Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S10 / 1

For a narrow, well-collimated photon beam, scattered photons are rejected and the uncollided intensity obeys the Beer-Lambert law, $I(x)=I(0)e^{-\mu x}$. With water mass density $\rho \simeq 1\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$, the linear attenuation coefficient is $\mu =\rho (\mu /\rho )$. The half-value layer is thereforeThus the annihilation photons associated with the fluorine-18 source require about $1.60$ times as much water for the same attenuation as the technetium-99m photons. The $511\mathrm{keV}$ here is the annihilation photon energy, not the variable kinetic energy of an emitted Positron. A broad-beam measurement admitting scattered photons would require a buildup correction and would not have this simple transmission interpretation.^[ref]

For photons, the total mass attenuation coefficient sums contributions from photoelectric absorption, Compton scattering, coherent Rayleigh scattering, and pair production in nuclear or electronic fields when above their thresholds. At sufficiently high photon energies, photonuclear reactions can also contribute. Attenuation counts removal from the incident narrow beam, including scattering without complete local deposition of the original energy.

In water at both specified energies, Compton scattering dominates. At $140\mathrm{keV}$, photoelectric absorption and coherent Rayleigh scattering make smaller contributions; at $511\mathrm{keV}$ these are still less important. Nuclear-field pair production is impossible below $2m_e{c}^2=1.022\mathrm{MeV}$, so it occurs in neither example. The positrons used to produce the fluorine-18 annihilation radiation do not change that threshold for interactions of its $511\mathrm{keV}$ photons in water.

Strictly, $\mu$ is a linear attenuation coefficient, whereas $\mu /\rho$ is a mass attenuation coefficient; the source's wording blurs these units. The mass energy-absorption coefficient ${\mu }_{\mathrm{en}}/\rho$ weights the fraction of incident photon energy transferred to charged particles and retained after radiative losses, rather than simply counting all removals. Coherent scattering contributes attenuation with negligible energy transfer. A photon deflected by Compton scattering can carry much of the original energy away. One often writes ${\mu }_{\mathrm{en}}=(1-g){\mu }_{\mathrm{tr}}$, with $g$ the fraction of secondary charged-particle kinetic energy lost radiatively. Under charged-particle equilibrium, multiplying photon energy fluence by ${\mu }_{\mathrm{en}}/\rho$ gives the absorbed dose; that equilibrium condition matters near interfaces and in buildup regions.^[ref]

A CT number is a dimensionless relative attenuation measure, conventionally $H=1000({\mu }_{\mathrm{CT}}-{\mu }_{\mathrm{water},\mathrm{CT}})/{\mu }_{\mathrm{water},\mathrm{CT}}$. The X-ray spectrum used in computed tomography has an effective energy far below $511\mathrm{keV}$ and is polychromatic. Its photoelectric absorption contribution depends strongly on composition, whereas water and soft-tissue attenuation at the positron emission tomography energy is mainly Compton scattering. Thus a CT number is neither a $511\mathrm{keV}$ linear attenuation coefficient nor an energy-independent mass attenuation coefficient.

The usual conversion calibrates a piecewise linear or bilinear relation between CT numbers and $\mu (511\mathrm{keV})$. The air–water branch models predominantly density differences in lung and soft tissue; a separate water–bone branch accounts for the different energy dependence of mineralized tissue. The calibration depends on the computed tomography spectrum. After geometric registration, the resulting linear attenuation coefficient map is forward projected along each coincidence line of response to obtain $\exp (\int {\mu }_{511}ds)$. If a mass coefficient is stored, its conversion to a linear coefficient must also include mass density. Unusual high-atomic-number materials can violate the simple tissue calibration, so the conversion is a physical material model, not a universal arithmetic rescaling.^[ref][ref]

In the ideal unscattered coincidence model, let a positron annihilation occur at position $s$ on a coincidence line of response connecting detectors $A$ and $B$. Its two oppositely directed photons must both survive. Their independent attenuation factors multiply:The source position cancels because the complementary path integrals span the whole chord. In a uniform slab of chord length $L$, the same argument is $e^{-\mu s}{e}^{-\mu (L-s)}=e^{-\mu L}$. Therefore attenuation correction requires the full linear attenuation coefficient map along the chord, but not the depth of the particular annihilation along that chord. This argument addresses attenuation of true PET coincidences; scatter, random coincidences and detector efficiency need their own corrections.