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Divergence of a/r^2

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the divergence and curl of [itex]\vec{E}[/itex]=α[itex]\frac{\vec{r}}{r^2}[/itex]


    2. Relevant equations

    Div([itex]\vec{E}[/itex])=[itex]\vec{∇}[/itex]°[itex]\vec{E}[/itex]
    Div([itex]\vec{E}[/itex])=[itex]\vec{∇}[/itex]x[itex]\vec{E}[/itex]

    Table of coordinate conversions, div, and curl:
    http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

    3. The attempt at a solution
    My confusion is stemming from the spherical coordinates. I believe I have the divergence down because the result is a scalar so the direction doesn't matter (phi & theta) so it's just a matter of the magnitude of the field at any given point which is dependent only on the distance from the origin. Or can I not make this assumption? I'm not sure how to bring phi and theta into the mix when dealing with the curl.

    For the divergence I know that [itex]\vec{r}[/itex]=r[itex]\hat{r}[/itex] so [itex]\vec{E}[/itex] simplifies to:

    α[itex]\frac{\hat{r}}{r}[/itex]

    Using the table linked above to find the form of the divergence in spherical coordinates I believe I can ignore the theta and phi 'contributions' because divergence will depend only on r.

    div(E)=[itex]\frac{α}{r^2}[/itex][itex]\frac{∂(r)}{∂r}[/itex]=[itex]\frac{α}{r^2}[/itex]

    I think the curl should be zero by observation but I don't know how to show this. Can I do the same as with divergence and ignore the theta and phi contributions because this is simply a function of r?

    Thank you!
     
  2. jcsd
  3. Sep 15, 2014 #2

    ZetaOfThree

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    Just use the formula in the table for the curl in spherical coordinates. Then just make sure you do the derivatives correctly. Easy.
     
  4. Sep 15, 2014 #3
    I'm sorry for not being more clear. Referencing the table in specific I don't know what A[itex]\theta[/itex] and A[itex]\phi[/itex] are. That's what is confusing me.
     
  5. Sep 15, 2014 #4

    ZetaOfThree

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    Those are the components of [itex]\vec{A}[/itex] in the [itex]\hat{\theta}[/itex] and [itex]\hat{\phi}[/itex] directions. In your case, [itex]\vec{A}=\vec{E}=a\frac{\vec{r}}{r^2}[/itex].
     
  6. Sep 15, 2014 #5
    edit: sorry, accidental double post
     
  7. Sep 15, 2014 #6
    Right, I understand that A is a placeholder for E in the table. It is the components of E in those directions that I don't know how to find.
     
  8. Sep 15, 2014 #7

    nrqed

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    In general, ##\vec{E} = E_r \hat{r} + E_\theta \hat{\theta} + E_\phi \hat{\phi} ##
     
  9. Sep 15, 2014 #8
    Right, I'm trying to find out what Er[itex]\hat{r}[/itex] and so on are functionally (starting from the value given for [itex]\vec{E}[/itex] in the problem). I can't take the partial derivative of the general form for E in the direction of theta without breaking E down into the relevant components first. I am given an actual value for E which is a function of r.
     
  10. Sep 15, 2014 #9

    ZetaOfThree

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    You should read up on spherical coordinates, it will be very helpful. The ##\theta## and ##\phi## components of ##\vec{E}## are zero because it only points in the radial direction and ##E_r=\frac{a}{r}##.
     
  11. Sep 15, 2014 #10

    nrqed

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    You are told that ## \vec{E}= a \frac{\vec{r}}{r^2} ##. By comparing with the general form, what do you conclude about ##E_\theta ## and ## E_\phi## ?
     
  12. Sep 15, 2014 #11
    Which is why I assumed by observation that the curl would be zero. I just wanted to verify that I didn't have to consider a general representation of phi and theta, to be complete, as they vary for any arbitrary r. For some reason I was thinking back to r in cartesian coordinates and projecting it onto some equivalent to x, y, and z for theta and phi. Sorry for the confusion!

    Thank you
     
  13. Sep 15, 2014 #12

    ZetaOfThree

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    This is wrong. Divergence absolutely depends of the direction of ##\vec{E}##.

    You really shouldn't intuit your way through these exercises, you need to do them nrqed's way to find the components of ##\vec{E}## and then plug them into your formulas from the table.


    Spherical coordinates is an orthogonal coordinate system, meaning that ## \hat{r}##, ##\hat{\theta}##, and ##\hat{\phi}## are perpendicular, so their projections onto each other are zero.
     
  14. Sep 15, 2014 #13
    This seems to contradict your earlier statement. In this case E only depends on r. I considered Er which is effectively the function they gave me for E and used the partial derivative equation from the table and the only non-zero component was the one dependent on r. In other words, the partial with respect to theta is zero because only r shows up in the equation which is not a function of theta and thus E is a constant with respect to theta (and phi). The partial of a constant is zero. The calculated divergence in this case is listed above in my original post (non-zero). Do you disagree with the result?

    I didn't -just- intuit my way through the problem but I did try to think about the problem and compare that to my results to make sure that I understand what is going on and that the results make sense.
     
    Last edited: Sep 15, 2014
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