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Divergence in spherical polar coordinates

  1. Aug 18, 2011 #1
    I took the divergence of the function 1/r2[itex]\widehat{r}[/itex] in spherical coordinate system and immediately got the answer as zero, but when I do it in cartesian coordiantes I get the answer as 5/r3.

    for [itex]\widehat{r}[/itex] I used (xi+yj+zk)/(x2+y2+z2)1/2

    what am i missing?
  2. jcsd
  3. Aug 18, 2011 #2
    What you're missing is the handling of the singularity at r=0.
  4. Aug 18, 2011 #3

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    Hi Idoubt! :smile:

    If I take the cartesian derivative with respect to x I get:
    [tex]{\partial \over \partial x}{x \over (x^2+y^2+z^2)^{3/2}} = {-2 x^2+y^2+z^2 \over (x^2+y^2+z^2)^{5/2}}[/tex]
    Of course the derivatives wrt to y and z have to be added yet...

    How did you get 5/r3?
  5. Aug 19, 2011 #4
    sorry I made a silly error in my calculation

    [itex]\frac{\partial}{\partial x}[/itex] x(x2+y2+z2)-[itex]\frac{3}{2}[/itex] =

    (x2+y2+z2)-[itex]\frac{3}{2}[/itex] - 3x2(x2+y2+z2)-[itex]\frac{5}{2}[/itex]

    When i add y & z parts it cancels and becomes zero, so the div of [itex]\frac{1}{r^2}[/itex][itex]\widehat{r}[/itex] is zero.

    The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0?
  6. Aug 19, 2011 #5

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    Not every electric field is generated by exactly 1 symmetrical charge.
    And in practice there is no singularity, since the charge would have to be a point charge for that.
    Last edited: Aug 19, 2011
  7. Aug 19, 2011 #6
    Isn't the field due to just one charged assumed to be a field created by a point charge? ( I mean it makes no difference if we make that assumption right? )

    But, how can an electric field have a non zero divergence then?
  8. Aug 19, 2011 #7

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    Yes. It would only make a difference when you get very close to where the point charge is supposed to be.

    Take for instance 2 point charges and you'll have non-zero divergence.
  9. Aug 19, 2011 #8
    But when we take two charges we no longer have the [itex]\frac{1}{r^2}[/itex] function ( unless we approximate at large distances)

    It is my understanding that even the electric field of a point charge has non zero divergence is it not so?
  10. Aug 19, 2011 #9

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    Right on both counts. :)

    My point about point charge is unrelated to divergence.
    Only that a real electric field has no singularity.
  11. Aug 19, 2011 #10
    my question is , how can a field which is scalar multiple of the function [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] have a divergence when i just proved that the divergence of this function is zero?
  12. Aug 19, 2011 #11


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    The divergence is not 0, it's [itex]4 \pi \delta^{(3)}(\vec{x})[/itex]. It's not too tricky to prove with help of Green's first integral theorem applied to the whole space with a little sphere around the origin taken out and then taking its radius to 0.
  13. Aug 19, 2011 #12
    This by definition is a singularity isn't it?
  14. Aug 19, 2011 #13
    In physics it's often modelled thusly AAMOI:


    I know not relevant but...
  15. Aug 19, 2011 #14

    greek to me :)
  16. Aug 19, 2011 #15
    meh a and b are just constants and in this case F(X) is obviously the Function of X.

    The Greek symbol that looks like an O with a hat means Tensor.

    E(X) I would assume means the Energy of the system in question in terms of the vector.
  17. Aug 19, 2011 #16
    greek with a few constants and a function of x now =)
  18. Aug 19, 2011 #17
    If you've done matrices you'll know a tensor is expressed as a matrix of a matrix.

    A vector in this case is simply a matrix with a scalar.



    So to steal more wiki imagery a tensor is:

    Last edited by a moderator: Apr 26, 2017
  19. Aug 19, 2011 #18
    hmm ok ok so thats a little advanced for me, but my question was about the divergence of the [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] function. In simpler terms can you tell me the reason for non-zero divergence in the electric field of a point charge?
  20. Aug 19, 2011 #19


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    No, it's a distribution, the socalled Dirac [itex]\delta[/itex] distribution. It's a very useful concepts. It's even so useful that the mathematicians built a whole new subfield of mathematics to define this concept rigorously, functional analysis!
  21. Aug 19, 2011 #20

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    I suspect you're referring to Maxwell's law that says:
    [tex]\textrm{div }\vec E = \frac {\rho} {\epsilon_0}[/tex]

    The divergence of a spherically symmetric charge is zero everywhere except at the place where the charge actually is.
    A point charge is a special case, since the charge density would be infinite (actually a Dirac delta-function).
    In practice the charge would take in a certain volume.
    The divergence of the electric field is not zero at the place where the charge density is not zero.
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