Does the dipole moment depend on the choice of origin
- if the total charge Q is not zero?
- 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.
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Emilio Pisanty
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4pie0
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No, why would it be? – my2cts Jul 21 '18 at 18:55
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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
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just editted as dipole moment – 4pie0 Jul 21 '18 at 19:34
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@my2cts Yes, why wouldn't it? – tparker Jul 21 '18 at 20:45
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If the total charge is non-zero, the dipole moment is ill defined in that its value depends on the choice of origin. For total charge zero distributions, the dipole moment does not depend on choice of origin.
mike stone
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How can i show them mathematically? And i need some drawings to visualize what it means "the choice of origin". – 4pie0 Jul 21 '18 at 20:03
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@4pieo. I assumed that you had already worked this out for yourself and just wanted reassurance. It follows immediately from the definition of dipole moment, after all. Emilo's answer says it all. – mike stone Jul 21 '18 at 22:45
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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.
For more on the subject see e.g. Why isn't there a centre of charge?, How does one prove that the lowest-order nonvanishing multipole moment of a charge distribution is independent of the origin, for arbitrary $\ell$?, Can one force the electric quadrupole moments of a neutral charge distribution to vanish using a suitable translation?, and links therein.
Emilio Pisanty
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Not sure if it's my bad English or my bad Physics, but "displacing the origin" in my understanding would mean $\int (\mathbf r - \mathbf r_0) \rho(\mathbf r - \mathbf r_0)\mathrm d\mathbf r$. To me this sounds more like "displacing the point of evaluation" or "choosing a different point of evaluation". How wrong am I? – Raphael J.F. Berger Feb 16 '23 at 12:29
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@RaphaelJ.F.Berger So long as $\rho(\boldsymbol{r})$ is under the integral it doesn't matter how you choose your coordinates the integral will be invariant of that (since $r_{0}$ is constant). ie the total charge doesn't change. – R. Rankin Apr 13 '23 at 08:20