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Curl in spherical polar coordinates

  1. Aug 16, 2011 #1
    Hey, I've been stuck on this question for quite a while now:

    1. The problem statement, all variables and given/known data

    1a. Write down an expression for the position vector r in spherical polar coordinates.

    1b. Show that for any function g(r) of r only, where r = |r|, the result [itex]\nabla[/itex] x [g(r)r] = 0 is true. Why does this imply that there is a potential function associated with any vector field g(r)r?

    2. Relevant equations

    3. The attempt at a solution

    So for (1a) I've written r = r[itex]\hat{e}[/itex][itex]_{r}[/itex]
    But for (1b) I really don't know what I'm doing, I know how to take the curl but not which function to use. So could anyone give me a clue as to where to start?
    For the potential function bit I've written about the vector field being path-independent.

    Thanks
     
    Last edited: Aug 16, 2011
  2. jcsd
  3. Aug 16, 2011 #2

    vela

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    What is [itex]\nabla \times \mathbf{A}[/itex] equal to when expressed in spherical coordinates?
     
  4. Aug 16, 2011 #3
  5. Aug 16, 2011 #4

    vela

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    OK, so just identify what the components of the vector field are equal to when the vector field is of the form g(r)r and plug it into that expression. You should find you can evaluate it without knowing the specific form of g(r).
     
  6. Aug 16, 2011 #5
    I don't really know how to do that. Am I supposed to use the answer from part (a)?
     
  7. Aug 16, 2011 #6

    vela

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    In Cartesian coordinates, you have the three basis vectors (for R3), [itex]\hat{e}_x[/itex], [itex]\hat{e}_y[/itex], and [itex]\hat{e}_z[/itex], and the vector assigned to the point (x,y,z) can be expressed in terms of these basis vectors:
    [tex]\mathbf{A} = A_x(x,y,z) \hat{e}_x + A_y(x,y,z) \hat{e}_y + A_z(x,y,z) \hat{e}_z[/tex]
    To calculate the curl in Cartesian coordinates, you need Ax, Ay, and Az, which you simply identify as the coefficients of the basis vectors.

    In spherical coordinates, it's convenient to use the three basis vectors [itex]\hat{e}_r[/itex], [itex]\hat{e}_\theta[/itex], and [itex]\hat{e}_\phi[/itex], and you can write
    [tex]\mathbf{A} = A_r(r,\theta,\phi) \hat{e}_r + A_\theta(r,\theta,\phi) \hat{e}_\theta + A_\phi(r,\theta,\phi) \hat{e}_\phi[/tex]
    Here you do the same thing as before. To find Ar, for instance, you just find the coefficient of the basis vector [itex]\hat{e}_r[/itex].

    In this problem, you're told the vector field is equal to [itex]\mathbf{A} = g(r)\mathbf{r}[/itex]. Using your answer to part (a), you can express this in terms of the basis vectors so you can identify what the various components are.
     
  8. Aug 16, 2011 #7
    So, I get out that Ar is g(r)r, so just a function of r. Does this mean that there are no [itex]\theta[/itex] or [itex]\phi[/itex] terms, and when calculating the curl you never get [itex]\delta[/itex]/[itex]\delta[/itex]r of an r function, so it's zero?
     
  9. Aug 16, 2011 #8

    vela

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    Yup, that's right.
     
  10. Aug 17, 2011 #9
    Thanks a lot for your help :smile:
    I wonder if you could help me with the next bit. It says:



    I imagine that's completely wrong!
     
    Last edited: Aug 17, 2011
  11. Aug 17, 2011 #10

    vela

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    Your expression for work is for a constant force. Here, the force varies with position, so you must use the general expression[tex]W = \int_C \mathbf{F}\cdot d\mathbf{r}[/tex] where C is the path the object follows.
     
  12. Aug 17, 2011 #11
    Ooh, okay, I thought there should be an integral somewhere! Thanks again :smile:
     
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