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We are taught that Gravitational force exerted by an object is a two-step process:

  1. The object creates a field around it.
  2. The field exerts a force on bodies present in the field.

Now, since we know that an object any object of certain mass creates its own field, is it correct or incorrect to say that 'Field-field interactions exerts force on the bodies'.
Can a similar doubt be raised in case of Electric fields created by charges?

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  • $\begingroup$ Is your question within classical physics ? $\endgroup$
    – anna v
    Commented Dec 23, 2020 at 5:27
  • $\begingroup$ Yes, because I am yet learn about quantum physics. $\endgroup$ Commented Dec 23, 2020 at 6:31
  • $\begingroup$ Fields dont interact. They just add. Its linear. Electric fields add the same way but produce the opposite force $\endgroup$
    – R. Emery
    Commented Dec 23, 2020 at 7:09
  • $\begingroup$ @R. Emery that is true only for free fields. $\endgroup$ Commented Dec 23, 2020 at 7:44

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The idea of a field can be conceived of in multiple ways. I will describe the one that I prefer.

I need to point out first: that which can be actually measured is motion. What we observe is that under particular circumstances the motion of objects is affected (acceleration). We can attribute that affect to a field, but that field cannot be measured directly. In that sense we cannot assert the existence of the field with the same certainty as we have for the observations of actual motion.

The concept of a field is used because it affords an economy of thought. Throughout the history of physics it has often proved to be a good policy to proceed in a direction that affords economy of thought,


The concept of a field is not as if a particle casts a net out towards the universe.

Rather, the field is conceived as an entity that exists anyway, but in the absence of an interaction source it is in a uniform state.

In the case of the electrostatic interaction: presence of electric charge causes a distortion from uniform state in the space adjacent to that, and that distortion transfers on to space adjacent to that, and so on.

The usual visualization for that is a sheet of elastic material. If you depress that material at a single point you get an extended depressed area because every tiny area of the elastic material transfers stress to adjacent areas.

With a fast oscillation you can induce traveling ripples in that elastic sheet. The ripples propagate precisely because the sheet is elastic.

In the particular case of electromagnetism we have that our technology is inducing traveling electromagnetic ripples everywhere. Radio waves are electromagnetic waves.

These traveling ripples keep traveling, independent of whether the original source is still there or not. That is of course very strong corroborating evidence that the electromagnetic field actually exists.


In the case of gravity:
We have the option of thinking of gravity as mediated by a field. This field exist anyway, but in the absence of a source of gravitational interaction this field is thought of as uniform.

A source of gravitational interaction induces a distortion away from that uniform state. This stressed state has an effect on the motion of matter moving through it.

This effect is thought of as a local effect. An object moving through a gravitational field is interacting with the field at the location of that object.


As far as we know, all types of interaction are mutual interactions. When you have two sources of gravity they each contribute to an overall gravitational field, and each object is affected by the gravitational field of the other object.

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  • $\begingroup$ so do you mean that an object is not affected by its own gravity ? $\endgroup$
    – Ankit
    Commented Dec 23, 2020 at 9:56
  • $\begingroup$ @Ankit That question is still an open question. We have that the motions that are observed are consistent with a theory of gravity in which particles do not interact with their own field. But that doesn't exclude the possibility. Likewise, in the theory of electromagnetism it is still an open question whether interaction of a particle with its own field should be incorporated in the theory. No evidence exists that makes such incorporation necessary, but that doesn't exclude the possibility. $\endgroup$
    – Cleonis
    Commented Dec 23, 2020 at 10:08
  • $\begingroup$ also we explain the formation of black holes by saying that the star collapsed in its own gravity.. that's why I asked.. $\endgroup$
    – Ankit
    Commented Dec 23, 2020 at 10:10
  • $\begingroup$ @Ankit It's not clear what you are talking about. A star consists of particles, each individual particle adds its own contribution to the overall gravitational field. Collapse to a denser state happens because all of the parts of the star are gravitationally attracting each other. $\endgroup$
    – Cleonis
    Commented Dec 23, 2020 at 10:14
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Here is the wikipedia article on classical fields that will help clarify:

A classical field theory is a physical theory that predicts how one or more physical fields interact with matter through field equations. The term 'classical field theory' is commonly reserved for describing those physical theories that describe electromagnetism and gravitation, two of the fundamental forces of nature. Theories that incorporate quantum mechanics are called quantum field theories.

Bold mine, to stress that a field in classical field theory interacts with matter not with another field.

The word interaction occurs once here:

The first field theory of gravity was Newton's theory of gravitation in which the mutual interaction between two masses obeys an inverse square law. This was very useful for predicting the motion of planets around the Sun.

and a second time:

Maxwell's theory of electromagnetism describes the interaction of charged matter with the electromagnetic field.

So fields are additive, with the algebra on which they are defined, vector or tensor afaik in classical physics. They do not interact with each other.

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  • $\begingroup$ It is incorrect to say that fields don't interact with themselves classically. One can always add a potential which contains self-interaction terms, although the notion of "self-interaction" cannot be made sense of classically. $\endgroup$ Commented Dec 23, 2020 at 7:42
  • $\begingroup$ @thunderbolt You seem to contradict yourself, ""self-interaction" cannot be made sense of classically. – thunderbolt 2 mins ago". I (the article in fact) am answering withint the classical framework $\endgroup$
    – anna v
    Commented Dec 23, 2020 at 7:46
  • $\begingroup$ Let me clarify. Nothing stops me from adding a $\phi ^n$ term to a classical scalar Lagrangian. This is exactly what one would call a self-interaction term. Of course, one cannot interpret that term as a vertex in the Feynman diagram (as done in QFT.) Nevertheless, your statement about the fields being additive won't be true for a classical $\phi ^4$ theory since the equations of motion will be non linear. $\endgroup$ Commented Dec 23, 2020 at 9:31
  • $\begingroup$ @thunderbolt classical theoriew describe and fit data, that is why they are validated. One can imagine a lot of mathematics that has no connection with data. $\endgroup$
    – anna v
    Commented Dec 23, 2020 at 9:39
  • $\begingroup$ That is an odd argument to make. Anyway, we also have general relativity which is a completely classical theory whose equations are nonlinear. Its predictions are robust and have been experimentally tested to satisfaction. $\endgroup$ Commented Dec 23, 2020 at 10:17
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Gravity is neither a force or a field. It is the effect of matter, or energy, on the curvature of spacetime, which affects the matter or energy imbedded in that spacetime. The single least understood component of physics as it is currently understood. Many paradoxical aspects of gravity exist. Most notably, gravity at the quantum scale. It does not dovetail into either special or general relativity. A breakthrough is needed, and is highly sought after.

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