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I am wondering if the mass density profile $\rho(\vec{r})$ has been characterized for atomic particles such as quarks and electrons. I am currently taking an intro class in quantum mechanics, and I have run this question by several professors. It is my understanding from the viewpoint of quantum physics a particle's position is given by a probability density function $\Psi(\vec{r},t)$. I also understand that when books quote the "radius" of an electron they are typically referring to some approximate range into which an electron is "likely" to fall, say, one standard deviation from the expectation value of its position or maybe $10^{-15}$ meters.

However it is my impression that, in this viewpoint, wherever the particle "is" or even whether or not the particle "had" any position to begin with (via the Bell Inequalities), it is assumed that if it were (somehow) found, it would be a point mass. This has been verified by my professors and GSIs. I am wondering if its really true.

If the particle was truly a point mass then wherever it is, it would presumably have an infinite mass density. Wouldn't that make electrons and quarks indistinguishable from very tiny black holes? Is there any practical difference between saying that subatomic particles are black holes and that they are point masses? I am aware of such problems as Hawking Radiation although at the scales of the Schwarchild radius of an electron (back of the envelope calculation yields $\tilde{}10^{-57}$ meters), would it really make any more sense to use quantum mechanics as opposed to general relativity?

If anyone knows of an upper bound on the volume over which an electron/quark/gluon/anything else is distributed I would be interested to know. A quick Google Search has yielded nothing but the "classical" electron radius, which is not what I am referring to.

Thanks in advance; look forward to the responses.

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  • $\begingroup$ I think the density will not have symetric form. It will have veriety density according to its speed. And also it will have a singularity in the middle of the particle, hard to determine where is the middle of the particle. And more impotantly, we don't know how fermions are shaped. $\endgroup$
    – Xiang
    Commented Apr 19, 2012 at 3:22
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    $\begingroup$ Is there any reason to suspect that it wouldn't? Either way that is why I wrote it as $\rho(\vec{r})$ instead of $\rho(r)$; $\vec{r}=[x,y,z]$ and can take on any value (at least that's the convention we've been using). $\endgroup$
    – clevy
    Commented Apr 19, 2012 at 3:26
  • $\begingroup$ I didnt say it would not. I just said we dont know what fermions really seems like. You can assume It will strongly relate with atomic spin. $\endgroup$
    – Xiang
    Commented Apr 19, 2012 at 3:38
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    $\begingroup$ related: physics.stackexchange.com/q/7322 $\endgroup$
    – user4552
    Commented Jul 7, 2013 at 3:04
  • $\begingroup$ The 3-dimensional Dirac-Delta Function . $\endgroup$ Commented Aug 5, 2013 at 1:42

4 Answers 4

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Let me start by saying nothing is known about any possible substructure of the electron. There have been many experiments done to try to determine this, and so far all results are consistent with the electron being a point particle. The best reference I can find is this 1988 paper by Hans Dehmelt (which I unfortunately can't access right now) which sets an upper bound on the radius of $10^{-22}\text{ m}$.

The canonical reference for this sort of thing is the Particle Data Group's list of searches for lepton and quark compositeness. What they actually list in that reference is not exactly a bound on the electron's size in any sense, but rather the bounds on the energy scales at which it might be possible to detect any substructure that may exist within the electron. Currently, the minimum is on the order of $10\text{ TeV}$, which means that for any process occurring up to roughly that energy scale (i.e. everything on Earth except high-energy cosmic rays), an electron is effectively a point. This corresponds to a length scale on the order of $10^{-20}\text{ m}$, so it's not as strong a bound as the Dehmelt result.

Now, most physicists (who care about such things) probably suspect that the electron can't really be a point particle, precisely because of this problem with infinite mass density and the analogous problem with infinite charge density. For example, if we take our current theories at face value and assume that general relativity extends down to microscopic scales, an point-particle electron would actually be a black hole with a radius of $10^{-57}\text{ m}$. However, as the Wikipedia article explains, the electron's charge is larger than the theoretical allowed maximum charge of a black hole of that mass. This would mean that either the electron would be a very exotic naked singularity (which would be theoretically problematic), or general relativity has to break at some point before you get down to that scale. It's commonly believed that the latter is true, which is why so many people are occupied by searching for a quantum theory of gravity.

However, as I've mentioned, we do know that whatever spatial extent the electron may have cannot be larger than $10^{-22}\text{ m}$, and we're still two orders of magnitude away from probing that with the most powerful particle accelerator in the world. So for at least the foreseeable future, the electron will effectively be a point.

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    $\begingroup$ I read that the electron has a gyromagnetic ratio not consistent with a rotating ball of uniformly distributed charge and the mass and that one possible solution was a ball of uniform mass with charge on the surface. See physics.stackexchange.com/questions/224675/… $\endgroup$ Commented May 3, 2016 at 0:55
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    $\begingroup$ What if the electron is a naked singularity? What then? $\endgroup$
    – ikrase
    Commented Apr 3, 2021 at 1:25
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    $\begingroup$ Could the electron be banana-shaped, or have any other shape, as long as it is less than 10^-22 m? $\endgroup$
    – Juan Perez
    Commented Nov 14, 2021 at 0:42
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    $\begingroup$ @JuanPerez As far as we know, sure. If it were a very tiny banana, there's no experiment currently known that would allow us to distinguish it from a very tiny sphere. (BTW that $10^{-22}\ \mathrm{m}$ figure, or whatever the most up-to-date value is, is not a hard cutoff; there are a lot of probabilities involved.) $\endgroup$
    – David Z
    Commented Nov 14, 2021 at 2:20
  • $\begingroup$ Exactly, why cannot electron and quarks be naked singularities? $\endgroup$
    – Jus12
    Commented Jul 30, 2022 at 17:48
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David Zaslavsky has given a solid, relatively model-independent explanation of the empirical bounds on the size of an electron based on particle-physics experiments that probe short distance scales by using collisions at short wavelengths. There is also another way of getting at this question, which has been studied by people who have tried to model quarks and leptons as composites of more fundamental particles called preons. If the preons are confined to a space of linear size $x$, then the uncertainty principle says their mass-energy is at least about $\hbar/x $. But given even a relatively weak bound on $x$, this makes the mass-energy of the preons greater than the mass of the electron they supposedly make up. This is called the confinement problem. Various people (e.g., 't Hooft 1979) have worked out various possible ways to get around the confinement problem, but essentially the confinement problem makes these ideas unlikely to work out.

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An electron being a ball of uniform mass and charge is not consistent with its observed gyromagnetic ratio.

The charge must be pushed out and the mass must be pushed comparatively inwards to satisfy the existing ratio of about 2.

See Classical proof of the gyromagnetic ratio $g=2$

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  • $\begingroup$ This is modeling the electron classically, which doesn't work. People gave up on this sort of thing ca. 1905. $\endgroup$
    – user4552
    Commented Nov 9, 2019 at 1:11
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substitute into Hawking equation to establish Schwartzchild area for a unitary state thus: U_A=4Ghbar*c^2 and assume electron a single state particle. So, its Euclidean radius 1.4190e-14 meters, volume 1.19686e-41 m^3 and Euclidean density 7.6110e10 kg-m^-3

Just a stab in the dark.

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