References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercise 2.7.8 (1-3).
Note that the new coordinates depend only on the old coordinates and not on the velocities . We also assume that the transformation is invertible, so it’s possible to find the as functions of the .
First, we need to show that the Euler-Lagrange equations are invariant under such a transformation. Starting with the inverse equations
(we’re using unsubscripted variables to refer to the entire set, so that ), we have
Since the velocities are independent variables, this implies that, if we hold the coordinates constant,
since the derivative just picks out the one term containing in the sum 3. Now consider the Euler-Lagrange equations in the new coordinates. To do this, we write the Lagrangian in terms of the new coordinates and velocities, so that
Taking derivatives, we have
The second term on the RHS is zero since the velocities don’t depend on the coordinates (and vice versa), so we’re left with
Now for the other derivative
The first term on the RHS is zero (same reason as in the previous equation), and we can apply 4 to the second term to get
We can now take the derivative with respect to time and apply the Euler-Lagrange equation (which we know to be valid for the coordinates). We’re also assuming that the coordinates have no explicit time dependence. Thus
Comparing this with 7 we see that
That is, the Euler-Lagrange equations are valid for the coordinates as well.
We can use the Lagrangian to see how the momenta transform under the coordinate change. The definition of the canonical momentum is
If we write the Lagrangian in terms of the coordinates and velocities as in 5, then the momenta in the new coordinate system are
At this point, it’s worth noting that although and are different functions, they have the same value at each point in the configuration space. That is, if we choose some point that has the coordinates in the system and coordinates in the system, then, numerically at that one point, we must have . Because of this, we can write
That is, if we’re keeping constant, the derivative of with respect to must be the same (numerically) no matter what coordinates we’re using to write . Therefore, we can use the latter form and then use the chain rule to write out the derivative:
Because the coordinates don’t depend on the velocities , the first term on the RHS is zero. We can use 4 in the second term, and we have
where we used the definition of in the last line.
If we review the derivation of Hamilton’s equations, we see that nowhere did we make any assumptions about the particular coordinate system that was being used in the Lagrangian. All that is required for Hamilton’s equations to be valid is that the momenta are defined as in 14, and that the Euler-Lagrange equations are satisfied. Therefore, in any such system, Hamilton’s equations are valid:
is called a point transformation.
In the -dimensional phase space of the Hamiltonian formalism, where and are the variables rather than the and used in the Lagrangian, we can envision a more general transformation in which
In such a general transformation, there’s no guarantee that 18 is satisfied, so such transformations need not be point transformations (though they could be). There’s also no guarantee that the momenta are related to the Lagrangian by 14, and thus Hamilton’s equations may not be satisfied.