Does the dipole moment depend on the choice of origin? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-17T20:37:16Z https://physics.stackexchange.com/feeds/question/418785 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/418785 1 Does the dipole moment depend on the choice of origin? 4pie0 https://physics.stackexchange.com/users/201646 2018-07-21T18:35:46Z 2018-07-21T21:00:27Z <p>Does the dipole moment depend on the choice of origin </p> <ol> <li>if the total charge Q is not zero? </li> <li>for a system of charges neutral overall?</li> </ol> <p>How can I show that mathematically? Also I need some drawings to visualize what "the choice of origin" means.</p> https://physics.stackexchange.com/questions/418785/-/418791#418791 2 Answer by mike stone for Does the dipole moment depend on the choice of origin? mike stone https://physics.stackexchange.com/users/80687 2018-07-21T19:03:44Z 2018-07-21T19:03:44Z <p>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. </p> https://physics.stackexchange.com/questions/418785/-/418810#418810 0 Answer by Emilio Pisanty for Does the dipole moment depend on the choice of origin? Emilio Pisanty https://physics.stackexchange.com/users/8563 2018-07-21T20:48:12Z 2018-07-21T20:48:12Z <p>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' &amp; = \int \mathbf r' \rho(\mathbf r)\mathrm d\mathbf r \\ &amp; = \int (\mathbf r - \mathbf r_0) \rho(\mathbf r)\mathrm d\mathbf r \\ &amp; = \int \mathbf r \rho(\mathbf r)\mathrm d\mathbf r - \mathbf r_0 \int \rho(\mathbf r)\mathrm d\mathbf r \\ &amp; = \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.</p> <p>For more on the subject see e.g. <a href="https://physics.stackexchange.com/questions/8221/why-isnt-there-a-centre-of-charge">Why isn&#39;t there a centre of charge?</a>, <a href="https://physics.stackexchange.com/questions/403652/how-does-one-prove-that-the-lowest-order-nonvanishing-multipole-moment-of-a-char">How does one prove that the lowest-order nonvanishing multipole moment of a charge distribution is independent of the origin, for arbitrary $\ell$?</a>, <a href="https://physics.stackexchange.com/questions/49745/can-one-force-the-electric-quadrupole-moments-of-a-neutral-charge-distribution-t">Can one force the electric quadrupole moments of a neutral charge distribution to vanish using a suitable translation?</a>, and links therein.</p>