## Maxwell-transport invariance between isolated observers We need a statement that does not smuggle “inertial frames” as a primitive. Define an **isolated observer** as an observer describing a region in which the energy-flow pattern is unforced and unperturbed, so that the net momentum flux through a large closed surface vanishes. In the flow-first language, this is the condition that there is no preferred direction of net $\mathbf{S}/c^2$ momentum throughput in that region. Now impose the only requirement we need: **Requirement (transport-law sameness).** Two isolated observers describing the same electromagnetic transport process must be able to write the same vacuum Maxwell transport law. In vacuum, Maxwell implies the wave operator $$ \mathcal{O}_c := \partial_t^2 - c^2 \nabla^2, $$ so transport-law sameness is the statement that the same process satisfies $$ \mathcal{O}_c \Phi = 0 \quad\Longleftrightarrow\quad \mathcal{O}'_c \Phi = 0, $$ for any field component $\Phi$, with the same constant $c$, when expressed in either observer’s coordinates. Equivalently: the wave operator is preserved up to an overall nonzero factor (which does not change the solution set of a linear homogeneous PDE): $$ \mathcal{O}_c = \lambda\,\mathcal{O}'_c,\qquad \lambda\neq 0. $$ This is not a statement about “space” having a metric. It is a statement about two descriptions of the same transport process agreeing on the transport PDE. ## Hyperbolic mixing is forced by operator invariance Assume the two isolated observers are in relative uniform translation along the $x$-axis. By homogeneity and straight-line preservation, the re-description is linear. By symmetry in the transverse directions, take $y'=y$, $z'=z$. Write the most general linear mixing of $t$ and $x$: $$ x' = a x + b t,\qquad t' = d x + e t, $$ with constants $a,b,d,e$ depending only on the relative translation rate. Compute how derivatives transform (chain rule): $$ \partial_x = a\,\partial_{x'} + d\,\partial_{t'},\qquad \partial_t = b\,\partial_{x'} + e\,\partial_{t'}. $$ Now impose wave-operator invariance in 1D: $$ \partial_t^2 - c^2 \partial_x^2 = \lambda\left(\partial_{t'}^2 - c^2 \partial_{x'}^2\right). $$ Expand the left-hand side: $$ (b\partial_{x'} + e\partial_{t'})^2 - c^2(a\partial_{x'} + d\partial_{t'})^2. $$ Collect coefficients of $\partial_{x'}^2$, $\partial_{x'}\partial_{t'}$, $\partial_{t'}^2$: 1) Mixed term must vanish (since the RHS has no $\partial_{x'}\partial_{t'}$): $$ 2(be - c^2 a d)=0 \quad\Longrightarrow\quad be = c^2 a d. $$ 2) Coefficients must match $\lambda(\partial_{t'}^2 - c^2\partial_{x'}^2)$: $$ e^2 - c^2 d^2 = \lambda, $$ $$ b^2 - c^2 a^2 = -\lambda c^2. $$ Now identify the relative translation rate $v$ by the motion of the primed origin $x'=0$: $$ 0 = a x + b t \quad\Rightarrow\quad x = -\frac{b}{a}t, \qquad v := \frac{dx}{dt} = -\frac{b}{a}. $$ So $b = -a v$. Use $be=c^2ad$: $$ (-av)e = c^2 a d \quad\Rightarrow\quad d = -\frac{v e}{c^2}. $$ Plug into $e^2 - c^2 d^2 = \lambda$: $$ e^2 - c^2\left(\frac{v^2 e^2}{c^4}\right)=\lambda \quad\Rightarrow\quad e^2\left(1-\frac{v^2}{c^2}\right)=\lambda. $$ Plug $b=-av$ into $b^2 - c^2 a^2 = -\lambda c^2$: $$ a^2 v^2 - c^2 a^2 = -\lambda c^2 \quad\Rightarrow\quad a^2\left(1-\frac{v^2}{c^2}\right)=\lambda. $$ Thus $a^2=e^2$. Choose the orientation-preserving branch $a=e$. Fix the overall scale by choosing $\lambda=1$ so that the inverse has the same form. Then $$ a=e=\gamma,\qquad \gamma:=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, $$ and therefore $$ b=-\gamma v,\qquad d=-\gamma\frac{v}{c^2}. $$ We have derived the unique linear mixing forced by $\mathcal{O}_c$ invariance: $$ x'=\gamma(x-vt),\qquad t'=\gamma\left(t-\frac{v}{c^2}x\right),\qquad y'=y,\quad z'=z. $$ This is the precise sense in which “hyperbolic mixing” is not assumed: it is forced by preserving the Maxwell transport operator. ## Hyperbolic velocity composition follows immediately (no Galilean c±v) Let a signal or feature move with $u=dx/dt$. Differentiate: $$ dx'=\gamma(dx-v\,dt), $$ $$ dt'=\gamma\left(dt-\frac{v}{c^2}dx\right). $$ Thus $$ u'=\frac{dx'}{dt'}=\frac{u-v}{1-\frac{uv}{c^2}}. $$ In particular, if $u=c$, then $u'=c$: $$ \frac{c-v}{1-\frac{v}{c}}=c. $$ This is the exact point where the classical ether-drift argument fails: it uses additive composition $u\mapsto u\pm v$, which contradicts the operator invariance of Maxwell transport. ## Virtual observer at the end of an arm: why v --- - [Preferred Frame Writing on GitHub.com](https://github.com/siran/writing) (built: 2026-03-19 17:43 EDT UTC-4)