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Potential from a simple Quadrupole expansion

  1. Dec 31, 2014 #1
    Hi everyone! I'm currently working on this problem for which I am getting inconsistencies depending on how I do it. I'm trying to find the potential due to the quadrupole moment of the following distribution:
    +q at (0,0,d), -2q at (0,0,0), and +q at (0,0,-2d)

    I am doing this using two different methods and they both get different answers:


    1) Using the general expansion Qij=sum[ql{3ril*rjl-rl^2deltaij)] and plugging into Vquad=1/(8*pi*epsilonor^3)sum[Qij*ni*nj]

    This method gives me some constant divided by r^3, with NO angular dependence whatsoever.


    2) Going back to the basics and using the very general potential by substituting in the addition theorem for spherical harmonics, etc to find the general potential for a general multipole moment: qlm=integral[rho*r^l*Ylm],
    and
    Vmulti=sum[1/(epsilono(r^(l+1)*(2l+1)*Ylm(theta,phi)*qlm]

    Notice how this answer definitely depends on theta for dipole moment and above (when the spherical harmonics introduce cos(theta)'s into them.)

    Doing it this method gives me the same constant divided by r^3 that I found earlier, except now it is multiplied by (3cos^2(theta)-1) which comes from exactly the Y20 spherical harmonic.


    These two methods SHOULD give the same results, but these are radically different... Any ideas?

    Thanks!

    PS sorry for the lack of Latex, but I figured most people should get the gist of it
     
  2. jcsd
  3. Dec 31, 2014 #2

    mfb

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    Where does the formula for the potential come from? It does not look right.

    What is "epsilonor"?
     
  4. Dec 31, 2014 #3
    You can see it derived here:
    http://physicspages.com/2012/04/03/quadrupole-moment/

    And sorry, epsilonor is just from my lack of latex understanding.. it is really supposed to be the vacuum constant times the magnitude of r:

    ε*|r|
     
  5. Dec 31, 2014 #4

    mfb

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    So ni*nj are the components of the vector r? Well, then your potential should depend on the direction of r.
    If it does not, please show your work.
     
  6. Dec 31, 2014 #5
    Yes that is correct. However, because all 3 charges are on the z axis, the only non-zero component of the quadrupole moment tensor is Q_zz. Every other one goes to 0.

    Then, only one term will be in the potential summation, the Q_zz term. This term corresponds to nz*nz (which are two unit vectors in the z direction) which boils down to 1.
     
  7. Dec 31, 2014 #6

    mfb

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    Those are not the unit vectors! The "unit" here refers to the whole vector r, its z component can be smaller.
     
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