Page 203 - Intro to Tensor Calculus

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is satisfied. Integrating the second term of this integral by parts we find

Z t 1 ∂L ∂L t 1 Z t 1 d ∂L
i
i
i
δI = η dt + η (t) − η (t) dt =0. (2.2.32)
∂x i ∂ ˙x i dt ∂ ˙x i
t 0 t 0 t 0
i
The end condition on η (t) makes the middle term in equation (2.2.32) vanish and we are left with the
integral
Z
t 1 ∂L d ∂L
i
δI = η (t) − dt =0, (2.2.33)
∂x i dt ∂ ˙x i
t 0
i
i
which must equal zero for all η (t). Since η (t) is arbitrary, the only way the integral in equation (2.2.33) can
i
be zero for all η (t) is for the term inside the brackets to vanish. This produces the result that the integral
of the Lagrangian is an extremum when the Euler-Lagrange equations

d ∂L ∂L
− =0, i =1,... ,N (2.2.34)
dt ∂ ˙x i ∂x i
are satisfied. This is a necessary condition for the integral I( ) to have a minimum value.
In general, any line integral of the form
Z t 1
i
i
I = φ(t, x , ˙x ) dt (2.2.35)
t 0
i
i
has an extremum value if the curve c defined by x = x (t),i =1,... ,N satisfies the Euler-Lagrange
equations

d ∂φ ∂φ
− =0, i =1,... ,N. (2.2.36)
dt ∂ ˙x i ∂x i
The above derivation is a special case of (2.2.36) when φ = L. Note that the equations of motion equations
(2.2.34) are just another form of the equations (2.2.24). Note also that

δT δ 1 i j i j i i
= mg ij v v = mg ij v f = mf i v = mf i ˙x
δt δt 2
∂V
and if we assume that the force Q i is derivable from a potential function V ,then mf i = Q i = − ,so
∂x i
δT i i ∂V i δV δ
that = mf i ˙x = Q i ˙x = − ˙ x = − or (T + V )= 0 or T + V = h = constant called the energy
δt ∂x i δt δt
constant of the system.
Action Integral
The equations of motion (2.2.34) or (2.2.24) are interpreted as describing geodesics in a space whose
line-element is
2
j
ds =2m(h − V )g jk dx dx k
where V is the potential function for the force system and T + V = h is the energy constant of the motion.
The integral of ds along a curve C between two points P 1 and P 2 is called an action integral and is

Z j k 1/2
√ P 2 dx dx
A = 2m (h − V )g jk dτ
dτ dτ
P 1