素粒子物理学２ 素粒子物理学序論B 2010年度講義第2回
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- ひろき ほうねん
- 5 years ago
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1 素粒子物理学２ 素粒子物理学序論B 2010年度講義第2回
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5 = J sec = MeV sec c = MeV m = c = c =1
6 1 m p = c(mev m) 938M ev = (m) 938 = (cm) 1 m p = (MeV sec) 938M ev = (sec) 938 = (sec)
7 G = m 3 kg 1 s 2 = c(gev/c 2 ) 2 G MM = Mc 2 r r = Mc M = c/g M = 1/G = (GeV )
8 Mc < 2GM c 2 M> 1 2 mc2 <G Mm r c 2G = 1 2 m pl r< 2GM c 2
9 m = E = t = MeV sec sec = MeV E t(= Γ τ) = τ = / E = /Γ = sec
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12 dω θ, φ n i /sec/cm 2 dσ dn = n i (θ, φ)dω dω dσ σ = dω dω
13 H 0 φ n = E n φ n, (H 0 + V (x, t))ψ = i ψ t V φ m φ nd 3 x = δ mn
14 遷移振幅 時刻 t=-t/2 から t=t/2 への遷移を考える (**)の解は ψ = an (t)φn (x)e ien t という形に表せるとする n daf = i dt (**)に入れて φf*をかけ体積vで積分 (*)を使う an (t) n ai ( T /2) = 1 相互作用前はH0の固有状態 daf = i dt af (t) = i Tf i af (T /2) = i T /2 T /2 φ f V φn d3 xei(ef En )t dt an ( T /2) = 0, (n = i) d3 xφ f V φi ei(ef Ei )t t dt T /2 d 3 xφ f V i(ef Ei )t φi e d3 x[φf (x)e ief t ] V (x, t)[φi (x)e iei t ] 14
15 フェルミの黄金律 ポテンシャルが時間に依存しない場合を考える Vf i Tf i = ivf i +T /2 T /2 d3 xφ f (x)v (x)φi (x) dtei(ef Ei )t = 2πiVf i δ(ef Ei ) 単位時間あたりの遷移確率 エネルギー保存 Δt = 2乗しても確率にならない Tf i 2 W = lim T T ρf 終状態密度 wf i = W def ρf (Ef ) ρf (Ef )def は Ef と Ef + def = 2π def ρf (Ef ) Vf i 2 δ(ef Ei ) wf i = 2π Vf i ρf (Ei ) の間の状態数 2 15
16 W fi = T fi 2 TV =(2π)4 δ(4) (p c + p D p A p B ) M 2 V 4 T fi = in A N B N C N D (2π) 4 δ (4) (p D + p C p B p A )M T fi = 2πiV fi δ(e f E i ) Vd 3 p C (2π) 3 2E C F = v A 2E A V Vd 3 p D (2π) 3 2E D 2E B V
17 dσ = V 2 1 v A 2E A 2E B V 4 M 2 (2π) 4 δ (4) (p C + p D p A p B ) Vd3 p C (2π) 3 2E C Vd 3 p D (2π) 3 2E D dσ = M 2 F dlips dlips =(2π) 4 δ (4) d 3 p C (p C + p D p A p B ) (2π) 3 2E C d 3 p D (2π) 3 2E D
18 ラザフォード散乱 非相対論的取り扱い Vf i = 3 d xφ f (x)his φi (x) φi = N eipi x, φf = N eipf x i(pi pf )x 2 e Ze d3 x HIS = 4πV r HIS Ze2 4π = 4πV q 2 1 Ze2 HIS = 4π r N = 1/ V iqx 4π e 3 =... = 2 I d x r q q pi pf q 2 = (pi pf )2 = 4p2 sin2 (θ/2) V d3 p V dp 2 V mpdω ρf = δ(ei Ef ) = p dω = 3 3 (2π) (2π) de 8π 3 入射粒子の速度をv=p/mとすると入射フラックスはv/V (σ=wfi V/v) dσ Ze2 2 4m2 Z 2 α2 m2 =( ) = dω 4π q 4p sin (θ/2) e2 1 α= 4π
19 dσ dω =(Ze2 4π )2 4m2 q 4 = Z2 α 2 m 2 4p 4 sin 4 (θ/2) H IS = 1 4π Ze 2 r
20 不変振幅 知りたい物理が不変振幅に含まれている Tf i = ina NB NC ND (2π)4 δ 4 (pd + pc pb pa )M 例として 仮想的に スピンを持たない電子とミューオンの 散乱を考える ゲージ対称性に基づくQEDより V = ie( µ Aµ + Aµ µ ) e2 A2 e pa jµ(e) e p 無視 C γ pb µ jµ(µ) pd µ 20
21 不変振幅 知りたい物理が不変振幅に含まれている Tf i = ina NB NC ND (2π)4 δ 4 (pd + pc pb pa )M 例として 仮想的に スピンを持たない電子とミューオンの 散乱を考える ゲージ対称性に基づくQEDより V = ie( µ Aµ + Aµ µ ) e2 A2 e pa jµ(e) e p C γ pb µ jµ(µ) pd µ 無視 電磁場Aμによる電子の散乱 Tf i = i φ f (x)v (x)φi (x)d4 x = i φ f ie(aµ µ + µ Aµ )φi d4 x = i jµ(e) Aµ d4 x jµ(e) = ena NC (pc + pa )µ ei(pc pa )x 20
22 不変振幅 続き e pa jµ(e) e p C γ pb µ jµ(µ) pd µ ミューオンが電磁場Aμを作ると考える 1 (µ) q = pd pb Aµ = 2 jµ q jµ(µ) = enb ND (pd + pb )µ ei(pd pb )x 21
23 p e A j (e) µ e p C γ p B j (µ) µ µ µ p D
24 不変振幅 続き = Tf i = i 1 (µ) (e) jµ (x)( 2 )jµ (x)d4 x q ina NB NC ND (2π)4 δ 4 (pd + pc pb pa )M gµν µ im = (ie(pa + pc ) )( i 2 )(ie(pb + pd )ν ) q e pa jµ(e) e p C γ pb µ jµ(µ) pd µ 21
25 不変振幅 続き = Tf i = i 1 (µ) (e) jµ (x)( 2 )jµ (x)d4 x q ina NB NC ND (2π)4 δ 4 (pd + pc pb pa )M gµν µ im = (ie(pa + pc ) )( i 2 )(ie(pb + pd )ν ) q 不変振幅 M はファインマンダイアグラムに対応している e pa jµ(e) e p C γ pb µ jµ(µ) pd µ 21
26 モット(Mott)散乱 QED クーロンポテンシャルに散乱されるスピン1/2の電子 ラザフォード散乱の相対論版 Hint = ej Aµ µ Ze Aµ = ( ; 0, 0, 0) 4πr j = ψγ ψ µ µ スピン状態を si, sf とすると Tf i = = = d4 xf Hint i d4 xpf, sf ej µ (x) pi, si Aµ (x)... ちょっと面倒な計算 スピノール があって dσ 4Z 2 α2 2 2 θ 2 = E (1 v sin ) 4 dω q 2 ラザフォード散乱に比べて m E θ 1 1 v sin ( ) スピンの効果
27 dγ = 1 2E C M 2 d 3 p A (2π) 3 2E A M Γ = M 2 F dlips d 3 p B (2π) 3 2E B (2π) 4 δ 4 (p C p A p B )
28 = c = c =1
素粒子物理学2 素粒子物理学序論B 2010年度講義第4回
素粒子物理学 素粒子物理学序論B 010年度講義第4回 レプトン数の保存 崩壊モード 寿命(sec) n e ν 890 崩壊比 100% Λ π.6 x 10-10 64% π + µ+ νµ.6 x 10-8 100% π + e+ νe 同上 1. x 10-4 Le +1 for νe, elμ +1 for νμ, μlτ +1 for ντ, τレプトン数はそれぞれの香りで独立に保存
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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
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9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) Ĥ0 ψ n (r) ω n Schrödinger Ĥ 0 ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ0 + Ĥint (
9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) 2. 2.1 Ĥ ψ n (r) ω n Schrödinger Ĥ ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ + Ĥint (t)] ψ (r, t), (2) Ĥ int (t) = eˆxe cos ωt ˆdE cos ωt, (3)
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. 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
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FPWS2018講義千代
千代勝実(山形大学) 素粒子物理学入門@FPWS2018 3つの究極の 宗教や神話 哲学や科学が行き着く人間にとって究極の問い 宇宙 世界 はどのように始まり どのように終わるのか 全てをつかさどる究極原理は何か 今日はこれを考えます 人類はどういう存在なのか Wikipediaより 4 /72 千代勝実(山形大学) 素粒子物理学入門@FPWS2018 電子レンジ 可視光では中が透け
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NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
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
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II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
![II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2](/thumbs/101/149159646.jpg "II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
0 ϕ ( ) (x) 0 ϕ (+) (x)ϕ d 3 ( ) (y) 0 pd 3 q (2π) 6 a p a qe ipx e iqy 0 2Ep 2Eq d 3 pd 3 q 0 (2π) 6 [a p, a q]e ipx e iqy 0 2Ep 2Eq d 3 pd 3 q (2π)
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( ) 2 S 3 ( ) ( ) 0 O 0 O ( ) O ϕ(x) ϕ (x) d 3 p (2π) 3 2Ep (a p e ipx + b pe +ipx ) ϕ (+) (x) + ϕ ( ) (x) d 3 p (2π) 3 2Ep (a pe +ipx + b p e ipx ) ϕ ( ) (x) + ϕ (+) (x) (px p 0 x 0 p x E p t p x, E p
July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i
July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac
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(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](/thumbs/104/162595168.jpg "(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
[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
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
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1 12 CP 12.1 SU(2) U(1) U(1) W ±,Z [ ] [ ] [ ] u c t d s b [ ] [ ] [ ] ν e ν µ ν τ e µ τ (12.1a) (12.1b) u d u d +W u s +W s u (udd) (Λ = uds)
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1 1 CP 1.1 SU() U(1) U(1) W ±,Z 1 [ ] [ ] [ ] u c t d s b [ ] [ ] [ ] ν e ν µ ν τ e µ τ (1.1a) (1.1b) u d u d +W u s +W s u (udd) (Λ = uds) n + e + ν e d u +W u + e + ν e (1.a) Λ + e + ν e s u +W u + e
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医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987
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8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J
26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK
Microsoft PowerPoint - qchem3-9
008 年度冬学期 量子化学 Ⅲ 章量子化学の応用 4.4. 相対論的効果 009 年 月 8 日 担当 : 常田貴夫准教授 相対性理論 A. Einstein 特殊相対論 (905 年 ) 相対性原理: ローレンツ変換に対して物理法則の形は不変 光速度不変 : 互いに等速運動する座標系で光速度は常に一定 ミンコフスキーの4 次元空間座標系 ( 等速系のみ ) 一般相対論 (96 年 ) 等価原理
ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx
i B5 7.8. p89 4. ψ x, tψx, t = ψ R x, t iψ I x, t ψ R x, t + iψ I x, t = ψ R x, t + ψ I x, t p 5.8 π π π F e ix + F e ix + F 3 e 3ix F e ix + F e ix + F 3 e 3ix dx πψ x πψx p39 7. AX = X A [ a b c d x
Radiation from moving charges#1 Liénard-Wiechert potential Yuji Chinone 1 Maxwell Maxwell MKS E (x, t) + B (x, t) t = 0 (1) B (x, t) = 0 (2) B (x, t)
Radiation from moving harges# Liénard-Wiehert potential Yuji Chinone Maxwell Maxwell MKS E x, t + B x, t = B x, t = B x, t E x, t = µ j x, t 3 E x, t = ε ρ x, t 4 ε µ ε µ = E B ρ j A x, t φ x, t A x, t
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
f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f
22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(x )
Techniques for Nuclear and Particle Physics Experiments Energy Loss by Radiation : Bremsstrahlung 制動放射によるエネルギー損失は σ r 2 e = (e 2 mc 2 ) 2 で表される為
Techniques for Nuclear and Particle Physics Experiments.. Energy Loss by Radiation : Bremsstrahlung 制動放射によるエネルギー損失は σ r e = (e mc ) で表される為 質量に大きく依存する Ex) 電子の次に質量の小さいミューオンの制動放射によるエネルギー損失 m e 0.5 MeV, m
総研大恒星進化概要.dvi
The Structure and Evolution of Stars I. Basic Equations. M r r =4πr2 ρ () P r = GM rρ. r 2 (2) r: M r : P and ρ: G: M r Lagrange r = M r 4πr 2 rho ( ) P = GM r M r 4πr. 4 (2 ) s(ρ, P ) s(ρ, P ) r L r T
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
Γ Ec Γ V BIAS THBV3_0401JA THBV3_0402JAa THBV3_0402JAb 1000 800 600 400 50 % 25 % 200 100 80 60 40 20 10 8 6 4 10 % 2.5 % 0.5 % 0.25 % 2 1.0 0.8 0.6 0.4 0.2 0.1 200 300 400 500 600 700 800 1000 1200 14001600
m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ
Microsoft Word - note02.doc
年度 物理化学 Ⅱ 講義ノート. 二原子分子の振動. 調和振動子近似 モデル 分子 = 理想的なバネでつながった原子 r : 核間距離, r e : 平衡核間距離, : 変位 ( = r r e ), k f : 力の定数ポテンシャルエネルギー ( ) k V = f (.) 古典運動方程式 [ 振動数 ] 3.3 d kf (.) dt μ : 換算質量 (m, m : 原子, の質量 ) mm
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Bethe-Bloch Bethe-Bloch (stopping range) Bethe-Bloch FNAL (Fermi National Accelerator Laboratory) - (SciBooNE ) SciBooNE Bethe-Bloch FNAL - (SciBooNE
21 2 27 Bethe-Bloch Bethe-Bloch (stopping range) Bethe-Bloch FNAL (Fermi National Accelerator Laboratory) - (SciBooNE ) SciBooNE Bethe-Bloch FNAL - (SciBooNE ) Bethe-Bloch 1 0.1..............................
simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =
II 6 ishimori@phys.titech.ac.jp 6.. 5.4.. f Rx = f Lx = fx fx + lim = lim x x + x x f c = f x + x < c < x x x + lim x x fx fx x x = lim x x f c = f x x < c < x cosmx cosxdx = {cosm x + cosm + x} dx = [
(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t
![(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t](/thumbs/67/56612994.jpg "(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t")
6 6.1 6.1 (1 Z ( X = e Z, Y = Im Z ( Z = X + iy, i = 1 (2 Z E[ e Z ] < E[ Im Z ] < Z E[Z] = E[e Z] + ie[im Z] 6.2 Z E[Z] E[ Z ] : E[ Z ] < e Z Z, Im Z Z E[Z] α = E[Z], Z = Z Z 1 {Z } E[Z] = α = α [ α ]
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)
Gmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r