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I'm studying basic physics. I'm using the text available at http://www.anselm.edu/internet/physics/cbphysics/downloadsI.html. It develops the universal law of gravitation by postulating the existence of a vector at each point of the form

$$ g_P = \sum G\frac{m_i}{|r|_i^3}r_i, $$

Where $m_i$ and $r_i$ are the mass of and separation vector from $P$ for all particles that aren't at the point $P$.

It examines the effect of one particle on another. If the separation vector is $r$ then from the above equation, we see that

$$ g = G\frac{m_1}{|r|^3}r $$

and that when a particle of mass $m_2$ is placed at the given point, the force will be

$$ F = G\frac{m_1m_2}{|r|^3}r $$

The authors then go on to claim that we can either repeat the development to see the effect of the second particle on the first one or apply Newton's third law.

How does Newton's third law apply through a field? If it's the field exerting the force, then Newton's third law would require a force on the field and not the object "generating" the field, correct?

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The third law is just about the conservation of the momentum. It follows from the translational symmetry of the laws of physics and it holds for any description, whether or not it contains fields. The description in terms of a force, action at a distance, has the same force with the opposite sign acting on both objects. If there are fields, the fields may carry some momentum, too. In quantum field theory, the force comes from virtual particles - virtual quanta of the fields - and they again carry the right momentum so that it is always conserved. So the law always holds. Does it answer "How"? –  Luboš Motl Apr 6 '11 at 12:17
    
BTW: All the everyday forces that you're familiar with---the ones for with you presumable accept the third law and we often call "contact" forces---are generated by fields. –  dmckee Apr 6 '11 at 14:18
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2 Answers

up vote 3 down vote accepted

In Newtonian physics, the field is not really something physical that has an independent existence. Particularly in Newtonian gravity, the gravitational force is really action at a distance with nothing mediating the force in between. The field $g$ defined here is simply for mathematical convenience and is not the usual field that you talk about in a fully relativistic classical field theory.

So as long as you're not doing any relativistic calculations and asking questions like "Does the force act instantaneously? In what frame?", it's perfectly fine to just treat the particles as obeying the Third Law, with equal and opposite forces between them that act at a distance, and use the field $g$ only as a convenience.

Of course, if you want a relativistic theory, you have to introduce a real physical field that can carry momentum and energy at every point in space. This is done in classical electrodynamics and general relativity, for example. You will see there that particles do not obey a simple Newton's Third Law since you have to take into account the dynamics of the field too. (There is no dynamics for the field itself in Newtonian physics, because like I said, the field is not something physical in Newtonian physics) This is what gives rise to electromagnetic and gravitational waves in those two field theories, respectively.

Summary: Don't take the "field" $g$ in a physical sense.

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Can a charge exert a force on the electromagnetic field? –  Physiks lover Oct 25 '11 at 22:10
    
"Force" is always defined as something that acts on a particle moving in a field. So it doesn't make sense to talk of "force acting on a field". The field can however however possess and transport energy and momentum. The force on particles measures how particles respond to this transfer of momentum and energy onto them by the field. –  dbrane Oct 26 '11 at 4:34
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The force field makes sense when it is an external force in a probe mass equation motion. It is too strong claim that the field exists "everywhere". In fact, it is a wrong claim.

If you have a classical source of light at a point A, it may emit a spherical wave that "exists" everywhere at a given R. But for a low-intensity, quantum source, this wave describes not the wave amplitude but the probability amplitude. Not all probe bodies placed around a quantum source can receive a single photon.

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