= Mean logarithmic energy decrement
{title2=$\xi=\mathbb E[\log(E/E')]$}
Consider a <nonrelativistic particle> of <mass> $m$ undergoing an <elastic collision> with a stationary free target of <mass> $M=Am$, where $A\geq1$. Assume the outgoing direction is uniformly distributed over solid angle in the <center of mass> frame; this assumption concerns the <scattering angle> distribution and need not hold for every target or <energy>. Let $\mu$ be the cosine of that frame's <scattering angle>. Transforming the outgoing <velocity> back to the laboratory gives
$$
x=\frac{E'}{E}=\frac{A^2+1+2A\mu}{(A+1)^2},\qquad
\alpha=\left(\frac{A-1}{A+1}\right)^2\leq x\leq1.
$$
The uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha,1]$. For $A>1$, integration yields
$$
\xi=\frac{1}{1-\alpha}\int_\alpha^1-\log x\,dx
=1+\frac{\alpha\log\alpha}{1-\alpha}.
$$
For equal <masses>, take the limit $\alpha\to0$: $\xi=1$. The <expected value> of the fractional <energy> after a collision is instead $\mathbb E[x]=(1+\alpha)/2$. These are different averages: averaging a <logarithm> does not give the <logarithm> of an average.
For independent collisions with the same distribution, the <logarithms> of successive <energy> ratios add. Thus
$$
\mathbb E[\log(E_n/E_0)]=-n\xi,\qquad
E_{\mathrm{geom}}=E_0e^{-n\xi},\qquad
n_{\mathrm{log}}=\frac{\log(E_0/E_*)}{\xi}.
$$
Here $E_{\mathrm{geom}}$ is the geometric-mean <energy>, and $n_{\mathrm{log}}$ is a characteristic collision count for reaching an <energy> scale $E_*$. It is not an exact first-passage <expected value>. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.
In an original chosen example, set $E_0=10^6\,\mathrm{eV}$ and $E_*=0.025\,\mathrm{eV}$. Approximate <hydrogen> by $A=1$ and <carbon> by $A=12$. Then
$$
\begin{array}{c|c|c}
A&\xi&n_{\mathrm{log}}\\
1&1&17.50\\
12&0.15777&110.95
\end{array}
$$
These are illustrative model values, not measured collision counts from 1934. The <energy> scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero <energy>.
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