Коткин Г.Л., Сербо В.Г., Черных А.И. Лекции по аналитической механике
Подождите немного. Документ загружается.
V(t) =
˙
R(t)
K
′
K
′
L
′
(r
′
,
˙
r
′
, t) =
m
2
(
˙
r
′
+ V(t))
2
− U(R(t) + r
′
) , (17.1)
K
′
m¨r
′
= −
∂U
∂r
′
− mW , (17.2)
W =
˙
V(t) K
′
K
′
(−mW)
“
“
K
′
(x
′
, y
′
, z
′
)
Ω = Ω(t)
K(x, y, z) x
′
= y
′
= z
′
= 0
x = y = z = 0 r
K
r
′
K
′
x, y, z x
′
, y
′
, z
′
K
′
K
v = [Ω, r] = [Ω, r
′
]
K
′
v
′
[Ω, r
′
]
r = r
′
, v = v
′
+ [Ω, r
′
] . (17.3)
L
′
(r
′
, v
′
, t)
L
′
(r
′
, v
′
, t) =
1
2
m (v
′
)
2
+mv
′
[Ω, r
′
]+
1
2
m [Ω, r
′
]
2
−U(r
′
) . (17.4)
p
′
=
∂L
′
∂v
′
= m (v
′
+ [Ω, r
′
]) , M
′
= r
′
× p
′
. (17.5)
p
′
M
′
K
p
′
= p , M
′
= M . (17.6)
E
′
K
′
E
′
= p
′
v
′
− L
′
=
1
2
m (v
′
)
2
+ U(r
′
) −
1
2
m [Ω, r
′
]
2
. (17.7)
E
′
−
1
2
m [Ω, r
′
]
2
,
E
′
E
K
E = pv − L = p
′
(v
′
+ [Ω, r
′
]) − L
′
= E
′
+ ΩM
′
. (17.8)
d
dt
∂L
′
∂v
′
=
∂L
′
∂r
′
.
m
dv
′
dt
= −
∂U
∂r
′
+ 2m[v
′
, Ω] + m[Ω, [r
′
, Ω]] + m[r
′
,
˙
Ω] . (17.9)
−∂U/∂r
′
2m[v
′
, Ω] ,
m[Ω, [r
′
, Ω]]
m [r
′
,
˙
Ω] ,
U(r)
B
L
K
L =
1
2
mv
2
− U(r) +
e
c
vA(r), (17.10)
e A(r)
A(r) =
1
2
[B, r]. (17.11)
K
′
Ω
K
r r
′
v v
′
K K
′
L
′
K
′
L
′
=
1
2
m (v
′
+ [Ω, r
′
])
2
− U(r
′
) +
e
2c
(v
′
+ [Ω, r
′
]) · [B, r
′
] .
L
′
= L
′
0
+ δL
′
, L
′
0
=
1
2
m(v
′
)
2
− U(r
′
) , (17.12)
δL
′
= mv
′
[Ω, r
′
] +
e
2c
v
′
[B, r
′
] +
1
2
m[Ω, r
′
]
2
+
e
2c
[Ω, r
′
] · [B, r
′
].
Ω
Ω
L
= −
eB
2mc
, (17.13)
B
δL
′
δL
′
= −
e
2
8mc
2
[B, r
′
]
2
. (17.14)
δL
′
K
′
U(r
′
)
U(r) = −α/r
B E < 0
K
′
K
B
x
U
L =
1
2
m ˙x
2
−
mgx
2
l
, (18.1)
m l
x
M k m
ω
0
=
p
k/M
m ≪ M
x
“
L
0
=
1
2
M ˙x
2
− kx
2
T =
ρ
2
l
Z
0
v
2
(ξ)dξ .
ρ = m/l l(t)
v( ξ) = ˙xξ/l
ξ
T = m ˙x
2
/6
ω =
s
k
M + (m/3)
≈
r
k
M
1 −
m
6M
. (18.2)
“
q
˙q
C L
C L
q(t)
˙q
L¨q +
q
C
= 0 . (18.3)
L(q, ˙q) =
1
2
L˙q
2
−
q
2
2C
(18.4)
q
E =
1
2
L˙q
2
+
q
2
2C
. (18.5)
q
˙q
“
“
xk
C L(x)
LC
L(x, q, ˙x, ˙q) =
m ˙x
2
2
+
L(x) ˙q
2
2
−
Cq
2
2
−
kx
2
2
(18.6)
q x
L(q, ˙q) = T (q, ˙q) − U(q) , T (q, ˙q) =
1
2
a(q) ˙q
2
≥ 0 . (19 .1 )
q
0
U(q) x = q − q
0
U(q) = U(q
0
) +
1
2
kx
2
,
dU
dq
|
q
0
= 0 ,
d
2
U
dq
2
|
q
0
= k > 0 (19.2)
k = 0
a(q) q
0
a(q) = m + O(x) , m = a(q
0
) . (19.3)