Яковенко Г.Н. Краткий курс теоретической механики
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ωωω
ωω
3
= ˙ϕe
3
ωωω
ωω
=
˙
ψi
3
+
˙
θn + ˙ϕe
3
. (26.6)
p q r
ωωω
ωω
= pe
1
+ qe
2
+ re
3
. (26.7)
p
e
1
p = (ωωω
ωω
, e
1
) =
˙
ψ(i
3
, e
1
) +
˙
θ(n, e
1
) + ˙ϕ (e
3
, e
1
).
p
q r
p =
˙
ψ sin θ sin ϕ +
˙
θ cos ϕ,
q =
˙
ψ sin θ cos ϕ −
˙
θ sin ϕ,
r =
˙
ψ cos θ + ˙ϕ.
(26.8)
˙
ψ = (p sin ϕ + q cos ϕ)
1
sin
θ
,
˙
θ = p cos ϕ − q sin ϕ,
˙ϕ = (p sin ϕ + q cos ϕ)ctgθ.
(26.9)
sin θ
i
1
, . . . , i
n
Λ =
n
P
k=1
λ
k
i
k
M =
n
P
l=1
µ
l
i
l
λ
k
µ
l
Λ ◦ M
Λ ◦ (M + N) = Λ ◦ M + Λ ◦ N
Λ ◦ M =
n
P
k, l=1
λ
k
µ
l
i
k
◦ i
l
i
k
◦ i
l
Λ = λ
0
i
0
+ λ
1
i
1
i
0
◦i
0
= i
0
i
0
◦i
1
= i
1
i
1
◦i
0
= i
1
i
1
◦ i
1
= −i
0
i
0
Λ◦i
0
= i
0
◦Λ = Λ
Λ = λ
0
+ λ
1
i
1
i
1
= i
i ◦ i = −1
i
0
i
1
i
2
i
3
i
0
Λ ◦ i
0
=
i
0
◦ Λ = Λ
Λ = λ
0
+
3
P
k=1
λ
k
i
k
k, l = 1, 3
i
k
◦ i
l
=
(
−1, k = l
[i
k
, i
l
], k 6= l
(27.1)
Λ = λ
0
+ λ
1
i
1
+ λ
2
i
2
+ λ
3
i
3
= λ
0
+ λλλ
λλ
(27.2)
λ
0
λλλ
λλ
= λ
1
i
1
+λ
2
i
2
+λ
3
i
3
Λ = λ
0
+ λ
1
i
1
+ λ
2
i
2
+ λ
3
i
3
= λ
0
+ λλλ
λλ
, M = µ
0
+ µ
1
i
1
+ µ
2
i
2
+ µ
3
i
3
= µ
0
+ µµµ
µµ
Λ ◦ M = λ
0
µ
0
− (λλλ
λλ
, µµµ
µµ
)
| {z }
+ λ
0
µµµ
µµ
+ µ
0
λλλ
λλ
+ [λλλ
λλ
, µµµ
µµ
]
| {z }
. (27.3)
Λ ◦ M − M ◦ Λ = 2[λλλ
λλ
, µµµ
µµ
]. (27.4)
Λ
2
= Λ ◦ Λ = λ
2
0
−
3
X
k=1
λ
2
k
+ 2λ
0
λλλ
λλ
. (27.5)
λ
0
= 0 µ
0
= 0
λλλ
λλ
◦µµµ
µµ
= −(λλλ
λλ
, µµµ
µµ
) + [λλλ
λλ
, µµµ
µµ
]. (27.6)
λλλ
λλ
◦λλλ
λλ
= −
3
X
k=1
λ
2
k
= −|λλλ
λλ
|
2
. (27.7)
|e| = 1
e ◦ e = −|e|
2
= −1. (27.8)
e
(Λ ◦ M) ◦ N = Λ ◦ (M ◦ N). (27.9)
Λ
2
+ 2bΛ + c = 0,
b c Λ
2
Λ
2λλλ
λλ
(λ
0
+ b) = 0, λ
2
0
− |λλλ
λλ
|
2
+ 2bλ
0
+ c = 0,
c = b
2
Λ = λ
0
= −b
c < b
2
Λ = λ
0
= −b ±
√
b
2
− c
c > b
2
λ
0
= −b
λλλ
λλ
|λλλ
λλ
|
2
= c − b
2
Λ = λ
0
+ λλλ
λλ
e
Λ =
λ
0
−λλλ
λλ
n
Λ = −
e
Λ
o
⇔
©
Λ = λλλ
λλ
(λ
0
= 0)
ª
. (27.10)
^
Λ
1
◦ ··· ◦ Λ
n
=
e
Λ
n
◦ ··· ◦
e
Λ
1
. (27.11)
kΛk = Λ ◦
e
Λ =
e
Λ ◦ Λ = λ
2
0
+ λ
2
1
+ λ
2
2
+ λ
2
3
, (27.12)
Λ 6= 0 Λ
−1
◦ Λ =
Λ ◦ Λ
−1
= 1
Λ
−1
=
1
kΛk
e
Λ. (27.13)
kΛ ◦ Mk = kΛkkMk. (27.14)
kΛk = Λ ◦
e
Λ =
e
Λ ◦ Λ = λ
2
0
+ λ
2
1
+ λ
2
2
+ λ
2
3
= 1. (27.15)
Λ
−1
=
e
Λ. (27.16)
|λ
0
| 6 1 λ
0
ϕ
λ
0
= cos
ϕ
2
,
λ
2
1
+ λ
2
2
+ λ
2
3
= 1 − cos
2
ϕ
2
= sin
2
ϕ
2
.
Λ = cos
ϕ
2
+ e sin
ϕ
2
, e . (27.17)
Λ ◦ r ◦
e
Λ = r
∗
(28.1)
r
e
Λ
r
∗
r e
ϕ
¤ rke
Λ ◦ r ◦
e
Λ = r ◦ Λ ◦
e
Λ = r [e, r] = b 6= 0 b = r sin γ
g
j
r
e
e
r
r*
a
b
c
a = [b, e] a = b = r sin γ a b e
r
r = a + r
e
r
∗
= r
e
+ c = r
e
+ a cos ϕ + b sin ϕ.
r
∗
= Λ ◦ r ◦
e
Λ
(27.4)
=
³
r ◦ Λ + 2[e sin
ϕ
2
, r]
´
◦
e
Λ = r ◦
1
z}|{
Λ ◦
e
Λ +2b ◦
e
Λ sin
ϕ
2
=
= r + 2b ◦
³
cos
ϕ
2
− e sin
ϕ
2
´
sin
ϕ
2
= r + b sin ϕ − a(1 − cos ϕ) =
= r − a
|{z}
r
e
+ a cos ϕ) + b sin ϕ
| {z }
c
.
b ◦ e = −(b, e) + [b, e] = −([e, r], e) + [b, e] = [b, e] = a.
¥
r
Λ
−→ r
∗
M
−→ r
∗∗
,
r
N=?
−−→ r
∗∗
.
Λ r r
∗
M r
∗
r
∗∗
N
r r
∗∗
r
∗
= Λ ◦r ◦
e
Λ
r
∗∗
= M ◦ r
∗
◦
f
M = M ◦ Λ ◦ r ◦
e
Λ ◦
f
M
(27.11)
= M ◦ Λ ◦ r ◦
^
M ◦ Λ = N ◦ r ◦
e
N,
N = M ◦ Λ
Λ M
N = M ◦ Λ
n
r
Λ
1
−→ ···
Λ
n
−→ r
∗
r
N
−→ r
∗
,
N = Λ
n
◦ ··· ◦ Λ
1
. (28.2)
Λ
1
, . . . , Λ
n
k β
¤ i
1
i
2
i
3
e
1
e
2
e
3
Λ
1
= cos
ψ + ϕ
2
+ i
3
sin
ψ + ϕ
2
, Λ
2
= cos
θ
2
+ n sin
θ
2
.
N
N = Λ
2
◦ Λ
1
= cos
θ
2
cos
ψ + ϕ
2
+
+i
3
cos
θ
2
sin
ψ + ϕ
2
+ n sin
θ
2
cos
ψ + ϕ
2
+ [n, i
3
] sin
θ
2
sin
ψ + ϕ
2
.
(28.3)
β
cos
β
2
= cos
θ
2
cos
ψ + ϕ
2
, (28.4)
k = i
3
cos
θ
2
sin
ψ + ϕ
2
+ n sin
θ
2
cos
ψ + ϕ
2
+ [n, i
3
] sin
θ
2
sin
ψ + ϕ
2
.
n i
1
i
2
i
3
n = i
1
cos ψ + i
2
sin ψ
N = Λ
2
◦ Λ
1
= cos
θ
2
cos
ψ + ϕ
2
+
+i
1
sin
θ
2
cos
ψ − ϕ
2
+ i
2
sin
θ
2
sin
ψ − ϕ
2
+ i
3
cos
θ
2
sin
ψ + ϕ
2
.
β
k = i
1
sin
θ
2
cos
ψ − ϕ
2
+ i
2
sin
θ
2
sin
ψ − ϕ
2
+ i
3
cos
θ
2
sin
ψ + ϕ
2
. (28.5)
¥
θ ω = ˙ϕ
ω =
˙
ψ
β
k
ϕ = ω t
ψ = ω t
θ = α + π/2 ω = Ω ω = Ω/ sin α
α = π/4
e
k
= Λ ◦ i
k
◦
e
Λ. (28.6)
e
Λ Λ
i
k
=
e
Λ ◦ e
k
◦ Λ. (28.7)
Λ = λ
0
+
3
X
k=1
λ
k
i
k
(28.8)
Λ
e
Λ
Λ ◦ Λ ◦
e
Λ
|{z}
1
= λ
0
Λ ◦
e
Λ
|{z}
1
+
3
X
k=1
λ
k
Λ ◦ i
k
◦
e
Λ
| {z }
e
k
Λ = λ
0
+
3
P
k=1
λ
k
e
k
Λ
t e
1
e
2
e
3
λ
0
λ
1
λ
2
λ
3
e
1
e
2
e
3
λ
0
(t)
λ
1
(t) λ
2
(t) λ
3
(t)
i
k
Λ=λ
0
+
3
P
k=1
λ
k
i
k
−−−−−−−−−→ e
k
M=µ
0
+
3
P
k=1
µ
k
e
k
−−−−−−−−−→ n
k
.
Λ i
k
M e
k
n
k
i
k
n
k
N = M ◦ Λ
Λ M
M i
k
N = M ◦ Λ =
Ã
µ
0
+
3
X
k=1
µ
k
e
k
!
◦ Λ =
Ã
µ
0
+
3
X
k=1
µ
k
Λ ◦ i
k
◦
e
Λ
!
◦ Λ =
= Λ ◦
M
∗
z }| {
Ã
µ
0
+
3
X
k=1
µ
k
i
k
!
◦
1
z}|{
e
Λ ◦ Λ = Λ ◦ M
∗
.
M
∗
M
M = µ
0
+
3
X
k=1
µ
k
e
k
, M
∗
= µ
0
+
3
X
k=1
µ
k
i
k
(28.9)
e
k
i
k
n
i
k
Λ
1
−→ ···
Λ
n
−→ n
k
i
k
N
−→ n
k
,
N = Λ
∗
1
◦ ··· ◦ Λ
∗
n
. (28.10)
Λ
1
, . . . , Λ
n
Λ
∗
j
= λ
0j
+
3
P
k=1
λ
kj
i
k
i
1
i
2
i
3
λ
0
(t) λ
1
(t) λ
2
(t) λ
3
(t) Λ(t)
t t Λ ◦
e
Λ = 1
˙
Λ ◦
e
Λ + Λ ◦
˙
e
Λ = 0
˙
Λ ◦
e
Λ = −Λ ◦
˙
e
Λ = −
]
˙
Λ ◦
e
Λ
˙
Λ ◦
e
Λ
b =
˙
Λ ◦
e
Λ = −Λ ◦
˙
e
Λ. (29.1)
t
˙
e
k
=
˙
Λ ◦ i
k
◦
e
Λ + Λ ◦ i
k
◦
˙
e
Λ
(28.7)
=
˙
Λ ◦
e
Λ
|{z}
b
◦e
k
◦ Λ ◦
e
Λ
|{z}
1
+ Λ ◦
e
Λ
|{z}
1
◦e
k
◦ Λ ◦
˙
e
Λ
|{z}
−b
=
= b ◦ e
k
− e
k
◦ b
(27.4)
= 2[b, e
k
] = [2b, e
k
].
(29.2)
ωωω
ωω
˙
e
k
= [ωωω
ωω
, e
k
]
Λ(t)
ωωω
ωω
= 2b = 2
˙
Λ ◦
e
Λ = −2Λ ◦
˙
e
Λ. (29.3)
Λ(t)
˙
Λ =
1
2
˙ϕ
³
−sin
ϕ
2
+ e cos
ϕ
2
´
+
˙
e sin
ϕ
2
,
e
Λ = cos
ϕ
2
− e sin
ϕ
2
,
ωωω
ωω
= e ˙ϕ +
˙
e sin ϕ + [e,
˙
e](1 − cos ϕ),
ϕ = 0
ωωω
ωω
˙ϕe
Λ(t)
Λ(t)
˙
Λ =
1
2
ωωω
ωω
◦ Λ. (29.3)
ωωω
ωω
ωωω
ωω
= pe
1
+ qe
2
+ re
3
.