# How is the potential in multipole expansion independent from the origin chosen?

Consider a charge distribution $\rho(x',y',z')$ and a point $P=(x,y,z)$ where we want to calculate the potential with multipole expansion.

Suppose also that the total charge is zero, that is $$Q=\mathrm{\int \int \int \rho(x',y',z') dx'dy'dz'=0}$$

Therefore the dipole moment of the distribution $\bf{p}$

$$\bf{p} = \mathrm{\int \int \int \rho(x',y',z') (x',y',z') dx'dy'dz'}$$

does not depend on the origin of the frame of reference chosen.

Nevertheless the potential is given by

$$V(x,y,z)=\frac{\bf{p} \cdot \mathrm{ (x,y,z)}}{\mathrm{4\pi \epsilon_0 \,\,\,(x^2+y^2+z^2)^{3/2}}}$$

And the vector $(x,y,z)$ does depend on which origin is chosen.

But the potential $V(x,y,z)$ should be indipendent from the choice of reference, so can this be? What am I missing?

• What is $\rho(x',y',z') (x',y',z')$? Nevertheless, a mean value over the whole space is naturally independent of the origin of the frame, whereas a punctual value of the field does depend on the origine of that frame. – user130529 Jan 6 '17 at 17:10

## 2 Answers

The dipole term in the potential $V(\vec{r})$ is not the exact potential, but is instead the second term in the multipole expansion: $$V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{Q}{r} + \frac{\vec{p} \cdot \hat{r}}{r^2} + \frac{Q_{ij} \hat{r}_i \hat{r}_j}{r^3} + \dots \right]$$ where $Q_{ij}$ is the quadrupole tensor; the dots indicate terms due to the octopole moment, hexadecupole moment, etc. All of the multipole moments depend on the choice of origin; and in particular, having zero net charge does not guarantee that the higher-order terms stay the same when you shift your origin. Even if you have a configuration that is a "pure dipole" relative to one origin (with $Q_{ij} = 0$ and similarly for higher moments), this configuration will have a non-zero quadrupole moment as measured relative to some other origin, and these additional multipole moments will "correct" the shifted dipole potential $V(\vec{r}')$ so that it agrees with the original potential $V(\vec{r})$.

To illustrate this, suppose we have two point charges $\pm q$ at $\vec{r} = \pm (d/2) \hat{z}$ respectively. With respect to this origin, the net charge is zero, the dipole moment is $\vec{p} = q d \hat{z}$, and the quadrupole moment is $$Q_{ij} = \sum_\alpha q_\alpha \left(r_{\alpha i} r_{\alpha j} - r_\alpha^2 \delta_{ij} \right),$$ where the sum runs over the two charges. This can easily be seen to vanish ($Q_{ij}=0$), since the quantity in brackets is the same for both charges and they have opposite sign.

Now let's look at the same charge configuration relative to a new origin, such that the charge $+q$ is at $d \hat{z}$ and the charge $-q$ is at the new origin. This is the same configuration, just shifted "up" by a vector $d \hat{z}/2$. Again, the net charge is zero, and the dipole moment is $\vec{p} = q d \hat{z}$. But the quadrupole moment is now non-zero; $\vec{r}_\alpha = 0$ for the negative charge, and the sum now contains a term only due to the positive charge: $$Q_{ij} = q (r_i r_j - d^2 \delta_{ij}) = \begin{bmatrix} - d^2 & 0 & 0 \\ 0 & -d^2 & 0 \\ 0 & 0 & 2d^2\end{bmatrix}.$$ Relative to this origin, the multipole expansion of the potential will contain not just a dipole term but also a quadrupole term (and presumably higher-order multipole terms as well.) Thus, the potential expressed in the new coordinates will have a different form than the potential expressed in the old coordinates; and both expressions would correspond to the same potential $V(\vec{r})$.

This is a general fact about multipole moments, by the way: the lowest non-zero multipole moment is always independent of the origin, but higher-order multipole moments will generally change when you change the origin. The statement with "monopole moments" is obvious: if we have a configuration with a net charge (i.e. monopole moment), then the charge is independent of the origin, but the dipole moment is generally not independent of the origin. If in the above sentence you replace "charge" with "dipole moment" and "dipole moment" with "quadrupole moment", it remains true.

Your potential ($V(x,y,z)$) is the product of a function ($\bf{p}$) which does not depend on the origin of the reference frame by a function (${\mathrm{ (x,y,z)}}/{\mathrm{(x^2+y^2+z^2)^{3/2}}}$) which does depend on the origin of the reference frame, hence it depends on the choice of origin of the reference frame.

Besides, if the potential $V(x,y,z)$ was independent from the choice of origin, it would be constant, which is very unusual for a potential.