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How do dipole moments work?

  1. Jan 23, 2012 #1
    Hi, I am sort of new to Physics forum but I am urgently in need of help. It would be very much appreciated if you could clear my confusion.
    Let me break them down into bullet points instead of one giant paragraph.

    1.Dipole moments are vector quantities. I can just add them up like I do to forces?
    2. Does dipole have to be composed of 1 positive and 1 negative charge? Can it be two positive charges?

    Right now I am trying to compute the dipole moment of a cluster of atoms in a specific shape discretely and then continuously.
    3.For the discrete part, I want to know whether if it's correct to treat each pair of atoms(in my case, they are ions+) with a dipole moment with a direction pointing somewhere, and then add them up like vectors of force?
    4.For continuous part, I must integrate it over a volume. I don't understand the equation
    P(r) = integral of p(r0) (r0 - r) dr^3.
    what is p(r0)*(r0 - r)
    r0 = r zero = r initial i think?

    What if my charge distribution is constant and uniform? Does that mean I can pull it out and times the volume of the shape?

    Thank you.
  2. jcsd
  3. Jan 23, 2012 #2

    Simon Bridge

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    1. dipole moments are pseudo-vectors like moment of inertia or torque, they add the same way.

    2. the electric dipole is defined in terms of + and - charge.

    3. depends on the situation. two positive ions won't have a dipole between them - their individual dipoles are usually magnetic and come from their electrons. In an electric field, though, the electrons will tend to hang about to one side creating an electric dipole.

    4. that sort of relation always comes with a diagram... look at the diagram.

    Looks like you have a specific problem in mind - how about posting it?
    Meantime - refresher:
    Last edited by a moderator: Sep 25, 2014
  4. Jan 23, 2012 #3


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    No, it's defined as a term in the multipole expansion of a general distribution of charge:

    Electric dipole moment (general case)

    Multipole expansion

    This is also covered in E&M textbooks like Griffiths, but I don't have my copy handy so I can't give a page reference.
  5. Jan 24, 2012 #4

    Simon Bridge

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    Fine - but how does that answer OPs question?
    You can have a dipole moment of 2 + charges?
    Last edited: Jan 24, 2012
  6. Jan 24, 2012 #5
    of course you cant have a dipole of 2 positive charges, which is precisely why simon gave an ok answer the first time. dipoles are little blobs of two clouds of charges, of opposite sign necessarily. the definition can be extended to more numbers of charges but the general description requires the net charge be zero, so that there should be no electric field in the domain, but only polarization fields appear.

    in the case of two neighbouring atoms, you find that the dipole exists as the oscillatory motion of the electrons in one shifts the orbitals on the other atoms' protons and electrons, thereby creating a dipole. the integral equation can be written, since the protons are very heavy, using the born-Oppenheimer approximation (feel free to google that one) we can assume they are pretty much stationary, and their charge distribution is therefore a Dirac delta function centred at the centre of mass in the continuous case. the equation then can describe the dipole moment matrix element describing the dipole interaction of an ion core on another identical nearby atom.
    Last edited: Jan 25, 2012
  7. Jan 25, 2012 #6


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    You can expand the long-distance fields of any charge distribution in terms of multipole moments.

    If you have two positive charges, there's of course a net monopole component with the sum of the charges in the origin, which is arbitrary somewhere near the charge distribution, and you should choose cleverly such that the calculation of the multipole moments is as easy as possible. The next corrections are then the dipole and quadrupole moments, etc. Only the first non-vanishing multipole moment is independent of the choice of the origin. In this case with a non-zero net charge only the monopole moment is independent of the choice of the origin.
  8. Jan 26, 2012 #7
    I wouldn't put it that way. The dipole is the second term in a multipole expansion. Any distribution of charge can be characterized as a monopole plus a dipole plus a quadrupole plus an octopole etc. The dipole moment is mathematically orthogonal to the monopole and to the higher multipoles as well. Now, if you calculate the dipole moment of a system of two positive charges about the centroid you will get zero, but not if you use a different origin.
  9. Jan 26, 2012 #8
    I was talking about something irrelevant, nevermind
    Last edited: Jan 26, 2012
  10. May 5, 2012 #9
    Hi frinds,
    I've a big confusion regarding the dipole.
    Why we take the direction of dipole moment from negative charge to the positive charge?
    Is it a convention only?
  11. May 5, 2012 #10

    Simon Bridge

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    Yes ... and no.
    It is a convention chosen to be consistent with other conventions. i.e. not an arbitrary convention. If you extrapolate backwards from the definition of the dipole, assuming the direction is the opposite way around, you get a lot of extra minus signs or you use a left-handed reference frame (iirc).
  12. Nov 12, 2012 #11
    Hi everyone. I have a question about dipole moments and I think it is closely related to the OP's posts so I would just ask it here: is the dipole moment just a measure of the tendency of the dipole to rotate under an external electric field?
  13. Nov 12, 2012 #12

    Simon Bridge

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    I think, historically, it came from investigations into how electric currents affect magnets. It is called a "moment" because a magnetic dipole is set up by a circulating electric current. In mechanics, a "moment" is a circular/twisting force - a "torque".

    However, the term has come to, also, be used to describe the property of being rotated by an external B field. Particularly of fundamental particles. I think it is useful to make a distinction between quantum mechanical magnetic moments and classical dipole moments.

    So - the answer to your question is, yes and no.
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