# 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

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 $V(\vec{r})$.
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.