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1 (Contents) 1. 2 2. (1) 22 3. 31 1. Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji SETO 22 3. Editorial Comments Tadashi YANO 31
Beginning of the Universe, Dark Energy and Dark Matter * 1 Noboru NAKANISHI * 2 1 *3 [1] 2 3 *1 *2 nbr-nak@trio.plala.or.jp *3 2
1 2 ( ) 1 20 4 g µν (x) g µν (x) *4 *4 3
20 *5 *6 40 10 5 2.7K 3 *7 η µν 4 (1, 1, 1, 1) x µ (µ = 0, 1, 2, 3) η µν x µ x ν x µ x ( ) φ A (x) A φ A (x) V V α β α α β 0 *5 *6 *7 c = 1, ħ h/2π = 1 4
*8 0 0 = 1 α α α V phys V phys V phys V 0 V phys H = V phys /V 0 *9 S = d 4 xl(x) L(x) φ A (x) 1 µ φ A (x) µ / x µ 2 S φ A (x) x µ L(x) S δs = 0 S S φ A (x) π A (x) = L(x)/ ( 0 φ A (x)) φ A (x) π B (y) y 0 = x 0 * 10 4 [φ A (x), φ B (y)] φ A (x) 0 V * 11 V [2] *8 *9 *10 [X, Y ] XY Y X X Y iδ(x 0 y 0 )[π A (x), φ A (y)] = δ 4 (x y) A 0 *11 5
2 S S 2 0 S S S S δs S 4 S J λ (x) λ J λ (x) = 0 0 3 Q = d 3 xj 0 (x) φ A (x) φ A (x) iϵ[q, φ A (x)] ϵ 6
P µ M µν P 0 P 0 0 P µ 0 = 0, M µν 0 = 0 P µ M µν 1 Q 0 = 0 0 [Q, χ(x)] 0 = 0 χ(x) 0 χ(x) 0 0 Q P µ * 12 χ(x) NG 2008 BCS ϕ(x) = ϕ 1 (x) + iϕ 2 (x) ϕ j (x) ϕ j = ϕ j L G = µ ϕ µ ϕ V (ϕ) (4.1) (x) * 13 V (ϕ) V (ϕ) = uϕ ϕ + λ 4 (ϕ ϕ) 2 (4.2) 2 λ L G ϕ e iθ ϕ ϕ = iϵϕ 0 ϕ(x) 0 ϕ 0 0 ϕ(x) 0 = v/ 2 v *12 Q P µ Q 0 = 0 *13 α X β = β X α 7
µ ϕ dv ( v / 2)/d v = 0 ( u + 1 4 λ v 2) v = 0 (4.3) u > 0 v = 0 u < 0 v = 0 v 2 = 4u/λ v 0 arg v arg v ϕ 2 ϕ 2 NG u ϕ 2 0 5 s s = 0, 1/2, 1 1/2 * 14 2 6 2 1 u d 2/3 1/3 uud udd 6 3 1 1 0 *14 8
3 3 SU(3) SU(3) µ SU(3) 2 SU(2) 1 U(1) 4 3 SU(2) U(1) U(1) * 15 4 3 NG 1 CERN * 16 δs = 0 B * 17 2 FP FP * 18 *15 *16 *17 NL *18 FP FP FP BRS B 9
BRS * 19 BRS FP BRS BRS Q B BRS Q B phys = 0 (5.1) V phys = { phys } U(1) NG NG NG NG * 20 NG 6 LHC *19 BRST *20 [3] 10
NG (5.1) NG * 21 SU(3) Q a (a = 1, 2,, 8) (5.1) Q a phys = 0 (6.1) [4] Q a Q B U(1) Q Q phys = 0 2 2 SU(3) ψ α (x) ψ α (x) 3 ψ α (y)ψ α (x) 0 (6.2) α=1 *21 1 1 11
x µ = y µ x µ y µ (6.1) * 22 [5] 3 1 2 1:2 u d 1:2 2!! 7 2 1 2 g µν (x) 2 *22 (6.1) 12
4 ds ds 2 = g µν (x)dx µ dx ν (7.1) * 23 R µν 1 2 g µνr = κt µν (7.2) R µν R g µν κ G 8π/c 4 T µν S M S M g µν S E 1 2κ d 4 x gr (7.3) g detg µν det g d 4 x g S M δ(s E + S M ) = 0 * 24 g R µν 1 2 g µνr + Λg µν = κt µν (7.4) 3 g µν (x) (7.1) g µν (x) * 25 *23 *24 R 2 *25 13
* 26 S E + S M S M * 27 S M [7] * 28 *26 [6] *27 *28 14
* 29 g µν (x) κ 0 η µν g µν (x) κ 0 8 10 [8] [ ] g µν (x) x µ x µ 4 1 4 1 f t f 1 t [9] *29 BRS 15
S E FP BRS g g hµ a (x) hµ a (x) 16 g µν (x) = ξ ab h a µ (x)h b ν (x) (8.1) ξ ab 2 +1 1 h = det hµ a ξ ab 1 1 ξ ab = ±η ab hµ a 6 BRS * 30 GL(4) x 0, x 1, x 2, x 3 1 x 0 GL(4) 1/2 1/2 A µ = hµ a A a x µ 3 5 4 0 P µ ˆM µ ν M ab (= M ba ) *30 B FP FP (4 4 =)16 IOSp(8, 8) 8+8 16
i[p µ, h c λ (x)] = µ h c λ (x), (8.2) i[ ˆM µ ν, h c λ (x)] = x µ ν h c λ (x) + δ µ λ h c ν (x), (8.3) i[m ab, h c λ (x)] = η ac h b λ (x) η bc h a λ (x). (8.4) 1 0 h c λ (x) 0 u c λ (8.5) ν u c λ = 0 (8.3) (8.4) i 0 [ ˆM µ ν, h c λ (x)] 0 = δ µ λ u c ν, (8.6) i 0 [M ab, h c λ (x)] 0 = η ac u b λ η bc u a λ (8.7) u c λ (8.5) (8.6) ˆM µ ν M ab SU(2) U(1) U(1) U(1) GL(4) x µ α, β, γ, u α γ v γ β u α γ v γ β = δα β M αβ ( η βγ u α δ v ϵ γ η αγ u β δ vϵ γ) ˆM δ ϵ + M αβ (8.8) (8.6)(8.7) 0 [ M αβ, h c λ (x)] 0 = 0 (8.9) M αβ 1/2 x α u α β xβ 2 NG 17
* 31 9 21 Ia 4 ϕ(x) 0 ϕ(x) 0 = 0 (9.1) g(x) 120 [10] *31 18
κ * 32 0 [2] g µν (x) 0 g µν (0) (x) (= η ab h (0) µ a (x)h (0) ν b (x)) g (0) (y) 4 4 F (z µν ) 0 F (g (0) µν (x)) 0 = F ( 0 g (0) µν (x) 0 ) (9.2) V 0 g (0) µν (x) 0 * 33 (9.2) ρλ 10 10 5 2.7K φ(x) *32 κ *33 19
* 34 * 35 [11] 8 3 11 10 *34 V (φ) *35 20
[1] http://ja.wikipedia.org/ [2] N. Nakanishi, Prog. Theor. Phys. 111 301 (2004). [3] N. Nakanishi, Mod. Phys. Letters A 17 89 (2002). [4] N. Nakanishi and I. Ojima, Prog. Theor. Phys. 71 1359 (1984); 72 1197 (1984). [5] 116 148 (2008); 21 No.2 (2016). [6] 2008), 192-193. [7] D. N. Page and C. D. Geilker, Phys. Rev. Letters 47 979 (1981). [8] 1524 (2006), 50. [9] N. Nakanishi, Intern. J. Mod. Phys. A 29 No.6 (2014) [DOI: 10.1142/S0217751X14500341] [10] 19 No.1 (2015). [11] 116 188 (2008). 21
(1) Problem of Heat Exchanger (1) Kenji Seto 1 ( 8 3 ) ( 77 10 84 4 ) 1 100 A 0 B 2 100 A 0 B 0 B 1 2 2 50 2 2 2 A 1, A 2, B 1, B 2 (1) A 1 B 1 50 (2) A 1 B 2 25 (3) A 2 B 1 75 (4) A 2 B 2 50 (5) 2 A (25 + 50)/2 = 37.5 B (75 + 50)/2 = 62.5 (6) A 50 B 50 B E-mail: seto@pony.ocn.ne.jp 22
2 2 2.1 A B N B B 1, B 2,, B N A B 1, A B 2, A B 3,, A B N A T 0 = 100, B 0 A B n 1 A B n 1 T n 1 A B n T n A T n 1 T n B n 0 T n A B n B n A 1/N T n 1 T n = 1 N T n, T n = ( 1 + 1 N ) 1Tn 1 (2.1) T n A T N ( T N = 1 + 1 ) N T0 (2.2) N T N N N 3 lim T N = e 1 T 0 = 36.787 (2.3) N B T 1, T 2,, T N A, B A B N B 100 36.787 = 63.212 A, B 2 B 62.5 A 2.2 A, B A M B N 3 B A A 1, A 2,, A M B 23
B 1, B 2,, B N A T 1,0, T 2,0,, T M,0 B T 0,1, T 0,2,, T 0,N T 1,0 = T 2,0 = = T M,0 = T 0 (= 100 ), T 0,1 = T 0,2 = = T 0,N = 0 (2.4) 3 A 1 B 1, A 1 B 2, A 1 B N, A 2 B 1, A 2 B 2, A 2 B N, A M B N 3 A 1 B 1 T 1,1, A 1 B 2 T 1,2 A 1, A 2,, A M T 1,N, T 2,N,, T M,N B 1, B 2,, B N T M,1, T M,2,, T M,N A m B n 1 A m B n A m T m,n 1 T m,n B n T m 1,n T m,n A m B n A m, B n 1/M, 1/N 1 ( ) 1 ( ) Tm,n 1 T m,n = Tm,n T m 1,n M N 2 (2.4) (2.5) T m,n = αt m 1,n + βt m,n 1, α = M M + N, β = N M + N, α + β = 1 (2.6) α, β 2 3 n m m 1 1 T m,n = α 2 T m 2,n + αβt m 1,n 1 + βt m,n 1 (2.7) 24
(2.6) m 2 m 1 T m,n = α 3 T m 3,n + β T m,n = α m T 0,n + β 3 α r 1 T m r+1,n 1 (2.8) m α r 1 T m r+1,n 1 (2.9) (2.4) T 0,n = 0 1 T m,n = β m α r 1 T m r+1,n 1 (2.10) n n 1 T m,n = β 2 m m r+1 r =1 α r+r 2 T m r r +2,n 2 = β 2 m k=1 l=1 k α k 1 T m k+1,n 2 (2.11) r l, r + r 1 k l l k k r T m,n = β 2 m rα r 1 T m r+1,n 2 (2.12) (2.10) n 2 T m,n = β 3 m m r+1 r =1 rα r+r 2 T m r r +2,n 3 = β 3 m k=1 ( k l=1 ) l α k 1 T m k+1,n 3 (2.13) r l, r + r 1 k l k(k + 1)/2 k r T m,n = β 3 T m,n = β 4 m m r+1 r =1 r(r + 1) 2 m α r+r 2 T m r r +2,n 4 = β 4 r(r + 1) α r 1 T m r+1,n 3 (2.14) 2 m k=1 ( k l k(k + 1)(k + 2)/6 k r T m,n = β 4 m r(r + 1)(r + 2) 6 α r 1 T m r+1,n 4 = β 4 T m,n = β 5 m m l=1 l(l + 1) ) α k 1 T m k+1,n 4 (2.15) 2 (r + 2)! 3!(r 1)! αr 1 T m r+1,n 4 (2.16) (r + 3)! 4!(r 1)! αr 1 T m r+1,n 5 (2.17) 25
p 1 T m,n = β p m * 1 (r + p 2)! (p 1)!(r 1)! αr 1 T m r+1,n p (2.18) r r =1 (r + p 2)! (r + p 1)! (p 1)!(r = 1)! p!(r 1)! (2.19) (2.18) p = n (2.4) T m r+1,0 = T 0 T m,n = β n[ m (r + n 2)! (n 1)!(r 1)! αr 1] T 0 (2.20) *2 A 1, A 2,, A M T 1,N, T 2,N,, T M,N A A M,N A M,N = 1 M M m=1 T m,n = βn T 0 M M m m=1 (r + N 2)! (N 1)!(r 1)! αr 1 (2.21) r m M m M M = m=1 m=r (2.22) m m A M,N = βn T 0 M M (M r + 1)(r + N 2)! α r 1 (2.23) (N 1)!(r 1)! r M N ( β N = N M + N ) N N e M (2.24) (r + N 2)! ( M ) r 1 α r 1 N = (r + N 2)(r + N 3) (N) M r 1 (2.25) (N 1)! M + N (2.23) N A M, = e M T 0 M M (M r + 1)M r 1 (r 1)! (2.26) *1 II p.5, 3 r r =1 *2 M = N 78 4 r +p 2C p 1 = r+p 1 C p 26
r M (M r + 1)M r 1 (r 1)! = M M r (r 1)M r 1 (r 1)! = M M r M (r 1)! M r 1 (r 2)! = M M (M 1)! r=2 (2.27) A M, A M, = e M M M 1 T 0 (2.28) (M 1)! A B T M,1, T M,2,, T M,N A B B M,N B M,N = T 0 A M,N (2.29) N (2.28) B M, = (1 e M M M 1 ) T 0 (2.30) (M 1)! M = 1, 2, 3, 4 3 B 1, = (1 e 1 )T 0 = 63.212, B 2, = (1 2e 2 )T 0 = 72.932, ( B 3, = 1 9 ( 2 e 3) T 0 = 77.595, B 4, = 1 32 3 e 4) T 0 = 80.463 (2.31) B A M Stirling M (M 1)! M M e M 2π M (2.32) (2.30) B M, ( 1 1 )T 0 (2.33) 2πM M 2 B, = T 0 (2.34) B A A B 3 A B A B A, B M A, M B (2.5) M A M ( ) M B ( ) Tm,n 1 T m,n = Tm,n T m 1,n N (3.1) 27
(2.6) T m,n = αt m 1,n + βt m,n 1, α = λm λm + N, β = N λm + N, λ = M B M A (3.2) α, β λ B A α, β (2.23) (2.24) (2.25) ( β N = N λm + N (r + N 2)! ( λm α r 1 = (r + N 2)(r + N 3) (N) (N 1)! λm + N (2.26) A M, = e λm T 0 M ) N N e λm (3.3) M (M r + 1)(λM) r 1 (r 1)! ) r 1 N (λm) r 1 (3.4) λ = 1 r A M, λ 1 (3.5) 1 M λ A M, = e λm T 0 (λm) r 1 (r 1)! (3.6) 2 λ 2 A λm (λm)m 1 M, = Me (M 1)! T 0 (3.7) M (2.32) Stirling 2 λ 2 A M, T 0 λ ( λe 1 λ ) M M (3.8) 2π λe 1 λ 1, λ = 1 M 2 λ 2 A, = 2 0 λ < 1, 0, 0 λ < 1, λ = 1 0, λ > 1 (3.9) 1 < λ A, λ 1 λ = 0 A, = T 0, λ = 1 A, = 0 λ (3.5) A M, > 0, (3.6) ( / λ)a M, < 0 λ = 1 A, = 0 λ > 1 A, = 0 A, λ A, = { (1 λ)t 0, 0 λ < 1 0, 1 λ (3.10) λ = 1 (3.9) A, λ 2 2 λ 2 A, = T 0 δ(λ 1) (3.11) 28
Dirac A T 0 A, T 0 A, M A (T 0 A, ) B B 0 M A (T 0 A, )/M B B B, B, = 1 λ (T 0 A, ) (3.12) B, = { T 0, 0 λ < 1 T 0 /λ, 1 λ (3.13) 4 1 2 4 A B 2 A B B 4 A B 29
[ ] 3 A B 30
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