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As far as I know this was never experimentally undertaken with a high precision measurement, probably by measuring the $E$ interaction field around the monopole charge of an isolated electron (i.e. using a penning trap?).

Opposite to general public science news articles in the media reporting the "perfect sphericity" or "roundness" of the electron charge measured for example by the 2018 ACMEII eEDM electron electric dipole moment upper limit experiment result at $|de|<1.1 x 10^{-29}$ e•cm thus $1.76 x 10^{-50}$ C•m or $17.6 x 10^{-49}$ C•cm upper limit eEDM (i.e. $1 e•cm = 1.602 x 10^{-21}$ C•m),

these type of eEDM experiments, all measure the sphericity (i.e. absence of electric dipole moment) of the electron charge $|e|$ at a single direction colinear to the spin axis of the electron (i.e. intrinsic magnetic dipole moment axis) but as far as I know, not $360°$ all around the electron monopole charge.

So far there is no charge anisotropy found (i.e. no eEDM) by any experiment on the spin axis of the electron. But what if the anisotropy of the electron charge is not colinear to the spin axis (i.e. there is no anisotropy found between the two poles of the spin axis) but instead observed at an other location around the electron charge compared to its spin axis? For example there could be an anisotropy observed between the equator and the poles location of its spin.

I am wondering what experiment if possible could accomplish such measurement of the total spatial isotropy of an isolated electron monopole charge deciding the sphericity of its monopole interaction E-field all around using similar precision with the ACMEII experiment? Could for example the ACME or JILA experiments be extended in the future to measure eEDM on other directions around the electron?

And regarding the electron, if its total all around sphericity as described above was found to be not perfect, say for example by an order of $10^{-12}$ e•cm or more, what implications would that have for particle/atomic physics and the Standard Model? Would this suggest new physics beyond the standard model?

electron monopole charge

image source: https://en.wikipedia.org/wiki/Electric_charge

Below a pedagogical illustration (not to be scaled) of what is asked in this thread:

toroidal moment

An intrinsic toroidal moment could possible exist on the electron charge manifold that although it does not generate an eEDM between the two poles (i.e. poles although deformed from the rest of the charge manifold, are both P-parity symmetric and identical, therefore no eEDM is generated between the two poles on the spin axis).

However, this toroidal moment illustrated above would possible create an anisotropy on the monopole electron charge E-field vectors, most prominent observed at the equator of the manifold relative to each pole in 3D space?

p.s. According to the Standard Model the electron is a dimensionless point charge and massive particle therefore its charge and mass cannot have any shape (i.e. manifold) and consequently any physical size when interacting with its environment and their origin is only intrinsic in nature. Nevertheless, the term intrinsic as in the case describing its other physical properties like spin can be used also for describing effectively a possible charge manifold for the purpose of analysis under the framework of effective geometry theory.

Qmechanic
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    Does this answer your question? Do electrons have shape? – John Rennie Jul 29 '22 at 11:20
  • Two remarks. (1) "All the way around" a sphere is a solid angle of $4\pi$ steradian, not a linear angle of $2\pi\text{ radian}=360⁰$. (2) As discussed at the linked duplicate, the Wigber-Eckart theorem requires a permanent EDM to align or anti-align with the spin axis. – rob Jul 29 '22 at 13:17
  • @JohnRennie Finding an eEDM much larger from the current SM prediction limit de<10^-38 e cm will by definition prove that the electron charge has an intrinsic shape or else can posses a certain intrinsic charge flux manifold. But this is not what I am asking here. – Markoul11 Jul 29 '22 at 13:29
  • @rob Thanks for your input. The "360⁰ all around" term in the question thread is for understanding purposes and was not intended to be epistemological. I think the readers here understand the difference between a 2D circle and a 3D sphere so there is no need to make explanations more complicated. quote: "Wigber-Eckart theorem requires a permanent EDM to align or anti-align with the spin axis". That is true but not so much of importance if there is no eEDM on the spin axis. – Markoul11 Jul 29 '22 at 13:43
  • But the theorem shows that there can be no EDM which isn't parallel to the spin axis. That's why the duplicate answers your question. – rob Jul 29 '22 at 14:29
  • The anisotropy could be symmetrical on the spin axis (i.e. two poles of spin axis could be identical, no CP violation) and therefore no eEDM on the spin axis will be ever discovered, although each pole could be anisotropic compared to the rest of the spherical charge. – Markoul11 Jul 29 '22 at 18:04
  • Using representation theory, the electron charge manifold could be similar to the Earth. A tiny wider on the equator generating an E monopole interaction field with its environment that is not perfect spherical. – Markoul11 Jul 29 '22 at 18:37
  • A "wide equator" corresponds to an oblate spheroid. That's a quadrupole moment, not a dipole moment. To have nonzero quadrupole moment, a quantum system must have spin $\hbar$ or larger. – rob Jul 29 '22 at 21:17
  • "A "wide equator" corresponds to an oblate spheroid. That's a quadrupole moment, not a dipole moment. To have nonzero quadrupole moment, a quantum system must have spin ℏ or larger" Not if the the spheroid is simply connected to its two poles. https://en.wikipedia.org/wiki/Simply_connected_space – Markoul11 Jul 30 '22 at 12:45
  • like this for example: https://tinyurl.com/249wss4j – Markoul11 Jul 30 '22 at 12:57
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    Oh, I see. Your spiral torus thing is related to the "anapole moment," which is sometimes used in discussions of parity violation. You are also misusing the term "simply connected." – rob Jul 30 '22 at 14:13
  • In my previous post I meant "Non-Simply Connected" and not "Simply Connected". Sorry. – Markoul11 Jul 31 '22 at 18:28

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