Chapter 1: X-ray CT and Projection

Q: What is the Radon transform?

The Radon transform maps a function $f(x,y)$ to the collection of its line integrals along every line. Specifying a line by its orientation (angle $\theta$) and position (distance $s$), the line integral of $f$ along that line $L_{s,\theta}$, viewed as a function of $(s,\theta)$, is the Radon transform $\mathcal{R}f$: $\mathcal{R}f(s,\theta)=\displaystyle\int_{L_{s,\theta}} f(x,y)\,dl$. In X-ray CT, this is the projection data measured from each direction. The explicit formula is treated in Chapter 2.

Q: Can CT be done with visible light?

Yes. All that reconstruction requires is that the line integral along a line be measurable at many angles; it does not depend on wavelength. As long as the wave travels straight and its intensity follows the accumulation of absorption along the path (the Beer-Lambert law), CT works with visible light too for nearly transparent samples (OPT and the optical CT gel dosimeter are real examples). The human body needs X-rays because visible light scatters and does not travel straight in tissue, and refraction bends it, breaking the straight-line premise.

As the starting point of the Radon transform—the computational principle behind X-ray CT—we study the projection (the line integral of the attenuation coefficient).

1.1 Historical Background

The Radon transform was first formulated by the Austrian mathematician Johann Radon (1887–1956) in a paper published in 1917.

Historical Facts

1917: Radon published the paper "Über die Bestimmung von Funktionen durch ihre Integralwerte längs gewisser Mannigfaltigkeiten" (On the determination of functions from their integral values along certain manifolds)
1963: Allan Cormack independently rediscovered the medical application
1971: Godfrey Hounsfield developed the first CT scanner
1979: Cormack and Hounsfield received the Nobel Prize in Physiology or Medicine

Originating as pure mathematics, the Radon transform was put to practical use about 50 years later in the form of medical imaging, bringing a revolution to modern medicine.

1.2 Intuitive Understanding

The essence of the Radon transform is to "infer the shape from its shadow."

Figure 1: Projection by parallel X-rays. When the X-rays passing through a uniform object are measured by the detector, the central part—where the path through the object is longest—is attenuated the most and becomes darkest (grayscale bar). The graph on the right is the X-ray intensity $I(s)$ reaching the detector: outside the object there is no attenuation, so it stays at the source intensity $I_0$, and it becomes smaller toward the object's center.

An everyday example: When you undergo an X-ray CT scan at a hospital, X-rays pass through your body and are absorbed in different amounts by bones and organs. Here $I_0$ is the source intensity (the intensity that would reach the detector if there were no object), and $I(s)$ is the X-ray intensity that passes through the object and reaches the detector. The larger the total absorption accumulated along the path, the smaller $I(s)$ becomes (when the object is homogeneous, as in Figure 1, this shows up as being darker where the path through the object is longer).

From measured intensity to a line integral: how absorption accumulates

All the detector can measure is the intensity $I(s)$. First, we want to relate this to the absorption inside the object along the path the X-ray traveled (extracting the absorptivity at each individual location—image reconstruction—is treated in Chapter 3). The key is that, as the X-ray passes through the object, absorption accumulates by multiplication, and taking the logarithm turns that multiplication into addition. Below, we first check this with a concrete numerical example, then generalize using thin layers.

First, the intuition (discrete version): think of the object as a stack of thin layers. Each time the X-ray passes through a layer, only the "transmittance" (the fraction that gets through) of the intensity remains, so the overall transmittance is the product of the layer transmittances. For example, if three layers have transmittances $0.7,\ 0.5,\ 0.6$, then

$$\frac{I}{I_0} = 0.7 \times 0.5 \times 0.6 = 0.21$$

A 3D figure of an X-ray passing from left to right through three semi-transparent layers of different transmittance (0.7, 0.5, 0.6 from the left), dimming at each layer until it finally becomes 0.21 times the original intensity — Figure 2: An X-ray (orange) of intensity $I_0$ passes from left to right through three layers of different transmittance ($0.7,\ 0.5,\ 0.6$ from the left). Since the X-ray intensity is multiplied by the transmittance at each layer, the intensity that finally emerges is $I = 0.7\times0.5\times0.6\,I_0 = 0.21\,I_0$, and the X-ray weakens each time it passes through a layer. The more transparent a layer (the closer its transmittance is to $1$), the more X-rays it lets through.

Taking the logarithm of both sides, the property $\ln(ab)=\ln a+\ln b$ turns the multiplication into addition:

$$\ln\frac{I}{I_0} = \ln 0.7 + \ln 0.5 + \ln 0.6 = \ln 0.21$$

Since transmittance is at most $1$ ($0\le\text{transmittance}\le 1$), each $\ln$ is at most $0$, and so is $\ln(I/I_0)$. Leaving it negative makes the later handling cumbersome, so using the sign-flipped $-\ln$ makes each layer's contribution a non-negative quantity that is easier to read:

$$-\ln\frac{I}{I_0} = (-\ln 0.7) + (-\ln 0.5) + (-\ln 0.6) \;\ge\; 0$$

The transmittance (the fraction that passes through the object) and the absorptance (the fraction absorbed by the object) are complementary, summing to $1$.

$$\text{transmittance} \;=\; 1 - \text{absorptance} \qquad(\text{e.g. transmittance }0.7 \;\Leftrightarrow\; \text{absorptance }0.3)$$ $$-\ln(\text{transmittance}) \;=\; -\ln(1 - \text{absorptance})$$

This $-\ln(\text{transmittance})$ is called the amount of absorption (optical depth). Since the product of transmittances became a sum under the logarithm (the equation above), $-\ln(\text{transmittance})$ is a quantity that can simply be added up.

Writing the absorptance as $x$, the following Taylor expansion centered at $x = 0$ holds.

$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots$$

If $x$ is small ($x\ll1$), the higher-order terms from $x^2/2$ on are negligibly small, so it can be approximated by the first-order term alone.

$$-\ln(1-x) \approx x \qquad (x \ll 1) \tag{1}$$

The absorptivity at each location, expressed as a quantity per unit length, is called the attenuation coefficient, denoted here by $f$ (in medical imaging the attenuation coefficient is often written $\mu$). In a thin layer of small thickness $dl$, the absorptance $x$ is approximately the product of the attenuation coefficient $f$ and the thickness $dl$.

$$x \;\approx\; f\,dl \qquad(\text{absorptance} \;\approx\; \text{attenuation coefficient}\,\times\,\text{thickness})$$

Three easily confused quantities

Attenuation coefficient $f$: the absorptivity per unit length at a location (a material property). Units: "1/length."
Absorptance $x$: the fraction actually absorbed in that layer. $x\approx f\,dl$ (dimensionless).
Amount of absorption (optical depth): $-\ln(\text{transmittance})$. A quantity that accumulates additively; in a thin layer, $\approx x$ (the absorptance).

Only $f$ is "per unit length"; the other two are fractions "for that layer" (so $x\ne f$, and $x\approx f\,dl$).

Let us apply this to a continuous object. Divide the path $L$ into thin layers of thickness $dl$. Since a general object is not homogeneous, each layer has its own attenuation coefficient $f_i$ ($i$ is the layer index). Passing through the $i$-th layer, the X-ray loses the absorptance $x_i\approx f_i\,dl$, and the intensity drops to $1-f_i\,dl$ times its value.

So over the whole path, the result is the successive product of the layer transmittances $1-f_i\,dl$:

$$I = I_0\,\underbrace{(1-f_1\,dl)(1-f_2\,dl)\cdots}_{\text{each layer along the path }L}$$

Flipping the sign of equation (1),

$$\ln(1-x) \approx -x$$

Apply $\exp$ to both sides (exponentiate). The left side is $e^{\ln(1-x)} = 1-x$, so

$$1-x \approx e^{-x}$$

Finally, setting $x = f_i\,dl$, each factor takes exponential form:

$$1 - f_i\,dl \;\approx\; e^{-f_i\,dl}$$

Substituting this, the product of exponentials collapses into an addition in the exponents:

$$I \;\approx\; I_0\,e^{-f_1\,dl}\,e^{-f_2\,dl}\cdots \;=\; I_0\,e^{-(f_1+f_2+\cdots)\,dl} \;=\; I_0\,e^{-\sum_i f_i\,dl}$$

Finally, making the layers infinitely thin ($dl\to0$), the sum $\displaystyle\sum_i f_i\,dl$ converges, as a Riemann sum, to the line integral $\displaystyle\int_L f(x,y)\,dl$. What we obtain is the Beer–Lambert law.

The Beer–Lambert law

The transmitted intensity $I(s)$ is the incident intensity $I_0$ attenuated exponentially by the line integral of the attenuation coefficient along the path $L$.

$$I(s) = I_0\,e^{-\int_L f(x,y)\,dl} \tag{2}$$

A 3D figure replacing the three discrete layers with a single continuous medium of attenuation coefficient f. The X-ray attenuates continuously (exponentially) as it advances through the medium, with the incident intensity I0 becoming I = I0 exp(-integral of f dl). — Figure 3: The three discrete layers of Figure 2 replaced by a single **continuous medium**. Here the attenuation coefficient $f$ varies with position and is **largest near the center** (nearly transparent at the ends), so the X-ray attenuates sharply through the dense center and gently at the ends, emerging as $I = I_0\,e^{-\int_L f(x,y)\,dl}$. This is the limit of infinitely many layers of small thickness $dl$ stacked together, where the product of the layer transmittances $e^{-f\,dl}$ collapses into the exponential $e^{-\int_L f(x,y)\,dl}$.

The detector can measure only the intensity $I(s)$, but taking $-\ln$ of both sides of equation (2) turns the multiplication (product/exponential) back into addition (a line integral), letting us extract from the measured intensity the line integral of absorption along the path. This is called the projection $p(s)$.

Projection

The projection $p(s)$ is the Beer–Lambert law solved for the line integral; it is computed from the measured value $I(s)$.

$$p(s) = -\ln\frac{I(s)}{I_0} = \int_L f(x,y)\,dl$$

This projection $p(s)$ is what image reconstruction is built on.

Line integral and the Radon transform

In this way, from the detected intensity we have extracted the line integral $\displaystyle\int_L f(x,y)\,dl$ along a single line $L$. A line is determined by two things: its orientation (the angle $\theta$) and its position (the distance $s$). Fixing one orientation and varying $s$, the collected $p(s)$ is merely one projection for that orientation. Varying the orientation $\theta$ as well, and collecting over all lines $(s,\theta)$, gives the Radon transform $p(s,\theta)$ of the function $f$.

$$p(s,\theta) = \mathcal{R}f(s,\theta) = \displaystyle\int_{L_{s,\theta}} f(x,y) \, dl$$

The precise formulation (the parametric representation of a line, the line integral, and the explicit form of the equation above) is treated in Chapter 2.

Why measure $-\ln(I/I_0)$ rather than the intensity $I$ itself?

What the detector directly measures is the transmitted intensity $I$. Even so, CT uses $-\ln(I/I_0)$ as the projection value because absorption accumulates by multiplication (exponentially) at each layer. With $I$ as is, each location's contribution sits inside a product, and you cannot add them up to extract the "total along the path." Taking $-\ln$ turns the multiplication into addition, giving the line integral $-\ln(I/I_0)=\int_L f(x,y)\,dl$—a "manageable quantity" that is simply the sum of each location's contribution.

Can visible light do CT? (the Radon transform is wavelength-independent)

We have written "X-ray" so far, but all the reconstruction mathematics requires is that the line integral $p(s)=\int_L f(x,y)\,dl$ along a line $L$ be measurable at many angles; wavelength appears nowhere. As long as the wave travels in a straight line and its intensity is determined by the accumulation of absorption along the path (the Beer–Lambert law), the same equations make the Radon transform and its inverse hold for visible light too. The "transparent layers" in the figures above are exactly the idealization of pure absorption with no scattering or refraction—this very wavelength-independent situation.

Indeed, for nearly transparent samples, CT with visible light is a real technology. Representative examples are OPT (Optical Projection Tomography), which performs projection tomography of cleared embryos and organoids with visible light, and the optical CT gel dosimeter, which reads out a transparent gel whose turbidity changes with radiation; a translucent body like agar is an ideal subject.

So why does the human body require X-rays? There are two reasons visible light fails in biological tissue. (1) Scattering: tissue is a "turbid medium," and visible and near-infrared photons scatter many times and do not travel straight (the same effect as the blurry red glow when you hold a hand over a flashlight). The premise of a straight line $L$ breaks down, so the simple Radon transform cannot be used. (2) Refraction: when the refractive index varies, the rays bend (a lens effect). This is why OPT index-matches (clears) the sample. For strongly scattering tissue there is a different method, diffuse optical tomography (DOT), but that solves a difficult inverse problem of the diffusion equation rather than the Radon transform, and its resolution is coarse.

In short, X-rays are chosen not because they are "mathematically special" but because they are one of the few means that can travel straight through and penetrate the human body. Given a transparent subject, CT is fundamentally a matter of pure geometry (line integrals) that does not care about wavelength.

1.3 Applications

Medical Imaging (CT, PET)

X-ray CT scans, PET scans, and the like acquire cross-sectional images of the body non-invasively. A foundational technology of modern medical diagnosis.

Electron-Microscope Tomography

Imaging a specimen with an electron beam while rotating it to reconstruct its three-dimensional structure. Used for the structural analysis of proteins and viruses.

Seismic Tomography

Estimating the internal structure of the Earth from seismic-wave propagation data. Contributes to elucidating the structure of the mantle and tectonic plates.

Synthetic Aperture Radar (SAR)

Processing radar observation data from satellites. Generates high-resolution images of the ground surface.

1.4 Exercises

Problem 1

An X-ray passes in succession through two layers of transmittance $0.8$ and $0.5$. By what factor of the incident intensity $I_0$ does it emerge? Also, express its projection value $-\ln(I/I_0)$ as the sum of the contributions of the individual layers.

Solution

Transmittance accumulates multiplicatively, so $I/I_0 = 0.8 \times 0.5 = 0.4$ (a factor of $0.4$). The projection value is

$$-\ln(I/I_0) = -\ln 0.4 = (-\ln 0.8) + (-\ln 0.5) \approx 0.223 + 0.693 = 0.916$$

which is the sum of each layer's contribution ($-\ln$).

Frequently Asked Questions

What is the Radon transform?

The Radon transform is the operation that maps a function $f(x,y)$ to the collection of "line integrals along every line." Specifying a line by its orientation (angle $\theta$) and position (distance $s$), the line integral of $f$ along that line $L_{s,\theta}$, viewed as a function of $(s,\theta)$, is the Radon transform $\mathcal{R}f$: $\mathcal{R}f(s,\theta)=\displaystyle\int_{L_{s,\theta}} f(x,y)\,dl$. In X-ray CT, this is the projection data measured from each direction. The explicit formula is treated in Chapter 2.

What applications does the Radon transform have?

Medical imaging (X-ray CT, PET, SPECT, MRI reconstruction) is the largest field of application. It also appears in seismic tomography (estimating the Earth's interior), electron-microscope tomography, non-destructive testing, astronomy (estimating the density distribution of celestial objects), radar imaging, and in every inverse problem of recovering internal structure from projection data.

Can the original image be recovered from the projection data?

Yes. A single projection in one direction does not carry enough information, but combining projections measured from many directions (the Radon transform) lets you recover the original function $f(x,y)$. This is the principle of CT (computed tomography). The concrete method of recovery (the inverse Radon transform) is treated in a later chapter.

Can CT be done with visible light?

Yes. All that reconstruction requires is that "the line integral along a line be measurable at many angles"; it does not depend on wavelength. As long as the wave travels straight and its intensity is determined by the accumulation of absorption along the path (the Beer–Lambert law), CT works with visible light too for nearly transparent samples (OPT, optical projection tomography, and the optical CT gel dosimeter are real examples). The human body needs X-rays because, in biological tissue, visible light scatters and does not travel straight, and refraction bends it, so the premise of a "straight-line integral" breaks down.