= Spectrum-weighted water-equivalence ratio
A <spectrum-weighted water-equivalence ratio> compares a specified material response with the corresponding <water> response after accounting for the radiation's energy distribution. Its definition must identify the interaction coefficient and the spectral weighting. An attenuation comparison and an energy-absorption comparison are different quantities.
For an energy-absorption comparison, let $\Psi_E(E)\,dE$ be the <photon energy fluence> in an energy interval, and write $a_j(E)=(\mu_{\rm en}/\rho)_j(E)$ for material $j$. Define the coefficient-weighted response and its ratio by
$$
A_j=\int\Psi_E(E)a_j(E)\,dE,\qquad Q_\Psi=\frac{A_m}{A_w},
$$
with $A_w>0$. The same prescribed local <photon energy fluence> is used for both materials in this comparison. NIST distinguishes the <mass energy-absorption coefficient> from the <mass attenuation coefficient> and relates energy fluence to photon fluence through the additional photon-energy factor.<Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients by John H. Hubbell and Stephen M. Seltzer>[§3, energy fluence relation and equation (13), mass energy absorption]{ref}
The ratio is a weighted mean of the monoenergetic ratios $q(E)=a_m(E)/a_w(E)$, but its normalized weights are
$$
w(E)=\frac{\Psi_E(E)a_w(E)}{A_w},\qquad Q_\Psi=\int w(E)q(E)\,dE.
$$
Thus the weights depend on the water response as well as the incident spectrum. Averaging the ratios with energy-fluence weights alone generally gives a different result.
For an original two-line calculation, use the NIST <water> and <PMMA> energy-absorption data: at $30\,\mathrm{keV}$ the coefficients are $0.1557$ and $0.09645\,\mathrm{cm^2/g}$ respectively; at $100\,\mathrm{keV}$ they are $0.02546$ and $0.02368\,\mathrm{cm^2/g}$.<Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients by John H. Hubbell and Stephen M. Seltzer>[Table 4, Water, Liquid and Polymethyl Methacrylate; 0.03 MeV and 0.1 MeV rows, energy-absorption column]{ref} Equal energy fluences in the two lines give
$$
Q_\Psi=\frac{0.09645+0.02368}{0.1557+0.02546}\approx0.66312.
$$
Equal photon counts instead give energy-fluence weights proportional to $30$ and $100$, producing $Q_\Psi\approx0.72904$. The different answers reflect different physical spectra, not a discrepancy in the tabulated coefficients.
An <absorbed dose> interpretation needs additional conditions such as <charged-particle equilibrium>. A finite object can also change the local spectrum through <beam hardening> and scattering. The ratio above therefore specifies a local coefficient-weighted comparison; it is not a universal conversion factor for every phantom size or irradiation geometry. A physical validation should characterize the local spectrum and detector energy response rather than assume that equal total fluence means equal spectral weighting.
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