# 202. Nested Transport and Hyperbolic Composition The double-curl transport closure of chapter 7 determines the local transport cone. Reapplying $$ \nabla\times(\nabla\times\mathbf{F}) $$ acts again on field structure and raises the spatial operator. Nested transport is a different question. It belongs to kinematics: how do successive bounded transport increments compose when they occur in the same approximately uniform region and therefore share the same local transport speed $k$? To keep the discussion in one lab, consider motion along one spatial direction $x$. Let $u$ denote the speed produced from rest by one standard transport pulse in that region. If a body is already moving at speed $v$, let $$ v \oplus u $$ denote the speed measured in the same lab after applying that same standard pulse again. Because $k$ is the local transport speed singled out by the electromagnetic closure, no transport process native to that region can push a mode outside the admissible interval $|v|0$. Therefore the ratio transforms multiplicatively: $$ R \mapsto \frac{a}{b}R. $$ If one standard pulse sends rest to speed $u$, then since rest has $R(0)=1$, that same pulse has $$ R(u)=\frac{a}{b}. $$ Applying it to a state already moving at speed $v$ gives $$ R(v\oplus u)=R(u)\,R(v). $$ Therefore $$ \frac{k+v\oplus u}{k-v\oplus u} = \frac{k+u}{k-u}\cdot\frac{k+v}{k-v}. $$ Solving this relation gives $$ \boxed{ v\oplus u = \frac{v+u}{1+vu/k^2} }. $$ This is the hyperbolic composition law forced by bounded momentum-flux transport. The same result can be written geometrically from the transport cone. In the same approximately uniform region, the local transport speed $k$ picks out the lines $$ x = \pm kt, $$ which bound the local transport cone. Writing the corresponding null coordinates $$ \xi = t + \frac{x}{k}, \qquad \chi = t - \frac{x}{k}, $$ any orientation-preserving linear map fixing those two directions takes the form $$ \xi' = a\,\xi, \qquad \chi' = b\,\chi, $$ with $a,b>0$. For speed composition only the ratio matters, so write $$ \Lambda^2:=\frac{a}{b}. $$ Along a line of constant speed $v$, the null-coordinate ratio is $$ \frac{\xi}{\chi} = \frac{t+x/k}{t-x/k} = \frac{1+v/k}{1-v/k} = \frac{k+v}{k-v}. $$ So the momentum-flux ratio $R(v)$ is exactly the null-coordinate ratio. A cone-preserving map rescales it by $\Lambda^2$, which is the same multiplicative law obtained above from channel rebalancing. Only after this step is it useful to introduce an additive parameter. Taking the logarithm of $R(v)$ gives $$ \eta(v):=\frac12\ln\!\frac{k+v}{k-v}. $$ Then $$ \eta(v\oplus u)=\eta(v)+\eta(u). $$ Equivalently, $$ \eta(v)=\operatorname{artanh}\!\left(\frac{v}{k}\right), \qquad v=k\tanh\eta. $$ So successive identical pulses add linearly in $\eta$, not in $v$. If one pulse contributes $\eta_0$, then after $n$ identical pulses the lab speed is $$ v_n=k\tanh(n\eta_0). $$ So the distinction is exact: - double curl organizes source-free transport locally - repeated double curl changes field structure - nested transport composes successive transport increments that preserve the same local bound $k$ - preserving that cone forces hyperbolic composition The train-and-passenger image is therefore valid, but only at the kinematic level. One transport process may be nested inside another. The resulting composition is hyperbolic because the same local transport speed $k$ is preserved at each step.