On an electrodynamic origin of quantum fluctuations

Abstract

We use the Liénard–Wiechert potential to show that very violent fluctuations are experienced by an electromagnetic charged extended particle when it is perturbed from its rest state. The feedback interaction of Coulombian and radiative fields among different charged parts of the particle makes uniform motion unstable. Then, we show that radiative fields and radiation reaction produce dissipative and antidamping effects, triggering a self-oscillation. Finally, we compute the self-potential, which in addition to rest and kinetic energy, gives rise to a new contribution that shares features with the quantum potential. We suggest that this contribution to self-energy produces a symmetry breaking of the Lorentz group, bridging classical electromagnetism and quantum mechanics.

Appendix

The following lines are devoted to obtain a power series relating the size of the particle d and the magnitude of the delay r/c. This relation allows us to approximate the distance l between the dumbbell’s position at time t and at the delayed time $t_{r}$, as a function of the mass center velocity, its derivatives and the particle’s size [20, 21]. We begin with the relation

$$\begin{aligned} d=r \sqrt{1-\left( \frac{l}{r} \right) ^2}=r \left( 1-\frac{z^2}{2}-\frac{z^4}{8}-\ldots \right) , \end{aligned}$$

(52)

where the variable $z=l/r$ has been introduced. On the other hand, Eq. (6) can be rewritten as

$$\begin{aligned} z=\frac{l}{r}=\beta +\frac{a}{2 c^2}r+\frac{\dot{a}}{6 c^3}r^2+\frac{\ddot{a}}{12 c^4}r^3+\frac{\dddot{a}}{120 c^5}r^4\ldots \end{aligned}$$

(53)

The square of z can then be computed. If we disregard the terms of the third order and higher orders as well, we obtain

$$\begin{aligned} z^2=\beta ^2+\frac{a}{c^2}\beta r+\frac{a^2}{4 c^4}r^2+\frac{\dot{a}}{3 c^3}\beta r^2+O(r^3). \end{aligned}$$

(54)

Concerning the fourth power of z we can write

$$\begin{aligned} z^4= & {} \beta ^4+\frac{2a}{c^2}\beta ^3 r+\frac{3 a^2}{c^4}\beta ^2 r^2\nonumber \\&\quad +\frac{2 \dot{a}}{3c^3}\beta ^3 r^2+O(r^3). \end{aligned}$$

(55)

to the same approximation as before.

Substitution of Eqs. (54) and (55) into Eq. (52), after gathering terms, yields

$$\begin{aligned}&d=\left( 1-\frac{\beta ^2}{2} -\frac{\beta ^4}{8} \right) r-\frac{a}{2c^2}\beta \left( 1+\frac{\beta ^2}{2}\right) r^2-\nonumber \\&\quad \left( \frac{a^2}{8 c^4}\left( 1+\frac{3\beta ^2}{2}\right) \right. \nonumber \\&\quad \left. \quad +\frac{\dot{a} \beta }{6 c^3} \left( 1+\frac{\beta ^2}{2} \right) \right) r^3+O(r^4). \end{aligned}$$

(56)

If we consider the non-relativistic limit, by just keeping terms of the first order in $\beta$, we arrive at the approximated relation

$$\begin{aligned} d=r-\frac{a}{2c^2}\beta r^2-\left( \frac{a^2}{8 c^4}+\frac{\dot{a}}{6 c^3} \beta \right) r^3. \end{aligned}$$

(57)