ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 1

Solution for assignment 5.

1. The Lagrangian is

L = \frac{1}{2} ẋ^{2} − \frac{1}{2} x^{2} .

Thus, the Euler-Lagrange equation becomes

0 = \frac{∂L}{∂x} − \frac{d}{dt} \frac{∂L}{∂ẋ} = −x − ẍ

and the general solution is given by x(t) = A sin(t) + B cos(t). The integration constants are determined by the initial conditions

x(0) = B = 0 and x(π/2) = A = 1

so that the exact path is

x(t) = sin(t) .

For this path the action is given by

S [x(t)] = \frac{1}{2} ∫_{0}^{π/2} dt [cos^{2}(t) − sin^{2}(t)] = \frac{1}{2} ∫_{0}^{π/2} dt cos(2t) = 0 .

2. For a linear path x(t) = a + b t subject to the boundary conditions:

x(0) = a = 0 and x(π/2) = b \frac{π}{2} = 1 ⇒ b = \frac{2}{π}

so that this path is

x(t) = \frac{2}{π} t .

For this path the action is given by

S [x(t)] = \frac{1}{2} ∫_{0}^{π/2} dt [(\frac{2}{π} )^{2} − (\frac{2}{π} )^{2} t^{2}] = \frac{2}{π^{2}} ∫_{0}^{π/2} dt (1 − t^{2})
= \frac{2}{π^{2}} [\frac{π}{2} − \frac{1}{3} (\frac{2}{π} )^{3}] = \frac{1}{π} (1 − \frac{π^{2}}{12}) = \frac{1}{π} − \frac{π}{12} = 0.0565 ….

3. Assuming that the previous result is in J · s, we find for the large number

S/h̄ ≈ 5.37 × 10^{32} .