ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 3

Solution for assignment 12:

With

H = (∑_{j} q̇_{j} \frac{∂L}{∂q̇_{j}} − L)

we find

dH = ∑_{j} [d (\frac{∂L}{∂q̇_{j}} ) q̇_{j} + \frac{∂L}{∂q̇_{j}} dq̇_{j} − \frac{∂L}{∂q̇_{j}} dq̇_{j} − \frac{∂L}{∂q_{j}} dq_{j}] .

The two central terms cancel out. Using Euler-Lagrange and the definition of the generalized momentum, we have

\frac{∂L}{∂q_{j}} = \frac{d}{dt} \frac{∂L}{∂q̇_{j}} = ṗ_{j}

and, therefore,

dH = ∑_{j} (q̇_{j} dp_{j} − ṗ_{j} dq_{j}) .

From this we read off Hamilton’s equations:

\frac{∂H}{∂p_{j}} = q̇_{j} and \frac{∂H}{∂q_{j}} = −ṗ_{j} .