Does the dipole moment depend on the choice of origin?

Does the dipole moment depend on the choice of origin

1. if the total charge Q is not zero?
2. for a system of charges neutral overall?

How can I show that mathematically? Also I need some drawings to visualize what "the choice of origin" means.

• No, why would it be? – my2cts Jul 21 '18 at 18:55
• Is your question about dipole moment or dipole potential (which is potential generated by dipole at some point in space)? – V.F. Jul 21 '18 at 19:26
• just editted as dipole moment – 4pie0 Jul 21 '18 at 19:34
• @my2cts Yes, why wouldn't it? – tparker Jul 21 '18 at 20:45

For a system with charge density $\rho(\mathbf r)$ (which might be volumetric, but which could also include point, line or surface charges by including suitable delta-function terms into $\rho(\mathbf r)$), the dipole moment is always defined to be $$\mathbf d = \int \mathbf r \rho(\mathbf r)\mathrm d\mathbf r,$$ where the integral is taken over all of space. This means that if you displace your origin by $\mathbf r_0$, then the new dipole moment will be given by \begin{align} \mathbf d' & = \int \mathbf r' \rho(\mathbf r)\mathrm d\mathbf r \\ & = \int (\mathbf r - \mathbf r_0) \rho(\mathbf r)\mathrm d\mathbf r \\ & = \int \mathbf r \rho(\mathbf r)\mathrm d\mathbf r - \mathbf r_0 \int \rho(\mathbf r)\mathrm d\mathbf r \\ & = \mathbf d - \mathbf r_0 Q, \end{align} i.e. it will change by the product of the coordinate translation and the total charge $Q =\int \rho(\mathbf r)\mathrm d\mathbf r$ of the system. This means that the dipole moment is origin-independent if the system is globally neutral, and it does depend on the coordinate origin if the global charge is nonzero.