Some arguments for the wave equation in quantum theory 4

Proof. As all the particles in a simple system have zero acceleration, using the calculation of the Lienard-Wiechert potentials for a single particle, and the fact that all the acceleration fields vanish, we have that; \(lim_{r\rightarrow\infty}\int_{B(0,r)}div(\overline{E}_{t}\times\overline{B}_{t})d\overline{x}=0\), for the causal fields \((\overline{E},\overline{B})\). When we transform between frames, the particles still move with constant but differing velocities and again all the acceleration fields vanish, so the result holds for the causal fields in all frames. Thermal equilibrium in the base frame follows from the fact that \(|{\overline{J}\over \rho}|=d\), as all the particles have the same constant speed. For the second claim, as the velocity \(\overline{\lambda}\) is constant, and the paths are parallel, we must have that, \(\overline{J}=\overline{\lambda}\rho\). In order to satisfy the continuity equation, we must have that; \[{\partial \rho\over \partial t}=-div(\overline{J})=-div(\overline{\lambda}\rho)=-\overline{\lambda}\centerdot \bigtriangledown(\rho).\]

This is a transport equation with solution \(\rho(\overline{x},t)=g(\overline{x}-\overline{\lambda}t)\), where \(g\in C^{\infty}(\mathcal{R}^{3})\). For the third claim, in a divergent system, with the intersection point centred at the origin, we must have that; \(\overline{J}=d\rho {\overline{x}\over |\overline{x}|}\), \((V)\). By the continuity equation, we obtain that;
\[{\partial \rho \over \partial t}=-div(\overline{J})=-div(d\rho {\overline{x}\over |\overline{x}|})=-d\bigtriangledown(\rho)\centerdot {\overline{x}\over |\overline{x}|}-d\rho div({\overline{x}\over |\overline{x}|})=-d\bigtriangledown(\rho)\centerdot {\overline{x}\over |\overline{x}|}-{2d\rho\over |\overline{x}}=-{d\over |\overline{x}|}(\bigtriangledown(\rho)\centerdot \overline{x}-2\rho)(A)\]

If \((\overline{E},\overline{J})=0\) for the causal fields such that \((\rho,\overline{J},\overline{E},\overline{B})\) satisfies Maxwell’s equations, then, for any volume \(V\subset \mathcal{R}^{3}\), we would have that \({dW\over dt}=0\), for the total mechanical energy \(W\) of the charge distribution in \(V\). In particular, as all the particles are travelling with equal speed, we must have that;
\[{dQ\over dt}={d\over dt}\int_{V}\rho d\overline{x}=\int_{V}{\partial \rho\over \partial t}d\overline{x}=0,\]

so that, as the volume \(V\) was arbitrary, \({\partial \rho\over \partial t}=\) and \(\rho\) is time independent. From \((A)\), we then obtain that \(\bigtriangledown(\rho)\centerdot \overline{x}=2\rho\). However;
if \(\rho={1\over 2}(\rho_{x}x+\rho_{y}y+\rho_{z}z)\), then \(\rho_{x}={1\over 2}(\rho_{xx}x+\rho_{x}+\rho_{xy}y+\rho_{xz}z)\) and \(\bigtriangledown(\rho_{x})\centerdot\overline{x}=\rho_{x}\). Similarly, it follows that; \(\bigtriangledown(\rho_{y})\centerdot\overline{x}=\rho_{y}\) and \(\bigtriangledown(\rho_{z})\centerdot\overline{x}=\rho_{z}\).

Repeating the argument, we obtain that; \(\bigtriangledown(\rho_{xx})\centerdot\overline{x}=\bigtriangledown(\rho_{xy})\centerdot\overline{x}=\bigtriangledown(\rho_{xz})\centerdot\overline{x}=\bigtriangledown(\rho_{yy})\centerdot\overline{x}\)
\(\bigtriangledown(\rho_{yz})\centerdot\overline{x}=\bigtriangledown(\rho_{zz})\centerdot\overline{x}=0\)
It follows that: \[\rho_{rxx}=\rho_{rxy}=\rho_{rxz}=\rho_{ryy}=\rho_{ryz}=\rho_{rzz}=0,\rho_{rx}=C_{1}\rho_{ry}=C_{2}\rho_{rz}=C_{3}, \rho_{r}=C_{1}x+C_{2}y+C_{3}z.\]

By the divergence theorem and the continuity equation, we have that;

\[\int_{B(\overline{0},r)}{\partial \rho\over \partial t} d\overline{x}=-\int_{B(\overline{0},r)}div(\overline{J})d\overline{x}=-\int_{S(\overline{0},r)}\overline{J}\centerdot d\overline{S}=-\int_{S(\overline{0},r)}d\rho {\overline{x}\over |\overline{x}|}\centerdot {\overline{x}\over |\overline{x}|}dS=-d\int_{S(\overline{0},r)}\rho dS=0\]

so that \(\int_{S(\overline{0},r)}\rho dS=0\). It follows, by the Reynolds transport theorem, that;
\[{d\over dr}\int_{B(\overline{0},r)}\rho d\overline{x}=\int_{B(\overline{0},r)}{\partial \rho\over \partial r}d\overline{x}+\int_{S(\overline{0},r)}\rho dS=\int_{B(\overline{0},r)}\overline{C}\centerdot \overline{x}d\overline{x}=0\]

and;

\[\int_{B(\overline{0},r)}\rho d\overline{x}=0\;\; \textrm{for}\;\; r\in\mathcal{R} (E)\]

As \(\rho\geq 0\), the condition \((E)\) then implies that \(\rho=0\), so that \(\overline{J}=\overline{0}\) as well. We can therefore assume that for the causal fields \((\overline{E},\overline{B})\) in the based frame, that \((\overline{E},\overline{J})\neq 0\). As the set of causal fields in the frames \(S_{\overline{v}}\), where \(|\overline{v}|<c\) is definable, and the condition \((\overline{E},\overline{J})=0\) defines a closed set, we can assume that generically in \(S_{\overline{v}}\), \((\overline{E}_{\overline{v},causal},\overline{J}_{\overline{v}})\neq 0\). We can then apply the arguments in [2], together with the classically non-radiating property proved above, to conclude that there is a transfer of mechanical energy between two volumes \(\{S,S_{\kappa}\}\). As this occurs over a finite time interval \((t_{0}=\epsilon,t_{0}+\epsilon)\) we can construct corresponding volumes \(\{T,T_{\kappa}\}\) in the base frame. Generically, either the energy change in \(T\) is positive and the energy change in \(T_{\kappa}\) is negative ot the energy change in \(T\) is negative and the energy change in \(T_{\kappa}\) is positive, both of which contradict thermal equilibrium in the base frame. Alternatively, the energy change in \(\{T,T_{\kappa}\}\) is of the same sign, in which case, as the process is reversible, we can assume that both energy changes are negative. In this case, we can assume that energy is transferred into the field, again a contradiction. By the main result of [2], we can conclude that \((\rho,\overline{J})\) must satisfy the wave equations \(\square^{2}\rho=0\), \(\square^{2}\overline{J}=\overline{0}\) in the base frame, as well as the connecting relation \(\bigtriangledown(\rho)+{1\over c^{2}}{\partial \overline{J}\over \partial t}=\overline{0}\). We have, by the definition of \(\overline{J}\), that; \(\bigtriangledown(\rho)+{1\over c^{2}}{\partial \overline{J}\over \partial t}=\overline{0}\) iff \(\bigtriangledown(\rho)+{1\over c^{2}}d{\overline{x}\over |\overline{x}|}{\partial \rho\over \partial t}=\overline{0}\).

In particularly, \(\bigtriangledown(\rho)\) is parallel to \(\overline{x}\), so that \(\rho\) is constant on spheres \(S(\overline{0},r)\), for \(r\in\mathcal{R}_{>0}\). We have that \(\square^{2}(\rho)=0\), so writing the Laplacian \(\bigtriangledown^{2}\) in polar form, we obtain that; \({1\over r}{\partial^{2}\over \partial r^{2}}(r\rho)-{1\over c^{2}}{\partial^{2}\rho\over \partial t^{2}}=0\) so that \(r\rho(r,t)\) satisfies the \(1\)-dimensional wave equation \(\square_{r}^{2}(r\rho)=0\), \((K)\), with speed \(c\). Similarly, \(\square_{r}^{2}(r{\partial\rho\over \partial t})=0\). From \((A)\), we have that;
\({\partial \rho\over \partial t}=-{d\over |\overline{x}|}(\bigtriangledown(\rho)\centerdot \overline{x}-2\rho)\) \((A)\)
\(=-d{\partial \rho\over \partial r}-{2d\rho\over r}\)
so that; \(r{\partial \rho\over \partial t}=-dr{\partial \rho\over \partial r}-2d\rho\) and, as \(\square_{r}^{2}(r{\partial \rho\over \partial t})=0\), we have that; \(\square_{r}^{2}(-dr{\partial \rho\over \partial r}-2d\rho)=0\), \((F)\). Also; \(d{\partial (r\rho)\over \partial r}=dr{\partial \rho\over \partial r}+d\rho\) so that as \(d\square_{r}^{2}({\partial (r\rho)\over \partial r})=0\), we have that; \(\square_{r}^{2}(dr{\partial \rho\over \partial r}+d\rho)=0\) \((G)\).

Combining \((F),(G)\) gives that \(\square^{2}_{r}(\rho)=0\). It follows from this and \((K)\) that;
\({\partial^{2 }\rho\over \partial^{2}r}={1\over c^{2}}{\partial^{2}\rho\over \partial^{2}t}\)
\[{\partial^{2}(r\rho)\over \partial r^{2}}={\partial (\rho+r{\partial \rho\over \partial r})\over \partial r}=2{\partial \rho\over \partial r}+r{\partial^{2}\rho\over \partial^{2}r}={1\over c^{2}}{\partial^{2}\rho\over \partial^{2}t}\]

so that; \((1-r){\partial^{2}\rho\over \partial^{2}r}=2{\partial \rho\over \partial r}\). Letting \(g={\partial \rho\over \partial r}\), we have that; \((1-r){\partial g\over \partial r}=2g\), so that \(g=A(t)e^{-2ln(1-r)}=A(t)(1-r)^{-2}={A(t)\over (1-r)^{2}}\) and \(\rho={A(t)\over (1-r)}+B(t)\). It follows from \(\square^{2}_{r}(\rho)=0\), that; \({1\over c^{2}}({A''(t)\over (1-r)}+B''(t))={2A(t)\over (1-r)^{3}}\), so that \(A''(t)=B''(t)=A(t)=0\), and \(\rho=B(t)=\alpha t+\beta\). It follows that \(\overline{J}=d(\alpha t+\beta){\overline{x}\over |\overline{x}|}\) and clearly, we cannot have that \(\square^{2}\overline{J}=\overline{0}\). ◻

Proof. As the acceleration is zero, the velocity \(\lambda\) is constant, so that, \(\overline{J}=\overline{\lambda}\rho\), \((A)\). By the continuity equation and the connecting relation, we have that; \[{\partial \rho\over \partial t}=-div(\overline{J})=-div(\overline{\lambda}\rho)=-\overline{\lambda}\centerdot \bigtriangledown(\rho)=\overline{\lambda}\centerdot (-{1\over c^{2}}{\partial\overline{J}\over \partial t})=-{\overline{\lambda}\over c^{2}}\centerdot {\partial\overline{J}\over\partial t}={\partial \over \partial t}({\overline{\lambda}\over c^{2}}\centerdot \overline{J})\]

so that, by the FTC;

\[\rho={\overline{\lambda}\over c^{2}}\centerdot\overline{J}+d(\overline{x})={\overline{\lambda}\over c^{2}}\centerdot \overline{\lambda}\rho+d(\overline{x})={|\lambda^{2}|\over c^{2}}\rho+d(\overline{x})\]

so that either \(\rho\) is time independent or \(|\lambda|=c\) and \(d(\overline{x})=0\). In the first case, as \(\square^{2}\rho=0\), we have that \(\bigtriangledown^{2}(\rho)=0\) and \(\rho\) is harmonic. Then, as \(\overline{J}=\overline{\lambda}\rho\), we have that, as \(\square^{2}\overline{J}=\overline{0}\), and the components of \(\overline{J}\) are time independent, that the components of \(\overline{J}\) are harmonic. In the second case, we obtain that \(\rho={\overline{\lambda}\over c^{2}}\centerdot\overline{J}\), \((B)\). The penultimate claim is clear.
For the final claim concerning thermal equilibrium, we have that, \(\overline{J}=\overline{\lambda}\rho\), with \(|\overline{\lambda}|=d\), so that \({\partial \overline{\lambda}\over \partial t}\centerdot \overline{\lambda}=0\). As above, we have that;
\({\partial \rho\over \partial t}=-div(\overline{J})=-div(\rho \overline{\lambda})=-\overline{\lambda}\centerdot \bigtriangledown(\rho)-\rho div(\overline{\lambda})(U)=-\overline{\lambda}\centerdot(-{1\over c^{2}}{\partial \overline{J}\over \partial t})-\rho div(\overline{\lambda})={\overline{\lambda}\over c^{2}}\centerdot {\partial \overline{J}\over \partial t}-\rho div(\overline{\lambda})={\partial \over \partial t}({\overline{\lambda}\centerdot \overline{J}\over c^{2}})-{1\over c^{2}}{\partial \overline{\lambda}\over \partial t}\centerdot \overline{J}-\rho div(\overline{\lambda})={\partial \over \partial t}({\overline{\lambda}\centerdot \overline{\lambda}\rho\over c^{2}})-{\rho\over c^{2}}{\partial \overline{\lambda}\over \partial t}\centerdot \overline{\lambda}-\rho div(\overline{\lambda})={\partial \over \partial t}{d^{2}\rho\over c^{2}}-\rho div(\overline{\lambda})\).

If \(d=c\), we obtain \(\rho div(\lambda)=0\), so that with analyticity assumptions, \(\rho\neq 0\), we obtain that \(div(\lambda)=0\). If \(d\neq c\), letting \(\epsilon=1-{d^{2}\over c^{2}}\), we have that; \(\epsilon{\partial \rho\over \partial t}=-\rho div(\overline{\lambda})\) iff \(\epsilon(div(\overline{J}))=\rho div(\overline{\lambda})\).

Again, we have that; \[div(\overline{J})=\bigtriangledown(\rho)\centerdot\overline{\lambda}+\rho div(\overline{\lambda})\] which implies that; \({\rho\over \epsilon}div(\overline{\lambda})=\bigtriangledown(\rho)\centerdot\overline{\lambda}+\rho div(\overline{\lambda})\), so that; \(\rho div(\overline{\lambda})({1\over \epsilon}-1)=\bigtriangledown(\rho)\centerdot\overline{\lambda}\) and, from \((U)\); \({\partial \rho\over \partial t}=-\overline{\lambda}\centerdot \bigtriangledown(\rho)-{\epsilon\over \epsilon-1}\overline{\lambda}\centerdot \bigtriangledown(\rho)={1-2\epsilon\over \epsilon-1}\overline{\lambda}\centerdot \bigtriangledown(\rho)={1-2\epsilon\over \epsilon-1}\overline{\lambda}\centerdot -{1\over c^{2}}{\partial \overline{J}\over \partial t}={1-2\epsilon\over \epsilon-1}\overline{\lambda}\centerdot -{1\over c^{2}}{\partial\over \partial t}(\rho \overline{\lambda})=-{1-2\epsilon\over \epsilon-1}{1\over c^{2}}(d^{2}){\partial \rho\over \partial t}\), so that; \((1+{1-2\epsilon\over \epsilon-1}{1\over c^{2}}(d^{2})){\partial \rho\over \partial t}=0\) iff \(2(1-{d^{2}\over c^{2}}){\partial \rho\over \partial t}=0\)
so that \(\rho\) is time independent. The conclusion that \(\rho\) is harmonic follows again from \(\square^{2}(\rho)=0\). ◻

Proof. As the system is parallel, by the proof of Lemma 1, we have that; \(\overline{J}=\overline{\lambda}\rho\) and \({\partial \rho\over \partial t}=-\overline{\lambda}\centerdot \bigtriangledown(\rho)\) \((A)\). The connecting relation implies that; \[\bigtriangledown(\rho)+{1\over c^{2}}{\partial\overline{J}\over \partial t}=\bigtriangledown(\rho)+{1\over c^{2}}{\partial(\overline{\lambda}\rho)\over \partial t}=\bigtriangledown(\rho)+{\overline{\lambda}\over c^{2}}{\partial \rho \over \partial t}=\overline{0},\] so that \(\bigtriangledown(\rho)=-{\overline{\lambda}\over c^{2}}{\partial \rho \over \partial t}\) \((B)\).

Conversely, if \(\rho\) satisfies \((A),(B)\), with \(\overline{J}=\overline{\lambda}\rho\), then we obtain a parallel system with the additional connecting relation. By \((A)\), we have that \(\rho(\overline{x},t)=g(\overline{x}-\overline{\lambda}t)\), \((C)\). In the case that \(\overline{\lambda}=(c,0,0)\), we have from \((B),(C)\), that we require; \((g_{x},g_{y},g_{z})=-{1\over c^{2}}(c,0,0){\partial\rho \over \partial t}\), so that, in particular, \(g_{y}=g_{z}=0\), \(g(\overline{x})=h(x)\) and then; \(\rho(\overline{x},t)=g(\overline{x}-\overline{\lambda}t)=h(x-ct)\). Observe that if \(\bigtriangledown(\rho)=\overline{\lambda} \delta(\overline{x},t)\), then, as \(|\lambda|=c\); \[{\partial \rho\over \partial t}=-\overline{\lambda}\centerdot \bigtriangledown(\rho)=-\overline{\lambda}\centerdot \lambda \delta(\overline{x},t)=-c^{2}\delta(\overline{x},t)\] and \(\delta(\overline{x},t)=-{1\over c^{2}}{\partial\rho\over\partial t}\), so that; \(\bigtriangledown(\rho)=-{\overline{\lambda}\over c^{2}}{\partial \rho\over \partial t}\), which is \((B)\), \((*)\).

We have that \(\rho_{x}=g_{x}\), \(\rho_{xx}=g_{xx}\), \(\rho_{y}=g_{y}=0\), \(\rho_{yy}=g_{yy}=0\), \(\rho_{z}=g_{z}=0\), \(\rho_{zz}=g_{zz}=0\), \(\rho_{t}=h_{x}(-c)\), \(\rho_{tt}=h_{xx}c^{2}\), so that \(\bigtriangledown^{2}(\rho)-{1\over c^{2}}{\partial^{2}\rho\over \partial t^{2}}=h_{xx}-h_{xx}=0\) and \(\square^{2}(\rho)=0\), so that \(\square^{2}(\overline{J})=\overline{0}\). In the case of an arbitrary \(\overline{\lambda}\) with \(|\overline{\lambda}|=c\), choose an orthogonal matrix \(U\) with \(U(1,0,0)={\overline{\lambda}\over c}\). We have found \(\rho(\overline{x},t)\) such that; \(\bigtriangledown(\rho)(\overline{x},t)=-{1\over c^{2}}(1,0,0)m(\overline{x}.t)\), so that, applying \(U\) to both sides, we have that;

\[U(\bigtriangledown(\rho))(\overline{x},t)=\bigtriangledown(\rho)(U^{-1}(\overline{x}),t)=-{1\over c^{2}}{\overline{\lambda}\over c}m(\overline{x}.t)\]

We have that;

\[\rho(U^{-1}(\overline{x}),t)=g(U^{-1}(\overline{x})-\overline{\lambda}t)=h(U^{-1}(\overline{x})_{1}-ct)=h({\lambda_{1}\over c}x+{\lambda_{2}\over c}y+{\lambda_{3}\over c}z-ct).\]

Observe that if \(g(\overline{x})=h({\lambda_{1}\over c}x+{\lambda_{2}\over c}y+{\lambda_{3}\over c}z)\), then; \(g(\overline{x}-\overline{\lambda}t)=h({\lambda_{1}\over c}x+{\lambda_{2}\over c}y+{\lambda_{3}\over c}z-{\lambda_{1}^{2}-\lambda_{2}^{2}-\lambda_{3}^{2}\over c}t)=h({\lambda_{1}\over c}x+{\lambda_{2}\over c}y+{\lambda_{3}\over c}z-ct)\), so that \((A)\) is still satisfied, and, by the above remark \((*)\), \((B)\) is satisfied as well, as \(\bigtriangledown(\rho\circ U^{-1})\) is parallel to \(\lambda\). Again \(\rho\circ U^{-1}\) satisfies the wave equation \(\square^{2}(\rho\circ U^{-1})=\bigtriangledown^{2}(\rho\circ U^{-1})-{1\over c^{2}}{\partial^{2}\rho\circ U^{-1}\over \partial t^{2}}=0\). Defining \(\overline{J}=\overline{\lambda}\rho\circ U^{-1}\), we have that \(\square^{2}(\overline{J})=\overline{0}\) as well. ◻

Proof. For the first claim, we have that \({\overline{J}\over \rho}\centerdot \overline{x}=0\), in particular \(\overline{J}\centerdot \overline{x}=0\). As \(\square^{2}(\overline{J})=\overline{0}\), and the continuity equations holds, we have that; \[\begin{aligned} \square^{2}(\overline{J}\centerdot \overline{x})&=({\partial^{2}\over \partial x^{2}}+{\partial^{2}\over \partial y^{2}}+{\partial^{2}\over \partial z^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})(j_{1}x+j_{2}y+j_{3}z)\\ &={\partial^{2}\over \partial x^{2}}(j_{1}x)+{\partial^{2}\over \partial y^{2}}(j_{2}y)+{\partial^{2}\over \partial z^{2}}(j_{3}z)+(({\partial^{2}\over \partial y^{2}}+{\partial^{2}\over \partial z^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{1})x+(({\partial^{2}\over \partial y^{x}}+{\partial^{2}\over \partial z^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{2})y\\ &+(({\partial^{2}\over \partial x^{2}}+{\partial^{2}\over \partial y^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{3})z\\ &=2{\partial j_{1}\over \partial x}+2{\partial j_{2}\over \partial y}+2{\partial j_{3}\over \partial z}+({\partial^{2}\over \partial x^{2}}j_{1})x+({\partial^{2}\over \partial y^{2}}j_{2})y+({\partial^{2}\over \partial z^{2}}j_{3})z+(({\partial^{2}\over \partial y^{2}}+{\partial^{2}\over \partial z^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{1})x\\ &+(({\partial^{2}\over \partial y^{x}}+{\partial^{2}\over \partial z^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{2})y+(({\partial^{2}\over \partial x^{2}}+{\partial^{2}\over \partial y^{2}}-{1\over c^{2}}{\partial^{2}\over \partial t^{2}})j_{3})z\\ &=2div(\overline{J})+(\square^{2}j_{1})x+(\square^{2}j_{2})y+(\square^{2}j_{3})z=2div(\overline{J})=-2{\partial \rho\over \partial t}=0, \end{aligned}\]

so that \(\rho\) is time independent and, as \(\square^{2}(\rho)=0\), that \(\rho\) is harmonic. As \(\bigtriangledown(\rho)\) is time independent, by the connecting relation \(\bigtriangledown(\rho)+{1\over c^{2}}{\partial\overline{J}\over \partial t}=\overline{0}\), we have that \({\partial \overline{J}\over \partial t}\) is time independent and \({\partial^{2}\overline{J}\over \partial t^{2}}=\overline{0}\), so that, as \(\square^{2}\overline{J}=\overline{0}\), the the components of \(\overline{J}_{t}\) are harmonic for \(t\in\mathcal{R}\). For the second claim, we have that, if \(\overline{\gamma}\) is a trajectory;
\[\begin{aligned} {d\rho\over ds}(\overline{\gamma}(s),t)&=\bigtriangledown(\rho)|_{\overline{\gamma}(s),t}\centerdot \overline{\gamma}'(s)=-{1\over c^{2}}{\partial\overline{J}\over \partial t}|_{\overline{\gamma}(s),t}\centerdot {\overline{J}\over \rho}|_{\overline{\gamma}(s),t}\\ &=-{1\over c^{2}\rho}|_{\overline{\gamma}(s),t}({\partial\overline{J}\over \partial t}\centerdot \overline{J})|_{\overline{\gamma}(s),t}=-{1\over 2c^{2}\rho}|_{\overline{\gamma}(s),t}{\partial \over \partial t}(\overline{J}\centerdot \overline{J})|_{\overline{\gamma}(s),t}\\ &=-{1\over 2c^{2}\rho}|_{\overline{\gamma}(s),t}{\partial \over \partial t}d^{2}\rho^{2}|_{\overline{\gamma}(s),t}=-{d^{2}\over c^{2}\rho}|_{\overline{\gamma}(s),t}\rho|_{\overline{\gamma}(s),t}{\partial \rho\over \partial t}|_{\overline{\gamma}(s),t}\\ &=-{d^{2}\over c^{2}}{\partial \rho\over \partial t}|_{\overline{\gamma}(s),t}. \end{aligned}\] If a closed trajectory \(\overline{\gamma}\) is disjoint from the locus of \({\partial \rho\over \partial t}_{t_{0}}\), with \(\overline{\gamma}(a)=\overline{\gamma}(b)\), for some \(t_{0}\in\mathcal{R}\), then either \({\partial \rho\over \partial t}|_{W}>0\) or \({\partial \rho\over \partial t}|_{W}<0\). We have that;
\[\int_{W}\bigtriangledown(\rho)\centerdot d\overline{\gamma}=\int_{a}^{b}{d\rho\over ds}(\overline{\gamma}(s),t)ds=-{d^{2}\over c^{2}}\int_{a}^{b}{\partial \rho\over \partial t}(\gamma(s),t)ds=\overline{\gamma}(b)-\overline{\gamma}(a)=0\]

which is a contradiction. For the next claim, we calculate, if \(\overline{\gamma}\) is a flowline; \[\begin{aligned} {d\rho\over ds}(\overline{\gamma}(s),s)&=\bigtriangledown(\rho)|_{\overline{\gamma}(s),s}\centerdot \overline{\gamma}'(s)+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}=-{1\over c^{2}}{\partial\overline{J}\over \partial t}|_{\overline{\gamma}(s),s}\centerdot {\overline{J}\over \rho}|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}\\ &=-{1\over c^{2}\rho}|_{\overline{\gamma}(s),s}({\partial\overline{J}\over \partial t}\centerdot \overline{J})|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}=-{1\over 2c^{2}\rho}|_{\overline{\gamma}(s),s}{\partial \over \partial t}(\overline{J}\centerdot \overline{J})|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}\\ &=-{1\over 2c^{2}\rho}|_{\overline{\gamma}(s),s}{\partial \over \partial t}d^{2}\rho^{2}|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}=-{d^{2}\over c^{2}\rho}|_{\overline{\gamma}(s),s}\rho|_{\overline{\gamma}(s),s}{\partial \rho\over \partial t}|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}\\ &=-{d^{2}\over c^{2}}{\partial \rho\over \partial t}|_{\overline{\gamma}(s),s}+{\partial \rho \over \partial t}|_{\overline{\gamma}(s),s}=0, \end{aligned}\] as \(d=c\), by Lemma 2. If \(\overline{\gamma}\) is a flowline, then it cannot intersect the locus of \(\overline{J}_{t}=\overline{0}\) at any time \(t\), as we have thermal equilibrium \(\overline{c}\), with \(|\overline{c}|=c\). It follows that the flowline cannot have any equilibrium points in its closure, and if it is bounded, must be a closed loop or spirals into a closed loop. The first case can be excluded, by O-minimality of the real closed field with Pfaffian functions, using the fact that for a point \(\overline{x}\) on the loop \(Range(\overline{\gamma})\), we can define;
\(\{t\in \mathcal{R}_{>0}:\overline{\gamma}(t)=\overline{x}\}\)
which cannot be a finite union of points and intervals, as the set is unbounded and discrete, see [8] and [9]. More specifically, we can approximate \({\overline{J}\over \rho}\) by a polynomial vector field \(\overline{W}\), using the Stone-Weierstrass approximation theorem, on an open ball \(B(\overline{0},r_{0})\) containing the loop, in such a way that a flowline \(\overline{\gamma}_{1}\) for \(\overline{W}\) is still a bounded closed loop. This follows as the cycle maps \(\theta_{t_{0},t}\) for \({\overline{J}\over \rho}\), defined by;
\(\theta_{t_{0},t}(\overline{x})=\overline{\gamma}(t_{0}+t)\)
where \(\overline{\gamma}(t_{0})=\overline{x}\) are invertible, by reversing the flow, letting;
\({\overline{J}\over \rho}^{rev}(\overline{x},s)=-{\overline{J}\over \rho}(\overline{x},t_{0}+t-s)\), \(0\leq s\leq t\)
and using the cycle map \(\theta^{rev}_{0,t}\) for the corresponding flowline. It follows, by continuity, that the cycle maps \(\theta_{t_{0},t}\) are proper and the deviation map \(\phi_{t_{0},t}(\overline{x})=\theta_{t_{0},t}(\overline{x})-\overline{x}\) is proper. As \(\overline{\gamma}\) forms a closed loop, we have that \(\overline{0}\in \phi_{t_{0},t}(B^{\circ}(\overline{0},r_{0}))\) and not a boundary point. By continuity, when we construct \(\overline{W}\), we can still obtain a closed loop. In the second case, we could find a limit point \(\overline{x}\in Range(\overline{\gamma})\), which is not an equilibrium point and which, wlog, is not on the trajectory, \((*)\). This follows as if it lies on the trajectory twice, we would form a closed loop. By the argument \((X)\) below, we can exclude the case that \(\overline{x}\) is an endpoint of \(\overline{\gamma}\). It follows that it must be part of a compact \(\omega\)-limit set with no fixed points. By the Poincare-Bendixson Theorem, applied to planar projections of the system in \(\mathcal{R}^{3}\), we would obtain a periodic orbit again, and we can repeat the argument to obtain a contradiction. We can, therefore assume that the system is open.

If the system is not parallel or divergent, then there exists a ball \(B(\overline{x},r)=\bigcup_{w\in I}\{\overline{\gamma}_{w}(t):0\leq t\leq t_(w)\}\), where \(I\) is a closed interval and \(\{\overline{\gamma}_{w}:w\in I\}\) is a set of flowlines, with the properties that;

(i). \(\overline{\gamma}_{w}(0)\) and \(\overline{\gamma}_{w}(t_{w})\) are the unique points of \(\overline{\gamma}_{w}([0,t_{w}])\) lying on \(\delta B(\overline{x},r)\).

\((ii)\). We have that \(-\overline{c}(\overline\gamma_{w}(0))\centerdot \hat{\overline{n}}>\overline{c}(\overline\gamma_{w}(t_{w}))\centerdot \hat{\overline{n}}\)

The property \((i)\) follows from the fact that the flowlines are unbounded and the existence of flowlines through a given point, the last property being a consequence of Peano’s existence theorem. The property \((ii)\) follows from the asymmetry of a path which is not straight and continuity. By the above, we have that \(\rho\) is constant along the flowline in the sense that \(\rho(\overline{\gamma}_{w}(s),s)=f(w)\) for some smooth function \(f\), \((S)\). By the continuity equation and divergence theorem, we have that;

\[\int_{B(\overline{x},r)}{\partial \rho\over \partial t}dB=-\int_{\delta B(\overline{x},r)}\overline{J}\centerdot \hat{\overline{n}}dS=-\int_{\delta B(\overline{x},r)}\rho \overline{c}\centerdot \hat{\overline{n}}dS(T)\]

so that by \((S),(T)\) and the properties \((i),(ii)\), the charge \(\rho\) would monotonically increase or decrease inside the ball \(B(\overline{x},r)\). We can then either use the fact that \(\rho\geq 0\) or the fact that \(\square^{2}(\rho)=0\), together with Kirchoff’s formula for \(\rho\), with initial conditions in \(S(\mathcal{R}^{3})\), so that \(|\rho_{t}|\leq {M\over t}\), where \(M\) is independent of \(t\), to get a contradiction. The addition of a constant doesn’t effect the argument.

We show that every trajectory \(\overline{\gamma}:[0,\infty)\rightarrow \mathcal{R}\) doesn’t have an endpoint, \((X)\). Suppose that a trajectory has an endpoint \(\overline{x}\). Then, by thermal equilibrium, we may suppose that \({\overline{J}\over \rho}(\overline{x})=\overline{v}\neq \overline{0}\). Without loss of generality, assume that \(v_{2}=v_{3}=0\). By continuity, we may suppose that \(|{j_{1}\over \rho}|\geq |{v_{1}\over 2}|\), for \(\overline{x}\in B(\overline{x},r)\), with \(r<{|v_{1}|\over 2}\), and, there exists \(t_{0}\in\mathcal{R}_{>0}\), with \(|\overline{\gamma}(t)-\overline{x}|<r<{|v_{1}|\over 2}\) \((A)\), for \(t\geq t_{0}\). By the intermediate value theorem, we have that; \[|(\gamma_{\overline{v}}(t_{0}+s))_{1}-(\gamma_{\overline{v}}(t_{0}))_{1}|=|s(\gamma'_{\overline{v}}(t_{0}+s_{0}))_{1}|\geq sv_{1}\over 2\] with \(s_{0}\in (0,s)\), so that if \(s>3\), we obtain a contradiction with \((A)\), as;

\(|(\gamma_{\overline{v}}(t_{0}+s))-(\gamma_{\overline{v}}(t_{0}))|\geq |(\gamma_{\overline{v}}(t_{0}+s))_{1}-(\gamma_{\overline{v}}(t_{0}))_{1}|> {v_{1}}\)

and, by \((A)\);

\(|(\gamma_{\overline{v}}(t_{0}+s))-(\gamma_{\overline{v}}(t_{0}))|\leq |(\gamma_{\overline{v}}(t_{0}+s))-\overline{x}|+|(\gamma_{\overline{v}}(t_{0}))-\overline{x}|\leq 2{v_{1}\over 2}=v_{1}\) \((X)\)
which is a contradiction. ◻