I'm considering on cotangent bundle of a two-dimensional variety where in canonical coordinates $(x^i,y_i)$ is defined in the 1-form $$\theta=y_{1}dx^1+y_2dx^2$$ let new coordinate: $$x^1=q^1\cosh(q^2)$$ $$x^2=q^1\sinh(q^2)$$ I have to determinate:
1) the conjugate momenta $(p_{\alpha})$ concerning to $(q^{\alpha})$
2) the components of $\theta(q^{\alpha},p_{\alpha})$.
I think I can replace:
$$dx^1=\cosh(q^2)dq^1+q^1\sinh(q^2)dq^2$$ $$dx^2=\sinh(q^2)dq^1+q^1\cosh(q^2)dq^2$$
Unfortunately I know the definition of conjugate momenta only in Hamiltonian mechanics $p_{\mu}=\frac{\partial L}{\partial u^{\mu}}$ and I do not know how to go on here. I looked for a definition but I did not find it.
In Hamiltonian mechanics, the conjugate momenta are the cotangent vectors. When the tautological one-form $\theta$ is written as $$\theta=\sum_i y_idx_i$$ in canonical coordinates $(x^i,y_i)$, the canonical momenta are precisely $y_i$.
You may refer to equation $(4)$ of the following answer on a physics stackexchange post to see how conjugate momenta transforms between coordinates: https://physics.stackexchange.com/questions/176555/momentum-is-a-cotangent-vector/176578#176578 . Remember that there is a hidden summation over all $i$ on the right-hand-side of that equation.
The $q^i$ in that answer corresponds to $x^i$ in your question; the $p_i$ in that answer corresponds to $y_i$ in your question; the $q'^j$ in that answer corresponds to $q^j$ in your question; the $p_j'$ in that question corresponds to $p_j$ in your question.