3

According to Quantum Electrodynamics, electrically charged particles interact by exchanging virtual photons. This answer gives a concise explanation of this process, which describes how the exchange of virtual photons creates repulsion and attraction between charged particles.

This got me thinking, if at any given time countless of photons are being exchanged between particles, why doesn't everything constantly shine? I know virtual photons exist for a brief moment and their only task is (instantly) traveling between charged particles to produce the electromagnetic field between them. But wouldn't charged particles be constantly emitting some sort of radiation? If so, what's the wavelength of this radiation?

Agnius Vasiliauskas
  • 13,467
Gabe12
  • 229
  • 1
    Indeed, a charged object does have an effect on your eye; for example, it pulls on your cells' electrons and neutrons. But to excite the actual photoreceptors, to see something, you need light of a given frequency, which only comes from charges shaking at that frequency. – knzhou Oct 30 '16 at 01:36
  • BTW charged particles are shining. We live in a world of thermic radiation. All the time every body around us receive and emit EM radiation. Only a body with temperature 0 Kelvin would not radiate, but reach this zero point is impossible. More than this, uncharged particles emit and receive EM radiation too. – HolgerFiedler Oct 30 '16 at 07:48
  • More on the idea of photon exchange and its actual meaning: http://physics.stackexchange.com/q/2244/50583, http://physics.stackexchange.com/q/142159/50583 and their linked questions. – ACuriousMind Oct 30 '16 at 15:14

2 Answers2

1

Well, one thing to be clear. The exchanged particles are virtual particles (in your case the virtual photon). Since they are not real, they shouldn't be detected.One way to know that there are indeed virtual photons is by studying the Feynman propagator: for free particle case, various virtual photons are excited at all directions. If you would like to know the influence of this at some other point, you do integration to get the probability. In fact the particle does excite virtual photons at all time. But these are not the photons in our everyday life, with, which you can indeed detect them. We are affected by these virtual photons but the method to analyze this process is through the method called QED(of cause, you know it).

ZHANG Juenjie
  • 1,041
  • A particle like electron does radiate some sort of virtual photons (and maybe others). The frequencies of these virtual particles depend on their energies, thus this is determined by the specific process you want to look at. I think there is no need for us to know the frequencies of these virtual particles. If you like you can simply calculate it. This is the same thing as saying that electrons have frequencies, but why do we need to know the exact value? As for constant, this is a time scale question. – ZHANG Juenjie Oct 30 '16 at 02:02
  • You can say that one particle is emitting virtual photons all the time. But it doesn't make any sense if is has not interacted with others. If it is indeed interacting with others, you can say, well there is one virtual particle or there are many virtual particles. The point is that this expression is simply an integration. You can regard this propagator as one particle, or you can say it is a composition of many particles. It seems that scientists prefer the former one to make this scenario easier. – ZHANG Juenjie Oct 30 '16 at 02:09
  • 2
    If you want to add something to your answer, please use the edit button instead of adding comments. – ACuriousMind Oct 30 '16 at 15:16
0

Charged particles exchange virtual photons. These are not the same as usual photons in electromagnetic spectrum, so they can't be detected directly. However, virtual photons "sea" can be detected indirectly, because quantum vacuum energy density is solely due to quantum vacuum fluctuations, aka. virtual photons. And these virtual photons are the main cause of Casimir force,- attraction or repulsion between close parallel plates. This force arises, because virtual photons makes a standing waves between those two plates. No any other serious explanation of Casimir force exist, so this force can be considered as a good proof to existence of virtual photons.

Agnius Vasiliauskas
  • 13,467