Kindly refer to the Multipole Expansion section (chapter 3) of David Griffith's Introduction to Electrodynamics.

Let us first discuss about why multipole expansion is needed. As far as I understand, if we have a extended charge distribution (not at the origin), then very far away from this distribution, the potential $V(\mathbf r)$ will be same as that due to a point charge (as if the whole charge of the extended charge distribution has accumulated at a single point). [$\mathbf r$ is the position vector of the field point, $\mathbf r'$ is the coordinates of the source points.] As we come closer and closer to the charge distribution while moving along the line joining the origin and the field point, the shape, size and configuration of the charge distribution will have its impact on the potential and we will have dipole, quadrupole, and so on, terms added to the point charge potential. Let us focus on the dipole term. The dipole moment is defined as $\sum_{i=1}^nq_i\mathbf r_i'$.

Now let us apply this to a actual single point charge $Q$ situated not at the origin. Let its position be $\mathbf R'$ with respect to origin. Now by applying the above formula, Griffiths says that it will have a dipole moment $Q\mathbf R'$.

My question is as follows. The coordinate origin is just a mathematical construction. It simply decides the direction in which the field point is approaching the point charge. The shape, size and configuration of a point charge is isotropic. From every direction you look at it, it always looks the same. Then how can one have a dipole moment of a point charge? To add to this, Griffiths further writes that if the point charge is situated at the origin, then the dipole moment is zero. How can a physically measurable quantity (the dipole moment in this case) be dependent on a purely mathematical construction which looks redundant since the configuration of the point charge looks same from all directions?


1 Answer 1


Quite simply, the dipole moment of a charged system depends on the coordinate origin. There is nothing particularly surprising or unphysical about this, and there are plenty of other quantities (such as orbital angular momentum) with that property. The dipole moment of a charged system is not a quantity that is defined for the system itself; it is only defined for the system in relation to a given coordinate system.

  • $\begingroup$ The dipole moment of a physical dipole (+q and -q separated by a distance d) is always charge times distance of separation between them (q*d) and is always independent of the coordinate system. Why is it so? $\endgroup$
    – user103515
    Apr 29, 2016 at 10:34
  • $\begingroup$ Because the system is neutral, in which case (as opposed to a system with nonzero total charge) the dipole moment is indeed independent of the coordinate origin. That's a simple calculation which you should be able to perform. $\endgroup$ Apr 29, 2016 at 11:45
  • $\begingroup$ If the moments up to order $n-1$ vanish, the moment of order $n$ is independent of coordinate system. Perhaps you should add "in general" to the first sentence? $\endgroup$
    – auxsvr
    Apr 29, 2016 at 13:17
  • $\begingroup$ @auxsvr Hence the careful qualifying of the entire post to apply only to charged systems where relevant. $\endgroup$ Apr 29, 2016 at 13:36
  • $\begingroup$ Kindly refer to en.wikipedia.org/wiki/Dipole and see "Quantum mechanical dipole operator" See the expression for CHARGED DIPOLE. If we apply this formula to a single electron or any single point charge the the dipole moment p becomes 0. What is your explanation on that? $\endgroup$
    – user103515
    May 17, 2016 at 12:37

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