|
|
|
- しのぶ ふじつぐ
- 7 years ago
- Views:
Transcription
1 BaHfO 3 PLD GdBa 2 Cu 3 O 7 δ
2 REBaCuO IBAD PLD MOD i
3 SQUID SQUID J c -B J c -B F p -B BHO BZO ii
4 iii
5 1.1 B c L d dx ( ) ( ) c K J c -B K J c -B K J c -B K U K F p -B K F p -B P1 20 K E J H1 20 K E J Z1 20 K E J iv
6 Leiden Kamerlingh-Onnes K 1957 Bardeen Cooper Schriffer BCS ( T c ) 30 K Johames G.Bednorz Karl Alex Müller (La-Ba-Cu-O) 30 K (77 K) Y-Ba-Cu-O Bi-Sr-Ca-Cu-O T c (Meissner ) Meissner B c T c B c1 Meissner B c2 B c2 1
7 1.2 B c2 B c Lorentz F L F L J B F L = J B F L v E = B v F L F p JB = J B JB F p J c J c = F p /B T c B c2 J c 1.3 NbTi Nb 3 Sn T c K T c (CuO 2 ) c ab c ab J c 2
8 Lorentz B c2 1.1 B c2 J c 0 J c = 0 J c 0 B i 1.1 J c B c2 3
9 [1] 1.2 Lorentz Lorentz ( A C) B U U 4
10 T k B T (k B Boltzmann ) U Arrhenius exp( U/k B T ) a f a f a f a f ν 0 Lorentz v + ( v + = a f ν 0 exp U ) k B T nu 0 (1.1) ν 0 = ζρ fj c0 2πa f B (1.2) ζ ζ 2π a f ζ = 4 ρ f J c0 Lorentz v ( v = a f ν 0 [exp U ) exp ( U )] k B T k B T (1.3) U Lorentz E = B v ( E = Ba f ν 0 [exp U ) exp ( U )] k B T k B T (1.4) E J c0 ) m ( J c0 = A (1 TTc B γ 1 1 B ) 2 (1.5) B c2 A m γ x 1.2 5
11 F (x) = U 0 2 sin(kx) fx (1.6) f V Lorentz f = JBV U 0 /2 k = 2π/a f x = x 0 x = x 0 0 F (x) = 0 x 0 = a ( ) f faf 2π cos 1 U 0 π U U = F (x 0 ) F ( x 0 ) [ ( )] U = U 0 sin cos 1 faf U 0 π { ( ) } = U 0 2f 1 U 0 k fa f π (1.7) ( ) faf cos 1 U 0 π ) (1.8) 2f ( 2f U 0 k cos 1 U 0 k sin(cos 1 x) = 1 x 2 U = 0 2f/U 0 k = 2J c0 BV/U 0 k = 1 J = J c0 2f U 0 k = J J c0 j (1.9) j (1.8) k = 2π/a f (1.9) U(j) = U 0 [(1 j 2 ) 1 2 j cos 1 j] (1.10) U (j) U + fa f = U + πu 0 j (1.11) (1.4) ( E = Ba f ν 0 exp U(j) ) [ ( 1 exp πu )] 0j k B T k B T (1.12) 6
12 Lorentz 1.2 Lorentz 1.3 Lorentz J B Lorentz Lorentz δ = v/ v J B δf p0 = 0 (1.13) F p0 J c0 J = F p0 /B = J c0 J > J c0 7
13 (1.13) J B δf p0 B ϕ 0 ηv = 0 (1.14) ϕ 0 η (1.14) J c0 = F p /B E = B v J J = J c0 + E ρ f (1.15) ρ f = Bϕ 0 /η (1.15) E E = ρ f (J J c0 ) (1.16) U 0 Û0 V U 0 = Û0V (1.17) Û0 Labusch α L d i Û 0 = 1 2 α Ld 2 i (1.18) F p F p = J c0 B = α L d i (1.19) d i ζ a f d i = a f ζ (1.20) (1.18) 1.28) U 0 = 1 2ζ J c0ba f V (1.21) 8
14 [2] (1.21) U 0 a f ϕ 0 a f = (2ϕ 0 / 3B) 1/2 R L R L (1.21) U R a f R = ga f (1.22) g 2 J c g 2 = g 2 e [ 5kB T 2U e ( Baf ν 0 log E c )] 4 3 (1.23) [3] g 2 e g2 U e g 2 = g 2 e U 0 g 2 e g 2 e = C 0 66 C 0 66 = B c 2 B 4µB c2 C πJ c0 Ba f (1.24) ( 1 B B c2 2 [3] B c L L = ( C44 α L ) 1 2 = ( Baf ζµ 0 J c0 ) ) 1 2 (1.25) (1.26) C 44 = B 2 /µ 0 L d L < d L > d 9
15 L > d L V V = R 2 L U 0 U 0 = 0.835g2 k B J 1/2 c0 (1.27) ζ 3/2 B 1/4 L > d 2 d V = R 2 d U 0 U 0 = 4.23g2 k B J c0 d ζb 1/2 (1.28) (1/2)(2/ 3) 7/4 (ϕ 7 0/µ 2 0) 1/ k B (1/2)(2/ 3) 3/2 ϕ 3/ k B U 0 k B T E cp E ff j 1 ( E cr = Ba f ν 0 exp U(j) ) [ ( 1 exp πu )] 0j k B T k B T (1.29) E ff = 0 (1.30) 1.4 L d 10
16 j > 1 ( E cr = Ba f ν 0 [1 exp πu )] 0 k B T (1.31) E ff = ρ f (J J c0 ) (1.32) E E = (E 2 cr + E 2 ff) 1/2 (1.33) (1.5) A f(a) = K exp [ (log A log A m) 2 ] 2σ 2 (1.34) K σ 2 A m A A E(J) = 0 E f(a)da (1.35) (1.35) E-J (0 x 2d) z (0 x d) x y J B = µ 0 (H e Jx) (1.36) x = 0 Maxwell B E = d B t = µ 0d 2 2 J t 11 (1.37)
17 (1.4) U U J U U (1.4) 2 U J U = U 0 sj U 0 J 0 s = U 0 /J c0 U = U 0 ( 1 J ) J c0 (1.38) [ J t = 2Ba fν 0 µ 0 d 2 exp U 0 k B T ( 1 J J c0 t = 0 J = J c0 J = 1 k ( BT 2Baf ν 0 U ) 0 t J c0 U0 log µ 0 d 2 J c0 k B T d ( ) J = k BT d(logt) J c0 U0 )] (1.39) (1.40) (1.41) U0 U0 U 0 U 0 [4] [5] U0 1.6 REBaCuO 1 REBaCuO(REBCO) (RE: ) J c RE Gd GdBCO GdBa 2 Cu 3 O 7 δ (δ ) GdBCO 12
18 REBCO (RE: ) J c c F p [6] 1.7 c REBCO Y 2 BaCuO 3 Y 2 O 3 BaZrO 3 (BZO) BaSnO 3 (BSO) PLD(Pulsed Laser Deposition) c J c MOD(Metal Organic Deposition) (ISTEC-SRL) Hf( ) BaHfO 3 PLD GdBCO IBAD IBAD(Ion Beam Assisted Deposition) [7] REBCO Ar + 13
19 IBAD MgO 2 CeO 2 IBAD PLD PLD(Pulsed Laser Deposition) GdBCO PLD PLD MOD c J c MOD MOD(Metal Organic Deposition) (TFA) MOD PLD c J c 1.10 GdBCO J c Superconducting Magnetic Energy Storage/ (SMES) ( 10 T) B ab J c BZO BSO 14
20 J c T c J c BHO GdBCO SQUID J c ( BZO ) 15
21 2 2.1 (ISTEC- SRL) GdBa 2 Cu 3 O 7 δ (GdBCO) Hastelloy IBAD(Ion Beam Assisted Deposition) MgO PLD(Pulsed Laser Deposition) CeO 2 PLD GdBCO Ag Ag/GdBa 2 Cu 3 O 7 δ /PLD-CeO 2 /IBAD-MgO/Hastelloy SQUID(Superconducting Quantum Interference Device) 4 2 mm 16
![2.1 2.1.3 2.1 BHO BZO 1.0[µm] 2.5[µm] G 2.2.2 2.1 Process Thickness d[µm] T c [K] G P1 GdBCO 1.1 90.7 1.00 10 3 H1 GdBCO+BHO(3.5mol%) 1.0 90.5 7.](/docs-images/82/85528011/images/22-0.jpg "75 10 4 Z1 GdBCO+BZO(3.5mol%) 1.1 89.2 8.56 10 4 P2 GdBCO 2.6 91.4 2.03 10 3 H2 GdBCO+BHO(3.5mol%) 2.5 89.7 1.99 10 3 Z2 GdBCO+BZO(3.5mol%) 2.4 89.8 2.")
22 BHO BZO 1.0[µm] 2.5[µm] G Process Thickness d[µm] T c [K] G P1 GdBCO H1 GdBCO+BHO(3.5mol%) Z1 GdBCO+BZO(3.5mol%) P2 GdBCO H2 GdBCO+BHO(3.5mol%) Z2 GdBCO+BZO(3.5mol%)
23 SQUID c 0 T 7 T 7 T 0 T 0 T M[emu] (J c -B) l w (l > w) x y z Bean 2.2 dx di c z dz di c = J c dxdz dx S S x ( S = 4x x + l w ) 2 = 4x 2 + 2x(l w) (2.1) dm = SdI c m = dm = S(x)J c dxdz = J c d S(x)dx (2.2) d m = J cw 2 (3l w)d (2.3) 12 18
24 2.2 dx 2.3 M m (2.3) m = J cw 2 (3l w)d (2.4) 12 m M = J cw (3l w) (2.5) 12l J c = 6l M (2.6) w(3l w) SQUID [emu] SI M[A/m] = M[emu] 10 3 (2.7) SQUID 19
25 B y l w x 2.3 ( ) ( ) Maxwell E-J V/m J (2.3) J = 12m w 2 d(3l w) (2.8) 2.4 Φ Φ = wlb e + µ 0m d (2.9) Faraday d w l E 1 E = 2(l + w) dφ dt (2.10) 20
26 B 0 Be 2.4 c E G (2.9) (2.10) E E = µ 0G 2d(l + w) dm dt (2.11) G d l L 1 = (µ 0 l/2) log(8l/d) d, l L 2 = πµ 0 l 2 /4d G = L 1 /L 2 (2.8), (2.11) SQUID E-J 21
27 3 3.1 J c -B J c [A/m 2 ] T=77.3 K GdBCO GdBCO+BHO GdBCO+BZO d 1.0 µm d 2.5 µm P1 H1 Z1 P2 H2 Z B [T] K J c -B K J c -B 0 T BHO J c J c BHO BZO J c 0.3 T BHO BZO J c BHO, 22
28 J c d J c BZO 3.2 3T J c J c Z2 3.2 J c -B 20, 77.3 K J c -B 3.2, K J c -B BHO J c 2 T BZO 2 J c BZO J c BHO J c H2 J c = A/m T 20 K J c -B BHO J c 20 K BHO BZO J c BHO J c BZO J c [A/m 2 ] T=77.3 K GdBCO GdBCO+BHO GdBCO+BZO P1 H1 Z1 d 1.0 µm d 2.5 µm P2 H2 Z2 J c [A/m 2 ] T=20 K d 1.0 µm d 2.5 µm GdBCO GdBCO+BHO GdBCO+BZO #1 #2 #3 #4 #5 # B [T] B [T] K J c -B K J c -B 23
29 J c 1.5 J c J c K 20 K 2 2 J c J c 3.3 SMES U0 SMES 20 K 1 6 T (1.41) U BHO U0 U 0 4 T BHO U 0 BHO 4 T U 0 BZO 2 T BHO H2 U 0 H1 0.1 GdBCO GdBCO+BHO GdBCO+BZO d 1.0 µm d 2.5µm P1 H1 Z1 P2 H2 Z2 U 0 * [ev] T = 20K B [T] K U 0 24
30 H1 H2 BHO 3.2 J c SMES 3.4 F p -B 20 K, 77.3 K F p -B K F p -B BHO BZO 4 GN/m K F p -B 20 K BHO F p H1 280 GN/m 3 BHO BZO F p [GN/m 3 ] 4 T=77.3 K GdBCO GdBCO+BHO GdBCO+BZO d 1.0 µm d 2.5 µm #1 #2 #3 #4 #5 #6 F p [GN/m 3 ] T=20 K B [T] GdBCO GdBCO+BHO GdBCO+BZO d 1.0 µm #1 #2 #3 #4 #5 #6 d 2.5 µm B [T] K F p -B K F p -B 25
31 BZO A m σ 2 γ g 2 (2.8) (2.11) 20 K E-J 20 K m A m σ 2 γ g 2 P H Z A m H1 σ 2 H1, Z1 P1 P1 26
32 E [V/m] T 1T 2T 3T 4T 5T 6T P1 E [V/m] T 2T 3T 4T 5T 6T H T=20 K GdBCO J [A/m 2 ] T=20 K GdBCO+BHO J [A/m 2 ] P1 20 K E J 4.2 H1 20 K E J E [V/m] T 2T 3T 4T 5T 6T Z T=20 K GdBCO+BZO J [A/m 2 ] Z1 20 K E J 27
33 γ H1 H1 g BHO BZO 3.4 BHO BZO BHO BZO 3.5 mol% c GdBCO BHO T c 0.2 K BZO 1.3 K 0 T BZO J c BHO J c J c T c TEM(Transmission Electron Microscope) c GdBCO BHO 5.9%, BZO 6.5% T c BHO BZO J c T c 1.7 T J c [8] [9] J c BHO BZO BHO BZO 28
34 5 BHO BHO BZO GdBCO J c -B 77.3 K BHO J c H2 J c H1 H1 H2 J c 0 T J c BHO BZO J c 20 K J c -B BHO J c BHO 2 J c J c 20 K U0 -B BHO 4 T U0 BHO 20 K J c -B SMES F p -B 77.3, 20 K BHO BZO 20 K E-J BHO A m, γ BHO BZO 29
35 BHO BZO BZO 30
36 6 ISTEC-SRL 31
37 [1] K. Yamafuji, T. Fujiyoshi, K. Toko and T.Matsushita: Physica C 159 (1989) 743 [2] (1998) [3] T. Matsushita, Physica C 217 (1993) 461 [4] (2009) [5] (2009) [6] M. Miura, at al.: Applied Physics Express 2 (2009) [7] RE123, 115, p.46-54, (2008) [8] T. Matsushita, M. Kiuchi, T. Haraguchi, T. Imada, K. Okamura, S. Okayasu, S. Uchida, J. Shimoyama, K. Kishio, Supercond. Sci. T echnol., 19, (2006). [9] M. Namba, S. Awaji, K. Watanabe, T. Nojima, S. Okayasu, P hysica C, 468, (2008). 32
Ni PLD GdBa 2 Cu 3 O 7 x 2 6
Ni PLD GdBa 2 Cu 3 O 7 x 2 6 1 1 1.1.................................. 1 1.2................................ 2 1.2.1......................... 3 1.3 RE 1 Ba 2 Cu 3 O 7 x...................... 3 1.3.1...............................
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
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
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(
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..................................
微分積分 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. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 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. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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
. 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.........
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......................
量子力学 問題
3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,
[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
S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
2008/02/18 08:40-10:10, 12:50-14:20 14:30-16:00, 16:10-17:40,
008/0/18 08:40-10:10, 1:50-14:0 14:30-16:00, 16:10-17:40, 1pt A 1911 Leiden Heike Kammelingh-Onnes H.Kammelingh Onnes 1907 He 1 4. K H H c T c T H c Hg:40 mt, Pb:80 mt, Sn:30 mt 100 mt I c H c H c H
n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
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....................................
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)
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....
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C602E646F63>
スピントロニクスの基礎 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/077461 このサンプルページの内容は, 初版 1 刷発行時のものです. i 1 2 ii 3 5 4 AMR (anisotropic magnetoresistance effect) GMR (giant magnetoresistance
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
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
: 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(
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
物理化学I-第12回(13).ppt
I- 12-1 11 11.1 2Mg(s) + O 2 (g) 2MgO(s) [Mg 2+ O 2 ] Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) - 2Mg(s) 2Mg 2+ (s) + 4e +) O 2 (g) + 4e 2O 2 (s) 2Mg(s) + O 2 (g) 2MgO(s) Zn(s) Zn 2+ (aq) + 2e +) Cu 2+ (aq)
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. 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.
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
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
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................
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
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
2 1 7 - TALK ABOUT 21 μ TALK ABOUT 21 Ag As Se 2. 2. 2. Ag As Se 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 1 Sb Ga Te 2. Sb 2. Ga 2. Te 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 1 1 2 3 4
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
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
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
( ) 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
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
(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
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
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/
1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (
![18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (](/thumbs/100/143999768.jpg "18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (")
8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin
(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z
B 4 24 7 9 ( ) :,..,,.,. 4 4. f(z): D C: D a C, 2πi C f(z) dz = f(a). z a a C, ( ). (ii), a D, a U a,r D f. f(z) = A n (z a) n, z U a,r, n= A n := 2πi C f(ζ) dζ, n =,,..., (ζ a) n+, C a D. (iii) U a,r
Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e
7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z
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
all.dvi
72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G
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 =
limit&derivative
- - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................
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
(yx4) 1887-1945 741936 50 1995 1 31 http://kenboushoten.web.fc.com/ OCR TeX 50 yx4 e-mail: [email protected] i Jacobi 1751 1 3 Euler Fagnano 187 9 0 Abel iii 1 1...................................
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
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
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
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
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
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
³ÎΨÏÀ
2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p
I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
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.............................
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
II 2 II
II 2 II 2005 [email protected] 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................
(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b)
(5) 74 Re, bondar laer (Prandtl) Re z ω z = x (5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 76 l V x ) 1/ 1 ( 1 1 1 δ δ = x Re x p V x t V l l (1-1) 1/ 1 δ δ δ δ = x Re p V x t V
C el = 3 2 Nk B (2.14) c el = 3k B C el = 3 2 Nk B
I [email protected] 217 11 14 4 4.1 2 2.4 C el = 3 2 Nk B (2.14) c el = 3k B 2 3 3.15 C el = 3 2 Nk B 3.15 39 2 1925 (Wolfgang Pauli) (Pauli exclusion principle) T E = p2 2m p T N 4 Pauli Sommerfeld
V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H
199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)
Z: Q: R: C:
0 Z: Q: R: C: 3 4 4 4................................ 4 4.................................. 7 5 3 5...................... 3 5......................... 40 5.3 snz) z)........................... 4 6 46 x
.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T
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
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
φ 4 Minimal subtraction scheme 2-loop ε 2008 (University of Tokyo) (Atsuo Kuniba) version 21/Apr/ Formulas Γ( n + ɛ) = ( 1)n (1 n! ɛ + ψ(n + 1)
φ 4 Minimal subtraction scheme 2-loop ε 28 University of Tokyo Atsuo Kuniba version 2/Apr/28 Formulas Γ n + ɛ = n n! ɛ + ψn + + Oɛ n =,, 2, ψn + = + 2 + + γ, 2 n ψ = γ =.5772... Euler const, log + ax x