Goldstein 2-16 (3rd ed. 2.17) The kinetic energy T and potential

Goldstein 2-16 (3rd ed. 2.17) The kinetic energy T and potential energy V for a system of n degrees of freedom are expressed as

T =

X

·2

fi (qi )qi , V =

i

X

Vi (qi ) ,

(1)

i

·

where qi and qi are the ith generalized coordinate and velocity, respectively, and the functions fi (qi ) and Vi (qi ) each depend on only a single generalized coordinate. The Lagrangian for this system is L=T −V =

X ·2 [fi (qi )qi − Vi (qi )] .

(2)

i

The Lagrange EOM are defined as ∂L d ∂L − =0 . dt ∂ q·j ∂qj

(3)

Note that it is highly advisable to use a dummy index (j) that differs from the one (i) used in defining L. To get a more explicit form for the EOM we need the partial derivatives ∂L · ∂ qj

·

= 2fj (qj )q j ,

(4)

and ∂Vj (qj ) ∂L · 2 ∂fj (qj ) = qj − . ∂qj ∂qj ∂qj

(5)

Please note that in Eqs.(4) and (5) there is only one term from Eq.(2) that survives the differentiation: the one for which the index i equals the index j. Next take the time derivative of Eq.(4) to obtain d ∂L ·· · 2 ∂fj (qj ) = 2fj (qj )q j + 2q j . dt ∂ q·j ∂qj

(6)

Now combine Eqs.(5) and (6) to obtain the explicit form of the EOM ··

· 2 ∂fj (qj )

2fj (qj )q j + q j

∂qj

1

+

∂Vj (qj ) =0 . ∂qj

(7)

It is obvious that the EOM are fully decoupled because each one depends only on the single generalized coordinate and generalized velocity specified by the index j. To find the solution of this EOM as a quadrature, we multiply Eq.(7) · by q j dt, and make use of the following properties ·· · q j q j dt

1 · 2 ∂fj (qj ) · ∂fj (qj ) = dq j , q j dt = dqj = d fj (qj ) , 2 ∂qj ∂qj

(8)

(Vj satisfies a similar relation as fj ) to obtain ·2

·2

·2

fj (qj )dq j + q j d fj (qj ) + dVj (qj ) = d[fj (qj )q j + Vj (qj )] = 0 .

(9)

·2

From this result, we see that the first integral, fj (qj )q j + Vj (qj ), is a constant of the motion, which we will set equal to the constant Ej , the energy of the jth degree of freedom, ·2

fj (qj )q j + Vj (qj ) = Ej .

(10)

·

Next, we solve Eq.(10) for q j to obtain s

·

qj = ±

Ej − Vj (qj ) . fj (qj )

(11)

Since the right hand side of Eq.(11) is only a function of qj , we can find a formally integrable expression by rearranging Eq.(11) as s fj (qj ) dqj . (12) dt = ± Ej − Vj (qj ) The quadrature solution is t − t0 = ±

Z

qj qj0

s

fj (q) dq , Ej − Vj (q)

where t0 and qj0 are initial values. You can choose the + or − branch of the ·0

solution based on the sign of the initial value q j using Eq.(11).

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