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I've asked this question How much electrons can absorb a single photon? , but guys contradicted each-other, in the question about how much virtual photons does rest electron emits. One of them said infinite, another say one.

First, as I understood virtual photons are not an energy, so, when electron emits them, it doesn't lose energy. Hence virtual photons, do not obey the statement, that the energy emits only by quants.

Let's consider the classic electric field of fixed charge. It creates forces around, in all direction. But what did I understand, when nothing interacts with it, there is no field, and forces, correct me, if I wrong. When something interacts with electrons, the field is happen, but only that part that interacts with something. To prove, that the electron emits more than one virtual photon, consider the case below

enter image description here

Positron(or electron too, no matter), electron and electron, situated at one straight line. According to Coulomb's law the result will be, that the electron and positron will attract, and electron and electron will repel instantly, which means, that there are two virtual photons, because one virtual photon can't move in two opposite directions.

So is it correct, that a rest particle emits two and more virtual photons?

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    $\begingroup$ Sorry, but you've been misled by popsci. The question you're asking simply doesn't make any sense in terms of the actual mathematics of quantum field theory. The people who answered your previous question don't know the math either. Everybody there is just kind of making stuff up on the spot, the blind leading the blind. $\endgroup$ – knzhou Sep 11 '18 at 8:58
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    $\begingroup$ If anybody ever tries to tell you anything about virtual particles, but doesn't precede it with at least three paragraphs about how the whole notion is fraught with peril, don't trust them! If they can't think of any equations besides the uncertainty principle, just walk away! $\endgroup$ – knzhou Sep 11 '18 at 8:58
  • $\begingroup$ @knzhou Maybe You can explain then? Even, if there are a lot of hard math or something else $\endgroup$ – user205695 Sep 11 '18 at 9:00
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    $\begingroup$ You cannot really count virtual particles like this. A proper treatment of problems like these involves either writing down a perturbation series of Feynman diagrams and doing the math, or treating the EM field as continuous and doing the math. Even in the QED treatment of Feynman diagrams, I don't think there is any proper way to define anything like a 'virtual photon number'. $\endgroup$ – Stijn B. Sep 11 '18 at 9:48
  • $\begingroup$ @StijnB., but I don't want to count them. I just asked is there only one virtual photon per time, or more $\endgroup$ – user205695 Sep 11 '18 at 11:53
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First, virtual photons are not real, they are a mathematical method of describing the (in your case) stationary electric field around the electron.

This field exist around the electron. It is something that QM/QFT tries to describe mathematically, because in real particles terms, they are not describable.

Virtual photons do not obey some laws, for example they do not obey the speed limit c.

Virtual photons are a way of describing the interaction between the electron (charge) field and the other particles' fields, for example a proton's field (strong force field because of quarks make up the proton). The proton field interacts with the electron field by exchanging virtual photons. This is how protons and electrons attract.

But this interaction, and the exchange of virtual photons is what confuses you. This is not a real particle exchange. There are no real particles in this case exchanged.

Now this interaction is described in math as the exchange of virtual photons. But we do not know whether something is actually exchanged between a proton and the electron.

We do know that the proton and the electron do interact electromagnetically (and gravitationally too).

Now this EM interaction between the proton and the electron is what causes them to attract. We describe this interaction with the exchange of virtual photons. These are not real photons. They are off mass shell. They do not obey the speed limit c.

Now you are asking how many virtual photons are exchanged between the proton and the electron. You are confused because you learned that EM energy is quantized. The smallest piece of this EM energy is the real photon. Not the virtual photon.

Both the proton and the electron have a static EM field around them. Your question is whether this static EM field is quantized.

Now even if you would try to measure this static field around the charge, and then divide it into pieces, and call them virtual photons, this would not be real. In reality, the electron or the proton, neither of them are emitting anything in this case.

The static EM field around the charges is not created by emitting real particles. We cannot count them, because no real particle is emitted.

What we do know, is that the static EM field around the charge is somewhat similar to gravity. Let's see how similar.

Now stress-energy is the source of gravity. This creates a gravitational field around a mass. How many virtual gravitons does the Sun emit instantly, per time? Similarly, the Sun does not emit virtual gravitons. This is just a theoretical (and experimentally not yet proven) way to describe the gravitational effects of stress-energy (mass).

So how do we explain that gravity (stress-energy) interacts with other particle's fields? We say that gravity bends spacetime. This is the way we describe it instead of virtual gravitons. But the reality is, that gravity has an effect on the fabric of spacetime, so that anything that interacts with the gravitational field, will be affected by it.

Similarly, when there is a charge, like an electron in your case, and it has a static EM field around it, anything that interacts electromagnetically, will be affected by it. Do we say similarly that EM bends spacetime? Actually we do. But it is indirect. What we can say is that the charge has an effect on the fabric of spacetime and anything that interacts with the EM field will be affected by it.

So just as stress-energy (or the mass of the Sun) in reality does not emit virtual gravitons, and we cannot count the virtual gravitons, similarly, we say that in reality the electron in your case does not emit virtual photons, and we cannot count them.

So both the electron and the proton create a field that has an effect on the fabric of spacetime, and these fields interact with each other. This is how they attract. Mathematically we describe them as an exchange of virtual photons. In reality, no real particle is exchanged.

The EM field is quantized, but virtual photons are not the real quantization of the static EM field around the electron and the proton.

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You may want a simpler explanation, depending on the level of depth you are looking for. To be able to say how many photons are emitted per time, you need to to be able to count the photons. Counting means that there is a formula you can use that tells you how many particles there are. In QFT one counts the number of particles by applying the so-called number operator on the state that describes the particle. Now, states describe external particles; the virtual particles that you talk about are a theoretical concept and they have no corresponding states. Virtual particles are a name given to a concept that originates from perturbation theory; they have no corresponding state. As a result you cannot count them because there is nothing the number operator can act upon. This answer glosses over many details and assumptions, maybe most importantly the question ``what is a particle?'' But this is a whole other can of worms.

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  • $\begingroup$ Я вот не понимаю вообще, как электрон, который покоится, и на которого ничто не влияет, начинает притягивать\отталкивать другие покоящиеся частицы? Другая частица влияет на электрон? Но что повлияло на неё, чтобы она повлияла на электрон? Это бессмысленно $\endgroup$ – user205695 Sep 12 '18 at 15:45
  • $\begingroup$ Artur, your questions are good. To answer them you need to realise that these phenomena occur at microscopic scales and humans have not necessarily evolved (neither our senses nor our brains) to have an intuitive understanding of them. We therefore use words and concepts that we are familiar with in our macroscopic world (particle, attract, influence etc) to describe them. But these are only associations that have no deeper meaning. The language we have to describe these phenomena is mathematics. How electrons interact is described by mathematics of field theory. That's really all there is! $\endgroup$ – Oбжорoв Sep 13 '18 at 20:00
  • $\begingroup$ The structure of electric and magnetic fields and of photons is not under consideration. I’ve found a model explaining the interaction between charges. A dry written foundation about Complex one-dimensional structures of space and Are photons composed particles. $\endgroup$ – HolgerFiedler Dec 20 '18 at 19:05
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A single electron (or a charge) in empty space with its Coulomb field is an abstraction. The field makes sense if it stands in the equation of motion of some other charge, as an external field (force).

Generally the virtual photons are associated with so called near field - it is a retarded field that doesn not propagate to infinity. It decays with distance as $1/R^2$. It may create bound states of charges if it is attractive. Note, each charge is a source of its own field, which is external for the other charges.

Now, this retarded near field of an electron can be decomposed into harmonics with different frequencies and those harmonics may be called "virtual photons". Even for Coulomb field $\vec{E}(\vec{R})\propto 1/R^2$ there are many virtual photons, all with the frequency zero, but with different wave vectors $\vec{k}$. For a uniformly moving electron the frequencies are not zero due to time dependence of the electron field.

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  • $\begingroup$ Can you explain the last sentence? If frequency of photon is not zero, hence energy non-zero, but as I know, the uniformly moving electron doesn't lose its energy. $\endgroup$ – user205695 Sep 11 '18 at 8:53
  • $\begingroup$ The near field is not radiated; it follows the charge; thus "virtual" instead of real photons. $\endgroup$ – Vladimir Kalitvianski Sep 11 '18 at 9:43
  • $\begingroup$ How it can "follows" charge, if anything can propagates only maximum with finite $c$ speed? $\endgroup$ – user205695 Sep 11 '18 at 11:49
  • $\begingroup$ The near field is "attached" to a charge and moves with the charge with the charge velocity if there is no acceleration. Read Electrodynamics textbooks before asking how many virtual photons.... $\endgroup$ – Vladimir Kalitvianski Sep 11 '18 at 13:38

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