Are the directions of charge flow (=current?) absolute?

[Dale]

I'm not sure if a charge can accelerate in isolation. If it is not even a physical possibility, this may as well be a non-issue.

There are physically possible scenarios. E.g. A single charge accelerating due to a EM wave. In fact, in QM all EM interactions are built up from a series of this basic interaction.

In any case, did you understand my point above about the magnetic field being a component of the EM field tensor?

 

[kmarinas86]

I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.

The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.

I'm not sure if a charge can accelerate in isolation. If it is not even a physical possibility, this may as well be a non-issue.

There are physically possible scenarios. E.g. A single charge accelerating due to a EM wave. In fact, in QM all EM interactions are built up from a series of this basic interaction.

The keyword was isolation.

Also, I never saw a way to interact with EM waves that didn't involve a mass on the receiving end. This mass invariably has internal charges, which is also true even if the receiver is a neutron. Is there a way to take a measurement of a magnetic field [err... I mean an EM field] by a means that doesn't involve a mass? I don't think so.

In any case, did you understand my point above about the magnetic field being a component of the EM field tensor?

Yes I did.

 

[Dale]

The keyword was isolation.

That was your word, not mine. I just said a single charge, I never said that it was isolated, just that there wasn't another charge.

 

[kmarinas86]

That was your word, not mine. I just said a single charge, I never said that it was isolated, just that there wasn't another charge.

How is there not going to be another charge when we observe a charge?

I said:

I don't think it is possible to even observe a charge's magnetic field without there being another charge to act as a receiver. So it would seem to me that a magnetic field is product of two or more charges.

You replied:

The first sentence is true, but the second one leads to the non-conservation of momentum in the case of a single accelerating charge.

 

[Dale]

How is there not going to be another charge when we observe a charge?

I said:
You replied:

Yes, I agreed with that first sentence. Did you not see that it was only the second sentence I was objecting to?

 

[kmarinas86]

Yes, I agreed with that first sentence. Did you not see that it was only the second sentence I was objecting to?

Yes, and that was:

So it would seem to me that a magnetic field is product of two or more charges.

When we observe the EM field (going back to the proper terminology), there is always something to receive it. We never saw a EM field of a charge in isolation. Also there must be accounting for the energy of the EM field. Wouldn't that take a form of energy that treads neither on the E nor B lines? How come I have rarely been exposed to, if ever, a depiction of these lines?

 

[Dale]

When we observe the EM field (going back to the proper terminology), there is always something to receive it.

Yes, for the third time, I agree with this part.

Also there must be accounting for the energy of the EM field.

You are correct, and this goes along with my objection to the second sentence. If we say that magnetic fields only exist when there are two or more charges then in the case of a single charge you will get non-conservation of momentum as I mentioned and also non-conservation of energy as you mentioned. That is why I don't think it is good to say "that a magnetic field is product of two or more charges" even though I agree that you cannot observe an EM field without a charge.

 

[kmarinas86]

Yes, for the third time, I agree with this part.

You are correct, and this goes along with my objection to the second sentence. If we say that magnetic fields only exist when there are two or more charges then in the case of a single charge you will get non-conservation of momentum as I mentioned and also non-conservation of energy as you mentioned. That is why I don't think it is good to say "that a magnetic field is product of two or more charges" even though I agree that you cannot observe an EM field without a charge.

I am not very clear as to why you say we would get "non-conservation of momentum" "in the case of a single charge" if "magnetic fields only exist when there are two or more charges". How does one define the conserved momentum of a particle where there is nothing but a stand alone charge? Space is arbitrary, and in the case of a stand alone charge, there is not even another particle of some relative velocity for an EM field to have any real physical effect on anything. Thus, such a field may as well not exist (unobservable, unphysical).

 

[Dale]

I am not very clear as to why you say we would get "non-conservation of momentum" "in the case of a single charge" if "magnetic fields only exist when there are two or more charges".

Consider an electron scattered by an EM wave. The electron changes momentum, and so does the field. If there is no EM field to carry momentum then the momentum of the charge changes without an equal and opposite change in momentum in anything else. So if EM fields only exist if there are at least two charges then momentum is not conserved for a single charge.

 

[kmarinas86]

Consider an electron scattered by an EM wave. The electron changes momentum, and so does the field.

Doesn't that simply mean that the charge does not have a constant momentum, but the combination of the charge and the EM field do?

If there is no EM field to carry momentum then the momentum of the charge changes without an equal and opposite change in momentum in anything else. So if EM fields only exist if there are at least two charges then momentum is not conserved for a single charge.

Why not? A "single charge" does not experience any force (i.e. any momentum changes over time) until you bring other charges or EM fields into the picture. Odds would tell you that all EM fields are produced by charges, unless you believe that EM fields were created ex nihilo from the Big Bang (which I do not). The EM force, as you recalled, is mediated by EM radiation events, so there is no EM radiation event if there is no EM force! There is no reason why it should violate momentum conservation. A "single charge" without anything external to it has a constant momentum. Add an EM field, and they simply exchange momentum.

I think what I'm trying to get at is this: Does the constraint "EM fields only exist if there are at least two charges" contradict experimental observations?

 

[Dale]

Doesn't that simply mean that the charge does not have a constant momentum, but the combination of the charge and the EM field do?

Yes, exactly. Which is why momentum would not be conserved without the field.

I think what I'm trying to get at is this: Does the constraint "EM fields only exist if there are at least two charges" contradict experimental observations?

Not that I know of. Of course, I don't know of any experimental observations which confirm it either. The constraint has no experimental evidence for or against it, but sound theoretical reasons against it.

 

[Dale]

Why not? A "single charge" does not experience any force (i.e. any momentum changes over time) until you bring other charges or EM fields into the picture.

Yes, charges OR fields. The point is that it doesn't have to be both.

Odds would tell you that all EM fields are produced by charges, unless you believe that EM fields were created ex nihilo from the Big Bang (which I do not).

EM waves are vacuum solutions to Maxwells equations. They don't require charges in principle, and in practice you cannot ever uniquely solve for a charge distribution given only an EM wave.

On a related note, I don't understand the motivation for this line of thought. Maxwells equations and QED work in all situations where they have been tested, and they predict that an isolated charge has an EM field. This prediction cannot be tested. So why would you want to make a more complicated theory that differs from the standard theory only in such a non-verifiable way?

 

[Dale]

For each link there is a pair of opposite forces.

This is not correct in general in EM. If there were always a pair of equal-and-opposite forces on each pair of charges then the mechanical momentum of the system of charges would always be conserved. Because the forces are not always equal and opposite you need to consider also the momentum of the field in order to have conservation of momentum.

 

[kmarinas86]

For each link there is a pair of opposite forces.

This is not correct in general in EM. If there were always a pair of equal-and-opposite forces on each pair of charges then the mechanical momentum of the system of charges would always be conserved. Because the forces are not always equal and opposite you need to consider also the momentum of the field in order to have conservation of momentum.

Of course there is propagation delay, but the forces are 1) equal and 2) opposite. I didn't say that the charges act on each other instantaneously. I don't see how you can read that from the post you quoted. By the way, that post quoted was deleted by bcrowell because it was deemed to be misinformation.

 

[Dale]

Of course there is propagation delay, but the forces are 1) equal and 2) opposite.

Not in general, no. Even allowing for the propagation delay.

 

[kmarinas86]

Not in general, no. Even allowing for the propagation delay.

You said:

Because the forces are not always equal and opposite you need to consider also the momentum of the field in order to have conservation of momentum.

I said:

Of course there is propagation delay, but the forces are 1) equal and 2) opposite.

$$F_1=d\mathbf{p}/dt$$
$$F_2=-d\mathbf{p}/dt$$
$$F_1+F_2=0$$

The field carries the opposite force and the opposite momentum simultaneously. Equal and opposite forces are implied by the conservation of momentum.

 

[kmarinas86]

I am speaking only of classical EM.

Equal and opposite forces were implied by conservation of momentum. Your comments seem to disagree with that. When is that not the case in classical EM?

BTW, classical EM is not the "general" case.

 

[Dale]

The field carries the opposite force and the opposite momentum simultaneously. Equal and opposite forces are implied by the conservation of momentum.

The total momentum of the two charges and the field is conserved. That in no way implies that the force on one charge is equal and opposite to the force on the other charge. In fact, unless the second charge receives all of the momentum of the field the conservation of momentum guarantees that the two forces are not equal and opposite. I already said all this in post 43.

 

[kmarinas86]

That in no way implies that the force on one charge is equal and opposite to the force on the other charge. In fact, unless the second charge receives all of the momentum of the field the conservation of momentum guarantees that the two forces are not equal and opposite. I already said all this in post 43.

I was never claiming that the forces are direct.

There may as well be four forces. Two between the first charge and the field, and two between the second charge and the field. F_1 and F_2 cancel. F_3 and F_4 cancel. Neither F_1 nor F_2 need to cancel F_3 or F_4. These pairs do not have to be collinear, and nor do they need to be equal in magnitude (e.g. (3-3) + (1-1) is still 0). You cannot truly model this system as two forces, but you can model them as pairs of equal, but opposite forces.

 

[Dale]

I was never claiming that the forces are direct.

And I never said that you did.

There may as well be four forces. Two between the first charge and the field, and two between the second charge and the field. F_1 and F_2 cancel. F_3 and F_4 cancel. Neither F_1 nor F_2 need to cancel F_3 or F_4. These pairs do not have to be collinear, and nor do they need to be equal in magnitude (e.g. (3-3) + (1-1) is still 0). You cannot truly model this system as two forces, but you can model them as pairs of equal, but opposite forces.

Sure, if you want to call the change of momentum of the field a force then you could certainly say that. It is not common, but I have no problem with it. It seems that you now understand the point I have been making since post 43.

 

[kmarinas86]

Sure, if you want to call the change of momentum of the field a force then you could certainly say that. It is not common, but I have no problem with it. It seems that you now understand the point I have been making since post 43.

It's not common? Really?

EM fields are regarded as mediators of momentum, but you say they are usually not regarded carriers of force. Contrarily, what I read about EM implies that the EM field consists of force. In fact, EM is one of the "fundamental forces" known to physics.

The definition of force is quite clear from Newton's formulation. If a system has momentum being exchanged, then there is force.

 

[Dale]

It's not common? Really?

Really, really.

 

[bcrowell]

Closing this thread because it has drifted away from the original topic. Kmarinas86, please feel free to open a new thread if you wish on the topic of Newton's third law in the context of electromagnetism.