatives, ## \partial U / \partial x ##, ## \partial U / \partial y ##, ## \partial U / \partial z ##.
     
  16. Oct 7, 2015 #15

    ehild

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    Yes.
     
  17. Oct 7, 2015 #16
    So I have a long mess of an answer that I won't put up cos it seems pointless. I'm fairly sure it was all differentiated correctly though. For the spherical part do I just start again and change ##z## to ##rcos\theta##?
     
  18. Oct 7, 2015 #17

    ehild

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    Yes.
     
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