Timeline for Detailed form of light waves in vacuum and how to test it experimentally?
Current License: CC BY-SA 3.0
30 events
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| Jun 3, 2022 at 6:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 24, 2022 at 22:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 18, 2021 at 20:47 | answer | added | Claudio Saspinski | timeline score: 1 | |
| Dec 18, 2021 at 18:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 12, 2021 at 15:17 | answer | added | barry | timeline score: 0 | |
| Jan 15, 2018 at 9:46 | comment | added | MJC | Must it be a plane wave? How to determine that by measurement? This is not possible. Plane waves are non-physical mathematical simplifications. | |
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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| Jan 11, 2016 at 15:05 | comment | added | Julia | But an LED or light bulb is not a dipole!? | |
| Jan 11, 2016 at 14:53 | comment | added | ProfRob | Just look up the standard treatment of an oscillating electric dipole. At large distances (compared with the wavelength) these are approximately plane waves. But not close to the dipole. | |
| Jan 11, 2016 at 14:44 | comment | added | Julia | @RobJeffries Do you have a reference of how the EM-field of a light bulb or LED or something like that look like (formulas and/or plots)? | |
| Jan 11, 2016 at 14:42 | comment | added | ProfRob | Light from a point source cannot be a plane wave. That E and B are perpendicular is only a property of plane wave solutions to Maxwell's equations. Other solutions are possible - as you seem to know. | |
| Jan 11, 2016 at 14:15 | comment | added | Julia | @RobJeffries I am talking about light waves in vacuum. Consider a LED or light bulb in vaccuum or the light of a star... | |
| Dec 20, 2015 at 3:25 | history | tweeted | twitter.com/StackPhysics/status/678415927197704193 | ||
| Dec 14, 2015 at 18:23 | comment | added | 299792458 | @Julia - Yes, you are right. I am sorry, somehow I totally messed up such a simple fact. Ok, the question makes sense and now I will vote on this post! | |
| Dec 14, 2015 at 17:40 | comment | added | ProfRob | You seem to be talking about plane wave solutions to Maxwell's equations. These do necessarily have the attributes you describe. But no, they are not the only possible solutions to Maxwell's equations. For example E and B are not in phase for an EM wave in a medium with conductivity. An oscillating electric dipole does not emit plane waves. | |
| Dec 14, 2015 at 16:36 | comment | added | HolgerFiedler | Go trough academia.edu/12172263/… and read about photons and about radio waves | |
| Dec 14, 2015 at 15:16 | history | edited | Julia | CC BY-SA 3.0 |
added 327 characters in body
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| Dec 14, 2015 at 14:59 | review | Close votes | |||
| Dec 14, 2015 at 20:08 | |||||
| Dec 14, 2015 at 14:41 | comment | added | Carl Witthoft | This was just asked last week -- as usual I can't find the question... here it is: physics.stackexchange.com/questions/219978/… | |
| Dec 14, 2015 at 14:18 | comment | added | Julia | @TheDarkSide Nothing the curl and time derivative operators are linear and applied to the added constant term they yield zero. | |
| Dec 14, 2015 at 14:05 | answer | added | John Duffield | timeline score: -5 | |
| Dec 14, 2015 at 13:44 | comment | added | 299792458 | @Julia - What happens when we substitute your modified ${\vec E}$ into the second and fourth Maxwell's equations (for the curls)? | |
| Dec 14, 2015 at 12:45 | comment | added | Julia | @TheDarkSide Why should it be modified? Without modification it leads to a solution of the Maxwell equations | |
| Dec 14, 2015 at 12:31 | comment | added | 299792458 | @Julia - And why will the ${\vec B}$ stand preserved in that case, and not get modified in accordance with ${\vec B} = \frac{1}{\omega} \left( {\vec k} \times {\vec E} \right)$ ? | |
| Dec 14, 2015 at 12:13 | comment | added | Julia | @TheDarkSide: Do you agree that I can add a vector field (constant in space and time) to a solution to the maxwell equations in vacuum and get a solution of it, too? Then suppose you have a plane wave solution where $\vec{E}$ and $\vec{B}$ are orthogonal. Then just add a constant $\vec{E_0}$ which is oblique to $\vec{E}$ to the field $\vec{E}$ and you may get a solution where $\vec{E'}$ and $\vec{B}$ are no more orthogonal. | |
| S Dec 14, 2015 at 12:09 | history | suggested | Stefan Bischof | CC BY-SA 3.0 |
clarified question according to comment.
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| Dec 14, 2015 at 11:43 | comment | added | 299792458 | Why do you say, in the first bullet, "Since you can add constants to a solution to Maxwell's equation it doesn't seem necessary from theory" | |
| Dec 14, 2015 at 11:32 | review | Suggested edits | |||
| S Dec 14, 2015 at 12:09 | |||||
| Dec 14, 2015 at 10:31 | comment | added | physicopath | I guess one of the B's must be E in the second bullet point | |
| Dec 14, 2015 at 10:04 | history | asked | Julia | CC BY-SA 3.0 |