Ratbay Myrzakulov
Ratbay Myrzakulov
Ratbay Myrzakulov
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As is well known, for our dynamical system, the Euler-Lagrange equations read as<br />
( )<br />
d ∂L37<br />
− ∂L 37<br />
= 0, (4.9)<br />
dt ∂Ṙ ∂R<br />
( )<br />
d ∂L37<br />
dt ∂T<br />
˙ − ∂L 37<br />
= 0, (4.10)<br />
∂T<br />
( )<br />
d ∂L37<br />
− ∂L 37<br />
= 0. (4.11)<br />
dt ∂ȧ ∂a<br />
Hence, using the expressions<br />
we get<br />
∂L 37<br />
∂Ṙ = −6ɛ 1F RR a 2 ȧ, (4.12)<br />
∂L 37<br />
∂T<br />
˙<br />
∂L 37<br />
∂ȧ<br />
respectively. Here<br />
= −6ɛ 1 F RT a 2 ȧ, (4.13)<br />
= −12(ɛ 1 F R − ɛ 2 F T )aȧ − 6ɛ 1 (F RR Ṙ + F RT ˙ T + F Rψ ˙ψ)a 2 + a 3 F T vȧ + a 3 F R uȧ, (4.14)<br />
a 3 F TT<br />
(<br />
T − v − 6ɛ 2<br />
ȧ 2<br />
a 2 )<br />
(<br />
)<br />
a 3 F RR R − u − 6ɛ 1 (ä<br />
a + ȧ2<br />
a 2 )<br />
= 0, (4.15)<br />
= 0, (4.16)<br />
U + B 2 F TT + B 1 F T + C 2 F RRT + C 1 F RTT + C 0 F RT + MF + 6ɛ 2 a 2 p = 0, (4.17)<br />
U = A 3 F RRR + A 2 F RR + A 1 F R , (4.18)<br />
A 3 = −6ɛ 1 Ṙ 2 a 2 , (4.19)<br />
A 2 = −12ɛ 1 Ṙaȧ − 6ɛ 1 ¨Ra 2 + a 3 Ṙuȧ, (4.20)<br />
A 1 = 12ɛ 1 ȧ 2 + 6ɛ 1 aä + 3a 2 ȧuȧ + a 3 ˙uȧ − a 3 u a , (4.21)<br />
B 2 = 12ɛ 2 Taȧ ˙ + a 3 Tvȧ, ˙<br />
(4.22)<br />
B 1 = 24ɛ 2 ȧ 2 + 12ɛ 2 aä + 3a 2 ȧvȧ + a 3 ˙vȧ − a 3 v a , (4.23)<br />
C 2 = −12ɛ 1 a 2 ṘT, ˙<br />
(4.24)<br />
C 1 = −6ɛ 1 a 2 T˙<br />
2 , (4.25)<br />
C 0 = −12ɛ 1 Taȧ ˙ + 12ɛ 2 Ṙaȧ − 6ɛ 1 a 2 ¨T + a 3 Ṙvȧ + a 3 Tuȧ, ˙<br />
(4.26)<br />
M = −3a 2 . (4.27)<br />
If F RR ≠ 0,<br />
F TT ≠ 0, from Eqs. (4.17) and (4.17), it is easy to find that<br />
R = u + 6ɛ 1 (Ḣ + 2H2 ), T = v + 6ɛ 2 H 2 , (4.28)<br />
so that the relations (4.1) are recovered. Generally, these equations are the Euler constraints of<br />
the dynamics. Here a,R,T are the generalized coordinates of the configuration space. On the<br />
other hand, it is also well known that the total energy (Hamiltonian) corresponding to Lagrangian<br />
L 37 is given by<br />
H 37 = ∂L 37<br />
∂ȧ ȧ + ∂L 37<br />
Ṙ + ∂L 37<br />
∂Ṙ ∂T<br />
˙ T˙<br />
− L 37 . (4.29)<br />
Hence using (4.12)-(4.14) we obtain<br />
H 37 = [−12(ɛ 1 F R − ɛ 2 F T )aȧ − 6ɛ 1 (F RR Ṙ + F RT ˙ T + F Rψ ˙ψ)a 2 + a 3 F T vȧ + a 3 F R uȧ]ȧ<br />
−6ɛ 1 F RR a 2 ȧṘ − 6ɛ 1F RT a 2 ȧ ˙ T − [a 3 (F − TF T − RF R + vF T + uF R + L m )−<br />
6(ɛ 1 F R − ɛ 2 F T )aȧ 2 − 6ɛ 1 (F RR Ṙ + F RT ˙ T + F Rψ ˙ψ)a 2 ȧ]. (4.30)<br />
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