# On the momentum of fields

1. May 9, 2010

### JK423

We know that when a charge interacts with an electric field, the latter makes the charge move. That means that the field gives the particle momentum.
However when the field's source is a charge, we notice that the field's strength doesnt decrease so it's able to give infinite momentum to particles.
For example, if we place infinite charges near a charge, they will all move ==> The field has given infinite momentum to all those charges, but still its strength/energy stays the same.
That's not what happens for example with electromagnetic waves.. When EM waves interact with a particle they lose energy/momentum...

So, im forced to say that a charge is a source of infinite energy..!
Is this thought valid?

Also, the field of a charge/source will move a 2nd charge.
Lets consider that the second charge interacts with the electric field of the charge/source first. (That means that the charge/source will feel the field of the 2nd charge an hour later).
If we are to apply conservation of momentum on the 2nd charge, it will have zero momentum at first, and some momentum during interaction. If there is to be no contradiction, we must say that the electric field must carry momentum, or else where did the 2nd charge find its momentum, from nowhere?

But what momentum does the electric field carry? The poynting vector is zero before the interaction (cause B=0) !!

2. May 9, 2010

### Staff: Mentor

You are correct, EM fields carry both energy and momentum. When you include the energy and the momentum of the fields then you always have that both are conserved. The conservation laws follow directly from Maxwell's equations, so the details of the scenario are not important.

Here is my favorite page on the conservation of energy in EM:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

And nearby is my favorite page on the conservation of momentum in EM:
http://farside.ph.utexas.edu/teaching/em/lectures/node91.html

3. May 9, 2010

### JK423

The "Momentum of fields" enters the game when you have a bunch of charges which interact in such a way that Newton`s third law no longer holds, and you are forced to assign a momentum to those fields.
But this kind of momentum has nothing to do with the fact that an electric field can make a charge move.
Let me give you an example in order to make myself clear:

Suppose that you have two charges in some distance.
If you let them free to interact, the electric force that they will exert to eachother
will be exact and opposite in direction ==> Momentum's conserved.
But lets disregard one of the charges and focus on the interaction between the system "Charge 1 + Electric Field of charge 2".
a) Initial momentum = zero
Charge 1 is still, and there is no momentum in the fields because there is no magnetic field. (Remember that the momentum of fields is expressed by the poynting vector which is ExB, so its zero in the above situation)
b) Momentum after interaction = NON zero
Because Charge 1 moved due to the interaction with the electric field of charge 2.

So, we see that momentum conservation in the system "Charge 1 + Electric Field of 2" cannot hold unless the electric field has also momentum, which is something different from the usual 'Momentum of Fields' which is described by the Poynting vector which is zero in that case!

Update:
I just noticed thats its wrong to say that the momentum of interaction is non zero because the motion of the charge creates also a magnetic field which combined with the eletriic field, gives opposite momentum to the electromagnetic field..
Interesting, because it actually shows why an accelerating charge gives momentum to the electromagnetic field, aka electromagnetic waves

Anyway, i tried to make an example in order to show that the electric field must carry momentum and it seems that i failed.
I still cannot understand how and electric field can move a charge...

If you have any ideas let me know!

4. May 9, 2010

### kcdodd

what do you mean?

5. May 10, 2010

### Staff: Mentor

I am glad that you thought through it on your own and understood the resolution of the problem.

If you will go back to the links that I posted you can see that the derivation is directly from Maxwell's equations. The reason that is important is that it means that the details of the example are not important. If you follow Maxwell's equations you will necessarily get conservation of energy and momentum.

6. May 10, 2010

### JK423

Well, i'm not trying to prove the theory wrong.. Im just trying to understand how it works and thats why the details are important. If you cant figure something out you havent understood it..

7. May 10, 2010

### planck42

Ummm, electric field strength falls off according to the inverse square law. How does the field's strength not decrease?

8. May 10, 2010

### Staff: Mentor

Well, it looks like you understood it correctly. You replied to yourself exactly with what I would have said in your "update" section.

Last edited: May 10, 2010