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I understand that forces are applied to charges when they are in an electric field, but what I cannot wrap my head around is how they know they are in the presence of a field. (I understand that they don't actually "know" anything). In mechanics, its relatively easy to see how forces are exchanged. For example, someone pushes a box, and the box accelerates. But the force is transmitted through the physical touch. In the case of electric charges, the there is no physical contact other than with the field itself. But the field is massless.

I like to think about two electrons in an isolated system. The electrons will experience an acceleration due to the electric field. Since electrons do have mass, they carry kinetic energy. Where does this kinetic energy come from? And how is it transferred without contact of any sort other than electric field, from one charge to the other?

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  • $\begingroup$ fields have their own energy. first they are not massless, but previously thought and they contain energy to accelerate electrons as i know. $\endgroup$ Commented Aug 23, 2020 at 3:35
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    $\begingroup$ I'm curious how you push a box without the electric field interacting with electrons. See here for more information about that. $\endgroup$
    – Sandejo
    Commented Aug 23, 2020 at 4:05
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    $\begingroup$ See my answer to In what medium are non-mechanical waves a disturbance? The aether?. You are right to think it is mysterious. But physics doesn't answer why questions like "Why does an electron feel a force?". It just says how the universe behaves. $\endgroup$
    – mmesser314
    Commented Aug 23, 2020 at 5:18
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    $\begingroup$ How do you know you’re in a gravitational field? One way is by the force acting on you due to that field. Same thing with the electron. The electric field causes a force on it, and that’s how it knows the field is there. $\endgroup$
    – The Photon
    Commented Aug 23, 2020 at 5:28
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    $\begingroup$ According to QED it’s constantly interacting with photons. And there are more photons where the field is stronger. $\endgroup$ Commented Aug 23, 2020 at 6:03

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As you said, they actually don't know anything. In the context of QED, one can interpret the fact that electrons trajectories are affected by the electric field as a consequence of the way the electron/positron field is coupled with the photon field. In fact, this last is what we call a gauge field: it is the field that tells us what will be the trajectories of particles in it. Since particles are excitation of fields, electrons are special excitations of the electron/positron field and, because of this coupling between our two fields, we can say that electrons only sense electric field via acceleration if they are stopped from their natural motion, just as Trula said. This is a rather rough explanation but you can think of this motion analogously as water current regarding whatever is in the water, but don't let your mind lie too much on it. This may be an unsatisfying answer because, just like Mmesser314 pointed out, physics does not explain the way the universe works, it describes how it works. We will probably never know anything more fundamental that "photon field is the gauge field where excitations of electron/positron field lie". From a mathematical point of view, here is what I said:

Let $\psi$ be a bispinor Dirac field and $\overline{\psi}$ its Dirac conjugate. Let $A_\mu$ be the components of the photon four-vector field. The free Dirac equation is \begin{equation} (i\gamma^\mu \partial_\mu-m)\psi=0. \end{equation} With $\gamma^\mu$ the gamma matrices. Now let's introduce de covariant gauge derivative associated to the $U(1)$ symmetry: $D_\mu=\partial_\mu+ieA_\mu$. Then the Dirac equation becomes: \begin{equation} (i\gamma^\mu D_\mu-m)\psi=0. \end{equation} This means that if we are in the gauge field, electrons will seem to be in free motion. Now we can rewrite this equation as: \begin{equation} (i\gamma^\mu \partial_\mu-m)\psi=e\gamma^\mu A_\mu \psi \end{equation} and this means that if we are outside the gauge field, electrons will not be in free motion. If electrons were at rest from our referential, then they will acquire kinetic energy. See that our coupling is a sort of abstract "contact" between our two fields. Not a physical one but more like a conceptual link.

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if an apple rests on a tree, you say it "feels" the gravitational field by the pull the tree excertes, but when it falls, there is no material connection, it still is accelerated, the same for you, if you jump down. what we call field ist just, that this force exists, regardless of matter.

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