Geometric Sensitivity of a Pinhole Collimator

Abstract

Geometric sensitivity for single photon emission computerized tomography (SPECT) is given by a double integral over the detection plane. It would be useful to be able to explicitly evaluate this quantity. This paper shows that the inner integral can be evaluated in the situation where there is no gamma ray penetration of the material surrounding the pinhole aperture. This is done by converting the integral to an integral in the complex plane and using Cauchy's theorem to replace it by one which can be evaluated in terms of elliptic functions.

1. Introduction

Nuclear-medicine imaging provides images that assess how the body is functioning [1, 2], as opposed to anatomical modalities (e.g., X-ray computed tomography, commonly known as a CT or “CAT” scans) that provide little or no information about function, but great detail of the body's structure. Nuclear-medicine images the biodistribution of radiolabeled molecules that are typically intravenously injected into the patient in tracer (i.e., nonpharmacological) quantities. The compounds may have different biochemical properties that affect the biodistribution and, hence, the choice of pharmaceuticals used to assess the disease state of a patient.

Two common nuclear-medicine techniques are single photon emission computed tomography (SPECT) and positron emission tomography (PET). Molecules labeled with a SPECT tracer emit a single photon. PET tracers emit a positron, which annihilates with a nearby electron to produce two nearly back-to-back photons at 511 keV, the rest energy of an electron. The line of these two photons contains the emission point of the positron. Many photon pairs are detected in coincidence and reconstructed into a three-dimensional (3D) image of the tracer's distribution.

Since SPECT tracers emit only one photon, the detection of the photon itself gives little information about the location of the emission since neither the direction nor origin of the emission is known. Consequently, SPECT uses collimation that allows only photons along certain lines to be detected. The origin is still unknown, but the 3D image may be reconstructed when photons from numerous angles are detected [4, 5]. Pinhole collimation is often used when a small organ (e.g., the thyroid) is imaged by a large detector with a SPECT tracer. Figure 1 is a diagram of the apparatus. It is not hard to see that the pinhole collimation's geometric efficiency, namely, the fraction of emitted photons that pass through the circular aperture of the collimator, is approximately given by

Figure 1 

Perspective View of the Knife-edge Pinhole Collimator. A point source at 𝑧=ℎ and angle 𝜃 emits a photon that intersects the 𝑧=0 plane at (𝜌cos𝛽,𝜌sin𝛽,0) with an incident angle 𝜃𝑎. The acceptance angle of the collimator is 𝛼 and the diameter of the aperture is 𝑑. Δ𝐿 is the path length of the photon in the attenuating medium. Reproduced from [3] with permission ( 2001 IEEE).

𝑔=𝑑²sin³𝜃16ℎ²,(1.1) where 𝑑 is the diameter of the pinhole, ℎ is the perpendicular distance, and 𝜃 is the incidence angle of the photon on the aperture plane at the center of aperture. This formula becomes exact in the limit that 𝑑 goes to zero. Also, an exact formula is easily obtained when 𝜃=𝜋/2

𝑔𝜃=𝜋2=12−211+2𝑑ℎ²^−1/2≈12−121−212𝑑ℎ²=1𝑑6ℎ²².(1.2) We derive stronger versions of these two results in the next section. But the main goal of this paper is to simplify the integral representation for the geometric sensitivity without taking 𝑑 or 𝜃−𝜋/2 to be small. We are concerned with the case where there is no penetration in the sense that not all gamma rays are stopped by the attenuating material surrounding the pinhole aperture. We take 𝜃≠𝜋/2. The geometric sensitivity, as defined above, is given by the integral over the aperture of the flux times the sine of the incidence angle at that point on the aperture:

𝑔=14𝜋ℎ²^0𝑑/2𝜌𝑑𝜌^02𝜋𝑑𝛽sin³𝜃_𝑎,(1.3) where 𝜌 and 𝛽 are polar coordinates on the aperture plane and 𝜃_𝑎 is the incidence angle of the photon at a particular value of 𝜌 and 𝛽, and is given by [3]

cot²𝜃_𝑎=cot²𝜃−2𝜌ℎcot𝜃cos(𝛽−𝜙)+ℎ𝜌²²,(1.4) where 𝜙 is the third coordinate of the point source. For integrals over 2𝜋 the difference 𝛽−𝜙 can be replaced with only 𝛽 without loss of generality. This yields

sin²𝜃_𝑎=csc²𝜃−2𝜌ℎcot𝜃cos𝛽+ℎ𝜌²²⁻¹.(1.5)

Applying this to (1.3),

𝑔=14𝜋ℎ²^0𝑑/2𝜌𝑑𝜌^02𝜋𝑑𝛽csc²𝜃−2𝜌ℎcot𝜃cos𝛽+ℎ𝜌²²^−3/2.(1.6)

When there is penetration the above would be modified to

𝑔=14𝜋ℎ²^∞0𝜌𝑑𝜌^02𝜋𝑑𝛽csc²𝜃−2𝜌ℎcot𝜃cos𝛽+ℎ𝜌²²^−3/2𝑒^{−𝜇Δ𝐿},(1.7) where the path length through attenuating material, Δ𝐿, is zero for 𝜌<𝑑/2 and is otherwise given by (11) in [3].

Thus, (1.7) can be rewritten as

𝑔=𝑔_NP+14𝜋ℎ²^∞𝑑/2𝜌𝑑𝜌^02𝜋𝑑𝛽csc²𝜃−2𝜌ℎcot𝜃cos𝛽+ℎ𝜌²²^−3/2𝑒^{−𝜇Δ𝐿}=𝑔_NP+𝑔_PEN,(1.8) where 𝑔_PEN is the efficiency due to penetration and 𝑔_NP, the efficiency under the assumption of no penetration, is given by (1.6).

In this paper we use complex variable methods to calculate the 𝛽 integral in (1.6).

2. Two Approximations for 𝑔_𝑁𝑃

The special form of 𝑔_NP makes it easy to derive approximations for 𝑑/ℎ small and for 𝜃−𝜋/2 small which generalize (1.1) and (1.2). Before doing this, we observe that only even powers of 𝑑 and of 𝜃−𝜋/2 can occur in the expansion of 𝑔_NP. We see this by writing the integral as the sum of integrals from 0 to 𝜋 and from 𝜋 to 2𝜋.

2.1. The Expansion for 𝑑 Small

Let us, in this section, denote 𝑔_NP by 𝑔. Upon substituting 𝑟=𝑑/2ℎ and 𝜂=𝜌/ℎ into (1.6) we obtain

𝑔(𝑟,𝜃)=14𝜋^02𝜋^𝑟0csc²=1𝜃−2𝜂cot𝜃cos𝛽+𝜂²^−3/2𝜂𝑑𝜂𝑑𝛽,4𝜋^02𝜋^𝑟0𝜂𝜙(𝛽,𝜂,𝜃)𝑑𝜂𝑑𝛽.(2.1) Clearly, 𝑔(0,𝜃)=0. Since

𝑔_𝑟(𝑟,𝜃)=14𝜋^02𝜋𝑟𝜙(𝛽,𝑟,𝜃)𝑑𝛽,(2.2) we have 𝑔_𝑟(0,𝜃)=0. Differentiating (2.12) three more times with respect to 𝑟 we obtain

𝑔_𝑟𝑟(𝑟,𝜃)=14𝜋^02𝜋𝑟𝜙_𝑟𝑔(𝛽,0,𝜃)+𝜙(𝛽,𝑟,𝜃)𝑑𝛽,_𝑟𝑟𝑟(𝑟,𝜃)=14𝜋^02𝜋𝑟𝜙_𝑟𝑟𝑔(𝛽,0,𝜃)+2𝜙_𝑟(𝛽,𝑟,𝜃)𝑑𝛽,_{𝑟𝑟𝑟𝑟}(𝑟,𝜃)=14𝜋^02𝜋𝑟𝜙_𝑟𝑟𝑟(𝛽,0,𝜃)+3𝜙_𝑟𝑟(𝛽,𝑟,𝜃)𝑑𝛽,(2.3) and then setting 𝑟=0 in each of the resulting equations we obtain

𝑔_𝑟𝑟(0,𝜃)=14𝜋^02𝜋𝑔𝜙(𝛽,0,𝜃)𝑑𝛽,_𝑟𝑟𝑟(0,𝜃)=12𝜋^02𝜋𝑔𝜙_𝑟(𝛽,0,𝜃)𝑑𝛽,_{𝑟𝑟𝑟𝑟}(0,𝜃)=34𝜋^02𝜋𝜙_𝑟𝑟(𝛽,0,𝜃)𝑑𝛽.(2.4) For convenience let us write

𝑢=csc²𝜃,𝑣=cot𝜃,(2.5) so 𝜙 becomes

𝜙(𝛽,𝑟,𝜃)=𝑢−2𝑣𝑟cos𝛽+𝑟²^−3/2.(2.6) Then

𝜙_𝑟𝜙(𝛽,0,𝜃)=𝑢^−3/2,𝜙(𝛽,0,𝜃)=3𝑢^−5/2𝑣cos𝛽,_𝑟𝑟(𝛽,0,𝜃)=15𝑢^−7/2𝑣²cos²𝛽−3𝑢^−5/2.(2.7) We substitute into the Taylor expansion

𝑔(𝑟,𝜃)=𝑔(0,𝜃)+𝑔_𝑟(0,𝜃)𝑟+12!𝑔^𝑟𝑟(0,𝜃)𝑟²+13!𝑔^𝑟𝑟𝑟(0,𝜃)𝑟³+14!𝑔^{𝑟𝑟𝑟𝑟}(0,𝜃)𝑟⁴+⋯(2.8) and obtain

𝑔(𝑟,𝜃)=1241𝜋^02𝜋𝑢^−3/2𝑑𝛽𝑟²+13!23𝜋^02𝜋6𝑢^−5/2𝑣cos𝛽𝑑𝛽𝑟³+14!41𝜋^02𝜋315𝑢^−7/2𝑣²cos²𝛽−3𝑢^−5/2𝑑𝛽𝑟⁴+⋯.(2.9) Since

^02𝜋cos𝛽𝑑𝛽=0,^02𝜋cos²𝛽𝑑𝛽=𝜋(2.10) and 𝑢⁻¹=sin²𝜃 we obtain

𝑔(𝑟,𝜃)=𝑟²4sin³𝜃+3325sin⁵𝜃cos²𝜃−2sin⁵𝜃𝑟⁴+⋯.(2.11) Finally we replace cos²𝜃 by 1−sin²𝜃 and 𝑟 by 𝑑/2ℎ and obtain

𝑔_NP=116sin³𝜃𝑑ℎ²+3512sin⁵𝜃3−5sin²𝜃𝑑ℎ⁴+𝒪𝑑⁶.(2.12)

2.2. The Expansion for 𝜃−𝜋/𝟐 Small

To start we let

𝑓(𝜃,𝜂,𝛽)=csc²𝜃−2𝜂cot𝜃cos𝛽+𝜂²^−3/2(2.13) and compute its Taylor expansion at 𝜃=𝜋/2,

𝑓(𝜃,𝜂,𝛽)=1+𝜂²^−3/2−31+𝜂²^−5/2𝜂cos𝛽𝜃−𝜋+122!151+𝜂²^−7/2𝜂²cos²𝛽−31+𝜂²^−5/2𝜃−𝜋2²+𝒪𝜃−𝜋2³.(2.14)

Since

𝑔_NP=14𝜋^02𝜋^0𝑑/2ℎ𝑓(𝜃,𝜂,𝛽)𝜂𝑑𝜂𝑑𝛽(2.15) and since we know that only even powers of 𝜃−𝜋/2 occur, we obtain

𝑔_NP=121−√1+1(𝑑/2ℎ)²−83(𝑑/2ℎ)²1+(𝑑/2ℎ)²^5/2𝜃−𝜋2²+𝒪𝜃−𝜋2⁴.(2.16)

3. The Contour Integral

We have

𝐼=^02𝜋𝑑𝛽csc²𝜃−2𝜌ℎcot𝜃cos𝛽+ℎ𝜌²²^−3/2𝑔,(3.1)_NP=14𝜋ℎ²^0𝑑/2𝐼𝜌𝑑𝜌.(3.2)𝐼 may be recast as a contour integral by setting 𝑧=𝑒^𝑖𝛽: 𝐼=_Γ𝑑𝑧𝑖𝑧−2𝑧𝑏𝑧²−2𝑎𝑏𝑧+1^−3/2=2𝑏^3/21𝑖_Γ(𝑧−𝑥⁰)(𝑥₁−𝑧)𝑧^−3/2𝑑𝑧𝑧,(3.3) where

Γ𝑎is=thcescu₂nitcircle,𝜃+𝜌²ℎ²𝑏=2𝜌,𝑥ℎcot𝜃,₀=𝑎𝑏1−1−𝑏²𝑥₁𝑎²,=𝑎𝑏1+1−𝑏²𝑎².(3.4) We note that 𝑎>𝑏 and so 0<𝑥₀<1<𝑥₁.

In this section we take 𝜌,𝜃, and ℎ to be constant. We write

𝐹(𝑧)=𝑧𝑧−𝑥₀𝑥₁−𝑧((𝑧−𝑥₀)𝑧(𝑥₁−𝑧))^1/2=11𝑧𝑧−𝑥₀𝑥₁−𝑧(𝑧−𝑥₀)𝑧(𝑥₁−𝑧)^1/2𝐼=2,𝑏^3/21𝑖_Γ𝐹(𝑧)𝑑𝑧.(3.5) We need to be careful because of the square roots. We use the usual convention that

lim_𝑦→0±√−1+𝑖𝑦=±𝑖.(3.6) We wish to show that 𝐹(𝑧) is holomorphic (and single valued) in

{𝑧∶|𝑧|<1}−𝑥∶0<𝑥<𝑥₀.(3.7) To isolate the square root factor in 𝐹 we set

𝑔(𝑧)=𝑧𝑧−𝑥₀𝑥₁−𝑧.(3.8) We take |𝑧|≤1. If in addition Re𝑧<0 and |Im𝑧|≪|𝑧−𝑥₀| then Re(𝑧−𝑥₀)(𝑥₁−𝑧)<0. This implies that 𝑔(𝑧) never takes on negative real values and so has a single valued square root in Re𝑧<0. Similarly, if |𝑧|≤1, and Re𝑧>𝑥₀ (no condition on Im𝑧 now) then Re(𝑧−𝑥₀)(𝑥₁−𝑧)>0 and 𝑔(𝑧) has a single valued square root in Re𝑧>𝑥₀.

It follows from Cauchy's theorem that the integral of 𝐹(𝑧) over the unit circle Γ is equal to the integral over the “ barbell” 𝛾 around 0 and 𝑥₀. More explicitly, 𝛾 is the boundary of

𝐵=𝐵_𝜖(0)∪𝐵_𝜖𝑥₀∪𝑅,(3.9) where 𝐵_𝜖(𝑥) is the ball of radius 𝜖 and centered at 𝑥 and 𝑅 is the rectangle

𝑅=𝑧=𝑥+𝑖𝑦∶0≤𝑥≤𝑥₀,−𝛿≤𝑦≤𝛿(3.10) with 𝛿≪𝜖, see Figure 2.

Figure 2 

The curve 𝛾.

The curve 𝛾 decomposes into smooth pieces which we describe using the angle 𝛼=arcsin(𝛿/𝜖)

𝛾_𝑟=𝑧𝛾∶||𝑧−𝑥₀||=𝜖,−𝜋+𝛼≤arg𝑧−𝑥₀≤𝜋−𝛼_𝑙,𝛾={𝑧∶|𝑧|=𝜖,𝛼≤arg(𝑧)≤2𝜋−𝛼},_±=𝑧∶𝑧=𝑥±𝑖𝛿,𝜖cos𝛼≤𝑥≤𝑥₀−𝜖cos𝛼.(3.11)

To bound the integral over 𝛾_𝑙 we set 𝑧=𝜖𝑒^𝑖𝜃

||||_𝛾𝑙𝐹(𝑧)𝑑𝑧||||==𝒪||||𝜖^{𝛼2𝜋−𝛼}_3/2𝐹𝜖𝑒^𝑖𝜃𝜖𝑒^𝑖𝜃𝑑𝜃||||.(3.12) To evaluate ∫_𝛾𝑟𝐹(𝑧)𝑑𝑧 we set 𝑧=𝑥₀+𝜖𝑒^𝑖𝜃. Then

𝐹𝑥₀+𝜖𝑒^𝑖𝜃=𝐶𝜖^−3/2𝑒^{−(3/2)𝑖𝜃}+𝒪𝜖^−1/2(3.13) with

𝐶=𝑥^01/2𝑥₁−𝑥₀^3/2,(3.14) and so

_𝛾𝑟𝐹(𝑧)𝑑𝑧==^{𝜋−𝛼−𝜋+𝛼}𝜖^−3/2𝐶𝑒^{−(3/2)𝑖𝜃}+𝒪𝜖^−1/2𝜖𝑖𝑒^𝑖𝜃(1+𝒪(𝜖))𝑑𝜃_{𝜋−𝛼−𝜋+𝛼}𝐶𝑖𝜖^−1/2𝑒^{−(1/2)𝑖𝜃}+𝒪𝜖^1/2𝑑𝜃.(3.15) Since 𝛼=𝑂(𝜖), we may set 𝛼=0 and absorb the error in the 𝒪(𝜖^1/2) term. We then do the integration and obtain

_𝛾𝑟𝐹(𝑧)𝑑𝑧=4𝑖𝐶√𝜖+𝒪𝜖^1/2.(3.16) We want to compute the integral of 𝐹 over 𝛾₊. We need to be careful about the square root. Recall that we are assuming

𝜖cos𝛼<𝑥<𝑥₀−𝜖cos𝛼,𝛿≪𝜖.(3.17)

Lemma 3.1. On 𝛾₊𝐹(𝑧)=𝑖𝑥₀−𝑥𝑥₁−𝑥^−3/2𝑥^1/2+𝒪(𝛿)(3.18) and on 𝛾₋𝐹(𝑧)=−𝑖𝑥₀−𝑥𝑥₁−𝑥^−3/2𝑥^1/2+𝒪(𝛿).(3.19)

Proof. This is a consequence of the branching of the square root function. Let 𝐻(𝑧)=𝑧𝑧−𝑥₀𝑥₁−𝑧=𝑧𝑧−𝑥₀𝑥₁−𝑧||𝑧−𝑥₀||²||𝑥₁−𝑧||²ℎ(𝑧)=𝑧,𝑧−𝑥₀𝑥₁−𝑧.(3.20) We see that ℎ(𝑥+𝑖𝛿)=𝑥𝑥−𝑥₀𝑥₁−𝑥+𝑖𝛿𝑥²−1+𝒪𝛿².(3.21) Recall that 𝛿 is replaced by −𝛿 on 𝛾₋ and that 0<𝑥<𝑥₀<𝑥₁. So Re𝐻(𝑧)<0 on both 𝛾₊ and 𝛾₋ while Im𝐻(𝑧)<0 on 𝛾₊ and Im𝐻(𝑧)>0 on 𝛾₋. Thus on 𝛾₊𝐻(𝑧)^1/2=−𝑖||||𝑥(𝑥−𝑥₀)(𝑥₁−𝑥)||||^1/2=−𝑖𝑥+𝒪(𝛿)(𝑥₀−𝑥)(𝑥₁−𝑥)^1/2+𝒪(𝛿),(3.22) while on 𝛾₋𝐻(𝑧)^1/2=𝑖𝑥(𝑥₀−𝑥)(𝑥₁−𝑥)^1/2+𝒪(𝛿).(3.23) Since 𝐹(𝑧)=𝐹(𝑥)+𝒪(𝛿)=1𝑥−𝑥₀𝑥₁−𝑥𝐻(𝑧)^1/2=−1+𝒪(𝛿)𝑥₀−𝑥𝑥₁−𝑥𝐻(𝑧)^1/2=−1+𝒪(𝛿)𝑥₀−𝑥𝑥₁−𝑥−𝑖(𝑥₀−𝑥)𝑥(𝑥₁−𝑥)^1/2+𝒪(𝛿),(3.24) the lemma follows.

We also need to be careful about the orientation. The curve 𝛾 is oriented counter-clockwise. This means that the integration over 𝛾₊ goes in the direction of 𝑥₀ to 0. We need to reverse its sign in order to write it in the conventional way

_𝛾+𝐹(𝑧)𝑑𝑧==−−^{𝑥0−𝜖cos𝛼𝜖cos𝛼}_{𝑥0−𝜖cos𝛼𝜖cos𝛼}𝐹(𝑥+𝑖𝛿)𝑑𝑥𝑖𝑥^1/2𝑑𝑥𝑥₀−𝑥^3/2𝑥₁−𝑥^3/2+𝒪(𝜖).(3.25) We let 𝑥=𝑥₀𝑡 and Δ=𝜖𝑥⁰⁻¹cos𝛼 and use 𝛿≪𝜖 to obtain

_𝛾+𝐹(𝑧)𝑑𝑧=−^Δ1−Δ𝑖𝑥_03/2𝑡^1/2𝑑𝑥𝑥₀−𝑥₀𝑡^3/2𝑥₁−𝑥₀𝑡^3/2+𝒪(𝜖).(3.26) Now let

𝜏=𝑥¹𝑥₀>1.(3.27) We have

_𝛾+𝐹(𝑧)𝑑𝑧=−𝑖𝑥_03/2^Δ1−Δ𝑡^1/2𝑑𝑡(𝜏−𝑡)^3/2(1−𝑡)^3/2+𝒪(𝜖).(3.28) We may let Δ→0 in the lower limit, but not in the upper limit. So we now have

_𝛾+𝐹(𝑧)𝑑𝑧=−𝑖𝑥_03/2^01−Δ𝑡^1/2𝑑𝑡(𝜏−𝑡)^3/2(1−𝑡)^3/2+𝒪(𝜖).(3.29) Here Δ=𝜖𝑥₀₋₁cos𝛼. We may let 𝛿→0 and redefine Δ to be 𝜖𝑥₀₋₁.

We get the same value for the integral over 𝛾₋. This is because the integrand is the negative of that in (3.29) but the orientation is reversed and so the two negatives cancel.

For any function 𝑔(𝑡), continuously differentiable on the interval [0,1], we have

^01−Δ𝑔(𝑡)(1−𝑡)^3/2𝑑𝑡=2𝑔(1)Δ^1/2−2𝑔(0)−2¹⁰(1−𝑡)^−1/2𝑔^(𝑡)𝑑𝑡+𝑂Δ^1/2(3.30) as can easily be seen by an integration by parts.

We apply this to 𝑔(𝑡)=𝑡^1/2/(𝜏−𝑡)^3/2 as in (3.29). This yields

_𝛾+𝐹(𝑧)𝑑𝑧=−2𝑖(𝜏−1)^3/2𝑥^03/2Δ^1/2+𝑖𝑥^03/2¹⁰𝑡^−1/2(𝜏−𝑡)^−3/2+3𝑡^1/2(𝜏−𝑡)^−5/2𝑑𝑡+𝒪𝜖^1/2.(3.31) We now take twice this quantity and add to it the results of (3.14) and (3.16),

_|𝑧|=1𝐹(𝑧)𝑑𝑧==4_𝛾𝑖𝐶𝐹(𝑧)𝑑𝑧√𝜖+2−2𝑖√𝑥₀+𝑖𝑥₁−𝑥₀^−3/2√𝜖𝑥_03/2¹⁰+𝒪𝜖_1/2(1−𝑡)^−1/2(𝜏−𝑡)^−3/2𝑡^−1/2+3𝑡^1/2(𝜏−𝑡)^−5/2𝑑𝑡.(3.32) We substitute the value for 𝐶 from (3.14) and let 𝜖→0

_|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖𝑥_03/2¹⁰(1−𝑡)^−1/2(𝜏−𝑡)^−3/2𝑡^−1/2+3𝑡^1/2(𝜏−𝑡)^−5/2𝑑𝑡.(3.33) Finally, we evaluate this integral using Mathematica and obtain

_|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖𝑥_03/22(2𝜏𝐸[1/𝜏(]𝜏−−(1−)1²√+𝜏𝜏)𝐾[1/𝜏]),(3.34) where 𝐾 and 𝐸 are the complete elliptic functions of the first and second kinds

𝐾𝐸[[𝑚𝑚]]==¹⁰1−𝑚𝑡²^−1/21−𝑡²^−1/2𝑑𝑡,₁₀1−𝑚𝑡²^1/21−𝑡²^−1/2𝑑𝑡.(3.35)

We now have

𝐼=2𝑏^3/21𝑖_Γ=2𝐹(𝑧)𝑑𝑧𝑏^3/22𝑥_03/22(2𝜏𝐸(1/𝜏)−(𝜏−1)𝐾(1/𝜏))=2(𝜏−1)²√𝜏𝑏^3/24𝜏^1/4(𝜏−1)²𝑔(2𝜏𝐸(1/𝜏)−(𝜏−1)𝐾(1/𝜏)),(3.36)_NP=14𝜋ℎ²^0𝑑/2𝜌𝐼𝑑𝜌(3.37) with

𝑏=2𝜌ℎcot𝜃,𝑎=csc²𝜃+𝜌²ℎ²𝜏=𝑥₁,(3.38)𝑥₀=1𝑥₀₂𝑥₀,=𝑎𝑏1−1−𝑏²𝑥₁𝑎²,=𝑎𝑏1+1−𝑏²𝑎².(3.39)

4. Limiting Cases

We validate (3.36) by considering two limiting cases.

4.1. Expansion for Small 𝑑

We look for an expansion of 𝑔_NP under the assumption that 𝑑/ℎ is small and rederive (2.12). The dependence of 𝑥₀ on 𝜌 is given in (3.39). The two-term Taylor expansion of 𝑥₀ as a function of 𝜌 is

𝑥₀=𝐴𝜌+𝐵𝜌³+𝒪𝜌⁵(4.1) with

𝐴=cos𝜃sin𝜃ℎ,𝐵=𝐴³−𝐴ℎ²csc²𝜃.(4.2) Since 𝜏=1𝑥₀₂,(4.3) we have 1𝜏=𝐴²𝜌²+𝒪𝜌⁴,(4.4)1(𝜏−1)²=1𝜏²+2𝜏³+𝒪𝜌⁸.(4.5) It is convenient to work with the quantity 𝜌²𝜏, so we note, using (4.1),

𝜌²𝜏=1𝐴²−2𝐵𝐴³𝜌²+𝒪𝜌⁴.(4.6) We start with

𝐼=2𝑏^3/24𝜏^1/4(𝜏−1)²2𝜏𝐸𝜏1−(𝜏−1)𝐾𝜏1.(4.7) Substituting the expression for 𝑏 from (3.38) and for (𝜏−1)² from (4.5) we obtain, after some manipulation

𝐼=ℎcot𝜃^3/24𝜌²𝜏^3/41𝜏+𝒪𝜌⁴2𝜏𝐸𝜏1−(𝜏−1)𝐾𝜏1.(4.8)

We now use the expansions (as given by Mathematica)

𝐸(𝑡)=𝜋𝐾(𝑡)=𝜋2−𝜋8𝑡+𝒪𝑡²,2+𝜋8𝑡+𝒪𝑡².(4.9) So

𝐼=ℎcot𝜃^3/24𝜌²𝜏^3/4𝜋2+8𝜋𝜏+𝜋𝜏+𝒪𝜌⁴.(4.10) From the expression for 𝜌²𝜏 in (4.6), we derive

1𝜌²𝜏^3/4=𝐴^3/21+32𝐵𝐴𝜌²+𝒪𝜌⁴,(4.11) and this leads to

𝐼=4ℎcot𝜃^3/2𝐴^3/2𝜋2+8𝜋𝜏+𝜋𝜏+3𝐵𝜌²𝜋4𝐴+𝒪𝜌⁴.(4.12) We now substitute for 𝐴 and 𝐵 using (4.2) and also for 𝜏⁻¹ using (4.4)

𝐼=sin³𝜃2𝜋+3𝜋23sin²𝜃−5sin⁴𝜃𝜌²ℎ²+𝒪𝜌⁴.(4.13) Substituting this into (3.37) and integrating, we again derive (2.12).

4.2. An Expansion for 𝜃 Near 𝜋/𝟐

We show how (4.7) also gives the expansion of Section 2.2. From (3.38) we see that

𝑎=1+𝜌²ℎ²+𝜃−𝜋2²+𝒪𝜃−𝜋2⁴𝑏=−2𝜌,(4.14)ℎ𝜃−𝜋2+𝒪𝜃−𝜋2³.(4.15) So from (3.39) we have

𝑥₀=𝑏2𝑎1+4𝑏𝑎²²+𝒪𝜃−𝜋2³.(4.16) Thus

𝜏=4𝑎²𝑏²−2+𝒪𝜃−𝜋2²1,(4.17)(𝜏−1)²=11𝜏²1+𝜏2+𝒪𝜃−𝜋2⁴,𝜏2𝜏𝐸𝜏1−(𝜏−1)𝐾𝜏1=𝜋2+98𝜋𝜏+𝒪𝜃−𝜋2⁴.(4.18) These expansions let us write (4.7) as

𝐼=2𝑏^3/24=2_7/2𝜏^3/41+𝜏2𝜋2+98𝜋𝜏+𝒪𝜃−𝜋2⁴𝑏²𝜏^3/41+𝜏2𝜋2+98𝜋𝜏+𝒪𝜃−𝜋2⁴.(4.19) Using (4.15) and (4.17) we obtain a substitute for (4.1)

1𝑏²𝜏^3/4=4^−3/4𝑎^−3/21+83𝑎𝑏²²+𝒪𝜃−𝜋2⁴.(4.20) This gives, in light of (4.14),

𝐼=2𝜋𝑎^−3/2+158𝜋𝑏²𝑎^−7/2+𝒪𝜃−𝜋2⁴=2𝜋1+𝜌²ℎ²^−3/2−3𝜋1+𝜌²ℎ²^−5/2𝜃−𝜋2²+152𝜋ℎ𝜌²²1+ℎ𝜌²²^−7/2𝜃−𝜋2²+𝒪𝜃−𝜋2⁴.(4.21) Finally, we integrate term by term.

𝑔=14𝜋ℎ²^0𝑑/2=1𝜌𝐼(𝜌)𝑑𝜌4𝜋ℎ²−2𝜋3ℎ𝜋²ℎ₂1−1+2𝑑ℎ²^−1/2131−1+2𝑑ℎ²^−3/2𝜃−𝜋2²+15+𝒪𝜃−𝜋2𝜋ℎ²125−2+51(𝑑5/2ℎ)²1+2𝑑ℎ²^−5/2𝜃−𝜋2²2⁴=121−1+2𝑑ℎ²^−1/2−381+2𝑑ℎ²^−5/22𝑑ℎ²𝜃−𝜋2²+𝒪𝜃−𝜋2⁴.(4.22)

5. Numerical Confirmation

The result in (3.36) is further validated in this section with numerical integration of (3.1). In this original formulation, the integral depends on the ratio of 𝜌/ℎ and on the value of 𝜃. We analytically considered one limiting case for each of these quantities in the previous section. We consider these and other values for these quantities in this section using numerical methods. For fixed values of 𝜌/ℎ, the numerical integration (solid lines) is shown with the results of the residue calculation (open circles) in Figure 3 as a function of 𝜃. Figure 4 shows the numerical integration and residue results for fixed values of 𝜃 as a function of 𝜌/ℎ. All corresponding values of the residue calculation and the numerical integration agree.

Figure 3 

Numerical integration of (3.1) (solid lines) and residue calculation of (3.36) (open circles) as a function of 𝜃 for fixed values of 𝜌/ℎ: 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0.

Figure 4 

Numerical integration of (3.1) (solid lines) and residue calculation of (3.36) (open circles) as a function of 𝜌/ℎ for fixed values of 𝜃: 50, 60, 70, 80, and 90 degrees.

6. Conclusions

The accuracy of collimator efficiency is important for design and for quantitative reconstruction programs. Analytic forms obviate the need for numerical models, which do not readily offer insight into the interplay of collimator parameters. This paper approaches the problem by recasting one of the integrals in terms of contour integration. This integration was completed, but the authors were unable to execute the second integral to yield a complete analytic solution. The result was studied for special cases with known outcomes for validation. Future efforts may be able to build on this validated result to find a complete analytic solution.

Acknowledgment

Scott D. Metzler was supported by the National Institute for Biomedical Imaging and Bioengineering of the National Institutes of Health under Grant R01-EB-6558.