# 218. Reconstruction of the Maxwell Pair from Energy Density and Flux The main text derives the transport core first and identifies the two complementary aspects $\mathbf E$ and $\mathbf B$ with the two transporting aspects $\mathbf F_+$ and $\mathbf F_-$. This appendix does not replace that derivation. It proves only the local converse: once a local energy density and energy flux are given, and once they obey the transport bound, one can reconstruct at least one Maxwell pair that reproduces them. This is a representation theorem, not a dynamical one. It shows what the local observables $(u,\mathbf S)$ are sufficient to encode. It does not by itself determine how the fields evolve. That still belongs to Maxwellian transport. ## 218.1 Problem Suppose a pointwise energy density and flux are given: $$ u(\mathbf r,t) > 0, \qquad \mathbf S(\mathbf r,t). $$ Assume the transport bound $$ |\mathbf S| \le c\,u. $$ We ask whether there exist vectors $\mathbf E$ and $\mathbf B$ such that $$ u = \frac{\varepsilon_0}{2}|\mathbf E|^2 + \frac{1}{2\mu_0}|\mathbf B|^2, $$ and $$ \mathbf S = \frac{1}{\mu_0}\,\mathbf E\times\mathbf B. $$ The answer is yes. ## 218.2 Reconstruction Theorem For every pair $(u,\mathbf S)$ with $u>0$ and $|\mathbf S|\le cu$, there exists at least one pair $(\mathbf E,\mathbf B)$ satisfying $$ u = \frac{\varepsilon_0}{2}|\mathbf E|^2 + \frac{1}{2\mu_0}|\mathbf B|^2, $$ $$ \mathbf S = \frac{1}{\mu_0}\,\mathbf E\times\mathbf B. $$ Moreover, the reconstruction is not unique. ## 218.3 Construction If $\mathbf S\neq 0$, let $$ \hat{\mathbf s}:=\frac{\mathbf S}{|\mathbf S|}. $$ If $\mathbf S=0$, choose any unit vector $\hat{\mathbf s}$. Choose any unit vector $\hat{\mathbf e}$ orthogonal to $\hat{\mathbf s}$, and define $$ \hat{\mathbf b}:=\hat{\mathbf s}\times\hat{\mathbf e}. $$ Then $(\hat{\mathbf e},\hat{\mathbf b},\hat{\mathbf s})$ is a right-handed orthonormal frame. Now define $$ \mathbf E := E\,\hat{\mathbf e}, \qquad \mathbf B := B\,\hat{\mathbf b}, $$ with scalars $E,B\ge 0$ to be chosen. Because $$ \hat{\mathbf e}\times\hat{\mathbf b}=\hat{\mathbf s}, $$ we have $$ \mathbf E\times\mathbf B = EB\,\hat{\mathbf s}. $$ So the flux condition becomes $$ \frac{1}{\mu_0}EB = |\mathbf S|. $$ The energy condition becomes $$ u = \frac{\varepsilon_0}{2}E^2 + \frac{1}{2\mu_0}B^2. $$ It is convenient to rescale: $$ X:=\sqrt{\varepsilon_0}\,E, \qquad Y:=\frac{B}{\sqrt{\mu_0}}. $$ Then the two conditions become $$ u=\frac{X^2+Y^2}{2}, $$ and, since $$ c^2=\frac{1}{\mu_0\varepsilon_0}, $$ also $$ |\mathbf S| = c\,X\,Y. $$ So it is enough to find nonnegative $X,Y$ such that $$ X^2+Y^2=2u, \qquad XY=\frac{|\mathbf S|}{c}. $$ ## 218.4 Existence Set $$ \rho := \frac{|\mathbf S|}{c\,u}. $$ By assumption, $$ 0\le \rho \le 1. $$ Choose an angle $\theta\in[0,\pi/4]$ such that $$ \sin(2\theta)=\rho. $$ Now define $$ X:=\sqrt{2u}\,\cos\theta, \qquad Y:=\sqrt{2u}\,\sin\theta. $$ Then $$ \frac{X^2+Y^2}{2} = \frac{2u(\cos^2\theta+\sin^2\theta)}{2} = u, $$ and $$ cXY = c(2u\cos\theta\sin\theta) = cu\sin(2\theta) = |\mathbf S|. $$ So the reconstructed pair $$ \mathbf E = \frac{X}{\sqrt{\varepsilon_0}}\,\hat{\mathbf e}, \qquad \mathbf B = \sqrt{\mu_0}\,Y\,\hat{\mathbf b} $$ satisfies the required relations exactly. Thus the reconstruction exists for every $(u,\mathbf S)$ satisfying $|\mathbf S|\le cu$. ## 218.5 Why the Bound Is Sharp The same inequality is also necessary. For any vectors $\mathbf E,\mathbf B$, $$ |\mathbf S| = \frac{1}{\mu_0}|\mathbf E\times\mathbf B| \le \frac{1}{\mu_0}|\mathbf E|\,|\mathbf B|. $$ Using the arithmetic-geometric mean inequality, $$ \frac{1}{\mu_0}|\mathbf E|\,|\mathbf B| \le \frac{c}{2}\left( \varepsilon_0|\mathbf E|^2+\frac{1}{\mu_0}|\mathbf B|^2 \right) = cu. $$ So any Maxwell pair must satisfy $$ |\mathbf S|\le cu. $$ The bound is therefore not an extra decoration. It is exactly the condition under which the local observables admit Maxwell representation. ## 218.6 Non-Uniqueness The reconstruction is not unique. First, the choice of transverse unit vector $\hat{\mathbf e}$ is arbitrary up to rotation in the plane orthogonal to $\hat{\mathbf s}$. That already gives a continuous family of solutions. Second, even after one pair $(\mathbf E,\mathbf B)$ is fixed, the duality rotation $$ \mathbf E'=\mathbf E\cos\phi + c\,\mathbf B\sin\phi, $$ $$ c\,\mathbf B'=-\mathbf E\sin\phi + c\,\mathbf B\cos\phi $$ leaves both $u$ and $\mathbf S$ unchanged. So $(u,\mathbf S)$ do not determine $(\mathbf E,\mathbf B)$ uniquely. The underdetermination is exactly the local polarization or duality freedom of the two-aspect transport. ## 218.7 What This Does and Does Not Determine The pair $(u,\mathbf S)$ determines: - whether a Maxwell reconstruction exists, - one local transport direction $\hat{\mathbf s}$ when $\mathbf S\neq 0$, - and the scalar load split consistent with the transport bound. It does not determine: - a unique polarization frame, - a unique duality phase, - or the dynamical evolution of the reconstructed fields. Those missing pieces are not flaws in the reconstruction. They are precisely what requires the full Maxwellian transport closure of chapter 7. ## 218.8 Final Statement The Maxwell pair is not an arbitrary superstructure placed on top of energy density and flux. Its primary origin in this book is still the double-curl transport closure. The present appendix says only that whenever the local observables satisfy $$ |\mathbf S|\le cu, $$ they already admit at least one exact Maxwell representation. the already-derived two-aspect transport admits at least one local Maxwell representation whose scalar and vector observables are $u$ and $\mathbf S$. What the representation alone cannot do is fix the evolution. That is the role of Maxwellian transport.