Timeline for What's the meaning of linear medium in electromagnetism?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2015 at 22:07 | comment | added | Brian Moths | @JohnM You got it backwards here and in a comment to your answer. $D$ gives you $\rho_f$ and $E$ of course gives you the total $\rho$. Since $\rho_f$ is zero, $\nabla \cdot D$ must be zero, but since we are linear, $\nabla \cdot P$ must be zero as well. | |
| S Aug 26, 2015 at 20:28 | history | edited | Sepideh Abadpour | CC BY-SA 3.0 |
fixed a typo and added a dielectric tag
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| S Aug 26, 2015 at 20:28 | history | suggested | John M | CC BY-SA 3.0 |
fixed a typo and added a dielectric tag
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| Aug 26, 2015 at 20:23 | review | Suggested edits | |||
| S Aug 26, 2015 at 20:28 | |||||
| Aug 26, 2015 at 20:08 | comment | added | John M | $\nabla \cdot {\boldsymbol P} = \rho_b$ is for bound volume charges. Free charges are slightly different -- total field responds to $\nabla \cdot {\boldsymbol E} = \rho_f$ | |
| Aug 26, 2015 at 20:06 | comment | added | Brian Moths | It is not clear to me that the absence of source charges implies that $\vec{M} = \vec{P} = \vec{0}$. In fact, I would have thought that the electromagnetic wave would disturb the medium and cause a polarization to form. Similarly, a magnetic field would induce a magnetization. However, with no free sources I would expect $\vec{\nabla} \cdot \vec{P} = \vec{0}$. | |
| Aug 26, 2015 at 20:01 | answer | added | John M | timeline score: 3 | |
| Aug 26, 2015 at 19:14 | history | edited | Sepideh Abadpour | CC BY-SA 3.0 |
added 105 characters in body
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| Aug 26, 2015 at 19:06 | history | edited | Sepideh Abadpour | CC BY-SA 3.0 |
added 436 characters in body; edited tags
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| Aug 26, 2015 at 18:55 | history | edited | Sepideh Abadpour | CC BY-SA 3.0 |
added 436 characters in body; edited tags
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| Aug 26, 2015 at 18:24 | history | asked | Sepideh Abadpour | CC BY-SA 3.0 |