The 3rd law of Newton with electrostatic fields

[Mad_Eye]

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..

 

[Andrew Mason]

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..

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

 

[Dale]

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..

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.

 

[rcgldr]

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..

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 electrical or gravitational field.

 

[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.

 

[Andrew Mason]

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.

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:

Richard Feynman: http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html"
Well, it seemed to me quite evident that the idea that a particle acts on itself, that the electrical force acts on the same particle that generates it, is not a necessary one - it is a sort of a silly one, as a matter of fact. And, so I suggested to myself, that electrons cannot act on themselves, they can only act on other electrons. That means there is no field at all. You see, if all charges contribute to making a single common field, and if that common field acts back on all the charges, then each charge must act back on itself. Well, that was where the mistake was, there was no field. It was just that when you shook one charge, another would shake later. There was a direct interaction between charges, albeit with a delay. The law of force connecting the motion of one charge with another would just involve a delay. Shake this one, that one shakes later. The sun atom shakes; my eye electron shakes eight minutes later, because of a direct interaction across.

Now, this has the attractive feature that it solves both problems at once. First, I can say immediately, I don't let the electron act on itself, I just let this act on that, hence, no self-energy! Secondly, there is not an infinite number of degrees of freedom in the field. There is no field at all; or if you insist on thinking in terms of ideas like that of a field, this field is always completely determined by the action of the particles which produce it. You shake this particle, it shakes that one, but if you want to think in a field way, the field, if it's there, would be entirely determined by the matter which generates it, and therefore, the field does not have any independent degrees of freedom and the infinities from the degrees of freedom would then be removed. As a matter of fact, when we look out anywhere and see light, we can always "see" some matter as the source of the light.

AM

 

[kcdodd]

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.

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?

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

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. ;)

 

[Andrew Mason]

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?

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

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. ;)

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

 

[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.

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..

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.

 

[kcdodd]

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

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.

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

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.

 

[Dale]

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.

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.