April 26 2004


Finite size scaling of triangular lattices f=1/49 at T=0.005


The labels of k1 and k2 in this page are the orthogonal coordinates in the structure function space, not the wave numbers based on the reciprocal lattice b=(b1,b2).

E=0.1

The simulation results are at April 18 2004 .

Labeling of peaks

The peaks are labeled as shown in the following picture:

Label k1 k2 dk=sqrt(k1*k1 + k2*k2) dk / d1
1 0 1/3Pi (1.0364567796) 1.0364567796 1
2 0 2/3Pi (2.0729135591) 2.0729135591 2
3 0 Pi (3.1093703387) 3.1093703387 3
4 2/7Pi (0.89759790103) 1/6Pi (0.51822838978) 1.0364567796 1
5 4/7Pi (1.7951958021) 1/3Pi 2.0729135592 2
6 6/7Pi (2.6927937031) 1/2Pi (1.5546851693) 3.1093703387 3
7 4/7Pi 0 1.7951958021 12/7
8 2/7Pi 1/2Pi 1.7951958021 12/7
11 6/7Pi 1/6Pi 2.7422068834 18/7

Log-Log plot of peak heights

The peak heights are fit by H = H0 L ^ ( 2 - Ak).

The fitted values are:

Parameter Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7
H0 0.0442603 0.300286 1.76738 0.0762866 0.688684 1.78499 0.295279
Ak 0.281669 1.02133 1.80014 0.462303 1.3944 1.98836 1.11211

To evaluate the deviation of Ak from equilibrium float solid case, we compute B1 = Ak ( d1 / dk )2.

Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7
B1 0.281669 0.255333 0.200016 0.462303 0.348600 0.220929 0.378426

Plot of fit parameter Ak

The abover parameter values Ak have been fitted by Ak=A0dkr. The values for point 1,2,3 are A0=0.316204, r=1.54119. Those for point 4,5,6 are A0=0.542514, r=1.16643.


E=0.2

The simulation results are at May 01 2004. The second run of L=112 has been used.

Log-Log plot of peak heights

The peak heights are fit by H = H0 L ^ ( 2 - Ak).

The fitted values are:

Parameter Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7
H0 0.0168089 0.0151078 0.95807 0.137418 0.553069 1.97377 0.488833
Ak 0.0597787 0.316067 1.55833 0.596852 1.34369 1.99347 1.22793

Plot of fit parameter Ak

The abover parameter values Ak have been fitted by Ak=A0dkr. The values for point 1,2,3 are A0=0.0191003, r=3.87958. Those for point 4,5,6 are A0=0.602414, r=1.06167.


E=0.28

The simulation results are at May 02 2004.

Log-Log plot of peak heights

The peak heights are fit by H = H0 L ^ ( 2 - Ak).

The fitted values are:

Parameter Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7
H0 0.0164558 0.0131147 0.0267984 0.0628677 0.0622811 0.350681 0.103151
Ak 0.0073707 0.0980165 0.389463 0.398437 0.76976 1.48844 0.833524

Plot of fit parameter Ak

The abover parameter values Ak have been fitted by Ak=A0dkr. The values for point 1,2,3 are A0=0.00809728, r=3.41445. Those for point 4,5,6 are A0=0.31348, r=1.35722.


E=0.4

The simulation results are at May 04 2004.

Log-Log plot of peak heights

The peak heights are fit by H = H0 L ^ ( 2 - Ak).

The fitted values are:

Parameter Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7
H0 0.0203877 0.0203877 0.0203877 0.0147965 0.0151877 0.0585157 0.0152019
Ak 0 0 0 0.0520818 0.333823 0.883244 0.333722

Plot of fit parameter Ak

The abover parameter values Ak have been fitted by Ak=A0dkr. The values for point 1,2,3 are A0=0, r=0. Those for point 4,5,6 are A0=0.0558114, r=2.43543.


E=2.0

The simulation results are at May 05 2004.

Log-Log plot of peak heights

The peak heights are fit by H = H0 L ^ ( 2 - Ak).

The fitted values are:

Parameter Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7 Point 8 Point 11
H0 0.0204082 0.0204082 0.0204082 0.0110128 0.00968585 0.0474475 0.00968585 0.0110128 0.0474475
Ak 0 0 0 -0.0100777 0.236641 0.835942 0.236641 -0.0100777 0.835942

To evaluate the deviation of Ak from equilibrium float solid case, we compute B1 = Ak ( d1 / dk )2.

Point 1 Point 2 Point 3 Point 4 Point 5 Point 6 Point 7 Point 8 Point 11
B1 0 0 0 -0.0100777 0.05916025 0.092882444444 0.080523673611 -0.0034292173611 0.12642332716