Timeline for Harmonic function in infinite domain in $\mathbb{R}^3$, constant on the boundary and decaying as $1/r^2$
Current License: CC BY-SA 4.0
20 events
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| Oct 8, 2022 at 11:06 | 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. | |
| Jun 10, 2022 at 10:07 | 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. | |
| Feb 10, 2022 at 9:06 | 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. | |
| Oct 13, 2021 at 9:04 | 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. | |
| Sep 14, 2021 at 21:45 | comment | added | Michael Engelhardt | ... so this is inconsistent with your specification that $f$ falls off faster than that. It is likewise inconsistent with your specification that the original conductor with cavity and charges is overall neutral. So we're left with $f_0 =0$, and therefore $f\equiv 0$. | |
| Sep 14, 2021 at 21:44 | comment | added | Michael Engelhardt | Ah, I see. I would argue like this (and you've said most of this already): All that matters for the electric field outside is the constant $f_0 =f|_{\partial \Omega }$, and the question boils down to what we can say about $f_0 $. Let's show that $f_0 \neq 0$ is inconsistent with your specifications. $f_0 \neq 0$ would be equally produced by a charged solid (without the cavity) conductor - an uncharged one would (uniquely) produce $f_0 =0$. But a charged solid conductor would produce asymptotic behavior $f\sim 1/r$ (continued ...) | |
| Sep 14, 2021 at 11:52 | comment | added | asv | @MichaelEngelhardt : I am trying to prove the following. Assume that in a cavity of a conductor there are charges with total charge 0. Then outside of the conductor there is no electric field. | |
| Sep 13, 2021 at 17:56 | comment | added | Michael Engelhardt | Thinking about it some more, I may finally be understanding your question: Perhaps the uniqueness theorem you seek is this one: "The electric field of a collection of conductors is uniquely determined if the total charge on each conductor is given." This is shown, e.g., in Griffiths' "Introduction to Electrodynamics" (section 3.1.6 in the 3rd edition). Does that help? | |
| Sep 13, 2021 at 17:37 | comment | added | Michael Engelhardt | Hmm, no, it can't be $1/r^2 $ either, that's the behavior of a dipole. For a (connected) conductor, your only options are: It is charged, then the leading behavior at infinity is $\sim 1/r$; or it is neutral, then the potential $f\equiv 0$. If you want a dipole, you need to have at least two conductors disconnected from one another. | |
| Sep 13, 2021 at 8:21 | comment | added | asv | @MichaelEngelhardt : You are right: it should be $1/r^2$. Corrected. | |
| Sep 13, 2021 at 8:19 | history | edited | asv | CC BY-SA 4.0 |
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| Sep 12, 2021 at 22:32 | answer | added | Michael Renardy | timeline score: 0 | |
| Sep 12, 2021 at 15:30 | comment | added | Michael Engelhardt | Just a remark (this also applies to the previous question you refer to): Your physical motivation does not match your mathematical question. The potential you describe (constant on the boundary, decaying as $1/r^3 $) does not come from a conductor. | |
| Sep 12, 2021 at 15:05 | comment | added | asv | @LeechLattice: Looks like the final answer. | |
| Sep 12, 2021 at 15:01 | comment | added | LeechLattice | @makt Even if $\Omega$ is not an Euclidean ball, if $f$ is nonzero at its boundary, $g$ will be bounded away from zero at its boundary (though not constant); in this case it's impossible to have $g(0)=0$ as $g$ is harmonic. | |
| Sep 12, 2021 at 14:31 | comment | added | asv | @LeechLattice: If $\Omega$ is not a Euclidean ball then $g$ is not constant on the boundary. But in the Euclidean ball this answers the question. | |
| Sep 12, 2021 at 13:59 | comment | added | LeechLattice | And that $g(x)=O(|x|^2)$ for $x \approx 0$ indicates that there is no singularity at $x=0$, hence $g$ is harmonic with constant boundary values. | |
| Sep 12, 2021 at 13:53 | comment | added | Giorgio Metafune | $g(x)=|x|^{2-n} f(x/|x|^2)$ in $R^n$. | |
| Sep 12, 2021 at 12:28 | history | edited | asv | CC BY-SA 4.0 |
added 4 characters in body
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| Sep 12, 2021 at 11:29 | history | asked | asv | CC BY-SA 4.0 |