ABSTRACT
We then have ∂L/∂t = 0. We can now replace ∂L/∂xk from the Euler equations to give
dL___ dt
∂L___∂x˙k dx˙k___ dt
+ d__ dt
(∂L___∂x˙k ) x˙k = d__ dt
∂L___∂x˙k
x˙k) (15.5) or
d__ dt
∂L___∂x˙k x˙k − L ) = 0 (15.6)
pk = ∂L___∂x˙k
(15.7)
we have
d__ dt
pk x˙k − L ) = 0 (15.8) The Hamiltonian H is defi ned by
H = ∑ k pk x˙k − L (15.9)
We can then conclude that the Hamiltonian does not depend on time and therefore is constant. Equation 15.8 expresses the law of conservation of energy, since the Hamiltonian corresponds to the energy E of the system.3,4 We then have
pkx˙k − L = E (15.10)
where E is a constant. In this case, it is possible to reduce the number of dimensions from four to three by eliminating time.1 Consider x1 and x2 as functions of x3, that is, x1 = x1(x3) and x2 = x2(x3), so that
x˙1 = dx1___ dx3
x˙3 x˙2 = dx2___ dx3
x˙3 (15.11)
and thus
L(x1, x2, x3, x˙1, x˙2, x˙3) = L (x1, x2, x3, dx1___dx3, dx2___ dx3
, x˙3) (15.12) Accordingly,
∂L___∂x˙k = fk (x1, x2, x3, dx1___dx3,
dx2___ dx3
, x˙3) (15.13) and considering Equation 15.7, Equation 15.10 can be written as
fk(x1, x2, x3, dx1___dx3, dx2___ dx3
, x˙3) x˙k − L (x1, x2, x3, dx1___dx3, dx2___ dx3
, x˙3) = E (15.14) Equation 15.14 can now be solved for x˙3 to give
x˙ 3 = Φ (x1, x2, x3, dx1___dx3, dx2___ dx3
,E ) (15.15)
x˙1 = dx1____ dx3
Φ(x1, x2, x3, dx1___dx3,
dx2___ dx3
,E ) x˙2 = dx2____dx3Φ (x1, x2, x3, dx1___ dx3
, dx2___ dx3
,E ) (15.16) Consider all the paths for which the system has some given constant energy E. We then compare only varied paths of the same energy as the real path. From Equation 15.10, we can write Equation 15.1 as1
pk x˙k − E ) dt = ∫ t1t2 ∑ k=1
pk x˙k dt − ∫ t1 t2
E dt = 0 (15.17)
or
pk x˙k dt = 0 (15.18)
since ∫E dt = 0 because E is a constant. Considering Equation 15.7, the integral in Equation 15.18 can be rewritten as
∂L___∂x˙k
(∂L___∂x˙1 dx1___ dx3
+ ∂L___∂x˙2
dx2___ dx3
+∂L___∂x˙3 ) dx3 (15.19)
since we are now considering xk = xk(x3), and therefore
x˙k dt = dxk___ dt
dt = dxk___ dx3
dx3___ dt
dt = dxk___ dx3
dx3 (15.20)
Now making
F(x1, x2, x3, x˙1, x˙2, x˙ 3) = ∂L___∂x˙1
dx1___ dx3
+ ∂L___∂x˙2
dx2___ dx3
+∂L___∂x˙3 (15.21)
and replacing for x˙1, x˙2, and x˙3 from Equations 15.15 and 15.16, we have6
F(x1, x2, x3, dx1___dx3, dx2___ dx3
,E ) dx3 = 0 (15.22) or since the energy E is given to be constant
F(x1, x2, x3, x′ 1, x′ 2) dx3 = 0 (15.23)
This is the Maupertuis’ principle of least action. Equation 15.10 enables the elimination of time derivatives x˙k =dxk/dt, which can now be expressed as geometrical derivatives, given by x′k = dxk/dx3 (k = 1, 2), where x3 is now independent variable. Equation 15.23 is purely geometrical and describes the
the canonical equations,1 which are obtained directly from Equations 15.1, 15.7, and 15.9 as
dxk___ dt
=∂H___∂pk
dpk___ dt
= −∂H___∂xk k = 1, 2, 3 (15.24)
The differential equations derived from Equation 15.23 have the form of the Euler equations
d___ dx3
(∂F____∂ x′ 1)= ∂F___∂x1
d___ dx3
(∂F____∂ x′ 2)= ∂F___∂x2
(15.25)
The integral in Equation 15.18 can also be written by using Equation 15.10 as
∑ k pk x˙k dt = ∫ P1
If T is the kinetic energy of the system and V the potential energy, then the total energy is E = T + V. Also, L = T − V.3,4 Replacing this in Equation 15.10, we have
pk x˙k = 2T (15.27)
Replacing this in expression 15.26, we have7,8
T dt = 0 (15.28)
Maupertuis’ principle of least action is also presented in classical mechanics textbooks as3,6
∆ ∫ ∑ k pk x˙k dt = 0 (15.29)
The variation appearing in expression 15.29 for the principle of least action is the ∆ variation. The δ variation corresponds to displacements in which the time is held fi xed and the coordinates of the system are varied subject to the constraints imposed on the system. In contrast, the ∆ variation deals with displacements in which, not only the coordinates of the system are varied, but it also involves a change in time. We are only considering cases, however, in which the energy is constant. In this case, we have seen that the dependence on time can be eliminated from the integral of expression 15.29. In this case, the ∆ and δ variations can be made identical and therefore expression 15.29 can be written as expression 15.26.6
n(x1, x2, x3 ) √ __________
2 dx3 (15.30)
L(x1, x2, x3, x′ 1, x′ 2 )dx3 = 0
Euler equations (Equation 15.25) are also the same as those found in optics. Maupertuis’ principle is, therefore, the equivalent in mechanics of Fermat’s principle of optics.
