δf = δn I [ ( FI (N I ) N I ) T,V δn I [ ( FI N I ( ) F N T,V ( ) FII (N N I ) + N I ) ( ) FII T,V N II T,V T,V ] ] = 0 = 0 (8.2) = µ (8.3) G

Size: px
Start display at page:

Download "δf = δn I [ ( FI (N I ) N I ) T,V δn I [ ( FI N I ( ) F N T,V ( ) FII (N N I ) + N I ) ( ) FII T,V N II T,V T,V ] ] = 0 = 0 (8.2) = µ (8.3) G"

Transcription

1 8 ( ) 8. 1 ( ) F F = F I (N I, T, V I ) + F II (N II, T, V II ) (8.1) F

2 δf = δn I [ ( FI (N I ) N I ) T,V δn I [ ( FI N I ( ) F N T,V ( ) FII (N N I ) + N I ) ( ) FII T,V N II T,V T,V ] ] = 0 = 0 (8.2) = µ (8.3) G F G(N, T, p) = F (N, T, V ) + pv (8.4) T, p N ( ) ( ) ( ) ( ) ( ) G F F V V = + + p N T,p N T,p V N,T N T,p N T,p (8.5) p = ( F/ V ) N,T ( ) G = µ (8.6) N T,p µ G = µn (8.7) 1 T, p N 2 G 2 G(αN, T, p) = αg(n, T, p) (8.8)

3 α 1 G(αN, T, p) G(αN, T, p) = αn G(N, T, p) α αn = N (8.9) α=1 α N µn G (8.7) i G = i µ i N i (8.10) N i x i G = i µ i x i (8.11) G 8. 2 T p n m f = 2 m + n (8.12) mn 1 m 2 + mn m (8.13) i

4 j µ (j) i µ (1) 1 = µ (2) 1 = = µ (m) 1 = µ 1 µ (1) 2 = µ (2) 2 = = µ (m) 2 = µ 2 µ (1) n = µ (2) n = = µ (m) n = µ n (8.14) n(m 1) f f = 2 + mn m n(m 1) = 2 m + n (8.15) f = 0 1 m + n 8. 3 I II (E I 0 < E II 0 ) (K I > K II ) 8.1 I II I II (phase transition) T tr G I (T tr ) = G II (T tr ) (8.16)

5 Phase I Free Energy Ttr Phase II Temperature 8.1 (I) (II) G G = H T tr S = 0 (8.17) H II H I = T tr (S II S I ) (8.18) S II S I > 0 I II II I (latent heat) 1 (first-order transition) 2 (second-order transition) II

6 Fe bcc fcc Debye 8. 4 C p H T H(T ) = H 0 + C p (T )dt (8.19) T 0 H 0 T 0 T 0 S T C p (T ) S(T ) = S 0 + dt (8.20) T 0 T S f 2ncal/K/mol 8.4nJ/K/mol Richards rule n 1 NaCl n = 2 (9 11J/K/n-mol) ( 14J/K/n-mol) ( 30 J/K/n-mol)

7 I (undercooled or supercooled liquid) (metastable equilibrium) Fe-Fe 3 C Tm[I] Ttr Tm[II] 8.2 T tr I II T m[i] I T m[ii] II 8. 6 H 2 O 8.3 H,O 2 H 2 + 1/2O 2 H 2 O

8 [MPa] 611 [Pa] H 2O - (p T diagram) f = 3 m (8.21) 1 m = 1 f = f = 3 2 = m = 3 f = 0 3 (triple point) 2 3 (critical point) A,B 2 2

9 (Clausius-Clapeyron) dp dt = S V (8.22) dg = SdT + V dp (8.23) 2 S A dt + V A dp = S B dt + V B dp (8.24) dp/dt p T p T (homogeneous nucleation) σ (driving force) r (embryo) (droplet model) G = G v 4πr 3 /3 + 4πr 2 σ (8.25) G v (II ) (I ) G v = G I G II = H T S (8.26)

10 e -10 G * r * 5e -07 Gsurface r -4e -10 Gtotal Gvolume 8.4 H S T H tr = H II H I = H T = T tr G = 0 S = H tr T tr (8.27) S, H tr G v = H tr + T H tr T tr = H tr T T tr (8.28) 8.4 (critical radius) r dg/dr = 0 r = 2σ G v = 2σT tr H tr T (8.29) (8.28) (energy barrier or activation barrier G ) G = 16π 3 σ 3 G 2 v = 16π 3 σ 3 T 2 tr (H tr T ) 2 (8.30) Cu Cu erg/cm erg/cm K

11 (over heating) I Z N n I = N nz (8.31) ) Nn exp ( G kt (8.32) G d ( ) I exp ( G + G d ) kt (8.33) exp ( 1/T T 2) exp ( 1/T ) TTT (Time- Temperature-Transformation diagram) 8.5 (amorphous) (inhomogeneous nucleation) 8.6 (substrate:s) (crystal:c) (liquid:l) (contact angle)θ

12 Temperature: T T m time: t TTT T m (melting temperature) liquid σ ls σ lc θ crystal σ cs substrate 8.6 σ ls = σ cs + cos θσ lc (8.34) G hetero = G homo f(θ) = ( G v 4πr 3 /3 + 4πr 2 σ lc ) 2 3 cos θ + cos 3 θ 4 (8.35) (Appendix )

13 θ (wet) (Jackson model) smooth surface facet rough surface non-facet Jackson N N A one layer Z S ( N NA N ) Z S (8.36) N A ɛ ( H = N A 1 N ) A Z s ɛ (8.37) N N N A W = N! N A!(N N A )! (8.38) S = k B ln W (Stirling s approximation) ln N! = N ln N N γ = N A /N S = k B N {(1 γ) ln(1 γ) + γ ln γ} (8.39) Z c L 0

14 α= α= γ α=2 α=1 8.7 L 0 = Z c ɛ (8.40) G Nk B T tr = αγ(1 γ) + {(1 γ) ln(1 γ) + γ ln γ} (8.41) α = L 0 k B T tr Z s Z c (8.42) α 8.7 α 2 KurzFisher) W. Kurz and D. J. Fisher, Fundamentals of Solidification, Trans Tech Publications, 1984, Switzerland. Chalmers) Bruce Chalmers, Principles of Solidification, John Wiley & Sons, Inc., 1964, New York Flemings) Merton C. Flemings, Solidification Processing, McGraw-Hill, 1974, New York.

15 A G interface = A lc σ lc + A cs σ cs A cs σ ls (43) G interface = A lc σ lc + πr 2 (σ cs σ ls ) (44) R = r sin θ σ lc = σ cs + σ ls cos θ (45) G interface = A lc σ lc πr 2 cos θσ ls (46) G interface = G volume + G interface = v c G v + (A lc πr 2 cos θ)σ ls v c (47) v c = πr3 (2 3 cos θ + cos 3 θ) 3 (48) A lc = 2πr 2 (1 cos θ) (49) G hetero = G homo f(θ) = ( G v 4πr 3 /3 + 4πr 2 σ lc ) 2 3 cos θ + cos 3 θ 4 (50) f(θ) = 2 3 cos θ + cos3 θ 4 = (2 + cos θ)(1 cos θ)2 4 (51)

(Jackson model) Ziman) (fluidity) (viscosity) (Free v

(Jackson model) Ziman) (fluidity) (viscosity) (Free v 1) 16 6 10 1) e-mail: nishitani@ksc.kwansei.ac.jp 0. 1 2 0. 1. 1 2 0. 1. 2 3 0. 1. 3 4 0. 1. 4 5 0. 1. 5 6 0. 1. 6 (Jackson model) 8 0. 1. 7 10. 1 10 0. 1 0. 1. 1 Ziman) (fluidity) (viscosity) (Free volume)(

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

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

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

untitled

untitled D nucleation 3 3D nucleation Glucose isomerase 10 V / nm s -1 5 0 0 5 10 C - C e / mg ml -1 kinetics µ R K kt kinetics kinetics kinetics r β π µ π r a r s + a s : β: µ πβ µ β s c s c a a r, & exp exp

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

ver.1 / c /(13)

ver.1 / c /(13) 1 -- 11 1 c 2010 1/(13) 1 -- 11 -- 1 1--1 1--1--1 2009 3 t R x R n 1 ẋ = f(t, x) f = ( f 1,, f n ) f x(t) = ϕ(x 0, t) x(0) = x 0 n f f t 1--1--2 2009 3 q = (q 1,..., q m ), p = (p 1,..., p m ) x = (q,

More information

[ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 i,j S i S j (4.39) i, j z 5 2 z = 4 z = 6 3

[ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 i,j S i S j (4.39) i, j z 5 2 z = 4 z = 6 3 4.2 4.2.1 [ ] (Ising model) 2 i S i S i = 1 (up spin : ) = 1 (down spin : ) (4.38) s z = ±1 4 H 0 = J zn/2 S i S j (4.39) i, j z 5 2 z = 4 z = 6 3 z = 6 z = 8 zn/2 1 2 N i z nearest neighbors of i j=1

More information

2007 5 iii 1 1 1.1.................... 1 2 5 2.1 (shear stress) (shear strain)...... 5 2.1.1...................... 6 2.1.2.................... 6 2.2....................... 7 2.2.1........................

More information

B

B B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T

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

30

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

More information

// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du

// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du 2 2.1 1 [ 1 ]< 33, 34 > 1 (the first law of thermodynamics) U du = δw + δq (1) (internal energy)u (work)w δw rev = PdV (2) P (heat)q 1 1. U ( U ) 2. 1 (perpetuum mobile) 3. du 21 // //( ) (Helmholtz, Hermann

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

Note.tex 2008/09/19( )

Note.tex 2008/09/19( ) 1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................

More information

untitled

untitled (a) (b) (c) (d) Wunderlich 2.5.1 = = =90 2 1 (hkl) {hkl} [hkl] L tan 2θ = r L nλ = 2dsinθ dhkl ( ) = 1 2 2 2 h k l + + a b c c l=2 l=1 l=0 Polanyi nλ = I sinφ I: B A a 110 B c 110 b b 110 µ a 110

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

December 28, 2018

December 28, 2018 e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................

More information

4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k

4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k 4.6 (E i = ε, ε + ) T Z F Z = e ε + e (ε+ ) = e ε ( + e ) F = kt log Z = kt loge ε ( + e ) = ε kt ln( + e ) (4.8) F (T ) S = T = k = k ln( + e ) + kt e + e kt 2 + e ln( + e ) + kt (4.20) /kt T 0 = /k (4.20)

More information

2000年度『数学展望 I』講義録

2000年度『数学展望 I』講義録 2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53

More information

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a ... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c

More information

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

More information

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 =

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 = 5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =

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

201711grade1ouyou.pdf

201711grade1ouyou.pdf 2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2

More information

70 : 20 : A B (20 ) (30 ) 50 1

70 : 20 : A B (20 ) (30 ) 50 1 70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................

More information

薄膜結晶成長の基礎2.dvi

薄膜結晶成長の基礎2.dvi 2 464-8602 1 2 2 2 N ΔμN ( N 2/3 ) N - (seed) (nucleation) 1.4 2 2.1 1 Makio Uwaha. E-mail:uwaha@nagoya-u.jp; http://slab.phys.nagoya-u.ac.jp/uwaha/ 2 [1] [2] [3](e) 3 2.1: [1] 2.1 ( ) 1 (cluster) ( N

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

[Ver. 0.2] 1 2 3 4 5 6 7 1 1.1 1.2 1.3 1.4 1.5 1 1.1 1 1.2 1. (elasticity) 2. (plasticity) 3. (strength) 4. 5. (toughness) 6. 1 1.2 1. (elasticity) } 1 1.2 2. (plasticity), 1 1.2 3. (strength) a < b F

More information

( ) ,

( ) , II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00

More information

I ( ) 2019

I ( ) 2019 I ( ) 2019 i 1 I,, III,, 1,,,, III,,,, (1 ) (,,, ), :...,, : NHK... NHK, (YouTube ),!!, manaba http://pen.envr.tsukuba.ac.jp/lec/physics/,, Richard Feynman Lectures on Physics Addison-Wesley,,,, x χ,

More information

70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain) (grain boundary) 5. 1

70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain) (grain boundary) 5. 1 5 0 1 2 3 (Carnot) (Clausius) 2 5. 1 ( ) ( ) ( ) ( ) 5. 1. 1 (system) 1) 70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain)

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

LLG-R8.Nisus.pdf

LLG-R8.Nisus.pdf d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =

More information

3章 問題・略解

3章 問題・略解 S S W R S O( l) O( ) c Jg g J Jg S R J 7. K.9 JK S W S R S JK S S R J 7. K.9JK 4 (a) -Tice 7.K T ice T N 77 K S R.9 JK 4. JK T T ice N.6JK S W S R S JK S S.6JK R (b) S R JK S.6 JK T T ice N 6 O( c) O(

More information

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t

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

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)

More information

main.dvi

main.dvi SGC - 70 2, 3 23 ɛ-δ 2.12.8 3 2.92.13 4 2 3 1 2.1 2.102.12 [8][14] [1],[2] [4][7] 2 [4] 1 2009 8 1 1 1.1... 1 1.2... 4 1.3 1... 8 1.4 2... 9 1.5... 12 1.6 1... 16 1.7... 18 1.8... 21 1.9... 23 2 27 2.1

More information

P F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2

P F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2 1 1 2 2 2 1 1 P F ext 1: F ext P F ext (Count Rumford, 1753 1814) 0 100 H 2 O H 2 O 2 F ext F ext N 2 O 2 2 P F S F = P S (1) ( 1 ) F ext x W ext W ext = F ext x (2) F ext P S W ext = P S x (3) S x V V

More information

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) = [

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) = [ 3 3. 3.. H H = H + V (t), V (t) = gµ B α B e e iωt i t Ψ(t) = [H + V (t)]ψ(t) Φ(t) Ψ(t) = e iht Φ(t) H e iht Φ(t) + ie iht t Φ(t) = [H + V (t)]e iht Φ(t) Φ(t) i t Φ(t) = V H(t)Φ(t), V H (t) = e iht V (t)e

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

The Physics of Atmospheres CAPTER :

The Physics of Atmospheres CAPTER : The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(

More information

3 3.1 R r r + R R r Rr [ ] ˆn(r) = ˆn(r + R) (3.1) R R = r ˆn(r) = ˆn(0) r 0 R = r C nn (r, r ) = C nn (r + R, r + R) = C nn (r r, 0) (3.2) ( 2.2 ) C

3 3.1 R r r + R R r Rr [ ] ˆn(r) = ˆn(r + R) (3.1) R R = r ˆn(r) = ˆn(0) r 0 R = r C nn (r, r ) = C nn (r + R, r + R) = C nn (r r, 0) (3.2) ( 2.2 ) C 3 3.1 R r r + R R r Rr [ ] ˆn(r) = ˆn(r + R) (3.1) R R = r ˆn(r) = ˆn(0) r 0 R = r C nn (r, r ) = C nn (r + R, r + R) = C nn (r r, 0) (3.2) ( 2.2 ) C nn (r r ) = C nn (R(r r )) [2 ] 2 g(r, r ) ˆn(r) ˆn(r

More information

main.dvi

main.dvi 固体物理の基礎 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/066581 このサンプルページの内容は, 初版 1 刷発行当時のものです. http://www.morikita.co.jp/support ( 03 3817 5670 FAX03 3815 8199) i 1 2 5 6

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

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

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

20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................

More information

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 = #A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/

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

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ = 1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A

More information

現代物理化学 1-1(4)16.ppt

現代物理化学 1-1(4)16.ppt (pdf) pdf pdf http://www1.doshisha.ac.jp/~bukka/lecture/index.html http://www.doshisha.ac.jp/ Duet -1-1-1 2-a. 1-1-2 EU E = K E + P E + U ΔE K E = 0P E ΔE = ΔU U U = εn ΔU ΔU = Q + W, du = d 'Q + d 'W

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

3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3) P = U V S,N,, µ = U N. (4) S,V,, ( ) ds = 1 T du + P T dv µ dn +, (5) T 1 T = P U V,N,, T

3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3) P = U V S,N,, µ = U N. (4) S,V,, ( ) ds = 1 T du + P T dv µ dn +, (5) T 1 T = P U V,N,, T 3 3.1 [ ]< 85, 86 > ( ) ds > 0. (1) dt ds dt =0, S = S max. (2) ( δq 1 = TdS 1 =0) (δw 1 < 0) (du 1 < 0) (δq 2 > 0) (ds = ds 2 = TδQ 2 > 0) 39 3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3)

More information

SO(2)

SO(2) TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

More information

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =

More information

1 2 2 (Dielecrics) Maxwell ( ) D H

1 2 2 (Dielecrics) Maxwell ( ) D H 2003.02.13 1 2 2 (Dielecrics) 4 2.1... 4 2.2... 5 2.3... 6 2.4... 6 3 Maxwell ( ) 9 3.1... 9 3.2 D H... 11 3.3... 13 4 14 4.1... 14 4.2... 14 4.3... 17 4.4... 19 5 22 6 THz 24 6.1... 24 6.2... 25 7 26

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................

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

知能科学:ニューラルネットワーク

知能科学:ニューラルネットワーク 2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,

More information

知能科学:ニューラルネットワーク

知能科学:ニューラルネットワーク 2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,

More information

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 + ( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n

More information

42 3 u = (37) MeV/c 2 (3.4) [1] u amu m p m n [1] m H [2] m p = (4) MeV/c 2 = (13) u m n = (4) MeV/c 2 =

42 3 u = (37) MeV/c 2 (3.4) [1] u amu m p m n [1] m H [2] m p = (4) MeV/c 2 = (13) u m n = (4) MeV/c 2 = 3 3.1 3.1.1 kg m s J = kg m 2 s 2 MeV MeV [1] 1MeV=1 6 ev = 1.62 176 462 (63) 1 13 J (3.1) [1] 1MeV/c 2 =1.782 661 731 (7) 1 3 kg (3.2) c =1 MeV (atomic mass unit) 12 C u = 1 12 M(12 C) (3.3) 41 42 3 u

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

第86回日本感染症学会総会学術集会後抄録(I)

第86回日本感染症学会総会学術集会後抄録(I) κ κ κ κ κ κ μ μ β β β γ α α β β γ α β α α α γ α β β γ μ β β μ μ α ββ β β β β β β β β β β β β β β β β β β γ β μ μ μ μμ μ μ μ μ β β μ μ μ μ μ μ μ μ μ μ μ μ μ μ β

More information

main.dvi

main.dvi 5 IIR IIR z 5.1 5.1.1 1. 2. IIR(Infinite Impulse Response) FIR(Finite Impulse Response) 3. 4. 5. 5.1.2 IIR FIR 5.1 5.1 5.2 104 5. IIR 5.1 IIR FIR IIR FIR H(z) = a 0 +a 1 z 1 +a 2 z 2 1+b 1 z 1 +b 2 z 2

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

positron 1930 Dirac 1933 Anderson m 22Na(hl=2.6years), 58Co(hl=71days), 64Cu(hl=12hour) 68Ge(hl=288days) MeV : thermalization m psec 100

positron 1930 Dirac 1933 Anderson m 22Na(hl=2.6years), 58Co(hl=71days), 64Cu(hl=12hour) 68Ge(hl=288days) MeV : thermalization m psec 100 positron 1930 Dirac 1933 Anderson m 22Na(hl=2.6years), 58Co(hl=71days), 64Cu(hl=12hour) 68Ge(hl=288days) 0.5 1.5MeV : thermalization 10 100 m psec 100psec nsec E total = 2mc 2 + E e + + E e Ee+ Ee-c mc

More information

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

More information

PDF

PDF 1 1 1 1-1 1 1-9 1-3 1-1 13-17 -3 6-4 6 3 3-1 35 3-37 3-3 38 4 4-1 39 4- Fe C TEM 41 4-3 C TEM 44 4-4 Fe TEM 46 4-5 5 4-6 5 5 51 6 5 1 1-1 1991 1,1 multiwall nanotube 1993 singlewall nanotube ( 1,) sp 7.4eV

More information

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a

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

数学Ⅱ演習(足助・09夏)

数学Ⅱ演習(足助・09夏) II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w

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

現代物理化学 2-1(9)16.ppt

現代物理化学 2-1(9)16.ppt --- S A, G U S S ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r S -- ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r d Q r e = P e = P ΔS d 'Q / e (d'q / e ) --3,e Q W Q (> 0),e e ΔU = Q + W = (Q + Q ) + W = 0

More information

Microsoft Word - 章末問題

Microsoft Word - 章末問題 1906 R n m 1 = =1 1 R R= 8h ICP s p s HeNeArXe 1 ns 1 1 1 1 1 17 NaCl 1.3 nm 10nm 3s CuAuAg NaCl CaF - - HeNeAr 1.7(b) 2 2 2d = a + a = 2a d = 2a 2 1 1 N = 8 + 6 = 4 8 2 4 4 2a 3 4 π N πr 3 3 4 ρ = = =

More information

数学の基礎訓練I

数学の基礎訓練I I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

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

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

More information

H21環境地球化学6_雲と雨_ ppt

H21環境地球化学6_雲と雨_ ppt 1 2 3 40 13 (0.001%) 71 24,000 (1.7%) 385 425 111 1,350,000 (97%) 125 (0.009%) 40 10,000 (0.7%) 25 (0.002%) 10 3 km 3 10 3 km 3 /y 4 +1.3 +5.8 (21) () ( ) 5 HNO 3, SO 2 etc 6 7 2009年度 環境地球化学 大河内 10種雲形と発生高度

More information

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) ( 6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

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

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

Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence

Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence Hanbury-Brown Twiss (ver. 2.) 25 4 4 1 2 2 2 2.1 van Cittert - Zernike..................................... 2 2.2 mutual coherence................................. 4 3 Hanbury-Brown Twiss ( ) 5 3.1............................................

More information

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k 7 b f n f} d = b f n f d,. 5,. [ ] ɛ >, n ɛ + + n < ɛ. m. n m log + < n m. n lim sin kπ sin kπ } k π sin = n n n. k= 4 f, y = r + s, y = rs f rs = f + r + sf y + rsf yy + f y. f = f =, f = sin. 5 f f =.

More information

23 1 Section ( ) ( ) ( 46 ) , 238( 235,238 U) 232( 232 Th) 40( 40 K, % ) (Rn) (Ra). 7( 7 Be) 14( 14 C) 22( 22 Na) (1 ) (2 ) 1 µ 2 4

23 1 Section ( ) ( ) ( 46 ) , 238( 235,238 U) 232( 232 Th) 40( 40 K, % ) (Rn) (Ra). 7( 7 Be) 14( 14 C) 22( 22 Na) (1 ) (2 ) 1 µ 2 4 23 1 Section 1.1 1 ( ) ( ) ( 46 ) 2 3 235, 238( 235,238 U) 232( 232 Th) 40( 40 K, 0.0118% ) (Rn) (Ra). 7( 7 Be) 14( 14 C) 22( 22 Na) (1 ) (2 ) 1 µ 2 4 2 ( )2 4( 4 He) 12 3 16 12 56( 56 Fe) 4 56( 56 Ni)

More information

A

A A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................

More information

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i 1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

More information

I

I I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

More information

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

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 1 1 1 + 1 4 + + 1 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 a n < ε 1 1. ε = 10 1 N m, n N a m a n < ε = 10 1 N

More information

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2 1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a

More information

keisoku01.dvi

keisoku01.dvi 2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.

More information

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3

More information

( )

( ) 7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................

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

Mathematical Logic I 12 Contents I Zorn

Mathematical Logic I 12 Contents I Zorn Mathematical Logic I 12 Contents I 2 1 3 1.1............................. 3 1.2.......................... 5 1.3 Zorn.................. 5 2 6 2.1.............................. 6 2.2..............................

More information