I apologize for seeming to return on a same question, but I don't have the feeling that the things are clear.

In a former question, "where is the potential energy stored", the conclusion was that this energy is stored in the field of mutual interaction between two objects. But it seems to me that this conclusion doesn't arrange well with other facts.

To shorten the story, my question is : if an asteroid approaches the Earth, does it find the gravitational field of the Earth there, ready, or the field appears between the two objects?

The thing is analogous with a big electric charge A. When a point charge B is brought in the neighborhood of A, does B find the field of A ready or the field appears between the two bodies?

One should not let oneself eluded by the fact that the Earth is a big and the asteroid is small. If the Earth has around itself a field, so has the asteroid.

And, if possible, what kind of matter is the static field? A travelling field is photons, gravitons, neutrinos, other particles. But the static field? It contains energy, the energy is stored in matter.


2 Answers 2


This question can be rephrased like the following. Suppose we created a big electric charge somewhere, and after a second passed, we created a small test charge in two light-seconds from the first one. Now the question would be: what time has to pass before the test charge begins accelerating?

If the field appeared only because of both bodies, then it'd mean that the test charge wouldn't start accelerating until the big charge "knows" the test charge is there. This means four seconds in our setup — two for the "message" to reach the big charge and another two for "response" to reach the test charge.

But actually, according to theory of electromagnetism, the test charge will accelerate just after one second, because the field of big charge has already travelled one light-second when we created our test charge, and had only to travel yet another light second to reach the test charge.

Moreover, if, when we created the test charge, we simultaneously removed our big charge, the test charge would still begin accelerating after one second as if the big charge was still there, and only after another one second would the acceleration end because the change of electrostatic field due to removal of its source has now reached the test charge.

So, the field is already there when the test charge probes it, not created due to both charges being present.

See also an illustration for evolution of static field of a uniformly moving charge in this answer to another question.

Having said this, I have to note that this is the most straightforward interpretation of the theory of electromagnetism. The other is Wheeler-Feynman absorber theory, in which the field is indeed created by both emitter and absorber, but this is actually somewhat tricky and experimentally indistinguishable from what I've said above.

  • $\begingroup$ it's plausible what you say. But, do you say implicitly that the charges radiate all the time? (Both of them, because big or small, they are on the same footing from the point of view of fundamental properties). It's a continuous loss of energy. Where from all this energy, and why don't we feel in some way that there is a loss of energy? $\endgroup$
    – Sofia
    Jan 22, 2015 at 18:23
  • 1
    $\begingroup$ @Sofia well charges don't actually emit if they don't accelerate. They instead sustain the field, don't let it change "unexpectedly". If they ever change their velocity (or disappear), this is what causes a radiation pulse, and it's this pulse, which has the energy flowing away from the charge. Likewise, the fact of creation of the charge is what initially puts the energy into the field, after the creation the pulse of energy spreads away by itself — as governed by Maxwell's equations, leaving static field around the charge. $\endgroup$
    – Ruslan
    Jan 22, 2015 at 18:50
  • $\begingroup$ how much energy was in that initial pulse? For the pulse to advance in space and leave a static field behind it should also leave energy in that field ~ $4 \pi \int E^2 r^2 dr$. Even if $E$ ~ $r^{-2}$, but immediately near a charge $E \to \infty$. $\endgroup$
    – Sofia
    Jan 22, 2015 at 19:11
  • $\begingroup$ @Sofia actually, for a point charge it's infinite (see also discussion if this issue under equation $(606)$ in that link). $\endgroup$
    – Ruslan
    Jan 22, 2015 at 19:37

I used to consider fields as mere mathematical convenience until recently a teacher asked me to consider the case of electromagnetic waves. They travel to far distances entirely due to the interaction between fields caused by a source.

Wikipedia defines field as a physical quantity. Physical quantities have to be measured, which requires interaction, but if you are willing to discard field as having meaning of its own because you need to measure its effects to feel its presence, the same should go true to other physical quantities as well.

In my opinion, it is up to interpretation. Technology has no meaning for a physical quantity unless measured. Although from a theoretical physics point of view, you can guarantee them existence which is exposed only when you measure it via an interaction.

Length, for example, is a quantity which all macroscopic bodies can be said to possess. You don't know it until you interact with it using a ruler. But the property exists and may be relevant to certain phenomena. Although, some interaction would be needed to observe any phenomena which brings us back to where we started.

  • $\begingroup$ @Ruslan I see you talk with me from San Petersburg. Isn't midnight in your town? Until which our can you be on the Internet? I would be glad to chat a bit, I mean, I have some thoughts still in the direction that around the charges there a virtual particles, or more exactly, dipoles as in a dielectric. Can you afford a chat, or it's bed time for you? $\endgroup$
    – Sofia
    Jan 22, 2015 at 20:47

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