electromagnetism - Why is a dipole moment called a dipole moment?

The General Formula for a moment is the following one:

$$\vec{M} = \vec{r} \times \vec{F}.$$ However the formula for a dipole moment is this one: $$\vec{p} = Q \vec{d}.$$ How comes this is still called a "moment"? A charge is not the same as a force, so if you strictly follow the definition it is not a moment, or am I mistaken?

Answer

A 'moment' is quite a general term, and its use ranges from electrostatics (e.g. dipole and other multipole moments) to mechanics (moment of force but also moment of inertia) to huge stretches of statistics. The general intuition is that you have some amount of 'stuff' (charge, force, mass, probability) with some distribution function $s(x)$, and the various moments of this distribution capture, usually very well and in a compact fashion, the information of how the stuff is distributed.

On one dimension, moments are usually defined as

$$S_n=\int x^n s(x)\,\text dx.$$ The zeroth moment tells you about the 'amount of stuff' that's present, which is just the integral of $s(x)$. The first moment gives you a good approximation of where the stuff is centered at, and the second moment gives you information of how wide the distribution is. Third moments give you information about how skewed this distribution is about the center, and if you go on you can keep extracting information about the distribution. In fact a question called the moment problem asks whether it's possible to reconstruct $s(x)$ from the moments, and it turns out that you can indeed do this as long as the 'stuff' doesn't change sign.

In more dimensions, and if your 'stuff' is a vector quantity (like a force) then there are many more choices of how to do this, but the general idea of multiplying $s(x)$ with some position-dependent function $M_n(x)$ (which may be matrix-valued! this is the case for the moment of a force distribution, as

$$\vec r\times \vec F=\begin{pmatrix} 0 &-x & y\\x &0&-z \\-y &z&0\end{pmatrix} \begin{pmatrix}F_x\\F_y\\F_z\end{pmatrix}$$ is a linear operator on $\vec F$), and then integrating. So, the 'most general' definition of a moment is along the lines of $$S_n=\int M_n(x) s(x)\,\text dx,$$ where depending on the situation the interpretation of the different objects can be complicated. In general, though, in physics, there is the expectation that the kernel functions $M_n(x)$ be some sort of homogeneous $n$-th order polynomial in the position, which can nicely be characterized as $$M_n(\lambda x)=\lambda^n M_n(x).$$ This expresses the commonality between the moment of force and the dipole moment, which are both linear in the relevant distance.

Finally, one ought to worry about how these fancy formulas, with all those integrals, connect with the two simple cases you stated above. In general:

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  for a given charge distribution $\rho(\vec x)$, the dipole moment is $\vec d =\int \vec x \rho(\vec x)\,\text d\vec x,$ and

  
  


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  for a force distribution $\vec F(\vec x)$ acting on a given object the moment of force (i.e. the net torque on the object about the chosen origin) is $\vec \tau=\int \vec x\times \vec F(\vec x)\,\text d\vec x$.

  
  

You can see that these fit the general pattern above, and that you can recover your original, simpler cases by taking a pair of point charges or a force acting at a single point.

For a simple take-home message, though, you can try something along the lines of "the first moment of something is the quantity times some sort of distance".