Laves A-B AB 2 MgCu 2 (C14) MgZn 2 (C15) MgNi 2 (C36) Laves VASP ZrCr 2 Laves VASP(Vienna Ab-initio Simulation Package) Laves Energy-Volume Quasi-Harm

Size: px
Start display at page:

Download "Laves A-B AB 2 MgCu 2 (C14) MgZn 2 (C15) MgNi 2 (C36) Laves VASP ZrCr 2 Laves VASP(Vienna Ab-initio Simulation Package) Laves Energy-Volume Quasi-Harm"

Transcription

1 ZrCr 2 Laves

2 Laves A-B AB 2 MgCu 2 (C14) MgZn 2 (C15) MgNi 2 (C36) Laves VASP ZrCr 2 Laves VASP(Vienna Ab-initio Simulation Package) Laves Energy-Volume Quasi-Harmonic Energy-Volume Phonon-DOS Phonon-DOS Energy-Volume Phonon-DOS 2 Phonon-DOS

3 MedeA Phonon-DOS Morse B(bulk modulus) D(θ D ) F Energy-Volume Phonon-DOS F

4 1 Laves A-B AB 2 Laves 200 Laves MgCu 2 (C14) MgZn 2 (C15) MgNi 2 (C36) FrankKasper [2] 1.1 C14 AB C36 ABAC A C15 ABC C14 12 C15 C36 24 [3] C15 C14 [4] Laves 1.1: C14 C15 C36 prototype MgCu 2 MgZn 2 MgNi AB ABC ABAC 2

5 1.1: ZrCr 2 Laves 3

6 2 2.1 VASP(Vienna Ab-initio Simulation Package) PAW VASP 2.2 MedeA VASP MedeA MedeA 1 Windows Phonon-DOS Phonon MedeA Phonon 4

7 Morse Moruzzi[5] quasi-harmonic a b c λ E(r) = a + be λr + ce λr (2.1) Morse E(r) = A 2De λ(r r 0) + De 2λ(r r 0) (2.2) a = A (2.3) b c = 2e λr 0 (2.4) D = b2 4c (2.5) B(bulk modulus) 2.5 B(bulk modulus) x = e λr (2.6) 2.1 E(x) = a + bx + cx 2 (2.7) 5

8 P P = de (2.8) dv V P = de/dx dv/dx (2.9) V = 4π 3 r3 (2.10) 2.6 r V r = ln x λ (2.11) V = 4π 3 ( ln x λ )3 (2.12) x 2.9 P P = xλ3 (b + 2cx) (2.13) 4π(ln x) 2 B P B = V dp dv B = dp/dx dv/dx (2.14) (2.15) B = xλ3 2 [(b + 4cx) (b + 2cx)] (2.16) 12π ln x ln x B θ D 6

9 2.6 v s g(ω) ω = v s k (2.17) g(ω) = V k2 2π 2 dk dω = V ω2 2π 2 v 3 s (2.18) v s ω k 1/v 3 s g(ω) = V ω2 2π 2 ( 1 v 3 l + 2 ) (2.19) vt 3 longitudinaltransverse v t v l 1 v 3 s = 1 v 3 l + 2 v 3 t (2.20) v s ω D ωd g(ω)dω = 3 (2.21) 0 ( V 1 6π 2 vl v 3 l 3 v 3 = 2 v 3 t ) + 1 v 3 l ω 3 D = 3 (2.22) (2.23) 7

10 2.22 V 6π 2 v 3 ωd 3 = 1 (2.24) v ρ B ρ v = B ρ ρ = M V (2.25) (2.26) V V = 4 3 πr3 (2.27) v = πr3 B (2.28) ω 3 D = 6π 2 3 4πr 3 v3 = 6π 2 ( 4 3 πr3 ) 1 ( 4πr3 3 ) 3 2 ( B M ) 3 2 (2.29) ω D = (6π 2 ) ( 3 πr3 ) 1 B 6 M (2.30) k B θ D = h 2π ω D (2.31) k B ω D k B Boltzmann s constants h Plank s cpnstants ω D Debye frequency 2.30 θ D = h (6π 2 ) ( 2πk B 3 πr3 ) 1 B 6 M rb θ D = M v t v l ρ v t = S ρ (2.32) (2.33) (2.34) 8

11 v l = L ρ (2.35) Molzzi S = 0.30B L = 1.42B B v = (2.36) ρ v 2.30 θ D (θ D ) 0 = r0 B M (2.37) r 0 [a.u.] M B r 0 [kbar] 2.7 D(θ D ) D(y) = 3 y 3 y 0 e x x 4 dx (2.38) (e x 1) 2 D(y) y 0 D 1 y = θ D 2.8 F E(r, T ) = E(r) + E D (r, T ) T S D (r, T ) (2.39) T E D S D E D (r, T ) = E 0 + 3k B T D ( ) θd [ ( ) ] 4 S D (r, T ) = 3k B 3 D θd ln(1 e θ DT ) T T (2.40) (2.41) E 0 = 9 8 k Bθ D (2.42) 9

12 E(r, T ) = E(r) k B T [ D ( θd T ) 3 ln(1 e θ DT ) ] k Bθ D (2.43) 2.9 1K β = α 3 (2.44) 1/3 r 0 [a.u.] T[K] α(t ) = 1 r 0 dr 0 dt (2.45) 10

13 3 ZrCr 2 C14 C15 C36 3 Laves VASP C14 C15 C36 Energy-Volume Phonon-DOS Energy-Volume Moruzzi 3.1 Energy-Volume ZrCr 2 Energy-Volume E-V x y y () x y Zr1 Cr // 2 5 //C14 infile1:=[[0.925, ], [0.95, ], [0.975, ], [1.0, ], [1.025, ]] //C15 11

14 infile2:=[[0.95, ], [0.975, ], [1.0, ], [1.025, ], [1.05, ]] //C36 infile3:=[[1.0, ], [1.025, ], [1.05, ], [1.075, ], [1.1, ]] //MedeA C14 V1:=Vector([ , , ]): V2:=Vector([ , , ]): V3:=Vector([ , , ]): //C15 V4:=Vector([ , , ]): V5:=Vector([ , , ]): V6:=Vector([ , , ]): //C36 V7:=Vector([ , , ]): V8:=Vector([ , , ]): V9:=Vector([ , , ]): // ; V1.(CrossProduct(V2,V3)); V4.CrossProduct(V5,V6); V7.CrossProduct(V8,V9); //3 ^3* 0 *3 for i from 1 to nops(infile1) do infile1[i,1]:=infile1[i,1]^3* /12*3; end do: for i from 1 to nops(infile2) do infile2[i,1]:=infile2[i,1]^3* /6*3; 12

15 end do: for i from 1 to nops(infile3) do infile3[i,1]:=infile3[i,1]^3* /24*3; end do: for j from 1 to nops(infile1) do infile1[j,2]:=infile1[j,2]/12*3; end do: for j from 1 to nops(infile2) do infile2[j,2]:=infile2[j,2]/6*3; end do: for j from 1 to nops(infile3) do infile3[j,2]:=infile3[j,2]/24*3; end do: //4 fitting data1:=convert(transpose(convert(infile1,array)),listlist): fit1:=fit[leastsquare[[x,y], y=c0+c4*x+c1*x^2+c2*x^3+c3*x^4, {c0,c1,c2,c3,c4,c5}]](data1): fit_c14:=unapply(rhs(fit1),x); data2:=convert(transpose(convert(infile2,array)),listlist): fit2:=fit[leastsquare[[x,y], y=c0+c4*x+c1*x^2+c2*x^3+c3*x^4, {c0,c1,c2,c3,c4,c5}]](data2): fit_c15:=unapply(rhs(fit2),x); data3:=convert(transpose(convert(infile3,array)),listlist): fit3:=fit[leastsquare[[x,y], y=c0+c4*x+c1*x^2+c2*x^3+c3*x^4, {c0,c1,c2,c3,c4,c5}]](data3): fit_c36:=unapply(rhs(fit3),x); // d1:=display(p1,plot(fit_c14(x),x=30..60,color=red)); d2:=display(p3,plot(fit_c15(x),x=30..60,color=blue)); d3:=display(p5,plot(fit_c36(x),x=30..60,color=green)); display(d1,d2,d3,view=[40..52, ]); 13

16 3.1: E-V fitting 3.1 C15 C36 C14 Pavlu [4] 3.2 Phonon-DOS Phonon-DOS Phonon-DOS VASP A(T) Phonon-DOS C14 Laves

17 Cv : vibrational heat capacity at constant volume. E(T)-E(0) : the change in vibrational internal energy from 0 K. E(0) is the zero point energy (ZPE). S(T) : the vibrational entropy at temperature T. -(A(T)-E(0)) : the change in the vibrational Helmholtz free energy from 0 K. E(T) : the electronic plus vibrational energy of formation. A(T) : the electronic plus vibrational Helmholtz free energy. 3.1: C14 Phonon-DOS T Cv E(T)-E(0) S(T) -(A(T)-E(0)) E(T) A(T) K J/K/mol kj/mol J/K/mol kj/mol kj/mol kj/mol

18 // pho_c14:=[[0., ], [1., ], [10., ], [100., ], [200., ], [300., ], [400., ], [500., ], [600., ], [700., ], [800., ], [900., ], [1000., ], [1100., ], [1200., ], [1300., ], [1400., ], [1500., ], [1600., ], [1700., ], [1800., ], [1900., ], [2000., ]]; pho_c15:=[[0., ], [1., ], [10., ], [100., ], [200., ], [300., ], [400., ], [500., ], [600., ], [700., ], [800., ], [900., ], [1000., ], [1100., ], [1200., ], [1300., ], [1400., ], [1500., ], [1600., ], [1700., ], [1800., ], [1900., ], [2000., ]]; pho_c36:=[[0., ], [1., ], [10., ], [100., ], [200., ], [300., ], [400., ], [500., ], [600., ], [700., ], [800., ], [900., ], [1000., ], [1100., ], [1200., ], [1300., ], [1400., ], [1500., ], [1600., ], [1700., ], [1800., ], [1900., ], [2000., ]]; //0 Zr2 Cr1 //[Ry] [ev] for i from 1 to nops(pho_c14) do pho_c14[i,2]:=(pho_c14[i,2] )*1000/4.1855/23060; pho_c15[i,2]:=(pho_c15[i,2] )*1000/4.1855/23060; pho_c36[i,2]:=(pho_c36[i,2] )*1000/4.1855/23060; end do: pho_c14; pho_c15; pho_c36; // poi1:=pointplot(pho_c14,color=red,legend="c14_phonon",symbol=box): 16

19 p1:=pointplot(pho_c14,color=red,connect=true): poi2:=pointplot(pho_c15,color=blue,legend="c15_phonon",symbol=circle): p2:=pointplot(pho_c15,color=blue,connect=true): poi3:=pointplot(pho_c36,color=green,legend="c36_phonon",symbol=cross): p3:=pointplot(pho_c36,color=green,connect=true): display(p1,p2,p3,poi1,poi2,poi3,view= , labels=["temperature[k]","free Energy[eV/ZrCr2 atoms]"], labeldirections=[horizontal,vertical]); 3.2: Phonon-DOS Phonon-DOS 0K 2000K 100K 3.2 C15 C K Pavlu 17

20 3.3 Energy-Volume ZrCr 2 C14 C15 C36 3 Laves 0K 2000K 100K r C14 C15 C36 // // 5 //C14 infile1:=[[0.925, ], [0.95, ], [0.975, ], [1.0, ], [1.025, ]] //C15 infile2:=[[0.95, ], [0.975, ], [1.0, ], [1.025, ], [1.05, ]] //C36 infile3:=[[1.0, ], [1.025, ], [1.05, ], [1.075, ], [1.1, ]] //C14 ; V1.(CrossProduct(V2,V3)); //C15 ; V4.CrossProduct(V5,V6); //C36 ; V7.CrossProduct(V8,V9); //MedeA 18

21 r_c14:=3.031; r_c15:=2.910; r_c36:=2.817; // C14 r [ ] [a.u.] for i from 1 to nops(infile1) do infile1[i,1]:=infile1[i,1]*r_c14*1.89: end do: //C14 E (3 )[ev] [Ry] for j from 1 to nops(infile1) do infile1[j,2]:=infile1[j,2]/12*3/ ; end do: //4 fitting data1:=convert(transpose(convert(infile1,array)),listlist): fit1:=fit[leastsquare[[x,y], y=c0+c4*x+c1*x^2+c2*x^3+c3*x^4, {c0,c1,c2,c3,c4}]](data1): fit_c14:=unapply(rhs(fit1),x): fit_c14(x); // p1:=pointplot(infile1,color=red,legend="c14",symbol=box): p2:=pointplot(infile1,color=red,connect=true): display(p1,p2); d1:=display(p1,plot(fit_c14(x),x= ),color=black): display(d1,view= ,labels=["[a.u.]"," [Ry]"], labeldirections=[horizontal,vertical]); C15 C36 Fitting 19

22 3.3: C14 Laves 3.4: C15 Laves 20

23 3.5: C36 Laves [KBar] [Ry/(a.u.) 3 ] [ev/ 3 ] [GPa] [kbar] [1] //r0( ) r0:=fsolve(diff(fit_c14(x),x)=0,x= ); //r1(medea 0 r[ ] r[a.u.] ) r1:=3.031*1.89; //MedeA 3 V1:= /12*3*1.89^3; //V0 V //r 1/3 //dv V0:=(r0/r1)^3*V1: 21

24 V:=r->r^3*V0/(r0)^3: r:=v->(v/ )^(1/3): dv:=unapply(1/(diff(v(r),r)),r); // B(C14):=unapply(V(r)*dV(r)*diff(dV(r)*diff(fit_C14(r),r),r) * /( )^3* ,r); // B1:=display(plot(B(C14)(x),x=1..5),color=black): display(b1); C15 C // B(C14)(r0); B(C15)(r0); B(C36)(r0);

25 3.6: //ZrCr2 M Zr1 Cr2 M= *2; // thetad_c14:=unapply(41.63*(r2*b(c14)(r2)/m)^(1/2),r2): thetad_c15:=unapply(41.63*(r2*b(c15)(r2)/m)^(1/2),r2): thetad_c36:=unapply(41.63*(r2*b(c36)(r2)/m)^(1/2),r2): // B10:=display(plot(thetaD_C14(r2),r2= ,color=red,legend="C14")): B11:=display(plot(thetaD_C15(r2),r2= ,color=blue,legend="C15")): B12:=display(plot(thetaD_C36(r2),r2= ,color=green,legend="C36")): display(b10,b11,b12,labels=[" [a.u.]"," [K]"], 23

26 labeldirections=[horizontal,vertical]); 3.7: 24

27 // f1:=unapply(exp(x)*x^4/(exp(x)-1)^2,x): Debye:=unapply(3/y^3*Int(f1(x),x=0..y),y): Df_C14:=unapply(Re(evalf(Debye(thetaD_C14(r)/T))),r,T): Df_C15:=unapply(Re(evalf(Debye(thetaD_C15(r)/T))),r,T): Df_C36:=unapply(Re(evalf(Debye(thetaD_C36(r)/T))),r,T): // B100:=display(plot(Df_C14(r,300),r= ,color=red,legend="C14")): B110:=display(plot(Df_C15(r,300),r= ,color=blue,legend="C15")): B120:=display(plot(Df_C36(r,300),r= ,color=green,legend="C36")): display(b100,b110,b120,view= ,labels=[" [a.u.]"," D( D)"], labeldirections=[horizontal,vertical]); 25

28 3.8: F E-V 2.43 //(kb/ry) Ry // Ry ev >FreeEnergy_C14:=unapply((fit_C14(r)-(8.617*10^(-5)/ ) *T*(Df(r,T)-3*ln(1-exp(-thetaD(r)/T))) +(9/8)*(8.617*10^(-5)/ )*thetaD(r))* ,r,T): // [K] p11:=plot(freeenergy(r5,1),r5= ): p12:=plot(freeenergy(r5,100),r5= ,color=blue): p13:=plot(freeenergy(r5,200),r5= ,color=green): p14:=plot(freeenergy(r5,400),r5= ,color=black): p15:=plot(freeenergy(r5,800),r5= ,color=yellow): 26

29 // p1:=evalf(freeenergy(r0,1)); p2:=evalf(freeenergy(r0,100)); p3:=evalf(freeenergy(r0,200)); p4:=evalf(freeenergy(r0,400)); p5:=evalf(freeenergy(r0,800)); p30:=pointplot([evalf(fsolve(diff(freeenergy(x,1),x)=0,x= )),p1],symbolsize=20); p31:=pointplot([evalf(fsolve(diff(freeenergy(x,100),x)=0,x= )),p2],symbolsize=20); p32:=pointplot([evalf(fsolve(diff(freeenergy(x,200),x)=0,x= )),p3],symbolsize=20); p33:=pointplot([evalf(fsolve(diff(freeenergy(x,400),x)=0,x= )),p4],symbolsize=20); p34:=pointplot([evalf(fsolve(diff(freeenergy(x,800),x)=0,x= )),p5],symbolsize=20); display(p11,p12,p13,p14,p15,p30,p31,p32,p33,p34,labels=[" [a.u.]"," [ev/zrcr2 atoms]"],labeldirections=[horizontal,vertical]); 3.9 T = [K] C

30 3.9: T = [K] C14 [ev/zrcr 2 atoms] 3.2: C14 T [K] r 0 [a.u.] F [ev/zrcr 2 atoms]

31 C15 C36 T = [K] 3.10: T = [K] C15 [ev/zrcr 2 atoms] 3.3: C15 T [K] r 0 [a.u.] F [ev/zrcr 2 atoms]

32 3.11: T = [K] C36 [ev/zrcr 2 atoms] 3.4: C36 T [K] r 0 [a.u.] F [ev/zrcr 2 atoms]

33 //0K 2000K 100K tmp:=[]: ttmp:=[]: tttmp:=[]: for i from 0 to 2000 by 100 do tmp:=[op(tmp),[i,0]]: ttmp:=[op(ttmp),[i,0]]: tttmp:=[op(tttmp),[i,0]]: end do: //0K tmp[1,2]:= ; ttmp[1,2]:= ; tttmp[1,2]:= ; for j from 2 to nops(tmp) do tmp[j,2]:= ((tmp_c14[1,2]-tmp_c14[j,2])); ttmp[j,2]:= ((tmp_c15[1,2]-tmp_c15[j,2])); tttmp[j,2]:= ((tmp_c36[1,2]-tmp_c36[j,2])); end do: // ppp1:=listplot(tmp,color=red,legend="c14"); ppp2:=listplot(ttmp,color=blue,legend="c15"); ppp3:=listplot(tttmp,color=green,legend="c36"); display(ppp1,ppp2,ppp3,labels=[" [K]"," [ev/zrcr2 atoms]"], labeldirections=[horizontal,vertical]); Energy-Volume 3.12 Phonon-DOS K 31

34 3.12: 3.4 0K 1500K 100K ZrCr 2 Laves //0[K] 1500[K] 100[K] tmp2:=[[0,evalf(fsolve(diff(freeenergy(x,1),x)=0,x= ))]]: for i from 100 to 1500 by 100 do tmp2:=[op(tmp2),[i,evalf(fsolve(diff(freeenergy(x,i),x)=0,x= ))]]; end do: //0[K] 1500[K] 100[K] tmp:=[]: for i from 0 to 1500 by 100 do tmp:=[op(tmp),[i,0]]: end do: 32

35 ///100[K] for j from 2 to nops(tmp) do tmp[j,2]:=(((tmp2[j,2])/tmp2[j-1,2])^3-1)/100; end do: //5 fitting data11:=convert(transpose(convert(tmp,array)),listlist): fit11:=fit[leastsquare[[x,y], y=c0+c1*x+c2*x^2+c3*x^3+c4*x^4+c5*x^5]](data11): fit_b1:=unapply(rhs(fit11),x): // pp:=plot(fit_b1(x),x= ,labels=[" [K]"," [10^-5/K]"], labeldirections=[horizontal,vertical]); 3.13: 33

36 3.14: Kellow C C36 C15 C14 Kellow C [6] 34

37 4 ZrCr 2 Laves Phonon-DOS Quasi- Harmonic Phonon-DOS Energy-Volume Phonon-DOS 35

38 [1] 2006 [2] 1975 [3] F. Stein, M. Palm, G. Sauthoff, Structure and stability of Laves phases, Part I. Critical assessment of factors controlling Laves phase stability, Intermetallics 12, (2004), [4] J. Pavlu, J. Vrest al, M.Sob, Stability of Laves Phases in the CrZr System, CALPHAD-COMPUTER COUPLING OF PHASE DIAGRAMS AND THERMOCHEMISTRY, [5] V. L. Moruzzi, J. F. Janak, Calculated thermal properties of metals, The American Phisical Society, [6] A. Kellow, T. Grosdidier, C. Coddet, H. Aourag, Theoretical study of structural, electronic, and thermal properties of Cr2(Zr,Nb) Laves alloys, Acta Meterialia Inc,

39 37

phonopy phonon-dispersion phonon-dos MedeA phonopy phonopy, phonopy MedeA phonopy MedeA Moruzzi Quasi-harmonic MedeA,Quasi-harmonic phonopy phonopy.

phonopy phonon-dispersion phonon-dos MedeA phonopy phonopy, phonopy MedeA phonopy MedeA Moruzzi Quasi-harmonic MedeA,Quasi-harmonic phonopy phonopy. phonopy phonon 1513 2015 3 phonopy phonon-dispersion phonon-dos MedeA phonopy phonopy, phonopy MedeA phonopy MedeA Moruzzi Quasi-harmonic MedeA,Quasi-harmonic phonopy phonopy. 1 3 1.1 phonopy.....................................

More information

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)

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) 1 9 v..1 c (216/1/7) Minoru Suzuki 1 1 9.1 9.1.1 T µ 1 (7.18) f(e ) = 1 e β(e µ) 1 E 1 f(e ) (Bose-Einstein distribution function) *1 (8.1) (9.1) E E µ = E f(e ) E µ (9.1) µ (9.2) µ 1 e β(e µ) 1 f(e )

More information

(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

(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 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

iBookBob:Users:bob:Documents:CurrentData:flMŠÍ…e…L…X…g:Statistics.dvi

iBookBob:Users:bob:Documents:CurrentData:flMŠÍ…e…L…X…g:Statistics.dvi 4 4 9............................................... 3.3......................... 4.4................. 5.5............................ 7 9..................... 9.............................3................................4..........................5.............................6...........................

More information

() 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)

() 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) 0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

More information

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

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 II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =

More information

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 ( 1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

More information

Gmech08.dvi

Gmech08.dvi 145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

More information

08-Note2-web

08-Note2-web r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)

More information

/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat

/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat / Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiation and the Continuing Failure of the Bilinear Formalism,

More information

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5. A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

More information

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E

5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 2, S 1 N 1 = S 2 N 2 2 (chemical potential) µ S N

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

More information

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

SFGÇÃÉXÉyÉNÉgÉãå`.pdf SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180

More information

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

More information

untitled

untitled 1 Physical Chemistry I (Basic Chemical Thermodynamics) [I] [II] [III] [IV] Introduction Energy(The First Law of Thermodynamics) Work Heat Capacity C p and C v Adiabatic Change Exact(=Perfect) Differential

More information

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2) 3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)

More information

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

More information

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) 4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7

More information

I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co

I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co 16 I ( ) (1) I-1 I-2 I-3 (2) I-1 ( ) (100 ) 2l x x = 0 y t y(x, t) y(±l, t) = 0 m T g y(x, t) l y(x, t) c = 2 y(x, t) c 2 2 y(x, t) = g (A) t 2 x 2 T/m (1) y 0 (x) y 0 (x) = g c 2 (l2 x 2 ) (B) (2) (1)

More information

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H 199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

More information

( ) ( )

( ) ( ) 20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))

More information

chap03.dvi

chap03.dvi 99 3 (Coriolis) cm m (free surface wave) 3.1 Φ 2.5 (2.25) Φ 100 3 r =(x, y, z) x y z F (x, y, z, t) =0 ( DF ) Dt = t + Φ F =0 onf =0. (3.1) n = F/ F (3.1) F n Φ = Φ n = 1 F F t Vn on F = 0 (3.2) Φ (3.1)

More information

master.dvi

master.dvi 4 Maxwell- Boltzmann N 1 4.1 T R R 5 R (Heat Reservor) S E R 20 E 4.2 E E R E t = E + E R E R Ω R (E R ) S R (E R ) Ω R (E R ) = exp[s R (E R )/k] E, E E, E E t E E t E exps R (E t E) exp S R (E t E )

More information

Contents 1 Jeans (

Contents 1 Jeans ( Contents 1 Jeans 2 1.1....................................... 2 1.2................................. 2 1.3............................... 3 2 3 2.1 ( )................................ 4 2.2 WKB........................

More information

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a

More information

Untitled

Untitled II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (

2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( ( (. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2

More information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

¼§À�ÍýÏÀ – Ê×ÎòÅŻҼ§À�¤È¥¹¥Ô¥ó¤æ¤é¤® - No.7, No.8, No.9

¼§À�ÍýÏÀ – Ê×ÎòÅŻҼ§À�¤È¥¹¥Ô¥ó¤æ¤é¤® - No.7, No.8, No.9 No.7, No.8, No.9 email: takahash@sci.u-hyogo.ac.jp Spring semester, 2012 Introduction (Critical Behavior) SCR ( b > 0) Arrott 2 Total Amplitude Conservation (TAC) Global Consistency (GC) TAC 2 / 25 Experimental

More information

Yuzo Nakamura, Kagoshima Univ., Dept Mech Engr. perfect crystal imperfect crystal point defect vacancy self-interstitial atom substitutional impurity

Yuzo Nakamura, Kagoshima Univ., Dept Mech Engr. perfect crystal imperfect crystal point defect vacancy self-interstitial atom substitutional impurity perfect crystal imperfect crystal point defect vacancy self-interstitial atom substitutional impurity atom interstitial impurity atom line defect dislocation planar defect surface grain boundary interface

More information

スライド 1

スライド 1 Matsuura Laboratory SiC SiC 13 2004 10 21 22 H-SiC ( C-SiC HOY Matsuura Laboratory n E C E D ( E F E T Matsuura Laboratory Matsuura Laboratory DLTS Osaka Electro-Communication University Unoped n 3C-SiC

More information

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

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 19 σ = P/A o σ B Maximum tensile strength σ 0. 0.% 0.% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional limit ε p = 0.% ε e = σ 0. /E plastic strain ε = ε e

More information

1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................

More information

4 2 Rutherford 89 Rydberg λ = R ( n 2 ) n 2 n = n +,n +2, n = Lyman n =2 Balmer n =3 Paschen R Rydberg R = cm 896 Zeeman Zeeman Zeeman Lorentz

4 2 Rutherford 89 Rydberg λ = R ( n 2 ) n 2 n = n +,n +2, n = Lyman n =2 Balmer n =3 Paschen R Rydberg R = cm 896 Zeeman Zeeman Zeeman Lorentz 2 Rutherford 2. Rutherford N. Bohr Rutherford 859 Kirchhoff Bunsen 86 Maxwell Maxwell 885 Balmer λ Balmer λ = 364.56 n 2 n 2 4 Lyman, Paschen 3 nm, n =3, 4, 5, 4 2 Rutherford 89 Rydberg λ = R ( n 2 ) n

More information

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P 6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

More information

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)

More information

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9

More information

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e 7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z

More information

Mott散乱によるParity対称性の破れを検証

Mott散乱によるParity対称性の破れを検証 Mott Parity P2 Mott target Mott Parity Parity Γ = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 t P P ),,, ( 3 2 1 0 1 γ γ γ γ γ γ ν ν µ µ = = Γ 1 : : : Γ P P P P x x P ν ν µ µ vector axial vector ν ν µ µ γ γ Γ ν γ

More information

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

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,. 24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)

More information

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

More information

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4 1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev

More information

SiC Si Si Si 3 SiC Si SiC Metastable Solvent Epitaxy MSE MSE SiC MSE SiC SiC 3C,4H,6H-SiC 4H-SiC(11-20) 4H-SiC (0001) Si Si C 3 C 1

SiC Si Si Si 3 SiC Si SiC Metastable Solvent Epitaxy MSE MSE SiC MSE SiC SiC 3C,4H,6H-SiC 4H-SiC(11-20) 4H-SiC (0001) Si Si C 3 C 1 SiC 4718 20 2 21 SiC Si Si Si 3 SiC Si SiC Metastable Solvent Epitaxy MSE MSE SiC MSE SiC SiC 3C,4H,6H-SiC 4H-SiC(11-20) 4H-SiC (0001) Si Si C 3 C 1 1 4 1.1.................... 4 1.2 SiC........................

More information

d > 2 α B(y) y (5.1) s 2 = c z = x d 1+α dx ln u 1 ] 2u ψ(u) c z y 1 d 2 + α c z y t y y t- s 2 2 s 2 > d > 2 T c y T c y = T t c = T c /T 1 (3.

d > 2 α B(y) y (5.1) s 2 = c z = x d 1+α dx ln u 1 ] 2u ψ(u) c z y 1 d 2 + α c z y t y y t- s 2 2 s 2 > d > 2 T c y T c y = T t c = T c /T 1 (3. 5 S 2 tot = S 2 T (y, t) + S 2 (y) = const. Z 2 (4.22) σ 2 /4 y = y z y t = T/T 1 2 (3.9) (3.15) s 2 = A(y, t) B(y) (5.1) A(y, t) = x d 1+α dx ln u 1 ] 2u ψ(u), u = x(y + x 2 )/t s 2 T A 3T d S 2 tot S

More information

( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1

( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1 (3.5 3.8) 03032s 2006.7.0 n (n = 0,,...) n n = δ nn n n = I n=0 ψ = n C n n () C n = n ψ α = e 2 α 2 n=0 α, β α n n (2) β α = e 2 α 2 2 β 2 n=0 =0 = e 2 α 2 β n α 2 β 2 n=0 = e 2 α 2 2 β 2 +β α β n α!

More information

dynamics-solution2.dvi

dynamics-solution2.dvi 1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj

More information

n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)

n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1) D d dx 1 1.1 n d n y a 0 dx n + a d n 1 y 1 dx n 1 +... + a dy n 1 dx + a ny = f(x)...(1) dk y dx k = y (k) a 0 y (n) + a 1 y (n 1) +... + a n 1 y + a n y = f(x)...(2) (2) (2) f(x) 0 a 0 y (n) + a 1 y

More information

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t) 338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x

More information

nm (T = K, p = kP a (1atm( )), 1bar = 10 5 P a = atm) 1 ( ) m / m

nm (T = K, p = kP a (1atm( )), 1bar = 10 5 P a = atm) 1 ( ) m / m .1 1nm (T = 73.15K, p = 101.35kP a (1atm( )), 1bar = 10 5 P a = 0.9863atm) 1 ( ).413968 10 3 m 3 1 37. 1/3 3.34.414 10 3 m 3 6.0 10 3 = 3.7 (109 ) 3 (nm) 3 10 6 = 3.7 10 1 (nm) 3 = (3.34nm) 3 ( P = nrt,

More information

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0 79 4 4.1 4.1.1 x i (t) x j (t) O O r 0 + r r r 0 x i (0) r 0 x i (0) 4.1 L. van. Hove 1954 space-time correlation function V N 4.1 ρ 0 = N/V i t 80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t

More information

B 1 B.1.......................... 1 B.1.1................. 1 B.1.2................. 2 B.2........................... 5 B.2.1.......................... 5 B.2.2.................. 6 B.2.3..................

More information

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2 filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin

More information

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

More information

85 4

85 4 85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V

More information

29

29 9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n

More information

K E N Z OU

K E N Z OU K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................

More information

215 11 13 1 2 1.1....................... 2 1.2.................... 2 1.3..................... 2 1.4...................... 3 1.5............... 3 1.6........................... 4 1.7.................. 4

More information

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C 8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

最新耐震構造解析 ( 第 3 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 3 版 1 刷発行時のものです.

最新耐震構造解析 ( 第 3 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 3 版 1 刷発行時のものです. 最新耐震構造解析 ( 第 3 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. http://www.morikita.co.jp/books/mid/052093 このサンプルページの内容は, 第 3 版 1 刷発行時のものです. i 3 10 3 2000 2007 26 8 2 SI SI 20 1996 2000 SI 15 3 ii 1 56 6

More information

1 nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC

1   nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC 1 http://www.gem.aoyama.ac.jp/ nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC r 1 A B B C C A (1),(2),, (8) A, B, C A,B,C 2 1 ABC

More information

Si SiO 2. Si 1 VASP Si 1,, Si-Si 0.28Å Si Si-Si 0.19Å Si 166 Si Å Si

Si SiO 2. Si 1 VASP Si 1,, Si-Si 0.28Å Si Si-Si 0.19Å Si 166 Si Å Si 8615 2012 3 Si SiO 2. Si 1 VASP Si 1,, Si-Si 0.28Å Si 160 1 2 Si-Si 0.19Å Si 166 Si 4 0.85Å Si 118 2 4 1 3 1.1 SiO 2... 3 1.2 Si... 3 1.3 Cz... 4 1.4 SiO 2... 4 1.5... 5 2 6 2.1... 6 2.2 VASP(Vienna Ab-initio

More information

1 1.1,,,.. (, ),..,. (Fig. 1.1). Macro theory (e.g. Continuum mechanics) Consideration under the simple concept (e.g. ionic radius, bond valence) Stru

1 1.1,,,.. (, ),..,. (Fig. 1.1). Macro theory (e.g. Continuum mechanics) Consideration under the simple concept (e.g. ionic radius, bond valence) Stru 1. 1-1. 1-. 1-3.. MD -1. -. -3. MD 1 1 1.1,,,.. (, ),..,. (Fig. 1.1). Macro theory (e.g. Continuum mechanics) Consideration under the simple concept (e.g. ionic radius, bond valence) Structural relaxation

More information

2009 2 26 1 3 1.1.................................................. 3 1.2..................................................... 3 1.3...................................................... 3 1.4.....................................................

More information

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π . 4cm 6 cm 4cm cm 8 cm λ()=a [kg/m] A 4cm A 4cm cm h h Y a G.38h a b () y = h.38h G b h X () S() = π() a,b, h,π V = ρ M = ρv G = M h S() 3 d a,b, h 4 G = 5 h a b a b = 6 ω() s v m θ() m v () θ() ω() dθ()

More information

30

30 3 ............................................2 2...........................................2....................................2.2...................................2.3..............................

More information

, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )

, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( ) 81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,.,

More information

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2.

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2. A A 1 A 5 A 6 1 2 3 4 5 6 7 1 1.1 1.1 (). Hausdorff M R m M M {U α } U α R m E α ϕ α : U α E α U α U β = ϕ α (ϕ β ϕβ (U α U β )) 1 : ϕ β (U α U β ) ϕ α (U α U β ) C M a m dim M a U α ϕ α {x i, 1 i m} {U,

More information

C el = 3 2 Nk B (2.14) c el = 3k B C el = 3 2 Nk B

C el = 3 2 Nk B (2.14) c el = 3k B C el = 3 2 Nk B I ino@hiroshima-u.ac.jp 217 11 14 4 4.1 2 2.4 C el = 3 2 Nk B (2.14) c el = 3k B 2 3 3.15 C el = 3 2 Nk B 3.15 39 2 1925 (Wolfgang Pauli) (Pauli exclusion principle) T E = p2 2m p T N 4 Pauli Sommerfeld

More information

QMII_10.dvi

QMII_10.dvi 65 1 1.1 1.1.1 1.1 H H () = E (), (1.1) H ν () = E ν () ν (). (1.) () () = δ, (1.3) μ () ν () = δ(μ ν). (1.4) E E ν () E () H 1.1: H α(t) = c (t) () + dνc ν (t) ν (), (1.5) H () () + dν ν () ν () = 1 (1.6)

More information

IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................

More information

September 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1

September 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1 September 25, 2017 1 1.1 1.2 p = nr = 273.15 + t : : K : 1.3 1.3.1 : e.g. 1.3.2 : 1 intensive variable e.g. extensive variable e.g. 1.3.3 Equation of State e.g. p = nr X = A 2 2.1 2.1.1 Quantity of Heat

More information

物性物理学I_2.pptx

物性物理学I_2.pptx phonon U r U = nαi U ( r nαi + u nαi ) = U ( r nαi ) + () nαi,β j := nαi β j U r nαi r β j > U r nαi r u nαiuβ j + β j β j u β j n α i () nαi,β juβj 調和振動子近似の復習 極 小 値近傍で Tylor展開すると U ( x) = U ( x ) + (

More information

OHO.dvi

OHO.dvi 1 Coil D-shaped electrodes ( [1] ) Vacuum chamber Ion source Oscillator 1.1 m e v B F = evb (1) r m v2 = evb r v = erb (2) m r T = 2πr v = 2πm (3) eb v

More information

untitled

untitled 1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β

More information

( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................

More information

II 2 II

II 2 II II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................

More information

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

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π 4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

More information

d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r

d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r 2.4 ( ) U(r) ( ) ( ) U F(r) = x, U y, U = U(r) (2.4.1) z 2 1 K = mv 2 /2 dk = d ( ) 1 2 mv2 = mv dv = v (ma) (2.4.2) ( ) U(r(t)) r(t) r(t) + dr(t) du du = U(r(t) + dr(t)) U(r(t)) = U x = U(r(t)) dr(t)

More information

0201

0201 2018 10 17 2019 9 19 SI J cal 1mL 1ºC 1999 cal nutrition facts label calories cal kcal 1 cal = 4.184 J heat capacity 1 K 1 J K 1 mol molar heat capacity J K mol (specific heat specific heat capacity) 1

More information

gr09.dvi

gr09.dvi .1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {

More information

(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )

More information

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A .. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.

More information

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq 49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

More information

08 p Boltzmann I P ( ) principle of equal probability P ( ) g ( )g ( 0 ) (4 89) (4 88) eq II 0 g ( 0 ) 0 eq Taylor eq (4 90) g P ( ) g ( ) g ( 0

08 p Boltzmann I P ( ) principle of equal probability P ( ) g ( )g ( 0 ) (4 89) (4 88) eq II 0 g ( 0 ) 0 eq Taylor eq (4 90) g P ( ) g ( ) g ( 0 08 p. 8 4 k B log g() S() k B : Boltzmann T T S k B g g heat bath, thermal reservoir... 4. I II II System I System II II I I 0 + 0 const. (4 85) g( 0 ) g ( )g ( ) g ( )g ( 0 ) (4 86) g ( )g ( 0 ) 0 (4

More information

K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................

More information

. 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

. 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 003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........

More information

CVMに基づくNi-Al合金の

CVMに基づくNi-Al合金の CV N-A (-' by T.Koyama ennard-jones fcc α, β, γ, δ β α γ δ = or α, β. γ, δ α β γ ( αβγ w = = k k k ( αβγ w = ( αβγ ( αβγ w = w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( βγδ w = = k k k ( αγδ

More information

pdf

pdf http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

More information

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1

More information

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c)   yoshioka/education-09.html pdf 1 2009 1 ( ) ( 40 )+( 60 ) 1 1. 2. Schrödinger 3. (a) (b) (c) http://goofy.phys.nara-wu.ac.jp/ yoshioka/education-09.html pdf 1 1. ( photon) ν λ = c ν (c = 3.0 108 /m : ) ɛ = hν (1) p = hν/c = h/λ (2) h

More information

SiC SiC QMAS(Quantum MAterials Simulator) VASP(Vienna Ab-initio Simulation Package) SiC 3C, 4H, 6H-SiC EV VASP VASP 3C, 4H, 6H-SiC (0001) (11 20) (1 1

SiC SiC QMAS(Quantum MAterials Simulator) VASP(Vienna Ab-initio Simulation Package) SiC 3C, 4H, 6H-SiC EV VASP VASP 3C, 4H, 6H-SiC (0001) (11 20) (1 1 QMAS SiC 7661 24 2 28 SiC SiC QMAS(Quantum MAterials Simulator) VASP(Vienna Ab-initio Simulation Package) SiC 3C, 4H, 6H-SiC EV VASP VASP 3C, 4H, 6H-SiC (0001) (11 20) (1 100) MedeA SiC QMAS - C Si (0001)

More information

液晶の物理1:連続体理論(弾性,粘性)

液晶の物理1:連続体理論(弾性,粘性) The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers

More information