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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$\begingroup$ No, why would it be? $\endgroup$– my2ctsCommented Jul 21, 2018 at 18:55
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$\begingroup$ Is your question about dipole moment or dipole potential (which is potential generated by dipole at some point in space)? $\endgroup$– V.F.Commented Jul 21, 2018 at 19:26
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$\begingroup$ just editted as dipole moment $\endgroup$– 4pie0Commented Jul 21, 2018 at 19:34
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2$\begingroup$ @my2cts Yes, why wouldn't it? $\endgroup$– tparkerCommented Jul 21, 2018 at 20:45
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2 Answers
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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.
answered Jul 21, 2018 at 20:48
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$\begingroup$ 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? $\endgroup$ Commented Feb 16, 2023 at 12:29
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$\begingroup$ @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. $\endgroup$ Commented Apr 13, 2023 at 8:20
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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.
answered Jul 21, 2018 at 19:03
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$\begingroup$ How can i show them mathematically? And i need some drawings to visualize what it means "the choice of origin". $\endgroup$– 4pie0Commented Jul 21, 2018 at 20:03
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$\begingroup$ @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. $\endgroup$ Commented Jul 21, 2018 at 22:45