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I Derivation of a formula

  1. Dec 3, 2017 #1
    Would someone explain the last step in eq.(2.34) in Heald and Marion?
    Much thanks ahead.
     

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  3. Dec 3, 2017 #2

    haruspex

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    You've a far better chance of getting assistance if you upload the relevant page. Not everyone will have that book.
     
  4. Dec 4, 2017 #3

    vanhees71

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    Well, that's simply working out the derivatives. It's just a bit of algebra!
     
  5. Dec 26, 2017 #4
    If you're asking about the second equal sign, it's just simple algebra working with derivatives.
    If you're asking about the first equal sign, that may be the process of finding the quadrupole component of an electric potential of a multipole.
    I do not have the book nor do I have the PDF file, so I'll just explain from scratch.
    Well, let us put an infinitesimal charge dq reside on a "source point", located on the position
    x = (x1,x2,x3) relative to the origin. (Vector notation in boldface)
    Then we can get the exact electric potential at position r as`

    φ(r) = 1/4πε0 ∫ [dq/(| r - x |)]

    and the infinitesimal charge dq as

    dq = ρ(x)dx1'dx2'dx3'

    If we perform a Taylor series expansion on the function 1/| r - x | = 1/[(x1 - x1')2+(x2 - x2')2+(x3 - x3')2] about x = 0, the third term after using the summation convention will be the term inside the summation operator multiplied by 1/2. I don't get why suddenly there's a 1/6 in front of the operator, so if you have more parts of the book I suggest you upload it. :)
     
  6. Dec 26, 2017 #5

    Charles Link

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    That equation looks a lot like equation (12) from @vanhees71 recent Insights article: https://www.physicsforums.com/insights/homopolar-generator-analytical-example/ I worked through the calculus of that and verified it was correct. It may not be the exact same thing, but I think it is likely the formulas leading up to it are quite similar. It's hard to tell for sure, because the OP is rather incomplete, but they do like quite similar. Also, I'm not sure what the (4) in ## \Phi^{(4)} ## represents.
     
    Last edited: Dec 26, 2017
  7. Dec 27, 2017 #6

    vanhees71

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    It's the quadrupole contribution to the electrostatic potential (or, if you use a magnetostatic potential in regions of space, where there are no currents the magnetostatic quadrupole contribution to this potential).

    In statics it's always derived from the (free) Green's function of the Laplace operator,
    $$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|},$$
    fulfilling
    $$\Delta_x G(\vec{x},\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}').$$
    You can most easily get the first few multipole moments from assuming ##r=|\vec{x}|\gg r'=|\vec{x}'|## and then doing an expansion in powers of ##r/r'##. Then you get the multipole moments in Cartesian coordinates, but that method becomes quite inconvenient at higher orders, and it is more convenient to use the mutlipole expansion in spherical coordinates, i.e., the representation theory of the rotation group SO(3), leading in a very beautiful way to the spherical harmonics. That's why I believe that didactically it would be much better to teach QM 1 first and then classical electrodynamics, because then you can teach the mathematical methods needed in field theory on hand of the relatively simple Schrödinger equation rather than right away jump into Maxwell theory with it's many vector fields :-).
     
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