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Let's say we have $(q, \dot{q})$ as the generalised coordinate and generalised velocity. If we have a Lagrangian given by $$L=Aq\dot{q}+Bq$$ where $A$ and $B$ are constants that give the right units to the Lagrangian, then the canonical momentum $p$ we use to perform the Legendre transformation to get the Hamiltonian is

$$p=\frac{\partial L}{\partial \dot{q}}=Aq.$$ But then $p$ and $q$ are not independent variables as $$\frac{\partial p}{\partial q}=\frac{\partial}{\partial q} Aq=A\neq0.$$

What am I doing wrong here?

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  1. OP's Lagrangian $$L(q,v,t)~=~Aqv+Bq\tag{1}$$ has EL equation $$B~\approx~ 0,\tag{2}$$ so the theory only has classical solutions if $B=0$. In the latter case, the Lagrangian is a total derivative, so the Euler-Lagrange (EL) equation is trivially satisfied. Either way, OP's Lagrangian leads to a rather singular theory.

  2. Concerning the corresponding Hamiltonian formulation, OP's Lagrangian is an example where the Legendre transformation is singular, i.e. we cannot invert the relation $$p~=~\frac{\partial L}{\partial v}~=~Aq\tag{3}$$ to find $v$ as a function of $p$. Instead eq. (3) is a primary constraint, i.e. $q$ and $p$ are indeed not independent variables, and one would have to perform a Dirac-Bergmann analysis. A secondary constraint is eq. (2). The Hamiltonian becomes $$H~=~\lambda(p-Aq),\tag{4}$$ where $\lambda$ is a Lagrange multiplier.

  3. For regular Lagrangians, see e.g. this related Phys.SE post.

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