l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r
|
|
|
- いちえい おうじ
- 6 years ago
- Views:
Transcription
1 2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2 w 2 1 γ α + iβ β 0 l 2 0 λ i 1 (7a)(7b) l 2 0 Re(γ) 1 r λ w + w i( w w ) 1 + w w w ( x w, y w, z w ), w ( x w, y w, z w ) λ 1 2 x w yz i( w zy 2 w ) 2 x w x r λ 2 y r λ 3 z r 1
2 l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r 2 2mxy 1 2my2 2myz 2mxz 2myz 1 2mz2 ds 2 (dx 0 ) 2 (dx) 2 2m r (dx0 + xdx + ydy + zdz ) 2 r m γ γ 1 r [(x a 1) 2 + (y a 2 ) 2 + (z a 3 ) 2 ] 1/2 a i const (1) a a γ γ 1 r [x2 + y 2 + (z ia) 2 ] 1/2 γ γ l 2 0 λ i (7a)(7b) w w + iσ [x 2 + y 2 + (z ia) 2 ] 1/2 (r 2 a 2 2iaz) 1/2 (2) r 2 x 2 + y 2 + z 2 σ σ 2
3 ( + iσ) 2 2 σ 2 + 2iσ r 2 a 2 2iaz 2 σ 2 r 2 a 2 σ az 4 2 (r 2 a 2 ) a 2 z r2 a 2 ± (r 2 a 2 ) 2 + 4a 2 z 2 2 r2 a 2 2 ± ( (r2 a 2 ) a 2 z 2 ) 1/2 ± + r a 2 r2 2 + r2 2 r2 r (2) a + iσ (r 2 a 2 2iaz) 1/2 r l0 2 γ γ 1 + iσ iσ ( + iσ)( iσ) 2 + σ 2 iσ 2 + σ 2 l 2 0 Re(γ) 2 + σ a2 z a 2 z 2 3
4 (7a) λ w w (2) w (r 2 a 2 2iaz) 1/2 ( x(r 2 a 2 2iaz) 1/2, y(r 2 a 2 2iaz) 1/2, (z ia)(r 2 a 2 2iaz) 1/2) r iak w w r + iak w r r (x, y, z) k k (0, 0, 1) z (7a) w w r2 + a 2 ww r2 + a 2 w 2, w w ( 2iya w 2, 2ixa w 2, 0) λ w + w i( w w ) 1 + w w w (r iak) + w(r + iak) + 2a(r k) w 2 + r 2 + a 2 ( iσ)(r iak) + ( + iσ)(r + iak) + 2a(r k) w 2 + r 2 + a 2 2[r aσk + a(r k)] w 2 + r 2 + a 2 w w 2 (r 2 a 2 2iaz) 1/2 (r 2 a 2 + 2iaz) 1/2 ((r 2 a 2 ) 2 + 4a 2 z 2 ) 1/2 (( 2 σ 2 ) 2 + 4σ 2 2 ) 1/2 (( 2 + σ 2 ) 2 ) 1/2 2 2 r 2 a 2 + 2a 2 λ λ 2[r aσk + a(r k)] w 2 + r 2 + a 2 [r + ak az + a(r k)] ( 2 + a 2 ) 2 + a 2 [r + a2 z 2 k + a (r k)] λ 4
5 λ 1 λ 2 λ 3 x + ay 2 + a 2 y ax 2 + a a 2 [z + a2 z 2 ] z r a l µ l 0 (1, x + ay y ax 2, + a2 2 + a 2, z ), 3 l a 2 z 2 g µν η µν 2ml µ l ν g m a 2 z 2, g m 3 + ay 4 + a 2 (x z2 2 + a 2 )2 g m 3 ax 4 + a 2 (y z2 2 + a 2 )2, g m a 2 z 2 (z )2 g 01 2m 3 x + ay 4 + a 2 z a 2, g 02 2m 3 y ax 4 + a 2 z a 2 g 03 2m 3 z 4 + a 2 z 2, g 12 2m 3 (x + ay)(y ax) 4 + a 2 z 2 ( 2 + a 2 ) 2 ds 2 g 13 2m 3 z(x + ay) 4 + a 2 z 2 ( 2 + a 2 ), g 23 2m 3 z(y ax) 4 + a 2 z 2 ( 2 + a 2 ) ds 2 (dx 0 ) 2 dx 2 2m3 4 + a 2 z 2 [(dx0 ) 2 x + ay + ( 2 + a 2 )2 dx 2 y ax + ( 2 + a 2 )2 dy 2 + ( z )2 dz 2 + 2(x + ay) 2 + a 2 dx 0 dx + 2(x + ay)(y ax) + ( 2 + a 2 ) 2 dxdy + 2(y ax) 2 + a 2 dx 0 dy + 2z dx0 dz 2z(x + ay) 2z(y ax) ( 2 + a 2 dxdz + ) ( 2 + a 2 ) dydz] 2m 3 /( 4 + a 2 z 2 ) (a + b + c + d) 2 a 2 + b 2 + c 2 + d 2 + 2ab + 2ac + 2ad + 2bc + 2cd + 2bd 5
6 [ ds 2 (dx 0 ) 2 dx 2 2m3 4 + a 2 z 2 dx 0 + (dx 0 ) 2 dx 2 2m3 4 + a 2 z 2 x + ay y ax 2 dx + + a2 2 + a 2 dy + z ] 2 dz [ dx 0 (xdx + ydy) a(ydx xdy) a a 2 + z dz ] 2 (3) (Kerr) θ cos θ z φ x + iy ( sin θ) cos φ + i( sin θ) sin φ sin θ(cos φ + i sin φ) ( sin θ)e iφ ia ( ia) sin θe iφ x + iy φ u x 0 + (3) dz 2 dz 2 (cos θd sin θdθ) 2 dx 2 + dy 2 dx 2 + dy 2 d(x + iy) 2 [sin θe iφ d + ( ia) cos θe iφ dθ + i( ia) sin θe iφ dφ] [sin θe iφ d + ( + ia) cos θe iφ dθ i( + ia) sin θe iφ dφ] sin 2 θd cos 2 θdθ 2 + a 2 cos 2 θdθ sin 2 θdφ 2 + a 2 sin 2 θdφ sin θ cos θddθ + 2a sin 2 θddφ (sin θd + cos θdθ + a sin θdφ) 2 + ( sin θdφ a cos θdθ) 2 6
7 xdx + ydy xdx + ydy 1 2 (2xdx + 2ydy) 1 2 d(x2 + y 2 ) 1 d x + iy d[(2 + a 2 ) sin 2 θ] sin 2 θd + ( 2 + a 2 ) sin θ cos θdθ xdy ydx xdy ydx Im[(x + iy)d(x iy)] Im[( ia)e iφ sin θd{( + ia)e iφ sin θ}] Im[( ia)e iφ sin θ{sin θe iφ d + ( + ia)e iφ cos θdθ i( + ia)e iφ sin θdφ}] Im[( ia) sin 2 θd i( 2 + a 2 ) sin 2 θdφ + ( 2 + a 2 ) sin θ cos θdθ] ( 2 + a 2 ) sin 2 θdφ + a sin 2 θd zdz zdz cos θ(cos θd sin θdθ) cos 2 θd 2 cos θ sin θdθ xdy ydx 2m a 2 z 2 2m a 2 2 cos 2 θ 2m 2 + a 2 cos 2 θ dx 0 du d (x ) 7
8 [ ds 2 (dx 0 ) 2 dx 2 2m3 4 + a 2 z 2 dx 0 (xdx + ydy) a(xdy ydx) a a 2 + z ] 2 dz (du 2 + d 2 2dud) [(sin θd + cos θdθ + a sin θdφ) 2 + ( sin θdφ a cos θdθ) 2 + (cos θd sin θdθ) 2 ] [ 2m 2 + a 2 cos 2 du d + ( sin2 θd + ( 2 + a 2 ) sin θ cos θdθ) θ 2 + a 2 + a{(2 + a 2 ) sin 2 θdφ + a sin 2 θd} 2 + a 2 + cos2 θd 2 ] 2 cos θ sin θdθ (du 2 + d 2 2dud) (d dθ 2 + (a ) sin 2 θdφ 2 + 2a sin 2 θddφ + a 2 cos 2 θdθ 2 ) 2m [ du d + d + a sin a 2 cos 2 θdφ ] 2 θ (du 2 2dud) ( 2 dθ 2 + (a ) sin 2 θdφ 2 + 2a sin 2 θddφ + a 2 cos 2 θdθ 2 ) (1 2m 2 + a 2 cos 2 θ [du2 + a 2 sin 4 θdφ 2 + 2a sin 2 θdφdu] 2m 2 + a 2 cos 2 θ )du2 ( 2 + a 2 cos 2 θ)dθ 2 [(a ) sin 2 θ + 2ma2 sin 4 θ 2 + a 2 cos 2 θ ]dφ2 2dud 4ma sin2 θ 2 + a 2 cos 2 θ dφdu 2a sin2 θddφ (4) φ (4) ω ds 2 (c 2 ω 2 r 2 )dt 2 (dr 2 + r 2 dφ 2 + 2ωr 2 dφdt + dz 2 ) dφdt (4) ddφ, dud ds 2 g 00 du 2 + g 22 dθ 2 + g 33 dφ 2 + 2g 03 dudφ + 2g 01 dud + 2g 13 ddφ (5) cˆt u A(), du cdˆt + A d ˆφ φ B(), dφ d ˆφ + B d A B (11) 8
9 ds 2 g 00 du 2 + g 22 dθ 2 + g 33 dφ 2 + 2g 03 dudφ + 2g 01 dud + 2g 13 ddφ g 00 (cdˆt + A d) 2 + g 22 dθ 2 + g 33 (d ˆφ + B d) 2 + 2g 03 (cdˆt + A d)(d ˆφ + B d) + 2g 01 (cdˆt + A d)d + 2g 13 (d ˆφ + B d)d g 00 [(cdˆt) 2 + A 2 d 2 + 2A cdˆtd] + g 22 dθ 2 + g 33 (d ˆφ 2 + B 2 d 2 + 2B d ˆφd) + 2g 03 (cdˆtd ˆφ + A dd ˆφ + A B d 2 + B cdˆtd) + 2g 01 (cdˆt + A d)d + 2g 13 (d ˆφ + B d)d g 00 (cdˆt) 2 + (g 00 A 2 + g 33 B 2 + 2g 03 A B + 2g 01 A + 2g 13 B )d 2 + g 22 dθ 2 + g 33 d ˆφ 2 + 2g 03 cdˆtd ˆφ + 2(g 33 B + g 03 A + g 13 )d ˆφd + 2(g 00 A + g 03 B + g 01 )cdˆtd (6) d ˆφd, cdˆtd 0 A B g 33 B + g 03 A + g 13 0 g 00 A + g 03 B + g 01 0 { A g03b +g 01 g 00 B g 03A +g 13 g 33 B g 13g 00 g 03 g 01 g 2 03 g 00g 33 A g 01g 33 g 03 g 13 g 2 03 g 33g 00 g µν A g 00 g 33 (1 2m 2 + a 2 cos 2 θ )[(a2 + 2 ) sin 2 θ + 2ma2 sin 4 θ 2 + a 2 cos 2 θ ] (a ) sin 2 θ 2ma2 sin 4 θ 2 + a 2 cos 2 θ + (a2 + 2 ) sin 2 2m θ 2 + a 2 cos 2 θ + (2m)2 a 2 sin 4 θ ( 2 + a 2 cos 2 θ) 2 g 2 03 (2ma sin2 θ) 2 ( 2 + a 2 cos 2 θ) 2 9
10 g 2 03 g 00 g 33 (a ) sin 2 θ + 2ma2 sin 4 θ 2 + a 2 cos 2 θ (a2 + 2 ) sin 2 2m θ 2 + a 2 cos 2 θ (2m)2 a 2 sin 4 θ ( 2 + a 2 cos 2 θ) 2 + (2ma sin2 θ) 2 ( 2 + a 2 cos 2 θ) 2 (a2 + 2 )( 2 + a 2 cos 2 θ) sin 2 θ + 2ma 2 sin 4 θ 2m(a ) sin 2 θ 2 + a 2 cos 2 θ A g 33 g 01 [(a ) sin 2 θ + 2ma2 sin 4 θ 2 + a 2 cos 2 θ ] g 03 g 13 a sin 2 θ 2ma sin2 θ 2 + a 2 cos 2 θ g 33 g 01 g 03 g 13 (a2 + 2 )( 2 + a 2 cos 2 θ) sin 2 θ + 2ma 2 sin 4 θ 2ma 2 sin 4 θ 2 + a 2 cos 2 θ (a2 + 2 )( 2 + a 2 cos 2 θ) sin 2 θ 2 + a 2 cos 2 θ A A (a )( 2 + a 2 cos 2 θ) sin 2 θ (a )( 2 + a 2 cos 2 θ) sin 2 θ + 2ma 2 sin 4 θ 2m(a ) sin 2 θ (a )( 2 + a 2 cos 2 θ) (a )( 2 + a 2 cos 2 θ) + 2ma 2 (1 cos 2 θ) 2m(a ) (a )( 2 + a 2 cos 2 θ) ( 2 + a 2 cos 2 θ)[(a ) 2m( 2 + a 2 cos 2 θ)/( 2 + a 2 cos 2 θ)] a a m B g 00 g 13 a sin 2 θ(1 2m 2 + a 2 cos 2 θ ) g 03 g 01 2ma sin2 θ 2 + a 2 cos 2 θ 10
11 g 00 g 13 g 03 g 01 a(2 + a 2 cos 2 θ) sin 2 θ 2 + a 2 cos 2 θ B a( 2 + a 2 cos 2 θ) sin 2 θ (a )( 2 + a 2 cos 2 θ) sin 2 θ + 2ma 2 sin 4 θ 2m(a ) sin 2 θ a( 2 + a 2 cos 2 θ) (a )( 2 + a 2 cos 2 θ) + 2ma 2 (1 cos 2 θ) 2m(a ) a a m (6) d 2 g 00 A 2 + g 33 B 2 + 2g 03 A B + 2g 01 A + 2g 13 B A (g 00 A + g 03 B + g 01 ) + B (g 03 A + g 33 B + g 13 ) + g 01 A + g 13 B d ˆφd, cdˆtd ( ) 0 d 2 g 01 A + g 13 B A B (4) (6) g 01 A a a m, g 13B a 2 sin 2 θ a m a2 a 2 cos 2 θ a m ds 2 g 00 (cdˆt) 2 + (g 01 A + g 13 B )d 2 + g 22 dθ 2 + g 33 d ˆφ 2 + 2g 03 cdˆtd ˆφ (1 2m 2 + a 2 cos 2 θ )(cdˆt) a 2 cos 2 θ a m d2 ( 2 + a 2 cos 2 θ)dθ 2 [(a ) sin 2 θ + 2ma2 sin 4 θ 2 + a 2 cos 2 θ ]d ˆφ2 2 2ma sin2 θ 2 + a 2 cos 2 θ cdˆtd ˆφ (Boyer-Lindquist) a 0 e 0 11
12 dˆtd ˆφ dˆt, d ˆφ dˆtd ˆφ ( ) ds 2 a2 sin 2 θ (cdt) 2 Σ Σ d2 (2 + a 2 ) 2 a 2 sin 2 θ Σ sin 2 θdφ 2 Σdθ 2 2 2ma sin2 θ cdtdφ Σ ( Σ 2 + a 2 cos 2 θ 2 + a 2 2m 12
notekiso1_09.dvi
39 3 3.1 2 Ax 1,y 1 Bx 2,y 2 x y fx, y z fx, y x 1,y 1, 0 x 1,y 1,fx 1,y 1 x 2,y 2, 0 x 2,y 2,fx 2,y 2 A s I fx, yds lim fx i,y i Δs. 3.1.1 Δs 0 x i,y i N Δs 1 I lim Δx 2 +Δy 2 0 x 1 fx i,y i Δx i 2 +Δy
20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................
TOP URL 1
TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................
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....................................
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
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
( ) 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
『共形場理論』
T (z) SL(2, C) T (z) SU(2) S 1 /Z 2 SU(2) (ŜU(2) k ŜU(2) 1)/ŜU(2) k+1 ŜU(2)/Û(1) G H N =1 N =1 N =1 N =1 N =2 N =2 N =2 N =2 ĉ>1 N =2 N =2 N =4 N =4 1 2 2 z=x 1 +ix 2 z f(z) f(z) 1 1 4 4 N =4 1 = = 1.3
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
II 2 II
II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2
9 E B 9.1 9.1.1 Ampère Ampère Ampère s law B S µ 0 B ds = µ 0 j ds (9.1) S rot B = µ 0 j (9.2) S Ampère Biot-Savart oulomb Gauss Ampère rot B 0 Ampère µ 0 9.1 (a) (b) I B ds = µ 0 I. I 1 I 2 B ds = µ 0
() 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 ()
( : December 27, 2015) CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x) f (x) y = f(x) x ϕ(r) (gradient) ϕ(r) (gradϕ(r) ) ( ) ϕ(r)
( : December 27, 215 CONTENTS I. 1 II. 2 III. 2 IV. 3 V. 5 VI. 6 VII. 7 VIII. 9 I. 1 f(x f (x y f(x x ϕ(r (gradient ϕ(r (gradϕ(r ( ϕ(r r ϕ r xi + yj + zk ϕ(r ϕ(r x i + ϕ(r y j + ϕ(r z k (1.1 ϕ(r ϕ(r i
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (
II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )
( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =
1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =
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 ),
.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
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)
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
) ] [ 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
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )
2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))
応力とひずみ.ppt
in yukawa@numse.nagoya-u.ac.jp 2 3 4 5 x 2 6 Continuum) 7 8 9 F F 10 F L L F L 1 L F L F L F 11 F L F F L F L L L 1 L 2 12 F L F! A A! S! = F S 13 F L L F F n = F " cos# F t = F " sin# S $ = S cos# S S
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
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)
(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
II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n
= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k
7 b f n f} d = b f n f d,. 5,. [ ] ɛ >, n ɛ + + n < ɛ. m. n m log + < n m. n lim sin kπ sin kπ } k π sin = n n n. k= 4 f, y = r + s, y = rs f rs = f + r + sf y + rsf yy + f y. f = f =, f = sin. 5 f f =.
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
..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A
![..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A](/thumbs/93/112175219.jpg "..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A")
.. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.
DVIOUT-HYOU
() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.
21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........
Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
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
K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
Gmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
Quiz x y i, j, k 3 A A i A j A k x y z A x A y A z x y z A A A A A A x y z P (x, y,z) r x i y j zk P r r r r r r x y z P ( x 1, y 1, z 1 )
Quiz x y i, j, k 3 A A i A j A k x y z A x A y A z x y z A A A A A A x y z P (x, y,z) x i y j zk P x y z P ( x 1, y 1, z 1 ) Q ( x, y, z ) 1 OP x1i y1 j z1k OQ x i y j z k 1 P Q PQ 1 PQ x x y y z z 1 1
sec13.dvi
13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:
Acrobat Distiller, Job 128
(2 ) 2 < > ( ) f x (x, y) 2x 3+y f y (x, y) x 2y +2 f(3, 2) f x (3, 2) 5 f y (3, 2) L y 2 z 5x 5 ` x 3 z y 2 2 2 < > (2 ) f(, 2) 7 f x (x, y) 2x y f x (, 2),f y (x, y) x +4y,f y (, 2) 7 z (x ) + 7(y 2)
1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx
1 1 1 1 1. U(x, t) U(x, t) + c t x c, κ. (1). κ U(x, t) x. (1) 1, f(x).. U(x, t) U(x, t) + c κ U(x, t), t x x : U(, t) U(1, t) ( x 1), () : U(x, ) f(x). (3) U(x, t). [ U(x, t) Re u k (t) exp(πkx). (4)
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
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
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
(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {
![(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {](/thumbs/94/118399854.jpg "(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {")
7 4.., ], ], ydy, ], 3], y + y dy 3, ], ], + y + ydy 4, ], ], y ydy ydy y y ] 3 3 ] 3 y + y dy y + 3 y3 5 + 9 3 ] 3 + y + ydy 5 6 3 + 9 ] 3 73 6 y + y + y ] 3 + 3 + 3 3 + 3 + 3 ] 4 y y dy y ] 3 y3 83 3
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.............................
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
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
) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
chap9.dvi
9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =
δ ij δ ij ˆx ˆx ŷ ŷ ẑ ẑ 0, ˆx ŷ ŷ ˆx ẑ, ŷ ẑ ẑ ŷ ẑ, ẑ ˆx ˆx ẑ ŷ, a b a x ˆx + a y ŷ + a z ẑ b x ˆx + b
23 2 2.1 n n r x, y, z ˆx ŷ ẑ 1 a a x ˆx + a y ŷ + a z ẑ 2.1.1 3 a iˆx i. 2.1.2 i1 i j k e x e y e z 3 a b a i b i i 1, 2, 3 x y z ˆx i ˆx j δ ij, 2.1.3 n a b a i b i a i b i a x b x + a y b y + a z b
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,
untitled
20010916 22;1017;23;20020108;15;20; 1 N = {1, 2, } Z + = {0, 1, 2, } Z = {0, ±1, ±2, } Q = { p p Z, q N} R = { lim a q n n a n Q, n N; sup a n < } R + = {x R x 0} n = {a + b 1 a, b R} u, v 1 R 2 2 R 3
Z: Q: R: C: sin 6 5 ζ a, b
Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,
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................................
6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P
6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t), y(t), z(t)), a t b.
2009 7 9 1 2 2 2 3 6 4 9 5 14 6 18 7 23 8 25 9 26 10 29 11 32 12 35 A 37 1 B 38 1 (x, y), (x, y, z) (x 1, x 2 ) (x 1, x 2, x 3 ) 2 : x 2 + y 2 = 1. (parameter) x = cos t, y = sin t. y = f(x) r(t) = (x(t),
mugensho.dvi
1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim
.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
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
Z: Q: R: C: 3. Green Cauchy
7 Z: Q: R: C: 3. Green.............................. 3.............................. 5.3................................. 6.4 Cauchy..................... 6.5 Taylor..........................6...............................
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
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +
( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n
1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h
IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r
( ) ( )
20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1
1/5 ( ) Taylor ( 7.1) (x, y) f(x, y) f(x, y) x + y, xy, e x y,... 1 R {(x, y) x, y R} f(x, y) x y,xy e y log x,... R {(x, y, z) (x, y),z f(x, y)} R 3 z 1 (x + y ) z ax + by + c x 1 z ax + by + c y x +
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,
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
tomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.
tomocci 18 7 5...,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. M F (M), X(F (M)).. T M p e i = e µ i µ. a a = a i
1 2 1 No p. 111 p , 4, 2, f (x, y) = x2 y x 4 + y. 2 (1) y = mx (x, y) (0, 0) f (x, y). m. (2) y = ax 2 (x, y) (0, 0) f (x,
No... p. p. 3, 4,, 5.... f (, y) y 4 + y. () y m (, y) (, ) f (, y). m. () y a (, y) (, ) f (, y). a. (3) lim f (, y). (,y) (,)... (, y) (, ). () f (, y) a + by, a, b. + y () f (, y) 4 + y + y 3 + y..3.
(yx4) 1887-1945 741936 50 1995 1 31 http://kenboushoten.web.fc.com/ OCR TeX 50 yx4 e-mail: yx4.aydx5@gmail.com i Jacobi 1751 1 3 Euler Fagnano 187 9 0 Abel iii 1 1...................................
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
KENZOU
KENZOU 2008 8 9 5 1 2 3 4 2 5 6 2 6.1......................................... 2 6.2......................................... 2 6.3......................................... 4 7 5 8 6 8.1.................................................
Lebesgue Fubini L p Banach, Hilbert Höld
II (Analysis II) Lebesgue (Applications of Lebesgue Integral Theory) 1 (Seiji HIABA) 1 ( ),,, ( ) 1 1 1.1 1 Lebesgue........................ 1 1.2 2 Fubini...................... 2 2 L p 5 2.1 Banach, Hilbert..............................
M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -
M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................
1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )
A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P
1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A
第10章 アイソパラメトリック要素
June 5, 2019 1 / 26 10.1 ( ) 2 / 26 10.2 8 2 3 4 3 4 6 10.1 4 2 3 4 3 (a) 4 (b) 2 3 (c) 2 4 10.1: 3 / 26 8.3 3 5.1 4 10.4 Gauss 10.1 Ω i 2 3 4 Ξ 3 4 6 Ξ ( ) Ξ 5.1 Gauss ˆx : Ξ Ω i ˆx h u 4 / 26 10.2.1
5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i
i j ij i j ii,, i j ij ij ij (, P P P P θ N θ P P cosθ N F N P cosθ F Psinθ P P F P P θ N P cos θ cos θ cosθ F P sinθ cosθ sinθ cosθ sinθ 5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6
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
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
. 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.........
all.dvi
38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
O x y z O ( O ) O (O ) 3 x y z O O x v t = t = 0 ( 1 ) O t = 0 c t r = ct P (x, y, z) r 2 = x 2 + y 2 + z 2 (t, x, y, z) (ct) 2 x 2 y 2 z 2 = 0
9 O y O ( O ) O (O ) 3 y O O v t = t = 0 ( ) O t = 0 t r = t P (, y, ) r = + y + (t,, y, ) (t) y = 0 () ( )O O t (t ) y = 0 () (t) y = (t ) y = 0 (3) O O v O O v O O O y y O O v P(, y,, t) t (, y,, t )
N cos s s cos ψ e e e e 3 3 e e 3 e 3 e
3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >
Microsoft Word - 表紙.docx
黒住英司 [ 著 ] サピエンティア 計量経済学 訂正および練習問題解答 (206/2/2 版 ) 訂正 練習問題解答 3 .69, 3.8 4 (X i X)U i i i (X i μ x )U i ( X μx ) U i. i E [ ] (X i μ x )U i i E[(X i μ x )]E[U i ]0. i V [ ] (X i μ x )U i i 2 i j E [(X i
( ) ) ) ) 5) 1 J = σe 2 6) ) 9) 1955 Statistical-Mechanical Theory of Irreversible Processes )
( 3 7 4 ) 2 2 ) 8 2 954 2) 955 3) 5) J = σe 2 6) 955 7) 9) 955 Statistical-Mechanical Theory of Irreversible Processes 957 ) 3 4 2 A B H (t) = Ae iωt B(t) = B(ω)e iωt B(ω) = [ Φ R (ω) Φ R () ] iω Φ R (t)
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,
第1章 微分方程式と近似解法
April 12, 2018 1 / 52 1.1 ( ) 2 / 52 1.2 1.1 1.1: 3 / 52 1.3 Poisson Poisson Poisson 1 d {2, 3} 4 / 52 1 1.3.1 1 u,b b(t,x) u(t,x) x=0 1.1: 1 a x=l 1.1 1 (0, t T ) (0, l) 1 a b : (0, t T ) (0, l) R, u
( )
7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................
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
: 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(