The X-ray transform in 2 dimensions

Joey Zou

April 13, 2020Stanford Kiddie Colloquium

Beer’s Law

Consider an X-ray traveling through a medium on a trajectory x(t).Beer’s Law gives the rate of decay of the intensity of the X-ray:

d

dt(I (x(t))) = −f (x(t))I (x(t)).

Here f (x) is the “absorption coefficient” which depends on thecomposition of the material at the point x .Solving the ODE gives If

I0= exp(−

∫f (x(t)) dt), i.e.∫

f (x(t)) dt = − log(IfI0

).

Recovering the absorption coefficient

We suppose all X-ray trajectories are straight lines (this is mostlytrue).Suppose we have a 2-dimensional object, and we’re allowed to sendX-rays through the object along all possible lines. Can we get backthe absorption coefficient f (x) (which might tell us thecomposition of the object)?Mathematically, for f : R2 → R and a line ` in R2, letRf (`) =

∫` f (x) dH1(x). If we know Rf (`) for all lines `, do we

know f ?(R is called the “X-ray transform” or “Radon transform”. Firststudied by Johann Radon in 1917.)

Parametrizing lines

Note that every line in R2 can be described as

x ∈ R2 : x · ω = s

for some s ∈ R and ω ∈ S1. (ω is a normal vector to the line, ands gives the corresponding signed distance to the origin.)This gives a 2-to-1 correspondence between Rs × S1

ω to the spaceof lines (note (s, ω) and (−s,−ω) parametrize the same line).Thus, we consider R as an operator from functions on R2 tofunctions on the cylinder R× S1.

Example: disk

Suppose f (x) = χB(0,1)(x). Then

Rf (s, ω) =

2√

1− s2 −1 ≤ s ≤ 10 otherwise

Figure: X-ray transform of χB(0,1)

Example: disk

In general, suppose f (x) = χB(x0,r)(x). Then

Rf (s, ω) =

2√

r2 − (s − x0 · ω)2 |s − x0 · ω| ≤ r0 otherwise

Figure: X-ray transform of χB(x0,r) for x0 = (−1, 1), r = 2 and forx0 = (2, 0.5), r = 0.5

Other examples

See the YouTube channel of Samuli Siltanen:https://www.youtube.com/watch?v=5DUGTXd26nA

https://www.youtube.com/watch?v=q7Rt_OY_7tU

etc.

What are some interesting things we can say?

Theorem (Fourier slice)

The Fourier transform of Rf in s

Fs(Rf )(σ, ω) :=

∫Re−iσsRf (s, ω) ds

equals f (σω), and hence one can obtain Rf as

Rf (s, ω) =1

2π

∫Re isσ f (σω) dσ

i.e. by taking the inverse Fourier transform of f “over a slice”.

Proof.In the Fourier transform, decompose R2 into hyperplanes (i.e.lines) of the form x · ω = s:

f (σω) =

∫R2

e−iσω·x f (x) dx

=

∫R

∫x ·ω=s

e−iσs f (x) dH1(x) ds

=

∫Re−iσsRf (s, ω) ds

= FsRf (σ, ω).

Corollary

If Rf ≡ 0 and f ∈ L2, then f ≡ 0. In other words, R is injective onL2.

Adjoint

We consider the adjoint for R. So for f : R2 → R andg : R× S1 → R we have

〈R∗g , f 〉L2(R2) = 〈g ,Rf 〉L2(R×S1)

=

∫S1

∫Rg(s, ω)

(∫x ·ω=s

f (x) dH1(x)

)ds dω

=

∫S1

∫R2

g(x · ω, ω)f (x) dx dω.

Thus R∗g(x) =∫S1 g(x · ω, ω) dω.

(R integrates a function over points on a line; R∗ integrates afunction over lines through a point.)

Claim: The operator 14πR

∗(−∆s)1/2 gives a left inverse for R,

where Fs(−∆s)1/2g(σ, ω) = |σ|Fsg(σ, ω).Proof: For f , g : R2 → R,

〈R∗(−∆s)1/2Rf , g〉L2(R) = 〈(−∆s)1/2Rf ,Rg〉L2(R×S1)

=1

2π〈Fs(−∆s)1/2Rf ,FsRg〉L2(R×S1)

=1

2π

∫S1

∫R|σ|f (σω)g(σω) dσ dω

=1

π

∫R2

f (ξ)g(ξ) dξ

= 4π〈f , g〉L2(R2).

(Note the use of polar coordinates.) Thus, 14πR

∗(−∆s)1/2Rf = f .

The formula 14πR

∗(−∆s)1/2Rf = f is known as the FilteredBackprojection Formula. It was discovered by Radon in 1917 andre-discovered by Cormack in the late ’60s (en route to him sharingthe 1979 Nobel Prize in Physiology or Medicine for helping todevelop actual CT scanners).

Some examples:https://www.youtube.com/watch?v=7jWC0T5C7a4

https://www.youtube.com/watch?v=tRD58IO1FKw

Qualitative behavior

We now ask if there are any qualitative features we can state (e.g.where the singularities are) without necessarily computing thefiltered back-projection.First, how does it act on smooth functions?

LemmaWrite ω = (cosφ, sinφ). We have

∂s(Rf )(s, ω) = R(ω · ∇f )(s, ω)

∂φ(Rf )(s, ω) = R((x1∂x2 − x2∂x1)f )(s, ω).

In particular, if f ∈ C∞c (R2), then Rf ∈ C∞c (R× S1).

Note that ω · ∇ and x1∂x2 − x2∂x1 are the vector fields generatingtranslation in the direction of ω and counterclockwise rotation,respectively.

Qualitative behavior for nonsmooth functions

We focus on χΩ where Ω is an opendomain with smooth boundary.We now ask: for which (s, ω) is RχΩ

smooth at (s, ω)?Claim: if the line x · ω = s intersects∂Ω transversely, then RχΩ is smooth at(s, ω).Idea: rotate until ω is horizontal (i.e.the line is vertical). Then transverseintersection implies ∂Ω near the line is aunion of segments which are graphs ofsmooth functions.

Ω

On the other hand, if the line intersects ∂Ωtangentially, then you may get a(non-differentiable) singularity.Idea: take the model case where ω = (1, 0) ishorizontal and Ω = x2 > x2

1 near theintersection (0, 0). Then

RχΩ(s, (1, 0)) =

2√s s ≥ 0

0 s < 0for s near 0.

Ω

Thus, singularities in the sinogram space correspond to tangentlines to the boundary (i.e. if RχΩ is singular at (s, ω) then the linex · ω = s is tangent to ∂Ω.)Moreover, if the singularity is a smooth curve, then the curve’slocal behavior pinpoints where the tangent line intersects ∂Ω.Idea: if x(t) is a path in ∂Ω, ω(t) the corresponding unit normal, ands(t) = x(t) · ω(t), then

s(t) = x(t) · ω(t) + x(t) · ω(t) = x(t) · ω(t)

Since ω(t) = ω⊥(t)φ(t) where ω⊥ = (−ω2, ω1) and φ is the angle,

s(t)/φ(t) = x(t) · ω⊥(t).

The LHS is the slope of the singularity curve. Since s(t) = x(t) · ω(t),

knowledge of s and the slope s/φ determines x since ω, ω⊥ forms a

basis.

A more analytic way to analyze singularities

DefinitionSuppose u is a distribution on Rn. For (x0, ξ0) ∈ T ∗Rn = Rn × Rn

with ξ0 6= 0, we say that u is “microlocally smooth” at (x0, ξ0) ifthere exists a cutoff ϕ ∈ C∞c (Rn) with ϕ(x0) 6= 0 and a conicalneighborhood1 C of ξ0 such that for every N, there exists CN suchthat

|F(ϕu)(ξ)| ≤ CN |ξ|−N for all ξ ∈ C, |ξ| ≥ 1.

Define the wavefront set WF (u) of u to be the set of(x , ξ) ∈ Rn × (Rn\0) where u is not microlocally smooth.The wavefront set captures not only where u is singular, but alsoin which (co)directions it is singular.

1i.e. C =ξ :

∣∣∣ ξ|ξ| −

ξ0|ξ0|

∣∣∣ < ε

Example: Heaviside function

Consider u(x) = H(x1) = χx1≥0(x) on R2 (H the Heavisidefunction).If x1 6= 0, then (x , ξ) 6∈WF (u) for any ξ 6= 0 (just localize awayfrom x1 = 0). In general, consider ϕ(x) = ϕ1(x1)ϕ2(x2),ϕi ∈ C∞c (R). Then

F(ϕu)(ξ) = F(ϕ1H)(ξ1)F(ϕ2)(ξ2).

Note F(ϕ2)(ξ2) decays super quickly as |ξ2| → ∞, while F(ϕ1H)is bounded. Hence, if ξ2 6= 0, then we can find a conicalneighborhood where |ξ2| > ε|ξ|, and leverage the rapid decay ofF(ϕ2) to conclude (x , ξ) 6∈WF (u). Thus, WF (u) is contained in

(x , ξ) ∈ T ∗R2\0 : x1 = 0, ξ2 = 0 = (0, x2; ξ1, 0) | ξ1 6= 0.

Notice that this is precisely N∗(x1 = 0) minus the zero section.

Conormal Distributions

More generally, we can consider distributions that are conormalw.r.t. some set S (think a submanifold or union of submanifolds).These are distributions u satisfying

V1V2 . . .VNu ∈ L2 if V1, . . . ,VN are vector fields tangent to S .

Example: If ∂Ω is smooth, then χΩ is conormal to ∂Ω. If we canwrite ∂Ω = f = 0 where df does not vanish on ∂Ω, then δ(f ) isconormal to ∂Ω.

TheoremIf S ⊂ Rn is a smooth submanifold, and u is conormal to S , then

WF (u) ⊂ N∗(S) = (x , ξ) ∈ T ∗Rn : x ∈ S , ξ|TxS ≡ 0.

N∗S is the “conormal bundle” of S . Note if S = f = 0 with dfnon-vanishing, then N∗x (S) = span dfx for all x ∈ S .

Schwartz kernel

Schwartz Kernel Theorem: Continuous linear mapsD(Y )→ D′(X ) are in one-to-one correspondence withdistributions in D′(X × Y ) (i.e. distributions on the productX × Y ). Informally, to a distribution K (x , y) (the “Schwartzkernel”) we associate the operator

Tf (x) =

∫YK (x , y)f (y) dy .

For the X-ray transform we have

Rf (s, ω) =

∫x ·ω=s

f (x) dH1(x)

=

∫R2

δ(s − x · ω)f (x) dx

so the Schwartz kernel of R is δ(s − x · ω).

Hormander-Sato Lemma

TheoremLet T : D(Y )→ D′(X ) be a continuous linear map, with Schwartzkernel K ∈ D′(X × Y ). Then

WF (Tu) ⊂ (x ; ξ) : ∃(y ; η) ∈WF (u) s.t. (x , y ; ξ,−η) ∈WF (K )= WF ′(K ) WF (u)

where WF ′(K ) = ((x ; ξ), (y ;−η)) : (x , y ; ξ, η) ∈WF (K )(regarded as a relation).

RemarkTechnically we also need to assume that the projections ofWF (K ) ⊂ T ∗(X × Y )\o onto T ∗X and T ∗Y do not meet the zerosections (this is necessary in part to make sure Tu is even well-definedwhen u is merely a distribution).

Applying Hormander-Sato to X-ray transform

For f = χΩ, we are looking for tuples (s, φ, x ;σ, η, ξ) such that

s = x · ω and σ ds + η dφ− ξ · dx is parallel to d(s − x · ω)

and (x ; ξ) ∈WF (f ).

Since

d(s − x · ω) = ds + (x1 sinφ− x2 cosφ)dφ− cosφ dx1 − sinφ dx2

it follows that ξ must be parallel to (cosφ, sinφ).But for f = χΩ we have (x ; ξ) ∈WF (f ) ⇐⇒ ξ is parallel to the(co)normal vector to ∂Ω at x . So φ is the angle of the conormal to∂Ω! Then s can be computed correspondingly.

An X-ray picture

Figure: A dental X-ray slice. Picture fromhttps://www.exxim-cc.com/metal_artifact_reduction.html

So, why are there streaking lines?

Recall how we got the data: we should have

If (s, ω)

I0(s, ω)= exp(−Rf (s, ω)) =⇒ Rf (s, ω) = − ln

(If (s, ω)

I0(s, ω)

)so by applying the filtered backprojection 1

4πR∗(−∆s)1/2 we

should get

f =1

4πR∗(−∆s)1/2

(− ln

IfI0

).

In reality, there is an issue: implicitly we assumed the X-rays arerays of photons with the same energy; in practice this is not quitetrue. Some materials (e.g. metals) have absorption coefficientsthat vary strongly with energy, i.e. f = fE depends on the energyE .

Thus the collected data IfI0

is actually an average:

IfI0

=

∫ E+

E−

η(E ) exp(−RfE ) dE .

If we assume fE (x) = fE0(x) + α(E − E0)χD(x) (e.g. D is ametallic object), and η(E ) = 1

2εχ[E0−ε,E0+ε](E ), then

IfI0

=

∫ E0+ε

E0−ε

1

2εexp(−RfE0) exp(−α(E − E0)RχD) dE

= exp(−RfE0)sinh(αεRχD)

αεRχD

Thus the image that we actually compute is

1

4πR∗(−∆s)1/2

(− ln

IfI0

)=

1

4πR∗(−∆s)1/2

(− ln

(exp(−RfE0)

sinh(αεRχD)

αεRχD

))= fE0 −

1

4πR∗(−∆s)1/2 ln

(sinh(αεRχD)

αεRχD

).

We have an additional artifact term fMA = 14πR

∗(−∆s)1/2F (RχD)

where F (x) = − ln(

sinh(αεx)αεx

). Could this cause the streaking

lines?

Theorem (Palacios, Uhlmann, Wang ’17)

Suppose D = ∪kj=1Di , where Dj are disjoint, and each ∂Dj issmooth and strictly convex. Then

WF (fMA) ⊂ N∗D ∪

(⋃`∈L

N∗(`)

),

where

L = lines tangent to both Di and Dj for some i 6= j.

What’s the difference between one object and manyobjects?

Fact: If u ∈ L∞ and is conormal to S , and F ∈ C∞(R), then F (u)is also conormal to S .If D is one strictly convex region, then the sinogram singularitycurve S is the union of two non-intersecting smooth curves, soWF (F (RχD)) ⊂ N∗S .However, if D is a union of such regions, then S is a union ofintersecting curves. F (RχD

) is still “conormal to S”, however thebest we can say is

S =k⋃

j=1

Sj , u conormal to S

=⇒ WF (u) ⊂

⋃i 6=j

N∗(Si\Sj)

∪⋃

i 6=j

N∗(Si ∩ Sj)

The singularities in

⋃i 6=j N

∗(Si ∩ Sj) are mapped under the filteredbackprojection to

⋃`∈LN

∗(`).

More recent results

Theorem (Wang, Z. ’20)

Suppose D = ∪kj=1Di , where Dj are disjoint, and each ∂Dj issmooth (not nec. strictly convex). Then

WF (fMA) ⊂ N∗D ∪

( ⋃`∈L∪L′

N∗(`)

),

where

L = lines tangent to both Di and Dj for some i 6= j,L′ = lines tangent to an inflection point on ∂D.

Thanks for your attention!