Using thymine-18 for enhancing dose delivery and localizing the Bragg peak in proton-beam therapy
UUsing thymine-18 for enhancing dose delivery and localizing the Bragg peak inproton-beam therapy
William Parke
The George Washington University, Washington, DC
Dalong Pang
Georgetown Medical School, Washington, DC (Dated: February 19, 2021 v0.7)Therapeutic protons acting on O18-substituted thymidine increase cytotoxicity in radio-resistanthuman cancer cells. We consider here the physics behind the irradiation during proton beam therapyand diagnosis using O18-enriched thymine in DNA, with attention to the effect of the presence ofthymine-18 on cancer cell death.
I. INTRODUCTION
This technical report presents the physics background behind a proposal for proton-beam therapy (PT) using dopedDNA in which one of the oxygen atoms in the thymine bases have been replaced by oxygen-18 (a stable isotope), forthe purpose of enhanced dose to tumors and for the side benefit of localization of the peak-dose delivery region.
FIG. 1. Thymine-18. O is attached to the carbon between the two nitrogens. (This carbon is atom
Proton therapy delivers radiation to cancers to cause malignant-cell death, principally by radical formation causingunrepairable damage to their DNA and thereby slowing or stopping tumor growth. Therapeutic protons, in passingthrough tissue, will scatter electrons and generate ions and radicals, such as hydroxyl. Through chemical reactions,the radicals can cause single-strand breaks (SSB) in DNA. Two such nearby breaks can generate a double-strandbreak (DSB). With a much lower probability, a direct DSB can be caused by a beam proton passing through theDNA helix itself, producing a ‘clustered’ pair of SSBs. Beam protons can also activate nuclei, such as those of oxygen,carbon and nitrogen atoms. These, in turn, can release, through strong-interaction decay, gammas and fast-movingsecondary particles such as protons, neutrons, deuterons, tritium and alpha particles, and through weak-interactiondecay, electrons, positrons, and neutrinos. Those secondaries with a charge can also produce localized damage to thefunctioning of tumor cells.A recent project [33]) at the Georgetown University Hospital initiated a study of the lethality enhancement ofirradiated cells due to modified DNA, replacing the O16 with O18 in thymine (see Figure 1). Upon proton irradiation,some O18 is transformed to F18, producing a DNA breakage mechanism and a positron-emission tomography (PET)signal, which can be monitored to help track where the proton beam Bragg peak occurred in the doped tissue. TheGeorgetown group exposed SQ20-B squamous carcinoma cells to physiologic O-thymidine concentration of 5 µ Mfor 48 hours followed by 1 to 9 Gy graded doses of proton radiation given 24 hours later. Survival analysis showed adose modification factor (
DM F ) of 1.2 due to the substituted thymidine.Here we wish to estimate the rate of O18 conversion to F18 for a given proton beam and fixed amount of dopedDNA, and the subsequent resulting DNA breakage induced. Also, we want to assess the use of the PT-produced F18to better localize the PT beam region of maximum dose via a post-PT PET scanning. a r X i v : . [ phy s i c s . m e d - ph ] F e b II. PHYSICS ANALYSISA. Proton beams in tissue
When an energetic proton (initial kinetic energy K from 50 to 250 MeV) enters tissue, the atoms and moleculesin the tissue will slow the beam protons, largely through scattering and ionization of tissue electrons. Because theproton is 1836 times more massive than the electron, the direction of the proton is not significantly changed byelectron encounters. (Appendix B 2 shows that for each encounter of a proton with a quasi-free electron, the protondeviates from a straight line no more than arcsin ( m e /m p ) = 0 . m e is the mass of the electron and m p is the mass of the proton.) The scattered electrons (‘delta’ electrons) spread the beam ionization over nearby tissue,but not more than about 2 mm for 200 MeV protons. The ‘stopping power’ (loss of beam-proton kinetic energy withdistance, − dK/dz ) depends on the density of the tissue, largely because of the electron density. Fluences in the rangeof 10 protons per square centimeter are typical for proton-beam therapy. The fluence rate d Φ /dt of protons in thebeam decreases as protons are taken from the beam, with interaction of beam protons with nuclei in the tissue a largecontributor to the fluence loss. These nuclear events are relatively rare because of the small size of the nuclei, but canwidely deflect beam protons by nuclear-Coulomb elastic and inelastic scattering. Nonelastic nuclear reactions alsooccur when the beam protons have sufficient energy to penetrate the nuclear Coulomb barrier, making fast-movingreaction products, generally moving away from the proton beam axis. The physical dose delivered by the beam within some small volume dV of tissue, i.e. the energy dE dep deposited inthat volume per unit mass of tissue D = dE dep dm = 1 ρ dE dep dV , (1)comes from electron ionizations (producing chemical transformations and radical formation), but is also affected bythe energy deposited by the products of nuclear reactions. For a proton beam with an initial energy of 90 MeV, 5 %of the beam energy loss from the beam comes from inelastic nuclear reactions ([36]).The ‘biologically effective dose’ ( BED ), D BE , with units sievert (Sv), defined as the physical dose times a unitlessfactor called the ‘relative biological effectiveness’ ( RBE ), attempts to make the
BED have the same biologicallydamaging effect as photons with about the same energy. The overall
RBE of proton beams is around 1 .
1, while forcarbon ions, its about 2 . RBE exceeds 3. However, the
RBE of a proton beam depends on itsenergy, reaching values as much as 4 or 5 when the beam’s rate of energy loss is about 100 keV per micrometer ([36])because of a larger number of atomic electron excitation and ionization, but also having contributions from a greaternumber proton-nuclei reactions making ionizing secondaries. A
BED over 1 Sv will frequently be fatal to living cells(lethality is 50 % with a
BED of 4 to 5 Sv delivered over a few minutes).In tissue, the released neutrons are likely to bounce around among the nearby nuclei until absorbed by a nucleusor thermalized. The energy loss is slow relative to protons, because there is no Coulomb interaction, so the neutronmust get within about a fermi of a nucleus to scatter. For neutrons in the 1 to 20 MeV energy range, scatteringagainst a high Z nucleus can cause that nucleus to be excited, but this has a lower probability than elastic scattering.At lower energies, the neutron is likely to be absorbed, even by hydrogen protons making deuterium. The neutronattenuation coefficient depends on the inverse of its relative speed for energies below about 1 eV. The time for diffusivethermalization ranges from about 200 microseconds (water) to 20 milliseconds (graphite). A thermalized neutron isthen likely to be absorbed by a nearby nucleus. Without a nuclear-absorption event, the neutron will beta decay(with a half-life time of 10 . .
782 MeV in kineticenergy. The neutrinos exit with practically no interaction with matter on their path. (A 1 MeV electron anti-neutrinohas a mean-free path through water of 60 light-years!)The fluence of secondary neutrons produced by proton beams in tissue is low [7], but because these neutrons areeasily spread, and their relative biological effectiveness can be 10 to 20 times higher than the protons, their presencecannot be ignored. However, with proper strategies, the neutron-generated dose to healthy tissue is still low withproton beam therapy when compared to the dose to healthy tissue under photon beam therapy[35]. We will take elastic collision to mean that the incoming and outgoing particles are the same, and that all the initial kinetic energy isreturned to the outgoing particles. A quasi-elastic collision occurs if the incoming particle knocks out a particle in a bound state of atarget, with little energy being transferred to the other particles in the target. In an inelastic collision, some of the initial kinetic energyis converted into internal energy in one or more of the outgoing particles. For a noneleastic collision, the set of outgoing particles differsin internal structure from the incoming ones.
B. Using Oxygen-18
Among the many reactions of an energetic proton-beam with a thymine-18 doped target, the reaction O( p, n ) Fhas a relatively large cross section (hundreds of millibarns) in the 4 to 14 MeV proton kinetic energy range. Thepresence of this reaction can be monitored because of the subsequent beta decay of fluorine-18. The beta decayproduces a positron, which may scatter with local nuclei and electrons, and then annihilate with an electron, makingtwo oppositely-directed 0 .
511 MeV gamma rays. Using the techniques of a PET scan, the location near (within afraction of a millimeter with high probability) where the reacted thymine resided can be found. We will see laterthat the location for maximal production of F18 is typically a fraction of a millimeter behind a pristine Bragg peaklocation. (See end of Appendix H.)The beta decay of the fluorine, ( F) → ( O) − + e + + ν e , (2)is the dominant (97%) cause of the finite lifetime of F ( τ / = 109 .
77 min), with about 3% occurrence of electroncapture (also via a weak interaction). (The parenthesis in the displayed reaction ‘equation’ indicates the atom ratherthan just the nucleus.)Both the reaction (2) and the electron capture reaction( F) + + ( e − ) → ( O) + ν e (3)are transitions of F[ 1 + ] directly to the nuclear ground state of O[ 0 + ] (so there will be no subsequent gammaemission). The bracketed notation shows the spin-parity of the nucleus. For the beta-decay reaction, with the nuclearangular momentum L = 0 in the final state, the positron-neutrino system will be in a triplet (negative parity)spin state, so the nuclear transition must be mediated by an odd parity operator, making this reaction an allowedGamow-Teller transition.Generally, a nuclear beta decay can be represented as a reaction: N ∗ ( A, Z ) → N ( A, Z ∓
1) + e ± + ν (cid:48) e , (4)where, for positron emission, ν (cid:48) e = ν e , i.e. a left-handed neutrino with lepton number 1, while for electron emission, ν (cid:48) e = ν e , i.e. a right-handed anti-neutrino with lepton number − N ∗ ( A, Z ) is the ‘parent’ nucleus with mass number A (the number of protons plus the number of neutrons; used as an isotope label) and nuclear charge Z | e | , while N isthe daughter nucleus. If the daughter nucleus is produced in an excited state, a ’prompt’ gamma ray is also likely tobe emitted from the daughter, although other channels for the fast nuclear decay of an excited nucleus are possible,such as α -particle emission, proton or neutron emission, or even fission. (These daughter decays into nuclear particlesare referred to as ‘beta-delayed’ nuclear decays.) Daughters may also undergo another beta decay.In the case of F18 decay, the experimentally-measured maximum positron kinetic energy is K max = 0 .
634 MeV,with an average of about a third of this maximum. This kinetic energy can be calculated using energy-momentumconservation, as shown in Appendix F. Including the Coulomb repulsion that the positron feels on exiting the daughternucleus, the number of positrons N ( E ) dE produced in the weak decay in a given energy interval dE , when the positronenergies have energy centered on E = K + m e c is given by (Wu and Moszkowski [41]): N ( E ) dE = gF ( Z, E ) pE ( E max − E ) dE . (5)Here, g is a coupling constant, and F ( Z, E ) is the Coulomb Fermi function (coming from the magnitude squared ofthe overlap of the wave functions for the two leptons), with E max = K max + m e c . Figure 2 shows the probability perunit energy, N ( E ), for the emission of a positron.Figure 3 shows simulated positron tracks from a series of beta decays at one localized point. At the end of each track,the positron combines with a local electron to make two gamma rays (oppositely directed) of energy m e c = 0 .
511 MeVeach. These gamma rays have an inverse attenuation coefficient of about 10 cm (Kinahan et al. [22]) in water and softtissue, so, if produced in a body, they will likely exit and can be simultaneously detected in a PET scanner.After the proton-beam induces O18 (in a double-covalent bond with carbon
FIG. 2. Positron number per unit with energy (Levin and Hoffman [23, p.786])FIG. 3. Positron tracks (from Levin and Hoffman [23, p.792])
With a F18 half-life of 109 .
77 min, the electrons around the new fluorine nucleus have plenty of time to settle intostable orbits. Later, the fluorine nucleus beta decays, with the positron quickly leaving the scene. Now the atomicelectrons find themselves in orbit around the new oxygen nucleus, with one more electron than needed to make theatomic states of neutral oxygen, and with transient energies lower than those for the new stable oxygen orbits, sincethe number of protons at the core of the atom is one less. As the electrons settle up, they will have to release theirexcess energy to adjacent atoms, directly or by radiation (in the eV range). From the complete ionization energiesof fluorine and oxygen (106,434 keV and 84.078 keV) and by estimating the binding energy of the last electron in FIG. 4. Positron spread around “Line of Response” (from Levin and Hoffman [23, p.793]). negatively charged oxygen (electron affinity of about 1.46 eV), there will be about 22 keV released by the new chargedoxygen in becoming neutral.
III. NUCLEAR REACTIONS ALONG THE PROTON BEAM
Protons and secondary neutrons in the PT beam can undergo nuclear interaction with the tissue nuclei via elastic,inelastic, and nonelastic collisions. For proton beams entering biological tissue, collisions with H1, O16, C12, and N14will dominate, as these are the most abundant nuclides. At the highest beam energies ( ∼
250 MeV), a beam particlehas time to interact with only one or two nuclei before reaching its range in the tissue. Because of Lorentz contraction,the beam particle acts as if the nucleus were flattened (perpendicular to the beam-particle’s momentum), and theparticles within as hardly moving and weakly held small-target dots. Bremsstrahlung photon emission is negligiblefor therapeutic proton beams. At intermediate energies (tens of MeV), the beam protons and secondary neutrons canknock out protons, neutrons, and alpha particles, and multiple scattering within the nucleus can occur. Even nuclearfission can be induced.The cross sections for proton-proton scattering and nuclear reactions produced by the proton beam are muchsmaller than the electron-proton scattering cross sections, and even more so when the proton energy is much abovethe resonance region of the nuclear cross sections. However, when the protons are slowed into the lower MeV range,i.e. in front of the Bragg peak, nuclear scattering and transmutations occur with higher probability. Near the endof the beam-protons range, the nuclear reactions largely produce ‘compound’ nuclides, wherein the proton energy isshared among a number of nucleons in the nucleus before that nucleus decays. Excitation of collective motion of thenucleons is possible, producing ‘giant resonances’ in the cross sections.The proton fluence drops with depth in the tissue because of these reactions. In addition, proton-proton scatteringcauses a spread of the beam. The radial spread has a standard deviation of about 2 % of the beam range, causing aspread in dose delivery. Beam intensities can produce over 1 Gy/min within the Bragg peak. Because protons at thefar end of their range lose energy faster than at any other location, uncertainties in the range of the proton beam cancause a large uncertainty (estimated to be about ± . p + O O → N ∗ +secondaries:beam proton secondary particles N ∗ energy p n d H He He recoils250 MeV 0 .
66 0 .
21 0 .
005 0 .
002 0 .
001 0 .
019 0 . .
57 0 . .
106 0 .
002 0 .
002 0 .
029 0 . .
36 0 .
073 0 .
051 0 .
001 0 .
003 0 .
098 0 . .
17 0 . . . . .
11 0 . Q values in Table TABLE III.1. Important nonelastic reactionsNuclear Reaction Q (MeV) Decay of N ∗ Half-life E EC (MeV) E maxe (MeV) p + O → F +0 . O + β + .
49 sec 2 . . p + O → n + F − . O + p prompt ( Q (cid:48) = +0 .
536 MeV) p + O → n + p + O − . N + β + .
04 min 2 . . p + O → p + p + N − . stablep + O → n + n + p + O − . N + β + .
61 sec 5 . . p + O → α + N − . C + β + .
965 min 2 . . p + O → p + α + C − . stablep + O → d + O − . stablep + O → F +7 . stablep + O → n + F − . O + β + .
77 min 1 . . p + O → n + p + O − . stablep + O → p + p + N − . O + β − .
173 sec 8 . p + O → n + n + p + O − . stablep + O → α + N +3 . stablep + O → d + O − . stablen + O → O +4 . stablen + O → p + N − . O + β − .
13 sec 10 . n + O → p + n + N − . stablen + O → n + n + O − . N + β + .
04 min 2 . . n + O → p + p + C − . N + β − .
45 sec 9 . n + O → O +3 . N + β − .
91 sec 4 . n + O → p + N − . O + β −
624 msec 4 . . n + O → p + n + N − . O + β − .
173 sec 8 . n + O → n + n + O − . stablen + O → p + p + C − . O + 2 β − . . III.1, i.e. the energy released from mass, are given using atomic masses, not fully ionized isotopes. (See Appendix D.The Q values come from Wang et al. [39] as reported by [32].) Any of these reactions may involve the release of agamma ray, but gammas are unlikely to ionize locally.The production of tritium and helium-3 occurs at a much lower rate, as the nucleons in these products do not formtight clusters in heavier nuclei, and so these clusters are not as likely to be knocked out as a particle. Slow neutronswith tissue hydrogen will also make deuterium, but this will occur most often outside the target tissue. As the tableshows, many of the daughter nuclei ( N ∗ ) after the ‘strong’ nuclear interaction are radioactive. Note the striking difference in the reaction Q values between the list of O16 vs O18 reactions shown in TableIII.1. Corresponding reactions of the proton with O18 show far less energy loss in the release of the reaction products.Two of the O18 reactions are strongly exoergic and these are among the largest in cross section, while for the proton-O16 reactions, only the direct capture is exoergic, but with only 600 keV released compared to almost 8 MeV forO18. Overall, far more energy will be available for kinetic energy of ionizing particles in the case of p + O
18 than for p + O
16. The root cause is the tighter binding of O16, being a doubly ‘magic’ nucleus, with the lowest lying nuclearstates for neutrons and protons being fully occupied, and with an energy gap to the next level.The reactions that produce radionuclides that are positron emitters may add 0 .
511 MeV gamma emissions during aPET scan, indistinguishable from the F18 decay gammas. (See Appendix K.) Beside the ones listed above, there areother nuclear reactions in tissue that contribute to positron production. Important ones are shown in Table III.2.
TABLE III.2. Other contributing positron-producing reactionsNuclear Reaction Decay of N ∗ N ∗ Half-life p + O → d + O N + β + .
04 min p + O → p + n + O N + β + .
04 min p + O → d + α + C B + β + . p + O → p + n + α + C B + β + . p + C → n + N α + β + . p + C → d + C B + β + . p + C → p + n + C B + β + . n + C → n + n + C B + β + . p + N → d + N C + β + .
965 min p + N → p + n + N C + β + .
965 min p + N → α + C B + β + .
39 min p + Na → d + Na Ne + β + .
602 yrs p + Na → p + n + Na Ne + β + .
602 yrs n + Na → n + n + Na Ne + β + .
602 yrs p + F → d + F O + β + . p + F → p + n + F O + β + . n + F → n + n + F O + β + . When a tumor is doped with O18, the reaction O18( p, n )F18 will occur in PT therapy. The produced F18 half-life is 109 . β + -induced gamma emissions from theradionuclides in the above table will have died down except for the F18 produced from fluorine in the tissue, and fromNa22. The concentration of natural sodium Na23 in tissue is low (0 .
037 % by atoms or 3 . × ions/cm ), and Na22has a long half-life (2 . . × atoms/cm . As shown in Table A.5, the number density of F18 after a full PTis about 8 to 24 × per cubic centimeter. Moreover, the cross section for F19( p, d )F18 is about 100 times lowerthan O18( p, n )F18 (see Fig. 8). The reaction F19( n, nn )F18 contributes even less, since the neutron flux is lower thanthe proton beam flux. Thus, the PET signal from fully exposed thymine-18 should be far larger than that from theF18 produced by proton-converted naturally-occurring F19. IV. CROSS SECTION FOR O18 ( n, p ) F18
The cross section for O18( p, n )F18 is shown in the two graphs below. For comparison, Fig. 7 shows the total
FIG. 5. Cross section for O18( p, n )F18 2 to 20 MeV (From Cyclotron Produced Radionuclides: Principles and Practice, (2008)) cross section for protons impinging on oxygen-16. Among all the reactions shown in Table III.1, the cross section forO18( p, n )F19 is large due to a number of resonances in the proton energy region of 4 to 12 MeV.
FIG. 6. Cross section for O18( p, n )F18 0 to 30 MeV, from Hess et al. [20, p361]FIG. 7. Total cross section for O( p, X ) from Ulmer and Matsinos [38] FIG. 8. Cross section for F19( p, np ) F
18 (plotted by EXFOR [44] from Marquez [25], Yule and Turkevich [43]) V. ESTIMATING THE INCREASE IN DOSE WITH O18 SUBSTITUTED FOR 16O IN DNA THYMINE
To estimate the increase in dose due to inelastic nuclear reactions with O18 instead of O16, one can first calculatethe LET due to the reaction O18( p, X ) and compare to O16( p, X ). This is an integration over energy for each p + N reaction cross section as a function of energy times an average energy given to charged-particles in that reaction timesthe number density of the reactant times the flux rate of the beam protons at the reaction location.There are several important factors that must be considered: • The energy and flux of protons in the beam at the location of the reaction; • The cross section for the reactions O18( p, X ); • The density of O18 in the exposed tissue; • The energies of the released charged particles.The consequent LET per unit area per unit time per length along the beam has the general form: dKdAdt dz = − J (cid:88) i ρ i (cid:88) j (cid:90) Ω j dσ ij d Ω j (cid:15) ij ( K, Ω j ) d Ω j , (6)where σ ij ( K ) is the cross section for the reaction with reactant index i and exiting particle j at beam energy K ; (cid:15) ij ( K, Ω j ) is the kinetic energy of a released charged particle labeled by j and exiting into solid angle Ω j . The quantity ρ i is the number density of reactants in the tissue and J is the number current density of protons in the beam, bothat the location where the beam has energy K . The expression Eq. (6) can be handled in a computer calculation,drawing on a database of reactions and cross sections, such as GEANT4 [1]. VI. SUMMARY
The substitution of O18 for O16 in tissue thymine causes a greater PT dose to be delivered to cancer cells, aconclusion one can draw from the fact that the important inelastic nuclear reactions with O16 are far more endoergicthan the corresponding reactions with O18. The delayed beta decay of F18 adds more to the dose delivered, andmakes it possible to know, within about a millimeter, where the delivery occurred, using PET. Given these facts, afull calculation of the numerical
RBE with O18 is justified.
Appendix A: Some Relevant Data
Some selected physical masses are given in Table A.1 in daltons and energy units of MeV (with time units to make c = 1). The column labeled ”Complete Ionization” comes from the CRC “Ionization energies of atoms and atomicions” data [24], converted to eV. The values for the bare isotopes of O16, O18 and F18 are calculated from thecomplete ionization energy and the atomic mass-energy using m bare = m atom − Z m e + I , (A1)where Z is the number of protons in the nucleus and I is the complete ionization energy of the atom (energy to removeall the atom’s electrons).The daughter nucleus of oxygen-18 is created with an extra atomic electron close-by. As such, that electron willhave an attractive Coulomb force acting on it. The extra electron will likely take energy from other atomic electrons asthey all shuffle to new stable oxygen-18 orbitals, higher in energy than the fluorine-18 orbitals they initially occupied.In addition, the following facts are relevant:The isotope O18 occurs naturally with a number ratio O18/O16 = 2 . × − ([17]) that varies from (1 .
88 to2 . × − . The greater amount is in seawater due to slower evaporation of water containing O18 compared to O16.Thus, the percent of O18 in a patient depends on the diet of that patient. A vegetable diet would favor an atmosphericabundance in tissue about 1 . × − for O18/O16, while a dominance of sea food will make 2 . × − for O18/O16.The density of O16 in soft tissue is dominated by that in the tissue water. Specific element mass densities in tissueare given in Table A.4.2 TABLE A.1. Selected Physical Masses (Coursey et al. [11])Mass Mc Complete(Daltons) (MeV) Ionization (eV) m p m e The F18 nucleus is a spin-parity J π = 1 + system; the O18 nucleus is a spin-parity J π = 0 + system. These meanthat the weak β + decay proceeds via an allowed Gamov-Teller transition, with the transition probability proportionalto the magnitude square of the nuclear transition matrix element, a sum over all the nucleons, giving (cid:12)(cid:12)(cid:80) κ (cid:104) Ψ O | (cid:126)σ N ( κ ) τ − N ( κ ) | Ψ F (cid:105) · χ † ν (cid:126)σ L ( κ ) χ e (cid:12)(cid:12) , with the nucleon spin operator (cid:126)σ N , lepton spin operator (cid:126)σ L and the nucleon isospin lowering operator τ − N . The symbol κ labels individual nucleons in the nucleus. The χ are lepton spin states. After summing over the unobserved spinstates of the leptons, the transition probability is proportional to (cid:12)(cid:12) (cid:104) Ψ O | (cid:80) κ (cid:126)σ N ( κ ) τ − N ( κ ) | Ψ F (cid:105) (cid:12)(cid:12) , where a vector dot-product is implied between the two matrix-element factors. We see that this beta decay will changea proton into a neutron, and, given the spin-parity of O18 and F18, will flip its spin.The lifetime of F18 is 109 . TABLE A.2. Some properties of beta-emitters used in PET (from Conti and Eriksson [10])Isotope Half-life Branching ( β + ) Kmax (MeV) Kmean (MeV) Rmax (mm) R mean (mm) C 20.4 min 99.8 % 0.960 0.386 4.2 1.2 N 10.0 min 99.8 % 1.199 0.492 5.5 1.8 O 2 min 99.9 % 1.732 0.735 8.4 3.0 F 110 min 96.9 % 0.634 0.250 2.4 0.6 Cu 12.7 h 17.5 % 0.653 0.278 2.5 0.7 Zr 78.4 h 22.7 % 0.902 0.396 3.8 1.3
Table A.3 shows the most important positron emitters created by a proton beam in tissue.3
TABLE A.3. Positron emitters created by proton beams in tissue (Studenski and Xiao [37])Reaction Threshold energy Half-life Positron energyMeV min MeV O (p,pn) O 16.79 2.037 1.72 O (p, α ) N 5.66 9.965 1.19 N(p,pn) N 11.44 9.965 1.19 C(p,pn) C 20.61 20.390 0.96 N(p, α ) C 3.22 20.390 0.96 O(p, α pn) C 59.64 20.390 0.96 TABLE A.4. Percent by mass of principal elements in tissue and ionization energy (H¨unemohr et al. [21, p.5])Material ρ [g/cm ] H · C · N · O · Ca P · I [eV]Lung deflated 0.26 10.4 10.6 3.1 75.7 0 0.2 74.54Yellow marrow 0.98 11.5 64.6 0.7 23.2 0 0 63.72Mammary gland1 0.99 10.9 50.8 2.3 35.9 0 0.1 66.73Mammary gland2 1.02 10.6 33.3 3 52.9 0 0.1 70.05Mammary gland3 1.06 10.2 15.9 3.7 70.1 0 0.1 73.78Red marrow 1.03 10.6 41.7 3.4 44.2 0 0.1 68.7Brain Cerebrospinal fluid 1.01 11.2 0 0 88.8 0 0 75.3Adrenal gland 1.03 10.7 28.5 2.6 58.1 0 0.1 70.92Smallintestine wall 1.03 10.7 11.6 2.2 75.5 0 0.1 73.98Urine 1.02 11.1 0.5 1 87.2 0 0.1 75.21Gall bladder bile 1.03 10.9 6.1 0.1 82.9 0 0 74.81Lymph 1.03 10.9 4.1 1.1 83.9 0 0 75Pancreas 1.04 10.7 17 2.2 69.9 0 0.2 73Brain white matter 1.04 10.7 19.6 2.5 66.8 0 0.4 72.5Prostate 1.04 10.6 9 2.5 77.9 0 0.1 74.58Testis 1.04 10.7 10 2 77.2 0 0.1 74.23Brain gray matter 1.04 10.8 9.6 1.8 77.5 0 0.3 74.17Muscle skeletal1 1.05 10.2 17.3 3.6 68.7 0 0.2 73.69Muscle skeletal2 1.05 10.3 14.4 3.4 71.6 0 0.2 74.03Muscle skeletal3 1.05 10.3 11.3 3 75.2 0 0.2 74.66Heart1 1.05 10.4 17.6 3.1 68.6 0 0.2 73.32Heart2 1.05 10.5 14 2.9 72.4 0 0.2 73.8Heart3 1.05 10.5 10.4 2.7 76.2 0 0.2 74.49Heart blood filled 1.06 10.4 12.2 3.2 74.1 0 0.1 74.22Blood whole 1.06 10.3 11.1 3.3 75.2 0 0.1 74.61Kidney1 1.05 10.3 16.1 3.4 69.9 0.1 0.2 73.79Kidney2 1.05 10.4 13.3 3 73 0.1 0.2 74.16Kidney3 1.05 10.5 10.7 2.7 75.8 0.1 0.2 74.48Stomach 1.05 10.5 14 2.9 72.5 0 0.1 73.81Thyroid 1.05 10.5 12 2.4 75 0 0.1 74.24Liver1 1.05 10.4 15.8 2.7 70.8 0 0.3 73.72Liver2 1.06 10.3 14 3 72.3 0 0.3 74.18Liver3 1.07 10.2 12.7 3.3 73.4 0 0.3 74.58Aorta 1.05 10 14.8 4.2 70.2 0.4 0.4 74.78Ovary 1.05 10.6 9.4 2.4 77.4 0 0.2 74.52Eye lens 1.07 9.6 19.6 5.7 64.9 0 0.1 73.97Spleen 1.06 10.4 11.4 3.2 74.7 0 0.3 74.47Trachea 1.06 10.2 14 3.3 72 0 0.4 74.38Skin1 1.09 10.1 25.2 4.6 59.9 0 0.1 72.25Skin2 1.09 10.1 20.6 4.2 65 0 0.1 73.17Skin3 1.09 10.2 15.9 3.7 70.1 0 0.1 73.89Connective tissue 1.12 9.5 21 6.3 63.1 0 0 73.79Cartilage 1.1 9.8 10.1 2.2 75.7 0 2.2 76.96Sternum 1.25 7.8 31.8 3.7 44.1 8.6 4 81.97Sacrummale 1.29 7.4 30.4 3.7 44.1 9.9 4.5 84.19Femur conical trochanter 1.36 6.9 36.7 2.7 34.8 12.9 5.9 86.69Sacrum female 1.39 6.6 27.3 3.8 43.8 12.6 5.8 89.11Humerus whole specimen 1.39 6.7 35.3 2.8 35.3 13.6 6.2 88.06Ribs 2nd to 6th 1.41 6.4 26.5 3.9 43.9 13.2 6 90.28Vert colC4 excl cartilage 1.42 6.3 26.3 3.9 43.9 13.4 6.1 90.78Femur total bone 1.42 6.3 33.4 2.9 36.3 14.4 6.6 90.24Femur whole specism 1.43 6.3 33.2 2.9 36.4 14.5 6.6 90.34Innominate female 1.46 6 25.2 3.9 43.8 14.4 6.6 92.76Humerus total bone 1.46 6 31.5 3.1 37 15.3 7 92.23Clavicle scapula 1.46 6 31.4 3.1 37.1 15.3 7 92.26Humerus cylindrical shaft 1.49 5.8 30.3 3.2 37.6 15.9 7.2 93.56Ribs10th 1.52 5.6 23.7 4 43.7 15.7 7.3 95.42Cranium 1.61 5 21.3 4 43.8 17.7 8.1 99.69Mandible 1.68 4.6 20 4.1 43.8 18.8 8.7 102.35Femur cylindrical shaft 1.75 4.2 20.5 3.8 41.8 20.3 9.4 105.13Cortical bone 1.92 3.4 15.6 4.2 43.8 22.6 10.4 111.63 TABLE A.5. Thymine-18 in doped tissue ([30],[29])Mass of DNA male 6.41 pgram, female 6.51 pgramTodal DNA length male: 205.0 cm, female 208.23 cmTotal base pairs male: 6.27 × , female 6.37 × A-T base pairs in DNA 59 . . × . × = 3 . × Cell diameter/nucleus diameter ∼ . ρ cell = 1 to 3 billion cells per cm Density of thymine in tissue ρ Thy = 4 to 12 × per cm Density of O18 in dopped tissue ρ O = 4 to 12 × per cm Cross section each O18 σ O < − cm Extinction length λ ≡ / ( ρ O σ O ) >
84 cm Appendix B: Maximum energy loss of a proton scattering from an electron and the largest scattering angle
The proton beam loses energy traversing tissue largely by Coulomb interactions with material electrons. Considerthe Coulomb scattering of a relativistic proton with electrons in tissue. As electrons bound to atoms and moleculeshave energies much lower than the initial protons in a proton beam, they will recoil much like free electrons until thebeam proton energies are much below 1 MeV. In this quasi-free region (ahead of the Bragg peak), the upper limit tothe scattered electron’s energy is determined by energy-momentum conservation. Denote the energy and momentumof the initial and final particles as ( E , (cid:126)p ) , ( m, , ( E , (cid:126)p ) , ( √ k + m , (cid:126)k ) and the mass of the proton as M .
1. Maximum delta-electron energy
We will find the maximum electron energy after a collision with a beam proton. Label the incoming particles withthe indices 1 ,
2, and outgoing particles 3 ,
4. We will distinguish the electron magnitude of its 3-mommentum with thesymbol k .Conservation of energy and momentum reads E + E = E + E (B1) −→ p + −→ p = −→ p + −→ p (B2)The electron receives its greatest energy for head-on collisions. In the ‘Lab’ frame, −→ p = −→ (cid:126)k = (cid:126)p for the electron. Then p = p + k (B3)and (cid:113) p + M + m = (cid:113) ( p − k ) + M + (cid:112) k + m (B4)Solving for k , and dropping the k = 0 solution, we have k maxe = 2 m (cid:112) p + M + mM + m + 2 m (cid:112) p + M p . (B5)There follows the maximum electron total energy E maxe = (cid:118)(cid:117)(cid:117)(cid:116)(cid:32) m (cid:112) p + M + mM + m + 2 m (cid:112) p + M p (cid:33) + m (B6)= m (cid:118)(cid:117)(cid:117)(cid:117)(cid:116) (cid:0) mM + K M (cid:1) (cid:16)(cid:0) mM (cid:1) + 2 mM K M (cid:17) (cid:18) K M (cid:19) K M . (B7)The maximum energy loss of a proton by electron scattering will by the maximum electron kinetic energy given tothe electron. For 85 MeV incoming protons, Eq. (B7) gives K maxe = E maxe − m = 193 . . K maxe , and so a shorter range, making theirexcursion from the proton beam track measurable in the tenths of a millimeter.
2. Maximum beam proton deflection in scattering by electrons
Now let’s find the maximum proton deflection angle, θ max by one collision with an electron. We have, from energy-momentum conservation E + m = E + E e (B9) −→ p = −→ p + −→ k (B10)7 FIG. 9. Electron range in water. (Plante and Cucinotta [31]) so p + p − p p cos θ = k (B11)cos θ = p + p − k p p (B12) E = ( E + m − E e ) (B13) p = (cid:113) E − M (B14)= (cid:113) ( E + m − E e ) − M (B15)Thus cos θ = E − M + ( E + m − E e ) − M − (cid:0) E e − m (cid:1) (cid:112) E − M (cid:113) ( E + m − E e ) − M . (B16)This cosine function is 1 ( θ = 0) when there is no scattering ( E e = m ) or at K θ =0 e = mp M + m ( E + (1 / m ) (B17)with scattering. It drops below one as K e increases.As an aside, if m = M (as for proton-proton scattering), or m > M the cosine function drops to zero at K θ =90 o e = p ( E + m ) . (B18)If m < M , cos θ rises back to one at a finite electron kinetic energy given by K θ =90 o e = 2 2 M + K M + 2 mM + m + 2 mK mK , (B19)= 4 m K M (cid:0) mM (cid:1) + 2 mM K M K M . (B20)8We can find the energy where the greatest deflection occurs by finding where the slope of cos θ vanishes withchanging E e . Setting ddE e E − M + ( E + m − E e ) − M − (cid:0) E e − m (cid:1) (cid:112) E − M (cid:113) ( E + m − E e ) − M = 0 , (B21)or E m + E m − M E e − E mE e = 0 , (B22)which gives E θ − max e = m m + E E m + M E , (B23)so the kinetic energy of the electrons when θ is maximum as K θ − max e = m m + M + K ( M + K ) m + M ( M + K ) − m (B24)= m M + K M + M m + mK K . (B25)= 2 m K M mM (1 + K M K ) (B26)For an 85 MeV beam, this is K θ − max e = 2 × (cid:18) . (cid:19) × . × (1 + (cid:0) . . (cid:1) × (cid:0) (cid:1) ×
85 (B27)= 0 . . E θ − maxe from Eq. (B23) in the formula Eq. (B16) for cos θ gives a remarkably simple result , and onethat does not depend on the proton beam energy:cos θ max = (cid:114) − (cid:16) mM (cid:17) , (B29)sin θ max = (cid:16) mM (cid:17) , (B30) θ max = arcsin (cid:16) mM (cid:17) = 180 π × arcsin (cid:18) . (cid:19) = 3 . × − deg . (B31)Each time a beam proton interacts with a medium electron, the proton cannot be deflected more than 0 . FIG. 10. “A curious fact” (Gottschalk [15])
Appendix C: Energy and momentum of emitted particles after a particle beta-decay
To simplify indices, we will use here (0,1,2,3) for the parent, daughter, beta, and neutrino. A notation on top ofan energy or momentum symbol indicates a frame of reference. Zero will be used for the frame in which the parentnucleus is at rest. (This is ‘the center of momentum (cm) frame’.)Energy-momentum conservation gives (for the 4-vectors) p = p + p + p . (C1)In the rest frame of the parent particle, the three vector-momenta must lie in a plane. The orientation of this planedetermines two of the independent kinematic variables measurable in the reaction. Orientation of one of the outgoingmomenta along some axis in the cm-reaction plane determines another. Momentum conservation means one of thethree momenta can be written in terms of the other two. Energy conservation gives the magnitude of one of theoutgoing momenta in terms of the other two. Of the nine variables in the three momenta, the constraints mentionleaves 9-2-1-3-1=2 variables not determined by kinematics. These two might be taken as the angle between particles(1,2) and the energy of particle 2. A more universal choice, a choice that does not depend on a selection of referenceframe, is to use kinematical variables that are relativistic invariants. These are often taken as a pair of Mandelstamvariables, defined by s = ( p + p ) , (C2) s = ( p + p ) , (C3) s = ( p + p ) . (C4)These three are not independent of each other, as can be seen from s + s + s = m + m + m + m . (C5)To preserve symmetry among the three variables in the plot of the kinematically allowed values of ( s , s , s ) , Dalitz first plotted the possible values of ( s , s , s ) in three dimensions. The constraint Eq. (C5) forces the valuesto lie in a plane oriented with a normal having positive values on each axis. This plane cuts the (1,2), (2,3), and(3,1) axis-planes, producing a triangular surface. The Dalitz plot is made on that surface. The density of values as0points within the triangle are made proportional to the physical probabilities for detecting particles produced withthe kinematics of the point.Now the question arises: What are the kinematical limits within the Dalitz triangle, or, equivalently, what are thelimits on the measured values of momentum and energy of the outgoing particles. Such physical limits come from theconstraints that the kinetic energies of the particles cannot be negative ( E i ≥ m i ), and that the angles between themomenta of the outgoing particles must be physical ( − ≤ cos θ ≤ s , we note that ( p · p ) , in the frame of reference of the parent particle, becomes p · p = m E (C6)But s = ( p + p ) = ( p − p ) = m + m − p · p (C7)giving E = 12 m (cid:0) m + m − s (cid:1) (C8)From E ≥ m , we have s ≤ ( m − m ) (C9)To get a lower limit on s , pick the rest frame for the pair of particles (2,3) (the ‘Jackson’ frame), so that J −→ p = − J −→ p (C10)Then s = ( p + p ) = (cid:18) J E + J E (cid:19) ≥ ( m + m ) (C11)So, for physically realizable energies, (by cyclically permuting of indices)( m + m ) ≤ s ≤ ( m − m ) (C12)( m + m ) ≤ s ≤ ( m − m ) (C13)( m + m ) ≤ s ≤ ( m − m ) (C14)When angular constraints are imposed, the s (cid:48) s may not be allowed to reach the above limits. In the ‘Jackson’ framefor particles 2 and 3’, in which the momentum of particles 2 and 3 satisfy J −→ p = J −−→ p . Furthermore: s = ( p − p ) (C15)= ( E − E ) − ( −→ p − −→ p ) (C16) s = ( E − E ) − ( −→ p + −→ p ) (C17)= (cid:18) J E − J E (cid:19) (C18) s = (cid:115)(cid:18) J −→ p (cid:19) + m − (cid:115)(cid:18) J −→ p (cid:19) + m . (C19)As a result, (cid:18) J −→ p (cid:19) = 14 (cid:16)(cid:0) m + m (cid:1) − s (cid:17) (cid:16)(cid:0) m − m (cid:1) − s (cid:17) s . (C20)1Now from s = ( p + p ) = (cid:18) J E + J E (cid:19) (C21)= (cid:115)(cid:18) J −→ p (cid:19) + m + (cid:115)(cid:18) J −→ p (cid:19) + m , (C22)we can solve for the momentum squared of particle 2 in the Jackson frame gives (cid:18) J −→ p (cid:19) = m + m + s − s m − s m − m m s , (C23)which is also the momentum squared for particle 3 in the Jackson frame. We now consider s = ( p + p ) (C24)= m + m + 2 ( p · p ) (C25)= m + m + 2 (cid:18) J E J E − (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) cos θ (cid:19) . (C26)Since, as we have shown, (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) and (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) can be expressed in terms of s (without s ), for fixed s , we have m + m + 2 (cid:18) J E J E − (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) ≤ s ≤ m + m + 2 (cid:18) J E J E + (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) J −→ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) (C27)with the lower limit in the case in which J −→ p is in the same direction as J −→ p and the upper limit in the case in which J −→ p is in the opposite direction as J −→ p . These limits determine the boundary in the Dalitz plot in the ( s , s ) plane.To get the maximum energy in the frame in which the parent particle is at rest (center-of-momentum frame), wenote that s = ( p − p ) = m + m − m E (C28)implies that E is determined by just one of the two independent Mandelstam variables, and that E reaches amaximum when s is a minimum. Solving for E , we have E = m + m − s m . (C29)Using the min of s above, E max1 = m + m − ( m + m ) m . (C30)It follows that p = (cid:118)(cid:117)(cid:117)(cid:116)(cid:32) m + m − ( m + m ) m (cid:33) − m (C31)= 12 m (cid:114)(cid:16) m − ( m − m − m ) (cid:17) (cid:16) m − ( m + m + m ) (cid:17) . (C32)By cyclic permutation, E max2 = m + m − ( m + m ) m , (C33)and E max3 = m + m − ( m + m ) m . (C34)2 Appendix D: Reaction ‘Q’ values and threshold energy in laboratory frame
Consider the reaction [1] + [2] → [3] + [4] + [5] + · · · , (D1)where [ i ] is particle labeled by index i .The “Q” value for this reaction is defined to be the energy released because of mass differences before and after thereaction: Q = m + m − (cid:88) f m f , (D2)the sum taken over the produced (final) particles.To find the threshold energy of particle [1] in the “lab” frame (where (cid:126)p = 0), consider the Mandelstam variable s = ( p + p ) = E cm , the total energy squared in the center-of-momentum frame of reference. From energy-momentumconservation, s = ( (cid:80) f p f ) (sum is over final particles), so we see that √ s ≥ (cid:80) f m f . Expressed in the laboratoryframe variables, s = m + m + 2 m E lab . Thus E lab ≥ ( (cid:80) f m f ) − m − m m , (D3)making the threshold kinetic energy of the incoming particle satisfy K th ≡ E th − m ≥ ( (cid:80) f m f ) − ( m + m ) m . (D4)In terms of Q , K th ≥ − (cid:80) i m i m Q , (D5)where now the sum is over all particles, in and out. If the reaction is exoergic,
Q > K th ≥
0. If endoergic,
Q <
0, the threshold kinetic energy for the incoming particle will be greater than zero, as given above.
Appendix E: Maximum energy of the recoiling O18 after beta decay of F18
The maximum oxygen recoil kinetic energy, from Eq. (C30), E O = m F − bare + m O − bare − m e m F − bare (E1) K O = m F − bare + m O − bare − m e m F − bare − m O − bare (E2)= ( m F − bare − m O − bare ) − m e m F − bare (E3)= 3 .
132 7 × − MeV = 31 . . (E4)The energy needed to cause a double-strand break in DNA is about 25 eV, so the recoil kinetic energy of the O18is sufficient to break both strands of its DNA, but this breakage depends on the O18 recoil direction. The electron,on the other hand, carries, on average, about 250 keV, which can leave a trail of ions by scattering from molecularelectrons along its path. For F18, the path is relatively short (average 0 .
24 mm in soft tissue (see Table A.2)).
Appendix F: Fate of the positron after the beta decay of F18
As shown above, in the particle-decay reaction P → P + P + P , the maximum recoil energy of the particle P after the parent decays can be found from energy and momentum conservation, giving E = m + m − ( m + m ) m . (F1)3For F18 decay, the positron maximum total energy and kinetic energy are therefore E e = m F − bare + m e − m O − bare m F − bare (F2)= 1 . , (F3) K e = E e − m e = 1 . − .
511 = 0 . . (F4) K e agrees with the experimental number 0 .
634 MeV, given in Table A.2.The positron energy is typically more than 200 times greater than the highest ionization energy of atoms andmolecules in tissue (see Table A.4). As the fast-moving positron (starting at 89% of the speed of light) travels throughtissue, it inelastically scatters from bound electrons and nuclei and slows down. Since the positron-electron crosssection depends inversely on the center-of-mass energy squared, the positron loses most of its energy in tissue at theend of its track, just as ions beams do. We should expect the emitted positron to scatter among the nearby molecularcharges, leading to molecular excitations, ionizations, and photons, before slowing to thermal energies. It can then bedirectly annihilated or captured by an electron. In water at 20 o C, the positron has about a 64% chance of undergoingdirect annihilation ([9]). If captured, positronium forms, either in a spin single state (ortho) or triplet state (para).With no preferred direction of the positron spin, 25% will be ortho-positronium, although interaction with nearbymolecules can flip one of the spins. If the positronium is created in an excited state with angular momentum, it willrelease UV and visible-light photons with discrete energies half those of atomic hydrogen, until the e + e − pair reachan l = 0 state (no angular momentum). At that point they annihilate each other. If the positronium is in a singletspin state, the annihilation has a lifetime of 0 .
12 ns and almost always two 0 .
511 MeV photons are emitted in oppositedirections. The spin triple state annihilates with a lifetime of 142 ns, with almost always the emission of three gammarays.
Appendix G: Threshold behavior of nuclear reaction cross sections
In 1948, Eugene Wigner showed ([40]) that under quite general circumstances, the low energy threshold behaviorof nuclear-reaction cross-sections have a definite dependence on the relative momentum (cid:126) k between the two incomingparticles, given by σ ∝ k exp ( − πµcZ Z α/ ( (cid:126) k )) for (+)(+) or (-)(-) charges (G1) σ ∝ k for (+)(-) charges , (G2)where µ is the reduced mass of the two particles, having charge numbers Z and Z , respectively, c is the speed oflight, and α is the fine-structure constant. A nuclear particle with a positive charge sent toward another one willbe inhibited from merging because of the Coulomb repulsive force acting. The effect is expressed by the exponentialfactor in Eq. (G1), used by George Gamow in 1928 to explain alpha-particle decay of nuclei, and is now called theGamow factor.Gamow realized [13] that if two particles are in a bound state held together by a strong short-range nuclear forceand repelled by a Coulomb force, they may still escape from each other by ‘tunneling’ through the potential barriercreated by the two forces. With this quantum idea, he was able to explain the wide range of lifetimes of alpha-particledecay of radioactive nuclei.Inversely, the probability of two nuclear particles, with masses m , m and charges Z | e | , Z | e | , approaching eachother and then overcome the Coulomb repulsion, is: P g ( K ) = exp (cid:32) − (cid:114) K g K (cid:33) (G3)where K is their total kinetic energy, K g = 2 µc ( παZ Z ) , with c the speed of light, µ ≡ m m / ( m + m ) (i.e.the reduced mass) and α ≡ e / ( (cid:126) c ) (the fine-structure constant). Note that as the kinetic energy K goes to zero, theprobability of penetration goes to zero. Eq. (G3) is a non-relativistic expression. Yoon and Wong [42] have given therelativistic form for the Gamow factor.By factoring out the strongly energy dependent factors P g and 1 /K from the cross section, the remaining ‘astro-physical factor S ( K )’ will have a more gentle low-energy behavior: σ = 1 K P g ( K ) S ( K ) . (G4)4The S ( K ) is proportional to the nuclear transition probability, so it contains all the nuclear physics of the process.Apart from nuclear resonances, one finds that S ( K ) is quite flat for K in the eV to keV range. Appendix H: Proton energy at given penetration depth
As protons from a PT beam enters tissue, a variety of mechanisms of interaction cause the protons to lose kineticenergy K , and to reduce their number (dropping beam fluence). The ionization of molecular electrons dominates,particularly above a few MeV. The inelastic collision with nuclei, although important because of the creation of nuclearfragments and because such collisions spread the beam, contributes only a small fraction to the proton energy-lossbecause of the small cross sections of nuclei. The energy loss of the ions in transversing through uniform tissue of agiven short length (Linear Energy Transfer, LET) follows closely to the Bethe-Bloch relation, with corrections, shownin Eq. (H1): (See Bethe [3],Fano [12], Ziegler [45].) dKdz = − πn e α z ( (cid:126) c ) m e c β (cid:18) ln (cid:18) m e c I (cid:19) + ln (cid:18) β − β (cid:19) − β − δ − Cz (cid:19) . (H1)Here, n e is the material electron density, z the number of charges on the beam ion, m e the electron mass, I theaverage ionization energy in the material, and β is the velocity of the beam ions divided by the speed of light. Theterm δ is a material electron density correction due to polarization of surround material as a relativistic ion passes,whose electric field spreads perpendicular to the ions velocity and shrinks parallel to that velocity for larger speeds.The term C/z is a “shell” correction needed when the beam particle slows to speed comparable to the speed of thebound electrons in the tissue molecules. For protons in energy range 1 MeV to 100 MeV, the shell correction can be asmuch as 6% [45]. In the derivation of Eq. (H1), the discrete nature of protons interacting with a medium is smoothed(“Continuous Slow-Down Approximation”, or CSDA for short), since, for most of its interaction with electrons, smalleffects rapidly occurring take place. These days, the discrete events can be handled by Monte Carlo methods (e.g.,using the GEANT4 code [8], [1], [2]). However, analytic tools involving continuous changes often can quite accuratelyaccount for discrete processes and many times the analytic arguments are simpler to understand and handle.To integrate the LET relation, we can use the relativistic connection between the kinetic energy and velocity of theprotons. We have, in units with c = 1 and with M the mass of the proton, K = M (cid:112) − β − M (H2)and, inversely β = (cid:112) − (1 + K/M ) − = (cid:112) K ( K + 2 M ) K + M , (H3)from which we can connect dK to d ( β ), facilitating the integration needed in Eq. (H1) to find z = z ( K ), which is animplicit solution for K = K ( z ). The connection is dK = M − β ) − / d ( β ) . (H4)When the proton speed becomes very small ( β (cid:28) lim v → (ln β ) /β = 0, so that we can see that thelarge value of the LET (causing the “Bragg peak”) comes from the first term with the inverse of β factor. This termgenerates an infinite peak for zero proton speed. However, we should not expect the Bethe formula to work when theprotons have slowed to below the average ionization energy I , which is typically found under 100 eV (see Table A.4).At proton kinetic energy of K = 100 eV, β ≈ . β is far above the thermal energy at bodytemperature T = 310 Kelvin, which is β ≈ (cid:112) k B T / (2 mc ) = 1 . × − . If we take I = 100 eV, then the protonspeed that makes Bethe’s dK/dz become negative is β ≈ . , which is below the range of validity of the formula.To simplify integrating, one can use β <
1, so an expansion of the second term in powers of β is justified: dKdz = − πn e α z ( (cid:126) c ) m e c β (cid:18) ln (cid:18) m e c I (cid:19) − ln (1 /β ) + 12 β + 13 β + 14 β + · · · (cid:19) . (H5)For an initial proton energy of K = 100 MeV, β = 0 . β = 0 . β ) = − . β ) have an even smaller contribution, and can be dropped.5 FIG. 11. Bragg peak (Gr¨un [16, p.9])
The proton beam Bragg peak is shown in Figure 11. By including a set of different initial proton energies in theincoming beam, the different ranges of the Bragg peaks can be made to produce a spread out Bragg peak (SOBP),designed to cover the width of a tumor. To reduce exposure of healthy tissue in front of the tumor for a given doseto the tumor and to work around structures needing avoidance, the beam angle can be rotated to a set of angles.The dose delivered along the proton beam path is well represented by Bortfield’s analytic expression derived fromthe Bethe-Block relation: (Bortfeld [5, eq.26], Newhauser and Zhang [27, eq.38]) D ( z ) = Φ e − ( z − R ) / (4 σ B ) Γ( q B + 1) √ πρ (1 + β B R ) (cid:18) σ B α B (cid:19) q (cid:20) σ B D − q B (( z − R ) /σ B ) + (cid:16) ( q B + γ B ) β B + ε B R (cid:17) D − q B − (( z − R ) /σ B ) (cid:21) , (H6)where z is the depth, Φ is the primary fluence, R is the range of the proton beam, σ B is the standard deviation ofthe Gaussian spread of the proton depth, ε is the fraction of low-energy proton fluence to the total proton fluence,and D y ( x ) is the parabolic cylinder function. The values of the material-dependent constants, q B and α B , are foundby fitting them using the classical LET Bragg-Kleeman (1905) formula: dKdz = − q B K − /q B α B . (H7)to experiment. Of course, the Bragg-Kleeman formula is far simpler than the Bethe-Bloch result, but it still workssurprisingly well. (The fitted constant q B is 0 . q B = 1 /
2, then the parabolic cylinder functions in Eq. (H6)become a Gaussian times a Hermite polynomial.) The parameters fitted by Bortfeld for a water target are given inTable H.1. The standard deviation of the range spectrum for an almost mono-energetic beam is denoted σ mono andthe standard deviation of the almost Gaussian part of the energy-spectrum at its peak E is given by σ E, . Thesemake up the full standard deviation of the range spectrum through σ B = σ mono + σ K (cid:18) dRdK (cid:19) = σ mono + σ E, (cid:18) α B q B K /q B − (cid:19) . (H8)An even better fit to the classical LET (Eq. (H7)) is a generalization to relativistic energies (see Ulmer and Matsinos[38]), taking a classical slow-down due to a frictional drag (‘damping’) proportional to the protons momentum p z alongthe beam axis to an inverse power: dp z dτ = − η/p qz , (H9)where τ is the relativistic proper-time dτ = (cid:112) − β dt . Integrating Eq. (H9) once gives p z = p o (1 − τ /τ R ) / ( q +1) , (H10)6 TABLE H.1. Dose-depth parameters for water (Bortfeld [5])Value Units q B α B . /q β B .
012 cm − γ B R α B K /q cm σ mono β (cid:48) R / − q cm β (cid:48) .
012 cm q − / σ E, ≈ . K MeV ε B ≈ . − . where p o is the initial proton beam momentum and τ R = p q +10 / (( q + 1) η ) is the proper time to stop the protons inthe tissue. We can integrate Eq. (H10), to get the proton distance in the tissue in terms of the proper time to reachthat distance, i.e. z = R (cid:34) − (cid:18) − ττ R (cid:19) q +2 q +1 (cid:35) , (H11)where R = (( q + 1) / ( q + 2)) τ R p /M is the range of the proton. Now with Eq. (H10) and (H11) we can express themomentum of the beam proton in terms of its distance of travel through the tissue: p z = p o (1 − z/R ) / ( q +2) . (H12)As a result, the kinetic energy of a beam proton follows K = c (cid:113) p (1 − z/R ) q +1 + M c − M c . (H13)Figure 12 shows how this energy behaves with the choice K = 85 MeV ( p = 408 MeV/c), R = 6 cm, and q = 1 . q coming from a fit of p = 1 + q/ FIG. 12. Proton kinetic energy vs z dKdz = − cp (2 + q ) R (1 − z/R ) − q/ ( q +2) (cid:113) p (1 − z/R ) / ( q +2) + M c . (H14)Figure 13 shows the corresponding slope − dK/dz , and exhibits the Bragg peak. FIG. 13. LET: Proton energy loss per unit distance (LET)
The O18 to F18 reaction occurs predominantly when K ≈ K = 85 MeV, the location for the O18( p, n )F18 nuclear reaction isgreatest at about 0 . /
100 of R . Appendix I: Chance of beam protons hitting tissue nuclei
The interaction of a proton beam passing into a material due to scattering from material electrons and nucleiis clearly an important topic for therapeutic proton beams. Besides the energy loss with penetration distance (asdescribed in Sec. H), the beam is spread by multiple scattering. A successful and widely used model describing thespreading and multiple scattering of the beam began with the work of Moliere [26]. The model gives the particle beamdistribution as a function of the angle away from the initial beam direction. Hans Bethe [4] simplified the derivation.Further refinements have been applied since 1953, including models for the relevant cross sections. These models canbe compared to brute force Monte Carlo calculations, which can incorporate all the physics describing a beam particlemultiply scattering through even an inhomogeneous material. For a review of such models, see Gottschalk [15]. Sincethe Monte Carlo calculations take much longer run times than current good models, proton beam planning typicallyuses such models.
Chance for a single proton to make a given number of hits to distributed nuclei
We want to estimate the probability that a proton in passing through a given thickness of tissue will undergo anumber of nuclear-scattering events. Let ρ N be the number density of nuclei in the tissue and ∆ z the thicknessthrough which the protons pass. Then the areal density of scatterers will be ρ N ∆ z. Let A B be the area that thebeam covers and σ N be the nuclear scattering cross-section. From a beam-proton’s perspective, if the areas aroundeach nucleus the size of their cross sections do not overlap, then the probability of a proton hitting within one8of the scattering-center areas is a ratio of all the center areas to the total area covered by the proton beam, i.e. P = ρ N A B ∆ z σ N /A B = ρ N σ N ∆ z . (In the frame of relativistic beam protons, the thickness ∆ z measures contracted,by a factor (cid:112) − ( v/c ) = m p c /E , so the scattering-center areas act as if they were all in a thin plane.)With a thick target, there will be a good chance that several nuclear-cross sections overlap when viewed from thebeam-proton frame of reference. To estimate the overlaps, let A represent the average area which covers one atomin the tissue (‘atomic areas’). Divide that area A into N ∼ A/σ ‘nuclear areas’. This number we expect to be quitelarge, as nuclear cross-sections are usually less than a barn ((10 − cm) ) while the atoms in tissue are most oftenseparated by 1 to 10 ˚Angstroms, making N of the order 10 or larger. The number of scattering nuclei behind oneatomic area A will be ρ N A ∆ z. With a fluence of 10 protons/cm = 10 − / ˚Angstrom , the number of protons passingthrough an atomic area at any given time is quite small. FIG. 14. Scattering tube for three atoms aligned with the proton beam. The atomic nuclei are projected onto the face of thetube. A beam proton happens to pass through one of the nuclear cross sections. The configuration of the depicted scatteringcenters we represented by a vector { , , , , · · · , } , here of dimension N = 8 = 64. We will assume no correlation in position of the scattering centers. A crystal will not have this property, but anamorphous mixture will, if thick enough. As a beam proton passes through an atomic area A , while traversing adistance ∆ z , the atomic volumes it passes through will make up a ‘scattering tube’ surrounding the path of the beamproton. The number of atomic volumes and therefore the number of nuclei in the tube will be n ∼ ∆ z/A / . We willassume that the n scattering centers in the scattering tube are spread randomly, so that the beam proton enteringany of the N nuclear areas has the same probability of being scattered as a proton entering any other such nucleararea.Given the equal a priori probabilities on entering any one of the nuclear areas, we can arrange all the nuclear areascovering the two dimensional atomic area A into a linear array of ‘cells’, symbolized by an N dimensional vector.Behind one of those nuclear areas there may be k > c . Because the position of thenuclear areas does not change the a priori probability for scattering, we can order the cell ‘occupation’ numbers k sothat, from left to right, k ≥ k ≥ k · · · ≥ k c .A single configuration for scattering centers will be called a ‘distribution’ of fixed values for the k i , with thedistribution ‘vector’ symbolized by an N dimensional vector (cid:126)k ≡ { k , k , · · · , k c , , , · · · } = { k , k , · · · , k N } . (I1)In the second expression, the k i values beyond i = c are all zero. Note that N (cid:88) i =1 k i = n . (I2)Evidently, the number of clusters is c ≡ N − number of empty cells (I3)and 1 ≤ c ≤ min ( n, N ). A distribution vector is then { k , k , · · · , k c , , , · · · , } . If a subset of non-zero k ’s have thesame value, we will call the number of such equal k ’s the ‘multiplicity’ ≡ m of that k .9After randomly distributing the n scattering centers in N cells, the number of configurations of scattering-centerclusters with a given distribution { k , k , k , · · · , k N } will be called the weight of the distribution, N c ( { k } ), since thisnumber is proportional to the probability of finding such a configuration after the scattering centers are randomlydistributed among the N cells. FIG. 15. Hand-calculation of number of configurations of five scattering centers in three cells. A single center (one ‘dot’) isdistributed first, and then probabilistic equivalence is used to represent the three possibilities as three equivalent ones like thefirst. In the figure for the distributions with four and five scattering centers, only the resulting configurations and their countare shown.
Figure 15 shows the case for N = 3 , n = 5. By filling three cells sequentially with five ‘dots’ and then counting,we find the following configurations and their probabilistic weights N c :config c m m N c { , , } { , , } { , , } { , , } { , , } N n = 3 = 243, a useful check that all configurationshave been found.To find a general expression for the weights N c ( { k } ), first note that if there are c clusters, there will be N ( N − · · · ( N − c + 1) = N ! / ( N − c )! ways to distribute distinct clusters. If there are m l cells with the same number k l > N ! / (( N − c )! (cid:81) m l !),since a cluster with label l has m l ! ways of equivalent re-arrangements. Now we must count the number of ways thescattering centers could have been placed in the given clusters. Starting with the left-most non-empty cluster whichcontains k scattering centers, there will have been (cid:0) nn − k (cid:1) ways to have selected the centers. But now we have n − k fewer centers to put in the second cluster, so there will have been (cid:0) n − k n − k − k (cid:1) ways to rearrange the centers in thesecond cluster, given the first. This continues until the last non-zero cluster, which has (cid:0) n − (cid:80) c − i =1 k i k c (cid:1) = (cid:0) k c k c (cid:1) = 1. As aresult, the number of distributions with vector { k , k , · · · k N } will be N c ( { k } ) = N !( N − c )! (cid:81) m l ! (cid:18) nk (cid:19)(cid:18) n − k k (cid:19)(cid:18) n − k − k k (cid:19) · · · (cid:18) n − (cid:80) c − k i k c − (cid:19)(cid:18) n − (cid:80) c − k i k c (cid:19) . (I4)0i.e. N c ( { k } ) = N !( N − c )! (cid:81) m l ! n ! (cid:81) k i ! . (I5)The counts N c ( { k } ) must satisfy (cid:88) c N c ( { k } ) = (cid:88) { k , ··· k c } N !( N − c )! (cid:81) m l ! n ! (cid:81) c k i ! = N n (I6)wherein the sum is over all k i that satisfy k ≥ k ≥ · · · k c and (cid:80) ci =1 k i = n. The identity Eq. (I6) can be derived from the following observation. The multinomial expansion is( a + a + a + · · · + a N ) n = N (cid:88) { k i =0 } n ! k ! k ! · · · k N ! a k a k · · · a k N N , (I7)where (cid:80) Ni =1 k i = n and now all the k ’s range from 0 to N . If we put all the a ’s to one, then we have N n = (cid:88) n ! k ! k ! · · · k N ! . (I8)In this sum, group all terms that have the same set of { k , k , · · · k N } . Consider the terms with c non-zero k ’s. Forthe case of distinct k ’s, there will be N ( N − · · · ( N − c ) such terms. But some terms with a given c may have adegeneracy, i.e. terms with m l identical k ’s. Then the number of distinct terms is N ( N − · · · ( N − c ) / (cid:81) m l !. Themultinomial for a sum of ones has become the sum of the N c ( { k } ).A useful alternative to the expression Eq. (I5) is N c = N !( N − c )! n ! (cid:81) Ll =1 m l !( κ l !) m l , (I9)where the κ l are all the distinct k ’s, with κ > κ · · · > κ L > L (cid:88) l =1 m l = c (I10) L (cid:88) l =1 m l κ l = n . (I11)With this notation, a given configuration of scattering centers can be denoted ( κ m , κ m , · · · , κ m L L ), where L ≤ c ≤ min ( n, N ).An even simpler expression results if we define m ≡ N − c = N − (cid:80) m l for the multiplicity of the empty cells.Then N c = (cid:18) Nm · · · m L (cid:19) (cid:18) nk · · · k c (cid:19) , (I12)= (cid:18) Nm · · · m N (cid:19) (cid:18) nk · · · k n (cid:19) (I13)where the two parenthetical expressions are multinomial coefficients, and we have taken advantage of the fact that (cid:80) L m l = m + (cid:80) L m l = N − c + c = N . In Eq. (I13), we assign zeros to the multiplicities m l for L < l < N , andthen use the fact that any number of m ’s or k ’s can be appended to the array in each multinomial, as long as theyare zero.The probability for a given configuration of { k , k , · · · k c } will be P c { k } = N c N n . (I14)Evidently, the configurations with the scattering centers spread out (having the k ’s close to the same values) willhave the larger probabilities, but, as seen in the example for N = 5 , n = 7 below, a configuration with fewer clusters1can have a larger probability than one with a more evenly spread k . (This is in contrast to the multinomial coefficientsthemselves, for which a more even spread of the k ’s always gives a larger coefficient.)The weight for the probability P > that two or more scattering events (‘hits’) occur for a given beam proton willbe a sum of the weight of a given cluster set times the probability p > of a hit in a cluster whose size k i is greaterthan one. This determination might best be seen in an example. Let N = 5 and n = 7. Then the possible clusterswith their weights N c are shown in Table I.1. For this example, the fraction P > of beam protons that hit two or TABLE I.1. Partitions of 7 centers in 5 cells, with the number of each partition given by N c and the probability of a hit on acell with more than one center given by P > . { k , k , · · · } N c × p > { , , , , } × / { , , , , } × / { , , , , } × / { , , , , } × / { , , , , } × / { , , , , } × / { , , , , } × / , , , ,
0) 4200 × / { , , , , } × / { , , , , } × / { , , , , } × / , , , ,
0) 12600 × / , , , ,
1) 12600 × / sum = 78125 33069 more scattering centers is 33069 / P > is given by P > = (cid:88) clusters N n n > N N !( N − c )! (cid:81) m l ! n ! (cid:81) k i ! (I15)Here, n > ≤ c is the number of k ’s bigger than one in the given distribution. Note that we have not required thatthe proton be deflected only to small angles. After each hit, the distribution of scattering centers is the same as thatpresented to the proton in the prior hit. This is a property following from the implicit assumption of homogeneityand isotropy of the tissue and of random distribution of scattering centers with cross-section σ among the possiblescattering tubes.A general expression for P ≥ is given by P ≥ = 1 N n (cid:88) clusters n ≥ N N ! m ! (cid:81) Nm l =1 m l ! n ! (cid:81) nk i =1 k i ! . (I16)Here, n ≥ ≤ c is the number of k ’s bigger than zero in the given cluster. Note that we have not required that theproton be deflected only to small angles. After each hit, the distribution of scattering centers is the same as thatpresented to the proton in the prior hit. This is a property following from the implicit assumption of homogeneityand isotropy of the tissue and of random distribution of scattering centers with cross-section σ among the possiblescattering tubes.The expression Eq. (I16) can be greatly simplified. First note that n ≥ is also the sum of the multiplicities m l except for m , and that m + n (cid:88) l =1 m l = N so that n ≥ = N − m . P ≥ = 1 N n (cid:32) (cid:88) clusters N ! m ! (cid:81) Nm l =1 m l ! n ! (cid:81) nk i =1 k i ! − (cid:88) clusters ( N − m − (cid:81) Nm l =1 m l ! n ! (cid:81) nk i =1 k i ! (cid:33) . (I17)(The m = 0 case is allowed in the second sum because 1 / (( − . ) We recognize each cluster sum to be anexpansion of a sum of ones to a power of n . Thus P ≥ = 1 N n ( N n − ( N − n ) = 1 − (1 − /N ) n ≈ − exp ( − n/N ) , (I18)where, in the last expression, we took N > n >>
1. Alternatively, this exponential behavior can be derived (asin Beer’s law for light scattering) from the assumption that the number of scattered protons is proportional to thenumber arriving into a given small volume of tissue and the density of scattering centers in that volume.To find the probability that a beam proton independently hits at least two scattering centers, we turn to P ≥ = 1 N n (cid:88) clusters n ≥ N N ! m ! (cid:81) Nm l =1 m l ! n ! (cid:81) nk i =1 k i ! . (I19)Now n ≥ = N − m − m . The evaluation of our expression Eq. (I19) is easier as a single multinomial sum. Thereresults P ≥ = 1 − (cid:18) − N (cid:19) n − N n +1 nN (cid:88) k ,...,k N (cid:18) n − k · · · k N , (cid:19) , (I20)where the k sum has been excluded. The factor N in front of the sum comes from the fact that there are N equivalentsuch sums. The ranges of the remaining k ’s go from 0 to n −
1, and they are constrained by (cid:80) Ni =2 k i = n −
1. Thesum in Eq. (I20) is just ( N − n − , resulting in P ≥ = 1 − (cid:18) n − N (cid:19) (cid:18) − N (cid:19) n − ≈ − (cid:18) n − N (cid:19) exp ( − ( n − /N ) . (I21)Note that if n = N (number of scattering centers is the same as the number of nuclear cells), then, for large n , P ≥ → − /e = 0 .
632 and P ≥ = 1 − /e = 0 . { k , k , · · · , k N } , calculates N c for a given distribution and then the probabilities P > , P > , · · · for a beam proton to be scattered once, twice,or any a number of times. Chance for a proton to undergo hits to O18 in tissue
In our applications, the number of nuclear scattering centers behind the area A is ρ N A ∆ z , while the number ofcells over which the scattering centers are distributed in N = A/σ , so n/N = ρ N σ ∆ z. As the proton beam slows, the events O18(p,n)F18 increase as the reaction cross section peaks at about 6 MeV (seeFig. 6). For a low density of parent nuclei, the probability that a proton will hit the cross-sectional area σ is given by P ≈ (∆ N/ ∆ V ) σ ∆ z , where (∆ N/ ∆ V ≡ ρ N ) is the number density of the parent nuclei and ∆ z is the distance theproton has traveled. When the area N σ becomes a significant fraction of the area A B = ∆ V / ∆ z , multiple independenthits are likely. According to Eq. (I18), the probability of at least two independent hits in the case 1 << n << N hasthe leading term P ≥ ≈ (cid:0) n − N (cid:1) , while P ≥ ≈ nN . The approximate expression for P ≥ would also follow from theassumption that the change in the flux of beam particles over a short distance drops in proportion to the flux itself,to the scatterer’s cross section and to their number density. The approximate expression for P ≥ would come aboutif the change of flux over a short distance dropped as the square of the distance the beam proton traveled, indicatingthat double scattering is required, just as the chance that a car is involved in two independent encounters with othercars is proportional to the density of cars squared.Table A.5 shows the relevant data for thymine-18 in the DNA of human tissue. One can see from the long ‘extinctionlength’ λ = 1 / ( ρ N σ ) for the O18(p,n)F18 reaction that inside a human, a beam proton is not likely to have more thanone encounter with O18 to make F18, assuming the O18 is uniformly spread out. However, the fact that the DNA is3compacted into a small volume inside each biologic cell increases the chance that one encounter will be followed by asecond, i.e. scattering events may be correlated.For scattering of beam protons off nuclei, we expect that the number of F18 isotopes that the proton beam producesin the doped tissue will be N F = JA B ∆ t (1 − exp ( − ρ N σ ∆ z )) ≈ JA B ρ N σ ∆ z ∆ t , (I22)where J is the number flux of protons in the beam, A B is the effective area of the beam, and ∆ t is the exposure time. Appendix J: Beam fluence loss
The loss of protons from a proton-therapy beam while it passes through tissue has a number of distinct causes: • Protons scattering from electrons (minimal loss) • Protons elastic scattering from other protons • Protons inelastically scattering from nuclei making excited nuclear states • Protons causing nuclear reactions (transmutation and fragmentation) • Protons becoming thermalizedThe loss is usually measured by the change in the beam fluence rate, or number flux, defined as the number of protonspassing through a unit area perpendicular to the beam per unit time. In relativity, this is the spatial part of theproton 4-current divided by the magnitude of the electron charge, i.e. the proton number density measured movingwith the beam, times the relativistic 4-velocity ( J µ = ρ o dx µ dτ ).Given the myriad of possible causes for diminishing proton flux as the beam passed through tissue, analytic ex-pressions for this loss as a function of penetration distance are hard to find. Rather, Monte Carlo techniques (e.g.GEANT4 code) are often employed. Even so, modeling and experiment show that the drop in proton flux with dis-tance is almost linear. The drop in proton number flux in water is 15 % before the end of the proton beam’s averagerange, where the flux tails off to zero within about 3 % of the beam’s range, ([34], [27]). Appendix K: Nuclear positron emitters produced by a proton beam
Reactions that produce positron-emitting radionuclides potentially could interfere with post-PT PET scans todetermine where the beam delivered its dose. It behooves us to look for the reactions that might produce significantamounts of positron nuclear decays in the time-frame of the measurable F18 decays.We will use (
A, Z ), to represent a nuclei. A proton is then represented by (
A, Z ) = (1 ,
1) , while a neutron is (1 , p, n, ( pp ) , ( pn ) , ( nn ) , α .These will be denoted (1 , , (1 , , (2 , , (2 , , (2 , , (4 , b, q )where b = 1 , q = 0 , ..., b , b = 4 , q = 2.Reactions are represented succinctly by(1 , z ) + ( A, Z ) = ( A ∗ , Z ∗ ) + ( b, q )with z = 1 , A, Z ) = ( A ∗ + b − , Z ∗ + q − z )The possible positron-emitting radionuclides that could interfere, and their half-lives, are4 β + emitter half-lifeC 11 = (11 ,
6) 20 .
334 minN 13 = (13 ,
7) 9 .
965 minO 15 = (15 ,
8) 2 . ,
9) 1 . ,
11) 2 .
602 yrsCu 64 = (64 ,
29) 12 . ,
31) 1 . ,
35) 6 .
46 minRb 82 = (82 ,
37) 1 .
273 minSr 83 = (83 ,
38) 32 .
41 hrsY 86 = (86 ,
39) 14 .
74 hrsZr 89 = (89 ,
40) 78 .
41 hrsI 124 = (124 ,
53) 4 .
176 daysExaminining each one up to copper, we have C
11 = (11 , A ∗ = 11 , Z ∗ = 6 half-life 20 .
334 minbeam proton ( z = 1) z b q A =11 + b − Z =6 + q − z name abundance (( percent )) reaction n
11 5 B .
01 1 1 p
11 6 C .
01 2 0 nn
12 5 B .
01 2 1 pn
12 6 C . C + p → C + n + p pp
12 7 N .
01 4 2 α
14 7 N . N + p → C + α ‘beam’ neutron ( z = 0) z b q A =11 + b − Z =6 + q − z name abundance ( percent ) reaction n
11 6 C .
00 1 1 p
11 7 N .
00 2 0 nn
12 6 C . C + n → C + n + n pn
12 7 N .
00 2 2 pp
12 8 O .
00 4 2 α
14 8 O . N
13 = (13 , A ∗ = 13 , Z ∗ = 7 half-life 9 .
965 minbeam proton ( z = 1) z b q A =13 + b − Z =7 + q − z name abundance ( percent ) reaction n
13 6 C .
01 1 1 p
13 7 N .
01 2 0 nn
14 6 C .
01 2 1 pn
14 7 N . N + p → N + n + p pp
14 8 O .
01 4 2 α
16 8 O . O + p → N + α ‘beam’ neutron ( z = 0)5 z b q A =13 + b − Z =7 + q − z name abundance ( percent ) reaction n
13 5 B .
00 1 1 p
13 6 C .
070 2 0 nn
14 7 N . N + n → N + n + n pn
14 8 O .
00 2 2 pp
14 9 F .
00 4 2 α
16 9 F . O
15 = (15 , A ∗ = 15 , Z ∗ = 8 half-life 2 . z = 1) z b q A =15 + b − Z =8 + q − z name abundance ( percent ) reaction n
15 7 N . p
15 8 O .
01 2 0 nn
16 7 N .
01 2 1 pn
16 8 O . O + p → O + n + p pp
16 9 F .
01 4 2 α
18 9 F . z = 0) z b q A =15 + b − Z =8 + q − z name abundance ( percent ) reaction n
15 8 O .
00 1 1 p
15 9 F .
00 2 0 nn
16 8 O . O + n → O + n + n pn
16 9 F .
00 2 2 pp
16 10
N e .
00 4 2 α
18 10
N e . F
18 = (18 , A ∗ = 18 , Z ∗ = 9 half-life 1 . hrs beam proton ( z = 1) z b q A =18 + b − Z =9 + q − z name abundance ( percent ) reaction n
18 8 O .
01 1 1 p
18 9 F .
01 2 0 nn
19 8 O .
01 2 1 pn
19 9 F . F + p → F + n + p pp
19 10
N e .
01 4 2 α
21 10
N e . z = 0) z b q A =18 + b − Z =9 + q − z name abundance ( percent ) reaction n
18 9 F .
00 1 1 p
18 10
N e .
00 2 0 nn
19 9 F . F + n → F + n + n pn
19 10
N e .
00 2 2 pp
19 11
N a .
00 4 2 α
21 11
N a . N a
22 = (22 , A ∗ = 22 , Z ∗ = 11 half-life 2 . yrs z = 1) z b q A =22 + b − Z =11 + q − z name abundance ( percent ) reaction n
22 12
M g .
01 1 1 p
22 13 Al .
01 2 0 nn
23 12
M g .
01 2 1 pn
23 13 Al .
01 2 2 pp
23 14
S i .
01 4 2 α
25 14
S i . z = 0) z b q A =22 + b − Z =11 + q − z name abundance ( percent ) reaction n
22 11
N a .
00 1 1 p
22 12
M g .
00 2 0 nn
23 11
N a . N a + n → N a + n + n pn
23 12
M g .
00 2 2 pp
23 13 Al .
00 4 2 α
25 13 Al . Appendix L: Mathematica expressions for proton scattering probabilities
In Appendix I, we derived a way of calculating the probability that a beam proton will scatter from a scatteringcenter in tissue at least a certain number of times. Here, we will give Mathematica expressions to perform this kind ofcalculation when the number of cells, N , and the number of scattering centers spread among those cells, are tractableby Mathematica. This limits N to be below about 100, and n to be below about 60, so the expressions given make auseful test-bed for examining the model, but not for cases in which N reaches billions or more.We distribute n scattering centers randomly over N ‘cells’. A given distribution of scattering centers in cells isdisplayed with the notation ( k , k , · , k N ), where the k ’s are the number of scattering centers which have ended in agiven cell. Since all cells have equivalent in a priori probability, the cell distributions can be ordered according to thesize of the k ’s as k ≥ k · · · ≥ k N . The k ’s are constrained by (cid:80) k i = n . The total number of distinct distributionsis some number c . For a given distribution, some of the k values may be the same. We say the k scattering centersform a ‘cluster’. If so, put these clusters in groups. Each such group will be a collection of a certain number ofcells. The number of such clusters we will call m l , the ‘multiplicity’ of the k value. The m l values are constrainedby (cid:80) m l = c and by (cid:80) m l k l = n . It is useful to extend the set of m l values with zeros, with the multiplicity ofzeros being m = N − c . The number of equivalent configurations with the same ( k , k , · , k N ) created after a randomdistribution of the n scattering centers among the N cells is called N c , and was shown in Appendix I to be N c = (cid:18) Nm · · · m N − (cid:19) (cid:18) nk · · · k n (cid:19) , (L1)where m i is the ‘multiplicity’ of the cells with the same number k of scattering centers. The m l are taken once forthe whole set of such cells. In Mathematica, after setting values for variables N N (for the value of N ) and n , thedistinct configurations can be listed usingc f = I n t e g e r P a r t i t i o n s [ n , { NN } , Range [ 0 , n ] ] ;We have put N → N N to avoid the Mathematica function N . A list of the values of N c for each distinct configurationwill result fromM u l t i n o m i a l @@@ ( T a l l y /@ c f ) [ [ All , All , 2 ] ] ∗ M u l t i n o m i a l @@@ c fNow, in order to get the probability of at least h hits by a beam proton, we definepos [ x ] := I f [ x > = h , 1 , 0 ]and use, sequentially, h = 0 , , , · · · . It follows thatTr [ F l a t t e n [ M u l t i n o m i a l @@@ ( T a l l y /@ c f ) [ [ All , All , 2 ] ] ∗ M u l t i n o m i a l @@@ c f ∗ Map [ pos , c f , { } ] ] ] / NNˆ ( n + 1 )7will give the probability that a beam proton will hit at least h scattering centers. The implied factor N c /N n is theprobability that a given configuration appears in the random distribution of scattering centers, and the implied factor n h /N gives the probability that a beam proton will hit a cell containing at least n h scattering centers. For n (cid:39) n taken N N ≡ k at a time is often called p ( n, k ), with p ( n, n ) ≡ p ( n )The number p ( n, k ) can be calculated in Mathematica for a given set of distinct distributions from the ‘length’ of cf ,i.e. p [ n , k ):= Length [ I n t e g e r P a r t i t i o n s [ n , k ] ]For the example, from Table I.1, with n = 7 and k = 5, p ( n, k ) = 13.In 1917, Hardy and Ramanujan [18, 19] gave an asymptotic expression for p ( n ): p ( n ) ≈ π e ν ( ν − √ ν + O (cid:18) e ν/ n (cid:19) (L2)where ν = (cid:112) (2 / π ( n − / n . Here, we give that of Brassesco and Meyroneic[6]: p ( n ) ≈ (cid:18) π √ (cid:19) exp ( r ( n ))(2 r ( n ) + 1) (cid:32) − K (cid:88) l =1 D ( l )(2 r ( n ) + 1) l + O (1 /n K/ ) (cid:33) (L3)where D ( l ) ≡ ( − l +1 ( l + 1)4 l l +1 (cid:88) k =0 (cid:18) lk (cid:19) ( − k ( l + 1 − k )! , (L4) r ( n ) ≡ (cid:115) π (cid:18) n − (cid:19) + 14 . (L5)The value of the integer K determines the error of the approximation. For example, values as little K = 3 give an p (71) within 0 . , , [1] Allison, J., Amako, K., Apostolakis, J., Araujo, H., Dubois, P. A., Asai, M., Barrand, G., Capra, R., Chauvie, S., Chytracek,R., et al. (2006). Geant4 developments and applications. IEEE Transactions on Nuclear Science , 53(1):270–278.[2] Allison, J., Amako, K., Apostolakis, J., Arce, P., Asai, M., Aso, T., Bagli, E., Bagulya, A., Banerjee, S., Barrand, G., et al.(2016). Recent developments in Geant4.
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