# Electrodynamics without fields?

1. Dec 15, 2012

### johne1618

Can one describe electrodynamics without any reference to fields?

I think you can.

Using the Heaviside-Feynman expression for the electromagnetic field due to an arbitrarily moving charge, together with the Lorentz force law, one can write down an expression for the electromagnetic force $\mathbf{F}$ on a charge $q_1$, that is instantaneously at rest in an inertial frame, due to an arbitrarily moving charge $q_2$ as:

$$\mathbf{F} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \left\{ \left[ \frac{\mathbf{\hat{r}}}{r^2} \right]_{ret} + \frac{\left[ r \right]_{ret}}{c} \frac{\partial}{\partial t}\left[\frac{\mathbf{\hat r}}{r^2}\right]_{ret} + \frac{1}{c^2} \frac{\partial^2 \left[ \mathbf{\hat r} \right]_{ret}}{\partial t^2} \right\} \\$$
where $[\mathbf{r}]_{ret}$ is the vector from the retarded position of $q_2$, at time $t - [r]_{ret}/c$, to $q_1$, at time $t$.

Does the above formula contain all of classical electromagnetism?

2. Dec 16, 2012

### Jano L.

In principle, one could use directly only the EM forces between the charged particles and then solve all the equations of motion. However, they are not given by the formula you wrote exactly - one should use the forces that are given by fields that are solutions of Maxwell's equations. You can find them in the book by Jackson or Landau&Lifgarbagez.

The problem with these formulae is that they are not unique - there are many formulae that give different forces but still are consistent with Maxwell's equations. Usually, however, the retarded form of the fields is assumed.

However, in macroscopic theory, one deals with enormous number of particles and it is much more tractable to use macroscopic fields. For example, the wave equation for medium can explain many things with fields - doing the same with the formula for retarded force is possible, but also much more difficult.

3. Dec 16, 2012

### johne1618

Do you mean the Electric and Magnetic field solutions of the Liénard–Wiechert potentials given in :

http://en.wikipedia.org/wiki/Liénard–Wiechert_potential ?

As far as I know the Heaviside-Feynman expression for the fields of a point charge is completely equivalent to the Liénard–Wiechert fields. Jackson on page 247,248 of his book derives the Heaviside-Feynman formula from Jefimenko's equations. He doesn't suggest that the Heaviside-Feynman formula is approximate.

4. Dec 16, 2012

### andrien

yes,of course.The formula you use in the expression of force is for only electric force.You can also write the formula for magnetic field also.They are just the generalized form of coulomb and biot-savart law.They hold for arbitrary motion of charge.It is very general.It is not approximate.

5. Dec 16, 2012

### andrien

if you want to describe electrodynamics as a theory in which action at a distance can be eliminated by the principle of locality(which holds for almost all physical accepted theories) then you will have to go with the field concept.

6. Dec 16, 2012

### johne1618

True.

But I can always eliminate the need for a magnetic field by using an inertial frame in which the receiving charge is at rest.

7. Dec 16, 2012

### johne1618

As far as I understand the Heaviside-Feynman formula is not an action-at-a-distance theory. It describes a "retarded" direct interaction between charges that only propagates at the speed of light.

8. Dec 16, 2012

### GeorgeRaetz

The fundamental problem here is that after you write up and solve the equations of motion for the charges, you will find the charges accelerating. The conventional particle-field description predicts loss of particle energy & momentum through radiation. So the equations of motion would seem to violate energy/momemtum conservation.

Last edited: Dec 16, 2012
9. Dec 16, 2012