Hamiltonian Mechanics - Deriving Hamilton's Equations

Deriving Hamilton's Equations

Hamilton's equations can be derived by looking at how the total differential of the Lagrangian depends on time, generalized positions and generalized velocities

$\mathrm{d} \mathcal{L} = \sum_i \left ( \frac{\partial \mathcal{L}}{\partial q_i} \mathrm{d} q_i + \frac{\partial \mathcal{L}}{\partial {\dot q_i}} \mathrm{d} {\dot q_i} \right ) + \frac{\partial \mathcal{L}}{\partial t} \mathrm{d}t \,.$

Now the generalized momenta were defined as and Lagrange's equations tell us that

$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathcal{L}}{\partial {\dot q_i}} - \frac{\partial \mathcal{L}}{\partial q_i} =0. \,$

We can rearrange this to get

$\frac{\partial \mathcal{L}}{\partial q_i} = {\dot p}_i \,$

and substitute the result into the total differential of the Lagrangian

$\mathrm{d} \mathcal{L} = \sum_i \left + \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t \,.$

We can rewrite this as

$\mathrm{d} \mathcal{L} = \sum_i \left + \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t \,$

and rearrange again to get

$\mathrm{d} \left ( \sum_i p_i {\dot q_i} - \mathcal{L} \right ) = \sum_i \left - \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t \,.$

The term on the left-hand side is just the Hamiltonian that we have defined before, so we find that

$\mathrm{d} \mathcal{H} = \sum_i \left - \frac{\partial \mathcal{L}}{\partial t}\mathrm{d}t = \sum_i \left [ \frac{\partial \mathcal{H}}{\partial q_i} \mathrm{d} q_i + \frac{\partial \mathcal{H}}{\partial p_i} \mathrm{d} p_i \right ] + \frac{\partial \mathcal{H}}{\partial t}\mathrm{d}t \,$

where the second equality holds because of the definition of the total differential of in terms of its partial derivatives. Associating terms from both sides of the equation above yields Hamilton's equations

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