H21環境地球化学6_雲と雨_ ppt
|
|
|
- えみ おえづか
- 4 years ago
- Views:
Transcription
1 1
2 2
3 (0.001%) 71 24,000 (1.7%) ,350,000 (97%) 125 (0.009%) 40 10,000 (0.7%) 25 (0.002%) 10 3 km km 3 /y
4 (21) () ( )
5 5 HNO 3, SO 2 etc
6 6
7 7 2009年度 環境地球化学 大河内 10種雲形と発生高度 15 km 巻雲 巻積雲 巻層雲 10 km 高積雲 5 km 高層雲 乱層雲 層積雲 0 km 層雲 上層雲 巻雲 巻積雲 対流圏上層の雲 ほとんど氷晶 中層雲 高積雲 上層 一部氷晶 下層雲を刺激 (シーダー) 下層雲 層雲 凝結核 多 水分量 高 雲水量(LWC) 大
8 8
9 9 PV γ = const γ= C p / C v cf. PV = const ( ) C p C v T 2 T 1 = V 1 V 2 γ- 1
10 10 P 2 T 2 V 2-47 q P 1 T 1 V 1
11 年度 環境地球化学 大河内 雨の生成機構 暖かい雨 水蒸気が大量に発生する低緯度 冷たい雨 中 高緯度
12 12 µ mm 0.1 mm 0.1 mm 3 mm
13 年度 環境地球化学 大河内 静止空気中を落下する水滴の形 13
14 S/V (cm -1 ) V t (cm sec -1 ) k g (cm sec -1 ) S/V = 6 D p k g V t D p V t = exp k g k g = K g D p D p V t ν 1 2 ν K g Diameter D p (mm) K g 14 ν
15 15 (Cloud Condensation Nuclei,CCN) (nucleation scavenging) CCN (Cloud Interstitial Aerosol,CIA) CCN
16 16 µm < 1 µm CCN Junge d 0.4 µm 0.4 µm d 2 µm d Whitby (fine) d 2 µm d 0.1 µm 0.1 µm d (coarse) 2 µm d 2 2 µm µm
17 17 1 % CCN 10 2 cm cm -3 S N C N c = CS k (0.2 < k <0.5) C=310, k=1/3 C=6000, k=2/5
18 18 CCN 1 % µm 0.1 µm 0.01 CCN 1 % % CCN
19 19
20 20 (Clausius-Clapeyron) 1 de s e dt = ΔH RT 2 s e s ΔH
21 21 r* ( ) e 2σ e = e exp s nktr * e s σ n k (1.381 x J deg -1 molecile -1 ) T (K) = (e e S )/e S 100
22 µm X 0.4%(X) 0.4 % ( µm) 0.5
23 23 () e e = f eʼ e f
24 24 m (M s ) r i(m / M s ) 4 3 πr3 ρ m / M w ρʼ M w i f = 4 3 πr3 ρ m / M w = πr3 ρ m / M w + im / M s imm w 4 M s 3 πr3 ρ m 1
25 25 i = C C =1+ (n 1)α Π = CRT i R C (mol/l)
26 26 r*eʼ e = exp 2 σ e s n ktr * 1+ ( ) eʼ e s imm w 4 M s 3 πr3 ρ m 1
27 % ( < 0.5 µm) 1 A 2 B ( ) %
28 28
29 29 20µm: ( ph CCN 1
30 30 interceptional collision inertial collision diffusional collision
31 31 (d 0.1 µm) (d 1 µm) 0.1 µm d µm 1 Greenfield gap
32 32
33 33 (Liquid Water Content, LWC) L (g of water m -3 ) (g of water m -3 ) (liquid water mixing ratio) w L w L (vol water/vol air) = L 10-6 (g m -3 )
34 34 C f A aq (mol L 1 air ) C g (mol L 1 air ) p g = C g RT C aq = [C A (aq)](mol L water 1 ) w L (L water L air 1 ) K H p g w L =10 6 K H p g L =
35 35 f A = K H p g w L RT p g = K H RTw = L 10 6 K H RTL X A aq = C aq C aq + C g = f A C g C g ( f A +1) = f A 1+ f A K H RTw L = 10 6 K H RTL X A aq = 1+ K H RTw L K H RTL
36 36 Moderately soluble Very soluble
37 37 SO 2 + H 2 O ( ) SO 2 H 2 O (aq) [SO 2 H 2 O] = K H (SO 2 ) p SO2 SO 2 H 2 O (aq) HSO H + (aq) K = [HSO ][H + ] 3 a1 [SO H O] 2 2 HSO 3 - SO H + (aq) K = [SO 2 ][H + ] 3 a 2 [HSO ] 3
38 38 [SO 2 (aq)] tot = [S(IV )] tot = [SO 2 H 2 O(aq)]+ [HSO 3 (aq)]+ [SO 3 2 (aq)] = [SO H O(aq)] 1+ K a1 2 2 [H + ] + K a1k a2 [H + ] 2 K (SO ) 1+ K a1 H 2 [H + ] + K a1k = a 2 [H + ] 2 p SO2 K eff (SO 2 )
39 39 K eff (SO 2 ) K eff (CO 2 ) ph 1 8 K eff (SO 2 ) K eff (CO 2 ) 44
40 40 SO 2 H 2 O HSO 3- SO 2-3 S SO 4 2- S 6 H 2 O 2 O 3 O 2 ( ) NO 2
41 41 H 2 O 2 (aq) + H 2 O ( ) HO 2- (aq) + H 3 O + (aq) HSO 3- (aq) + H 2 O 2 (aq) SO 2 OOH 2- (aq) + H 2 O ( ) SO 2 OOH 2- (aq) + H 3 O + (aq) H 2 SO 4 (aq) + H 2 O ( )
42 42 d[s(iv )] dt = k[h 3 O+ (aq)][h 2 O 2 ][HSO 3 (aq)] 1+ K[H 3 O + (aq)] k = (7.5 ±1.6) 10 7 M 1 s 1, K =13 M 1 ph ph S(IV)
43 43 ph 5.5 O 3 H 2 O 2 S(IV) + O 3 (aq) S(VI) + O 2 (aq) d[s(iv )] dt = {k 0 [SO 2 H 2 O(aq)] + k 1 [HSO 3 (aq)] +k 2 [SO 3 2 (aq)]}[o 3 (aq)] k 0 = (2.4 ±1.1) 10 4 M 1 s 1 k 1 = (3.7 ± 0.7) 10 5 M 1 s 1 k 2 = (1.5 ± 0.6) 10 9 M 1 s 1
44 44
45 45 1) ΔV = 0 2) ΔP = 0 3) ΔT = 0 4) P q = 0 V
46 46 (reversible process) P ext (irreversible process) P V 1 46 P = nrt V T=const V 2 w rev w irrev V
47 47 H = U + PV ΔH = q p H U P V ΔH = ΔU + PΔV ΔH = ΔU + Δ(PV) (sensible heat)
48 48 ( ) q = CΔT Q, ΔT
49 49 C = lim Δq ΔT 0 ΔT = d q dt C v C v = d q dt dv = 0 = d q v dt = du dt C p C p = d q dt dp= 0 = d q p dt = dh dt
50 50 C p = + nr C v C p C v R PV dʼq = du - dʼw = du - (-P ext dv) = du + P ext dv P ext = P gas = P dʼq = du + PdV PV=nRT d(pv) = P dv + V dp = nr dt P dv = nr dt - V dp dʼq = du + nr dt - V dp
51 51 C αn ncα C(1-α) Cʼ =C[1+(n-1)α]
52 52
H22環境地球化学4_化学平衡III_ ppt
1 2 3 2009年度 環境地球化学 大河内 温度上昇による炭酸水の発泡 気泡 温度が高くなると 溶けきれなくなった 二酸化炭素が気泡として出てくる 4 2009年度 環境地球化学 圧力上昇による炭酸水の発泡 栓を開けると 瓶の中の圧力が急激に 小さくなるので 発泡する 大河内 5 CO 2 K H CO 2 H 2 O K H + 1 HCO 3- K 2 H + CO 3 2- (M) [CO
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
09_organal2
4. (1) (a) I = 1/2 (I = 1/2) I 0 p ( ), n () I = 0 (p + n) I = (1/2, 3/2, 5/2 ) p ( ), n () I = (1, 2, 3 ) (b) (m) (I = 1/2) m = +1/2, 1/2 (I = 1/2) m = +1/2, 1/2 I m = +I, +(I 1), +(I 2) (I 1), I ( )
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
( ) ,
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
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
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
64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k
63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5
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
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
Microsoft Word - 11問題表紙（選択）.docx
A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx
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](/thumbs/92/109118076.jpg "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
/ 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,
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)
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,
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
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(
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...........................
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 =
30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
![II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re](/thumbs/94/118770263.jpg "II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re")
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
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..................................
// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du
![// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du](/thumbs/91/105254510.jpg "// //( ) (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
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
基礎からの冷凍空調 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/067311 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. 03-3817-5670FAX 03-3815-8199 i () () Q&A
追加演習問題
v (.5 v x (.98 3R k R. v v x B M m M v x v y v z v 3 v x m v mol M v 3 U U R ****************** 500 kcal 60 kg (.3 0 m. mgh t w t (500 kcal (4.84 0 J kcal (4 h (60 min h (60 s min (60 kg (9.80 m s (60
(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
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
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
![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](/thumbs/91/106215968.jpg "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")
8 ( ) 8. 1 ( ) F F = F I (N I, T, V I ) + F II (N II, T, V II ) (8.1) F δf = δn I [ ( FI (N I ) N I 8. 1 111 ) 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
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
2
th 37 ICh Theoretical Examination - - - - 5 - - : - ( ): ( ) - : 279 - - - - - - - G D L U C K 1 2 1 amu = 1.6605 10-27 kg N = 6.02 10 23 mol -1 k = 1.3806503 10-23 J K -1 e = 1.6022 10-19 C F = 9.6485
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
N/m f x x L dl U 1 du = T ds pdv + fdl (2.1)
23 2 2.1 10 5 6 N/m 2 2.1.1 f x x L dl U 1 du = T ds pdv + fdl (2.1) 24 2 dv = 0 dl ( ) U f = T L p,t ( ) S L p,t (2.2) 2 ( ) ( ) S f = L T p,t p,l (2.3) ( ) U f = L p,t + T ( ) f T p,l (2.4) 1 f e ( U/
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)
6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1
6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) (e) Γ (6.2) : Γ B A R (reversible) 6-1 (e) = Clausius 0 = B A: Γ B A: Γ d Q A + d Q (e) B: R d Q + S(A) S(B) (6.3) (e) // 6.2 B A: Γ d Q S(B) S(A) = S (6.4) (e) Γ (6.5)
m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2
3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1
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........................
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(
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
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 {
1
I II II 1 dw = pd = 0 1 U = Q (4.10) 1K (heat capacity) (mole heat capacity) ( dq / d ) = ( du d C = / ) (4.11) du = C d U = C d (4.1) 1 1 du = dq + dw dw = pd dq = du + pd (4.13) p dq = d( U + p ) p (4.14)
meiji_resume_1.PDF
β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E
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)
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..................
,., 5., ,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c
29 2 1 2.1 2.1.1.,., 5.,. 2.1.1,. 2.2,., x z. y,.,,,. du dt + α p x = 0 dw dt + α p z + g = 0 α dp dt + pγ dα dt = 0 α V dα dt = 0 (2.2.1), γ = c p /c v., V = (u, w), = ( / x, / z). 30 2.1.1: 31., U p(z),
i 18 2H 2 + O 2 2H 2 + ( ) 3K
i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................
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,
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](/thumbs/82/86676506.jpg "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)
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
Donald Carl J. Choi, β ( )
:: α β γ 200612296 20 10 17 1 3 2 α 3 2.1................................... 3 2.2................................... 4 2.3....................................... 6 2.4.......................................
2013 25 9 i 1 1 1.1................................... 1 1.2........................... 2 1.3..................................... 3 1.4..................................... 4 2 6 2.1.................................
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
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..............................
(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
: 2005 ( ρ t +dv j =0 r m m r = e E( r +e r B( r T 208 T = d E j 207 ρ t = = = e t δ( r r (t e r r δ( r r (t e r ( r δ( r r (t dv j =
72 Maxwell. Maxwell e r ( =,,N Maxwell rot E + B t = 0 rot H D t = j dv D = ρ dv B = 0 D = ɛ 0 E H = μ 0 B ρ( r = j( r = N e δ( r r = N e r δ( r r = : 2005 ( 2006.8.22 73 207 ρ t +dv j =0 r m m r = e E(
2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP
1. 1 213 1 6 1 3 1: ( ) 2: 3: SF 1 2 3 1: 3 2 A m 2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP
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
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
36 th IChO : - 3 ( ) , G O O D L U C K final 1
36 th ICh - - 5 - - : - 3 ( ) - 169 - -, - - - - - - - G D L U C K final 1 1 1.01 2 e 4.00 3 Li 6.94 4 Be 9.01 5 B 10.81 6 C 12.01 7 N 14.01 8 16.00 9 F 19.00 10 Ne 20.18 11 Na 22.99 12 Mg 24.31 Periodic
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
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
Sample function Re random process Flutter, Galloping, etc. ensemble (mean value) N 1 µ = lim xk( t1) N k = 1 N autocorrelation function N 1 R( t1, t1
Sample function Re random process Flutter, Galloping, etc. ensemble (mean value) µ = lim xk( k = autocorrelation function R( t, t + τ) = lim ( ) ( + τ) xk t xk t k = V p o o R p o, o V S M R realization
05Mar2001_tune.dvi
2001 3 5 COD 1 1.1 u d2 u + ku =0 (1) dt2 u = a exp(pt) (2) p = ± k (3) k>0k = ω 2 exp(±iωt) (4) k
KENZOU Karman) x
KENZO 8 8 31 8 1 3 4 5 6 Karman) 7 3 8 x 8 1 1.1.............................. 3 1............................................. 5 1.3................................... 5 1.4 /.........................
) ] [ 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
![) ] [ 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](/thumbs/92/107866707.jpg ") ] [ 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
006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................
1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ 2 2
1.500 m X Y 0.200 m 0.200 m 0.200 m 0.200 m 0.200 m 0.000 m 1.200 m m 0.150 m 0.150 m m m 2 24.5 N/ 3 18.0 N/ 3 30.0 0.60 ( ) qa 50.79 N/ 2 0.0 N/ 2 20.000 20.000 15.000 15.000 X(m) Y(m) (kn/m 2 ) 10.000
a a b a b c d e R c d e A a b e a b a b c d a b c d e f a M a b f d a M b a b a M b a M b M M M R M a M b M c a M a R b A a b b a CF a b c a b a M b a b M a M b c a A b a b M b a A b a M b C a M C a M
() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
001_046_物理化学_解答_責_2刷_Z06.indd
演習問題の解答例 ( 注 ) ( 注 ) 第 章.5 dt 7 d ( 7 dl l ( ) l ) ( ) 7 ( mi) l l. 5596,. 865 mi 7 4 d t. 865 mi l. 8 mi.6.7 R b R b R a + ( b) R ( b) f x l x.8 f f f f ( x ) f + x + x + + x!!! f f ( x) R a + ( b) R
素粒子物理学２ 素粒子物理学序論B 2010年度講義第2回
素粒子物理学２ 素粒子物理学序論B 2010年度講義第2回 =1.055 10 34 J sec =6.582 10 22 MeV sec c = 197.33 10 15 MeV m = c = c =1 1 m p = c(mev m) 938M ev = 197 10 15 (m) 938 =0.2 10 13 (cm) 1 m p = (MeV sec) 938M ev = 6.58
untitled
- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =
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
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)
閨75, 縺5 [ ｨ ﾁ573, 縺 ｨ ｨ
39ｨ 8 998 3. 753 68, 7 86 ﾀ7 9 9989769 438 縺48 縺55 3783645 ﾀ5 縺473 ﾀ7996495 ｨ 59754 8554473 9 8984473 3553 7. 95457357, 4.3. 639745 5883597547 6755887 67996499 ｨ 597545 4953473 9 857473 3553, 536583, 89573,
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
2.1: n = N/V ( ) k F = ( 3π 2 N ) 1/3 = ( 3π 2 n ) 1/3 V (2.5) [ ] a = h2 2m k2 F h2 2ma (1 27 ) (1 8 ) erg, (2.6) /k B 1 11 / K
![2.1: n = N/V ( ) k F = ( 3π 2 N ) 1/3 = ( 3π 2 n ) 1/3 V (2.5) [ ] a = h2 2m k2 F h2 2ma (1 27 ) (1 8 ) erg, (2.6) /k B 1 11 / K](/thumbs/86/94275608.jpg "2.1: n = N/V ( ) k F = ( 3π 2 N ) 1/3 = ( 3π 2 n ) 1/3 V (2.5) [ ] a = h2 2m k2 F h2 2ma (1 27 ) (1 8 ) erg, (2.6) /k B 1 11 / K")
2 2.1? [ ] L 1 ε(p) = 1 ( p 2 2m x + p 2 y + pz) 2 = h2 ( k 2 2m x + ky 2 + kz) 2 n x, n y, n z (2.1) (2.2) p = hk = h 2π L (n x, n y, n z ) (2.3) n k p 1 i (ε i ε i+1 )1 1 g = 2S + 1 2 1/2 g = 2 ( p F
0 = m 2p 1 p = 1/2 p y = 1 m = 1 2 d ( + 1)2 d ( + 1) 2 = d d ( + 1)2 = = 2( + 1) 2 g() 2 f() f() = [g()] 2 = g()g() f f () = [g()g()]
![0 = m 2p 1 p = 1/2 p y = 1 m = 1 2 d ( + 1)2 d ( + 1) 2 = d d ( + 1)2 = = 2( + 1) 2 g() 2 f() f() = [g()] 2 = g()g() f f () = [g()g()]](/thumbs/92/110082312.jpg "0 = m 2p 1 p = 1/2 p y = 1 m = 1 2 d ( + 1)2 d ( + 1) 2 = d d ( + 1)2 = = 2( + 1) 2 g() 2 f() f() = [g()] 2 = g()g() f f () = [g()g()]")
8. 2 1 2 1 2 ma,y u(, y) s.t. p + p y y = m u y y p p y y m u(, y) = y p + p y y = m y ( ) 1 y = (m p ) p y = m p y p p y 2 0 m/p U U() = m p y p p y 2 2 du() d = m p y 2p p y 1 0 = m 2p 1 p = 1/2 p y
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
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
= = = = = = 1, 000, 000, 000, = = 2 2 = = = = a,
2011 10 26 1. 1627 27, 682, 574, 402 = 2 7 7 7 7 7 7 7 7 7 7 7 7 2 2 7 2 7 7..., 100 100 99 2 100 100 two to the hundred n n, n 2, n 3,... n 10... 10 10 1 = 10 10 2 = 10 10 = 100 10 3 = 10 10 10 = 1000
( ) ( ) 20 11 11-13 15 20 21 ( ) 114cm 100cm 85cm () () 11 18 19 19-25 26 ( 1 ) 1 2 10 ( ) () 11 16 19 21 24-13 20 3 20 ( ) ( ) 14 15 19 20 23 29 13 20 4/15 600 400 5 7 1 8 5 7 20 3 1999 1000 100 86
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) = [](/thumbs/99/141438442.jpg "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
0.45m1.00m 1.00m 1.00m 0.33m 0.33m 0.33m 0.45m 1.00m 2
24 11 10 24 12 10 30 1 0.45m1.00m 1.00m 1.00m 0.33m 0.33m 0.33m 0.45m 1.00m 2 23% 29% 71% 67% 6% 4% n=1525 n=1137 6% +6% -4% -2% 21% 30% 5% 35% 6% 6% 11% 40% 37% 36 172 166 371 213 226 177 54 382 704 216
10 117 5 1 121841 4 15 12 7 27 12 6 31856 8 21 1983-2 - 321899 12 21656 2 45 9 2 131816 4 91812 11 20 1887 461971 11 3 2 161703 11 13 98 3 16201700-3 - 2 35 6 7 8 9 12 13 12 481973 12 2 571982 161703 11
From Evans Application Notes
3 From Evans Application Notes http://www.eaglabs.com From Evans Application Notes http://www.eaglabs.com XPS AES ISS SSIMS ATR-IR 1-10keV µ 1 V() r = kx 2 = 2π µν x mm 1 2 µ= m + m 1 2 1 k ν = OSC 2
x E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx
x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
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
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,
K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................