I am trying to understand the idea of a force carrier with the following example.

Let's say there are two charges $A$ and $B$ that are a fixed distance from each other. What is causing the force on $B$ by $A$?

Classically charge $A$ has an associated electric field which causes a force on $B$. From the standard model, photons are the force carrier for the electromagnetic force. With this view does it mean that $A$ is constantly emitting photons but in a way that the magnetic component cancels out? If that is the case then doesn't that mean that charge $A$ is constantly losing energy?


The notion of stationary electrically charged objects is a picture of classical electrostatics. So using the concepts of quantum mechanics like photons directly in this situation can lead to misconceptions.


The electromagnetic interaction takes place between the elementary particles that carry electric charge. If we treat the objects classically, then we should also treat their electromagnetic interaction classically in terms of classical concepts like forces (Coulomb law.)

To be accurate and consistent with nature, and quantum physics, one must take into account the fact that a classical charged object is a large congregation of electrically charged elementary particles, electrons for example, hosted by the classical objects A and B. The electrons move randomly inside the two objects A and B and obey the rules of quantum mechanics. So by no means they are stationary with respect to each other. One electron in object A can “sense” the existence of other electrons in object B, and it can of course sense the existence of the other electrons in object A. Similarly for electrons in object B

For any pair of electrically charged particles (electrons), one in object A and one in object B, we can calculate their interaction energy using the rules of quantum mechanics (Feynman diagrams etc,) and when we finish with all possible pairs of electrons we can add all these up together, and we end up with a classical force (the Coulomb law.) Therefore, the averaging of all exchanges replaces the quantum mechanical rules by the classical notion of the force field. So instead of exchanges of photons we are talking about forces, which are a classical concept in physics. The photons exchanged between pairs of electrons in objects A and B are normal photons, they have electric and magnetic field components but since the distance between the two objects is large, the number of photons is small compared with that when the objects are closer to each other.

None of the two objects A and B are loosing energy on average.

This is reflected, even in terms of classical physics, by the law of energy conservation

$ \frac{1}{2}mv^2_A+ \frac{1}{2}mv^2_B+\frac{1}{4\pi\epsilon_0}\frac{Q_AQ_B}{ r}=E_T$

which must constant.

  • 1
    $\begingroup$ Ok, so the objects actually have an uncertainty in position and momentum. This means photons do get randomly emitted and absorbed due to the acceleration of the charges. The calculation of all this would break down to Coulomb's Law. Is this the correct way of thinking about it? $\endgroup$
    – cspirou
    Mar 11 '13 at 1:24
  • $\begingroup$ @cspirou In many body problems such as this, developing effective theories, models, methods of calculation, is a good way. Such theories and methods are used very extensively in the condensed matter theory due to the extremely large number of particles involved. $\endgroup$
    – JKL
    Mar 11 '13 at 5:39
  • $\begingroup$ I'm not even really talking about many body problems. I'm really trying to understand what it means for the photon to be a force carrier in terms of Coulomb's Law. A and B could just as easily be a proton and electron in a hydrogen atom. $\endgroup$
    – cspirou
    Mar 12 '13 at 1:16
  • $\begingroup$ @cspirou I had the impression you were talking about two classical charged objects. In the case of the fundamental level interactions between two particles, we use the the QED Lagrangian which involves the four vector $A_\mu$ that represents properly the electromagnetic field. Then for special cases such as Coulomb field or just magnetic field $A_\mu$ reduces to the appropriate vector. We do this via the Schrodinger or Dirac's equation because the problem you set is a quantum mechanical since you are talking about interaction between fundamental particles. $\endgroup$
    – JKL
    Mar 13 '13 at 1:44

Force is given by the field.

Why carriers were needed is to explain WHY some forces are very short acting (strong force, electroweak) and some forces act in vast distances of space (electromagnetism,gravity). I guess the first person who introduced this idea for particle physics was Enrico Fermi.

Forces who have massive carrier can not travel far and forces who have massless carriers can travel infinitely far.

In QED (quantum electrodynamics) you can INDEED speak and write some mathematics which looks like creation and annihilation of the VIRTUAL photons. VIRTUAL means they are not subject to conservation of energy.

So, strictly speaking the idea of the FIELD is most profound and theoretically correct. Just think of some FIELDS as not traveling too far and you will be spared from confusion.

  • $\begingroup$ So charge A is emitting virtual photons with a magnetic components that cancels out? The whole energy issue is just side stepped virtual photons not having to conserve energy? $\endgroup$
    – cspirou
    Mar 6 '13 at 0:16
  • $\begingroup$ Yes, energy issue is not a problem. There are no photons with magnetic component cancelled out. They all have electric and magnetic field in interplay. That is why they can "carry" both electric and magnetic force. $\endgroup$
    – Asphir Dom
    Mar 6 '13 at 0:21
  • $\begingroup$ For static charges the field is purely electric. This means that the magnetic field component of a photon must cancel out or else a magnetic field is present in this example. I think you misinterpreted what I said. I didn't say photons don't have a magnetic field. Rather the sum of the magnetic fields from all the photons should cancel out. $\endgroup$
    – cspirou
    Mar 6 '13 at 0:27
  • $\begingroup$ Well - yes, something like that. $\endgroup$
    – Asphir Dom
    Mar 6 '13 at 0:39

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