= Solution
In the ultrarelativistic regime of <1/a>, retaining a constant mean fractional <energy> gain and ignoring losses gives, after $N$ individual crossings,
$$
\boxed{E_N\simeq E_0\left(1+\frac{u}{2c}\right)^N
\simeq E_0\exp\!\left(\frac{Nu}{2c}\right).}
$$
After $n$ complete return cycles, $N=2n$ and $E_n\simeq E_0(1+u/c)^n$, to the same first-order accuracy. These products approximate the mean amplification; individual <electron> crossing angles give a distribution around it.
The downstream <electron> distribution is nearly <isotropic>. Its crossing <particle flux> per unit area is $n_{\rm a}c/4$, where $n_{\rm a}$ is the accelerated-particle <number density>. The downstream flow permanently advects particles away at <particle flux> $n_{\rm a}u_2$. Their ratio estimates the escape probability after a downstream visit:
$$
P_{\rm esc}\simeq\frac{n_{\rm a}u_2}{n_{\rm a}c/4}
=\frac{4u_2}{c}=\frac uc,
\qquad \boxed{P_{\rm ret}\simeq1-\frac uc.}
$$
Upstream particles are swept back towards the <shock wave> in this idealized infinite-upstream model. Consequently this return probability applies once per \b[complete cycle], not once per individual crossing or local scattering event.
Let $A\simeq1+u/c$ be the amplification per complete cycle and $P=P_{\rm ret}$. A steady injection of $\mathcal N_0$ particles leaves $\mathcal N_0P^n$ surviving to reach $E=E_0A^n$. Eliminating $n$ gives the cumulative <power law> and differential <energy> distribution
$$
\mathcal N(>E)\propto E^{\ln P/\ln A},\qquad
\frac{d\mathcal N}{dE}\propto E^{-s},\qquad
s=1-\frac{\ln P}{\ln A}\simeq2.
$$
The escape distribution has the same differential exponent when $P$ is independent of <energy>. Thus the standard <Strong-shock Rankine-Hugoniot conditions> give
$$
\boxed{\mathcal N(>E)\propto E^{-1},\qquad d\mathcal N/dE\propto E^{-2}.}
$$
Equivalently the phase-space <one-particle distribution function> is $f(p)\propto p^{-4}$ for ultrarelativistic particles. Equal logarithmic intervals then contain equal particle <energy>. A finite source age, upstream escape, and radiative losses eventually introduce cutoffs or breaks.
A particle exponent is not automatically the <radiation spectral index>. For <optically thin> uncooled <synchrotron radiation> from $d\mathcal N/dE\propto E^{-s}$, the characteristic <frequency> scales as $\nu\propto E^2$ and the emitted <power> as $E^2$. Hence $L_\nu\propto E^{-s}E^2/(d\nu/dE)\propto\nu^{-(s-1)/2}$. With the convention $L_\nu\propto\nu^{-\alpha}$, this gives $\boxed{\alpha\simeq1/2}$ and $\nu L_\nu\propto\nu^{1/2}$. Strong <synchrotron cooling> steepens a continuously injected particle distribution by one exponent, giving $\alpha\simeq1$ above its cooling break. A <photon> number distribution $d\dot N_\gamma/d\varepsilon\propto\varepsilon^{-\Gamma}$ has $\Gamma=\alpha+1$, or $3/2$ in the uncooled case.
\b[Convention qualification.] If the printed $P=1-u/c$ is instead imposed independently at every individual crossing, its combination with $A_{\rm crossing}=1+u/(2c)$ gives $s\simeq3$ and uncooled $\alpha\simeq1$. For example $u/c=0.01$ gives $s=1-\ln(0.99)/\ln(1.005)\simeq3.015$, whereas the consistent cycle model gives $s\simeq2.010$. The standard $s=2$ estimate requires matching the probability and amplification to the same complete cycle.
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