8B (Classical Dynamics)

Use the first-order action

S [q, p] = \int_{t_{1}}^{t_{2}} (p \overset{q}{˙} - H (q, p, t)) d t .

Varying

q

with fixed endpoints and varying

p

freely gives

δ S = \int_{t_{1}}^{t_{2}} [(\overset{q}{˙} - H_{p}) δ p + (- \overset{p}{˙} - H_{q}) δ q] d t .

\overset{q}{˙} = H_{p}, \overset{p}{˙} = - H_{q} .

For

H = p^{2} + q^{- 2}

these become

\overset{q}{˙} = 2 p

and

\overset{p}{˙} = 2 q^{- 3}

H = E

is conserved and

\frac{d}{d t} (pq) = \overset{p}{˙} q + p \overset{q}{˙} = 2 q^{- 2} + 2 p^{2} = 2 H .

Thus

F = pq - c t H

is an integral of motion exactly when $c = 2$ .

Write its constant value as

F_{0}

pq - 2 tE = F_{0}

in extended

(q, p, t)

-space give

pq = F_{0} + 2 Et .

Using

E q^{2} = p^{2} q^{2} + 1

q (t) = \pm \frac{( F _{0} + 2 Et ) ^{2} + 1}{E}, p (t) = \frac{F _{0} + 2 Et}{q ( t )} .

The sign of

q

is fixed. In phase space, each energy orbit is one of the two branches

p^{2} + q^{- 2} = E, q = \pm (E - p^{2})^{- 1/2}, ∣ p ∣ < E .

Each branch approaches

p = - E

and

p = + E

∣ q ∣ \to \infty

and turns at

(p, q) = (0, \pm E^{- 1/2})

 Ancestors (9)

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