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- Timestamp:
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Dec 2, 2021, 3:48:35 PM (2 years ago)
- Author:
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gallmei
- Comment:
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Update infos on this page
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5 | 5 | First the target has to be adjusted to Deuterium: |
6 | 6 | {{{ |
7 | | $target |
8 | | target_Z=1 |
9 | | target_A=2 |
10 | | fermimotion=.true. |
11 | | $end |
| 7 | ! file: code/density/nucleus.f90 |
| 8 | &target |
| 9 | Target_Z = 1, Target_A = 2 |
| 10 | / |
12 | 11 | }}} |
13 | 12 | To distribute the nucleons in position and momentum space we can choose between two different wave function models: |
14 | 13 | {{{ |
15 | | $deuteriumFermi |
16 | | waveFunction_switch=2 |
17 | | ! 1=Bonn |
18 | | ! 2=Argonne |
19 | | $end |
| 14 | ! file: code/init/deuterium.f90 |
| 15 | &deuteriumFermi |
| 16 | waveFunction_switch=2 ! 1=Bonn, 2=Argonne |
| 17 | / |
20 | 18 | }}} |
21 | 19 | Next, we need to define a potential to bind the two nucleons. For this we can't use a mean field, because Deuterium represents a too small system. Instead we use a real two-body potential. Using the parallel ensemble technique, the potential ''V'' for each nucleon in the ''j''th ensemble is given by \( V=V_\text{2-body}(r_{1,j}-r_{2,j}) \) |
22 | 20 | where \( r_{i,j} \) is the position of the ''i''th nucleon in the ''j''th ensemble. For the full ensemble method, a Deuterium potential is not yet properly implemented. So we choose for the general input and the propagation routines the following switches: |
23 | 21 | {{{ |
| 22 | ! file: ./inputOutput/input.f90 |
24 | 23 | $input |
25 | | delta_T = 0.025 ! small time step sizes since the two-body potential is stiff and therefore the propagation is sensitive to too large time steps |
26 | | fullensemble=.false. ! => use parallel ensemble technique |
27 | | freezeRealParticles=.false. |
28 | | length_perturbative=1 ! We don't use perturbative particles, see comments below |
| 24 | delta_T = 0.025 ! small time step size since the two-body potential |
| 25 | ! is stiff and therefore the propagation is sensitive |
| 26 | ! to too large time steps |
| 27 | fullensemble=.false. ! => use parallel ensemble technique |
| 28 | freezeRealParticles=.false. |
29 | 29 | ... |
30 | | $end |
| 30 | / |
31 | 31 | |
32 | | $initDensity |
33 | | densitySwitch=1 |
34 | | splineExtraPolation=.true. !Switch for linear spline extrapolation for dynamically calculated density: Extrapolates density between |
35 | | gridPoints(1)=100 |
36 | | gridPoints(2)=100 |
37 | | gridPoints(3)=100 |
38 | | gridSize(1)=8. |
39 | | gridSize(2)=8. |
40 | | gridSize(3)=8. |
41 | | $end |
| 32 | ! file: code/density/density.f90 |
| 33 | &initDensity |
| 34 | densitySwitch = 2 ! 2=analytic |
| 35 | / |
42 | 36 | |
43 | | $propagation |
44 | | delta_P=0.01 ! Delta Momentum for derivatives |
45 | | DerivativeType=2 ! 1=first order Range-Kutta, 2=second order Range-Kutta |
46 | | predictorCorrector=.true. ! Whether to use a predictor/corrector algorithm to do the propagation |
47 | | $end |
| 37 | ! file: code/propagation/propagation.f90 |
| 38 | &propagation |
| 39 | RungeKuttaOrder=2 ! 2=second order Runge-Kutta |
| 40 | / |
48 | 41 | |
49 | | $baryonPotential |
50 | | EQS_Type=7 ! => Two body potential for deuterium |
51 | | DeltaPot=1 ! Switch for potential of spin=3/2 resonances |
52 | | ! 1=nucleon (spin=1/2) potential times 3/5 [according to ericson/Weise book] |
53 | | ! 2= 100 MeV *rho/rhoNull |
54 | | symmetriePotFlag=.false. ! Switch for the assymetry term in the nucleon potential |
55 | | $end |
56 | | |
57 | | $Yukawa |
58 | | yukawaFlag=.false. !decides whether Yukawa is switched off(.false.) or on (.true.) |
59 | | $end |
| 42 | ! file: code/potential/baryonPotential.f90 |
| 43 | &baryonPotential |
| 44 | EQS_Type=7 ! => Two body potential for deuterium |
| 45 | / |
60 | 46 | }}} |
61 | 47 | |
62 | 48 | |
63 | | Oliver prefers not to use perturbative particles with Deuterium, since there is no unperturbed nucleus left if there is a nuclear reaction in deuterium. So he chooses |
| 49 | Up to now is probably all you have to change. |
| 50 | |
| 51 | Oliver prefers not to use perturbative particles with Deuterium, since there is no unperturbed nucleus left if there is a nuclear reaction in deuterium. So he chooses for his special case |
64 | 52 | {{{ |
| 53 | ! file: ./inputOutput/input.f90 |
| 54 | $input |
| 55 | ... |
| 56 | length_perturbative=1 ! We don't use perturbative particles |
| 57 | ... |
| 58 | / |
| 59 | |
65 | 60 | $low_photo_induced |
66 | 61 | ... |
67 | | realRun=.true. ! => reaction products are set into real particle vector |
68 | | $end |
| 62 | realRun=.true. ! => reaction products are set into real particle vector |
| 63 | / |
69 | 64 | }}} |
| 65 | This may not work if you use another initialization. |
70 | 66 | |
71 | | |
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