Are there physical systems for which Hamiltonian is not defined?

Question

Consider the following Lagrangian : $$\mathcal{L}=\frac{1}{2}\dot{q}\sin^2q$$ It's easy to see for such a lagrangian we can't find the Hamiltonian since $$p=\partial_{\dot{q}}\mathcal{L}=\frac{1}{2}\sin^2q$$ We can't write $\dot{q}$ as function of $p$. Technically saying, Legendre transform doesn't exist. I'm asking if there can be systems which describe by such Lagrangians.

Answer 1

You can in fact calculate the Hamiltonian to your Lagrangian! But first let us take a look at the Lagrangian proposed by OP. What are the dynamics?

The EoM obtained by the EL-equations are $$ \frac{d}{dt} \frac 1 2 \sin^2 q - \dot q \sin q \cos q = 0\\ \sin q \cos q \dot q - \dot q \sin q \cos q = 0\\ 0 = 0 $$ Hence the Lagrangian seemingly describes no dynamics. Lets go back to your original question and check for the Hamiltonian. The Hamiltonian is given by $$ H = p \dot q - L = \frac 1 2 \sin^2 q \dot q - L = 0 $$ Again no dynamics. So one has to be sure that you have a propagating degree of freedom, hence you should be sure about the kinetic term in your theory. So here you dont have a kinetic term.

Answer 2

As a complement to @AlmostClueless's answer:

The reason your Lagrangean does not describe any dynamics is that it is a total time derivative, $$ \mathcal{L}=\frac{1}{2}\dot{q}\sin^2q=\frac14 \frac{\text{d}}{\text{d}t}\left(q-\sin q \cos q\right)\,. $$