|
|
|
- そう かたづ
- 5 years ago
- Views:
Transcription
1
2
3 ,,,,,,., uid, (uid dynamics) Newton. (geophysical uid).,.,. 2,,,. 1 2, II,.
4 % 40 % ,. 1., ( ) ( ),.,. 30.,,...,,", ". ( ) 2., (, 21),.,,.,. 3 ", (, " 3. G. K. Batchelor, An Introduction to Fluid Mechanics (Cambridge U. P.) 3
5 1.3. 5, Journal of Fluid Mechanics 1999.,. 4. L. D. Landau and E. M. Lifshitz, Fluid Mechanics (Pergamon Press) Landau - Lifshitz 1. Landau S. H. Lamb, Hydrodynamics (Cambridge U. P.) Sir ,.,. 2,. 2. J. Pedlosky, Geophysical Fluid Dynamics (Springer). 3. R. Salmon Lecture on Geophysical Fluid Dynamics (Oxford U. P.). Hamilton. Review Hamilton. 4., 5., : (, 13) 1., ( 2 ) ( ).
6 iwayamakobe-u.ac.jp 1.5, \ ",
7 (Cartesian coordinate).,, nonation 2.2.1, ~ A A x y z i j k e 1 e 2 e 3 e 1 = i e 2 = j e 3 = k (2.1). x y z. 1,.
8 r x. 2, r r = x i + y j + z k (2.2) x y z r x y z r =(x y z) (2.2). 3 (2.1) x = x 1 e 1 + x 2 e 2 + x 3 e 3 (2.3) v v = u i + v j + w k (2.4). u v w u x y z. 4 (2.1) v = v 1 e 1 + v 2 e 2 + v 3 e 3 (2.5) , (westerly: u > 0) (easterly: u < 0) (northerly: v<0) (southerly: v>0), (eastward: u >0) (westward: u <0) (southward: v<0) (northward: v>0),.
9 2.3. summation rules, Einstein's notation t x y z t x y z x 1 x 2 x r. r = i x + j y + k z (2.6) = i x + j y + k z (2.7) = e e e 3 3 (2.8) 2.3 summation rules, Einstein's notation 2, 1 3. x = x 1 e 1 + x 2 e 2 + x 3 e 3 = 3X i=1 x i e i = x i e i : (2.9) P. 1: 2 x i e i = x j e j = x k e k. 2: 2, 1 2. : v = v i e i : (2.10) r = e i i : (2.11) df(x 1 x 2 x 3 ) = dx 1 ( 1 f)+dx 2 ( 2 f)+dx 3 ( 3 f) = dx i ( i f): (2.12)
10 Kronecker 2 Kronecker : ij ( 0 (i j ): (2.13) 1 (i j ): Kronecker ij e i e j (2.14). ra = (e i i ) (A j e j ) = ij i A j = j A j : (2.15) : Kronecker,. 1. ii 2. ij A j 2.5 Eddington 3 Eddington : " ijk 8 >< >: 1 (i j k) =(1 2 3) (2 3 1) (3 1 2) : ;1 (i j k) =(3 2 1) (2 1 3) (1 3 2) : 0. (2.16) " ijk = ;" ikj : (2.17) " ijk =1 (i j k) =(1 2 3), " ijk = ;1 (i j k) = (1 2 3).
11 2.5. Eddington 11 A = 0 B a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 1 C A, Eddington deta = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = " ijk a 1i a 2j a 3k : (2.18) A B.. A B = ra = e 1 e 2 e 3 A 1 A 2 A 3 B 1 B 2 B 3 = " ijk e i A j B k : (2.19) e 1 e 2 e A 1 A 2 A 3 = " ijk e i j A k (2.20) 1. (2.18). 2., ij " ijk., ij " ijk. (a) B A = ;A B, (b) A (B C) = B (C A) =C (A B) 3. " ijk " ilm = jl km ; jm kl.
12 Gauss S V, 2. 5 I: v : ZZZ ZZ rvdv = v ds (2.21) V S, ds n, ds. II: Q : ZZ S Qn i ds = ZZZ, n i n i. V Q x i dv (2.22) I Gauss. : (2.22). 2.7 Stokes v, : I C v dr = ZZ S (rv) ds: (2.23) S C., C, S., S. n S. S, C,. S, ZZ. v. 5,. S (r v) ds =0 (2.24)
13 2.7. Stokes 13 n S C 2.1: Stokes C, C S,. Stokes. Stokes C., (2.23).,. Stokes,., ( ), ( ). ( ), Stokes ( )., Stokes.. ( ), (2.24), Stokes. (, ( ).) ( ), (2.24) ZZ S ( x v ; y u)dx dy (2.25)., S, x y. (.)
14
15 15 3, ( ),. 3.1, (, ),,..,.,., P, P v., P r, P (r) Nm (3.1) v., m,, N v. N v t.,, 2. L (v) 1=3 1. L K = L (3.2), K 1. K Knudsen. 1.
16 16 3 P ρ (r,t) P 3.1:. P, P. T 2. :., 0 o C 10 ;7 m 10 ;10 s L =1 10 km T 12 h ( ) L 10 3 km T 10 day. (uid parcel/uid particle)., v.,,,.,,,. 3.2, Newton,,., 2,.
17 ,,,.,.. (body force).,, Coliois,. (surface force),,.,,.. (stress). MKS Nm ;2. \ ", N. P, P S, S P ( n ), Tn. Tn S, n.,. Tn 2 1., 2 (tensor), 9, ij (i j =1 2 3),., ij. x j, x j x i ij.,. T;n = ;Tn (3.3), (tangential stress), (shear stress), (normal stress)., (pressure), (tension). 3.3,,, : \
18 18 3 τ 33 Τn τ 13 P τ : P z., P T n ,.",.,,,.,,.. \, 0,.", 0,. {[ ] { [ :] P, PAB,. PA, PB, AB p 1, p 2, p 3,., PA= l 1 PB = l 2, 6 A = 1 6 B = 2. : AB, p 1 l 1 sin 1 ; p 2 l 2 sin 2 =0: sin 1 l 2 = sin 2 l 1
19 , p 1 = p 2 : P. : O(l 2 ),., l,.,. : D'Alembert.,. {[ ] { P p 1 p 2 A α α 1 2 B p 3 3.3: P. 3.4,.
20 , (viscosity). 3 (inviscid/perfect/ideal uid),.,., ij = ;p ij : (3.4),. (viscous uid).,. Newton (Newtonian uid).. Newton e ij ; 13 e kk ij ij = ;p ij +2 e ij = 1 ui + u j 2 x j x i (3.5) (3.6). (coecient of viscosity). Newton. (3.5) 0, (3.4). : (3.5),.,. (3.5),,. 4,,.,, 5,, 3,,,,.. 4,,, 2,. 5. 1, 1.
21 , ,,.,.,,.. 6,,,., (compressible uid). (incompressible uid).. 7,,. 8 : 1. p = RT, T ds = c v dt + p d( 1 ), p p = C (Poisson )., T S c v R C,,,,, c v c p = c v + R, = c p =c v. 2., 300 K c =. (a) r p S. 6 c entropy,. c p (p=) S. 7,. 8 Bussinesq.,.. r p S
22 22 3 (b) c v = 5R c 2 p = 7R R = 287 J 2 K;1 kg ;1, c. (,.) v, E, S,... p, T,,. 9,, 2.,, , 5, 5,. 5,,.,,,. 3., ,, ,,,,,.,,, v,, T, p.. 10 d, d
23 Lagrange (Lagrange ),,., t = t 0 (a b c) t (x y z). x y z a b c t, x = f 1 (a b c t) y = f 2 (a b c t) (3.7) z = f 3 (a b c t). f 1 f 2 f 3,. t 0 t 0 =0. (3.7) (a b c), (a b c) material coorinates. (a b c),,, (a b c).. 11 N, i (i =1 2 :::N) t (x i y i z i ). i ( ). 2. i, (a b c).., ( ). Lagrange,, Lagrange, material dierentiation D Dt. Lagrange, F Lagrange DF F Dt = t a b c (3.8) 11 Lagrange naive Hamilton Hamilton.
24 24 3 (Lagrange ) ( ) (a b c) : i D Dt d dt 3.1: Lagrange. d dt. Euler (Euler ) t, (x y z) v T p :::.,.,,..,. Euler Lagrange, x y z Lagrange, Euler. Lagrange Euler F Euler F (x y z t). Lagrange Euler. t r = (x y z), t +t r + vt =(x + ut y + vt z + wt). F F, F = F (x + ut y + vt z + wt t +t) ; F (x y z t) F = t + u F x + v F y + w F ; t + O 2 (t) z DF Dt F = lim t!0 t = F t + u F x + v F y + w F z (3.9). O ((t) 2 ) t. F
25 , Lagrange Euler D Dt = t + u x + v y + w z = t + v r (3.10). (3.10) 2. v r rv r.,. v r, ( v r ), rv. 2. Langrange Euler a b c t x y z t x y z p T ::: u v w p T ::: Lagrange, D Dt t t + v r : 1. f(a b c), f=a, b c, a f. (f=a) b c,,,. Lagrange =t Euler =t. 2. Lagrange Euler (3.10) v r ,.,
26 26 3,,. g, N=m 3, N=m 2 g.,,,,.. 3.4, ( ds, dz)., ( ) ( ). 12 p(x y z +dz)ds, p(x y z)ds. g ds dz.,, ;p(x y z +dz)ds ; g dz ds + p(x y z)ds =0 (3.11). dz, p(x y z +dz) p(x y z +dz) =p(x y z)+ p z dz + O(dz2 ) (3.12) Taylor., O(dz 2 ) dz 2. (3.11) 1, ; p z + O(dz2 ) ; g dz ds =0: (3.13) dz ds, dz ds! 0., O(dz 2 ), ; p ; g =0 (3.14) z. (3.14) (hydrostatic balance), :. 13., Coriolis (geostrophic balance).
27 (3.14), p., x y ; p ; p. x y, ;rp.,, (pressure gradient force). p(z+dz) ds z+dz z p(z) ρg ds dz ds ds 3.4: ds, z z +dz., ds, dz 0 (, )., (3.11), ;p(x y z +dz)ds ; 0 g dz ds + p(x y z)ds (3.15). 1 3 g ds dz., (3.15) ( ; 0 )gds dz (3.16)., > 0,,, (3.16) > 0,.,
28 [ ], : (buoyancy force) \ ( ),,.., ( )"., (3.15) ;p(x y z +dz)ds + p(x y z)ds (3.17). (3.17), ;p(x y z +dz)ds + p(x y z)ds = gdsdz (3.18),..
29 29 4 (1): 3.5.1, v 2,,,,.,,. (equation of continuity). Euler Euler., Lagrange Lagrange, Euler. 4.1 Euler S. 1 S V. V, ZZZ V dv. V, d dt ZZZ V dv. V S V. ds, ds v n., v n v v n = v n ( 4.1 )., 1 Euler,,.
30 30 4 (1): S V ; ZZ S v n ds = ; ZZZ V r(v) dv. S n,. Gauss (2.6 ). d dt ZZZ V dv = ; ZZZ V r(v) dv:, ZZZ V t + r(v) dv =0. V, 0., + r(v) =0 (4.1) t.. Lagrange D Dt., D + rv =0 (4.2) Dt : 1. (2.3 ), (4.1). 2. Lagrange D, (4.1) (4.2) Dt. :,, D Dt =0, rv =0 (4.3).
31 4.2. Lagrange 31 n S ds V n ds v v.n 4.1: V ds. 4.2 Lagrange t = t 0 S 0 V 0,, t S V. Lagrange,. : ZZZ ZZZ 0 dx 0 dy 0 dz 0 = dx dy dz: V 0 V (a b c) t = t 0 (x 0 y 0 z 0 ), abc- : (x y z) (a b c),. = = ZZZ ZZZ V 0 0 dx 0 dy 0 dz 0 = V dx dy dz = ZZZ ZZZ V 0 0 da db dc V 0 (x y z) da db dc: (a b c), xyz abc Jacobian (x y z) (a b c) = 0 (4.4) Lagrange., 0 a b c. 2 2, 4.3.
32 32 4 (1): : = 0, (x y z) (a b c) =1 (4.5). 4.3 Euler Lagrange Euler Lagrange,. Lagrange (4.4) Jacobian, J (x y z) (a b c) (4.6)., (4.4) J = 0. (4.4) Lagrange, 2 chain rule DJ Dt ; = Dx y z Dt (a b c) = = ( ; Dx Dt y z (x y z) u x + v y + w z D Dt J + DJ Dt + ; x Dy Dt z (a b c) ; + x Dy z Dt (x y z) J =0: (4.7) ; + x y Dz Dt (a b c) ; + x y Dz Dt (x y z) ) (x y z) (a b c) = (rv) J (4.8). 3 (4.8) (4.7), Euler D + rv =0 Dt. Lagrange Lagrange Euler. 3 D=Dt Lagrange, a b c t (t ) a b c, x y z.
33 4.3. Euler Lagrange 33 Euler Lagrange, (4.2) Lagrange, Lagrange ( ) ( ). (4.8), rv = J ;1 DJ=Dt, (4.2) D Dt + J DJ Dt =0: D (J) =0 (4.9) Dt. J 0. Lagrange D=Dt =( t ) a b c, 0 a b c.
34
35 35 5 (2): \, ", Newton,. Euler Euler. Euler Lagrange, Euler (x y z) Lagrange (a b c), Lagrange. 5.1 Euler (2.3 ). S. R S V. V i v V i dv., V. d dt ZZZ V v i dv. i ij, ZZ S ij n j ds. i K i V i ZZZ V K i dv
36 36 5 (2):. S V. " ". Euler, ; ZZ S (v i ) v j n j ds. 1 V, V, S S., d dt ZZZ V v i dv = ZZZ V K i dv + ZZ S ij n j ds ; ZZ S (v i ) v j n j ds (5.1) 2.6 Gauss., ZZ ij n j ds ; S. ZZZ V ZZ S (v i ) v j n j ds = (5.1) t (v i) ; K i ; ZZZ x j ij + V x j ( ij ; v i v j )dv (v i v j ) dv =0 (5.2) x j. V,.,. t (v i)+ x j (v i v j )=K i +. (5.3) : t (v i) = v i t + v i t x j (v i v j ) = v j v i x j + v i x j ij (5.3) x j (v j ) : (4.1). (5.3) vi t + v j v i = ij + K i x j x j , V, v n ds., V, v n ds v.
37 5.1. Euler 37 v i t + v j v i x j = 1 ij + K i x j (5.4a).. Lagrange Dv i Dt = 1 ij + K i : x j (5.4b), ij = ;p ij (5.5), (5.4) v i t + v j v i x j. (5.6a) = ; 1 p x i + K i t + v r v = ; 1 rp + K (5.6b) (5.6a). (5.6) Euler (Euler's equation). Newton (5.4) v i t + v j v i x j (3.5) Newton, = ; 1 p + 2 v i + 1 x i x 2 j 3 vj + K i (5.7a) x i x j t + v r v = ; 1 rp + v + 1 r (rv)+k 3 (5.7b). = = (kinematic viscosity). (5.7) Navier-Stokes (Navier-Stokes equation),.
38 38 5 (2): 5.2,.. (5.4a) v i,, t v i v i t + v iv j v i x j = v i x j ij + v i K i 1 2 v iv i + x j 1 2 v iv i v j ; 1 2 v iv i., t + v j x j {z } ij v i = ( ijv i ) v i ; ij x j x j x j = ( ijv i ) ; ij e ij (5.8) x j ij ( ij = ji )., e ij = 1 ui + u i 2 x j x j (5.9)., T = 1 2 v iv i, (T ) t + (T v j) x j = ( ijv i ) x j + v i K i ; ij e ij (5.10).,, ( ) K ( ),.,. (5.10),., ij.
39 , ij = ;p ij, (5.8) ; (pv j) x j + pe jj., (5.10) (T ) t + (T + p) v j x j = v i K i + pe kk (5.11a). (T ) t + rf(t + p) vg = v K + prv (5.11b) Newton (3.5) Newton, (5.8), ;pv j +2 x j.,, (5.10) e ij v i ; 13 e kkv j ; ;pe jj +2 e ij e ij ; 13 e kke jj (e ij ; 1 3 e kk ij ) 2 = e ij e ij ; 2 3 e kk ij e kke kk ij {z} ij =3 (T ) t = e ij e ij ; 1 3 e kke jj : (5.12) + (T + p) v j ; 2 e ij v i ; 13 x e kkv j j = v i K i + pe kk ; (5.13) =2(e ij ; 1 3 e kk ij ) 2 (5.14) (dissipation function),....
40 40 5 (2): 5.3 Lagrange 5.1 Euler (5.6) Lagrange. (5.6) Lagrange, Dv Dt = ;1 rp + K (5.15). Euler, K p. Lagrange 2 r. ;rp Lagrange t a b c 2 (a b c). (5.6) x- a x, (5.6) y- a y, (5.6) z- a z rp : x p a x + y p a y + z p a z = a p,. Lagrange a-, (5.6) x- a x, (5.6) y- a y,(5.6) z- a z : ; 2 t x ;K x a x + ; 2 t y ;K y a y + ; 2 t z ;K z a z = ; 1 ap: b-, (5.6) x- b x, (5.6) y- b y, (5.6) z- b z, c-, (5.6) x- c x, (5.6) y- c y, (5.6) z- c z. : 2 x x t ;K 2 x a + 2 y y t ;K 2 y a + 2 z z t ;K 2 z a = ;1 p a (5.16) 2 x x y z p t ;K 2 x b + 2 y t ;K 2 y 2 x x t ;K 2 x c + 2 y t ;K 2 y b + 2 z t ;K 2 z y c + 2 z t ;K 2 z b z c = ; 1 = ; 1 b (5.17) p c : (5.18)
41 41 6 (3) \,,,,, ",., Euler. 6.1 S. S V. R V ( ) (T + U) dv V., T U. V d dt ZZZ V (T + U) dv.. V, ZZZ V v i K i dv + ZZ S v i ij n j ds. 1 S V, Euler, ; ZZ S f (T + U)g v j n j ds (6.1).,,. (thermal diusivity), 1 F, r, F r.
42 42 6 (3). S V. ZZ ; j n j ds (6.2) S. 2, J. V, V, S, S V,, d dt ZZZ V (T + U) dv = ZZZ ; + ZZ V S ZZZ v i K i dv + ZZ S v i ij n j ds [f (T + U)g v j + j ] n j ds, 2.6 Gauss ; ZZZ V V J dv (6.4) x j (T v j + Uv j ; v i ij + j )dv (6.5). V, t f (T + U)g + f (T + U) v j g = (v i ij ) ; j + v i K i + J (6.6) x j x j x j. (5.10) (6.6), t (U)+ f(u) v j g = ij e ij ; j + J (6.7) x j x j., (6.7) Lagrang. DU Dt = 1 ije ij ; 1 j x j + J (6.8) 2, Fourier. Fourier.. = ;rt (6.3)
43 , ij = ;p ij, (6.8). ij e ij = ;pe jj = ;p v j x j (6.9) 1 D Dt = ; v j x j (6.8) DU Dt + p D Dt 1 = ; 1 r + J (6.10)., (6.10), U + p( 1 )=Q, Lagrange D. Dt Newton Newton, (6.8). (6.8) DU Dt + p D Dt ij e ij = ;pe jj + (6.11) 1 =; 1 r + J (6.12),,,. (5.13).
44
45 ,, 3, 5,,, v, p, U 6. 1,. ( ), 2, 5., , : p = RT: (7.1) (7.1). n kmol PV = nr T (7.2)., R (R =8:314times10 ;3 JK ;1 kmol ;1 ). V m kg, M 1 Fourier T, 7. 2, Boussinesq.
46 46 7, (7.2) P = m V R M T (7.3). = m=v, (7.1) (7.3) R = R M (7.4). ( 28) 75.5 %, ( 32) 23.1 %, ( 40) 1.3 % 3, M =28 0: : :013 = 28:96. R = R =28:96 = 287J K ;1 kg ;1.,, R Boussinesq = f(p T ) T 0, p 0 Taylor, 1 f f = f(p T ) = f(p 0 T 0 )+ p, 1 0 = 0 + ( = 0 p T T p p T (p ; p 0 )+ T (p ; p 0 )+ T =0,, p T p (T ; T 0 ) (p ; p 0 )+ 1 0 T (T ; T 0 ) p (T ; T 0 ) = 0 [1 ; (T ; T 0 )] (7.5) ; 1 (7.6) 0 T 3 80 km. p )
47 (7.5) Boussinesq,,.,,. 4,, = 0 [1 ; (T ; T 0 )+(s ; s 0 )] (7.7) ; 1 (7.8) 0 s p T., s (salinity). s, 1kg.,,, (coecient of saline contraction). 7.3, v., p.,, = F(p) (7.9) f(p ) =0 (7.10),,,. (7.9) (7.10) (barotropic uid). (7.10).,. (baroclinic uid).,, barotropic uid ( ( ) ),, baroclinic (7.6),,., < 0.,,. 0. T p
48 D =0, (incompressible homogeneous Dt uid), = const: (7.11),. Newton,, P = p=. Dv Dt = ;rp + r2 v + K (7.12) rv =0 (7.13).,, Navier-Stokes ,. 1. T =const. p / (7.14) 7.3.3,. p,, S p = p 0 0 exp S ; S0 C v (7.15)
49 ,. C v C p = C p =C v. (S = S 0 ), (7.15). p p 0 = 0 (7.16) du = TdS; pd( ;1 ), p = RT, du = C v dt, (7.15). (7.16) Poisson. Poisson, T p, R=Cp T 0 T = p0 (7.17) p., T 0 p (7.17) p R=C p T = const (7.18), T 0, p 0,, T 0 (potential temperature),., z p(z), T (z) p 0 ( p 0 = 1000hPa ),, T 0, z.,., (z),.
50 50 7 S S = C p ln +const (7.19) , K, K = ;gk. 5, (5.7b), 1 rp = ;g k: (7.21), 0,., dp(z) dz = ;g: (7.22) (7.22) (7.1). 6 z, (geopotential), Z =g 0 (geopotential hight) (z) Z z 0 gdz: (7.23), g 0 g 0 =9:81ms ;2. z. 5, g. g g = GM (a + z) 2 : (7.20), G, M, a, z. 6 (7.22) z,.
51 (7.23) z p., (7.22) p T. Z p(z) (z) ; (0) = ;R Tdln p: (7.24) p(0) (7.24) (hypsometric equation). (7.24), p 1 p 2 (, (thickness)) 7 Z T Z(p 2 ) ; Z(p 1 )=; R g 0 Z p2 p 1 Tdln p (7.25). ( ),, (7.25),., p 1 p 2 ht i ht i R p2 p R 1 p2 p 1 Tdln p d ln p (7.26). Z T = ;H ln(p 2 =p 1 ) (7.27) H R ht i g 0 (7.28)., p(z) =p(0) e ;Z=H (7.29),. H (scale hight), e ;1. 255K 7km. 7 p 1 >p 2, p 1 p 2.
52
53 n, n., t =0,., k m, m d2 x dt 2 = ;kx (8.1) x(t) = A cos(!t)+b sin(!t)! = k=m: (8.2), A B t =0 x(0) _x(0).,. 1, 2 t 2 (x t) =c 2 2 x 2 (x t) (8.3) (x t) =G(x ; ct) +H(x + ct) (8.4). G H. 1,. 0 x L, (0 t)= (L t) =0, Fourier. Fourier, G(x ; ct), H(x + ct). 2, 2, 2, 2 ( (x 0) t (x 0)), 2 ( ). 3, 1979:,, 223 pp.
54 54 8 (8.4) (8.3). 4 (, ),,. Euler, Euler 1, t =0 v p.,. 8.2,. P v, P n.,,, v n =0 (8.5). 5 z = h(x y), r(h;z), (8.5). v r(h ; z) = Dh Dt ; Dz Dt =0 w = Dz Dt = Dh Dt (8.6) 8.3 2, ( 2 ). 4 : = x ; ct = x + ct,. t x 5 v 0, v 0, (8.5)., (v ; v 0 ) n =0.
55 F (x y z t) =0 (8.7). 2.,., P t v t, F (x + u t y + v t z + w t t +t) =0. F (x + u t y + v t z + w t t +t) = F (x y z t) {z } =0 DF Dt + DF Dt t +O(t2 )=0 = 0: (8.8) 8.4
56
57 57 9 (4) 9.1, 5.1 (5.6).,. 1??,, ( Coriolis )., dv dt d 0 rv, dt Coriolis ((??) )., d dt Lagrange D Dt. Euler v Lagrange : Dv Dt ;! Dv Dt +2 v + ( r): (9.1)??, prime,, r., Euler, Lagrange Euler v t + v rv +2 v = ;1 rp + K ; ( r) (9.2).,, (9.2). 1 Newton, K,., Landau, L. D. & Lifshitz, E. M., Fluid Mechanics. 2 nd. Ed. Pergamon Press p.48 { 49..,.
58 58 9 (4) r: () r: : () :, ( + = ) 2 : () : 9.1: ,,,,,.,. ( r), 3 (r ),., r,,,. r, a, z = r ;a.. e r (e ϕ)e λ e ϕ r (e (θ) θ ) ϕ (ϕ)λ 9.1:..
59 a, M., z (eective gravity) g. 3 a2 g ; (a + z) 2 g e r + 2 (a + z) cos e : (9.3) e r ( r), e ( z). e. g G. Ω e = cos e r ; sin e (9.4) g = GM a 2 (9.5) e ϕ g* ϕ δ g e r eρ - Ω XΩX r tangential plan 9.2:,,,. 9.2, (;e r ),, (, e r ) 2 Coriolis,. 3 gravitational force,, gravity force, gravity.
60 60 9 (4).., z =0. e r = cos ;1 g (;g e r ) g jgj g ; = cos ;1 2 a cos 2 p (g ; 2 a cos 2 ) 2 +( 2 a cos sin ) 2! (9.6). 4 a =6: m =7:3 10 ;5 s ;1 g =9:8ms ;2 (9.6), 45 =9:9 10 ;2 ( 6 ). 1km 1.7 m,.,,,, (gioid),,.,. 6 jgj (9.3),. 7, z a jgj. 8, 4 g (;g e r ) = ; ;g e r + 2 a cos e (;g e r ) = g 2 ; 2 a cos (cos e r ; sin e ) g e r = g 2 ; g 2 a cos 2 jgj = p (g ; 2 a cos 2 ) 2 +( 2 a cos sin ) 2 5 6: m, 6: m. 6,,. 7 (9.3) jgj 3:4 10 ;2 ms ;2. 9:78 m s ;2, 9:83ms ;2. (, M.L. Salby. Fundamentals of Atmospheric Physics. Academic Press, 1996,,,,, , O(10 2 )km. O(10)km., 6400 km.
61 , (g ' g 0 =9:806ms ;2 ). (9.6), sin cos O(10 ;1 ), 2 a=g 1, 2 a sin(2)=(2g ).. (Taylor.) 9.4,.,,. ( ),, g = ; a2 (a + z) 2 g 0 e r (9.7). ( ),, Coriolis. Euler (9.2) v t + v v t + v v r t + v 1 v r cos + v 1 v r cos + v 1 v r r cos + v 1 v r + v v r v v r + r r ; v v tan r {z } 1 ; 2 sin v +2 cos v {z } r = ; 1 y1 1 v r + v v r + 2 sin v = ; 1 1 v r r + v v r r v rv r + r 1 r r ; v2 1 r cos + v2 tan r {z } 2 p + K (9.8) p + K (9.9) + v2 {z r } 3 ;2 cos v {z } y2 = ; 1 p r ; g + K r (9.10). K =(K K K r ) K K. *, r,
62 62 9 (4) (curvature term) ,.,,. 0, 0. 10, (9.8) (9.10). 1.,,, x y z, x y z u v w.,, x = r cos( 0 )( ; 0 ) y = r( ; 0 ) z = r ; a (9.11) x = 1 r cos y = 1 r z = r (9.12) u = v v = v w = v r (9.13).,. 2.,, (9.8) (9.10) 1=r, 1=a., y1. z (, r ),., y1,, y2.,, Coriolis. y1 y2, y1 y2, Coriolis. 3., 11 (9.8) (9.10) (* ). 4. Coriolis 2 sin, Coriolis, f., 0 9 (9.8) (9.10),,. 10,. 11
63 , f 0 Taylor y 1. f = 2 sin ' 2 sin( 0 + y=a) ' f 0 + y (9.14) f 0 2 sin 0 (9.15) 2 cos 0 : (9.16) a, =0, f. 12,,., f = f k. u t + u u x + v u y + w u z ; p fv= ;1 x + K x (9.17) v t + u v x + v v y + w v p z + fu= ;1 y + K y (9.18) w t + u w x + v w y + w w p z = ;1 z + K z ; g (9.19) v t + v rv + f v = ;1 rp + g + K (9.20),, Coriolis (9.20).,, Coriolis (y ), (9.20) Coriolis f (9.14) (9.16). 2 4,,,.,,,.,,., 12 f 2 f = f 0 + y; 1 2 y2 2 sin 0 = a 2, [Yang, H., Wave Packets and Their Bifurcations in Geophysical Fluid Dynamics, Springer, 1990, 247pp.]
64 64 9 (4) L 1000 km = 10 6 m H 10 km = 10 4 m U 10 m s ;1 W 10 ;2 ms ;1 T 1day 10 5 s P 10 3 Pa 1kgm ;3 9.2:.,.,,,.,,.,,. 9.2,, (9.8) Coriolis 2 sin v, 2 cos v r. (9.9) Coriolis,.
5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1
4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1
D v D F v/d F v D F η v D (3.2) (a) F=0 (b) v=const. D F v Newtonian fluid σ ė σ = ηė (2.2) ė kl σ ij = D ijkl ė kl D ijkl (2.14) ė ij (3.3) µ η visco
post glacial rebound 3.1 Viscosity and Newtonian fluid f i = kx i σ ij e kl ideal fluid (1.9) irreversible process e ij u k strain rate tensor (3.1) v i u i / t e ij v F 23 D v D F v/d F v D F η v D (3.2)
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,
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
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
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
4 2016 3 8 2.,. 2. Arakawa Jacobin., 2 Adams-Bashforth. Re = 80, 90, 100.. h l, h/l, Kármán, h/l 0.28,, h/l.., (2010), 46.2., t = 100 t = 2000 46.2 < Re 46.5. 1 1 4 2 6 2.1............................
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
/ 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,
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
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..........
chap03.dvi
99 3 (Coriolis) cm m (free surface wave) 3.1 Φ 2.5 (2.25) Φ 100 3 r =(x, y, z) x y z F (x, y, z, t) =0 ( DF ) Dt = t + Φ F =0 onf =0. (3.1) n = F/ F (3.1) F n Φ = Φ n = 1 F F t Vn on F = 0 (3.2) Φ (3.1)
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
all.dvi
5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0
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..................................
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
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
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
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
( 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
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
![18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb](/thumbs/93/114079295.jpg "18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb")
r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
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
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....
note1.dvi
(1) 1996 11 7 1 (1) 1. 1 dx dy d x τ xx x x, stress x + dx x τ xx x+dx dyd x x τ xx x dyd y τ xx x τ xx x+dx d dx y x dy 1. dx dy d x τ xy x τ x ρdxdyd x dx dy d ρdxdyd u x t = τ xx x+dx dyd τ xx x dyd
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),
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(
A 99% MS-Free Presentation
A 99% MS-Free Presentation 2 Galactic Dynamics (Binney & Tremaine 1987, 2008) Dynamics of Galaxies (Bertin 2000) Dynamical Evolution of Globular Clusters (Spitzer 1987) The Gravitational Million-Body Problem
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........................
A
A05-132 2010 2 11 1 1 3 1.1.......................................... 3 1.2..................................... 3 1.3..................................... 3 2 4 2.1............................... 4 2.2
genron-3
" ( K p( pasals! ( kg / m 3 " ( K! v M V! M / V v V / M! 3 ( kg / m v ( v "! v p v # v v pd v ( J / kg p ( $ 3! % S $ ( pv" 3 ( ( 5 pv" pv R" p R!" R " ( K ( 6 ( 7 " pv pv % p % w ' p% S & $ p% v ( J /
1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O
![1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O](/thumbs/93/114442636.jpg "1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O")
: 2014 4 10 1 2 2 3 2.1...................................... 3 2.2....................................... 4 2.3....................................... 4 2.4................................ 5 2.5 Free-Body
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
2 Chapter 4 (f4a). 2. (f4cone) ( θ) () g M. 2. (f4b) T M L P a θ (f4eki) ρ H A a g. v ( ) 2. H(t) ( )
http://astr-www.kj.yamagata-u.ac.jp/~shibata f4a f4b 2 f4cone f4eki f4end 4 f5meanfp f6coin () f6a f7a f7b f7d f8a f8b f9a f9b f9c f9kep f0a f0bt version feqmo fvec4 fvec fvec6 fvec2 fvec3 f3a (-D) f3b
b3e2003.dvi
15 II 5 5.1 (1) p, q p = (x + 2y, xy, 1), q = (x 2 + 3y 2, xyz, ) (i) p rotq (ii) p gradq D (2) a, b rot(a b) div [11, p.75] (3) (i) f f grad f = 1 2 grad( f 2) (ii) f f gradf 1 2 grad ( f 2) rotf 5.2
IA
IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................
p = mv p x > h/4π λ = h p m v Ψ 2 Ψ
II p = mv p x > h/4π λ = h p m v Ψ 2 Ψ Ψ Ψ 2 0 x P'(x) m d 2 x = mω 2 x = kx = F(x) dt 2 x = cos(ωt + φ) mω 2 = k ω = m k v = dx = -ωsin(ωt + φ) dt = d 2 x dt 2 0 y v θ P(x,y) θ = ωt + φ ν = ω [Hz] 2π
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
C:/KENAR/0p1.dvi
2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x
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
v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i
1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [
1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0
![1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0](/thumbs/93/114340663.jpg "1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0")
: 2016 4 1 1 2 1.1......................................... 2 1.2................................... 2 2 2 2.1........................................ 2 2.2......................................... 3 2.3.........................................
dynamics-solution2.dvi
1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj
2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ
1 1 1.1 (Isaac Newton, 1642 1727) 1. : 2. ( ) F = ma 3. ; F a 2 t x(t) v(t) = x (t) v (t) = x (t) F 3 3 3 3 3 3 6 1 2 6 12 1 3 1 2 m 2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t)
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..............
構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
() 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 ()
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,
F S S S S S S S 32 S S S 32: S S rot F ds = F d l (63) S S S 0 F rot F ds = 0 S (63) S rot F S S S S S rot F F (63)
211 12 1 19 2.9 F 32 32: rot F d = F d l (63) F rot F d = 2.9.1 (63) rot F rot F F (63) 12 2 F F F (63) 33 33: (63) rot 2.9.2 (63) I = [, 1] [, 1] 12 3 34: = 1 2 1 2 1 1 = C 1 + C C 2 2 2 = C 2 + ( C )
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
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
1 4 1 ( ) ( ) ( ) ( ) () 1 4 2
7 1995, 2017 7 21 1 2 2 3 3 4 4 6 (1).................................... 6 (2)..................................... 6 (3) t................. 9 5 11 (1)......................................... 11 (2)
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....................
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..........................................
difgeo1.dvi
1 http://matlab0.hwe.oita-u.ac.jp/ matsuo/difgeo.pdf ver.1 8//001 1 1.1 a A. O 1 e 1 ; e ; e e 1 ; e ; e x 1 ;x ;x e 1 ; e ; e X x x x 1 ;x ;x X (x 1 ;x ;x ) 1 1 x x X e e 1 O e x x 1 x x = x 1 e 1 + x
OHP.dvi
t 0, X X t x t 0 t u u = x X (1) t t 0 u X x O 1 1 t 0 =0 X X +dx t x(x,t) x(x +dx,t). dx dx = x(x +dx,t) x(x,t) (2) dx, dx = F dx (3). F (deformation gradient tensor) t F t 0 dx dx X x O 2 2 F. (det F
構造と連続体の力学基礎
12 12.1? finite deformation infinitesimal deformation large deformation 1 [129] B Bernoulli-Euler [26] 1975 Northwestern Nemat-Nasser Continuum Mechanics 1980 [73] 2 1 2 What is the physical meaning? 583
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
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
8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J
26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK
all.dvi
29 4 Green-Lagrange,,.,,,,,,.,,,,,,,,,, E, σ, ε σ = Eε,,.. 4.1? l, l 1 (l 1 l) ε ε = l 1 l l (4.1) F l l 1 F 30 4 Green-Lagrange Δz Δδ γ = Δδ (4.2) Δz π/2 φ γ = π 2 φ (4.3) γ tan γ γ,sin γ γ ( π ) γ tan
Korteweg-de Vries
Korteweg-de Vries 2011 03 29 ,.,.,.,, Korteweg-de Vries,. 1 1 3 1.1 K-dV........................ 3 1.2.............................. 4 2 K-dV 5 2.1............................. 5 2.2..............................
215 11 13 1 2 1.1....................... 2 1.2.................... 2 1.3..................... 2 1.4...................... 3 1.5............... 3 1.6........................... 4 1.7.................. 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
24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x
![24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x](/thumbs/94/118744578.jpg "24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x")
24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),
, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )
81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,.,
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
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
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
() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
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.........................
1 B () Ver 2014 0 2014/10 2015/1 http://www-cr.scphys.kyoto-u.ac.jp/member/tsuru/lecture/... 1. ( ) 2. 3. 3 1 7 1.1..................................................... 7 1.2.............................................
I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
B ver B
B ver. 2017.02.24 B Contents 1 11 1.1....................... 11 1.1.1............. 11 1.1.2.......................... 12 1.2............................. 14 1.2.1................ 14 1.2.2.......................
( ) ,
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
23 7 28 i i 1 1 1.1................................... 2 1.2............................... 3 1.2.1.................................... 3 1.2.2............................... 4 1.2.3 SI..............................
m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d
m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
.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
. 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.........
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:
x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v
12 -- 1 4 2009 9 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8 4-9 4-10 c 2011 1/(13) 4--1 2009 9 3 x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2
2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+
R 3 R n C n V??,?? k, l K x, y, z K n, i x + y + z x + y + z iv x V, x + x o x V v kx + y kx + ky vi k + lx kx + lx vii klx klx viii x x ii x + y y + x, V iii o K n, x K n, x + o x iv x K n, x + x o x
X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2
,., 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),
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
第5章 偏微分方程式の境界値問題
October 5, 2018 1 / 113 4 ( ) 2 / 113 Poisson 5.1 Poisson ( A.7.1) Poisson Poisson 1 (A.6 ) Γ p p N u D Γ D b 5.1.1: = Γ D Γ N 3 / 113 Poisson 5.1.1 d {2, 3} Lipschitz (A.5 ) Γ D Γ N = \ Γ D Γ p Γ N Γ
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 +
phs.dvi
483F 3 6.........3... 6.4... 7 7.... 7.... 9.5 N (... 3.6 N (... 5.7... 5 3 6 3.... 6 3.... 7 3.3... 9 3.4... 3 4 7 4.... 7 4.... 9 4.3... 3 4.4... 34 4.4.... 34 4.4.... 35 4.5... 38 4.6... 39 5 4 5....
25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3
4 14 4 14 4 1 1 4 1.1................................................ 4 1............................................. 4 1.3................................................ 5 1.4 1............................................
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.............................
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................................
1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D
1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
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
I 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ
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
(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
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................