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[求助]
關(guān)于MEAM勢(shì)能的選擇問(wèn)題
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最近想做純鐵的模擬,用lammps自帶的library.meam中的參數(shù),發(fā)現(xiàn)結(jié)果存在嚴(yán)重錯(cuò)誤。 后搜到一組勢(shì)能參數(shù)來(lái)自于篇文章http://journals.aps.org/prb/abstract/10.1103/PhysRevB.89.094102(見(jiàn)附件) 做的是純Fe3C的勢(shì)能,相關(guān)參數(shù)文件如下: # Fe3C (Cementite) library MEAM file used in Phys. Rev. B 89, 094102 (2014) # Fe potential referenced from J. Phys.: Condens. Matter 24 225404 # C potential referenced from Phys. Rev. B 81, 104103 # Used together with "Fe3C.meam" for Fe-C alloy potential parameters # elt lat z ielement atwt # alpha b0 b1 b2 b3 alat esub asub # t0 t1 t2 t3 rozero ibar 'Fe' 'bcc' 8 26 55.847000 5.027000 3.800000 2.000000 0.900000 0.000000 2.851000 4.280000 0.585000 1.000000 -0.800000 12.300000 2.000000 1.000000 0.000000 'C' 'dia' 4. 6 12.0111 4.38 4.26 5.00 3.20 3.980 3.567 7.37 1.490 1.0 7.5 1.04 -1.01 1.0 0 來(lái)自:http://www.ctcms.nist.gov/potent ... _Liyanage_2014.meam # Fe-C alloy MEAM potential parameters from Phys. Rev. B 89, 094102 # Developed using April 2011 version of LAMMPS (lammps.sandia.gov) # Should work on recent release of LAMMPS # Contact information: lsiliyanage@gmail.com (Dr. Laalitha S. I. Liyanage) augt1=0 ialloy=1 nn2(1,1)=1 nn2(2,2)=1 nn2(1,2)=1 rc=4.500000 repuls(1,1)=0.300000 Cmin(1,1,1)=0.680000 Cmax(1,1,1)=1.900000 Cmin(2,2,2)=2.000000 Cmax(2,2,2)=2.800000 lattce(1,2)='b1' repuls(1,2)=0.125301 attrac(1,2)=0.125301 delta(1,2)=0.002348 alpha(1,2)=4.750000 re(1,2)=1.920000 rho0(1)=1.000000 rho0(2)=5.491125 Cmin(1,1,2)=0.059531 Cmax(1,1,2)=2.800000 Cmin(1,2,1)=2.000000 Cmax(1,2,1)=2.800000 Cmin(1,2,2)=2.000000 Cmax(1,2,2)=2.800000 Cmin(2,2,1)=0.507375 Cmax(2,2,1)=2.800000 來(lái)自:http://www.ctcms.nist.gov/potent ... _Liyanage_2014.meam 這兩個(gè)勢(shì)能參數(shù)文件是作者針對(duì)Fe3C的計(jì)算擬合結(jié)果。 如果用這兩個(gè)勢(shì)能參數(shù)文件來(lái)做純Fe的模擬是否合適? 如果是Fe-C合金呢?(成分在0~Fe3C之間) 希望大家多給意見(jiàn),更希望能得到高手的指點(diǎn)! |
找到一些相關(guān)的精華帖子,希望有用哦~
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