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The question is: How to calculate the energy of electrostatic field of a newly created particle.

The usual formula involves integration 'all over space'. According to GR I believe that 'all over space' means the interior volume with radius c*t_now, where t_now is the lifetime of the particle at the time of calculation and c is the light speed.

Thus the energy stored in the field is a quantity dependent on time evolution.

this is a clarification of the question of PS: conservation of energy and shrinking matter

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The energy of electrostatic field is not involved in any exchanges so it is a meaningless notion. –  Vladimir Kalitvianski Jan 22 '11 at 14:31

2 Answers 2

classical physics doesn't have any well-defined framework to describe the "creation of particles". So you have to talk about the real world - or, theoretical speaking, the world of quantum field theory (or something that contains it) - to be able to discuss the creation of particles.

However, in quantum field theory, you can't separate the energy coming from the field from the energy coming from the particles themselves in such a sharp way. In particular, even the lightest charged particle - the electron - has mass equal to 511 keV. That's big, and even if you compute the whole energy of the electrostatic field at the distance of the classical electron radius, $2.82 \times 10^{-15}$ meters, or shorter (not far from the radius of the proton, by the way), you will obtain a smaller energy than the rest energy of the electron.

If you want to say that a (virtual) electron existed at least for a little while, you need to assume that the lifetime was at least the Compton wavelength of the electron (divided by $c$). That's about 100 times longer a distance. And even with this distance, the uncertainty of the energy is comparable to those 511 keV.

This leads me to the key point which is the uncertainty principle for time and energy.

If you determine the timing of the places where you measure the energy with a much better accuracy than the Compton wavelength of the electron (over $c$), then the uncertainty of the energy you measure will inevitably be much greater than the whole electron rest mass (times $c^2$). And if you determine the the timing with a worse accuracy, then you will inevitably include almost the whole electric field of the electron, and you will get the electron mass (times $c^2$) of 511 keV with a big accuracy.

At any rate, you won't be able to show any violation of the total energy which holds exactly in any system with a time-translational symmetry. You will be extremely far from finding such a contradiction: even if you could attribute parts of energy to fields and particles at every moment, as you suggested and which you cannot because of the uncertainty principle, you couldn't measure it with the same precision, and even if you could measure it, the variable mass of the electron at the "center" of the field could always compensate any violation of the conservation law, anyway.

So the really important message is that the total energy is conserved, it can be measured as long as we have a long enough time to measure it, and attributing energy to small pieces of space around a particle is not the right way to proceed - neither theoretically nor experimentally. Instead, one should ask what are the probabilities of various processes.

External and internal particles will always show that energy is conserved. However, energy can only be measured accurately if it is measured for a long enough time - longer than $\hbar/\Delta E$ where $\Delta E$ is the required precision. If you think about these matters, you will find out that you are trying to measure the amount and localization of energy - in space and/or in time - more accurately than the uncertainty principle allows.

Cheers LM

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Luboš Motl: Thanks for your response (I follow your TRF;) I firmly believe that a particle is not separable from its exterior field and the total energy of the ensemble is constant. The problem is that if we accept that the field spreads out in space at the speed of light then the field 'outside', using a maximal radius by a suitable convention, has an energy content increased over time, implying that the energy located in what we call 'particle' is expected to decrease over time. Even working with probabilities I arrive to the same conclusion. –  Helder Velez Jan 19 '11 at 14:27

According to the usual laws of physics, charge cannot be created or destroyed. So it's impossible to create a particle with charge q without annihilating a particle with the same charge, or also creating a particle with charge -q, or some combination of these.

Suppose you annihilated a particle with the same charge. Then your problem might better be described as one of "which particle should you attribute the electrostatic energy to, the annihilated one or the created one?"

And if instead it was a case of creating two particles with charges +q and -q, then your problem could be described as one of having to separate the electrostatic energies of the two particles.

So I don't see that there is any real problem with defining the energy of the electrostatic field in a particle. It's a matter of semantics. I think the universe has no opinion as to which particle we attribute the electrostatic energy.

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We can invent a time varying charge density $\rho (t)$ to model appearing and disappearing charge and watch the filed evolution. It seems to me I saw somewhere a problem of filed propagation from suddenly created charge. Of course, such a phenomenon does not occur in reality. As I said previously, the electrostatic filed energy is not involved in any exchange so its "evolution" does not matter. –  Vladimir Kalitvianski Jan 23 '11 at 19:44
    
The electrostatic field (or gravitational) detector work by energy transfer from the environment (the field) and the matter of the detector. As energy got transferred we can assume that field has energy. –  Helder Velez Jan 24 '11 at 18:58
    
@all The particle (or the pair) spent all his existence impressing the environment (vacuum, space,..) with a spreading field. When the particle (or pair) got annihilated (converted to gamma-rays) the interaction time is very short. The far field can not be instantly unset or reverted to the source.Thus the energy of the field 'is lost' into the space and keeps spreading. Do you agree that 'all over space' has to be only the space interior to the sphere with radius c*t_now ? –  Helder Velez Jan 24 '11 at 19:23

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