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Note:

I acknowledge that this thought experiment is rather crude due to it's lack of quantum mechanical emphasis however I am really looking for a more principled explanation if possible. All I kindly ask is that you try to peel through my words in attempt to understand my point if I did not correctly represent the question; after all I am not really sure how else to ask it.

Intro:

Suppose there was a proton with a mass and charge denoted by $m_{p}$ and $e_{+}$ respectively and one could calculate the electric field energy of the proton by $\frac{\epsilon_{0}}{2}\int_{R}^{\infty}\left | \textbf{E}\right |^{2}dV = \frac{\epsilon_{0}}{2}\int_{R}^{\infty}\left | \frac{e+}{4\pi \epsilon_{0} r^{2}}\right |^{2}4\pi r^{2} dr = \frac{e_{+}^{2}}{8 \pi \epsilon_{0}R}$ and the rest mass energy by $m_{p}c^{2}$.

Question:

How exactly does the electric field energy of the proton relate to the mass-energy of the proton?

Motivation:

This question is motivated by the fact that if an electron can simply cancel a proton's electric field assuming they were separated by an infinitesimal distance (again crude), the total energy would just be $\left (m_{p} + m_{e} \right )c^{2}$ and yet such a reduction in electric field energy would seem to reduce the mass energy of both particles since the electric field energy is (assumed to be) encompassed within it.

*This zero reduction in mass-energy seems to contraction the notion that the electric field energy of some particle is contained the mass energy.

In reality I do understand that the information regarding the field's existence is not lost (which may ultimately answer my question) however I am unsure how to rectify the relationship between the electric field energy and the mass-energy in a straightforward way and any guidance would be much appreciated.

Thanks in advance you ridiculously amazing people.

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I think I have an answer to your question, assuming I understood what you are asking. Therefore I will express the question as I understand it and answer that question – and hopefully this will answer yours too. First though, I’m going to assume a static situation where charges are always at rest, or moving very slowly.

The set-up of your question uses reasoning along the following lines: the electric field can be shown to have an energy density given by u = (1/2)εoE2 Since this field takes up space, we can calculate the total energy contained in a volume of space. Since energy and mass have equivalence in relativity theory, the field associated with a charged particle should contribute to the mass of the particle.

So the question is, IF the electric field associated with any charged particle is reduced – say, by bringing an oppositely charged particle nearby, shouldn’t the mass of this new, 2-particle system be lowered as well? After all, hasn’t the total electric field, the vector superposition of the two constituent fields, now been reduced?

The mass of this 2-particle system is reduced, exactly according to your reasoning: that essentially, you have reduced the field, and since the field contains energy and therefore mass, less field means less mass.

But consider more precisely what you have done by allowing two oppositely charged particles to come together: You have created a bound system, a system that would require energy be added to it to separate the particles. The bound system therefore is at a lower energy and therefore possesses less mass than the sum of the separated constituents (This effect by the way, is measurable in nuclear reactions).

So you have reduced the field energy of the system in your scenario. In a sense, your question and thought process seems to perhaps try to separate the Newtonian mass – approximately what the charge particle would weight without the charge – and the field-energy mass. But I ask you to consider the opposite scenario: You start separating 2 oppositely charged particles that initially, were bound together. As you separate the particles, you are ‘creating electric field’ since the field cancellation mechanism decreases with separation. These fields possessed energy, and therefore mass, and this mass is appearing because you are doing work – by supplying energy to the particles. You are unbinding them, and the price of their freedom is energy, and this energy appears as mass.

But there are interesting implications here too. Far from say, a proton, there is an electric field. It seems this field is contributing to the proton’s mass, and it is distributed in space. My guess would be that we can’t peel away the field from the mass (or vice versa) in a proton or an electron, certainly not easily.

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  • $\begingroup$ Wow, I will need time for this to sink in deeper but you definatly elevated my understanding. I never considered that I need to put energy back into the system in order to get your electric field back! Very elegent. Great response, thankyou $\endgroup$ – spacetimeengineer May 22 '17 at 19:36
  • $\begingroup$ This doesn't look right to me. If this were correct, then it would take infinite energy to create an electron-antielectron pair, because the energy stored in the fields of these pointlike particles would be infinite. $\endgroup$ – Ben Crowell May 22 '17 at 21:37
  • $\begingroup$ @BenCrowell, it would take infinite energy if the Coulomb formula was valid all the way. But if it is valid only for distances greater than some limit $r_{min}$, and another dependence on distance is valid for distances smaller than that, the total energy needed to separate positive and negative electron may be finite. $\endgroup$ – Ján Lalinský Oct 3 '17 at 9:27
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I'm not sure whether this answers the question you're asking because I'm not 100% sure what you're asking. But if I guessed your question right, the apparent dilemma is resolved conceptually when you consider the effects of accelerating a charged particle.

Suppose you're a proton, and you feel the tug of attraction of a stationary electron. If someone now gives the electron a push, you will not notice it until light has had time to reach you, since no information travels faster than light. A proton closer to this electron will notice the acceleration before you do. This means that there is a period of time when you and another proton are being pulled in different directions, even though you're being pulled towards the same point particle. This means there must be a distortion in the field surrounding the electron. This distortion travels at the speed of light and is indeed an electromagnetic wave or a photon. Accelerating charges therefore generate EM radiation.

If we pretend an electron is going to fall towards a proton until it is within a defined distance, the electric field energy of both particles will propagate away as EM radiation as they accelerate towards each other. By the time they get within a defined radius all of the field energy outside that radius will have propagated away, and you'll be left only with the remaining mass energy. Supposing they don't touch, but form a dipole instead, the remaining field energy is a matter of how close you want to look around one of the charges. The "infinity" term in your integral is just for practical purposes, it designates whatever you're willing to quantify as being "far enough" from the field source. You will find field energy very close to one of the poles, but that energy is cancelled out by the time the radius you care about encompasses both charges... thus from the outside world, it just looks like a neutral object with the combined mass of a proton and electron.

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  • $\begingroup$ Actually I might be slightly wrong about that. Just as most of the mass of a proton or neutron is in the energy of the bonds of the quarks that form it, it seems the proton-electron dipole should have more mass due to the energy of their bond. However, I think this would be negligible since it couldn't be more energy than the ionization energy of hydrogen (13.6 eV), where the mass of the electron is equivalent to 511,000 eV. But if we don't have quantum to limit the ionization energy, I guess you'd have to add the energy it would take to separate the charges to their combined mass. $\endgroup$ – Ryan Franz Nov 28 '16 at 22:43
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You're making a seemingly reasonable assumption, but it turns out to be false. Although you start out by talking about a proton, and only later refer to electrons, the whole thing becomes clearer if you just talk about electrons. As far as we know, these are pointlike particles. Therefore the energy in an electron's electric field would be infinite. Special relativity says that energy has inertia, so the mass of an electron would therefore be infinite. This tells us that special relativity cannot self-consistently describe a pointlike charged particle.

In reality, when you get below a certain scale, you need to use quantum mechanics.

For more on this topic, read up on the classical electron radius.

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  • $\begingroup$ > Therefore the energy in an electron's electric field would be infinite. Only if you use the Poynting formula $E=\frac{\epsilon_0}{2}\int E^2dV$, which is not valid for point particles. For point particles, there is a consistent special relativistic theory where electromagnetic energy is given by another formula, one which gives 0 for EM energy of single point particle. Check my answer here: physics.stackexchange.com/questions/281739/… $\endgroup$ – Ján Lalinský Oct 3 '17 at 9:24
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For electron we can only estimate classical energy contribution from its electric field. Energy stored in the field of the electron is at least $\alpha m_e/2$, where $\alpha$ is fine structure constant (approximately equal $1/137$). We have integrated energy density around an electron from infinity up to the so called reduced Compton length of the electron ($386\ \mathrm{fm}$) i.e. to the limit of localisation of electron. So the answer is that minimum contribution of classical electromagnetic energy to the electron mass is $1/274$ of electron mass. Below the distance $386\ \mathrm{fm}$ (fermi) we have still divergences in calculations even on the field theory level. We cannot integrate classical energy formula to $r=0$.

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We know that energy stored in electromagnetic field is real. When we store energy in a capacitor that energy is $\frac12 E\cdot D\cdot V$, where $V$ is the volume of capacitor. For proton it is integral over $\frac12 E\cdot D$, where $E$ is electric field and $D$ is electric induction $= \epsilon_0E$. Energy of the electric field is real. We can discharge capacitor connecting it to the electric bulb and energy will be emitted by photons having certain mass equivalent energy. Now we treat proton as a uniformly charged ball with root mean square charge radius equal to $R= 0.87\ \mathrm{fm}$ (from scattering experiments). Performing integration from infinity to $r=0$ we obtain energy of the field $$W= \frac{3}{5} \frac{ k e^2} {R}, $$ where $ k=\frac{1}{4 \pi \epsilon_{0} } $. It gives $W = 1.94\ m_e c^2 $ in terms of electron mass. This is an approximate contribution of electric field to proton mass. We cannot do the same with electron! We were allowed to treat proton classically because the value $ \frac{\hbar}{m_p c} = 0.2\ \mathrm{fm}$ related to lower limit of localisation of proton is well below proton charge radius.

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