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

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1 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) h y 1 (x) u 1 (x) y 1 (x) y 0 (x) y u(x, t) y(x, t) y 0 (x) (3) (4) u(x, t) (5)

2 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) cot x B n n! xn B 3 = B 5 = B 7 = = 0 cot x = 1 x n=0 A n Bernoulli A n n! xn (4) cos(kx) x [ π, π] Fourier cos(kx) = C { } C n cos(nx) + S n sin(nx) n=1 C n S n

3 I-3 ( ) (100 ) 1, 2 φ 1 (x), φ 2 (x) φ 1, φ 2 2 s 1 s 2 ψ s (1) z, ψ s (2) z (1) S = s 1 + s 2 S S z Ψ S,Sz S = 1, S = 0 x 1 x 2 (2) P 1, P 0 P 1 Ψ 1,Sz = Ψ 1,Sz, P 1 Ψ 0,0 = 0, P 0 Ψ 0,0 = Ψ 0,0, P 0 Ψ 1,Sz = 0 P 0, P 1 P i = c i + d i s 1 s 2 c i, d i (3) H = e 2 /r 12 r 12 x 1 x 2 S = 0, S = 1 H = a + b s 1 s 2 S b b = 2 d 3 x 1 d 3 x 2 φ 1(x 2 )φ 2(x 1 ) e2 φ 1 (x 1 )φ 2 (x 2 ) r 12 s 1 s 2 S S s 1 s 2 a, b S (4) Ω φ 1(r)φ 2 (r) = 1 a q e iq r Ω b d 3 eiq r r lim d 3 eiq r ɛr r r ɛ +0 r (5) S = 1, S = 0 q

4 16 II (3 30 ) (1) II-1 II-2 II-3 1 (2) II-1 ( ) (100 ) N m k (N + 1) (1) n t x n (t) (2) x n (t) n N x(t) d 2 dt 2 x = Lx N L (3) x (p) n (t) = a (p) n cos ω p t (n = 1, 2,..., N) N p = 1, 2,..., N ω p a (p) n ω p = Ω sin θ p 2, a(p) n = A sin nθ p (A) Ω = 2 k/m θ p = pπ/(n + 1) A a (p) n = a sin(nθ + α) θ α (4) a (p) = t (a (p) 1, a (p) 2,..., a (p) n ) a (p) (A) A = 2/(N + 1) N n=1 a (p) n a (p ) n = δ pp (p, p = 1, 2,..., N) N δ pp sin 2 nθ p = (N + 1)/2 n=1

5 (5) 2mγẋ n γ > 0 F n (t) = mf n cos ω e t d 2 dt 2 x Lx + 2γ d dt x = f cos ω et (B) f = t (f 1, f 2,..., f N ) q p (t) f (p) = N n=1 a (p) n f n (6) q p (t) N n=1 a (p) n x n (t)

6 II-2 ( ) (100 ) (1) a b L C 0 ɛ 0 (2) C α α h 1 h 1 ( g ) α h 1 C θ ρ ( ) ( α) (3) C 1 h 1 C 1 ɛ ɛ 0 (4) V h 2 C 2 h 2 ρ (5) C 3 ɛ a, b, C 0, C 1, C 2, C 3, V, ɛ 0, θ α α' α θ C a b V L h1 h2 h1

7 II-3 (Ising ) (100 ) (1) V N µ H (+) ( ) σ i (i = 1,..., N) H +1 1 Hamiltonian H = J σ i σ j µh ij i σ i (A) ij i Hamiltonian H i σ i σ H i = Jz σ σ i µhσ i (B) z Jz σ = µh H eff = H + H (a) Z 1 H eff (b) Helmholtz F S U (c) M(= N µ σ ) V M = Nµ µh V tanh( kt + JzV M ) kt µn (d) (C) H = 0 M = 0 Curie T c = Jz/k k Boltzmann (2) N H (+) ( ) N + N Φ N + /N = (1 + Φ)/2 N /N = (1 Φ)/2 zn/2 +, N ++ N N + U (C) U = J(N ++ + N N + ) (D)

8 + p + = N + /N p = N /N U N ++ N ++ = 1 2 zn +p + = z N(1 + Φ)2 (E) 8 N + = zn + p = z 4 N(1 Φ2 ) N = 1 2 zn p = z N(1 Φ)2 (G) 8 (F) (a) N N + N Φ Stirling (log x! x log x x) (b) H Helmholtz F Φ (c) Φ (C) H = 0 Φ (d) C v T c

9 平成 16 年度大学院入学試験問題 III 3 時間 注意 (1) 問題は III-1 から III-8 まで 8 問ある これから 3 問選択せよ (2) 選択した問題の回答はそれぞれ別の用紙一枚に記入せよ 裏面を用いても よい (3) 各用紙ごとに 左上に問題番号 右上に受験番号と氏名を記入せよ III-1 (選択)(振り子の実験) (100 点) 剛体を水平な固定軸で支えた実体振り子で重力加速度 g を求める実験を行うも のとする Borda の振り子 この振り子は図1のように細い針金で吊られた半径 r 質量 M の金属球からなり 支持体のナイフエッジ K を支点として振動する 針金の長さを l 振り子が最大振幅になるときの角度を α 振動の周期を T とする と 重力加速度 g は次の式で求めることができる " 4π 2 (l + r) 2r2 g= 1 + T2 5(l + r)2 #" α # (A) K l L (1) 金属球の直径をノギス キャリパー を用いて測定したところ 図 2 のよう な表示だった 直径の値を誤差を含めて書け

10 (2) T = = = = = = = = = =209.9 T T T 4.32 = , 0.43 = K L ± 0.01 cm α 2 = (3) l (4) g (5) (4) g (6) l, r, T l r T g g g = L + r L r + 2 T T (B) g g

11 III-2 ( )( ) (100 ) (1) 10 cm 50 cm 100 A 10 T L µ 0 = 4π 10 7 kg m C 2 (2) Ω m 0.1 mm 1 m R 0 t=0 I 0 t B I 0, L, R L, R (3) (2) r

12 1 ppm(= ) r (4) (2) 0 t=0 I 0 =βt t B I 0, L, R, β β=0.1 A/sec L, R (5) 100A (S1) ( 1) (PSW) (D1,D2,R1 ) D1,D2 2 R Ω I D2 R1 D1 S1 60A 0.6V 1.2V V

13 III-3 ( )( ) (100 ) ( p 2 + ω 2 x 2) H = 1 2 x, p ω 1 a = (2ω) 1/2 (p iωx) a = (2ω) 1/2 (p + iωx) a a h = 1 (1) a a H ( H = ω a a + 1 ) 2 (2) 0 a 0 = 0 ϕ 0 (x) = x 0 π e ax2 dx = a (3) 1 = a 0 ϕ 1 (x) = x 1 H 0 H 0 = 1 2 ( p 2 x + p 2 y) ω2 ( x 2 + y 2) [ H = λ ( p e) 2 1 ] 2 (p2 x + p 2 y) p = (p x, p y ) e (cos θ, sin θ)

14 (4) θ = π/4 H H (5) θ = 0 λ > 0 λ < 0 ϕ 0 (x), ϕ 0 (y), ϕ 1 (x), ϕ 1 (y) (6) θ (7) (5) (6)

15 III-4 ( ) ( ) (100 ) µ 0 z x(> a) a x z ( ) (1) I 1 I 2 I 1 P H (2) (1) P ds = adθ x df x z df z (3) F π 0 dθ u t cos θ = π u2 t 2 (u > t) (4) I 1 = I sin ωt M e ( ) (5) I 1 = I x v 0 x(> a) I 2 R (6) (5) x = d 1 (> a) x = d 2 W Q

16 III-5 ( ) ( ) (100 ) Maxwell-Boltzmann ε(p) = 1 (A) 2m p2 m Maxwell-Boltzmann h k B (1) V T Z 1 Gauss : dx e x2 = π (2) N Z N ( ) (3) N µ Ξ (4) N = Ω µ Ω = pv = k BT ln Ξ µ = µ cl N T V µ cl = k B T ln [( ) ] N λ 3 T V λ T h λ T = 2 2πmk B T (5) m (A) ε(p) Ω = 2 V m 3/2 dε ε 3/2 ze βε (B) 3 2π2 3 1 gze βε 0

17 β = 1 k B T z = eβµ µ g +1 1 (6) (B) z N V T z = e βµ 1 (B) z N = Ω z N µ λ T V z N g Γ : Γ(z) = 0 dx x z 1 e x, Γ(z + 1) = zγ(z), Γ(1/2) = π (7) : z = e βµ 0 z (2) z = exp(βµ cl ) (8) (6) (7) z : x N V λ3 T (6) z x x (9) : pv Nk B T p

18

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20 III-7 ( )( : ) (100 ) M r p(r), ρ(r), v(r) (1) (2) v 1 dv 2a2 v dr = r GM r 2 v 2 a 2 (A) a p/ρ G (A) (critical point) r crit a (A) r (3) v r (A) (4) (3) (5) dyn cm 2 g erg deg 1 mol g cm K

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22 III-8 ( ) ( ) (100 ) ( ) 1940 Fe XIV Fe X Edlen ( 1) : Fe X Grotorian Fe X 1939 (1) λ R (2) Fe X Fe XIV 300eV 2 1eV= erg k= erg K m ( ) L1 L2 L3 F L1 C C 1 A 3 L1 L1 L2 L1 D ( ) (3) L2 1 A F A L1 D Fe XIV 5303Å

23 2 T Fe XIV Maxwell FeXIV v v + dv f(v)dv = ( ) m 1/2 e mv 2 /(2kT ) dv 2πkT m k (4) Fe XIV 1 kt 2 Fe XIV 5303Å Fe XIV 3: Fe XIV 5303Å Fe XIV λ 0 =5303Å (5) 1/e λ D c λ D = λ 0 c ( 2kT m (6) ( ) 6 ) 1/2

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

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