|
|
|
- たかよし なつ
- 5 years ago
- Views:
Transcription
1 K E N Z U HP M. 1
2
3 ( )
4 S 1 S 1 S S S 1 z S z S V t 0 y y V 4
5 S S V S t (t) = (t) V t (1.2.1) d (t) dt = d(t) dt V (1.2.2) S 1 S V S S S S 1 (1.2.2) t d 2 (t) dt 2 = d2 (t) dt 2 (1.2.3) S, S V S m d2 (t) dt 2 = f (1.2.4) f S S f f = f (1.2.3) m d2 (t) dt 2 = f (1.2.5) S (1.2.4) S S S S S (1.2.2) V (1.2.2) t d 2 dt 2 = d2 dt 2 dv dt a = a A (1.2.6) S 1 S 2 S S m ma = ma ma S ma = f ma (1.2.7) 2 3 5
6 f = f S f ma S S f = 0 (1.2.7) ma = ma S A S f = ma f 3 ( A θ S A S T θ T sin θ mg A S ma T mg θ 4 S A T m d2 = ma = T sin θ (1.2.8) dt2 0 m d2 = 0 = T sin θ ma (1.2.9) dt2 2 ma A S z ω t = 0 S, S P t S, S, { = cos ωt y sin ωt y = sin ωt + y cos ωt (1.2.10) S P f P S { mẍ = f mÿ = f y (1.2.11) 5 (1.2.10) ẍ = (ẍ 2ωy ω 2 ) cos ωt (ÿ + 2ωẋ y ω 2 ) sin ωt ÿ = (ẍ 2ω y ω 2 ) sin ωt + (ÿ + 2ωẋ y ω 2 ) cos ωt (1.2.12) f S S 6
7 (1.2.11), (1.2.12) { f = A cos ωt B sin ωt f y = A sin ωt + B cos ωt A = m(ẍ 2ω y ω 2 ) B = m(ÿ + 2ωẋ y ω 2 ) (1.2.13) A = f cos ωt + f y sin ωt B = f sin ωt + f y cos ωt (1.2.14) f, y { f = f cos ωt + f y sin ωt f y = f sin ωt + f y cos ωt (1.2.15) (1.2.14), (1.2.15) S P P S (1.2.11) S (1.2.16) S f f y mω 2, mω 2 y 2ω y, 2ωẋ 2 { m ẍ = f + mω 2 + 2mωy mÿ = f y + mω 2 y 2mωẋ (1.2.16) ω z ω z 0 v 0 f 0 o f 0 c ω = (0, 0, ω) (1.2.17) = (, y ), v = (ẋ, y ) f c f o y f c = mω 2 f o = 2mv ω (1.2.18) 6 v S m=1 P f = 1 P S (1.2.16) «6 a 2 a 3 a b = b 2 b 3, a3 a1 b 3 b 1, a1 a2 b 1 b 2 y f = 1 ω = 2 v 0 = 10t = 10 S P y y ωt ωt P y S 7
8 (1.2.16) f = 0, f y = 0 { m ẍ = cos ωt + ω 2 + 2ωy mÿ = sin ωt + ω 2 y 2ωẋ Mathematica ω = 2 v 0 = 10 t = 10 P S P (1.2.1) 8 S, S, t S S t t S z V t y S z 0 t = t (1.3.1) y V { = V t t = t (1.3.2) 9 S v V (1.2.2) v = v V (1.3.3)
9 (1.3.3) 0 = t, 1 =, 2 = y, 3 = z, V = (V 1, V 2, V 3 ) = V V V µ = G µν 2 V 4 G µν ν (1.3.4) ν= G A B C C G C = AB 2. G A AE = EA = A E G E 3. G A AA 1 = A 1 A = E A 1 G 4. 3 ABC A(BC) = (AB)C = ABC (1.3.2) G(V ) G(V ) : = V t, t = t (1.3.5) G(V ) : = V t, t = t, t t t G(V )G(V ) G(V + V ) = ( V t) V t = (V + V )t, t = t (1.3.6) G(V )G(V ) = G(V + V ) (1.3.7) G(0) G(V ) G( V ) t v = vt G(V )(G(V )G(V )) = G(V )G(V + V ) = G(V + V + V ), etc G(V )G( V ) = G( V )G(V ) = G(0) t v =tan α α 0 P 13 = 1 2 gt2 + v 0t + 0 9
10 = V t t = t (1.3.8) (, t) = (0, 0) t = 0 (1.3.8) = t = 0 t = (1/V ) (t, ) 1/V t t t φ t φ V : φ π/2) p p, tan φ = ( p p)/t p,... p = p t p tan φ = p V t p (1.3.10) 1 tp = t p cos φ t p = V = tan φ (1.3.9) t p cos φ = t p 1 + tan 2 φ = 1 + V 2 t p (1.3.11) (1.3.8) V 2 t t p 1 + V 2 cos φ = A B = 2 t p t p t p A t t φ φ t p B t t 1 + V 2 = t p cos φ,... t 1 + V 2 p = t p (1.3.12) t t t t 1 + V 2 P S v P S V S P P Q sin(α φ) = Q sin(π/2 α)... Q sin(α φ) = P Q cos α = Q cos α = tan α cos φ sin φ = cos φ(tan α tan φ) = Q = p, P Q = 1 + V 2 t p t p v V 1 + V 2 p φ p P P, Q P Q = 1 p 1 + V 2 t p = v 1 + V 2,... v = v V 10
11 t tan α = v P t φ t 1 + V 2 tan φ = V P α α α φ π/2 α Q, v ϕ(, t) = A cos(k ωt), (k = 2π/λ, ω = 2πν, T = 2π/ω) (1.3.13) k, ω ν T λ v = λν t = 0 = 0 t 1 = 2π/ω = 0 V V < v 2π/ω=1/ν A n A = n 1 ν = n ν (1.3.14) B A t 5 t 4 t 3 t 2 t 1 t α t t 1 + V 2 B λ v = λ 1/ν = tan α ν t A tan 1 v B tan 1 V λ v (tan 1 v tan 1 V ) B = n ν 1 + V 2 (1.3.15) 11
12 ... B sin AB = A B = sin BA sin AB A sin BA = sin(tan 1 v) cos(tan 1 V ) cos(tan 1 v) sin(tan 1 V )) sin(tan 1 v) sin α cos β cos α sin β = sin α = v V 1 v 1 + V 2 (1.3.16) tan 1 v = α cos α = tan 1 V = β cos β = 1, sin α = v 1 + v v 2 1, sin β = V 1 + V V 2 (1.3.16) (1.3.14) (1.3.15) A/B = ν ν V 2 (1.3.16) ν ν = v V v = 1 V v < 1 (1.3.17) V t 1/V t α t 1 + V 2 B B β tan 1 v = α tan 1 V = β A C β A α (π α β) C ν BC = n ν 1 + V 2 (1.3.18) AB BC = sin(α + β) sin α = sin α cos β + cos α sin β sin α = v + V v V 2 (1.3.19) AB = n/ν AB/BC = (n/ν ) 1 + V 2 (1.3.19) ν ν = v + V v = 1 + V v > 1 (1.3.20) 12
13 1.4 v ρ(, t) ( y z ) v 2 t 2 ρ(, t) = 0 (1.4.1) v v V v) S (1.4.1) 1 = V t, t = t = t = t t + t t + t t = = t V 1 14 ( V v 2 V 2 (1.4.2) ρ(, t) v t 1 2 ) v 2 V 2 t 2 ρ(, t ) = 0 (1.4.3) (1.4.1) V ( ) V E(, t) = 1 ε 0 B(, t) E(, t) = t B(, t) = 0 B(, t) = µ 0 J(, t) + 1 E(, t) c 2 t E B ( y z ) c 2 t 2 E(, t) = 0 ( y z ) c 2 t 2 B(, t) = 0 (1.4.4) (1.4.5) ( ) 15 c
14 V (1.4.3) v = c V V v +V v V 16 c + V c V 17 V V V B L M V B B M A L c + V M M C L c V M A B B A M C A B l A t 1 c + V c V t 1 = l c + V + l c V = C 2l/c 1 β 2, β = V c V l ct 2 /2 ct 2 /2 V t 2 (1.4.6) B B B t 2 ( ) 1 2 ( ) ct 2 = l V t 2,... t 2 = 2l c 2 V 2 = 2l/c 1 β 2 (1.4.7) L 2 β 1 18 ( ) 1 = c(t 2 t 1 ) = 2l 1 1 β 2 1 β 2 lβ 2 (1.4.8) V 1 β 2 16 V, V km/s 30km/s 18 c cm/sec V V cm/sec 14
15 A l/ 1 β 2 t 1 = 2(l/c)/ 1 β 2 t 1 = t 2 19 M E M 1 (1.4.4) (S ) V (S ) S S B = µ 0 J + 1 E c 2 t 1 (V E) c2 E = B t + (V B) B = 0, E = 1 ε 0 ρ (1.4.9) ( ) ( 2 c 2 t 2 φ(, t) = 0 u ) c 2 w 2 φ(u, w) = 0 (1.4.10) (, t) (u, w) u = A 21 { ( ) ( ) ( ) u = A + Bt u A B = w = C + Dt v C D t (1.4.11) = A 1 u, (A 1 0) ( ) A 1 = 1 D B A B, = = AD BC 0 C A C D ( ) ( ) ( ) (1.4.12)... = 1 D B u t C A w = (u, w), t = t(u, w) = u u + w w = A u + C w t = u t u + w t w = B u + D (1.4.13) w 19 A
16 (1.4.10) ( ) c 2 t 2 φ(, t) ) = (A 2 B2 2 c 2 u 2 ( D 2 c 2 C 2) 1 c 2 2 w 2 + A 2 B 2 /c 2 = 1 D 2 c 2 C 2 = 1 AC BD/c 2 = 0 ( AC BD ) c 2 u A, B, C, D (1.4.14) w (1.4.14) (1.4.15) c 2 t 2 u 2 1 c 2 w (1.4.16) (u, w) A, B, C, D (1.4.15) A B, C, D 1 B = c A 2 1, C = 1 c A 2 1, D = A (1.4.17) χ A = cosh χ 22 B = c sinh χ, C = 1 sinh χ, D = cosh χ (1.4.18) c B, C, D (1.4.11) u = cosh χ ct sinh χ w = t cosh χ c sinh χ (χ ) (1.4.19) 23 cosh χ = cos(iχ) = eχ + e χ sinh χ = i sin(iχ) 2 (1.4.20) (1.4.19) { u = cos(iχ) + ict sin(iχ) icw = sin(iχ) + ict cos(iχ) ( ) ( u cos θ = icw sin θ ) ( ) sin θ cos θ ict (1.4.21) θ = iχ ict iχ 2 (1.4.21) u + (icw) 2 = ( cos θ + ict sin θ) 2 + ( sin θ + ict cos θ) 2 = 2 + (ict) 2 (1.4.22) ict icw ict iχ icw u iχ u( ) 22 cosh 2 sinh 2 = 1 23 cosh θ = (e θ + e θ )/2, sinh θ = (e θ e θ )/2, cos θ = (e iθ + e iθ )/2, sin θ = (e iθ e iθ )/2i 16
17 2 + (ict) 2 ict iχ (ict) 2 3, y, z, ict 4 ict iχ = cos θ + 0 sin θ y = y z = z (1.4.23) ict = sin θ + ict cos θ ict cos θ sin θ 0 0 ict y = sin θ cos θ 0 0, θ = iχ (1.4.24) y z z (ict ) y 2 + z 2 = (ict) y 2 + z 2 (1.4.25) iχ cosh χ 1 sinh χ χ (1.4.26) (1.4.19) u = ctχ w = t c χ (1.4.27) χ = V/c (1.4.28) u = V t w = t V c 2 { = V t t = t (1.4.28) u w t c (1.4.28) 4 u w t 1. (t) = (t) V t, t = t 2. v = v V 17
18 3. ( y z ) c 2 t 2 φ(, t) = 0 (1.4.29) V (1.4.29) 1 ( V c 2 V 2 c t 1 2 ) c 2 V 2 t 2 φ(, t ) = 0 (1.4.30) 4. V 5. V 1 β 2 6. (1.4.29) ict iχ 7. 18
19 t t 2.2 S 1 S S(, t) S (, t ) 2 S S V 3 = A(V ) + B(V ) 0 y = K(V )y (2.2.1) z = K(V )z 0 = C(V ) + D(V ) 0 0 = ct, 0 = ct (2.2.2) S S V S, S 2 ( ) A, B, C, D, K S S V V = A( V ) + B( V ) 0, y = K( V )y z = K( V )z, 0 = C( V ) + D( V ) 0 (2.2.3)
20 2 c 2 t c 2 t 2 = 2 c 2 t 2 (2.2.4) A, B, C, D, K (2.2.1) (2.2.3) = {A( V )A(V ) + B( V )C(V )} + {A( V )B(V ) + B( V )D(V )} 0 y = K( V )K(V )y z = K( V )K(V )z (2.2.5) 0 = {C( V )B(V ) + ( V )D(V )} + {C( V )A(V ) + D( V )C(V )} 0, y, z, 0 A( V )A(V ) + B( V )C(V ) A( V )B(V ) + B( V )D(V ) = 0 K( V )K(V ) = 1 (2.2.6) C( V )B(V ) + ( V )D(V ) = 0 C( V )A(V ) + D( V )C(V ) = 1 (2.2.3) (2.2.5) 2 + y 2 + z = { A 2 (V ) C 2 (V ) } 2 + K 2 (V )(y 2 + z 2 ) { D 2 (V ) B 2 (V ) } {A(V )B(V ) C(V )D(V )} 0 (2.2.7) 2 + y 2 + z A, B, C, D, K A 2 (V ) C 2 (V ) D 2 (V ) B 2 (V ) K 2 (V ) = 1 A(V )B(V ) C(V )D(V ) = 0 (2.2.8) 5 K(V ) = 1 (2.2.9) (2.2.8) 2 (2.2.8) 1 C(V ) A(V ) = B(V ) β(v ) (2.2.10) D(V ) ( ) C 2 1 = 1 β 2 = 1 A A(V ) 2,.. 1. A(V ) = ± 1 β(v ) 2 ( ) B 2 1 = 1 β 2 1 = D D(V ) 2,.. 1. D(V ) = ± 1 β(v ) 2 (2.2.11) (2.2.8) V 0 (, 0 ) (, 0) 6 (2.2.1) A(V ) = 1 1 β(v ) 2, D(V ) = 1 1 β(v ) 2 (2.2.12) K 2 (V ) = 1 K(V ) = 1 6 V = 0 =, 0 = 0 20
21 (2.2.8) 1 C(V ) = β(v ), B(V ) = β(v ) 1 β(v ) 2 1 β(v ) 2 (2.2.13) (2.2.12), (2.2.13) (2.2.6) 2 A( V )B(V ) + B( V )D(V ) = 1 1 β( V ) 2 β(v ) + β( V ) 1 β(v ) 2 1 β( V ) β(v ) 2 = 0 (2.2.14)... β(v ) + β( V ) = 0 A, B, C, D 1 V β(v ) (2.2.1) = β(v ) 0 = + β(v ) 0 1 β(v ) 2 1 β(v ) 2 y = y z = z y = y z = z (2.2.15) 0 = 0 β(v ) 1 β(v ) 2 0 = 0 β(v ) 1 β(v ) 2 β(v ) S... d = d + d 0 = 0 1 d β(v ) d 0 1 β(v ) 2 1 β(v ) 2 d 0 = 0 d d = 1 d β(v ) 0 d 1 β(v ) 2 1 β(v ) 2 d d 0 = d β(v )d 0 d 0 β(v )d = d d 0 β(v ) 1 β(v ) d d 0 (2.2.16) S S V d/d 0 = 0, d /d 0 = d /cdt = V/c V c = β(v ) (2.2.17) β(v ) (2.2.15) β < 1 7 (2.2.15) (2.2.17) (2.2.15) V c β 1 (2.2.15) β 1 = ( β 0 )(1 β 2 ) 1/2 ( β 0 )( + β 2 /2) β 0 = V t 0 = ( 0 β)(1 β 2 ) 1/2 ( β 0 )( + β 2 /2) 0 β 0,... t = t (2.2.18)
22 (4 ) = cosh χ ct sinh χ = β(v ) 0 1 β(v ) 2 0 = t cosh χ sinh χ c 0 = 0 β(v ) 1 β(v ) 2 (2.2.19) cosh χ = 1, sinh χ = β, 1 β 2 1 β 2 tanh χ = β = V c (2.2.20) θ iχ cosh χ = cos θ = 1/ 1 β 2, sinh χ = i sin θ = β/ 1 β 2 { ( ) ( = cos θ + (i 0 ) sin θ cos θ i 0 = sin θ + (i 0) cos θ i = 0 sin θ sin θ cos θ ) ( ) i 0 (2.2.21) S S = γ( β 0 ), γ = 1/ 1 β 2 0 = γ( 0 β) (2.2.22) S 1 S = γ ( β 0), γ = 1/ 1 β 2 0 = γ ( 0 β ) (2.2.23) S S = γ ( β 0), γ = 1/ 1 β 2 = γγ { (1 + ββ ) (β + β ) 0 } = γγ (1 + ββ ) ( β + ) β 1 + ββ 0 (2.2.24) β = β + β 1 + ββ (2.2.25) 1 β 2 = (1 + ββ ) 2 (β + β ) 2 (1 + ββ ) 2 = (1 β2 )(1 β 2 ) (1 + ββ ) 2 =... γγ (1 + ββ ) = 1 1 β 2 1 {γγ (1 + ββ )} 2 (2.2.26) = 1 1 β 2 ( β 0 ) = γ ( β 0 ), γ = 1/ 1 β 2 (2.2.27) 22
23 S S 9 0 = γ ( 0 β ) (2.2.28) i i 0 0 i 0 cos θ 1 = γ 1, sin θ 1 = iγ 1 β 1 cos θ 2 = γ 2, sin θ 2 = iγ 2 β 2 (2.2.29) cos(θ 1 + θ 2 ) = γ 1 γ 2 + γ 1 β 1 γ 2 β 2 sin(θ 1 + θ 2 ) = i(γ 1 β 1 γ 2 + γ 1 γ 2 β 2 ) (2.2.30) θ 2 θ 1 S S S ( ) ( ) ( ) ( ) γ 1 iγ 1 β 1 γ 2 iγ 2 β 2 i 0 ( iγ 1 β 1 γ 1 iγ 2 β 2 γ 2 ) ( i 0 ) γ 1 γ 2 (1 + β 1 β 2 ) iγ 1 γ 2 (β 1 + β 2 ) = iγ 1 γ 2 (β 1 + β 2 ) γ 1 γ 2 (1 + ) i 0 ( ) β 1 β 2 ) ( = cos(θ 1 + θ 2 ) sin(θ 1 + θ 2 ) sin(θ 1 + θ 2 ) cos(θ 1 + θ 2 ) i 0 (2.2.31) γ = γ 1 γ 2 (1 + β 1 β 2 ) γ β = γ 1 γ 2 (β 1 + β 2 ) (2.2.32) (2.2.25) (2.2.33) (2.2.29) β = β 1 + β β 1 β 2 (2.2.33) γ 2 (1 β 2 ) = γ1(1 2 β1)γ 2 2(1 2 β2) 2 = 1... γ 1 = 1 β 2 (2.2.34) A B 2 2 W A W B A W B B W B W B 9 0 i 23
24 ( ) W A B W A 2 A B W A W B A t A B W B t B B A W A t A t B t A = t A t B t B = (t A + t A)/2 W A W B B W B t B (t A + t A )/2 t B W B B t B { } 1 t B = t B + = t B + 2 (t A + t A) t B = 1 2 (t A + t A) (2.3.1) ( ) S S t c 0 ( ct) V S, 0 S S (, 0 ) (, 0 ) (2.2.15) ( )2 A, B 2 1. S A B 2. S A B A B S S V S S A = γ( A V t A ), t A = γ(t A V A ) (... 0 = c t A, 0 = c t A) B = γ( B V t B ), t B = γ(t B V B ) (2.3.2) A = B, t A t B (2.3.2) B A = γv (t B t A ) t B t A = γ(t B t A ) (2.3.3) 1 B A 0 S 2 A,B 2 t B t A > t B t A 2 A B t B t A S S γ S S t B = t A, B A B A = γ( B A ) t B t A = γv ( B A )/c 2 (2.3.4) 24
25 S A B t B t A 0 A B A > B t B > t A S A B S A B A < B l 0 S V S S A, B l 0 = B A (2.3.5) S l l B A A, B S S V l 0 A B A B A = γ( A + βct A ), B = γ( B + βct B ) t A = t B B A l = B A = γ( B A) = γl 0 (2.3.6) γ 1 l < l 0 10 v l 0 D D A A l l 0 0 v D F D v B B C C C E l 0 β C l 0 θ l 0 θ F l 0 v E l 0 1 β 2 10 l 0 25
26 AB CD l 0 /c vl 0 /c = l 0 β ABC D C E l 0 1 β 2 β = θ θ V S S S W t S C S S t = t = 0 t t t t (2.2.15) t = t (β/c)v t 1 β 2 = t { 1 (V/c) 2} 1 β 2 = t 1 β 2 = t γ < t (... γ > 1) (2.3.7) S S V t t V t t t A B A B B A A B A B A B B A S t A A c B t B 26
27 ( A t A B t B B A = ±c(t B t A ) (2.3.8) S t B t A > 0 t A = γ(t A β A /c), t B = γ(t B β B /c)... t B t A = γ(t B t A β( B A )/c) = γ(t B t A )(1 β) (2.3.9) γ(1 β) t B t A > S n A ν λ ω k 11 ψ(, t) = A sin(ωt k ), k = (ω/v)n, ω = 2πν = 2π(v/λ) (2.4.1) S V S t = t = 0 S S y S y S V ωt k MEM-3 S S ω, k ωt k = ω t k (2.4.2) (2.2.15) k ω ωt k = ωt k k y y k z z = γω{t + (β/c) } γk { + (β/c)t } k y y k z z n θ k = γt {ω (β/c)k } γ {k (β/c)ω} k y y k z z = ω t k k yy k zz (2.4.3)... ω =γ{ω (β/c)k }, k = γ{k (β/c)ω}, k y = k y, k z = k z k, ω S 11 v v = ω/k c 27
28 k θ k θ (2.4.3) ν, ν k = k cos θ = (ω/c) cos θ = 2π(ν/c) cos θ k = k cos θ = (ω /c) cos θ = 2π(ν /c) cos θ (2.4.4) ν = γν (1 β cos θ), ν cos θ = γν (cos θ β) ν sin θ = ν sin θ (2.4.5) (2.4.5) 1 θ = 0 ν ν 1 β = ν (2.4.6) 1 + β ν ν 1 β ν = ν 1 + β β = ν β = β ν = ν 1 β 0 ν 0 ν (2.4.5) β β(= V/c) ν = ν 1 + β 1 β (2.4.7) (2.4.5) cos θ = cos θ β 1 β cos θ, sin θ = sin θ 1 β 2 1 β cos θ, tan θ = sin θ 1 β 2 cos θ β (2.4.8) θ θ θ = π/2 cos θ = β (2.4.9) V c (2.4.8) 1 θ θ θ V cos θ = (cos θ β)(1 β cos θ) 1 (cos θ β)(1 + β cos θ) cos θ β sin 2 θ (2.4.10) 28
29 θ = θ θ θ 0 cos θ cos θ θ sin θ, cos θ 1, sin θ θ (2.4.10) θ = β sin θ = V c sin θ (2.4.11) θ θ θ H H V t ct V H H S z S (2.4.5) θ = 90 ν 1 = ν 1 β 2 (2.4.12) 12 1 (2.4.9) cos θ = β (2.4.13) β = θ = θ 90 S ψ(, t) = A sin(ωt k), k = ω/c (2.4.14) S (, t ) (, t) = γ( + βct ), t = γ(t + (β/c) ) ψ(, t ) = A sin (γω(1 β)t γω c = A sin(ω t k ) (1 β) ) (2.4.15) ω, k S ω = γω(1 β), k = 1 γω(1 β) (2.4.16) c ν = γ(1 β)ν = ν 1 + β 1 + β (2.4.17) (2.4.5) θ = 0 S z S z z V ψ(t, z) = A sin(ωt kz), k = ω/c (2.4.18) t = γ(t + (β/c) ) k θ = 90 ψ(t,, z ) = A sin(ω t k z ) = A sin(γωt k(z γβ )) (2.4.19) ω = γω ν = γν (2.4.12) 12 ν α ν ν = 1 1 (v/v ) cos α ν α = 90 ν = ν 29
30 ( ) S v(v, v y.v z ) 13 S V S v (v, v y.v z) (v, v y, v z ) = ( d dt, dy dt, dz ) ( ) d, (v dt, v y, v z) = dt, dy dt, dz dt S S (2.2.15) (2.2.16) (2.5.1) d d 0 = 1 c v = d β(v )d 0 d 0 β(v )d =... v = v V 1 1 c 2 v V d d 0 β(v ) 1 β(v ) d d 0 = v c β 1 β c v (2.5.2) v y v z dy d 0... v y = = dy (d 0 β d)/ 1 β 2 = v y ( γ 1 1 c 2 v V ) v z = dy d 0 1 β 1 V 2 d c d 0 v z ( γ 1 1 c 2 v V v = v V 1 1 v = v + V c 2 v V c 2 v V v y v y = ( γ 1 1 ) v y v c 2 v y = ( V γ ) c 2 v V v z = v z ( γ 1 1 c 2 v V ) v z = ) v z ( γ c 2 v V ) (2.5.3) (2.5.4) (2.5.4) S v S v v = v + V v = v + V, v y = v y, v z = v z (2.5.5) 14 S v v y (2.5.4) c (2.5.5) 30
31 v z = 0 (2.5.4) v z = 0 v y v θ v θ { v = v cos θ v y = v sin θ { v = v cos θ v y = v sin θ v y y y S S v v θ θ v v y v S (2.5.4) v = v cos θ = v V 1 1 = c 2 v V v y = v sin θ = v y 1 (V/c) c 2 v V v cos θ V 1 V (v cos θ)/c 2 = v sin θ 1 (V/c) 2 1 V (v cos θ)/c 2 (2.5.6) θ V = 0.4c 0.1c 0.01c v = 0.9c 0 θ tan θ = v y v = v sin θ 1 (V/c) 2 (2.5.7) v cos θ V { v v 2 = v 2 + v y 2 2V (v cos θ) + V 2 { 1 (v sin θ/c) 2}} 1/2 = 1 V (v cos θ)/c 2 (2.5.8) θ θ v = 0.9c V v (2.5.8) θ = 0 v = v V 1 vv/c 2 (2.5.9) v V V, v v v = v + V 1 + v V/c 2 (2.5.10) S v S v S V S S S V v (, ct ) (, ct) v S (2.5.10) S V = k 2 c v = k 1 c, (0 < k 1, k 2 1) (2.5.10) S S S V v v v = k 1 + k k 1 k 2 c (2.5.11) k 1 + k 2 (1 + k 1 k 2 ) = (k 1 1)(k 2 1) 0 31
32 v c c (k 1 = k 2 = 1) v = c M E M 2 (2.2.20) tanh χ = β = V c (2.5.10) β = V c = tanh χ, β = v c = tanh χ, β = v c = tanh χ (2.5.12) β = β + β 1 + β β tanh χ = tanh χ + tanh χ 1 + tanh χ tanh χ = tanh(χ + χ)... χ = χ + χ (2.5.13) χ c n c/n V 2 1. c/n + V 2. α c/n + αv (α < 1) α = 1 (1/n 2 ) v = c n + ( 1 1 n 2 ) V (2.5.14) 2 (2.5.14) α (2.5.14) y n c/n S V S 32 c/n V
33 S c/n v (2.5.10) v = c/n + V ( c ) (1 1 + V/nc = n + V + V ) 1 ( c ) ( nc n + V 1 nc) V c ( n ) n 2 V (2.5.15) (2.5.14) 2 A B c S A v c 2c A v B u S u B A S v rel B d A dt A = v, d B dt B = u (2.5.16) A S S B v rel S A B v rel (2.5.2) 15 d A dt A = 0,... v rel = u + v d B dt B c 2 uv = v rel = d B dt B v 1 v c 2 d B dt B u + v = c 2 uv (2.5.17) A B c v rel u = v = c v rel = c (2.5.18) S a S a 16 (2.5.10) t dv dt = d ( v ) + V dt 1 + v V/c 2 = 1 (V/c)2 dv (1 + v V/c 2 ) 2 dt = 1 (V/c)2 dv dt (1 + v V/c 2 ) 2 dt dt (2.5.19) 15 (2.5.10) 16 S S V S S 33
34 (2.2.16) ) dt dt = γ (1 + β d d = γ ( 1 + v V /c 2), γ = 1/ 1 (V/c) 2 (2.5.20) 0 a = dv dt = 1 γ 1 (V/c) 2 dv (1 + v V/c 2 ) 3 dt = 1 dv {γ(1 + v V/c 2 )} 3 dt = 1 {γ(1 + v V/c 2 )} 3 a (2.5.21) S a (2.5.21) a a, V V, v v a = a {γ(1 vv/c 2 )} 3 (2.5.22) S S S t v t + t v + v t v S S t (v = 0) t v t v S S (2.5.10) v + v = t t v + v 1 + v v/c 2,... v = 1 (v/c)2 1 + v v/c 2 v (2.5.23) t = t 1 (v/c) 2 = γ p t, γ p = 1 1 (v/c) 2 (2.5.24) (2.5.23) (2.5.24) v t = t 0, v 0 γ 3 p 1 + v v/c 2 v t (2.5.25) dv dt = lim v t 0 t = dv dt γ3 p (2.5.26) v 0 dv/dt = a, dv /dt = a 0 a = a 0γ 3 p (2.5.27) S a S a 0 (2.5.21) v 0, V v, γ γ p, a a 0 a 0 = ξ = (2.5.25) ξ = dv 1 dt (1 v 2 /c 2 ) 3/2 (2.5.28) 34
35 t 0 = 0, v 0 = 0 1 (1 v 2 /c 2 dv = ξdt,.. v. ) 3/2 1 v 2 /c = ξ t (2.5.29) 2 v v = ξt 1 + (ξ t/c) 2 (2.5.30) 0 v t ( ) t = 0 = 0 v v = c t t ξt = vdt = dt 1 + (ξ t/c) 2 0 = c2 ξ 0 ( ) (2.5.31) 1 + ξ2 c 2 t2 1 ( + c 2 /ξ) 2 c 4 /ξ 2 (ct)2 c 4 /ξ 2 = 1 (2.5.32) t > = 0 ct = + c 2 /ξ ξt c ( c ξ 2 ) ξ 2 c 2 t2 1 = 1 2 ξt2 (2.5.33) = 1 2c 2 ξ(ct) 2 S (2.5.24) dt = 1 (v/c) 2 dt (2.5.34) t t ( ) t = 1 (v/c) 2 1 c ξ dt = dt = ξ2 t c 2 ξ sinh 1 c t 2... ξ ( ) (2.5.35) ξ c t = sinh c t t 17 ( ) v 0 S v 0 = 0 (2.5.36) α v 1 = α (2.5.37) 17 35
36 S 2 α S (2.5.10) 2 S 3 S v 3 = v 2 + α 1 + αv (2.5.39) 2 c 2 n S v n+1 = v n + α v 2 = v 1 + α 1 + αv (2.5.38) 1 c 2 v v = c. v 3 v 2 v 1 t 1 t 2 t 3 t αv n c 2 (n = 0, 1, 2, ; v 0 = 0) (2.5.40) c δ S n n + 1 t n = t n+1 t n = δ 1 v 2 n /c 2 (n = 0, 1, 2, ; t 0 = 0) (2.5.41) t S (2.5.30) v n = v n+1 v n = v n + α 1 v2 n 1 + αv v n n = α c 2 c α v n c 2.. v n. = α ( 1 v 2 n /c 2) 3/2 dv t n δ 1 + α v2 n dt = α ( 1 v 2 /c 2) 3/2 δ c α v2 c 2 ξ = α/δ α, δ 0 (2.5.42) dv dt = lim v n = ξ(1 v 2 /c 2 ) 3/2 (2.5.43) δ 0 t n α 0 ξ = a 0 (2.5.27) 2.6 S 1, 2 W 1, W 2 W 1, W t = t = 0 V 36
37 S W S 1 1 W 1 t 1 W τ W 2 t 2 W τ 2 W 1, W 2 W t 2 t 1 S S W V 1 2 W 1 W 2 τ 2 τ 1 S S S τ 2 τ 1 = 1 V 2 /c 2 (t 2 t 1 ) (2.6.1) τ 2 τ 1 τ V V = 2 1 t 2 t 1 (2.6.2) (2.6.1) 2 (2.6.2) ( τ) 2 = (1 V 2 /c 2 )(t 2 t 1 ) 2 = {1 1c ( 2 1 ) 2 } 2 (t 2 t 1 ) 2 (t 2 t 1 ) 2 = (t 2 t 1 ) 2 ( 2 1 ) 2 c 2 (2.6.3) S S 18 ( 2 1 ) 2 = {( 2 1 ) + U(t 2 t 1 )}2 1 U 2 /c 2 { (t (t 2 t 1 ) 2 2 t 1 ) + U c = 2 ( 2 1) 1 U 2 /c 2 } 2 (2.6.4) (2.6.3) ( τ) 2 = (t 2 t 1 ) 2 ( 2 1 ) 2 c 2 = (t 2 t 1) 2 ( 2 1 )2 c 2 (2.6.5) τ S S (2.6.6) (2.5.41) t2 τ 2 τ 1 = 1 v2 c 2 dt = t 1 t2 t 1 dt γ p (2.6.6) 3 18 S S U 37
38 2.7 0 = 0 β 1 β 2, = β 0 1 β 2 y = y z = z 0 = ct (2.7.1) 0 ( ct) 0 α 0 45 α y, z 0 ct 19 t ct 0 = 0 0 = 0 (2.7.1) 0 = β 0, 0 = β (2.7.2) 0, 0, α = tan 1 β 20 cos α = 1, sin α = β 1 + β β 2 β 1 α β = 1 (2.7.3) S S S S S 0 0 S S 0 S X 0 X { = X cos α + X 0 sin α 0 = X sin α + X 0 cos α (2.7.4) X = 1 + β 2 1 β β 2 X 0 = 0 1 β 2 (2.7.5) (2.2.15) = β 0 1 β 2 0 = 0 β 1 β 2 = + β 0 1 β 2 0 = 0 β 1 β 2 (2.7.5) (2.7.6) 0 α X 0 ( k 0 ) P (, 0) (X, X 0 ) α X ( k ) k = (1 + β 2 )/(1 β 2 ) X = k, X 0 = k 0 X k X 0 k 0 k 0 19 ict
39 S 2 A B (t A = t B = t AB ) S A(t AB, A ) B(t AB, B ) S A (t A, A ) B (t B, B ) A t A B t B A t B t AB 0 α t B t A 0 B B A A α B A A S X l 0 AB 21 AB X 0 ( k 0 ) 1 + β 2 π/2 2α AB = kl 0 = l 0 (2.7.7) B X ( k ) 1 β 2 A α A B D E C D E DE S l ABC 0 α AB sin ACB = AC sin ABC,.. cos 2α. AC = l = kl 0 cos α = l 0 1 β 2 (2.7.8) S S 2 1. S 0 = τ 2 A B 2 S V S 0 = τ A B 22 AB = τ, A B = kτ kτ = cos α cos 2α τ, (2.3.3)... τ = 1 k cos α cos 2α τ = 1 τ (2.7.9) 1 β 2 2. S S 0 = τ 2 A B S 2 0 = τ S τ 23 kτ cos α = τ,... τ = 1 1 β 2 τ (2.7.10) 21 S A B 22 A A X 23 A A 39
40 0 B τ A α kτ X 0 ( k 0 ) A B α B τ X ( k ) A π 2 α α kτ B A 0 B τ X 0 ( k 0 ) kτ B A α α B B τ α kτ A X ( k ) S v AB γ... tan δ = 0 = c t = c v,... v c AC sin η = BC sin ξ, AC = k, BC = kct AC BC = 1 c t = v c = sin η sin ξ v = c sin η sin ξ ξ, η ξ = δ α, (2.7) η = π 2 α δ V c = cot δ (2.7.11) = tan α (2.7.12) 0 α X 0 α A δ η ξ B X A ξ X 0 η C B X v = c... v = v V cos(δ + α) sin(δ α) = c 1 1 c 2 vv cot δ tan α 1 cot δ tan α = c (v/c) β 1 (v/c)(v/c) (2.7.13) (2.5.2) 2 A B A u B v A B u, v tan α = u/c, tan δ = v/c y A u v B 0 α A δ B y A v rel B 0 X 0 C α + δ α π/2 + α δ X A α B A B 40
41 A A B S A C 2 A C=kct B A B = k A B v rel (= /t ) A CB A B A C = 1 c t = 1 c v rel (2.7.14) A B A C = sin(α + δ) sin(α + δ) = sin(π/2 + α δ) cos(α δ) tan α + tan δ = 1 + tan α tan δ = 1 u + v c... v rel = u + v c 2 u v c 2 u v = 1 c v rel = sin α cos δ + cos α sin δ cos α cos δ + sin α sin δ (2.7.15) (2.5.17) V 45 1 ν c/ν V c/ν 0 A 3 A 2 A 1 A 1 X 0 ( k 0 ) A 3 A 2 α T = 1/ν X ( k ) A 1A 1 = π π 4 = 3 4 π, A 1A 1 = π (α+ 3 4 π) = 1 4 π α A 1 A 1 A 1 / sin(π/4 α) = A 1/ sin(3π/4) (2.7.16) A 1 = c/ν, A 1 = (c/ν )k 24 (2.7.3) ν 1 + β 2 1 β = νk(cos α sin α) = ν 1 β 2 = ν 1 β 1 + β β (2.7.17) (2.4.6) M E M 3 v ψ(t, ) = A sin(ωt k) = ωt k = 2πν(t /v) 2π ν(t /v) ν t = 0 P t 0 t(> t 0 ) P ν(t t 0 ) = ν(t /c) t P 24 X 0 41
42 P S PA BA=ct 0 = c(/c) =, PA= ct BA= ct PA/(c/ν) = ν(t /c) S BA B A, PA PA PA = ct, B A = c( /c) = PA PA /(c/ν ) = ν (t /c) S (2.7.6) (2.7.17) ν (ct ) = ν ( 0 ) = ν γ(1 + β)( 0 ), γ = 1/ 1 β 2 1 β 1 = ν 1 + β (1 + β)( 0 ) = ν(ct ) 1 β 2... ν (t /c) = ν(t /c) (2.7.18) t = 0 t = t 0 t = t S S P(, ct) P(, ct ) P ν(t t 0 ) t 0 = /c c/ν A B A B c/ν 42
43 i 0 = ict 1 = ( 1 + iβ 4 )/ 1 β 2 ( ) ( 1 1/ 1 β 2 iβ/ ) ( ) 1 β = ( 4 iβ 1 )/ = 1 β 2 iβ/ 1 β 2 1/ (3.1.1) 1 β a 11 = 1/ 1 β 2, a 14 = iβ/ 1 β 2 (3.1.1) a 41 = iβ/ 1 β 2, a 44 = 1/ 1 β 2 (3.1.2) 1 = a a 14 4 µ = 4 = a a 44 4 ν=1,4 a µν ν µ = a µν ν (3.1.3) (3.1.2) a 11 a 14 det(a µν ) = a 41 a 44 = 1 (3.1.4) µ=1,4 µ=1,4 µ=1,4 µ=1,4 (3.1.5) µ=1,4 a µ1 a µ1 = a µ1 a µ1 = a 11 a 11 + a 41 a 41 = 1 a µ4 a µ4 = a µ4 a µ4 = a 14 a 14 + a 44 a 44 = 1 a µ1 a µ4 = a µ1 a µ4 = a 11 a 14 + a 41 a 44 = 0 a µ4 a µ1 = a µ4 a µ1 = a 14 a 11 + a 44 a 41 = 0 a µν a µλ = a µν a µλ = a 1ν a 1λ + a 4ν a 4λ = δ νλ { ν = λ : 1 ν λ : 0 (3.1.5) (3.1.6) 43
44 µ=1,4 a νµ a λµ = a νµ a λµ = a ν1 a λ1 + a ν4 a λ4 = δ νλ { ν = λ : 1 ν λ : 0 (3.1.7) (3.1.3) µ µ = = (a a 14 4 )(a a 14 4 ) + (a a 44 4 )(a a 44 4 ) = (a 11 a 11 + a 41 a 41 ) (a 14 a 14 + a 44 a 44 ) (a 11 a 14 + a 41 a 44 ) 1 4 (3.1.8) = = µ µ 1 µ µ = a µν a µλ ν λ = δ νλ ν λ = ν ν (3.1.9) (= i 0 ) = (1) φ (3.1.3) φ φ = φ φ µ µ (3.1.9) (2) A 1 A 4 A A 4 A 4 = ia 0 (3.1.10) A 0 A 1, A 4 (3.1.3) A µ A µ = a µν A ν (3.1.11) (3) D µ = A µ + B µ (3.1.12) (4) 2 A µ B µ = A 1 B 1 + A 4 B 4 = A 1 B 1 A 0 B 0 (3.1.13) 1 µ, ν, λ, ρ A µ = (A 1, ia 0), B µ = (B 1, ib 0) 44
45 A µb µ = a µν a µλ A ν B λ = δ νλ A ν B λ = A ν B ν (3.1.14) A µ B µ 0 A µ B µ 3 A µ B µ = A 1 B 1 A 0 B 0 = 0 (3.1.15) A µ A µ = A 2 1 A2 0 > 0 A µ 2. A µ A µ = A 2 1 A2 0 < 0 A µ 3. A µ A µ = A 2 1 A2 0 = 0 A µ A A2 0 A µ 0 0 A 0 α 0 π 4 A 1 1 A 0 β 0 A 1 1 A µ A µ a a 2 = A µ A µ = A 2 0 A 2 1 (3.1.16) 5 A 21 + A20 = A 2 0 A2 1 A A2 1 A 2 0 A2 1 A 2 0 = a + A2 1 A 2 0 A2 1 A 2 0 A 2 1 = (A A 2 0) sin 2 (α 0 + π/4) A 2 1 = 1 2 (3.1.17) { A 2 0 A (A A 2 1) sin 2α 0 }... A 2 0 A 2 1 = (A A 2 1) sin 2α 0 (3.1.18) 3 A, B 90 (3.1.15) 4 null 5 45
46 (3.1.16) A A2 0 = a (3.1.19) sin 2α0 A 1, A 0 A 1 = a sin 2α0 cos(π/4 + α 0 ), A 0 = 2 A µ, B µ a sin 2α0 sin(π/4 + α 0 ) (3.1.20) A µ B µ = A 1 B 1 A 0 B 0 ab = {cos(π/4 + α 0 ) cos(π/4 + α 1 ) sin(π/4 + α 0 ) sin(π/4 + α 0 )} sin 2α0 sin 2α 1 = ab sin(α 0 + α 1 ) sin 2α0 sin 2α 1 (3.1.21) A µ A µ a 2 = A µ A µ = A 2 1 A 2 0 (3.1.22) A A2 0 = a A A2 1 A 2 1 A2 0 A 1, A 0 A 1 = a sin 2β0 cos(π/4 β 0 ), A 0 = A µ B µ = a sin 2β0 (3.1.23) a sin 2β0 sin(π/4 β 0 ) (3.1.24) A µ B µ = A 1 B 1 A 0 B 0 ab = {cos(π/4 + α 0 ) cos(π/4 beta 0 ) sin(π/4 + α 0 ) sin(π/4 β 0 )} sin 2α0 sin 2β 0 = ab sin(β 0 α 0 ) sin 2α0 sin 2β 0 (3.1.25) α 0 = β 0 A µ B µ A µ, B µ a, b A µ B µ D µ d d a + b (3.1.26) A + B A + B (3.1.26) A µ, B µ. A µ A µ = a 2, B µ B µ = b 2 (a, b > 0) (3.1.27) 46
47 D µ d 2 = D µ D µ d 2 = D µ D µ = (A µ + B µ )(A µ + B µ ) = a 2 + b 2 2A µ B µ (3.1.28) (3.1.21) A µ B µ = ab sin(α 0 + α 1 ) sin 2α0 sin 2α 1 ab (3.1.29) d 2 = a 2 + b 2 2A µ B µ a 2 + b 2 + 2ab = (a + b) 2 (3.1.30) ( ) 1, 2, 3 4 = i 0 = ict 4 µ (µ = 1, 2, 3, 4) 4 µ µ = = (3.1.31) µ = a µν ν, a µν a µλ = δ νλ { ν = λ : 1 ν λ : 0 (3.1.32) 1 a 11 a 12 a 13 a = a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a (3.1.32) (3.1.33) 6 (3.1.32) ν λ ν λ 10 7 a µν = 6 1, 2, 3 3 (V 1, V 2, V 3 ) φ (3.1.32) φ φ = φ (3.1.34) 6 a ij 7 n n n 2 n (n 2 n) (n 2 n)/2 = n(n 1)/2 n =
48 (3.1.32) A µ = a µν A ν (3.1.35) A µ = a νµ A ν (3.1.36) 8 (3.1.35) A µ = a νµ a νλ A λ = δ µλa λ = A µ (3.1.37) (3.1.36) A µ B µ = a νµ A νa λµ B λ = δ νλa νb λ = A νb ν (3.1.38) A µ 2 1. A µ A µ = A 2 1 A2 2 + A2 3 A2 0 > 0 A µ 2. A µ A µ = A 2 1 A2 2 + A2 3 A2 0 < 0 A µ 3. A µ A µ = A 2 1 A2 2 + A2 3 A2 0 = 0 A µ 0 A µ A µ A µ 1 2 S 2 P Q P = ( 1, y 1, z 1, ict 1 ), Q = ( 2, y 2, z 2, ict 2 ) S P = ( 1, y 1, z 1, ict 1 ), Q = ( 2, y 2, z 2, ict 2 ) 2 A = Q P A µ = ( 2 1, y 2 y 1, z 2 z 1, ic(t 2 t 1 )) A = Q P A µ = ( 2 1, y 2 y 1, z 2 z 1, ic(t 2 t 1)) S S s 12, s 12 s 2 12 = A µ A µ = c 2 (t 2 t 1 ) 2 + ( 2 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 (3.1.39) = c 2 t l 2 12 s 2 12 = A µa µ = c 2 (t 2 t 1) 2 + ( 2 1) 2 + (y 2 y 1) 2 + (z 2 z 1) 2 = c 2 t l a µν a νµ (3.1.40) 48
49 c 2 t l 2 12 = c 2 t l 2 12 (3.1.41) 1. 2 P, Q S S t 12 = 0 (3.1.41) s 2 12 = c 2 t l 2 12 = l 2 12 > 0 (3.1.42) A µ A µ > 0 S S 2 l 12 > ct 12 (3.1.43) 2 l 12 t 12 2 P, Q A µ 2. P,Q2 S S l 12 = 0 (3.1.41) S 2 s 2 12 = A µ A µ = c 2 t l 2 12 = c 2 t 2 12 < 0 (3.1.44) l 12 < ct 12 (3.1.45) 2 P, Q A µ 3. 2 P, Q 2 / ν ν = µ µ ν = a µλ µ a ρν ν λ = a µν ν µ a ρν = a ρν a µν ν µ = δ ρµ µ = ρ ρ µ µ = a µν ν (3.1.46) (3.1.35) / µ µ µ = i i = 2 1 c 2 2 t 2 (3.1.47)
50 2 1 c 2 2 t 2 / µ A µ A µ µ = (a νµ A A ν ρ A ν ν) = a µν µ = a µν a ρµ ρ µ ρ = δ νρ A ν ρ = A ν ν (3.1.48) A µ B µ 0 A µ B µ = 0 (3.1.49) 16 T µν (3.1.32) T µν = a µλ a νρ T λρ (3.1.50) T λρ 2 9 T µν = a λµ a ρν T λρ (3.1.51) T µν = T νµ (3.1.52) T µν = T νµ (3.1.53) 10 2 S µν A µν T µν = 1 2 (T µν + T νµ ) (T µν T νµ ) (3.1.54) S µν = S νµ = 1 2 (T µν + T νµ ), A µν = A νµ = 1 2 (T µν T νµ ) (3.1.55) 2 S µν A µν S µν A µν S 11 S 12 S 13 S 14 0 A 12 A 13 A 14 S 12 S 22 S 23 S 24 S µν = S 13 S 23 S 33 A 34 A A 12 0 A 23 A 24 µν = A 31 A 32 0 A 34 (3.1.56) S 14 S 24 S 34 S 44 A 14 A 24 A A µ B ν A µ = a µλ A λ, B ν = a νρ B ρ, A µb ν = a µλ a νρ A λ B ρ... T µν = a µλ a νρ T λρ, T µν = A µb ν, T λρ = A λ B ρ 9 µ ν = 16 n n 4 n
51 T µν A µ T µν A µ B ν = T µν A µ (3.1.57) T µν = a µλ a νρ T λρ, A µ = a µξ A ξ B ν = T µνa µ = a µλ a νρ T λρ a µξ A ξ = a νρ δ λξ T λρ A ξ = a νρ (T λρ A λ ) = a νρ B ρ A µ = T µν ν (3.1.58) A µ = T µν ν = a µλ a νρ T λρ ν = a µλ a νρ a τν T λρ τ = a µν a νρ T λρ τ = a µλ δ τρ T λρ τ τ ν = a µλ T λρ ρ = a µλ a νρ T λρ τ a τν (3.1.59) = a µλ A µ µ ν T µν T 11 T 12 T 13 T 14 T 12 T 22 T 23 T 24 T = T 13 T 23 T 33 T 34 T 14 T 24 T 34 T 44 ( ) T 11 + T 22 + T 33 + T 44 = tr T (3.1.60) T µν T µν A µ B ν 2 A µ B ν A 1 B 1 A 1 B 2 A 1 B 3 A 1 B 4 A 2 B 1 A 2 B 2 A 2 B 3 A 2 B 4 A µ B ν = A 3 B 1 A 3 B 2 A 3 B 3 A 3 B 4 A 4 B 1 A 4 B 2 A 4 B 3 A 4 B 4 ( ) A 1 B 1 + A 2 B 2 + A 3 B 3 + A 4 B 4 = A µ B µ (3.1.61) 51
52 3.1.3 S V S V = (V 1, V 2, V 3 ) S V V // ( ) β β // = β β = β β 2 β (3.1.62) = // // 1 = 1 β 0 1 β 2 2 = 2 3 = 3 0 = 0 β 1 1 β 2 // = // β 0 1 β 2 = 0 = 0 β // 1 β 2 (3.1.63) S / S V β = V /c 11 = // + (3.1.62) ( β = γ( // β 0 ) + = γ β 2 β + γ 1 ) β γ β 2 β γβ 0 = + 1 β 2 (γ 1)( β)β γβ 0 (3.1.64) = ( 1, 2, 3 ), β = (β 1, β 2, β 3 ) ( β)β = ( 1 β β β 3 )(β 1, β 2, β 3 )... {( β)β} 1 = ( 1 β β β 3 )β 1 = ( j β j )β 1 {( β)β} 2 = ( 1 β β β 3 )β 2 = ( j β j )β 2 (3.1.65) {( β)β} 3 = ( 1 β β β 3 )β 3 = ( j β j )β 3 (3.1.64) i = [δ ij + 1 ] β 2 (γ 1)β iβ j j γβ i 0 (3.1.66) 0 = γ( 0 β ) = γ( 0 β i i ) (3.1.67) i = [δ ij + 1 ] β 2 (γ 1)β iβ j j + γβ i 0 0 = γ( 0 + β i i) (3.1.66) (3.1.67) 0 γ γβ 1 γβ 2 γβ = γβ (γ 1)β1 2 /β2 (γ 1)β 1 β 2 /β 2 (γ 1)β 1 β 3 /β 2 γβ 2 (γ 1)β 1 β 2 /β (γ 1)β2 2/β2 (γ 1)β 2 β 3 /β 2 γβ 3 (γ 1)β 1 β 3 /β 2 (γ 1)β 2 β 3 /β (γ 1)β3 2β2 3 β 1 = β, β 2 = β 3 = (3.1.68) (3.1.69) 52
53 v V v // v v // = v V V 2 V, v = v v // (3.1.70) v // = v // V 1 v // V/c 2 = v // V 1 v V /c 2 v = v γ (1 v V /c 2 ) v = v // + v = γ(v // V ) + v v // γ(1 v V /c 2 ) = (γ 1)V (v V )/V 2 γv + v γ(1 v V /c 2 ) v 2 = v 2 // + v 2 = v // v // + v v = (v // V ) 2 + v 2 (1 V 2 /c 2 ) (1 v V /c 2 ) 2 = v2 2v // V + V 2 v 2 V 2 /c 2 (1 v V /c 2 ) 2 = v2 2v V + V 2 [v 2 V 2 (v V ) 2 ]/c 2 (1 v V /c 2 ) 2 = v2 + V 2 v 2 V 2 /c 2 c 2 + c 2 2v V + (v V ) 2 /c 2 (1 v V /c 2 ) 2 = (c2 v 2 )(1 V 2 /c 2 ) + c 2 (1 v V /c 2 ) 2 (1 v V /c 2 ) 2 = c 2 (c2 v 2 )(1 V 2 /c 2 ) (1 v V /c 2 ) 2 (3.1.71) (3.1.72) (3.1.73) v < c, V < c 2 v < c t( ) = (t) t t τ 1 = 1 (τ), 2 = 2 (τ), 3 = 3 (τ), 4 = 4 (τ), (3.2.1) 4 2 P,P 53
54 PP ds ds ( ) ds 0 B (ds) 2 = d µ d µ = c 2 (dt) 2 d i d i ( = c 2 (dt) ) d i d i c 2 dt dt (3.2.2) S v i = d i /dt ds = c dt 1 v2 c 2 (3.2.3) A ds P (ic(t + dt), + d) P(ict, ) 12 S v = 0 ds ds = cdt (3.2.4) dt dτ ds = cdτ dτ = ds c (3.2.5) τ (3.2.1) (3.2.5) (3.2.3) dt dτ = 1 ( v ) = γ p (3.2.6) 2 1 c τ = m t = τ 1 (v/c) 2 (3.2.7) 99.99% ,000 A B (B) (A) 2 B Earth A 0 Q B A P 12 v = c 0 54
55 B 2 B 0 A B B A B (3.2.5) B (3.2.5) v = 0 A τ A = τ B = Q Q dt dτ B = Q P P 1 1 c 2 v(t)2 < P P dt = t Q t P (3.2.8) Q dt = τ B (3.2.9) τ 1 (τ), 2 = 2 (τ), 3 = 3 (τ), 4 = 4 (τ), (2.2.15) τ 1 (τ) = γ{ 1(τ) + iβ 4 (τ)} 2 (τ) = 2(τ) 3 (τ) = 3(τ) 4 (τ) = γ{ 4(τ) iβ 1 (τ)} d u(τ) dτ µ(τ) = a µν ν (τ) (3.2.10) = a µν d ν (τ) dτ (3.2.11) dτ d ν 4 d ν (τ)/dτ 4 4 u µ u µ (τ) = d µ(τ) dτ (3.2.12) µ = 1, 2, 3 µ = 4 4 u µ (u 1, u 2, u 3, u 4 ) 3 v i (v 1, v 2, v 3 ) 13 u i (τ) = d i(τ) dτ u 4 = d 4(τ) dτ = = d i dt 1 1 (v/c) 2 = γ pv i c 1 (v/c) 2 = iγ pc (3.2.13)... u µ = (γ p v, iγ p c) = i 0, u 4 = iu 0 55
56 v c 4 v i 4 u µ u µ = u i u i u 0 u 0 = c 2 < 0 (i = 1, 2, 3) (3.2.14) 4 u 4 ( iu 0 ) u i 4 u u = u µ u µ = c (3.2.15), 4 u µ = (0, 0, 0, iγ p c) (3.2.16) u µ (3.2.10) u 1 = d 1 dτ = γ(u 1 + iβu 4 ) u 2 = d 2 dτ = d 2 dτ = u 2 u 3 = d 3 dτ = d 3 dτ = u 3 u 4 = d 4 dτ = γ(u 4 iβu 1 ) u µ = a µν u ν u 1 u 2 u 3 u 4 = u 1 = d 1 dτ = γ(u 1 iβu 4 ) u 2 = d 2 dτ = d 2 dτ = u 2 u 3 = d 3 dτ = d 3 dτ = u 3 u 4 = d 4 dτ = γ(u 4 + iβu 1 ) γ 0 0 iγβ iγβ 0 0 γ u 1 u 2 u 3 u 4 (3.2.17) (3.2.18) 4 4 (3.2.14) τ a µ = du µ dτ (3.2.19) u µ du µ dτ = 0 (3.2.20) p µ m 0 u µ (3.2.21) m 0 (3.2.13) p i = m 0 v i 1 (v/c) 2 = γ pm 0 v i p 4 = m 0 c 1 (v/c) 2 = iγ pm 0 c (3.2.22)... p µ = ( γ p m 0 v, iγ p m o c) 56
57 v c 4 p i = m 0 v i (3.2.20) m 0 ) dp µ 0 = u µ dτ = d i dp i dτ dτ d 0 dp 0 dτ dτ = γ2 p ( dp i v i dt cdp 0 dt (3.2.23) v i dp i /dt ( ) 14 E (3.2.23) t dp i v i dt = de dt de dt = cdp 0 dt (3.2.24) (3.2.25) E = cp 0 = γ p m 0 c 2 = m 0 c 2 1 v 2 /c 2,... p 0 = E c (3.2.26) E E 4 c p(p 1, p 2, p 3 ) 4 (3.2.22) p µ = (p, ie/c) (3.2.27) m(v) = m 0 1 v 2 /c 2 = γ pm 0 (3.2.28) (3.2.22),(3.2.26) p i = γ p m 0 v i = m(v)v i (3.2.29) p 0 = γ p m 0 c = m(v)c (3.2.30) E = γ p m 0 c 2 = m(v)c 2 (3.2.31) (3.2.31) m(v) 15 p µ (3.2.14) p µ p µ = m 2 0u µ u µ = p 2 p 2 0 = p 2 1 c 2 E2 = m 2 0c 2 (3.2.32) E = c p 2 + m 2 0 c2 (3.2.33) p 2 m 2 0 c2 ( ) 1/2 E = m 0 c p2 m 2 m 0 c p 2 (3.2.34) 0 c2 2m 0 2 (3.2.33) m 0 = 0 E = c p (3.2.35) ` 14 dw = mv2 d 1 15 dt dt 2 ( ) (Wiki) 57
58 (3.2.29) (3.2.31) v i = c 2 p i E (3.2.36) (3.2.33) p i (3.2.36) E p i = 4 p 1 = γ (p 1 + iβp 4 ) p 2 = p 2 p 3 = p 3 p 4 = γ (p 4 iβp 1 ) p µ = a µν p ν cp i p 2 + m 2 0 c2 = c2 p i E 2 (3.2.37) p 1 p 2 p 3 p 4 v i = E p i (3.2.38) = γ 0 0 iγβ iγβ 0 0 γ p 1 p 2 p 3 p 4 (3.2.39) (3.2.40) p F dp dt = F (3.2.41) 4 p i (3.2.41) f(f 1, f 2, f 3 ) dp i dt = f i dp i dτ = γ pf i (i = 1, 2, 3) (3.2.42) (3.2.24), (3.2.25) 4 F µ dp i v i dτ = v dp i dt i dt dτ = cdp 0 dt dt dτ = cdp 0 dτ = γ pv i f i (3.2.43) dp 0 dτ = γ p c v if i dp 4 dτ = i c γ pv i f i (3.2.44) F µ = ( γ p f, i ) c γ pv f (3.2.45) dp µ dτ = F µ (3.2.46) 58
59 4 γ 0 0 iγβ p 1 F µ = dp u dτ = d p 2 dτ p 3 iγβ 0 0 γ p 4 γ 0 0 iγβ dp 1 /dτ dp 2 /dτ = dp 3 /dτ iγβ 0 0 γ dp 4 /dτ (3.2.47)... F µ = a µν F ν (3.2.46) τ m 0 4 (3.2.50) d µ dτ = a d ν µν dτ µ = a µν ν (3.2.48) u µ = a µν u ν (3.2.49) dp µ dτ = a µνf ν (3.2.50) F µ = a µν F ν (3.2.51) dp µ dτ = F µ (3.2.52) S f β 0 f = dp dt = γ(dp βp 0 γ (dt βc ) = d dp dt β de 1 dt c 1 1 c β d dt f β = c f v 1 β v c (3.2.53) f v 4 F µ f y, f z f y f y = ( γ 1 β v ), f z = c ( γ f z 1 β v c ) (3.2.54) (3.2.46) 59
60 (3.2.55) dp i dτ = F i d (mv) = f (3.2.55) dt dp 4 dτ = F 4 c 2 dm dt = v f (... p 4 = ie/c, E = mc 2 ) (3.2.56) (3.2.56) dm dt v + mdv dt = f (3.2.57) m dv dt = f v f c 2 v (3.2.58) v c m m 0 2 v f 1. v f v f = 0 γ p m 0 dv dt = f (3.2.59) m 0 γ p m 0 2. v / f v = v(f/f) m dv dt = f vf c 2 v f f = f γp 2,... γpm 3 dv 0 dt = f (3.2.60) m 0 γ 3 pm 0 γ p m 0, γ 3 pm 0 γ p m 0 γ 3 pm α b t1 I = α ds = αc 1 v2 t1 a c 2 dt = L 0 dt (3.2.61) t 0 t 0 ds 17 L 0 L 0 = αc 1 v2 c 2 (3.2.62)
61 α (L = T = 1 2 mv2 ) v c 2 L 0 αc + 1 α 2 c v2 (3.2.63) 1 L 0 2 m 0 v 2 /2 L 0 α = m 0 c (3.2.64) L 0 = m 0 c 2 1 v2 c 2 (3.2.65) 0 I t1 t1 δi = δ L 0 dt = t 0 = L 0 δ i i t 0 t 2 t1 t t {L 0 ( i + δ i ) L 0 (δ i )}dt = i=1 d dt ( L0 i ) t1 δ i dt = t 0 d dt t2 L 0 dδ i t 1 i dt dt ( ) L0 δ i dt = 0 i (3.2.66) δ i t = t 1, t 2 0 δ i δi 0 ( ) d L0 = 0 (i = 1, 2, 3) (3.2.67) dt i d dt ( m 0 v i 1 v 2 /c 2 ) = 0 (3.2.68) p i = L 0 i p i = L 0 v i = (3.2.22) m 0 v i 1 v 2 /c 2 = γ pm 0 v i (i = 1, 2, 3) (3.2.69) H 0 = p i v i L 0 = γ p m 0 vi 2 + m 0c 2 = γ p m 0 c 2 = E γ p (3.2.70) = c p 2 + m 20 c2 H 0 p i = c 2 p i E = v i (3.2.71) (3.2.37) 61
62 f V L = L 0 V = m 0 c 2 1 v2 c 2 V ( 1, 2, 3 ) (3.2.72) ( ) d L L = 0 dt i i (i = 1, 2, 3) (3.2.73) L = 1 m 0 v 1 1 v 2 /c 2, L = V = f 1 (3.2.74) i 1 ( ) d m 0 v 1 = f 1 d dt 1 v 2 /c 2 dt ( m 0 v 1 v 2 /c 2 ) = f (3.2.75) p i = L = i m 0 v i 1 v 2 /c 2 = γ pm 0 v i = mv i (3.2.76) 62
63 4 4.1 E B S E(, t) = 1 ε 0 ρ(, t) (4.1.1) B(, t) = 0 (4.1.2) B(, t) E(, t) = t (4.1.3) B(, t) = µ 0 j + 1 E(, t) c 2 t (4.1.4) ρ j µ 0 ε 0 µ 0 = (ε 0 c 2 ) 1 S E(, t ) = 1 ε 0 ρ(, t ) (4.1.5) B(, t ) = 0 (4.1.6) E(, t ) = B(, t ) t (4.1.7) B(, t ) = µ 0 j + 1 c 2 E(, t ) t (4.1.8) t = γ (t βc ), = γ( V t), y = y, z = z t = γ (t + βc ) (4.1.9), = γ( + V t ), y = y, z = z S S ( t = t t t + t = γ t V ) (, t = γ t + V ) ( = + t t = γ β ) ( c t, = γ + β ) c t y = y, z = z (4.1.10) (4.1.11) (4.1.3) y E z z E y = t B (4.1.12) 63
64 y E z z E y = γ ( t V ) B (4.1.13) (4.1.2) B + y B y + z B z = γ... γ B = γβ c ( β c t B y B y z B z ) t B + y B y + z B z = 0 (4.1.14) (4.1.13)... y (E z + V B y ) z (E y V B z ) = 1 γ t B ( γ(ez y + V B y ) ) ( γ(ey z V B z ) ) = t B (4.1.15) (4.1.7) (4.1.15) (4.1.16) y z E E z + t B y = γ t (4.1.17) y E z z E y = t B (4.1.16) E z = γ(e z + V B y ), E y = γ(e y V B z ), B = B (4.1.17) γ (E z + V B y ) = (4.1.19) (4.1.18) ( B y + β ) c E z γ (E z + V B y ) + z E = 0 (4.1.18) z E E z = γ t E z (4.1.19) ( B y + β ) c E z (4.1.20) (4.1.7) ( ) E = E, B y = γ B y + β c E z (4.1.21) z ( ) B z = γ B z β c E y (4.1.22) E = E, E y = γ(e y V B z ), E z = γ(e z + V B y ) ( ) ( ) B = B, B y = γ B y + β c E z, B z = γ B z β c E y (4.1.23) 64
65 (4.1.5) E(, t ) = γ ( + β c t ) E + y {γ(e y V B z )} + z {γ(e z + V B y )} = γ E + γβ c t E γv ( B) = γ ρ + γβ ε 0 c t E γv ( B) (4.1.24) (4.1.4) ( B) = µ 0 J + 1 c 2 E t γβ c t E γv ( B) = γv µ 0 j (4.1.25) ( 1 ) E(, t ) γ ρ µ o V j = 1 ( γ ρ β ) ε 0 ε 0 c j = 1 ε ρ (4.1.26) cρ = γ(cρ βj ) (4.1.27) ( B ) = y B z z B y = { ( γ B z β )} y c E y { ( γ B y + β )} z c E z { = γ ( B) β ( c y E y + )} (4.1.28) z E z ( E t ) ( = γ t E + V ) E = γ t E + γcβ E (4.1.29) (4.1.27) { ( B ) = γ ( B) β ( c y E y + )} z E z = µ 0 j + γ ( c 2 t E + cβ ) E.. {. µ 0 j = γ ( B) 1 c 2 t E β } c E = µ 0 γ(j βcρ) (4.1.30) j y = j y, j z = j z (4.1.31) 1, y 2, z 3 j 4 = ij 0 = icρ (4.1.32) j 1 = γ(j i + iβj 4 ), j 2 = j 2, j 3 = j 3, j 4 = icρ = γ(j 4 iβj 1 ) (4.1.33) 65
66 j µ = a µν j ν j 1 j 2 j 3 j 4 = γ 0 0 iγβ iγβ 0 0 γ j 1 j 2 j 3 j 4 (4.1.34) j µ 4 j µ 4 j µ = (j 1, j 2, j 3, j 4 ) = (j, icρ) (4.1.35) φ A 1 E(, t) = φ(, t) A(, t) t (4.1.36) B(, t) = A(, t) (4.1.37) E(, t ) = φ(, t ) t A(, t ) (4.1.38) B(, t ) = A(, t ) (4.1.39) (4.1.36) E = E E = φ ( t A = γ β ) c t φ γ = {γ(φ cβa )} {γ(a t βc } φ = φ t A = E... φ c = γ ( φ c βa ), A = γ (A βc ) φ ( t cβ ) A (4.1.40) y E y = γ(e y V B z ) E y = y φ t A y = γ { y φ t } A y V ( A) y { = γ y φ ( t A y V A y )} y A { = γ ( y (φ V A ) t + V ) } A y (4.1.41) = γ y (φ V A ) t A y = y φ t A y.. ( ). φ φ c = γ c βa, A y = A y z A z = A z (4.1.42) 1 φ A HP 66
67 A µ = a µν A ν A 4 = ia 0 = iφ/c (4.1.43) A 1 = γ(a 1 + iβa 4 ) A 2 = A 2 A 3 = A 3 A 4 = γ(a 4 iβa 1 ) A 1 A 2 A 3 A 4 = γ 0 0 iγβ iγβ 0 0 γ A 1 A 2 A 3 A 4 (4.1.44) (4.1.45) A µ 4 A µ 4 A µ = (A 1, A 2, A 3, A 4 ) = (A 1, A 2, A 3, iφ/c) (4.1.46) µ (µ = 1, 2, 3, 4), 4 = i µ c t (4.1.47) c 2 t 2 = = µ µ (4.1.48) µ µ 4 4 F µν ρ t + j = 0 µj µ = 0 (4.1.49) A µ = (A 1, A 2, A 3, ia 0 ) (A 0 = φ/c) (4.1.50) F µν = µ A ν ν A µ (F µν = F νµ ) (4.1.51) 6 E(, t) = φ(, t) A(, t) t = ic( 1 A 4 4 A 1, 2 A 4 4 A 2, 3 A 4 4 A 3 )... ie c = ( 1A 4 4 A 1, 2 A 4 4 A 2, 3 A 4 4 A 3 ) = (F 14, F 24, F 34 ) B(, t) = A(, t) = ( 2 A 3 3 A 2, 3 A 1 1 A 3, 1 A 2 2 A 1 ) = (F 23, F 31, F 12 ) (4.1.52) 67
68 F µν E = E 1 = icf 14 B = B 1 = F 23 1 E = E y = E 2 = icf 24 E i = icf i4, B = B y = B 2 = F 31 B i = E z = E 3 = icf 2 ε ijkf jk 34 B z = B 3 = F 12 0 F 12 F 13 F 14 0 B 3 B 2 ie 1 /c F 12 0 F 23 F 24 F µν = F 13 F 23 0 F 34 = B 3 0 B 1 ie 2 /c B 2 B 1 0 ie 3 /c F 14 F 24 F 34 0 ie 1 /c ie 2 /c ie 3 /c 0 (4.1.53) (4.1.54) 2 F µν 1 i, j, k 123 F ij = ε ijk B k, ε ijk = 1 i, j, k 123 (4.1.55) 0 J = 1 2 F µνf µν = 1 2 (F 11F 11 + F 12 F 12 + F 13 F 13 + F 14 F 14 + F 21 F 21 + F 22 F 22 + F 23 F 23 + F 24 F 24 + F 31 F 31 + F 32 F 32 + F 33 F 33 + F 34 F 34 (4.1.56) J + F 41 F 41 + F 42 F 42 + F 43 F 43 + F 44 F 44 = B 2 E2 c 2 F µν = a µλ a νρ F λρ (4.1.57) F µν F 12, F 13, F 14, F 23, F 24, F 34 6 a µν γ 0 0 iγβ a µν = (4.1.58) iγβ 0 0 γ F µν (4.1.58) F 12 = a 1λ a 2ρ F λρ = (a 11 a 22 a 12 a 21 )F 12 + (a 11 a 23 a 13 a 21 )F 13 + (a 13 a 24 a 14 a 21 F 14 + (a 12 a 23 a 13 a 22 )F 23 + (a 12 a 24 a 14 a 22 )F 24 + ((a 13 a 24 a 14 a 23 )F 34 = a 11 a 22 F 12 a 14 a 22 F 24 = γ(f 12 iβf 24 ) (4.1.59) F 13 = γ(f 13 iβf 34 ), F 14 = F 14, F 23 = F 23 F 24 = γ(f 24 + iβf 12 ), F 34 = γ(f 34 + iβf 13 ) 68
69 0 γ(f 12 iβf 24 ) γ(f 13 iβf 34 ) F 14 F µν γ(f 12 iβf 24 ) 0 F 23 γ(f 24 + iβf 12 ) = γ(f 13 iβf 34 ) F 23 0 γ(f 34 + iβf 13 ) F 14 γ(f 24 + iβf 12 ) γ(f 34 + iβf 13 ) 0 (4.1.60) (4.1.54) (4.1.62) (4.1.23) B 2 = γ F µν ( B 2 + β ) ( c E 3, B 3 = γ B 3 β ) c E 2, E 2 = γ(e 2 V B 3 ), E 3 = γ(e 3 + V B 2 ) F 12 = γ(f 12 iβf 24 ) = γ{( 1 A 2 2 A 1 ) iβ( 2 A 4 4 A 2 )} ( = γ B 3 β ) c E 2 = B 3 F 13 = γ(f 13 iβf 34 ) = γ{( 1 A 3 3 A 1 ) iβ( 3 A 4 4 A 3 )} ( = γ B 2 + β ) c E 3 = B 2 (4.1.61) F 14 = 1 A 4 4 A 1 = i E 1 c, F 23 = B 1, F 24 = i E 2 c, F 34 = i E 3 c (4.1.54) (4.1.1) (4.1.4) E(, t) = 1 ρ(, t) ε 0 B(, t) = µ 0 j + 1 c 2 E(, t) t E 1 ε 0 ρ(, t) = 1 E E E 3 1 ε 0 ρ ( B) 1 µ 0 j 1 1 c 2 E 1 t µ F µν = µ 0 j ν (4.1.62) = ic( 1 F F F 34 + µ 0 j 4 ) = F F F 34 + µ 0 j 4 = 0 = 2 B 3 3 B 2 µ 0 j 1 1 c 2 E 1 t = 2 F F F 14 µ 0 j 1 = 0 (4.1.63) ( B) 2 µ 0 j 2 1 c 2 E 2 t = 3 B 1 1 B 3 µ 0 j 2 1 c 2 E 2 t = 3 F 23 1 F F 24 µ 0 j 2 = 0 (4.1.64) ( B) 3 µ 0 j 3 1 c 2 E 3 t = 1 B 2 2 B 1 µ 0 j 3 1 c 2 E 3 t = 1 F 31 2 F F 34 µ 0 j 4 = 0 69
70 B(, t) = 0 B(, t) E(, t) = t µ F νλ + ν F λµ + λ F µν = 0 (4.1.65) B = 1 B B B 3 = 1 F F F 12 = 0 (4.1.66) ( E) 1 + t B 1 = 2 E 3 3 E 2 + ic 4 B 1 = ic( 2 F F F 23 ) = 0 ( E) 2 + t B 2 = 3 E 1 1 E 3 + ic 4 B 2 = ic( 3 F 14 1 F F 31 ) = 0 ( E) 3 + t B 3 = 1 E 2 2 E 1 + ic 4 B 3 = ic( 1 F 24 2 F F 12 ) = 0 (4.1.67) 2 E(, t) = 1 ρ(, t) ε 0 B(, t) = 0 E(, t) = B(, t) t B(, t) = µ 0 j + 1 c 2 E(, t) t µ F µν = µ 0 j ν µ F νλ + ν F λµ + λ F µν = 0 (4.1.68) (4.1.51) F µν (4.1.68) 2 F µν = µ A ν ν A µ µ F νλ + ν F λµ + λ F µν = µ ( ν A λ λ A ν ) + ν ( λ A µ µ A λ ) + λ ( µ A ν ν A µ ) = 0 (4.1.69) 4 (4.1.50) F µν (4.1.51) (4.1.68) 2 2 (4.1.71) (4.1.70) µ F µν = µ 0 j ν (4.1.70) F µν = µ A ν ν A µ (4.1.71) µ F µν = µ ( µ A ν ν A µ ) = A ν ν µ A µ = µ 0 j ν (4.1.72) 4 j ν A µ (4.1.71) E B j ν 70
71 F µν F µν F µν = i 2 ε µνλρf λρ (4.1.73) ε µνλρ ε 1234 = 1 F µν 1 µ, ν, λ, ρ 1234 ε µνλρ = 1 µ, ν, λ, ρ (4.1.74) F 12 = i 2 ε 1234F 34 + i 2 ε 1243F 43 = i 2 ε 1234F 34 + i 2 ε 1234F 34 = if 34 (4.1.75) { F 12 = if 34 F 13 = if 24 F 14 = if 23 F 23 = if 14 F 24 = if 13 F 34 = if 21 (4.1.76) F i4 = i 2 ε ijkf jk ( i, j, k 1, 2, 3 ) (4.1.77) (4.1.53) 0 E 3 /c E 2 /c ib 1 E 3 /c 0 E 1 /c ib 2 F µν = E 2 /c E 1 /c 0 ib 3 ib 1 ib 2 ib 3 0 (4.1.78) (4.1.68) 2 F µν B(, t) E(, t)= t B(, t)=µ 0 j + 1 E(, t) c 2 t µf µν = 0 (4.1.79) ν = 1 1F F F F 41 = ( 2 E 3 3 E 2 ) t B 1 = 0 ν = 2 1F F F F 42 = ( 3 E 1 1 E 3 ) t B 2 = 0 ν = 3 1F F F F 43 = ( 1 E 2 2 E 1 ) t B 3 = 0 (4.1.80) ν = 4... E = B t (4.1.81) 1F F F F 44 = i( 1 B B 2 ) 3 B 3 = 0 (4.1.82)... B = 0 (4.1.83) 71
72 F µν F µν K = 1 4 F µν F µν = B E (4.1.84) K A µ Λ A µ A µ + µ Λ Ãµ (4.1.85) F µν E B Λ (4.1.85) Λ(t, ) A µ F µν µ F µν = µ ( µ A ν ν A µ ) = A ν ν µ A µ = µ 0 j ν (4.1.86) Λ() 2 A µ = µ Λ() (4.1.86) µ F µν = µ ( µ ν Λ() ν µ Λ()) = 0 (4.1.87) 0 A µ A µ + µ Λ() Λ() Λ() 2 A ν ν µ A µ = µ 0 j ν (4.1.88) 1 A old ν µ 0 µ A µ old () = s() (4.1.89) Λ() Λ() = s() (4.1.90) Λ() 1 A new µ () = A old ν + µ Λ() (4.1.91) 2 A µ Ludvig V alentin Lorenz (January 18, 1829 June9, 1891) 72
73 A new µ µ A µ new = µ A old ν + Λ() = s() s() = 0 (4.1.92) 3 (4.1.88) A new ν µ A µ new = 0 (4.1.93) A new ν = µ 0 j ν (4.1.94) { A ν = µ 0 j ν (4.1.95) µ A µ = 0 A ν (4.1.71) F µν (4.1.93) Λ() (4.1.95) q v f = q(e + v B) (4.2.1) (3.2.13) (3.2.45) ( u µ = γ p (v, ic), F µ = γ p f, i ) c v f (4.2.2) 4 F 1 = γ p f 1 = γ p q{e 1 + (v B) 1 } = q(f 12 u 2 + F 13 u 3 + F 14 u 4 ) F 2 = γ p f 2 = γ p q{e 2 + (v B) 2 } = q(f 21 u 1 + F 23 u 3 + F 24 u 4 ) F 3 = γ p f 3 = γ p q{e 3 + (v B) 3 } = q(f 31 u 1 + F 32 u 2 + F 34 u 4 ) (4.2.3)...F i = qf ij u j i F 4 = γ p c v f = γ i p c qe v = q(f 41u 1 + F 42 u 2 + F 43 u 3 ).. (4.2.4).F 4 = qf 4j u j 4 F µ = qf µν u ν (4.2.5) (3.2.46) dp µ dτ = qf µνu ν 3 Lorenz gauge A + 1 c A 0 = 0 or dp i dt = qf dτ µνu ν dt = qf µνv i (4.2.6) 73
74 E = (E, 0, 0) B = 0 t = 0 (4.2.6) d dt (mv) = m d 0 dt (γ pv) = 1 qf 1ν u ν = 1 q( ie 1 /c)(iγ p c) = qe γ p γ p ( ).. d. dt (γ pv) = d v = q v E dt 1 v 2 /c 2 m 0 1 v 2 /c = 2 v q m 0 Et (4.2.7) v = d dt = c qet/m 0 c 1 + (qet/m0 c),... = m 0c 2 2 qe ( 1 + (qet/m 0 c) 2 + 1) (4.2.8) qet/m 0 c 2 1 E = 0, B = (0, 0, B) f = q(v B) 4 v z = 0 F µ = γ p ( q(v B), 0 ) (4.2.9) d dt (mv) = m d 0 dt (γ pv) = q(v B) (4.2.10) dv dt = qb v y = ωv y, m 0 γ p dv y dt = qb v = ωv m 0 γ p dv z dt = 0 ω = 1 γ p qb m 0 (4.2.11) v = A cos ωt + B sin ωt, v y = B cos ωt A sin ωt.. (4.2.12). = (A/ω) sin ωt (B/ω) cos ωt + 0, y = (B/ω) sin ωt + (A/ω) cos ωt + y 0 ( 0 ) 2 + (y y 0 ) 2 = A2 + B 2 ω 2 = v2 ω 2 (4.2.13) 0, y 0 r = v/ω = γ p (m 0 v/qb) ω c = qb m 0 (4.2.14) ω = qb m 0 1 (v/c) 2 (4.2.15) 74
75 4.2.2 (3.2.61) I 0 = m 0 c ds = m 0 c 2 dτ (4.2.16) I 0 e A µ u µ dτ I = m 0 c 2 dτ e A µ u µ dτ (4.2.17) e A µ u µ dτ = e γ p (v A φ)dτ = (ev A eφ)dt (4.2.18) I = ( m 0c 2 γ p ) eφ + ev A dt (4.2.19) L = m 0c 2 γ p eφ + ev A (4.2.20) d dt ( L v i L = m 0 γ p v i + ea i = mv i + ea i v i L = e φ + e (v j A j ) = e φ A j + ev j i i i i d L = m dv i dt v i dt + eda i dt.. ( d L. dt v i ) L = m dv i i dt + e φ e A i i t = m dv i dt ee i ev j F ij ) L i = 0 (4.2.21) i ( ) = m dv i dt + e A i d j j dt + A i t ( A i + ev j A j j i = m dv i dt ee i eε ijk v j B k = 0 (... F ij = ε ijk B k ) m dv i dt = ee i + eε ijk v i B k m dv dt ) (4.2.22) = e(e + v B) (4.2.23) p i = L v i = m 0 γ p v i + ea i = mv i + ea i p = mv + ea (4.2.24) v = 1 m 0 γ p (p ea) = 1 m 0 (p ea) 1 (v/c) 2 (4.2.25) 75
76 2 v 2 v 2 = c 2 (p ea) 2 (m 0 c) 2 + (p ea) 2,... 1 = 1 v2 γ p c 2 = m 0 c (m0 c) 2 + (p ea) 2 (4.2.26) (4.2.25) v i = c(p i ea i ) (m0 c) 2 + (p ea) 2 (4.2.27) H = p i v i L = m 0 γ p v γ p m 0 c 2 + eφ = c (mc) 2 + (p ea) 2 + eφ (4.2.28) (4.2.23) e m 1 E (pe) = 1 2 m ξ 2 δ( ξ) j (pe) i = 1 2 mξ2 ξi δ( ξ) (4.3.1) E + j (pe) = j E (4.3.2) E = 1 2 ε 0E B 2 (4.3.3) 2 µ 0 E + 1 µ 0 (E B) = j E (4.3.4) 4 S = 1 µ 0 E B (4.3.5) (4.3.6) E + S = j E (4.3.6) (4.3.2) (4.3.6) ( E + E (pe)) { } 1 + (E B) + j (pe) = 0 (4.3.7) t µ 0 4 John Henry P oynting
77 4.3.2 (4.3.7) T µν = 1 (F µλ F λν + 1 µ 0 4 δ µνf λσ F λσ ) (4.3.8) (T µν = T νµ ) T µµ = 0 (4.3.9) 0 (4.1.56) F λσ F λσ = 2(B 2 E 2 /c 2 ) (4.3.10) T 44 = 1 µ 0 ( F 4λ F λ F λσf λσ ) = 1 2 ε 0E B 2 (4.3.11) 2 µ 0 E = T 44 = 1 2 ε 0E B 2 (4.3.12) 2 µ 0 T i4 T 14 = 1 F 1λ F λ4 = i µ 0 c (E 2B 3 E 3 B 2 ) = i c (E B) 1 T 24 = 1 F 2λ F λ4 = i µ 0 c (E 3B 1 E 1 B 3 ) = i c (E B) 2 (4.3.13) T 34 = 1 µ 0 F 3λ F λ4 = i c (E 1B 2 E 2 B 1 ) = i c (E B) 3 T i4 i i T i4 = 1 F i4 F 4j = i 1 (E B) i = i µ 0 c µ 0 c S i (4.3.14) T ij T 11 = 1 F 1λ F λ1 + 1 µ 0 4 F λσf λσ = ε 0 E 1 E ( 1 B 1 B 1 µ 0 2 ε 0E ) 1 B 2 2 µ 0 T 22 = 1 F 2λ F λ2 + 1 µ 0 4 F λσf λσ = ε 0 E 2 E ( 1 B 2 B 2 µ 0 2 ε 0E ) 1 B 2 2 µ 0 T 33 = 1 F 3λ F λ3 + 1 µ 0 4 F λσf λσ = ε 0 E 3 E ( 1 B 3 B 3 µ 0 2 ε 0E ) 1 B 2 2 µ 0 T 12 = 1 µ 0 F 1λ F λ2 = ε 0 E 1 E µ 0 B 1 B 2 (4.3.15) T 13 = 1 µ 0 F 1λ F λ3 = ε 0 E 1 E µ 0 B 1 B 3 T ij = ε 0 E i E j + 1 ( 1 B i B i δ ij µ 0 2 ε 0E ) 1 B 2 2 µ 0 (4.3.16) 77
78 j i T µν ( ) ( ) T ij T i4 T ij i T µν = = c S i T 4j T 44 i c S i E (4.3.8) ν 5 (4.1.86) ν F λν = µ 0 j λ ν T µν = 1 µ 0 ( ν F µλ F λν + F µλ ν F λν δ µν ν F λσ F λσ ) (4.3.17) = 1 µ 0 ( ν F µλ F λν + µ 0 F µλ j λ µf λσ F λσ ) (4.3.18) 1 3 ν F µλ F λν µf λσ F λσ = 1 2 νf µλ F λν νf µλ F λν µf λσ F λσ = 1 2 νf µλ F λν νf µλ F λν µf λσ F λσ (4.3.19) ν F µλ F λν = λ F µν F νλ = λ F νµ F λν (ν λ, λ ν) (4.3.20) (4.3.18) νf µλ F λν = 1 2 λf µν F νλ 1 2 µf λσ F λσ = 1 2 µf λν F λν (σ ν ) (4.3.21) (4.3.19) (4.1.65) 1 2 ( νf µλ + λ F µν + µ F λν )F λν = 0 (4.3.22) ν T µν = F µλ j λ (4.3.23) µ = 4 µ = 1, 2, 3 //
K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
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......................
( )
7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................
) 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
: 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(
TOP URL 1
TOP URL http://amonphys.web.fc2.com/ 1 30 3 30.1.............. 3 30.2........................... 4 30.3...................... 5 30.4........................ 6 30.5.................................. 8 30.6...............................
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.............................
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
016 9 3 6 0 016 1 0 1 10 1 1 17 1..,,..,. 1 1 c = h = G = ε 0 = 1. 1.1 L L T V 1.1. T, V. d dt L q i L q i = 0 1.. q i t L q i, q i, t L ϕ, ϕ, x µ x µ 1.3. ϕ x µ, L. S, L, L S = Ld 4 x 1.4 = Ld 3 xdt 1.5
(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x
![(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x](/thumbs/104/162595168.jpg "(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x")
Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k
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 )
[ ] 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 =
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..................................
( ) ( )
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))
Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x
7 7.1 7.1.1 Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x 3 )=(x 0, x )=(ct, x ) (7.3) E/c ct K = E mc 2 (7.4)
(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
2009 2 26 1 3 1.1.................................................. 3 1.2..................................................... 3 1.3...................................................... 3 1.4.....................................................
D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j
![D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j](/thumbs/103/158001152.jpg "D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j")
6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..
( ; ) 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 =
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(
() 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 ()
量子力学Ａ
c 1 1 1.1....................................... 1 1............................................ 4 1.3.............................. 6 10.1.................................. 10......................................
(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)
untitled
1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β
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 (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
08-Note2-web
r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)
( ) (ver )
ver.3.1 11 9 1 1. p1, 1.1 ψx, t,, E, p. = E, p ψx, t,. p, 1.8 p4, 1. t = t ρx, t = m [ψ ψ ψ ψ] ρx, t = mi [ψ ψ ψ ψ] p4, 1.1 = p6, 1.38 p6, 1.4 = fxδ ϵ x = fxδϵx = 1 π fxδ ϵ x dx = fxδ ϵ x dx = [ 1 fϵ π
meiji_resume_1.PDF
β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E
I 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
K E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
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
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..........
9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P
9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
) ] [ 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
() (, 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 =
chap10.dvi
. q {y j } I( ( L y j =Δy j = u j = C l ε j l = C(L ε j, {ε j } i.i.d.(,i q ( l= y O p ( {u j } q {C l } A l C l
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
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,
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
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
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..............
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
#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =
#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1.
1.1 1. 1.3.1..3.4 3.1 3. 3.3 4.1 4. 4.3 5.1 5. 5.3 6.1 6. 6.3 7.1 7. 7.3 1 1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N
( ) ) ) ) 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)
e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,
01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,
70 : 20 : A B (20 ) (30 ) 50 1
70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................
/Volumes/NO NAME/gakujututosho/chap1.tex i
2012 4 10 /Volumes/NO NAME/gakujututosho/chap1.tex i iii 1 7 1.1............................... 7 2 11 2.1........................................... 11 2.2................................... 18 2.3...................................
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,
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
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
IA
IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................
II 1 II 2012 II Gauss-Bonnet II
II 1 II 212 II Gauss-Bonnet II 1 1 1.1......................................... 1 1.2............................................ 2 1.3.................................. 3 1.4.............................................
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
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.........................
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....
( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................
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......................
4 5.............................................. 5............................................ 6.............................................. 7......................................... 8.3.................................................4.........................................4..............................................4................................................4.3...............................................
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
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
1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
![1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1](/thumbs/94/121802164.jpg "1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1")
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................
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
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 +
. 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.........
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),
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
.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
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x
. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +
4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
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
July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i
July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac
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)
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.............................
( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
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 =
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)
(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
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
2 1 x 2 x 2 = RT 3πηaN A t (1.2) R/N A N A N A = N A m n(z) = n exp ( ) m gz k B T (1.3) z n z = m = m ρgv k B = erg K 1 R =
1 1 1.1 1827 *1 195 *2 x 2 t x 2 = 2Dt D RT D = RT N A 1 6πaη (1.1) D N A a η 198 *3 ( a =.212µ) *1 Robert Brown (1773-1858. *2 Albert Einstein (1879-1955 *3 Jean Baptiste Perrin (187-1942 2 1 x 2 x 2
(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
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
/Volumes/NO NAME/gakujututosho/chap1.tex i
2010 4 8 /Volumes/NO NAME/gakujututosho/chap1.tex i iii 1 5 1.1............................... 5 2 9 2.1........................................... 9 2.2................................... 16 2.3...................................
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 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
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....................
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 ; >
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
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
( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e
![( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e](/thumbs/98/136160482.jpg "( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e")
( ) Note 3 19 12 13 8 8.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R, µ R, τ R (1a) L ( ) ) * 3) W Z 1/2 ( - )
1
GL (a) (b) Ph l P N P h l l Ph Ph Ph Ph l l l l P Ph l P N h l P l .9 αl B βlt D E. 5.5 L r..8 e g s e,e l l W l s l g W W s g l l W W e s g e s g r e l ( s ) l ( l s ) r e l ( s ) l ( l s ) e R e r
, 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,.,
液晶の物理１：連続体理論（弾性，粘性）
The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers
A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
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
7 π L int = gψ(x)ψ(x)φ(x) + (7.4) [ ] p ψ N = n (7.5) π (π +,π 0,π ) ψ (σ, σ, σ )ψ ( A) σ τ ( L int = gψψφ g N τ ) N π * ) (7.6) π π = (π, π, π ) π ±
![7 π L int = gψ(x)ψ(x)φ(x) + (7.4) [ ] p ψ N = n (7.5) π (π +,π 0,π ) ψ (σ, σ, σ )ψ ( A) σ τ ( L int = gψψφ g N τ ) N π * ) (7.6) π π = (π, π, π ) π ±](/thumbs/91/106462075.jpg "7 π L int = gψ(x)ψ(x)φ(x) + (7.4) [ ] p ψ N = n (7.5) π (π +,π 0,π ) ψ (σ, σ, σ )ψ ( A) σ τ ( L int = gψψφ g N τ ) N π * ) (7.6) π π = (π, π, π ) π ±")
7 7. ( ) SU() SU() 9 ( MeV) p 98.8 π + π 0 n 99.57 9.57 97.4 497.70 δm m 0.4%.% 0.% 0.8% π 9.57 4.96 Σ + Σ 0 Σ 89.6 9.46 K + K 0 49.67 (7.) p p = αp + βn, n n = γp + δn (7.a) [ ] p ψ ψ = Uψ, U = n [ α
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7