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Question in proof of Helmholtz Theorem

  1. Aug 9, 2013 #1
    I have a math question in this:http://en.wikipedia.org/wiki/Helmholtz_decomposition.

    If you look at the proof, what is
    [tex]\nabla_x\frac {1}{|\vec x-\vec x'|}\;and\; \nabla_x \times \frac {1}{|\vec x-\vec x'|}[/tex]

    My thinking the first one is just the gradient of ##\frac {1}{|\vec x-\vec x'|}##. But the second one is puzzling. A curl has to perform on a vector....that is ##\nabla\times\vec A##, not ##\nabla\times K## where ##K## is a scalar function. Last I check, ##|\vec x-\vec x'|## is a scalar function.

    Please help. Thanks
     
  2. jcsd
  3. Aug 10, 2013 #2

    mathman

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    I didn't see any ∇x?
     
  4. Aug 10, 2013 #3
    It's the 5th line of equation under "Proof".
     
  5. Aug 11, 2013 #4
    Well what you saying is true. But I think they meant this is the rule if you have some vector with this argument, because in the rest of proof there isn't this mistake.
     
    Last edited: Aug 11, 2013
  6. Aug 11, 2013 #5
    Thanks, that make me feel better.

    There is something about this Helmholtz theorem. Every single article I read had problems. The only one that is close is http://faculty.uml.edu/cbaird/95.657%282012%29/Helmholtz_Decomposition.pdf. Still it does not explain why the surface integral is zero at infinite distance. I have another thread asking this question:

    https://www.physicsforums.com/showthread.php?t=705015

    If you have time, please take a look at that. Thanks.

    Look at this video starting from 7:00.
    http://www.youtube.com/watch?v=OUOi-j_WzBM

    The derivation is all wrong, but magically he came to the right answer!!! He used the wrong product rule and come out with the right answer. I wrote a comment yesterday, he actually acknowledge the error today!!!
     
  7. Aug 11, 2013 #6

    lurflurf

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    Yes this means
    [tex]\nabla_x\frac {*}{|\vec x-\vec x'|}\;and\; \nabla_x \times \frac {*}{|\vec x-\vec x'|}[/tex]
    where * is some appropriate object.

    If you enjoy wrong proofs of Helmholtz, you will love this one that is beloved by physicists and engineers.
    $$-\nabla^2 F=-\nabla \nabla \cdot F+\nabla \times \nabla \times F \\
    (-\nabla^{-2})(-\nabla^2) F=(-\nabla^{-2})(-\nabla) \nabla \cdot F+(-\nabla^{-2})\nabla \times \nabla \times F \\
    F=-\nabla (-\nabla^{-2})\nabla \cdot F+\nabla \times (-\nabla^{-2})\nabla \times F \\
    \text{let} \\
    \varphi=(-\nabla^{-2})\nabla \cdot F \\
    A=(-\nabla^{-2})\nabla \times F \\
    \text{so} \\
    F=-\nabla \varphi+\nabla \times A \\
    \text{QED}$$
    I smile every time I see that.
     
  8. Aug 11, 2013 #7
    It would be a lot funnier if I have not stuck for a few days!!! I thumbed through quite a few articles, spending my time going through them and one place or the other, they all starting to have an aroma....and it's not a present one!!

    AND they all just said the surface integral goes to zero......like as if it is so simple that everyone supposed to know!!!! I dug through 7 or 8 EM books, Advanced Calculus, Vector calculus, PDE, ODE books...........none!!! How is:

    [tex]\int_{s'}\frac{\vec {F}(\vec {r})}{|\vec {r}-\vec {r'}|}\cdot d\vec {s'}=0\;\hbox { and }\;\int_{s'}\frac{\vec {F}(\vec {r})}{|\vec {r}-\vec {r'}|}\times d\vec {s'}=0[/tex]
    at large distance?
     
  9. Aug 11, 2013 #8

    lurflurf

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    The wiki link assumes we have "sufficiently smooth, rapid decay."
    So we have essentially
    $$\int f(r)/r \mathrm{d}r \rightarrow 0$$
    the 1/r is not enough to insure this say f is constant.
    In applications it is common to assume f is zero (or small) for large r.
    If we just assume f is zero we might have smoothness problems, but we can always smooth the function.
    This assumption is sometimes questionable but if we cannot assume the effects of distant objects is negligible we are in trouble. More generally we need boundary conditions that assure that the decomposition is unique. So that
    $$\varphi=(-\nabla^{-2})\nabla \cdot F \\
    A=(-\nabla^{-2})\nabla \times F $$
    are well defined. Without such conditions there could be many such functions.
     
  10. Aug 11, 2013 #9
    Thanks for the reply. I know!!! It is just common sense to know that the effect of a localized event at far away is zero. But math doesn't work with common sense!!! You have to proof it!!!..........Or.....am I becoming a math junky?!! But....I am not a student, I already have 30 years career as an engineer and manager of engineering. I am retired......common sense and practicality is not important anymore. I like to look at it in pure mathematical and logical sense.....it's better than crossword puzzle!!!!

    But thanks for your support.
     
    Last edited: Aug 11, 2013
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