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If one models the electron as a hollow spherical conductor with charge $e$ and radius $a$ then its electrostatic energy is given by:

$$E_{em}=\frac{1}{2}\frac{e^2}{4\pi\epsilon_0a}$$

However if one calculates the momentum in the field of a moving electron then one finds that the total mass in the field is given by:

$$m_{em}=\frac{2}{3}\frac{e^2}{4\pi\epsilon_0c^2a}$$

Therefore we have a discrepancy in energy given by:

$$E_p = \frac{1}{6}\frac{e^2}{4\pi\epsilon_0a}$$

Poincare hypothesised that there must be stresses holding the electron together against the electrostatic repulsion of the charge on its surface. Somehow it must take an energy $E_p$ in order to maintain these Poincare stresses.

Perhaps the electron can be modelled by a conducting spherical shell with a vacuum inside it. Presumably the vacuum would lead to a negative pressure on the charged shell due to the Casimir effect. This pressure must balance the electrostatic repulsion of the charged shell.

According to cosmological models, for example, the vacuum has an equation of state given by:

$$p = -\rho c^2$$

This expression can be justified on the grounds that the stress-energy tensor of the vacuum must be Lorentz invariant.

The outward pressure on the surface of the charged sphere due to Coulomb repulsion is given by:

$$p = \frac{1}{2}\epsilon_0\left(\frac{e}{4\pi\epsilon_0a^2}\right)^2$$

This pressure must be balanced by the negative pressure of the vacuum inside the shell. If we substitute into the above equation of state for the vacuum inside the sphere we find that its energy is given by: $$E_p=\frac{1}{6}\frac{e^2}{4\pi\epsilon_0a}$$

Thus we seem to have accounted for the energy discrepancy between the EM field of a static electron and a moving electron by including the energy of the vacuum inside the electron holding it together.

But in fact the inward pressure on the electron is due to the Casimir effect. This means that it is due to an excess of zero-point electromagnetic modes outside the conducting shell compared to the number of modes inside. Thus the extra energy $E_p$ associated with these extra modes is located outside the shell. This makes sense as we want to account for the discrepancy in the total mass/energy in the field outside the shell.

One can make the following close analogy with the case where one pulls a piston out of a cylinder that is surrounded by normal atmospheric pressure. One has to supply energy to do work against the outside atmosphere. The energy supplied is not stored in the vacuum created in the cylinder; instead it is located outside in the surrounding atmosphere.

Is this the correct way to think about Poincare stresses?

I have just found a very interesting paper that argues against my hypothesis (actually Casimir's hypothesis!) of a classical electron held together by zero-point energy. The author calculates the Casimir forces on a spherical conductor from first principles and concludes that they are repulsive rather than attractive.

But why does this result contradict the well-known vacuum equation of state $p=-\rho c^2$?

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    $\begingroup$ The assumption (problem with em. mass) is erroneous; cf. my answer here: physics.stackexchange.com/questions/160264/… $\endgroup$ – Ján Lalinský Mar 9 '15 at 4:28
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    $\begingroup$ Can you make explicit what exactly you believe the contradiction is? $\endgroup$ – Leandro M. Mar 9 '15 at 18:56
  • $\begingroup$ Using the vacuum equation of state $p=-\rho c^2$ one can derive the energy discrepancy $E_p$ correctly whereas when a careful calculation is done of the Casimir forces due to the vacuum inside the sphere one finds instead that they tend to expand the sphere instead of holding it together. $\endgroup$ – John Eastmond Mar 10 '15 at 7:59
  • $\begingroup$ To Jan Lalinsky: I would say that the discrepancy between the static field mass/energy and the moving field mass/energy still needs to be accounted for. I accept your point that one should not include the inertia in the em field if one's equation of motion only deals with forces that act on the body itself. $\endgroup$ – John Eastmond Mar 10 '15 at 8:22
  • $\begingroup$ Well, assuming that the equation of state is applicable (it's not obvious that it is but let's roll with it) the problem lies with the assumption that the vacuum pressure is responsible for stabilizing a charged shell. It clearly cannot be, because the Casimir force is repulsive. Therefore some other force must be responsible for equilibrating both the Coulomb repulsion and the vacuum pressure. $\endgroup$ – Leandro M. Mar 10 '15 at 15:19
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It is indeed the case that the Casimir force for a three dimensional sphere is repulsive.

http://arxiv.org/abs/hep-th/9406048

Neither myself nor the authors of this paper know an intuitive explanation of why this is the case. It is important to remark, however, that this result actually disproves the naive and very common "mode counting" argument that seemingly works in the Casimir effect for a parallel plate geometry.

The way you derive the pressure properly is to write down the zero point energy as a sum over the modes. This sum is not meaningful by itself because it's typically badly divergent. You have to regularize it, that is, extract a finite number from the divergent, infinite series. What you get after regularizing the sum might be positive or negative. There's no obvious way to tell.

For instance, in the parallel plate version of the Casimir effect you encounter the famous series

$$1+2+3+4+...=-\frac{1}{12}$$

This result may be obtained by zeta function regularization. The negative sign results in an attractive force.

However, for a related series, zeta function regularization will give

$$1+2^3 + 3^3+ 4^3+...=1+8+27+64+...= - \frac{B_{3+1}}{3+1}=\frac{1}{120}$$

If you found this series when calculating the Casimir effect for some geometry, you would find a positive pressure.

And if you found this other series

$$1+2^2+3^2+4^2+...=1+4+9+16+...=0$$

there would be no pressure at all.

The regularization involved in calculating the Casimir effect for a sphere is a little more involved but this is the general principle. Also see Terry Tao's take on it to get a little more intuition on what the regularized sums mean and why they're the correct answer.

In other words, there is no contradiction, because the mode counting argument fails for all but the simplest geometries.

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I don't think that would be possible since there is vacuum outside of the electron as well, everything between the electron and the nucleus is vacuum.

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  • $\begingroup$ Since free electrons are stable, I'm not sure what the nucleus has to do with anything. $\endgroup$ – Jon Custer Mar 8 '15 at 17:08
  • $\begingroup$ @JonCuster It doesn't, I didn't said it has. My point is that in this case, not only the vacuum at the center of the electron would hold it together, but the vacuum outside would be doing just the opposite. $\endgroup$ – Ivan Lerner Mar 8 '15 at 17:15
  • $\begingroup$ The idea is that only certain zero-point electromagnetic standing waves are allowed inside the electron whereas all possible waves are allowed outside. Thus the "true" vacuum outside the electron exerts a net pressure on it. $\endgroup$ – John Eastmond Mar 8 '15 at 19:08
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Influence of internal EM forces on the sphere cannot be calculated based on the field momentum outside the sphere - it is more complicated since the field is not simply moving with the sphere when the sphere moves in a general way.

Also, the electromagnetic defect is present, but it is explainable purely with mutual EM forces. No effect of zero-point field is necessary.

Also, Poincaré stresses are necessary to keep the sphere together, but not for explanation of the alleged problem with em. mass; the sphere does not need to be stable for the alleged problem to occur. It is sufficient that the charges have extremely large mechanical mass, so the sphere survives for long enough; then the Poynting energy and the Poynting momentum already lead to different EM mass.

I do not think there is any good reason to introduce the equation

$$ p = -\rho c^2 $$

for the space inside or outside the charged sphere. Besides the fact there is no apparent problem with mass of the sphere, the way this equation was arrived at in the gif file referenced above is dependent on a series of assumptions that are just off mark in the present context. For example, it is assumed that vacuum has stress four-tensor that is different from the EM tensor and this tensor has the form of standard tensor for perfect fluids (which was introduced for material fluid only, not vacuum), and that all components of the tensor are Lorentz-invariants. Why should we think that? There is just no reason whatsoever to assume such weird ideas in the context of charged sphere. And even if we do assume the above equation is valid, this brings nothing to the question of effective mass of the sphere.

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  • $\begingroup$ The vacuum equation of state $p=-\rho c^2$ can be justified on the grounds that it is Lorentz invariant so one does not require gravitational considerations. $\endgroup$ – John Eastmond Mar 9 '15 at 9:31
  • $\begingroup$ @JohnEastmond, I've edited my answer. $\endgroup$ – Ján Lalinský Mar 9 '15 at 18:42

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