# I Field-free formulation of ED: Conservation laws?

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1. Sep 24, 2016

### greypilgrim

Hi.

In the (mainstream) books of electrodynamics I know, the electric and magnetic fields are introduced as force fields normalized to a charged test particle of 1 C. This makes those fields appear as an unnecessary, but convenient mathematical tool. They cannot be measured in the absence of charged particles, but it's comfortable to assume they're actually there.

However, later those fields are ascribed properties like energy and momentum even in the complete absence of particles (e.g. EM waves). Energy and momentum are NOT conserved if only particles are taken into account.

I'm somehow missing the step from "mathematical tool" to "gizmo that, although massless, is capable of carrying energy and momentum". Especially with momentum which is, by it's definition p=mv, bound to massive particles.
If I were to reformulate ED sticking only to force laws and avoiding fields (if this is possible), would I generate a theory that violates conservation of energy and/or momentum?

2. Sep 24, 2016

### Staff: Mentor

Probably because the characterization as "mathematical tool" is bad in the first place.

3. Sep 24, 2016

### greypilgrim

Is the problem that the definition $\vec{E}:=\frac{\vec{F}}{Q}$ already uses a more general concept of force and momentum than somebody who only studied classical mechanics before is aware of? I immediately (and exclusively) associated the "$\vec{F}$" in this definition with $\vec{F}=\frac{d\vec{p}}{dt}$ where $\vec{p}=m\cdot\vec{v}$, which is only nonzero for massive particles.

But then again, even if one assumes a more general concept of momentum and force that is applicable to massless structures, how does the division by Q make sense? At least here I'm pretty sure only massive particles can carry charge.

4. Sep 24, 2016

### Staff: Mentor

I just mean that calling it only a "mathematical tool" is a mischaracterization in the first place. It is a mathematical model which accurately predicts the outcome of experiments.

If a model accurately predicts forces then it should be unsurprising that it accurately predicts momenta. Forces and momenta are closely related.

5. Sep 25, 2016

### greypilgrim

I think my problem comes down to the following: The definition $\vec{E}(P):=\vec{F}(P)/Q$ is about the hypothetical force that would act on a charged test particle placed at point $P$ in space. It's clearly implied that this force is not actually there in the absence of said particle.

I just re-read a chapter on the introduction of energy and momentum density, the Poynting vector and the Maxwell stress tensor. They all are motivated by computing how much hypothetical work and momentum the fields transfer would transfer to a charged test particle and it is then concluded that the energy and momentum transfer would come from the fields. However, all further treatment assumes that the fields carry this energy and momentum even in the abscence of said particle.

It's those different assumptions I don't understand: In the absence of the test particle, the forces of the fields do not exist, but energy and momentum do.

We need the concept of test charges to define fields, and we need test charges to define energy and momentum of fields. This seems to me that it should be possible to formulate the whole theory with only actual, real, massive charged particles and an analogy of Maxwells equations to describe the interactions between them. But obviously energy and momentum cannot be conserved both in this formulation AND in conventional ED since the first only takes into account massive particles, the second massive particles AND fields. This seems weird to me, conservation laws shouldn't depend on the mere formulation of a theory.

6. Sep 25, 2016

### Staff: Mentor

Why not?

Although that is the usual derivation, there is an alternative approach. One can start with knowledge of Noether's theorem.

We note that EM experiments done in London have the same results as identical experiments in New York. Therefore we know that the laws have a symmetry wrt spatial translation. Per Noether's theorem that implies a conserved momentum.

Similarly, we note that experiments done today have the same results as experiments done yesterday. Per Noether's theorem we know there is a conserved energy.

We look into the EM Lagrangian and find the appropriate symmetries and the corresponding conserved quantities. We note that they are the same as found above. So now we have more confidence in our labeling. Our fields carry energy and momentum because they have the required symmetries.

7. Sep 26, 2016

### greypilgrim

Okay, I guess different formulations could lead to different conserved quantities. But shouldn't symmetry wrt spatial translation lead to the same momentum formula $\vec{p}=m\cdot \vec{v}$ for massive particles in all formulations? And again: If $\sum_i m_i\cdot \vec{v}_i$ is conserved in one formulation, how can $\sum_i m_i\cdot \vec{v}_i+field\enspace contributions$ be conserved in the other formulation? It obviously could if the "field contributions" are conserved separately, but this case is not interesting.

Actually I think they secretly used that approach in that chapter I mentioned (and just re-read again). They introduce quantities like
$$\vec{j}=\frac{c}{4\pi}\left(\nabla\times\vec{H}-\frac{1}{c}\partial_t \vec{D}\right)$$
to show energy conservation
$$\partial_t u+\nabla\cdot \vec{S}+\vec{j}\cdot\vec{E}=0$$
without properly motivating the definition of $\vec{j}$. My guess is that this is some Noether current and the definition follows from the application of Noether's theorem?

Let's consider the simple system of two separated charges. It shouldn't be difficult to write the EM forces between those charges without fields, i.e. Coulomb forces plus magnetic forces. And I guess one could formulate a Lagrangian without fields as well. Since this simple system has all kind of symmetries, we should find conserved quantities. But those quantities must be properties of solely the two particles, for example total momentum $m_1\cdot \vec{v}_1+m_2\cdot \vec{v}_2$, since there's just nothing else there that could carry momentum in this formulation.
Suppose now one of the charges is (mechanically) made oscillating. We note that it loses energy. At some later point in time (since the speed of the EM wave is finite), the other charge starts resonating.
I'm interested in the time before the second charge starts moving. The conventional formulation of EM explains the energy loss of the first charge as an energy transfer to the EM wave, i.e. to the fields.
In a field-free formulation, we would see the first charge losing and the second gaining energy, but there's a time span (before the second charge starts moving) where the energy of the first charge seems to be lost (since we have no concept of fields that could carry this energy). Where am I wrong in this train of thought?