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Taking the gradient of 1/r (solid sherical harmonics?)

  1. Aug 1, 2009 #1
    Hi all, just an very elementary question, arising from the first study of generating harmonic solutions.

    How to get the gradient twice for
    [itex]
    1/r
    [/itex] in spherical coordinates?
     
    Last edited: Aug 1, 2009
  2. jcsd
  3. Aug 2, 2009 #2

    tiny-tim

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    Hi hanson! :smile:

    What's the difficulty?

    1/r obviously doesn't depend on θ or φ, so just use ∂/∂r. :wink:
     
  4. Aug 2, 2009 #3
    Hello. Yea,
    I think I can get the gradient of 1/r once, using the ∂/∂r, and that will gives me a vector.

    What confuses me is taking it another time, which will gives me a tensor. However, the answer I got is different from what is supposed to be.
     
  5. Aug 2, 2009 #4

    arildno

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    This is SO wrong!

    Let us do this properly (sort of!):

    First:
    [tex]\nabla\frac{1}{r}=-\frac{1}{r^{2}}\vec{i}_{r}[/tex]

    Now, [tex]\nabla^{2}\frac{1}{r}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{1}{r\sin\phi}\frac{\partial}{\partial\theta}+\vec{i}_{\phi}\frac{\partial}{r\partial\phi})\cdot(-\frac{1}{r^{2}}\vec{i}_{r})[/tex]

    Now, you are to apply each of these operators PRIOR to perform the dot products!

    Since the vector [itex]\vec{i}_{r}[/tex] depends upon both the angular variables, differentiating that vector will yield components PARALLELL to the direction of the angular variable.
    Thus, non-zero contributions to the full Laplacian of 1/r will appear from these differentiations.
     
    Last edited: Aug 2, 2009
  6. Aug 2, 2009 #5

    arildno

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    Okay, I'm generous today, so we use the identities:
    [tex]\frac{\partial\vec{i}_{r}}{\partial\phi}=\vec{i}_{\phi}[/tex]
    [tex]\frac{\partial\vec{i}_{r}}{\partial\theta}=\sin\phi\vec{i}_{\theta}[/tex]

    Thus, we get:
    [tex]\nabla^{2}\frac{1}{r}=\frac{2}{r^{3}}-\frac{1}{r^{3}}-\frac{1}{r^{3}}=0[/tex]

    As you should get.
     
  7. Aug 2, 2009 #6

    arildno

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    In general, when dealing with a function f(r), where r is the radial vector in spherical coordinates, we have the simple relation:
    [tex]\nabla^{2}f=\frac{1}{r^{2}}\frac{d}{dr}(r^{2}\frac{df}{dr})[/tex]
     
  8. Aug 2, 2009 #7
    Hi, arildno. Thanks for the detail reply.
    But what if it is not a dot product between [tex]\nabla[/tex] and [tex]\nabla \frac{1}{r}[/tex]? If it is a dot product, then we will get a Laplacian. However, if it is like a "dyadic product" (I am not sure if this is the right term to use, but seems to be), then we should get a tensor out of this, right? And the result should be something like [tex]\frac{\vec{\vec{I}}}{r^3} + 3 \frac{\vec{r}\vec{r}}{r^5}[/tex], where [tex]\vec{r} = r\vec{i}_{r}[/tex] and [tex]\vec{\vec{I}}[/tex] is the identity matrix. I am not sure how to get this. I kind of got the second term, but not the first one. Could you please help?

    p.s. I am sorry about the missing "arrows" above the identity matrix and the "r"s. They cannot display properly, since I am not familiar with latex typing here....
     
    Last edited: Aug 2, 2009
  9. Aug 2, 2009 #8

    arildno

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    OKay, I was a bit unsure if it was the Laplacian or the dyadic tensor you were after; it is, however, fairly trivial to set up in this case (remember that the Laplacian is the sum of the diagonal terms)!

    You get the tensor:
    [tex]f^{,,}\vec{i}_{r}\vec{i}_{r}+\frac{f^{,}}{r}\vec{i}_{\theta}\vec{i}_{\theta}+\frac{f^{,}}{r}\vec{i}_{\phi}\vec{i}_{\phi}[/tex]
     
  10. Aug 2, 2009 #9

    arildno

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    Note that this is rewritable as:
    [tex]\frac{3}{r^{3}}\vec{i}_{r}\vec{i}_{r}-\frac{I}{r^{3}}=\frac{3}{r^{5}}\vec{r}\vec{r}-\frac{I}{r^{3}}[/tex]
    where I is the identity matrix.
     
  11. Aug 2, 2009 #10
    Oh...the "f' here refers to a general function? So, you get this formula basically by performing similar steps below, right?

    I have done this is the most boring and lengthy way

    In spherical coordinates,
    [tex]
    \nabla = \vec{i}_r \frac{\partial}{\partial r}+ \vec{i}_{\phi} \frac{1}{r} \frac{\partial}{\partial \phi}+\vec{i}_\theta \frac{1}{r sin \phi} \frac{\partial}{\partial \theta}
    [/tex]

    So,
    [tex]
    \nabla \left(\frac{1}{r}\right) = \vec{i}_r \frac{-1}{r^2}
    [/tex]

    which is

    [tex]
    \nabla(\nabla(\frac{1}{r})) = \vec{i}_r \vec{i}_r \frac{2}{r^3} + \vec{i}_\phi \vec{i}_\phi \frac{-1}{r^3}+\vec{i}_\theta \vec{i}_\theta \frac{-1}{r^3}
    [/tex]

    And, it is rewritten as
    [tex]
    \vec{i}_r \vec{i}_r \frac{3}{r^3} +\vec{i}_r \vec{i}_r \frac{-1}{r^3} + \vec{i}_\phi \vec{i}_\phi \frac{-1}{r^3}+\vec{i}_\theta \vec{i}_\theta \frac{-1}{r^3},
    [/tex]

    which is
    [tex]
    \vec{i}_r \vec{i}_r \frac{3}{r^3} - \frac{\vec{\vec{I}}}{r^3}
    [/tex]

    p.s. I edited this 100 times but the latex still doesn't show up properly...i want to give up...

    But the above steps are basically correct, right?
     
    Last edited: Aug 2, 2009
  12. Aug 3, 2009 #11

    arildno

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    YEs, f was a general scalar function, only dependent upon the radial variable.

    What you have done above is correct, and you Latex seems just fine.

    If you want a similar tensorial form for a general f, you might rewrite post 8 as:
    [tex]r\frac{d}{dr}(\frac{f^{,}}{r})\vec{i}_{r}\vec{i}_{r}+\frac{f^{,}}{r}I[/tex]
    where I is, again, the identity matrix.
     
    Last edited: Aug 3, 2009
  13. Aug 3, 2009 #12
    Thanks arildno!
     
  14. Aug 4, 2009 #13

    tiny-tim

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    divergence theorem

    You can also find ∇2 of a spherically symmetric f(r) using the divergence theorem

    V2f(r)dV = ∫S f(r).dS = ∫S df/dr er.dS

    So if V is the volume between two concentric spheres of radius r and r + dr (so that S is the surface of the two spheres, and er.dS = ±dS),

    then 4πr22f(r) dr = d/dr(4πr2 df/dr) dr,

    so ∇2f(r) = (1/r2) d/dr(r2 df/dr) :wink:
     
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