Special relativity, de Broglie waves, dark energy and quantum mechanics

Abstract

In this paper, we connect quantum mechanics with the recent work of the author Hill (Z Angew Math Phys 69:133–145, 2018; Z Angew Math Phys 70:5–14, 2019), suggesting that dark energy arises from the conventional mechanical theory neglecting the work done in the direction of time and consequently neglecting the de Broglie wave energy. Using special relativity and validation through Lorentz invariance, Hill (2018, 2019) develops expressions for the de Broglie wave energy $\mathscr {E}$ by making a distinction between particle energy $e = mc^2$ and the total work done by the particle W, so that both momentum ${\mathbf{p} = m\mathbf{u}}$ and particle energy e contribute to the total work done $W = e + \mathscr {E}$. This formulation provides an extension of Newton’s second law that is invariant under the Lorentz group and gives work done expressions for $\mathscr {E}$ involving the $\log$ function, indicating that large energies might be generated even for slowing mechanical systems. Although inherent in Hill (2018, 2019), here we propose explicitly that the total work done W by a single particle comprises two contributions, namely particle energy e and wave energy $\mathscr {E}$; thus, $W = e + \mathscr {E}$. Since in any experiment either particles or de Broglie waves are reported, only one of e or $\mathscr {E}$ is physically measured, which leads to the expectation that particles appear for $e < \mathscr {E}$ and de Broglie waves occur for $\mathscr {E} \leqslant e$, but in either event, both a measurable energy and an unmeasurable energy exist, the latter registering its presence in the form of dark energy. In particular, in this formulation conventional quantum mechanics operates under circumstances such that the spatial physical force $\mathbf {f}$ vanishes, and the force g in the direction of time becomes pure imaginary. If both $\mathbf {f}$ and g are generated as the gradient of a potential, then the total particle energy is necessarily conserved in a conventional manner. The present paper makes a formal connection between special relativity and quantum mechanics, linking two new invariances of the Lorentz group of special relativity with the corresponding Lorentz invariant differential operators arising in quantum mechanics and the de Broglie particle and wave duality in Hill (2018, 2019) and giving rise to the Klein–Gordon equation of relativistic quantum mechanics.

Appendices

Appendices

Appendix A: Alternative derivation of Eqs. (4.9) and (4.10)

On taking the total time derivative ${\text {d}}/{\text {d}}t$ of the two invariants $\xi$ and $\eta$ given by (4.1) and making use of ${\text {d}}e/{\text {d}}t = u{\text {d}}p/{\text {d}}t$, we obtain

$$\begin{aligned} \frac{{\text {d}}\xi }{{\text {d}}t} = \frac{\eta }{\left( 1 - (u/c)^2\right) }\frac{{\text {d}}u}{{\text {d}}t}, \quad \quad c\frac{{\text {d}}\eta }{{\text {d}}t} = \frac{\xi }{c\left( 1 - (u/c)^2\right) }\frac{{\text {d}}u}{{\text {d}}t} - e_0c\left( 1 - (u/c)^2\right) ^{1/2}, \end{aligned}$$

on using ${\text {d}}p/{\text {d}}t = m_0\left( 1 - (u/c)^2\right) ^{-3/2}{\text {d}}u/{\text {d}}t$. These two equations simplify to give

$$\begin{aligned} \frac{{\text {d}}\xi }{{\text {d}}u} = \frac{\eta }{\left( 1 - (u/c)^2\right) }, \quad \quad c\frac{{\text {d}}\eta }{{\text {d}}u} = \frac{\xi }{c\left( 1 - (u/c)^2\right) } - e_0\frac{{\text {d}}s}{{\text {d}}u}, \end{aligned}$$

so that on introducing the substitution $u = c\sin \phi$ we have

$$\begin{aligned} \cos \phi \frac{{\text {d}}\xi }{{\text {d}}\phi } = c{\eta }, \quad \quad c\cos \phi \frac{{\text {d}}\eta }{{\text {d}}\phi } = {\xi } - e_0\cos \phi \frac{{\text {d}}s}{{\text {d}}\phi }. \end{aligned}$$

Now on introducing $\chi$ defined by

$$\begin{aligned} {\text {d}}\chi = \frac{{\text {d}}\phi }{\cos \phi } = \frac{{\text {d}}\phi }{\left( \cos ^2(\phi /2) - \sin ^2(\phi /2)\right) } = \frac{\sec ^2(\phi /2){\text {d}}\phi }{\left( 1 - \tan ^2(\phi /2)\right) }, \end{aligned}$$

we may readily deduce that

$$\begin{aligned} \chi = \log \left( \frac{1 + \tan (\phi /2)}{1 - \tan (\phi /2)}\right) = \frac{1}{2} \log \left( \frac{1 + \sin \phi }{1 - \sin \phi }\right) = \frac{1}{2} \log \left( \frac{1 + u/c}{1 - u/c}\right) = \theta , \end{aligned}$$

where $\theta$ is as defined in Eq. (3.5), and Eqs. (4.9) and (4.10) follow immediately.

Appendix B: Further differential relations in terms of angles

From the two relations $\xi ^2 - (c\eta )^2 = e_0^2(x^2 - (ct)^2)$ and $x^2 - (ct)^2 = \zeta ^2$, we may formally introduce angles $\Phi$ and $\Psi$ such that

$$\begin{aligned} \xi = e_0\zeta \cosh \Phi , \quad c\eta = e_0\zeta \sinh \Phi , \quad x = \zeta \cosh \Psi , \quad ct = \zeta \sinh \Psi , \end{aligned}$$

(B.1)

so that from relations (4.4) and using Eq. (4.6) we may deduce

$$\begin{aligned} e = e_0\cosh (\Phi + \Psi ) = e_0\cosh \theta , \quad \quad pc = e_0\sinh (\Phi + \Psi ) = e_0\sinh \theta , \end{aligned}$$

and therefore, we may conclude that simply $\theta = \Phi + \Psi$. On totally differentiating $\zeta ^2 = x^2 - (ct)^2$ with respect to time, we find that $\zeta {\text {d}}\zeta /{\text {d}}t = xu - c^2t = (xp - c^2et)/m$ from which we may deduce the equation

$$\begin{aligned} \zeta \frac{{\text {d}}\zeta }{{\text {d}}\theta } = \frac{c\eta }{f}, \end{aligned}$$

(B.2)

and the three relations Eqs. (4.7) and (B.2) constitute the three basic equations connecting the three variables $\xi$, $\eta$ and $\zeta$ as functions of $\theta = \Phi + \Psi$, where the angle $\theta$ relates to the velocity $u = c\tanh \theta$, and $\Phi$ and $\Psi$ connect, respectively, with $(\xi , c\eta )$ and (x, ct) through (B.1), noting that only one of the two relations (4.7) is independent. Indeed, we may show from Eqs. (4.7) and the basic definition $u = {\text {d}}x/{\text {d}}t = c\tanh \theta$ that all three relations give rise to the single condition

$$\begin{aligned} \frac{{\text {d}}\Phi }{{\text {d}}\theta } + \frac{e_0}{f\zeta }\cosh \Phi = 1, \end{aligned}$$

(B.3)

while (B.2) yields

$$\begin{aligned} \frac{{\text {d}}\zeta }{{\text {d}}\theta } = \frac{e_0\sinh \Phi }{f}, \end{aligned}$$

(B.4)

further noting that the physical force f needs to be specified (say gravitational or electrical) before these two key Eqs. ((B.3) and (B.4)) can be fully solved as two equations in the two unknowns $\zeta$ and $\Phi$. Again on using $e_0/f = {\text {d}}s/{\text {d}}\theta$, Eqs. (B.3) and (B.4)

$$\begin{aligned} \zeta \frac{{\text {d}}\Phi }{{\text {d}}s} + \cosh \Phi = \zeta \frac{{\text {d}}\theta }{{\text {d}}s}, \quad \quad \frac{{\text {d}}\zeta }{{\text {d}}s} = \sinh \Phi , \end{aligned}$$

so that on using $\theta = \Phi + \Psi$ we have simply

$$\begin{aligned} \zeta \frac{{\text {d}}\Psi }{{\text {d}}s} = \cosh \Phi , \quad \quad \frac{{\text {d}}\zeta }{{\text {d}}s} = \sinh \Phi , \end{aligned}$$

and therefore by division, we obtain the deceptively simple result

$$\begin{aligned} \frac{{\text {d}}\zeta }{{\text {d}}\Psi } = \zeta \tanh \Phi , \end{aligned}$$

connecting the three variables $\zeta = (x^2 - (ct)^2)^{1/2}$, $\Phi = \tanh ^{-1}(c\eta /\xi )$ and $\Psi = \tanh ^{-1}(ct/x)$.

Appendix C: Time-dependent Dirac equation for a free particle

In this appendix, for ease of reference, we state the details for the time-dependent Dirac equation for a free particle. The matrices ${\mathbf {A_x}}$, ${\mathbf {A_y}}$, ${\mathbf {A_z}}$ and $\mathbf {B}$ appearing in the time-dependent Dirac equation for a free particle are given, respectively, by

$$\begin{aligned} {\mathbf {A_x}} = \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}, \quad {\mathbf {A_y}} = \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 &{}\quad -i \\ 0 &{}\quad 0 &{}\quad i &{}\quad 0 \\ 0 &{}\quad -i &{}\quad 0 &{}\quad 0 \\ i &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}, \quad {\mathbf {A_z}} = \begin{pmatrix} 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad -1 &{}\quad 0 &{}\quad 0 \end{pmatrix}, \end{aligned}$$

and

$$\begin{aligned} \mathbf {B} = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad -1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 \end{pmatrix}, \end{aligned}$$

and the Dirac equation becomes

$$\begin{aligned} \begin{pmatrix} \frac{\partial }{\partial t} &{}\quad 0 &{}\quad \frac{\partial }{\partial z}&{}\quad \frac{\partial }{\partial x} - i\frac{\partial }{\partial y}\\ 0 &{}\quad \frac{\partial }{\partial t} &{}\quad \frac{\partial }{\partial x} + i\frac{\partial }{\partial y}&{}\quad -\frac{\partial }{\partial z} \\ \frac{\partial }{\partial z} &{}\quad \frac{\partial }{\partial x} - i\frac{\partial }{\partial y} &{}\quad \frac{\partial }{\partial t} &{}\quad 0 \\ \frac{\partial }{\partial x} + i\frac{\partial }{\partial y}&{}\quad - \frac{\partial }{\partial z} &{}\quad 0 &{}\quad \frac{\partial }{\partial t} \end{pmatrix} \begin{pmatrix} \psi _1 \\ \psi _2 \\ \psi _3 \\ \psi _4 \end{pmatrix} = \begin{pmatrix} -if_0\psi _1 \\ -if_0\psi _2 \\ if_0\psi _3 \\ if_0\psi _4 \end{pmatrix}, \end{aligned}$$

where $f_0 = e_0/\hbar$. Further, it can be shown that each component $\psi _{j} = \psi _{j}(x, y, z, t)$ for $j = 1, 2, 3, 4$ satisfies the three spatial dimensions Klein–Gordon equation, thus

$$\begin{aligned} \frac{\partial ^2 \psi _{j}}{\partial t^2} - {c^2}{\nabla ^2 \psi _{j}} = -\left( \frac{e_0}{\hbar }\right) ^{2}\psi _{j}. \end{aligned}$$

(C.1)