Similar documents
プログラム

CMP Technical Report No. 92 DOS Department of Computational Nanomaterials Design ISIR, Osaka University

Erased_PDF.pdf


本文/目次(裏白)

第1章 微分方程式と近似解法

: , 2.0, 3.0, 2.0, (%) ( 2.

プログラム

放射線専門医認定試験(2009・20回)/HOHS‐05(基礎二次)

Supplement to Osaka2k Department of Computational Nanomaterials Design ISIR, Osaka University

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

IV (2)

TOP URL 1

PDF

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

n 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m

BH BH BH BH Typeset by FoilTEX 2

Untitled

( ) ( )

4/15 No.

untitled

I

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

Note.tex 2008/09/19( )

医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.

untitled

_0212_68<5A66><4EBA><79D1>_<6821><4E86><FF08><30C8><30F3><30DC><306A><3057><FF09>.pdf

LLG-R8.Nisus.pdf

数学の基礎訓練I

a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552

1. ( ) 1.1 t + t [m]{ü(t + t)} + [c]{ u(t + t)} + [k]{u(t + t)} = {f(t + t)} (1) m ü f c u k u 1.2 Newmark β (1) (2) ( [m] + t ) 2 [c] + β( t)2

02-量子力学の復習

,, 2. Matlab Simulink 2018 PC Matlab Scilab 2

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

. (.8.). t + t m ü(t + t) + c u(t + t) + k u(t + t) = f(t + t) () m ü f. () c u k u t + t u Taylor t 3 u(t + t) = u(t) + t! u(t) + ( t)! = u(t) + t u(

untitled

analog-control-mod : 2007/2/4(8:44) 2 E8 P M () r e K P ( ) T I u K M T M K D E8.: DC PID K D E8. (E8.) P M () E8.2 K P D () ( T ) (E8.2) K M T M K, T

19 σ = P/A o σ B Maximum tensile strength σ % 0.2% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional

1 X X A, B X = A B A B A B X 1.1 R R I I a, b(a < b) I a x b = x I 1.2 R A 1.3 X : (1)X (2)X X (3)X A, B X = A B A B = 1.4 f : X Y X Y ( ) A Y A Y A f

ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.

AC Modeling and Control of AC Motors Seiji Kondo, Member 1. q q (1) PM (a) N d q Dept. of E&E, Nagaoka Unive


第86回日本感染症学会総会学術集会後抄録(I)

i

seminar0220a.dvi

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +



9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) Ĥ0 ψ n (r) ω n Schrödinger Ĥ 0 ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ0 + Ĥint (

日本内科学会雑誌第97巻第7号

日本内科学会雑誌第98巻第4号

Outline I. Introduction: II. Pr 2 Ir 2 O 7 Like-charge attraction III.

Part () () Γ Part ,


.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

スケーリング理論とはなにか? - --尺度を変えて見えること--

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n

さくらの個別指導 ( さくら教育研究所 ) a a n n A m n 1 a m a n = a m+n 2 (a m ) n = a mn 3 (ab) n = a n b n a n n = = 3 2, = 3 2+

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

τ τ

.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

1 9 v.0.1 c (2016/10/07) Minoru Suzuki T µ 1 (7.108) f(e ) = 1 e β(e µ) 1 E 1 f(e ) (Bose-Einstein distribution function) *1 (8.1) (9.1)

2 ( ) i

JFE.dvi

<B54CB5684E31A4E9C0CBA4E5AA6BC160BEE3B27AA544A5552E706466>

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0


福大紀要 02730816/教育科学 太田 氏家

Microsoft PowerPoint - 粉体特論2018 [互換モード]

総研大恒星進化概要.dvi

Numerical Analysis II, Exam End Term Spring 2017

EGunGPU

通信容量制約を考慮したフィードバック制御 - 電子情報通信学会 情報理論研究会(IT) 若手研究者のための講演会

9 8 7 (x-1.0)*(x-1.0) *(x-1.0) (a) f(a) (b) f(a) Figure 1: f(a) a =1.0 (1) a 1.0 f(1.0)

1 I

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)


December 28, 2018

Microsoft Word - 11問題表紙(選択).docx

多体問題

news

, , B 305, ,


D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j

H.Haken Synergetics 2nd (1978)

impulse_response.dvi

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (

‘¬”R.qx

keisoku01.dvi

薄膜結晶成長の基礎4.dvi

TOP URL 1

(1) 1 y = 2 = = b (2) 2 y = 2 = 2 = 2 + h B h h h< h 2 h

Powered by TCPDF ( Title 第 11 講 : フィッシャー統計学 II Sub Title Author 石川, 史郎 (Ishikawa, Shiro) Publisher Publication year 2018 Jtitle コペンハーゲン解

H 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [

外国語外国文学論集

1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc

Transcription:

CMP Technical Report No. 4 Department of Computational Nanomaterials Design ISIR, Osaka University

2 2................................. 2.2......................... 2 3 3 3................................ 3 3.2............................... 4 3.3.................................. 6 3.4............................... 8 4

MDS MDS u i (t) MDS Si 64 2 2. u i (t) u i (t) q u i (t) = q u i (q)e iω qt () u i (t) u i (q) u i (q) q u ( q) 2 = i u i (q) u i (q) (2) ū 2 ū 2 g(ω) ω ω + ω { 2 Mω2 ū(ω) 2 = n(ω) + } ωg(ω) (3) 2 3 n(ω) ū(ω) 3 ω

2.2 3 u i (t)u j () (4) J ij (ω) = u i (t)u j () e iωt dt (5) G(u i (t), u j ()) = i T [u i (t), u j ()] (6) E vib = D ij u i u j (7) 2 i,j { } E vib = 2 lim I D i,j G(u i (t), u j ()) (8) t + i,j 8 6 G(i, j; ω) ω G(ω) E vib = dω ω 2 I {G(ω)} (9) 2 π I {G(ω)} = {(n(ω) + ) + n(ω)} g(ω) () 2ω 9 3 4 5 G R (ω) J ij (ω) = 2[n(ω) + ]I { G R (ω) } () 2

ω k 3 Si 64 ==================== K-Space Setup ============================== k sampling point set Nkpts = No NM index in p in c A/gmin i/o star WTK GM / 2 / 2.. sum of wtk =. ==================== PW_Expansion ============================== Cutoff in the reciprocal space am =.74 (rel. units) kcut = 3.28776 (ab^-) with 2Pi Ecut =.893 (Ry) UNIT of K.362 (a.u.) Planewave expansion with NHDIM = 585 Name: GM / 2 Nstr= WTK=. NPW= 585 INV= Sum over WTK. 3. MDS MDS parameters ============================ time step :. (Rysec) = 4.8378 (fsec) max iterations : 25 Simulation time :.29 (psec) mds resume : NO way to give v : random initial T_atom : 2. (K) =.76 (Ry) Thermo control : OFF T at () = 2 [K] (2) T 6 K T span =.29 [ps] 3

MDS noctrl dt = 25 Etot.4.2 -.2-7.2-7.4-7.6-7.8-8 -8.2 Log E -.4 2 4 6 8-8.4.5.5 Epot.4.3.4.3 Ekin.2 2 4 6 8 ITER.2 : Time evolution of energies of Si 64. t=25 Rysec. ν = 27.28 [cm ] (3) t t =25 Rysec 2 t = Rysec 3 3.2 MDS T span t = 25 Rysec f = [/Rysec] 25 α/2 = 25 = [a B 25 274.74 ] = 27.28 [cm ] (4) 4

MDS noctrl dt = -6.4 Etot -. -6.6-6.8-7 Log E -7.2 -.2-7.4 5 5 2 25-55..5 Epot -55.2-55.3.4.3 Ekin -55.4.2 5 5 2 25 ITER 2: Time evolution of energies of Si 64. t= Rysec..2 (ps). u x -. -.2 5 5 2 25 STEP 3: Displacement of atom, u x in relative units. The black line shows data obained by setting t= Rysec, while the red line 25 Rysec. 5

<u(ω) - 2 > [a.u.^2] x -3 5 4 3 2 Frequency t=25 5 5 2 25 t= - ( x 27.28 [cm ]) 4: Fourier spectra of atomic displacements. The black line shows data of t=25, while the red line of t= Rysec. 4 27.28 cm 4 2 u i (ω) 4 MDS t 3.3 g(ω) 3 g(ω) T T at (t) T = 522 [K] (5) T span 2.4 ps ū 2 (ω) 5 6

<u(ω) - 2 > [a.u.^2] x -3 5 4 3 2 T=522 K 274 2 3 4 5 - Frequency ( x 3.64 [cm ]) 5: Power spectra at different temperatures. [a.u.]. T=522 K 274 g(ω) 2 3 4 5 6 ω 6: Phonon density of states g(ω) [cm -] 7

3 /ω g(ω) 6 g(ω) 52 cm TO 2 cm A 5 ū 2 g(ω) g(ω) 3.4 k i i i = lκα l κ α Si κ =, 2 u κα (kω) = l exp[ik R l ]u lκα (ω) (6) κ S α (kω) = κ u κα(kω)u κα (kω) (7) T = 522 K 2.4 ps 8

<u(ω) - 2 > [a.u.^2] x -4 k=(,,) (/2,,) (,,) 2 3 4 5 - Frequency ( x 3.64 [cm ]) 7: k dependent phonon peaks. Shown is the longitudinal branch along the line. k is expressed in 2π/a. x -4 <u(ω) - 2 > [a.u.^2] (,,) k=(,,) (/2,,) 2 3 4 5 - Frequency ( x 3.64 [cm ]) 8: The transverse branch along the line. k is expressed in 2π/a. 9

7 k = (,, ) x L X (u x (ω), u 2x (ω)) Γ x 39 {-.553, -.66386} {-.683262 +.5988 i,.685896 -.4392 i} LA LO X x 3 {.462358+.246 i, -.23872 +.29449 i} 8 k = (,, ) z T 7 8 k = (,, ) k /2 (2π/a ) 4 ω k