= X-ray spectral attenuation operator
{c}
{title2=$(A_xf)(E)=e^{-\mu(E)x}f(E)$}
For a fixed homogeneous absorber in a narrow-beam geometry, define the <X-ray spectral attenuation operator> on <L1 space> over a bounded positive <energy> interval. A physical input $f(E)\ge0$ is an <integrable function> giving incident <photon> rate density. Let its <linear attenuation coefficient> $\mu(E)\ge0$ be a <measurable function>, with $x\ge0$. The transmitted uncollided density is
$$
(A_xf)(E)=e^{-\mu(E)x}f(E),\qquad
S(x)=\int w(E)(A_xf)(E)\,dE.
$$
Here $w(E)\ge0$ is a fixed bounded detector response described by a <measurable function>; background has been subtracted. This is a positive <linear operator> on <L1 space>. Since $0\le e^{-\mu(E)x}\le1$, it cannot increase the integrated absolute incident rate. Multiplication of exponentials proves $A_{x+y}=A_xA_y$ for successive layers of the same material. Ignore secondary radiation entering the detector, saturation and changes to the source. Discrete <spectral lines> use the finite-sum version below instead of an ordinary rate density.
For a chosen filter, stipulate $\mu_\alpha=0.20\,\mathrm{mm^{-1}}$, $\mu_\beta=0.80\,\mathrm{mm^{-1}}$ and input line rates $(N_\alpha,N_\beta)=(4,1)$ in a common arbitrary rate unit. Assume equal calibrated detector weight for both lines. At $x=2.0\,\mathrm{mm}$,
$$
\begin{pmatrix}N_\alpha(x)\\N_\beta(x)\end{pmatrix}
=\begin{pmatrix}e^{-0.4}&0\\0&e^{-1.6}\end{pmatrix}
\begin{pmatrix}4\\1\end{pmatrix}
\simeq\begin{pmatrix}2.68128\\0.201897\end{pmatrix}.
$$
The selected $8.5\,\mathrm{keV}$ line's fraction rises from $4/5=0.800$ to $0.929974$, but its absolute rate falls by about $33.0\%$. Filtering improves purity at the cost of signal. It multiplies existing components and cannot create a new emitted <spectral line>.
\Image[/characteristic-x-ray-line-filtering-model.png]
{title=Shell differences set line <energies>; an absorber changes their weights}
{id=image-characteristic-x-ray-line-filtering-model}
{description=Original stipulated shell and filter model, not a measured Barkla <frequency spectrum>. Positive shell <binding energies> are plotted as negative bound-state <energies>. The two K lines are sticks with chosen input rates 4 and 1. The L line is outside the displayed filtering band. All attenuation coefficients are illustrative; they do not identify a real filter material.}
With just one integrated reading, $S=N_\alpha+N_\beta$, the two unknown line rates are underdetermined. If the source is unchanged between an empty-path reading and a filtered reading, the ideal <inverse problem> becomes
$$
\begin{pmatrix}S(0)\\S(2\,\mathrm{mm})\end{pmatrix}
=\begin{pmatrix}1&1\\e^{-0.4}&e^{-1.6}\end{pmatrix}
\begin{pmatrix}N_\alpha\\N_\beta\end{pmatrix}
=\begin{pmatrix}5\\2.88318\end{pmatrix}.
$$
The <determinant> $e^{-1.6}-e^{-0.4}\simeq-0.468424$ is nonzero, so these two ideal readings determine the two stipulated line rates uniquely. Identical coefficients would give zero <determinant> and destroy that separation. This small <linear algebra> example explains why absorber contrast supplies information, while also showing its limit: unknown scattered components or unknown detector weights add unknowns that two readings cannot determine.
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