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Multipole expansion of a line charge distribution

  1. Nov 17, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm very stuck trying to solve this problem, hopefully some of you can give me a clue about in which direction I should go:

    Determine the multipole expansion in two dimensions of the potential of a localized charge distribution ##\lambda(\vec{x})## until the quadrupole term.
    Hint: a point charge situated at the origin in 2D has the potential ##\phi(\vec{x}) = - k q \ln\vec{x}^2##.

    2. Relevant equations

    The multipole expansion formula I have (up to the quadrupole term): ##\phi(\vec{x}) = k \big(\frac{q}{|\vec{x}|} + \frac{\vec{x} \cdot \vec{p}}{|\vec{x}|^3} + \frac{1}{2} \sum Q_{ij} \frac{x_i x_j}{|\vec{x}|^5} \big) ##

    3. The attempt at a solution

    I am afraid I don't even know where to start. If I go cylindrical coordinates I can rewrite the given potential as ##\phi(\vec{r}) = -2kq \ln(r)##. But I don't really know how to link this equation with the multipole expansion formula I quoted above. I tried expanding ##\ln(r)## but I am not sure if that brings anything.

    Any hint?

    Thank you in advance for your suggestions.


    Julien.
     
  2. jcsd
  3. Nov 17, 2016 #2
    I also have those formulas for the components of the expansion:

    Monopole: ##q = \int d^3 x' \rho(\vec{x'})##
    Dipole moment: ##\vec{p} = \int d^3x' \vec{x'} \rho(\vec{x'})##
    Quadrupole moment: ##Q_{ij} = d^3 x' \rho(\vec{x'}) (3 x_i' x_j' - \vec{x}^2 \delta_{ij})##

    Maybe I'm looking at it in the wrong way the whole time. Should I just expand the given potential with a Taylor series for scalar fields?
     
  4. Nov 17, 2016 #3

    TSny

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    Hi, Julien. If you understand the derivation of this formula for 3D, then you can modify the derivation for 2D.
    Yes, that is part of what you need to do. But we would need to see the details of your work to know if you are going about it correctly.
     
  5. Nov 18, 2016 #4
    Hi @TSny and thank you for your answer. The first part of the derivation of the formula for 3D is about expanding the position term for a far field. I understand that more or less. I expanded ##\ln \vec{x}^2## and here is what I got:

    ##\ln (\vec{x} - \vec{x}')^2 \approx \ln \vec{x}^2 + (\vec{x}' \cdot \nabla_x) \ln \vec{x}^2 + \frac{1}{2} (-\vec{x}' \cdot \nabla_x)^2 \ln \vec{x}^2##
    ##= \ln \vec{x}^2 - \vec{x}' \cdot \frac{2 \vec{x}}{|\vec{x}|^2} + \frac{1}{2} (- \vec{x}')^2 \nabla_x \big(\frac{2 \vec{x}}{|\vec{x}|^2} \big)##
    ##= \ln \vec{x}^2 - \frac{2 \vec{x} \cdot \vec{x}'}{|\vec{x}|^2} + \frac{2 \vec{x}'^2 \cdot \vec{x}^2 - 2 (\vec{x} \cdot \vec{x}')^2}{|\vec{x}|^4}##

    I encountered a problem in my third term (it looks like zero!). Other students got ##\frac{\vec{x}^2 \vec{x}'^2 - 2 (\vec{x} \cdot \vec{x}')}{|\vec{x}|^4}## so I'd better detail my calculations here:

    ##\nabla \big(\frac{\vec{x}}{|\vec{x}|^2} \big) = \partial_{x_1} \big(\frac{x_1}{|\vec{x}|^2} \big) + \partial_{x_2} \big(\frac{x_2}{|\vec{x}|^2} \big)##
    ##= \big(\frac{1}{|\vec{x}|^2} - \frac{2 x_1^2}{|\vec{x}|^4} \big) + \big(\frac{1}{|\vec{x}|^2} - \frac{2 x_2^2}{|\vec{x}|^4} \big)##
    ##= \frac{2}{|\vec{x}|^2} - \frac{2 (x_1^2 + x_2^2)}{|\vec{x}|^4}##
    ##= \frac{2 |\vec{x}|^2 - 2 |\vec{x}|^2}{|\vec{x}|^4} \overset{!}{=} 0##

    I think I don't quite understand the notation really. Isn't ##\vec{x}^2 \vec{x}'^2## equivalent to ##(\vec{x} \vec{x}')^2##? Anyway once I get the correct expansion, can I just insert it in the given potential and get each term corresponding to the monopole, dipole and quadrupole moments? I struggle to understand the last part of the derivation for 3D.


    Thank you again for your answer, your help is very appreciated.


    Julien.
     
  6. Nov 18, 2016 #5
    I've rewritten my answer in terms of components and of integrals of the line charge density, I think that's how the monopole, dipole and quadrupole moments were defined in the derivation in 3D. I also assumed the other students got the right answer for the third term. I then get:

    ##\frac{1}{k} \phi(\vec{x}) = \ln \vec{x}^2 \int d^2x' \lambda (\vec{x}') - \frac{2 \vec{x}}{|\vec{x}|^2} \int d^2x' \vec{x}' \lambda (\vec{x}') + \sum_{i,j} \frac{x_i x_j}{|\vec{x}|^4} \int d^2x' \lambda (\vec{x}') (\vec{x}'^2 \delta_{ij} - 2 x_i' x_j')##

    Sorry I didn't write the whole development, I'm a bit short in time right now. But if it isn't clear to you how I get there, I'd very happily write the steps in detail this evening (most importantly I've inserted the Taylor expansion in the given potential and replaced ##q## by a surface integral instead of a volume integral for the 3D case). My assumption so far is that each integral is a (monopole, dipole, quadrupole) moment. Does that make sense?


    Thank you very much.


    Julien.
     
  7. Nov 18, 2016 #6

    TSny

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    Should the differentiation be with respect to the primed variables?

    Looks like a misprint in the second term of the numerator. Shouldn't ##(\vec{x} \cdot \vec{x}')## be squared?

    ##\vec{x}^2 \vec{x}'^2## ## = (x_1^2 + x_2^2)## ##({x'}_1^2+ {x'}_2^2)##

    ##(\vec{x} \vec{x}')^2## doesn't make sense notationally. Is that a scalar product of the two vectors inside the parentheses? If so, then a dot is required.
     
  8. Nov 18, 2016 #7

    TSny

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    Check the overall sign. Recall that for a point charge ##\phi(\vec{x}) = -kq \ln \vec{x}^2##

    In the second term on the right, did you mean to indicate a scalar product between ##\vec{x}## and the integral?

    Otherwise, your expression looks correct to me.

    Yes. I believe there are different conventions regarding the definitions of the quadrupole moments. For example, some people would define the quadrupole moments for the 2D case as ##Q_{ij} = \int d^2x' \lambda (\vec{x}') ( x_i' x_j' - \frac{1}{2}\vec{x}'^2 \delta_{ij})##, which differs by a factor of -1/2 from your integrals. They would then make up for this factor by including a factor of -2 for the ##x_i x_j## outside the integral.
     
  9. Nov 25, 2016 #8
    Hi @TSny and sorry for the late answer. I have been working on it again in class and it now seems much clearer. I just wanted to say thank you again for all your help. ;)


    Julien.
     
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