# In canonical transformation, how the new coordinate is the old momenta?

I am studying Canonical transformation using Goldstein (3ed), Ch.9.

I do understand everything he does in the first section and why we do need a generating function, $$F$$. The problem I am facing is that in some generating function, like $$F=F_1(q,Q,t).$$ In a trivial case we have $$Q_i=p_i,\quad P_i=-q_i$$ which basically means that the new coordinates $$\textbf{Q}$$ is the old momenta $$\textbf{p}$$, and vice versa.

I do not understand that, how a coordinate can be the same as a momenta? Each defines something different.

• You are setting the functions equal, you are not setting the physical interpretation of the functions equal. Remember that $p_i$ and $q_i$ are just manifold coordinates. – Charlie Mar 5 at 18:43

When you build a Lagrangian, you choose the generalized coordinates $$q^i$$ to describe the physical system, and those coordinates may have certain physical meaning, they can be distances, angles... Mathematically, what your are doing is selecting coordinates of the configuration space $$\mathbb{Q}$$, which has the mathematical structure of a differentiable manifold. In a Hamiltonian system, we are working on the phase space, which is the cotangent bundle $$\mathbb{T^*Q}$$ of the configuration space $$\mathbb{Q}$$. Any point in $$\mathbb{T^*Q}$$ can be labeled with $$2n$$ numbers $$(q^1,q^2,...,q^n,p_1,p_2,...,p_n)$$, where $$q^i$$ are generalized coordinates and $$p_i$$ are conjugate momenta. They may have an obvious physical meaning, for example when cartesian coordinates are chosen at the beginning, but it doesn't have to be that way.
When you perform a canonical transformation$$(q^1,...,q^n,p_1,...,p^n)\to(Q^1,...,Q^n,P_1,...,P_n),$$ you are just changing coordinates on that mathematical structure, to make equations easier or whatever. It is true that your new coordinates may lost the "intuitive" meaning, but you can always go back to your first coordinates once you have solved the problem.