2
$\begingroup$

The question of how $N$ electrons (seen as point charges) on a conducting sphere will arrange themselves in the electrostatic final state was first posed by J.J. Thomson in 1904--hence, aka the Thomson Problem. If these abstract point charge electrons are initially placed randomly, they will migrate along potential gradients to a state with a locally minimum potential energy. The Thomson Problem is seen as finding the geometrical arrangement of the N charges with the global minimum potential energy. Unfortunately there are usually a great number of local minima, e.g. on the order of $10^6$ for $N$ of several hundred, so numerical techniques don't necessarily produce the global minimum and analytic techniques to date have only solved the problem for some small values of $N$.

Notwithstanding the applications of the problem to many other practical phenomena, I have some general questions about the specific real case of electrons on a conducting sphere. If the sphere initially has a random distribution of excess charge ($N$ electrons), will they in fact somehow end up in the global minimum potential energy state, or will they as in a numerical simulation just find a local minimum and be stuck there? Is there any way to know that? If so, how do they do it?

Another question: is it valid to think of the electrons as ultimately stationary points on the vertices of some geometric arrangement on the sphere in the first place? I.e., given quantum effects, statistical considerations, etc.

$\endgroup$
  • 1
    $\begingroup$ To eliminate questions on quantum effects, you might try asking this with charges that are obviously classical, like charged bits of dust or plastic spheres. Whether the charges adopt a local minimum depends on initial conditions; for example, if placed directly in a local minimum, they are likely to stay there. Calculating this likelihood involves comparing the energy required to escape the local minimum to the kinetic energy the charges have when they enter the minimum. Also, in general, it shouldn't be surprising that nature sometimes does seemingly uncomputable things (e.g. turbulence). $\endgroup$ – probably_someone Jan 18 at 12:02
  • 1
    $\begingroup$ Ah, but the last thing I'd want to do is eliminate quantum or any other effects, because I'm interested in what real electrons will do, not some idealized abstraction of ball bearings rolling down a landscape or what have you! There is such interest in the problem with practical application to virus forms, fullerenes, etc., there is an implication I think that the geometry with the global minimum energy is somehow attained in real world systems. $\endgroup$ – Joe Knapp Jan 18 at 12:43
  • 1
    $\begingroup$ Well, I can tell you already that the global minimum energy is not always attained in real-world systems; otherwise, graphite wouldn't exist (diamond is a lower-energy configuration of carbon, but the conversion from graphite to diamond requires a large activation energy, and so only occurs under intense heat and pressure - therefore, most carbon remains in the "local minimum"). $\endgroup$ – probably_someone Jan 18 at 13:12
  • $\begingroup$ Good to keep in mind. Perhaps electrons behave differently. $\endgroup$ – Joe Knapp Jan 18 at 20:04
  • 1
    $\begingroup$ Another thing to keep in mind is that the Thomson problem doesn't really make sense if you start to include quantum effects. An electron doesn't have a well-determined location because its behavior is characterized by a wavefunction dispersed over the sphere, so it doesn't really make sense to talk about a geometrical arrangement in that case. Also, there's not really an "electrostatic final state" either, because the momentum of the electron is also never completely determined, and so it cannot ever be said to be totally stationary. $\endgroup$ – probably_someone Jan 20 at 13:39
1
$\begingroup$

If the sphere has no resistance, the initially distributed electrons will move in a complicated way and will never reach the local or global minima to stay - they will "oscillate".

In order to reach some minimum to stay, there should be losses of excess of energy.

If the resistance is high, the electrons may find local minima and stay there (the extreme case - an insulator).

If the resistance is weak enough, then reaching the global minimum is possible.

$\endgroup$
  • $\begingroup$ Yes, assuming a real case with non-zero resistance, say, copper. Would reaching the global minimum be assured? $\endgroup$ – Joe Knapp Jan 18 at 19:57
  • $\begingroup$ Yes, I think so. But in reality, electrons in a conuctor are like a liquid - they are present everywhere, so the final distribution will be homogenuous rather than "singular". $\endgroup$ – Vladimir Kalitvianski Jan 19 at 5:56
0
$\begingroup$

The question is a bit ambiguous.

On a real conductor (like a copper ball) there are lots of movable charges. All the movable charges – the original and the added ones – will move to a position minimizing potential energy. If the conductor was made of metal, it would not be possible to distinguish between the original and the added charges, so this would not give a solution to the Thomson problem. But even if you could mark the added charges in some way, their charge would be shielded by the movable charges of the conductor and they could move freely. So, this would not solve Thomson problem either.

What you might have thought of would be a sphere, in which the added charges can move freely, but in which other charges do not exist. To make the experiment complete, the sphere would never the less have to be constructed in a way, that the electrons can not leave it. If such a sphere could be constructed – might be for some other charged particles but electrons – these charges would tend to find local optima of the minimization problem.

Quantum effects would be relevant only, if the sphere is sufficiently small and the charges are sufficiently lightweight. Another practical problem using electrons or similar particles would be cooling. If you do not remove energy form the system, the particles will not move to the minimum anyway or not stay there. For larger particles like charged table tennis balls the problem might be solved however. Never the less this will be more difficult than solving the problem numerically and again it will find local optima only.

$\endgroup$
  • $\begingroup$ This answer is wrong. This is not how, I quote the OP, "electrons on a conducting sphere" behave. $\endgroup$ – my2cts Jan 18 at 17:57
  • $\begingroup$ Thanks--since these are electrons, seems like appreciable quantum effects are possible? For example, tunneling from one local minimum to another at a lower energy state without having to go over the hump. The thing is, the electrons would have to do so in a coordinated fashion as the potential energy is the sum of all interactions. $\endgroup$ – Joe Knapp Jan 18 at 20:03
  • $\begingroup$ @my2cts: I wondered whether my answer is already too long. So could you be a bit more specific, about what is wrong in it or what part of the behavior is not like that of real electrons? Thanks. $\endgroup$ – A.Degenhardt Jan 24 at 16:18
  • 1
    $\begingroup$ @JoeKnapp: The "coordinated" tunneling of multiple electrons would not be a problem. Multi particle systems are described by their coordinates in product space in QM (3 coordinates for each particle) without distinguishing between coordinates that originate from the same or from different particles. Regarding the quantum effects you would have to make sure that your sphere is not to small, in which case the electrons would be have more like electrons in a real atom. Main problem would be a constructing a sphere binding "your" electrons, without having extra electrons. $\endgroup$ – A.Degenhardt Jan 24 at 16:34
-3
$\begingroup$

If the sphere is a good conductor then the electrons will be all over the sphere. If it is an insulator then they will remain where you put them.

$\endgroup$
  • 1
    $\begingroup$ This doesn't really answer the question, as the OP was asking about a specific number of electrons, and their specific distribution "all over the sphere," which is the substance of the Thomson problem. It's definitely not enough to say that they will simply be "all over the sphere," given that context. $\endgroup$ – probably_someone Jan 18 at 14:46
  • $\begingroup$ @probably_someone You should read the OP's question again. My post constitutes a complete and correct answer. The question is not about classical charges but about " the specific real case of electrons on a conducting sphere." $\endgroup$ – my2cts Jan 18 at 17:55
  • $\begingroup$ The question is where the excess electrons (excess over the neutral state) are positioned, per the Thomson problem: en.wikipedia.org/wiki/Thomson_problem $\endgroup$ – Joe Knapp Jan 18 at 19:55
  • $\begingroup$ @Joe Knapp The answer is : spread out thinly over the sphere, as it is a conductor. $\endgroup$ – my2cts Jan 18 at 20:41
  • $\begingroup$ E.g., "Global minimum for Thomson's problem of charges on a sphere" "Using numerical arguments, we find that for N=306 a tetrahedral configuration (T(h)) and for N=542 a dihedral configuration (D5) are likely the global energy minimum for Thomson's problem of minimizing the energy of N unit charges on the surface of a unit conducting sphere." ncbi.nlm.nih.gov/pubmed/15903830 $\endgroup$ – Joe Knapp Jan 18 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.