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International Journal of Mathematics and Mathematical Sciences
Volume 2010, Article ID 915958, 18 pages
Research Article

Geometric Sensitivity of a Pinhole Collimator

1Department of Mathematical Sciences, Rutgers University, Camden, NJ 08102, USA
2Department of Radiology, University of Pennsylvania, Philadelphia, PA 19104, USA

Received 25 August 2009; Revised 7 December 2009; Accepted 19 February 2010

Academic Editor: Harvinder S. Sidhu

Copyright © 2010 Howard Jacobowitz and Scott D. Metzler. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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= 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).

𝑑𝑔=2sin3𝜃162,(1.1) where 𝑑 is the diameter of the pinhole, 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

𝑔𝜋𝜃=2=1212𝑑1+221/21212112𝑑22=𝑑2162.(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 𝜃𝜋/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 𝜃𝜋/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:

1𝑔=4𝜋20𝑑/2𝜌𝑑𝜌02𝜋𝑑𝛽sin3𝜃𝑎,(1.3) where 𝜌 and 𝛽 are polar coordinates on the aperture plane and 𝜃𝑎 is the incidence angle of the photon at a particular value of 𝜌 and 𝛽, and is given by [3]

cot2𝜃𝑎=cot2𝜌𝜃2𝜌cot𝜃cos(𝛽𝜙)+22,(1.4) where 𝜙 is the third coordinate of the point source. For integrals over 2𝜋 the difference 𝛽𝜙 can be replaced with only 𝛽 without loss of generality. This yields


Applying this to (1.3),


When there is penetration the above would be modified to

1𝑔=4𝜋20𝜌𝑑𝜌02𝜋𝑑𝛽csc2𝜌𝜃2𝜌cot𝜃cos𝛽+223/2𝑒𝜇Δ𝐿,(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=𝑔NP+14𝜋2𝑑/2𝜌𝑑𝜌02𝜋𝑑𝛽csc2𝜌𝜃2𝜌cot𝜃cos𝛽+223/2𝑒𝜇Δ𝐿=𝑔NP+𝑔PEN,(1.8) where 𝑔PEN is the efficiency due to penetration and 𝑔NP, 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 𝑔𝑁𝑃

The special form of 𝑔NP makes it easy to derive approximations for 𝑑/ small and for 𝜃𝜋/2 small which generalize (1.1) and (1.2). Before doing this, we observe that only even powers of 𝑑 and of 𝜃𝜋/2 can occur in the expansion of 𝑔NP. 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 𝑔NP by 𝑔. Upon substituting 𝑟=𝑑/2 and 𝜂=𝜌/ into (1.6) we obtain

1𝑔(𝑟,𝜃)=4𝜋02𝜋𝑟0csc2𝜃2𝜂cot𝜃cos𝛽+𝜂23/2=1𝜂𝑑𝜂𝑑𝛽,4𝜋02𝜋𝑟0𝜂𝜙(𝛽,𝜂,𝜃)𝑑𝜂𝑑𝛽.(2.1) Clearly, 𝑔(0,𝜃)=0. Since

𝑔𝑟1(𝑟,𝜃)=4𝜋02𝜋𝑟𝜙(𝛽,𝑟,𝜃)𝑑𝛽,(2.2) we have 𝑔𝑟(0,𝜃)=0. Differentiating (2.12) three more times with respect to 𝑟 we obtain

𝑔𝑟𝑟1(𝑟,𝜃)=4𝜋02𝜋𝑟𝜙𝑟𝑔(𝛽,0,𝜃)+𝜙(𝛽,𝑟,𝜃)𝑑𝛽,𝑟𝑟𝑟1(𝑟,𝜃)=4𝜋02𝜋𝑟𝜙𝑟𝑟(𝛽,0,𝜃)+2𝜙𝑟𝑔(𝛽,𝑟,𝜃)𝑑𝛽,𝑟𝑟𝑟𝑟1(𝑟,𝜃)=4𝜋02𝜋𝑟𝜙𝑟𝑟𝑟(𝛽,0,𝜃)+3𝜙𝑟𝑟(𝛽,𝑟,𝜃)𝑑𝛽,(2.3) and then setting 𝑟=0 in each of the resulting equations we obtain

𝑔𝑟𝑟1(0,𝜃)=4𝜋02𝜋𝑔𝜙(𝛽,0,𝜃)𝑑𝛽,𝑟𝑟𝑟1(0,𝜃)=2𝜋02𝜋𝜙𝑟𝑔(𝛽,0,𝜃)𝑑𝛽,𝑟𝑟𝑟𝑟3(0,𝜃)=4𝜋02𝜋𝜙𝑟𝑟(𝛽,0,𝜃)𝑑𝛽.(2.4) For convenience let us write

𝑢=csc2𝜃,𝑣=cot𝜃,(2.5) so 𝜙 becomes

𝜙(𝛽,𝑟,𝜃)=𝑢2𝑣𝑟cos𝛽+𝑟23/2.(2.6) Then

𝜙(𝛽,0,𝜃)=𝑢3/2,𝜙𝑟(𝛽,0,𝜃)=3𝑢5/2𝜙𝑣cos𝛽,𝑟𝑟(𝛽,0,𝜃)=15𝑢7/2𝑣2cos2𝛽3𝑢5/2.(2.7) We substitute into the Taylor expansion

𝑔(𝑟,𝜃)=𝑔(0,𝜃)+𝑔𝑟1(0,𝜃)𝑟+𝑔2!𝑟𝑟(0,𝜃)𝑟2+1𝑔3!𝑟𝑟𝑟(0,𝜃)𝑟3+1𝑔4!𝑟𝑟𝑟𝑟(0,𝜃)𝑟4+(2.8) and obtain

1𝑔(𝑟,𝜃)=214𝜋02𝜋𝑢3/2𝑟𝑑𝛽2+133!2𝜋02𝜋6𝑢5/2𝑟𝑣cos𝛽𝑑𝛽3+114!4𝜋02𝜋315𝑢7/2𝑣2cos2𝛽3𝑢5/2𝑟𝑑𝛽4+.(2.9) Since

02𝜋cos𝛽𝑑𝛽=0,02𝜋cos2𝛽𝑑𝛽=𝜋(2.10) and 𝑢1=sin2𝜃 we obtain

𝑟𝑔(𝑟,𝜃)=24sin33𝜃+325sin5𝜃cos2𝜃2sin5𝜃𝑟4+.(2.11) Finally we replace cos2𝜃 by 1sin2𝜃 and 𝑟 by 𝑑/2 and obtain


2.2. The Expansion for 𝜃𝜋/𝟐 Small

To start we let

𝑓(𝜃,𝜂,𝛽)=csc2𝜃2𝜂cot𝜃cos𝛽+𝜂23/2(2.13) and compute its Taylor expansion at 𝜃=𝜋/2,



𝑔NP=14𝜋02𝜋0𝑑/2𝑓(𝜃,𝜂,𝛽)𝜂𝑑𝜂𝑑𝛽(2.15) and since we know that only even powers of 𝜃𝜋/2 occur, we obtain


3. The Contour Integral

We have

𝐼=02𝜋𝑑𝛽csc2𝜌𝜃2𝜌cot𝜃cos𝛽+223/2,𝑔(3.1)NP=14𝜋20𝑑/2𝐼𝜌𝑑𝜌.(3.2)𝐼 may be recast as a contour integral by setting 𝑧=𝑒𝑖𝛽: 𝐼=Γ𝑑𝑧𝑖𝑧𝑏𝑧2𝑧2𝑎2𝑏𝑧+13/2=2𝑏3/21𝑖Γ(𝑧𝑥0)(𝑥1𝑧)𝑧3/2𝑑𝑧𝑧,(3.3) where

Γistheunitcircle,𝑎=csc2𝜌𝜃+22,𝜌𝑏=2𝑥cot𝜃,0=𝑎𝑏1𝑏12𝑎2,𝑥1=𝑎𝑏1+𝑏12𝑎2.(3.4) We note that 𝑎>𝑏 and so 0<𝑥0<1<𝑥1.

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

𝑧𝐹(𝑧)=𝑧𝑥0𝑥1𝑧𝑧((𝑧𝑥0)(𝑥1𝑧))1/21𝑧=1𝑧𝑥0𝑥1𝑧𝑧(𝑧𝑥0)(𝑥1𝑧)1/2,2𝐼=𝑏3/21𝑖Γ𝐹(𝑧)𝑑𝑧.(3.5) We need to be careful because of the square roots. We use the usual convention that

lim𝑦0±1+𝑖𝑦=±𝑖.(3.6) We wish to show that 𝐹(𝑧) is holomorphic (and single valued) in

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

𝑧𝑔(𝑧)=𝑧𝑥0𝑥1.𝑧(3.8) We take |𝑧|1. If in addition Re𝑧<0 and |Im𝑧||𝑧𝑥0| then Re(𝑧𝑥0)(𝑥1𝑧)<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(𝑧𝑥0)(𝑥1𝑧)>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=𝐵𝜖(0)𝐵𝜖𝑥0𝑅,(3.9) where 𝐵𝜖(𝑥) is the ball of radius 𝜖 and centered at 𝑥 and 𝑅 is the rectangle

𝑅=𝑧=𝑥+𝑖𝑦0𝑥𝑥0,𝛿𝑦𝛿(3.10) with 𝛿𝜖, 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)


To bound the integral over 𝛾𝑙 we set 𝑧=𝜖𝑒𝑖𝜃

||||𝛾𝑙𝐹||||=||||(𝑧)𝑑𝑧𝛼2𝜋𝛼𝐹𝜖𝑒𝑖𝜃𝜖𝑒𝑖𝜃||||𝜖𝑑𝜃=𝒪3/2.(3.12) To evaluate 𝛾𝑟𝐹(𝑧)𝑑𝑧 we set 𝑧=𝑥0+𝜖𝑒𝑖𝜃. Then

𝐹𝑥0+𝜖𝑒𝑖𝜃=𝐶𝜖3/2𝑒(3/2)𝑖𝜃𝜖+𝒪1/2(3.13) with

𝑥𝐶=01/2𝑥1𝑥03/2,(3.14) and so

𝛾𝑟𝐹(𝑧)𝑑𝑧=𝜋𝛼𝜋+𝛼𝜖3/2𝐶𝑒(3/2)𝑖𝜃𝜖+𝒪1/2𝜖𝑖𝑒𝑖𝜃=(1+𝒪(𝜖))𝑑𝜃𝜋𝛼𝜋+𝛼𝐶𝑖𝜖1/2𝑒(1/2)𝑖𝜃𝜖+𝒪1/2𝑑𝜃.(3.15) Since 𝛼=𝑂(𝜖), we may set 𝛼=0 and absorb the error in the 𝒪(𝜖1/2) term. We then do the integration and obtain

𝛾𝑟𝐹(𝑧)𝑑𝑧=4𝑖𝐶𝜖𝜖+𝒪1/2.(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


Lemma 3.1. On 𝛾+𝑥𝐹(𝑧)=𝑖0𝑥𝑥1𝑥3/2𝑥1/2+𝒪(𝛿)(3.18) and on 𝛾𝑥𝐹(𝑧)=𝑖0𝑥𝑥1𝑥3/2𝑥1/2+𝒪(𝛿).(3.19)

Proof. This is a consequence of the branching of the square root function. Let 𝑧𝐻(𝑧)=𝑧𝑥0𝑥1=𝑧𝑧𝑧𝑥0𝑥1𝑧||𝑧𝑥0||2||𝑥1||𝑧2,(𝑧)=𝑧𝑧𝑥0𝑥1𝑧.(3.20) We see that (𝑥+𝑖𝛿)=𝑥𝑥𝑥0𝑥1𝑥𝑥+𝑖𝛿2𝛿1+𝒪2.(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||||𝑥=𝑖(𝑥𝑥0)(𝑥1||||𝑥)1/2𝑥+𝒪(𝛿)=𝑖(𝑥0𝑥)(𝑥1𝑥)1/2+𝒪(𝛿),(3.22) while on 𝛾𝐻(𝑧)1/2𝑥=𝑖(𝑥0𝑥)(𝑥1𝑥)1/2+𝒪(𝛿).(3.23) Since 1𝐹(𝑧)=𝐹(𝑥)+𝒪(𝛿)=𝑥𝑥0𝑥1𝑥𝐻(𝑧)1/21+𝒪(𝛿)=𝑥0𝑥𝑥1𝐻𝑥(𝑧)1/21+𝒪(𝛿)=𝑥0𝑥𝑥1𝑥𝑥𝑖(𝑥0𝑥)(𝑥1𝑥)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=𝑥0𝜖cos𝛼𝜖cos𝛼𝐹(𝑥+𝑖𝛿)𝑑𝑥=𝑥0𝜖cos𝛼𝜖cos𝛼𝑖𝑥1/2𝑑𝑥𝑥0𝑥3/2𝑥1𝑥3/2+𝒪(𝜖).(3.25) We let 𝑥=𝑥0𝑡 and Δ=𝜖𝑥01cos𝛼 and use 𝛿𝜖 to obtain

𝛾+𝐹(𝑧)𝑑𝑧=Δ1Δ𝑖𝑥03/2𝑡1/2𝑑𝑥𝑥0𝑥0𝑡3/2𝑥1𝑥0𝑡3/2+𝒪(𝜖).(3.26) Now let

𝑥𝜏=1𝑥0>1.(3.27) We have

𝛾+𝑖𝐹(𝑧)𝑑𝑧=𝑥03/2Δ1Δ𝑡1/2𝑑𝑡(𝜏𝑡)3/2(1𝑡)3/2+𝒪(𝜖).(3.28) We may let Δ0 in the lower limit, but not in the upper limit. So we now have

𝛾+𝑖𝐹(𝑧)𝑑𝑧=𝑥03/201Δ𝑡1/2𝑑𝑡(𝜏𝑡)3/2(1𝑡)3/2+𝒪(𝜖).(3.29) Here Δ=𝜖𝑥01cos𝛼. We may let 𝛿0 and redefine Δ to be 𝜖𝑥01.

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

01Δ𝑔(𝑡)(1𝑡)3/2𝑑𝑡=2𝑔(1)Δ1/22𝑔(0)210(1𝑡)1/2𝑔Δ(𝑡)𝑑𝑡+𝑂1/2(3.30) as can easily be seen by an integration by parts.

We apply this to 𝑔(𝑡)=𝑡1/2/(𝜏𝑡)3/2 as in (3.29). This yields

𝛾+𝐹(𝑧)𝑑𝑧=2𝑖(𝜏1)3/2𝑥03/2Δ1/2+𝑖𝑥03/210𝑡1/2(𝜏𝑡)3/2+3𝑡1/2(𝜏𝑡)5/2𝜖𝑑𝑡+𝒪1/2.(3.31) We now take twice this quantity and add to it the results of (3.14) and (3.16),

|𝑧|=1𝐹(𝑧)𝑑𝑧=𝛾=𝐹(𝑧)𝑑𝑧4𝑖𝐶𝜖+22𝑖𝑥0𝑥1𝑥03/2𝜖+𝑖𝑥03/210(1𝑡)1/2(𝜏𝑡)3/2𝑡1/2+3𝑡1/2(𝜏𝑡)5/2𝜖𝑑𝑡+𝒪1/2.(3.32) We substitute the value for 𝐶 from (3.14) and let 𝜖0

|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖𝑥03/210(1𝑡)1/2(𝜏𝑡)3/2𝑡1/2+3𝑡1/2(𝜏𝑡)5/2𝑑𝑡.(3.33) Finally, we evaluate this integral using Mathematica and obtain

|𝑧|=1𝐹(𝑧)𝑑𝑧=2𝑖𝑥03/2[][])2(2𝜏𝐸1/𝜏(1+𝜏)𝐾1/𝜏(𝜏1)2𝜏,(3.34) where 𝐾 and 𝐸 are the complete elliptic functions of the first and second kinds


We now have

2𝐼=𝑏3/21𝑖Γ=2𝐹(𝑧)𝑑𝑧𝑏3/22𝑥03/22(2𝜏𝐸(1/𝜏)(𝜏1)𝐾(1/𝜏))(𝜏1)2𝜏=2𝑏3/24𝜏1/4(𝜏1)2𝑔(2𝜏𝐸(1/𝜏)(𝜏1)𝐾(1/𝜏)),(3.36)NP=14𝜋20𝑑/2𝜌𝐼𝑑𝜌(3.37) with


4. Limiting Cases

We validate (3.36) by considering two limiting cases.

4.1. Expansion for Small 𝑑

We look for an expansion of 𝑔NP under the assumption that 𝑑/ 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+𝐵𝜌3𝜌+𝒪5(4.1) with

𝐴=cos𝜃sin𝜃,𝐵=𝐴3𝐴2csc2𝜃.(4.2) Since 1𝜏=𝑥02,(4.3) we have 1𝜏=𝐴2𝜌2𝜌+𝒪4,(4.4)1(𝜏1)2=1𝜏2+2𝜏3𝜌+𝒪8.(4.5) It is convenient to work with the quantity 𝜌2𝜏, so we note, using (4.1),

𝜌21𝜏=𝐴22𝐵𝐴3𝜌2𝜌+𝒪4.(4.6) We start with

2𝐼=𝑏3/24𝜏1/4(𝜏1)212𝜏𝐸𝜏1(𝜏1)𝐾𝜏.(4.7) Substituting the expression for 𝑏 from (3.38) and for (𝜏1)2 from (4.5) we obtain, after some manipulation


We now use the expansions (as given by Mathematica)

𝜋𝐸(𝑡)=2𝜋𝑡8𝑡+𝒪2,𝐾𝜋(𝑡)=2+𝜋𝑡8𝑡+𝒪2.(4.9) So

𝐼=cot𝜃3/24𝜌2𝜏3/4𝜋2+𝜋+𝜋8𝜏𝜏𝜌+𝒪4.(4.10) From the expression for 𝜌2𝜏 in (4.6), we derive

1𝜌2𝜏3/4=𝐴3/231+2𝐵𝐴𝜌2𝜌+𝒪4,(4.11) and this leads to

𝐼=4cot𝜃3/2𝐴3/2𝜋2+𝜋+𝜋8𝜏𝜏+3𝐵𝜌2𝜋𝜌4𝐴+𝒪4.(4.12) We now substitute for 𝐴 and 𝐵 using (4.2) and also for 𝜏1 using (4.4)

𝐼=sin3𝜃2𝜋+3𝜋23sin2𝜃5sin4𝜃𝜌22𝜌+𝒪4.(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

𝜌𝑎=1+22+𝜋𝜃22𝜋+𝒪𝜃24,(4.14)𝑏=2𝜌𝜋𝜃2𝜋+𝒪𝜃23.(4.15) So from (3.39) we have

𝑥0=𝑏𝑏2𝑎1+24𝑎2𝜋+𝒪𝜃23.(4.16) Thus

𝜏=4𝑎2𝑏2𝜋2+𝒪𝜃22,1(4.17)(𝜏1)2=1𝜏221+𝜏𝜋+𝒪𝜃24,1𝜏12𝜏𝐸𝜏1(𝜏1)𝐾𝜏=𝜋2+9𝜋𝜋8𝜏+𝒪𝜃24.(4.18) These expansions let us write (4.7) as

2𝐼=𝑏3/24𝜏3/421+𝜏𝜋2+9𝜋𝜋8𝜏+𝒪𝜃24=27/2𝑏2𝜏3/421+𝜏𝜋2+9𝜋𝜋8𝜏+𝒪𝜃24.(4.19) Using (4.15) and (4.17) we obtain a substitute for (4.1)

1𝑏2𝜏3/4=43/4𝑎3/21+3𝑏28𝑎2𝜋+𝒪𝜃24.(4.20) This gives, in light of (4.14),

𝐼=2𝜋𝑎3/2+158𝜋𝑏2𝑎7/2𝜋+𝒪𝜃24𝜌=2𝜋1+223/2𝜌3𝜋1+225/2𝜋𝜃22+152𝜋𝜌22𝜌1+227/2𝜋𝜃22𝜋+𝒪𝜃24.(4.21) Finally, we integrate term by term.


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/ 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/, 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/. 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/: 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/ 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.


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.


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