Computational Physics - M. Jensen Episode 1 Part 5 pptx
Computational Physics - M. Jensen Episode 1 Part 1 ppsx
... . . . . . . . . . . 18 4 11 Monte Carlo methods in statistical physics 18 7 11 .1 Phase transitions in magnetic systems . . . . . . . . . . . . . . . . . . . . . . . 18 7 11 .1. 1 Theoretical background ... . 18 7 11 .1. 2 The Metropolis algorithm . . . . . . . . . . . . . . . . . . . . . . . . . 19 3 11 .2 Program example . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 2 pot
... 0.935762E -1 3 -0 .34 21 3 4E-04 10 0 40.0 0. 424 835E -1 7 -0 .2 21 0 33E+ 01 127 50.0 0 .19 28 75E- 21 -0 .833851E+05 15 5 60.0 0.875651E -2 6 -0 .850381E+09 17 1 70.0 0.397545E-30 NaN 17 1 80.0 0 .18 0485E-34 NaN 17 1 90.0 ... -0 .3066 811 1E-04 10 0 40.000000 0. 424 83543E -1 7 -0 . 316 57 319 E+ 01 127 50.000000 0 .19 28 7498E- 21 0 .11 0 729 33E+05 15 5...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 3 doc
... equation that (3 .18 ) (3 .19 ) and (3. 20) These equations have the solution (3. 21) and (3. 22) yielding To determine , we require in the last equation that (3. 23) (3. 24) 42 CHAPTER 3. NUMERICAL DIFFERENTIATION Assume ... e.g., (3 .11 ) We could also define five-points formulae by expanding to two steps on each side of . Using a Taylor expansion around in a region we have (3...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 4 pdf
... 20.0857 04 20.085539 20.085537 20.25 046 7 20.085537 4. 0 54. 643 6 64 54. 598605 54. 59 815 5 54. 59 815 1 54. 711 789 54. 59 815 0 5.0 14 8.536878 14 8. 41 4 396 14 8. 41 3 172 14 8. 41 3 1 61 150.635056 14 8. 41 3 159 Table 3 .1: Result ... 7.38905 610 0.00000003 3.0 20 .10 228 045 20.0897 217 6 20.08658307 20.08553692 0.00000009 4. 0 54. 643 66366 54. 60952560 54. 60099375 54. 5...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 5 pptx
... With respect to pointers this means that matr is pointer-to-a-pointer-to-an-integer which we can write matr. Furthermore matr is a-pointer-to-a-pointer of five integers. This interpretation is important ... 1 (by construction since the diagonals of equal 1) we can use the inverse of to obtain (5 .13 ) which yields the intermediate step (5 .14 ) and multiplying with on both sides we reob...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 6 doc
... that is (6. 20) It suffices in our case to study , which results in (6. 21) and with Eq. (6 .19 ) we obtain (6. 22) meaning . 6. 4 Newton-Raphson’s method Perhaps the most celebrated of all one-dimensional ... function with a second-order polynomial . The first and second derivatives are given by (8.8) and (8.9) 90 CHAPTER 6. NON-LINEAR EQUATIONS AND ROOTS OF POLYNOMIALS 1. (6 .15 )...
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Computational Physics - M. Jensen Episode 1 Part 7 docx
... the Gauss-Legendre method. 1 1.305 3.334 2 6 .74 7 7. 473 3 16 .030 10 .954 4 28.330 13 .463 5 42.556 14 .77 6 6 57. 444 14 .77 6 7 71 . 670 13 .463 8 83. 970 10 .954 9 93.253 7. 473 10 98.695 3.334 8.3.6 Applications ... Gauss-Legendre 10 1. 8 210 20 1. 214 025 0 .14 60448 20 0. 912 678 0.6098 97 0. 2 17 80 91 40 0. 478 456 0.333 71 4 0. 219 3834 10 0 0. 273 724 0.2 312...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 8 pdf
... ran1 ran2 ran3 0. 0-0 .1 1 013 9 91 9 38 10 47 0. 1- 0 .2 10 02 10 09 10 40 10 30 0. 2-0 .3 989 999 10 30 993 0. 3-0 .4 939 960 10 23 937 0. 4-0 .5 10 38 10 01 1002 992 0. 5-0 .6 10 37 10 47 10 09 10 09 0. 6-0 .7 10 05 989 10 03 ... 1. 59064E-02 10 00 7 .83 486 E- 01 5 .14 102E-03 10 000 7 .85 488 E- 01 1.60 311 E-03 10 0000 7 .85 009E- 01 5. 087 45E-04...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 1 Part 9 pptx
... thereafter (9 .10 8) by rewriting (9 .10 9) since (9 .11 0) Perform then a Monte Carlo sampling for (9 .11 1) with , 9. 5. IMPROVED MONTE CARLO INTEGRATION 15 3 where is a random number in the interval [0 ,1] . The algorithm ... These are (10 .8) implying that when we study the time-derivative , we obtain after integration by parts and using Eq. (10 .3) (10 .9) leading to (10...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 2 Part 1 ppsx
... spin [ j ] spin [ j +1] ; } 10 .4. THE METROPOLIS ALGORITHM AND DETAILED BALANCE 18 1 Figure 10 .5: Probability distribution for one walker after 10 , 10 0 and 10 00 steps. 10 .5. PHYSICS PROJECT: SIMULATION ... statistical physics at a given temperature, we need a probability distribution (11 .2) 18 7 10 .3. MICROSCOPIC DERIVATION OF THE DIFFUSION EQUATION 17 7 with the obv...
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Computational Physics - M. Jensen Episode 2 Part 2 doc
... two-dimensional lattice we obtain the following partition function (11 .22 ) and mean energy (11 .23 ) 20 4 CHAPTER 12. QUANTUM MONTE CARLO METHODS which can be rewritten as ( 12. 5) In general, the integrals involved ... the one-dimensional Ising model with spins with free ends (FE) and periodic boundary conditions (PBC). Number spins up Degeneracy Energy (FE) Energy (PBC) Magnetization 2...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 2 Part 3 pps
... 7.71 826 E-04 1 .20 000E+00 1.08667E+00 2. 233 89E-01 1.49462E- 03 1 .30 000E+00 1.17168E+00 4.78446E-01 2. 18 734 E- 03 1.40000E+00 1 .26 37 4E+00 8.55 524 E-01 2. 924 93E- 03 1.50000E+00 1 .38 897E+00 1 .30 720 E+00 3. 61553E- 03 ... of 2 was used in order to obtain an acceptance of . 5.00000E-01 2. 06479E+00 5.78 739 E+00 7.60749E- 03 6.00000E-01 1.50495E+00 2. 32 7 82E+00 4....
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 2 Part 4 ppsx
... 9.898985E-01 2. 949 052E+00 4. 86 622 3E+00 6.739916E+00 8.56 844 2E+00 100 9.9 748 93E-01 2. 98 744 2E+00 4. 96 727 7E+00 6.936913E+00 8.89 628 2E+00 20 0 9.993715E-01 2. 996864E+00 4. 991877E+00 6.9 843 35E+00 8.9 743 01E+00 40 0 ... the potential is symmetric with nm eV nm [eV] 86 42 0 -2 -4 -6 -8 0 -1 0 -2 0 -3 0 -4 0 -5 0 -6 0 Figure 12. 5: Plot of f...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 2 Part 5 pdf
... second-order equation goes typically like (14 .2) A well-known second-order equation is Newton’s second law (14.3) where is the force constant. ODE depend only on one variable, whereas 25 5 26 2 CHAPTER ... derivative in our Taylor expansion. We have then (14 .22 ) The second derivative can be rewritten as (14 .23 ) and we can rewrite Eq. (14.14) as (14 .24 ) 25 0 CHAPTER 13. EIGENSYS...
Ngày tải lên: 07/08/2014, 12:22
Computational Physics - M. Jensen Episode 2 Part 6 pptx
... errmax ; double yout [ 2] , y_h [ 2] , y_m [2] , y1 [ 2] , y2 [ 2 ] , delt a [ 2] , ys cal [ 2 ]; const double eps =1.0e 6; const double s a fet y =0.9; const double er rcon =6. 0e 4; const double ... phi and v / yout [0]= yin [ 0 ] +1.0 / 6 .0 ( k1 [0] +2 k2 [0] +2 k3 [0]+ k4 [ 0]) ; yout [1]= yin [ 1 ] +1.0 / 6 .0 ( k1 [1] +2 k2 [1] +2 k3 [1]+ k4 [ 1]) ; } 28 0 CHAPTER 1...
Ngày tải lên: 07/08/2014, 12:22