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A Divergence of 1/r^n...n>0

  1. Feb 29, 2016 #1
    I have read in Griffiths electrodynamics that divergence of 1/r^2 is delta function and I thought it was the only special case...I have understood the logic there.... but a question came in mind...what would happen in general if the function is 1/r^n ....where n is positive integer>0.....because the function blows up at origin..how to define divergence..found no book uptill now......I want answer from physics point of view...because I have heard mathematicians don't like delta function !!
     
  2. jcsd
  3. Feb 29, 2016 #2
    why people are viewing but not answering....have I asked something very silly ?...I am utterly confused
     
  4. Feb 29, 2016 #3
    Can you please explain how you take the divergence of a scalar?
     
  5. Feb 29, 2016 #4
    Respected Sir,I am not taking divergence of scaler .. I would never..unit vector in radial direction is assumed....I thought it to be pretty obvious and no one would ask me this....I was wrong however...from next time I would post total question.
     
    Last edited: Feb 29, 2016
  6. Feb 29, 2016 #5
    OK. Show us how you evaluate the divergence of ##\frac{\vec{i}_r}{r^n}##. I guess you're looking at spherical coordinates, right?
     
  7. Mar 2, 2016 #6
    Divergence of a vector field at any point is defined as the amount of flux per unit volume diverging from that point.
    Whereas in mathematical point of view, divergence have a different meaning with respect to sequences and series.
     
  8. Mar 2, 2016 #7
    He's clearly talking about the divergence of a vector, and there is a specific formula for evaluating the divergence of a vector in spherical coordinates. We weren't asking for a definition of "divergence."
     
  9. Mar 3, 2016 #8
    For a vector whose only non-zero component in spherical coordinates is in the radial direction, vr, the divergence is given by:

    $$\vec{\nabla} \centerdot \vec{v}=\frac{1}{r^2}\frac{\partial (r^2v_r)}{\partial r}$$

    I leave it up to you to do the differentiation.
     
  10. Mar 3, 2016 #9

    vanhees71

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    one should add that this is true in the points of the function where it is sufficiently smooth. You have to be carful with singularities. E.g., the Coulomb field
    $$\vec{E}(\vec{x})=q \frac{\vec{x}}{4 \pi |\vec{x}|^3}=\frac{q}{4 \pi r^2} \vec{e}_r$$
    obeys
    $$\vec{\nabla} \cdot \vec{E}(\vec{x})=\rho(\vec{x})=q \delta^{(3)}(\vec{x}).$$
    Of course, the formula in the previous posting is correct, since obviously for ##r>0## you have indeed
    $$\vec{\nabla} \cdot \vec{E}=0 \quad \text{for} \quad r>0.$$
    To find what's going on in singularities you have to analyze the situation going to the integral definition of the divergence
    $$\vec{\nabla} \cdot \vec{E}(\vec{x})=\lim_{\Delta V \rightarrow \infty} \int_{\partial \Delta V} \mathrm{d}^2 \vec{f} \cdot \vec{E},$$
    where ##\Delta V## is a small volume around the point ##\vec{x}## and ##\partial \Delta V## the closed boundary surface with the normal vectors pointing out of the volume.

    Of course, for the singularity in our case the limit diverges because of the point charge sitting in the origin. Here choose a sphere of radius ##a## around the origin for the integral:
    $$\int_{S_a} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \vec{e}_r \cdot \vec{E}=\frac{q}{4 \pi} \int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \int_0^{2 \pi} \mathrm{d} \varphi=q.$$
    No matter how small you make the sphere, you always get the charge inside sitting at the origin, as it must be due to Gauss's Law. Since you know from Chestermiller's formula that everywhere ##\vec{\nabla} \cdot \vec{E}=0##, you conclude that indeed the above formula with the Dirac-##\delta## distribution holds.
     
  11. Mar 11, 2016 #10
    @Chestermiller sir, I know the formula you have written but the formula holds for 'r' not equal to zero.....but if r=0 i.e at origin the function itself is undefined...So I guess spatial delta function should be employed....but I do not know how and @vanhees71 the singularity at origin for basic 1/r^2 field is taken care by spatial delta function...but i did not find any generalization anywhere for 1/r^n
     
  12. Mar 11, 2016 #11

    vanhees71

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    Look for multipole expansion. This gives an expansion of the electromagnetic field/potential in powers of ##1/r## for a charge-current distribution around the origin.
     
  13. Mar 11, 2016 #12
    @vanhees71 yes multipole expansion is used to find far fields of a certain current distribution....i.e to expand 1/R where R is distance between source point and observation point with the condition that distance of observation point is much larger than distance of source from origin.... @Chestermiller sir will correct me if I am wrong ..........but that is not related to the question I asked for
     
  14. Mar 11, 2016 #13

    vanhees71

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    It is precisely related to your question since it let's you determine the possible forms of fields and the corresponding charge-current distributions that have a singularity at the origin only.
     
  15. Mar 11, 2016 #14
    precisely what I am asking for is the divergence of 1/r^n (unit vector in radial direction) at the origin particularly .....because at all other points the function is differentiable..............now please elaborate your answer in your free time with the help of multipole expansion as you have suggested
     
  16. Mar 12, 2016 #15

    blue_leaf77

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    You don't need delta function if ##n\neq 2##. If ##n=2##, its divergence is undefined at the origin because you get an expression like ##0/0##, but outside this point the divergence is uniformly zero.
    On the other hand if ##n\neq 2##, the divergence will just be given by Chestermiller's formula above, and one thing to emphasize here is that the resulting function is a smooth function. Therefore you don't need a delta function. Consider a special case of ##n=5##, the divergence will be ##-3/r^6##, why do you think this function is necessarily expressible in terms of delta function? If this still doesn't convince you, consider a 1D function ##1/x^6##, can you express this function in terms of delta function?
     
  17. Mar 12, 2016 #16
    Sir, correct me if I am wrong..... at the origin the vector function has (whether n=2 or not) 1/0 form , not 0/0.......also for n=5 case which you have mentioned the divergence function(computed using Chestermiller's formula) has 1/0 form at origin........... and 1/0 is atleast not defined
     
  18. Mar 12, 2016 #17

    blue_leaf77

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    Consider the special case when ##n=2## and compute the divergence, then reconfirm whether you get 1/0 or 0/0.
    Yes, it's infinity at the origin, but what's wrong with that? What's wrong, for instance, with a function of the form ##1/x^2##?
    I have got the impression that you want to express the divergence for any value of ##n## in terms of delta function. Why do you think it's necessary? Back to the example of ##n=5##, why would one need to re-express ##-3/r^6## in terms of delta function if the very form we already have there suffices to represent the behavior of the function?
     
  19. Mar 12, 2016 #18
    But Sir ,for n=2 we are getting zero divergence everywhere....and when I mentioned that at origin the vector function has 1/0 form I meant the original function,not its divergence......and I don't have any problem with infinite divergence but what exactly does this physically mean?........and Sir one more thing... for n>2...while computing volume integral how do we deal with the lower limit i.e origin ?
     
  20. Mar 12, 2016 #19

    blue_leaf77

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    In
    and in
    you clearly said that you wanted to express the divergence no matter what ##n## is in terms of delta function.
    In that case the integral does not converge, but I doubt there is any physical system where you are required to do that, is there?
     
  21. Mar 12, 2016 #20
    okkk.....my limited knowledge does not know any physical situation as such........this problem encountered actually when I tried to find electric field of dipole using gauss' law ..and also found volume charge density of dipole using 2 delta functions.........but then I realized that due to lack of symmetry gauss' law can't be used here
     
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