# The 3rd law of Newton with electrostatic fields

1. Feb 27, 2010

Well, say there is an positive particle, and a negative particle.
Newton said one is pulling the other, and thus it is pulling back.

But we now know both particle actually creating fields, and the fields actually exerting the force on the other particle.

so basically, this is not action and reaction, each particle is creating an independent field, that exerting forces on particles around..

2. Feb 27, 2010

### Andrew Mason

Fields are a mathematical construct to deal with action at a distance. They do not necessarily have to exist as something physical. Indeed, a static massless energy field creates theoretical problems in physics. The theory of quantum electrodynamics (QED) uses the concept of virtual photons to explain the static electric field. A virtual photon, by definition, is not a real photon.

I am not sure what your question is, though. If you are simply observing that all interactions of masses are simply interactions of force fields (electromagnetic, gravitational, nuclear) you are correct. Ultimately, all matter consists of point particles and their associated fields (or actions at a distance).

AM

3. Feb 27, 2010

### Staff: Mentor

This is correct. Newton's 3rd law is a statement of the conservation of momentum. It turns out that EM fields also carry momentum and that when you include the field you find that momentum is conserved. If you accidentally neglect the momentum of the field then, as you suggest, you do think that momentum is not conserved.

4. Feb 27, 2010

### rcgldr

I don't know what Newton was thinking in his actual wording, but Newtons 3rd law as described today just states that all forces only exist in equal and opposing pairs, and charged particles do not violate the more generic version (equal and opposing forces).

One exception could be relativistic speeds between two objects and the propagation speed of an eletrical or gravitational field.

Last edited: Feb 27, 2010
5. Feb 28, 2010

### kcdodd

Actually, the EM field does have to be something physical. If you take a single charged particle, and there are no other charged particles anywhere, it will still feel a reaction force from it's own field if you accelerate it (above the inertia). In other words, the final kinetic energy is less than what you put in. That energy went into building up new field. So there is "something" there, it's not just action at a distance.

6. Feb 28, 2010

### Andrew Mason

First of all, there is only one way to accelerate a single charged particle if there are no charged particles anywhere and that is by gravity. But, perhaps remarkably, the acceleration of a charged particle by gravity does NOT result in the particle feeling a reaction force.

Second, in order to cause a charged particle to accelerate due to something other than gravity there has to be an electromagnetic interaction (eg a static electromagnetic field or em wave) which has to, ultimately, come from another charged particle somewhere. So the "reaction force" that the particle feels can also be explained as the interaction of the two particles at a distance.

This quote from Feynman's Nobel lecture might put things into perspective:
AM

Last edited by a moderator: May 4, 2017
7. Feb 28, 2010

### kcdodd

So, what you are saying is if I put a charged sphere on the top of a building and drop it, so the only acceleration is due to gravity, then it will not emit radiation?

Feynman ignores the issue of how the particle is "shook". The fact is that shaking the electron "here" emits radiation, whether there is an electron over "there", or not. Where did that energy go? If it is not "here", and it is not "there", can it be "anywhere"? I do not eat green eggs and ham. I do not like them, Sam-I-am. ;)

8. Feb 28, 2010

### Andrew Mason

That's right. It is a consequence of the principle of equivalence in GR.

I don't think Feynman ignored anything. His point was that you cannot "shake" an electron here except by having another electron over 'there". You can only shake an electron by applying an external electromagnetic force to it and all external electromagnetic forces must originate with other charges.

AM

9. Feb 28, 2010

### TurtleMeister

It is my understanding that the third law of motion will hold true regardless of the source of the force. The source (of the field) can be one particle, the other, or both. If particle A moves 1mm then particle B must also move 1mm (assuming they both have the same inertial mass), regardless of whether radiation occurs or not.
Yes, it is action and reaction. If it were not true then if particle A pulled harder on particle B than particle B pulled on particle A then the third law would be violated. As far as I know the third law has never been observed to be violated.

10. Mar 1, 2010

### kcdodd

I have heard that before. You can only say there is no radiation in the particle's free fall frame. If there is an electron supported in the lab frame, there is obviously no radiation, yet the particle is being accelerated in the free-fall frame. You would expect that if radiation was seen in the free fall frame in that case, you would also see radiation in the lab frame when the particle is in free fall. The only way out is to say the particle never radiates under uniform acceleration, which really has nothing to do with gravity.

And the one over there cannot shake without another electron somewhere else shaking. And so on and so fourth. Well, it is a convenient argument, but I know that I, being a neutrally charged person, can grab a heavily charged object and shake it around without creating any radiation myself.

11. Mar 1, 2010

### Staff: Mentor

I don't think that it is so obvious. Let's say that you are detecting the radiation with a simple but sensitive antenna. If the antenna is in free-fall and the charge is experiencing proper acceleration then the antenna will detect radiation regardless of if the proper acceleration is the result of being supported against gravity or being pushed in a rocket in deep space. Conversely, if the antenna and charge are both accelerating together then no radiation will be detected in either case. I really don't know the answers here, but I don't think that it is obvious and it appears to me that the existence of radiation is itself a frame-dependent thing.