### Instantaneous electromagnetic interactions

Newtonian gravity introduced a model of instantaneous direct interactions among massive particles. This model was latter replicated by Coulomb for charged particles. Those models have been traditional named the action-at-a-distance model; although, this name is misleading and generated unending polemics among physicists and philosophers about how a particle can act at a distance in a vacuum over distant particles. A better name for this model is direct-particle-interaction.

Maxwell electrodynamics and General Relativity introduced an alternative model of contact-action, where particles don't interact directly but by means of a mediator. The former theory uses fields, whereas the latter uses a concept of curved spacetime as mediator. In the contact-action model a particle emits a signal —e.g. a photon— which propagates through the media —e.g. the electromagnetic field— until reaching a far particle, which then feels the force or interaction of the first particle. Since the maximum possible speed is the speed of light, interactions are retarded in this model. The common belief during almost a century has been that the contact-action model is accurate and that instantaneous interactions have been disproved. Nothing more far from reality!

In fact, partially due to the defects of the contact-action model, partially due to set of modern experiments, the models of instantaneous interactions are seeing a renaissance in the specialized literature. It is worth to revisit this topic and clarify some misunderstandings in the literature. I will limit to electromagnetic interactions, but the material discussed here can easily ported to gravitation.

We will start with the original Coulomb potential at point $$x$$ 'generated' by another charge $$e$$ placed at $$r$$ $\phi(x, t) = K \frac{e}{|R(t)|} = K \frac{e}{|x - r(t)|} .$ This potential is instantaneous because depends on the position of the 'source' particle at present time $$t$$. Now we will expand the position of the 'source' charge around its position at some early time $$t_0$$ $r(t) = r(t_0) + v(t_0) (t-t_0) + a(t_0) (t-t_0)^2 / 2 + \cdots$ and assuming that this charge is non-accelerating at the initial time $$t_0 = (t - |R(t_0)|/c)$$ we obtain the next potential $\phi(x, t) = K \frac{e}{|R(t_0)| - v(t_0) R(t_0)/c} ,$ which is evidently the scalar Lienard & Wiechert potential. The vector Lienard & Wiechert potential can be obtained in the same way if we start from the instantaneous potential $A(x, t) = K \frac{ev}{|R(t)|} = K \frac{ev}{|x - r(t)|} .$ Note that the Lienard & Wiechert potentials have been derived under the approximation of particles being in inertial initial states, $$a(t_0) = 0$$. This means that Lienard & Wiechert potentials aren't complete and this explains why they have to be complemented by adding reaction-radiation potentials to the equations of motion for curing such issues like non-conservation of energy on accelerating particles. Not only those additional potentials are obtained ad hoc but the resulting improved equations of motion are still subjected to criticism due to non-physical behaviors.

Note that the $$a(t_0) = 0$$ approximation also explains why the Lienard & Wiechert potentials for a moving charge can be obtained from the potentials for a charge at rest $$A=0$$ and $$\phi = \phi_\mathrm{Coulomb}$$ applying the Lorentz transformations between the frame where the particle is at rest and the frame where the particle is moving with velocity $$v$$. The Lorentz transformations can be applied because the frames are inertial, i.e., the particle is non-accelerating. In fact some derivations explicitly assume that the charge is moving with "with uniform velocity $$v$$ through a frame $$S$$".

This same approximation is also the reason why the quantum field theory assumes as the only physicall admisible states those of free particles, i.e. non-accelerating. This is picturesquely described in Feynman diagrams
The diagram consider particles in free motion, until a virtual photon is emitted and absorbed and both electrons change their state of motion to a new inertial state. Quantum field theory only can rigorously describe the initial and the final states —before and after the interaction— but not cannot provide an accurate description of what happens during the interaction.

The Lienard & Wiechert potentials have been traditionally associated to retarded interactions because positions and velocities of the 'source' particles are evaluated at early time $$t_0$$. However, I have shown how they can be derived from instantaneous potentials that are function of positions and velocities at present time $$t$$. This urge to consider what is the origin of the myth of retarded interactions. For such goal I will start again with the Coulomb potential without any lost of generality, because the application to the vector potential $$A$$ is straightforward, $\phi(x, t) = K \frac{e}{|x - r(t)|}$ which will be rewritten as $\phi(x, t) = K \int \frac{\rho(y,t)}{|x - y|} \mathrm{d}y$ using a pure state electron density in position space $$\rho(y,t) = e\delta(y - r(t))$$. Now we can use the equation of motion to relate the present density to a previous density $\phi(x, t) = K \int \exp \Big[ L(t-t_0) \Big] \frac{\rho(y,t_0)}{|x - y|} \mathrm{d}y .$ $$L$$ in the above expression is the Liouvillian. We can now see clearly which is the equivalence between instantaneous and retarded potentials. A kernel $$1/|x - y|$$ evaluated at present time $$t$$ is identical to a modified kernel evaluated at retarded time $$t_0$$ $\left\{ \frac{1}{|x - y|}\right\}_t = \left\{ \exp[L(t-t_0)] \frac{1}{|x - y|}\right\}_{t_0} .$ If we replace the full Liouvillian by its free part $$L^\mathrm{free} = - v \nabla$$ we obtain the kernel of the Lienard & Wiechert potentials $\exp \Big[ L^\mathrm{free}(t-t_0) \Big] \frac{1}{|x - y|} = \frac{1}{\kappa |x - y|} ,$ with $$\kappa = 1 - v(x-y)/|x-y|c$$. Note that the denominator of the kernel being linear in space variables implies that the power series expansion of the exponential vanishes identically after the linear term in the Liouvillian. In the introduction we derive the Lienard & Wiechert potentials by neglecting acceleration and higher order terms. We can now confirm that the Lienard & Wiechert potentials are an exact consequence of the free component of the full Liouvillian, this free component of course describe inertial particles. The interaction Liovillian will introduce acceleration and higher-order corrections to the Lienard & Wiechert potentials. Corrections to the Lienard & Wiechert potentials will be explored elsewhere, now we want to identify some flaws have remained unnoticed in the electromagnetic literature during decades.

Let us start with the ordinary 'wave' equation for the scalar potential $\square \phi = -4\pi K \rho$ Note that the right-hand-side contains the instantaneous charge density, not the density at early times. Now the ordinary literature integrates the equation and obtains the approximated potential $\phi = K \int \rho(t',y) \frac{\delta(t-t'-|x-y|/c)}{|x-y|} \mathrm{d}y \, \mathrm{d}t'$ If we integrate first on position and then on time we obtain the Lienard & Wiechert potential. If we integrate on time then we would obtain $\phi(x, t) = K \int \exp \Big[ L^\mathrm{free}(t-t_0) \Big] \frac{\rho(y,t_0)}{|x - y|} \mathrm{d}y .$ However the standard literature gives the wrong result $\phi(x, t) = K \int \frac{\rho(y,t_0)}{|x - y|} \mathrm{d}y ,$ with a retarded density $$\rho(y,t_0)$$ which is the origin of the myth of retarded interactions. This discrepancy is due to the standard literature performing the integration of the delta function without careful analysis of the functional dependences of the argument of the delta function on the variable of integration. The correct integration is as follows. First we let $$s \equiv t' + |x-y|/c - t$$ be the new variable of integration. We have $$\mathrm{d}s / \mathrm{d}t' = 1 + \mathrm{d}|x-y|/c\mathrm{d}t'$$ and $\phi = K \int \rho(t',y) \frac{\delta(s)}{|x-y| (\mathrm{d}s / \mathrm{d}t')} \mathrm{d}y \, \mathrm{d}s$ Much care has to be taken on evaluating the term $$\mathrm{d}s / \mathrm{d}t'$$; on a first attempt we could assume that $$|x-y|$$ doesn't depend on time, which implies $$\mathrm{d}s / \mathrm{d}t' = 1$$ and recover the incorrect expression for $$\phi$$ with a retarded density. The subtle issue is that $$|x-y|$$ doesn't depend on time only outside the path of the 'source' particle, but in this trivial case the potential is identically zero. Within the particle path the term $$|x-y|$$ depends on time via the density $$\rho(y,t') = e\delta(y - r(t'))$$. Therefore, it is better to leave the term $$\mathrm{d}s / \mathrm{d}t'$$ in the integral without evaluating it when performing the integration on $$s$$ and use $$y = r(t')$$ at then end, when integrating on space coordinates. With this rigorous method we will obtain $\phi(x, t) = K \frac{e}{|x - r(t)|} ,$ in full agreement with the mechanical result.

A pair of final remarks. First, I have focused on retarded potentials but it is possible to obtain the advanced potentials when integrating the equation of motion taking some future time as baseline $$\rho(y,t) = \exp[L(t-t_F)] \rho(y,t_F)$$; there is no violation of causality because the equations of motion are deterministic and can be integrated both backward and forward in time. I have also focused on electromagnetic interactions but the same arguments can be applied to gravitation resulting on instantaneous potentials $$h_{\mu\nu}$$.