P HI T S Melt a snowman by proton beam Multi-Purpose Particle and Heavy Ion Transport code System title1 Aug revised

Purpose2 Purpose of this exercise Let us consider with current technology how realistic beam weapons in animations are by performing numerical exercise of proton beam melting a snowman Geometry setup Change of source Concept of normalization Lecture on Courtesy of D. Satoh

snowman.inp 3 Check Input File Basic setup Projectile ： Geometry ： Tally ： track.eps 100MeV proton (pencil beam with radius 1.0cm) Water sphere of 5cm radius at the origin [t-track] flux distribution [t-deposit] absorbed dose (Gy/source) in water sphere deposit.out … x: Serial Num. of Region y: Dose [Gy/source] p: xlin ylog afac(0.8) form(0.9) h: x n n y(all ),l3 n # num reg volume dose r.err E E x (Gy/source) Definition of volume of sphere is necessary at [volume] section

Flow chart of this exercise 4Procedure 1.Set geometry of a snowman 2.Set beam condition 3.Determine beam current and power to melt a snowman

Setup geometry of a snowman 5 Direction A simple structure of big and small ice balls with aluminum plate Material of the ice balls are 1g/cm 3 water (without temperature option*) The aluminum plate is placed on the small ice ball without gap \phits\utility\rotate3dshow Geometry to construct in this exercise *Temperature option influence only motion of low energy neutron

Step 1: Construct a big ice ball 6Step 1 Set a big ice ball around original 10cm sphere (radius 20cm at center z=0cm) Hint Geometry check can be done with icntl = 8 A spherical surface centering the origin is defined by ”so [radius]” Define the region of the big ice ball not to overlap with the region of the original sphere of 5cm radius (avoid double defined region)

Step 2: Construct a small ice ball 7Step 1 Set a small ice ball on top of the big ice ball (radius 15 cm at center z=-25cm) Hint A spherical surface centering on z axis is defined by ”sz [center z position] [radius]” Exclude region of big ice ball from small ice ball or vise versa, otherwise double defined region is created by two ice balls

Step3: Set aluminum plate 8Step 2 1.Define aluminum (27Al) at [material] section 2.Construct an aluminum plate of 10cm radius and 4cm thickness (-40cm < z < -36cm) Hint A cylindrical surface parallel to z axis is defined by “cz [radius]” A plane perpendicular to z axis is defined by ”pz [z position]” Density of aluminum is 2.7g/cm 3 (negative value for mass density) Exclude region of the aluminum plate from that of the small ice ball

Step 4: Set proton beam condition 9Step 3 Tune a proton beam energy so that the absorbed dose at central sphere is maximized (Tune the energy at the level of 1MeV) Hint Transport calculation can be executed with icntl = 0 Beam energy is given by e0 parameter at [source] section Absorbed dose at central sphere can be checked by deposit.out Proton absorbed dose is maximized at the Bragg peak

Default output of PHITS is normalized to per particle emitted from source Set totfact at [source] section to change the normalization 1A denotes status that 1C current is conducting in 1 second The electric charge of a proton is 1.6x C Step 5: Tune proton beam current 10Step 4 Calculate absorbed dose (Gy) in 1 second deposited by 10nA beam used in general proton therapy Hint Maxcas and maxbch control statistical uncertainty, and are not related to norm f.g. Absorbed dose for 100 protons is given by setting totfact=100.0

Answer 5 11Answer to Step 4 Check the scale change with totfact 1.Number of emitted protons for 1 ampere in 1 second is 1.0 / 1.6E-19 = 6.25E18 (particles) 2.Number of emitted protons for 10nA in 1 second is 6.25E18 x 10 x 1E-9 = 6.25E10 (particles) Absorbed dose with totfact = 6.25E10 is (Gy) (for 294MeV proton beam)

Step6: Calculate beam current to melt central sphere in the snow man 12Step 5 1.Assume the snowman is made of ice of -10  C 2.Calculate the proton beam current (in ampere) and the power (in Watt) necessary to heat up the ice to 0  C in 1 second Hint (assumption for simplicity) Specific heat of ice is 0.5 (cal/g/K) = 2.1 (J/g/K) Latent heat of ice (heat necessary for phase transition from ice to water) is （ J/g ） 1Gy = 1 （ J/kg ） = （ J/g ） Beam power (in MW) = Particle energy (in MeV)  Beam current (in A) By the way… The maximum power of the biggest power accelerator facilities in the world is the order of 1MW

Answer 6 13Answer to Step 5 You can solve the problem by answering the following questions step by step 1.How much is the absorbed dose in central sphere by 294 MeV proton beam with 10nA in 1 second ? 2.How much heat is needed to heat up the ice 10K and melt the ice ? 3.How much current is required to give that heat in 1 second ? 4.How much is the power at this beam current ? Melting a snowman is just what we can do with current technology Beam weapons in animation which destroy robots instantly are far beyond current technology ! (Of course, with longer time, we can melt metals) About 1MW beam power is required to melt a snowman (Details are given in answer\answer-exercise3-en.ppt)

14 Geometry of a snowman was constructed and PHITS simulation was conducted with tuned proton beam energy to melt it effectively Normally, tally outputs of PHITS are normalized to per particle Tally outputs have to be re-normalized by using totfact parameter to simulate actual conditions Summary