Answer to Question #97635 in Mechanics | Relativity for Bob

Question

Consider the system described by the Hamiltonian:

L(q,q') = 1/2*sum(q'i*q'j*Aij) + sum(bi(q)*q'i)-V(q)

(I've used ' to describe an overdot/derivative respect to time, and i and j are the indices being summed over. We are not given anything about b, other than it is summed over i, and is a function of q):

where the matrix A = Aij is symmetric and positive definite, with unique inverse C.
Find the corresponding Hamiltonian.

Answer 1

$L(q,\dot{q})=\frac{1}{2}\sum_{i,j}\dot{q}_i\dot{q}_jA_{ij}+\sum_ib_i(q)\dot{q}_i-V(q)\\ p_i=\frac{\partial L}{\partial\dot{q}_i} =\sum_j\dot{q}_jA_{ij}+b_i(q)\\ \vec{p}-\vec{b} = \vec{\dot{q}}\cdot A\\ \dot{q}_i = \sum_j (p_j-b_j)C_{ij}\\ H=\sum_i p_i\dot{q}_i-L = \sum_ip_i\left(\sum_j (p_j-b_j)C_{ij}\right) - \left(\frac{1}{2}\sum_{i,j}\dot{q}_i\dot{q}_jA_{ij}+\sum_ib_i(q)\dot{q}_i-V(q)\right) = \\ =\frac{1}{2}\sum_{i,j}\dot{q}_i\dot{q}_jA_{ij}+V(q) = \\ =\frac{1}{2}\sum_{i,j}\left(\sum_m (p_m-b_m)C_{im}\right)\left(\sum_k (p_k-b_k)C_{jk}\right)A_{ij}+V(q)$