Medical Imaging

Radioactive decay kinetics

Radioactive decay is a first-order process. The activity A = \lambda N falls exponentially with time, where \lambda is the decay constant, the probability per unit time that a nucleus decays. The half-life is T_{1/2} = \ln 2/\lambda, and the activity at time t is A(t) = A_0 e^{-\lambda t}. The activity after an interval depends only on \lambda and the elapsed time, not on the starting amount. Convert the elapsed time to seconds when \lambda is given per second, because leaving the interval in hours is a common error.

If the activity is known, the number of nuclei is N = A/\lambda, and the mass follows from m = NM/N_A, where M is the molar mass and N_A is Avogadro's number. Know the unit conversions: 1 Ci = 3.7\times10^{10} Bq, so 1 mCi = 3.7\times10^{7} Bq.

Two sources with different decay constants have activities that cross at a finite time. Setting A_{01}e^{-\lambda_1 t} = A_{02}e^{-\lambda_2 t} gives t = \ln(A_{01}/A_{02})/(\lambda_1-\lambda_2). If the two sources start with equal numbers of atoms, the source with the smaller decay constant starts with the larger activity. When one half-life is four times the other, the activities become equal at about 0.66\,T_{1/2} of the longer-lived nuclide. For sources with unequal initial activities, the same crossing formula applies, and the crossing time follows directly from the logarithm of the activity ratio divided by the difference of decay constants.

Ultrasound reflection and transmission

Acoustic impedance is Z = \rho c, where \rho is the density and c is the speed of sound. Soft tissues have similar sound speeds, so their impedances scale approximately with density.

At normal incidence, pressure is continuous across an interface. The pressure reflection coefficient is r = (Z_2 - Z_1)/(Z_2 + Z_1), and the transmitted pressure amplitude is t = 2Z_2/(Z_1 + Z_2). The sum of the incident and reflected amplitudes therefore equals the transmitted amplitude, P_0(1+r) = 2P_0 Z_2/(Z_1+Z_2).

For energy, the intensity reflection coefficient is R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 and the intensity transmission coefficient is T = \frac{4Z_1Z_2}{(Z_1+Z_2)^2}, with R + T = 1. Reflection depends on the impedance mismatch, so a tissue boundary with matched impedances reflects little. For example, if Z_1 = 1.5 and Z_2 = 4.5 in units of 10^6 rayl, then R = (3/6)^2 = 0.25, so a quarter of the incident intensity is reflected. Tissues with clearly different impedances, such as muscle and bone, reflect a substantial fraction of the energy.

For oblique incidence, refraction follows Snell's law, \sin\theta_t / \sin\theta_i = c_t / c_i, so the transmitted angle depends on the ratio of sound speeds. Impedance matching does not remove the need for this angle calculation.

Doppler ultrasound frequency shift

A moving reflector shifts the echo frequency. The Doppler shift is \Delta f = \frac{2 f_0 v \cos\theta}{c}, where f_0 is the transmitted frequency, v is the blood velocity, \theta is the angle between the beam and the flow, and c is the speed of sound. The factor of 2 arises because the moving blood cell acts first as a receiver and then as a source. The cosine factor means that the shift is zero for flow perpendicular to the beam and maximal for flow along it.

The velocity is obtained by rearranging, v = \frac{\Delta f\, c}{2 f_0 \cos\theta}. Use the speed of sound in blood or tissue consistently and convert \text{cm/s} to \text{m/s} before combining with c in \text{m/s}.

The transmitted frequency of a piezoelectric transducer is set by its thickness, f_0 = c/(2t), so the crystal thickness must be found before the Doppler shift can be computed.

MRI relaxation times T1 and T2

After a radiofrequency pulse, the nuclear magnetization relaxes through two independent processes. T1, the spin-lattice or longitudinal relaxation time, is the time constant for longitudinal magnetization to recover toward equilibrium by exchanging energy with the surroundings. T2, the spin-spin or transverse relaxation time, is the time constant for transverse magnetization to decay because of dephasing between spins. In general T_2 \le T_1.

Magnetization recovery is M_z(t) = M_0(1 - e^{-t/T_1}), and transverse decay is S(t) = S_0 e^{-t/T_2}. For example, a signal of 500 units with T_2 = 100 ms falls to 500e^{-0.5} \approx 303 units after 50 ms.

To obtain proton density weighting, the sequence must suppress both T1 and T2 contrast. A long repetition time, TR \gg T_1, lets longitudinal magnetization recover almost completely, and a short echo time, TE \ll T_2, limits transverse decay. The signal is then governed mainly by the number of hydrogen nuclei. A short TR gives T1 weighting, and a long TE gives T2 weighting.

Larmor frequency

The Larmor frequency is the precession frequency of a nuclear spin in a static field, f = \gamma B_0, where B_0 is the flux density and \gamma is the gyromagnetic ratio. Values are usually quoted as \gamma/2\pi in MHz per tesla, so the frequency is directly the product of this constant and the field.

• Gyromagnetic ratio — the constant of proportionality between magnetic moment and angular momentum; for protons \gamma/2\pi \approx 42.58 MHz/T. Example: at 3 T, protons precess at 42.58 \times 3 \approx 127.7 MHz.
• Nuclide-dependent value — different nuclei have different ratios, so the same field gives different frequencies. For sodium-23 the ratio is about 11.26 MHz/T, and for phosphorus-31 about 17.23 MHz/T.

For protons at 1.5 T the Larmor frequency is about 63.9 MHz, which is the value to use when the field strength is given in tesla.

Planar X-ray imaging

In planar radiography, a fast intensifying screen has a thick phosphor layer that absorbs more X-ray photons and reduces the dose required. The emitted light spreads laterally before it reaches the film or detector, which blurs the image and worsens spatial resolution. Speed and resolution therefore trade against each other.

The tube current determines the number of photons produced per unit time. Fewer photons reach the detector when the current is reduced. Because photon noise is approximately the square root of the count, a lower count lowers the signal to noise ratio.

When X-rays cross several tissue layers, each layer attenuates the beam independently. The transmitted fraction is the product of the layer factors, I/I_0 = \exp(-\sum_i \mu_i x_i), so the exponents add. For example, a 1 cm layer with \mu = 1 cm^{-1} followed by a 2 cm layer with \mu = 10 cm^{-1} transmits e^{-21} of the photons.

X-ray beam filtration

Low-energy X-rays are absorbed in the patient and contribute to dose but add little to the image. Metal filters, usually aluminium or copper, are placed in the beam to remove them. The filter removes low-energy photons preferentially because the attenuation coefficient falls steeply with energy.

Two filters of different materials give the same filtering if their linear attenuation coefficients times thickness are equal, \mu_1 x_1 = \mu_2 x_2. The linear coefficient is the mass attenuation coefficient multiplied by density. For example, if a 2 mm aluminium filter has \mu = 50 m^{-1} and copper has \mu = 600 m^{-1}, the equivalent copper thickness is 50 \times 2/600 \approx 0.17 mm.

Beam hardening occurs when the spectrum passes through the body. Low-energy photons are removed preferentially, so the mean energy of the transmitted beam increases while the low-energy part of the spectrum shrinks. The transmitted spectrum therefore has a higher mean energy than the incident spectrum.

Inverse square law in radiography

Radiation from a point source spreads over an area proportional to d^2, so the intensity at distance d is proportional to 1/d^2. In radiography, the exposure at a fixed tube voltage is proportional to \text{mAs}/d^2.

To keep exposure unchanged when the distance changes, the mAs must scale with the square of the distance, \text{mAs}_2 = \text{mAs}_1 (d_2/d_1)^2. For example, if the distance is increased from 1 m to 1.5 m, the mAs must rise from 20 to 20 \times 2.25 = 45 mAs.

The same law governs the dose from a point source. The dose rate is proportional to 1/r^2, so for a fixed dose the time of exposure is proportional to r^2. Doubling the distance from a source therefore requires four times the exposure time to receive the same dose, assuming no attenuation in air.

Half-value layer attenuation

For a monochromatic beam in a homogeneous medium, intensity falls exponentially, I = I_0 e^{-\mu x}, where \mu is the linear attenuation coefficient. The half-value layer (HVL) is the thickness that halves the intensity, x_{1/2} = \ln 2/\mu.

• Constant HVL for monochromatic beams — each successive halving requires the same thickness, because the exponential law is self-similar. The HVL depends on \mu, which depends on photon energy and material, but not on the incident intensity.
• Counting HVLs — transmission through a stack of layers is I/I_0 = (1/2)^{x/\text{HVL}} for each layer. For example, 6 cm of a material with a 2 cm HVL corresponds to 3 HVLs and transmits (1/2)^3 = 12.5\%.

For a multi-layer tissue path with different HVLs, the number of HVLs is the sum of the thickness divided by the HVL of each layer, and this total determines the transmitted fraction.

MRI slice selection

A slice is selected by applying a gradient along the slice direction while a radiofrequency pulse excites a band of frequencies. The RF bandwidth is BW = \gamma G \Delta z, where \gamma is the gyromagnetic ratio, G is the gradient strength in T/m, and \Delta z is the slice thickness. A stronger gradient or a thicker slice requires a wider bandwidth.

Check units carefully, because gradients are often given in gauss per centimetre, where 1 G = 10^{-4} T and 1 cm = 10^{-2} m. For example, a gradient of 0.03 T/m with a 4 mm slice and protons (\gamma = 42.58 MHz/T) requires BW = 42.58\times10^{6}\times0.03\times0.004 \approx 5.1 kHz.

The static field strength sets the centre frequency but does not appear in the bandwidth formula.

Ultrasound transducer resonance

A piezoelectric plate resonates most strongly at its fundamental frequency when its thickness is half a wavelength of the sound in the material. The thickness is t = \lambda/2 = v/(2f), where v is the sound speed in the crystal. Thickness and frequency are inversely related, so a higher operating frequency requires a thinner crystal.

For example, a crystal with v = 4000 m/s at 2 MHz has t = 4000/(2\times2\times10^{6}) = 1 mm.

CT scanner generations

Early CT scanners differed in how the source and detectors moved. A first-generation scanner uses a single pencil beam and a single detector. The assembly translates across the field of view to acquire each projection and then rotates by a small angular increment before the next translation.

The scan time is the number of projections multiplied by the time for one translation plus one rotation, T_{\text{scan}} = N_{\text{proj}}(t_{\text{trans}} + t_{\text{rot}}). The translation time is the field-of-view width divided by the translation speed. For example, if the translation takes 1 s and each rotation takes 0.5 s, then 360 projections require 360 \times 1.5 = 540 s.

Later generations used fan beams and detector arrays to avoid translation, which reduced scan times considerably.

Radon transform projection geometry

A projection is the set of line integrals of the attenuation coefficient along parallel rays. For a ray at perpendicular distance t from the centre of a circular object of radius R, the chord length in the object is L = 2\sqrt{R^2 - t^2}. Along this ray the Beer-Lambert law gives \ln(I_i/I_o) = \int \mu\, ds. For a uniform object with \mu constant, this reduces to \ln(I_i/I_o) = \mu L = 2\mu\sqrt{R^2 - t^2}.