8.1 Types of transformations

Point transformations

The simplest type of transformation, “point transformations” simply change the \(q_i\) to a new set of coordinates, \(Q_i\). We know from our liberty in choosing a set of generalized coordinates that the Lagrangian is invariant under point transformation. Naturally, Hamilton’s equations are also preserved, and point transformations of the type:

\begin{align} q_i=q_i(Q_i,t) \tag{8.2}\end{align}

are canonical.

General transformations

Consider the general invertible canonical transformation of the form:

\begin{align} q_i&=q_i(Q_i,P_i,t)\\ p_i&=p_i(Q_i,P_i,t) \tag{8.3}\end{align}

The equations of motion will be preserved if the integrand of the action (i.e. the Lagrangian) is modified at most by a total time differential of some function, \(F\):

\begin{align} \sum_i\dot q_ip_i-H=\sum_i \dot Q_iP_i-K+\frac{d}{dt}F(q_i,Q_i,t) \tag{8.4}\end{align}

where we have, arbitrarily, chosen that \(F\) depends on the pair of variables \(q_i, Q_i\). We could have just as well chosen any pair of variables that mixes the new and old coordinates, \(F(p_i, P_i)\), \(F(p_i, Q_i)\), or \(F(q_i, P_i)\). We assume (impose) that the transformation is canonical, so that:

\begin{align} \dot Q_i&=\die{K}{P_i}\\ \dot P_i&=-\die{K}{Q_i} \tag{8.5}\end{align}

We will require that Equation 8.4 is true so that we can make the transformation canonical. Consider the time derivative of \(F\):

\begin{align} \frac{dF}{dt} =\sum_i\left(\die{F}{q_i}\dot q_i+\die{F}{Q_i}\dot Q_i\right)+\die{F}{t} \tag{8.6}\end{align}

which we can substitute back into 8.4:

\begin{align} \sum_i\dot q_ip_i-H&=\sum_i \dot Q_iP_i-K+\sum_i\left(\die{F}{q_i}\dot q_i+\die{F}{Q_i}\dot Q_i\right)+\die{F}{t}\\ \sum_i\left(p_i-\die{F}{q_i}\right)\dot q_i-H&=\sum_i \left(P_i +\die{F}{Q_i}\right)\dot Q_i-K+\die{F}{t} \tag{8.7}\end{align}

We can guarantee the validity of this equation by setting the coefficients of \(\dot q_i\) and \(\dot Q_i\) to zero:

\begin{align} p_i&=\die{}{q_i}F(q_i,Q_i,t)\\ P_i&=-\die{}{Q_i}F(q_i,Q_i,t)\end{align}

and by requiring that the new Hamiltonian is then given by:

\begin{align} K(Q_i,P_i,t)=H(p_i,q_i,t)+\die{F}{t} \tag{8.9}\end{align}

We call \(F\) the “generator” of the canonical transformation, since it tells how to define the new Hamiltonian and the new coordinates. Given \(F\) and the above equations, one can always invert the transformation equations to get:

\begin{align} q_i&=q_i(Q_i,P_i,t)\\ p_i&=p_i(Q_i,P_i,t)\\ Q_i&=Q_i(q_i,p_i,t)\\ P_i&=P_i(q_i,p_i,t) \tag{8.10}\end{align}

and Hamilton’s canonical equations for \(K(Q_i,P_i,t)\).

We have four possible “types” of canonical transformations depending on the variables that \(F\) depends on:

\(F=F_1(q_i,Q_i,t)\) are transformations of the first type
\(F=F_2(q_i,P_i,t)-\sum_iQ_iP_i\) are transformations of the second type
\(F=F_3(p_i,Q_i,t)+\sum_ip_iq_i\) are transformations of the third type
\(F=F_4(p_i,P_i,t)+\sum_ip_iq_i-\sum_iQ_iP_i\) are transformations of the fourth type

These relations between the generating functions are similar to Legendre transformations, although it is not always true that, for a given situation, one can use any of the coordinate transformations (this may lead to expressions that are singular, ill-defined, etc.). It should also be noted that one can use generating functions that mix the above possibilities for different indices. For example, a valid generating function could be of one type for \(i=1\) and of another for \(i=2\).

One can use the same formalism as above to determine how the various generating functions give different coordinate transformations. These are summarized in table 8.1.

Type	Transformation equations
\(F=F_1(q_i,Q_i,t)\)	\(p_i=\die{F_1}{q_i}\) \(P_i=-\die{F_1}{Q_i}\)
\(F=F_2(q_i,P_i,t)-\sum_iQ_iP_i\)	\(p_i=\die{F_2}{q_i}\) \(Q_i=\die{F_2}{P_i}\)
\(F=F_3(p_i,Q_i,t)+\sum_ip_iq_i\)	\(q_i=-\die{F_3}{p_i}\) \(P_i=-\die{F_3}{Q_i}\)
\(F=F_4(p_i,P_i,t)+\sum_ip_iq_i-\sum_iQ_iP_i\)	\(q_i=-\die{F_4}{p_i}\) \(Q_i=\die{F_4}{P_i}\)

Table 8.1Summary of the four types of canonical transformations

It is interesting to note that one can choose an arbitrary function \(F(q_i, Q_i,t)\) and generate a canonical transformation, giving us a new set of arbitrary coordinates and Hamiltonian that will automatically satisfy the equations of motion. Being able to re-write the equations of motion in an arbitrary coordinate system is a powerful tool.

Example 8-1

Find the transformation equations \(Q_i(q_i,p_i,t)\) and \(P_i(q_i,p_i,t)\) for the generating function given by \(F=\sum_iQ_iq_i\)

For a canonical transformation of the first type, we have

\begin{align*} p_i&=\die{F}{q_i}=Q_i\nonumber\\ P_i&=-\die{F}{Q_i}=-q_i\nonumber\\ K&=H \end{align*}

These can be trivially inverted to obtain the new coordinates in terms of the old ones:

\begin{align*} Q_i&=p_i\nonumber\\ P_i&=-q_i \end{align*}

which had the net effect of reversing the coordinates with their conjugate momenta, highlighting how these should really be treated as independent in Hamiltonian mechanics.

Example 8-2

Find the transformation equations \(Q_i(q_i,p_i,t)\) and \(P_i(q_i,p_i,t)\) for the generating function given by \(F=\sum_iq_iP_i\)

For a canonical transformation of the second type, we have

\begin{align*} p_i&=\die{F}{q_i}=P_i\nonumber\\ Q_i&=\die{F}{P_i}=q_i \end{align*}

Again, these are trivially inverted

\begin{align*} Q_i&=q_i\nonumber\\ P_i&=p_i \end{align*}

which is the identity transformation.

Example 8-3

Show that the harmonic oscillator problem can be solved easily using transformations from the generating function \(F=\frac{m\omega}{2}q^2\cot Q\), where \(\omega=\sqrt{\frac{k}{m}}\) is the angular frequency.

The Hamiltonian for the simple harmonic oscillator is:

\begin{align*} H&=\frac{p^2}{2m}+\frac{k}{2}q^2\\ &=\frac{1}{2m}(p^2+m^2\omega^2q^2) \end{align*}

The transformation equation for the given generating function of the first kind are:

\begin{align*} p&=\die{F}{q}=m\omega q\cot Q \\ P&=-\die{F}{Q}=\frac{m\omega q^2}{2\sin^2 Q}\\ K&=H \end{align*}

Rearranging for \(p\) and \(q\):

\begin{align*} q^2&=\frac{2}{m\omega}P\sin^2Q\\ p^2&=2m\omega P\cos^2Q \end{align*}

which we can now substitute to get \(K\):

\begin{align*} K&=H=\frac{1}{2m}(p^2+m^2\omega^2q^2)\\ &=\frac{1}{2m}\left(2m\omega P\cos^2Q +m^2\omega^2\frac{2}{m\omega}P\sin^2Q \right)\\ &=\omega P(\cos^2Q+\sin^2Q)\\ &=\omega P\\ \end{align*}

and since \(Q\) is clearly cyclic, we have that \(P\) is constant. In this case, the total Hamiltonian is the energy, \(E\), so we can write:

\begin{align*} P=\frac{E}{\omega} \end{align*}

The canonical equation for \(Q\) gives:

\begin{align*} \dot Q&=\die{K}{P}=\omega\\ \therefore Q(t)&=\omega t +\beta \end{align*}

where \(\beta\) is an integration constant that depends on the initial conditions. We can now transform this back to the original variables:

\begin{align*} q(t)&=\sqrt{\frac{2P}{m\omega}}\sin Q\\ &=\sqrt{\frac{2E}{m\omega^2}}\sin(\omega t +\beta) \end{align*}

as expected.