Timeline for Polarization ellipse for an EM wave [duplicate]
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2023 at 8:50 | history | left closed in review |
Miyase Michael Seifert joseph h |
Original close reason(s) were not resolved | |
| Oct 8, 2023 at 4:10 | review | Reopen votes | |||
| Oct 11, 2023 at 8:50 | |||||
| Apr 3, 2019 at 21:05 | comment | added | Emilio Pisanty | (2) Once in that canonical form, it is clearer that the major axis of the ellipse is when both components add constructively (so with length $|E_+|+|E_-|$) and the minor axis happens when they interfere destructively (so it has length $|(|E_+|-|E_-|)|$). The identification of the major-to-minor-axis ratio as $(1+r)/(1-r)$ then follows easily. | |
| Apr 3, 2019 at 20:59 | comment | added | Emilio Pisanty | It appears my reopen vote has aged away, but there's little to add here to the given duplicate. In short: (1) the ellipse rotation can be seen from the fact that $R(\theta) \epsilon_\pm = e^{\mp i\theta} \epsilon_\pm$, i.e. the circular basis vectors are eigenvectors of rotations about $\epsilon_3$, so that a rotation by $\alpha/2$ can be used to nullify the phase difference between $E_+$ and $E_-$, i.e. rotating the system into the canonical form where both coefficients are positive. | |
| Mar 11, 2019 at 20:05 | review | Reopen votes | |||
| Mar 16, 2019 at 16:25 | |||||
| Mar 11, 2019 at 19:46 | comment | added | Charlie | It was marked as duplicated however I had already checked before the link provided and it didn't answer my question, hence why I opened a new one. I include additional information to justify this. | |
| Mar 11, 2019 at 19:45 | history | edited | Charlie | CC BY-SA 4.0 |
added 617 characters in body
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| Mar 11, 2019 at 12:43 | history | closed |
Emilio Pisanty GiorgioP-DoomsdayClockIsAt-90 Jon Custer Kyle Kanos Martin |
Duplicate of How can I get the axes of the polarization ellipse from the Jones vector of the light? | |
| Mar 10, 2019 at 2:05 | review | Close votes | |||
| Mar 11, 2019 at 12:43 | |||||
| S Mar 10, 2019 at 0:20 | history | suggested | xray0 | CC BY-SA 4.0 |
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| Mar 9, 2019 at 23:54 | review | Suggested edits | |||
| S Mar 10, 2019 at 0:20 | |||||
| Mar 9, 2019 at 22:50 | comment | added | Charlie | I didn't think about that but it seems a better approach, I'm going to try it. | |
| Mar 9, 2019 at 21:51 | comment | added | Mechanix | I have the feeling you are doing it much more complicated than necessary. If the ratio of the two amplitudes is $re^{i\alpha}$, the electric field can be given as: $E(\vec{r},t)=(E_{+}(\epsilon_{+}+re^{i\alpha}\epsilon_{-})e^{i(\vec{k}\vec{r}-wt)})$. Now you can callculate the norm of the term before the wave term and chose $\alpha$ such that it becomes maximal (semi-major axis) and minimal (semi minor axis). | |
| Mar 9, 2019 at 21:32 | history | edited | user137289 | CC BY-SA 4.0 |
TeX (sin, cos)
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| Mar 9, 2019 at 20:58 | history | asked | Charlie | CC BY-SA 4.0 |