---
title: The Physics of Energy Flow - Nested Transport and Hyperbolic Composition
date: 2026-03-13
---

# 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 $|v|<k$. So the composition law must preserve that bound.
This already rules out Galilean additivity as an exact transport law, since

$$
v+u
$$

can exceed $k$ even when both $|v|<k$ and $|u|<k$ separately hold. The problem
is therefore not whether the bound survives composition, but what exact shape
the composition law must take once that bound is respected.

The composition operation $\oplus$ should satisfy four basic requirements:

- identity: $v\oplus 0 = v$ and $0\oplus u = u$
- associativity: successive standard pulses can be grouped arbitrarily
- oddness: reversing both directions reverses the result
- boundedness: if $|v|<k$ and $|u|<k$, then $|v\oplus u|<k$

The same law can also be derived from momentum-flux bookkeeping, and this is
arguably the more physical route.

In one spatial direction, the local transport cone gives two causal channels:
forward and backward. Let $T_{++}\ge 0$ and $T_{--}\ge 0$ denote the
corresponding channel loads through a transverse cut. Then the total energy
density and flux are

$$
u = T_{++}+T_{--},
\qquad
S = k(T_{++}-T_{--}).
$$

For a coherent moving mode, the measured lab speed is therefore

$$
v=\frac{S}{u}
=
k\,\frac{T_{++}-T_{--}}{T_{++}+T_{--}}.
$$

So the speed is determined entirely by the ratio

$$
R(v):=\frac{T_{++}}{T_{--}}
=
\frac{k+v}{k-v}.
$$

A standard transport pulse preserving the same two causal channels cannot alter
the local bound $k$; it can only rebalance the forward and backward channel
loads. So such a pulse acts by

$$
T_{++}\mapsto a\,T_{++},
\qquad
T_{--}\mapsto b\,T_{--},
$$

with $a,b>0$. 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.