Timeline for Dipole Moment: what does it really represent?
Current License: CC BY-SA 4.0
4 events
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| Nov 9, 2020 at 17:02 | comment | added | Emilio Pisanty | On the other hand, for a globally-charged system, then the separation of $\rho(\mathbf r)$ into positive and negative parts is not necessary, and you can just define the center of charge directly from the total density. | |
| Nov 9, 2020 at 17:01 | comment | added | Emilio Pisanty | @ric.san The separate densities $\rho_+$ and $\rho_-$ are non-negative, so if they integrate to zero then they must be identically zero everywhere (i.e. there is no charge anywhere). Moreover, for a neutral system, you know that $Q_+=\int \rho_+(\mathbf r)d\mathbf r = \int \rho_-(\mathbf r)d\mathbf r=Q_-$ (or else you'd have $Q=Q_+-Q_- \neq 0$), so if one of them vanishes, then both do. | |
| Nov 9, 2020 at 16:55 | comment | added | ric.san | The problem is... what if one of them is zero? How can I define a 'separation' if there's only one type of distribution density'? | |
| Nov 9, 2020 at 15:51 | history | answered | Javier | CC BY-SA 4.0 |