6Canonical Evolution

What, then, is time? I know well enough what it is, provided that nobody asks me; but if I am asked what it is and try to explain, I am baffled. All the same I can confidently say that I know that if nothing passed, there would be no past time; if nothing were going to happen, there would be no future time; and if nothing were there would be no present time.

Augustine of Hippo, from Confessions, Book XI, Section 14. Translation by R.S. Pine-Coffin, 1961.

Time evolution generates a canonical transformation: if we consider all possible initial states of a Hamiltonian system and follow all the trajectories for the same time interval, then the map from the initial state to the final state of each trajectory is a canonical transformation. Hamilton–Jacobi theory gives a mixed-variable generating function that generates this time-evolution transformation. For the few integrable systems for which we can solve the Hamilton–Jacobi equation this transformation gets us action-angle coordinates, which form a starting point to study perturbations.

6.1   Hamilton–Jacobi Equation

If we could find a canonical transformation so that the transformed Hamiltonian was identically zero, then by Hamilton's equations the new coordinates and momenta would be constants. All of the time variation of the solution would be captured in the canonical transformation, and there would be nothing more to the solution. A mixed-variable generating function that does this job satisfies a partial differential equation called the Hamilton–Jacobi equation. In most cases, a Hamilton–Jacobi equation cannot be solved explicitly. When it can be solved, however, a Hamilton–Jacobi equation provides a means of reducing a problem to a useful simple form.

Recall the relations satisfied by an F₂-type generating function:

$$\begin{array}{cc}q\prime ={\partial }_2{F}_2\left(t,q,p\prime \right)& (6.1)\end{array}$$

$$\begin{array}{cc}p={\partial }_1{F}_2\left(t,q,p\prime \right)& \left(6.2\right)\end{array}$$

$$\begin{array}{cc}H\prime \left(t,q\prime ,p\prime \right)=H\left(t,q,p\right)+{\partial }_0{F}_2\left(t,q,p\prime \right).& \left(6.3\right)\end{array}$$

If we require the new Hamiltonian to be zero, then F₂ must satisfy the equation

$$\begin{array}{cc}0=H\left(t,q,{\partial }_1{F}_2\left(t,q,p\prime \right)\right)+{\partial }_0{F}_2\left(t,q,p\prime \right).& \left(6.4\right)\end{array}$$

So the solution of the problem is “reduced” to the problem of solving an n-dimensional partial differential equation for F₂ with unspecified new (constant) momenta p′. This is a Hamilton–Jacobi equation, and in some cases we can solve it.

We can also attempt a somewhat less drastic method of solution. Rather than try to find an F₂ that makes the new Hamiltonian identically zero, we can seek an F₂-shaped function W that gives a new Hamiltonian that is solely a function of the new momenta. A system described by this form of Hamiltonian is also easy to solve. So if we set

$$\begin{array}{lll}H″\left(t,q″,p″\right)\hfill & =H\left(t,q,{\partial }_{1}W\left(t,q,p″\right)\right)+{\partial }_{0}W\left(t,q,p″\right)\hfill & \hfill \\ \hfill & =E(p″)\hfill & \left(6.5\right)\hfill \end{array}$$

and are able to solve for W, then the problem is essentially solved. In this case, the primed momenta are all constant and the primed positions are linear in time. This is an alternate form of the Hamilton–Jacobi equation.

These forms are related. Suppose that we have a W that satisfies the second form of the Hamilton–Jacobi equation (6.5). Then the F₂ constructed from W

$$\begin{array}{cc}{F}_2\left(t,q,p\prime \right)=W\left(t,q,p\prime \right)-E(p\prime )t& \left(6.6\right)\end{array}$$

satisfies the first form of the Hamilton–Jacobi equation (6.4). Furthermore,

$$\begin{array}{cc}p={\partial }_1{F}_2\left(t,q,p\prime \right)={\partial }_{1}W\left(t,q,p\prime \right),& \left(6.7\right)\end{array}$$

so the primed momenta are the same in the two formulations. But

$$\begin{array}{lll}q\prime \hfill & ={\partial }_2{F}_2\left(t,q,p\prime \right)\hfill & \hfill \\ \hfill & ={\partial }_{2}W\left(t,q,p\prime \right)-DE(p\prime )t\hfill & \hfill \\ \hfill & =q″-DE(p\prime )t,\hfill & \left(6.8\right)\hfill \end{array}$$

so we see that the primed coordinates differ by a term that is linear in time—both $p\prime \left(t\right)=p_0^{\prime }$ and $q\prime \left(t\right)=q_0^{\prime }$ are constant. Thus we can use either W or F₂ as the generating function, depending on the form of the new Hamiltonian we want.

Note that if H is time independent then we can often find a time-independent W that does the job. For time-independent W the Hamilton–Jacobi equation simplifies to

$$\begin{array}{cc}E(p\prime )=H\left(t,q,{\partial }_{1}W\left(t,q,p\prime \right)\right).& \left(6.9\right)\end{array}$$

The corresponding F₂ is then linear in time. Notice that an implicit requirement is that the energy can be written as a function of the new momenta alone. This excludes the possibility that the transformed phase-space coordinates q′ and p′ are simply initial conditions for q and p.

It turns out that there is flexibility in the choice of the function E. With an appropriate choice the phase-space coordinates obtained through the transformation generated by W are action-angle coordinates.

Exercise 6.1: Hamilton–Jacobi with F₁

We have used an F₂-type generating function to carry out the Hamilton–Jacobi transformations. Carry out the equivalent transformations with an F₁-type generating function. Find the equations corresponding to equations (6.4), (6.5), and (6.9).

6.1.1 Harmonic Oscillator

Consider the familiar time-independent Hamiltonian

$$\begin{array}{cc}H\left(t,x,p\right)=\frac{p^2}{2m}+\frac{k{x}^2}{2}.& \left(6.10\right)\end{array}$$

We form the Hamilton–Jacobi equation for this problem:

$$\begin{array}{cc}0=H\left(t,x,{\partial }_1{F}_2\left(t,x,p\prime \right)\right)+{\partial }_0{F}_2\left(t,x,p\prime \right).& \left(6.11\right)\end{array}$$

Using F₂(t, x, p′) = W (t, x, p′) − E(p′)t, we find

$$\begin{array}{cc}E(p\prime )=H\left(t,x,{\partial }_{1}W\left(t,x,p\prime \right)\right).& \left(6.12\right)\end{array}$$

Writing this out explicitly yields

$$\begin{array}{cc}E(p\prime )=\frac{{\left({\partial }_{1}W\left(t,x,p\prime \right)\right)}^2}{2m}+\frac{k{x}^2}{2},& \left(6.13\right)\end{array}$$

and solving for ∂₁W gives

$$\begin{array}{cc}{\partial }_{1}W\left(t,x,p\prime \right)=\sqrt{2m\left(E(p\prime )-\frac{k{x}^2}{2}\right)}.& \left(6.14\right)\end{array}$$

Integrating gives the desired W:

$$\begin{array}{cc}W\left(t,x,p\prime \right)={\displaystyle {\int }^x}\sqrt{2m\left(E(p\prime )-\frac{k{z}^2}{2}\right)}dz.& \left(6.15\right)\end{array}$$

We can use either W or the corresponding F₂ as the generating function. First, take W to be the generating function. We obtain the coordinate transformation by differentiating:

$$\begin{array}{lll}x\prime \hfill & ={\partial }_{2}W\left(t,x,p\prime \right)\hfill & \hfill \\ \hfill & ={\displaystyle {\int }^x\frac{mDE(p\prime )}{\sqrt{2m\left(E(p\prime )-\frac{k{z}^2}{2}\right)}}dz}\hfill & \left(6.16\right)\hfill \end{array}$$

and then integrating to get

$$\begin{array}{cc}x\prime =\sqrt{\frac{m}{k}}DE(p\prime )\arcsin \left(\sqrt{\frac{k}{2E(p\prime )}x}\right)+C(p\prime ),& \left(6.17\right)\end{array}$$

with some integration constant C(p′). Inverting this, we get the unprimed coordinate in terms of the primed coordinate and momentum:

$$\begin{array}{cc}x=\sqrt{\frac{2E(p\prime )}{k}}\sin \left[\frac{1}{DE(p\prime )}\sqrt{\frac{k}{m}}\left(x\prime -C(p\prime )\right)\right].& \left(6.18\right)\end{array}$$

The new Hamiltonian H′ depends only on the momentum:

$$\begin{array}{cc}H\prime \left(t,x\prime ,p\prime \right)=E(p\prime ).& \left(6.19\right)\end{array}$$

The equations of motion are just

$$\begin{array}{ll}Dx\prime \left(t\right)={\partial }_{2}H\prime \left(t,x\prime \left(t\right),p\prime \left(t\right)\right)=DE(p\prime )\hfill & \hfill \\ Dp\prime \left(t\right)=-{\partial }_{1}H\prime \left(t,x\prime \left(t\right),p\prime \left(t\right)\right)=0,\hfill & \left(6.20\right)\hfill \end{array}$$

with solution

$$\begin{array}{ll}x\prime \left(t\right)=DE(p\prime )t+x_0^{\prime }\hfill & \hfill \\ p\prime \left(t\right)=p_0^{\prime }\hfill & \left(6.21\right)\hfill \end{array}$$

for initial conditions $x_0^{\prime }$ and $p_0^{\prime }$. If we plug these expressions for x′(t) and p′(t) into equation (6.18) we find

$$\begin{array}{lll}x\left(t\right)\hfill & =\sqrt{\frac{2E(p\prime )}{k}}\sin \left[\frac{1}{DE(p\prime )}\sqrt{\frac{k}{m}}(DE(p\prime )t+x_0^{\prime }-C(p\prime ))\right]\hfill & \hfill \\ \hfill & =\sqrt{\frac{2E(p\prime )}{k}}\sin \left[\sqrt{\frac{k}{m}}\left(t-t_0\right)\right]\hfill & \hfill \\ \hfill & =A\sin \left(\omega t+\phi \right),\hfill & \left(6.22\right)\hfill \end{array}$$

where the angular frequency is $\omega =\sqrt{k/m}$, the amplitude is $A=\sqrt{2E(p\prime )/k}$, and the phase is $\phi =-\omega t_0=\omega (x_0^{\prime }-C\left(p\prime \right))/DE\left(p\prime \right)$.

We can also use F₂ = W − Et as the generating function. The new Hamiltonian is zero, so both x′ and p′ are constant, but the relationship between the old and new variables is

$$\begin{array}{lll}x\prime \hfill & ={\partial }_2{F}_2\left(t,x,p\prime \right)\hfill & \hfill \\ \hfill & ={\partial }_{2}W\left(t,x,p\prime \right)-DE(p\prime )t\hfill & \hfill \\ \hfill & ={\displaystyle {\int }^x}\frac{mDE(p\prime )}{\sqrt{2m\left(E(p\prime )-\frac{k{z}^2}{2}\right)}}-DE(p\prime )t\hfill & \hfill \\ \hfill & =\sqrt{\frac{m}{k}}DE(p\prime )\arcsin \left(\sqrt{\frac{k}{2E(p\prime )}}x\right)+C(p\prime )-DE(p\prime )t.\hfill & \left(6.23\right)\hfill \end{array}$$

Plugging in the solution $x\prime =x_0^{\prime }$ and $p\prime =p_0^{\prime }$ and solving for x, we find equation (6.22). So once again we see that the two approaches are equivalent.

It is interesting to note that the solution depends upon the constants E(p′) and DE(p′), but otherwise the motion is not dependent in any essential way on what the function E actually is. The momentum p is constant and the values of the constants are set by the initial conditions. Given a particular function E, the initial conditions determine p′, but the solution can be obtained without further specifying the E function.

If we choose particular functions E we can get particular canonical transformations. For example, a convenient choice is simply

$$\begin{array}{cc}E(p\prime )=\alpha p\prime ,& \left(6.24\right)\end{array}$$

for some constant α that will be chosen later. We find

$$\begin{array}{cc}x=\sqrt{\frac{2\alpha p\prime }{k}}\sin \frac{\omega }{\alpha }x\prime .& \left(6.25\right)\end{array}$$

So we see that a convenient choice is $\alpha =\omega =\sqrt{k/m}$, so

$$\begin{array}{cc}x=\sqrt{\frac{2p\prime }{\beta }}\sin x\prime ,& \left(6.26\right)\end{array}$$

with $\beta =\sqrt{km}$. The new Hamiltonian is

$$\begin{array}{cc}H\prime \left(t,x\prime ,p\prime \right)=E(p\prime )=\omega p\prime .& \left(6.27\right)\end{array}$$

The solution is just $x\prime =\omega t+x_0^{\prime }$ and $p\prime =p_0^{\prime }$. Substituting the expression for x in terms of x′ and p′ into H(t, x, p) = H′(t, x′, p′), we derive

$$\begin{array}{l}\begin{array}{lll}p\hfill & ={\left[2m\left(p\prime \alpha -\frac{k}{2}{x}^2\right)\right]}^{1/2}\hfill & \hfill \\ \hfill & =\sqrt{2p\prime \beta }\cos x\prime .\hfill & \left(6.28\right)\hfill \end{array}\\ \end{array}$$

The two transformation equations (6.26) and (6.28) are what we have called the polar-canonical transformation (equation 5.29). We have already shown that this transformation is canonical and that it solves the harmonic oscillator, but it was not derived. Here we have derived this transformation as a particular case of the solution of the Hamilton–Jacobi equation.

We can also explore other choices for the E function. For example, we could choose

$$\begin{array}{cc}E(p\prime )=\frac{1}{2}\alpha p{\prime }^2.& \left(6.29\right)\end{array}$$

Following the same steps as before, we find

$$\begin{array}{cc}x=\sqrt{\frac{\alpha p{\prime }^2}{k}}\sin \frac{\omega }{\alpha }\frac{x\prime }{p\prime }.& \left(6.30\right)\end{array}$$

So a convenient choice is again α = ω, leaving

$$\begin{array}{ll}x=\frac{p\prime }{\beta }\sin \frac{x\prime }{p\prime }\hfill & \hfill \\ p=\beta p\prime \cos \frac{x\prime }{p\prime },\hfill & \left(6.31\right)\hfill \end{array}$$

with β = (km)^1/4. By construction, this transformation is also canonical and also brings the harmonic oscillator problem into an easily solvable form:

$$\begin{array}{cc}H\prime \left(t,x\prime ,p\prime \right)=\frac{1}{2}\omega p{\prime }^2.& \left(6.32\right)\end{array}$$

The harmonic oscillator Hamiltonian has been transformed to what looks a lot like the Hamiltonian for a free particle. This is very interesting. Notice that whereas Hamiltonian (6.27) does not have a well defined Legendre transform to an equivalent Lagrangian, the “free particle” harmonic oscillator has a well defined Legendre transform:

$$\begin{array}{cc}L\prime \left(t,x\prime ,\dot{x}\prime \right)=\frac{\dot{x}{\prime }^2}{2\omega }.& \left(6.33\right)\end{array}$$

Of course, there may be additional properties that make one choice more useful than others for particular applications.

Exercise 6.2: Pendulum

Formulate and solve a Hamilton–Jacobi equation for the pendulum; investigate both the circulating and oscillating regions of phase space. (Note: This is a long story and requires some knowledge of elliptic functions.)

6.1.2 Hamilton–Jacobi Solution of the Kepler Problem

We can use the Hamilton–Jacobi equation to find canonical coordinates that solve the Kepler problem. This is an essential first step in doing perturbation theory for orbital problems.

In rectangular coordinates (x, y, z), the Kepler Hamiltonian is

$$\begin{array}{cc}{H}_r\left(t;x,y,z;p_x,p_y,p_z\right)=\frac{p^2}{2m}-\frac{\mu }{r},& \left(6.34\right)\end{array}$$

where r² = x² + y² + z² and $p^2=p_x^2+p_y^2+p_z^2$.

We try a generating function of the form $W\left(t;x,y,z;p_x^{\prime },p_y^{\prime },p_z^{\prime }\right)$. The Hamilton–Jacobi equation is then¹

$$\begin{array}{l}\\ \begin{array}{lll}E(p\prime )\hfill & =\frac{1}{2m}[{\left({\partial }_{1,0}W\left(t;x,y,z;p_x^{\prime },p_y^{\prime },p_z^{\prime }\right)\right)}^2\hfill & \hfill \\ \hfill & +{\left({\partial }_{1,1}W\left(t;x,y,z;p_x^{\prime },p_y^{\prime },p_z^{\prime }\right)\right)}^2\hfill & \hfill \\ \hfill & +{\left({\partial }_{1,2}W\left(t;x,y,z;p_x^{\prime },p_y^{\prime },p_z^{\prime }\right)\right)}^2]-\frac{\mu }{r}.\hfill & \left(6.35\right)\hfill \end{array}\end{array}$$

This is a partial differential equation in the three partial derivatives of W. We stare at it a while and give up.

Next we try converting to spherical coordinates. This is motivated by the fact that the potential energy depends only on r. The Hamiltonian in spherical coordinates (r, θ, φ), where θ is the colatitude and φ is the longitude, is

$$\begin{array}{cc}{H}_s\left(t;r,\theta ,\phi ;p_r,p_{\theta },p_{\phi }\right)=\frac{1}{2m}\left[p_r^2+\frac{p_{\theta }^2}{r^2}+\frac{p_{\phi }^2}{r^2{(\sin \theta )}^2}\right]-\frac{\mu }{r}.& \left(6.36\right)\end{array}$$

The Hamilton–Jacobi equation is

$$\begin{array}{l}E(p_0^{\prime },p_1^{\prime },p_2^{\prime })\\ \begin{array}{lll}\hfill & =\frac{1}{2m}[{({\partial }_{1,0}W(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }))}^2\hfill & \hfill \\ \hfill & +\frac{1}{r^2}{({\partial }_{1,1}W(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }))}^2\hfill & \hfill \\ \hfill & +\frac{1}{r^2{(\sin \theta )}^2}{({\partial }_{1,2}W(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }))}^2]-\frac{\mu }{r}.\hfill & (6.37)\hfill \end{array}\end{array}$$

We can solve this Hamilton–Jacobi equation by successively isolating the dependence on the various variables. Looking first at the φ dependence, we see that, outside of W, φ appears only in one partial derivative. If we write

$$\begin{array}{cc}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)=f\left(r,\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)+p_2^{\prime }\phi ,& \left(6.38\right)\end{array}$$

then ${\partial }_{1,2}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)=p_2^{\prime }$, and then φ does not appear in the remaining equation for f:

$$\begin{array}{l}E(p_0^{\prime },p_1^{\prime },p_2^{\prime })\\ \begin{array}{lll}\hfill & =\frac{1}{2m}\{{({\partial }_{0}f(r,\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }))}^2\hfill & \hfill \\ \hfill & +\frac{1}{r^2}[{({\partial }_{1}f(r,\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }))}^2+\frac{{(p_2^{\prime })}^2}{{(\sin \theta )}^2}]\}-\frac{\mu }{r}.\hfill & (6.39)\hfill \end{array}\end{array}$$

Any function of the $p_i^{\prime }$ could have been used as the coefficient of φ in the generating function. This particular choice has the nice feature that $p_2^{\prime }$ is the z component of the angular momentum.

We can eliminate the θ dependence if we choose

$$\begin{array}{cc}f\left(r,\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)=R\left(r,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)+\Theta \left(\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)& \left(6.40\right)\end{array}$$

and require that Θ be a solution to

$$\begin{array}{cc}{\left({\partial }_0\Theta \left(\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\right)}^2+\frac{{\left(p_2^{\prime }\right)}^2}{{(\sin \theta )}^2}={\left(p_1^{\prime }\right)}^2.& \left(6.41\right)\end{array}$$

We are free to choose the right-hand side to be any function of the new momenta. This choice reflects the fact that the left-hand side is non-negative. It turns out that $p_1^{\prime }$ is the total angular momentum. This equation for Θ can be solved by quadrature.

The remaining equation that determines R is

$$\begin{array}{cc}E\left(p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)=\frac{1}{2m}\left[{\left({\partial }_{0}R\left(r,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\right)}^2+\frac{1}{r^2}{\left(p_1^{\prime }\right)}^2\right]-\frac{\mu }{r},& \left(6.42\right)\end{array}$$

which also can be solved by quadrature.

Altogether the solution of the Hamilton–Jacobi equation reads

$$\begin{array}{lll}W\left(r,\theta ,\phi ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\hfill & ={\displaystyle {\int }^r{\left(2mE\left(p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)+\frac{2m\mu }{r}-\frac{{\left(p_1^{\prime }\right)}^2}{r^2}\right)}^{1/2}dr}\hfill & \hfill \\ \hfill & +{\displaystyle {\int }^{\theta }}{\left({\left(p_1^{\prime }\right)}^2-\frac{{(p_2^{\prime })}^2}{{(\sin \theta )}^2}\right)}^{1/2}d\theta \hfill & \hfill \\ \hfill & +p_2^{\prime }\phi .\hfill & \left(6.43\right)\hfill \end{array}$$

It is interesting that our solution to the Hamilton–Jacobi partial differential equation is of the form

$$\begin{array}{l}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\\ \begin{array}{cc}=R\left(r,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)+\Theta \left(\theta ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)+\Phi \left(\phi ,p_0^{\prime },p_1^{\prime },p_2^{\prime }\right).& \left(6.44\right)\end{array}\\ \end{array}$$

Thus we have a separation-of-variables technique that involves writing the solution as a sum of functions of the individual variables. This might be contrasted with the separation-of-variables technique encountered in elementary quantum mechanics and classical electrodynamics, which uses products of functions of individual variables.

The coordinates q′ = (q′⁰, q′¹, q′²) conjugate to the momenta $p\prime =\left[p_0^{\prime },p_1^{\prime },p_2^{\prime }\right]$ are

$$\begin{array}{ll}q{\prime }^0\hfill & ={\partial }_{2,0}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\hfill \\ \hfill & =m{\partial }_{0}E(p\prime ){\displaystyle {\int }^r}{\left(2mE(p\prime )+\frac{2m\mu }{r}-\frac{{\left(p_1^{\prime }\right)}^2}{r^2}\right)}^{-1/2}dr\hfill \\ q{\prime }^1\hfill & ={\partial }_{2,1}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\hfill \\ \hfill & =p_1^{\prime }{\displaystyle {\int }^{\theta }}{\left({\left(p_1^{\prime }\right)}^2-\frac{{\left(p_2^{\prime }\right)}^2}{{(\sin \theta )}^2}\right)}^{-1/2}d\theta \hfill \\ \hfill & +{\displaystyle {\int }^r}\left(m{\partial }_{1}E(p\prime )-\frac{p_1^{\prime }}{r^2}\right){\left(2mE(p\prime )+\frac{2m\mu }{r}-\frac{{\left(p_1^{\prime }\right)}^2}{r^2}\right)}^{-1/2}dr\hfill \\ q{\prime }^2\hfill & ={\partial }_{2,2}W\left(t;r,\theta ,\phi ;p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)\hfill \\ \hfill & =\phi -\frac{p_2^{\prime }}{{(\sin \theta )}^2}{\displaystyle {\int }^{\theta }}{\left({\left(p_1^{\prime }\right)}^2-\frac{{\left(p_2^{\prime }\right)}^2}{{(\sin \theta )}^2}\right)}^{-1/2}d\theta \hfill \\ \hfill & +m{\partial }_{2}E(p\prime ){\displaystyle {\int }^r}{\left(2mE(p\prime )+\frac{2m\mu }{r}-\frac{{\left(p_1^{\prime }\right)}^2}{r^2}\right)}^{-1/2}dr.\hfill \\ \hfill & \hfill \end{array}$$

We are still free to choose the functional form of E. A convenient (and conventional) choice is

$$\begin{array}{cc}E\left(p_0^{\prime },p_1^{\prime },p_2^{\prime }\right)=-\frac{m{\mu }^2}{2{\left(p_0^{\prime }\right)}^2}.& \left(6.45\right)\end{array}$$

With this choice the momentum $p_0^{\prime }$ has dimensions of angular momentum, and the conjugate coordinate is an angle.

The Hamiltonian for the Kepler problem is reduced to

$$\begin{array}{cc}H\prime \left(t,q\prime ,p\prime \right)=E(p\prime )=-\frac{m{\mu }^2}{2\left(p_0^{\prime }\right)2}.& \left(6.46\right)\end{array}$$

Thus

$$\begin{array}{cc}q{\prime }^0=nt+{\beta }^0& \left(6.47\right)\end{array}$$

$$\begin{array}{cc}q{\prime }^1={\beta }^1& \left(6.48\right)\end{array}$$

$$\begin{array}{cc}q{\prime }^2={\beta }^2,& \left(6.49\right)\end{array}$$

where $n=m{\mu }^2/{\left(p_0^{\prime }\right)}^3$ and where β⁰, β¹, and β² are the initial values of the components of q′. Only one of the new variables changes with time.

The canonical phase-space coordinates can be written in terms of the parameters that specify an orbit. We merely summarize the results; for further explanation see [36] or [38].

Assume we have a bound orbit with semimajor axis a, eccentricity e, inclination i, longitude of ascending node Ω, argument of pericenter ω, and mean anomaly M. The three canonical momenta are $p_0^{\prime }=\sqrt{m\mu a},p_1^{\prime }=\sqrt{m\mu a\left(1-e^2\right)}$, and $p_2^{\prime }=\sqrt{m\mu a\left(1-e^2\right)}\cos i$. The first momentum is related to the energy, the second momentum is the total angular momentum, and the third momentum is the component of the angular momentum in the ẑ direction. The conjugate canonical coordinates are (q′)⁰ = M, (q′)¹ = ω, and (q′)² = Ω.

6.1.3 F₂ and the Lagrangian

The solution to the Hamilton–Jacobi equation, the mixed-variable generating function that generates time evolution, is related to the action used in the variational principle. In particular, the time derivative of the generating function along realizable paths has the same value as the Lagrangian.

Let ${\tilde{F}}_2\left(t\right)=F_2\left(t,q\left(t\right),p\prime \left(t\right)\right)$ be the value of F₂ along the paths q and p′ at time t. The derivative of ${\tilde{F}}_2$ is

$$\begin{array}{lll}D{\tilde{F}}_2\left(t\right)\hfill & ={\partial }_0{F}_2\left(t,q\left(t\right),p\prime (t)\right)\hfill & \hfill \\ \hfill & +{\partial }_1{F}_2\left(t,q\left(t\right),p\prime \left(t\right)\right)Dq\left(t\right)\hfill & \hfill \\ \hfill & +{\partial }_2{F}_2\left(t,q\left(t\right),p\prime \left(t\right)\right)Dq\prime \left(t\right).\hfill & \left(6.50\right)\hfill \end{array}$$

Using the Hamilton–Jacobi equation (6.4), this becomes

$$\begin{array}{lll}D{\tilde{F}}_2\left(t\right)\hfill & =-H\left(t,q\left(t\right),{\partial }_1{F}_2(t,q\left(t\right),p\prime (t))\right)\hfill & \hfill \\ \hfill & +{\partial }_1{F}_2\left(t,q\left(t\right),p\prime \left(t\right)\right)Dq\left(t\right)\hfill & \hfill \\ \hfill & +{\partial }_2{F}_2\left(t,q\left(t\right),p\prime \left(t\right)\right)Dq\prime \left(t\right).\hfill & \left(6.51\right)\hfill \end{array}$$

Now, using equation (6.2), we get

$$\begin{array}{l}\begin{array}{lll}D{\tilde{F}}_2\left(t\right)\hfill & =-H\left(t,q\left(t\right),p\left(t\right)\right)\hfill & \hfill \\ \hfill & +p\left(t\right)Dq\left(t\right)\hfill & \hfill \\ \hfill & +{\partial }_2{F}_2\left(t,q\left(t\right),p\prime \left(t\right)\right)Dp\prime \left(t\right).\hfill & \left(6.52\right)\hfill \end{array}\\ \end{array}$$

But $p\left(t\right)Dq\left(t\right)-H\left(t,q\left(t\right),p\left(t\right)\right)=L\left(t,q\left(t\right),Dq\left(t\right)\right)$, so

$$\begin{array}{cc}D{\tilde{F}}_2\left(t\right)=L\left(t,q\left(t\right),Dq\left(t\right)\right)+{\partial }_2{F}_2\left(t,q\left(t\right),p\prime (t)\right)Dp\prime \left(t\right).& \left(6.53\right)\end{array}$$

On realizable paths we have Dp′(t) = 0, so along realizable paths the time derivative of F₂ is the same as the Lagrangian along the path. The time integral of the Lagrangian along any path is the action along that path. This means that, up to an additive term that is constant on realizable paths but may be a function of the transformed phase-space coordinates q′ and p′, the F₂ that solves the Hamilton–Jacobi equation has the same value as the Lagrangian action for realizable paths.

The same conclusion follows for the Hamilton–Jacobi equation formulated in terms of F₁. Up to an additive term that is constant on realizable paths but may be a function of the transformed phase-space coordinates q′ and p′, the F₁ that solves the corresponding Hamilton–Jacobi equation has the same value as the Lagrangian action for realizable paths.

Recall that a transformation given by an F₂-type generating function is also given by an F₁-type generating function related to it by a Legendre transform (see equation 5.142):

$$\begin{array}{cc}{F}_1\left(t,q,q\prime \right)=F_2\left(t,q,p\prime \right)-q\prime p\prime ,& \left(6.54\right)\end{array}$$

provided the transformations are nonsingular. In this case, both q′ and p′ are constant on realizable paths, so the additive constants that make F₁ and F₂ equal to the Lagrangian action differ by q′p′.

Exercise 6.3: Harmonic oscillator

Let's check this for the harmonic oscillator (of course).

a. Finish the integral (6.15):

$W\left(t,x,p\prime \right)={\displaystyle {\int }^x}\sqrt{2m\left(E(p\prime )-\frac{k{z}^2}{2}\right)}dz.$

Write the result in terms of the amplitude $A=\sqrt{2E(p\prime )/k}$.

b. Check that this generating function gives the transformation

$x\prime ={\partial }_{2}W\left(t,x,p\prime \right)=\sqrt{\frac{m}{k}}DE(p\prime )\arcsin \left(\frac{x}{\sqrt{2E(p\prime )/k}}\right),$

which is the same as equation (6.17) for a particular choice of the integration constant. The other part of the transformation is

$p={\partial }_{1}W\left(t,x,p\prime \right)=\sqrt{mk}\sqrt{A^2-x^2},$

with the same definition of A as before.

c. Compute the time derivative of the associated F₂ along realizable paths (Dp′(t) = 0), and compare it to the Lagrangian along realizable paths.

6.1.4 The Action Generates Time Evolution

We define the function $\overline{F}\left(t_1,q_1,t_2,q_2\right)$ to be the value of the action for a realizable path q such that q(t₁) = q₁ and q(t₂) = q₂. So $\overline{F}$ satisfies

$$\begin{array}{cc}\overline{F}\left(t_1,q\left(t_1\right),t_{2,}q\left(t_2\right)\right)=S\left[q\right]\left(t_1,t_2\right)={\displaystyle {\int }_{t_1}^{t_2}L\circ \Gamma \left[q\right]}.& \left(6.55\right)\end{array}$$

For variations η that are not necessarily zero at the end times and for realizable paths q, the variation of the action is

$$\begin{array}{lll}{\delta }_{\eta }S\left[q\right]\left(t_1,t_2\right)\hfill & =\left({\partial }_{2}L\circ \Gamma \left[q\right]\right)\eta {|}_{t_1}^{t_2}\hfill & \hfill \\ \hfill & =p\left(t_2\right)\eta \left(t_2\right)-p\left(t_1\right)\eta \left(t_1\right).\hfill & \left(6.56\right)\hfill \end{array}$$

Alternatively, the variation of S[q] in equation (6.55) gives

$$\begin{array}{lll}{\delta }_{\eta }S\left[q\right]\left(t_1,t_2\right)\hfill & ={\partial }_1\overline{F}\left(t_1,q\left(t_1\right),t_2,q\left(t_2\right)\right)\eta \left(t_1\right)\hfill & \hfill \\ \hfill & +{\partial }_3\overline{F}\left(t_1,q\left(t_1\right),t_2,q\left(t_2\right)\right)\eta \left(t_2\right).\hfill & \left(6.57\right)\hfill \end{array}$$

Comparing equations (6.56) and (6.57) and using the fact that the variation η is arbitrary, we find

$$\begin{array}{ll}{\partial }_1\overline{F}\left(t_1,q\left(t_1\right),t_2,q\left(t_2\right)\right)=-p\left(t_1\right)\hfill & \hfill \\ {\partial }_3\overline{F}\left(t_1,q\left(t_1\right),t_2,q\left(t_2\right)\right)=p\left(t_2\right).\hfill & \left(6.58\right)\hfill \end{array}$$

The partial derivatives of $\overline{F}$ with respect to the coordinate arguments give the momenta. Abstracting off paths, we have

$$\begin{array}{ll}{\partial }_1\overline{F}\left(t_1,q_1,t_2,q_2\right)=-p_1\hfill & \hfill \\ {\partial }_3\overline{F}\left(t_1,q_1,t_2,q_2\right)=p_{2.}\hfill & \left(6.59\right)\hfill \end{array}$$

This looks a bit like the F₁-type generating function relations, but here there are two times. Solving equations (6.59) for q₂ and p₂ as functions of t₂ and the initial state t₁, q₁, p₁, we get the time evolution of the system in terms of $\overline{F}$. The function $\overline{F}$ generates time evolution.

If we vary the lower limit of the action integral we get

$$\begin{array}{cc}{\partial }_0\left(S\left[q\right]\right)\left(t_1,t_2\right)=-L\left(t_1,q\left(t_1\right),D_q\left(t_1\right)\right).& \left(6.60\right)\end{array}$$

Using equation (6.55), and given a realizable path q such that q(t₁) = q₁ and q(t₂) = q₂, we get the partial derivatives with respect to the time slots:

$$\begin{array}{lll}{\partial }_0\left(S\left[q\right]\right)\left(t_1,t_2\right)\hfill & ={\partial }_0\overline{F}\left(t_1,q_1,t_2,q_2\right)+{\partial }_1\overline{F}\left(t_1,q_1,t_2,q_2\right)Dq\left(t_1\right)\hfill & \hfill \\ \hfill & ={\partial }_0\overline{F}\left(t_1,q_1,t_2,q_2\right)-p\left(t_1\right)D_q\left(t_1\right).\hfill & \left(6.61\right)\hfill \end{array}$$

Rearranging the terms of equation (6.61) and using equation (6.60) we get

$$\begin{array}{lll}{\partial }_0\overline{F}\left(t_1,q_1,t_2,q_2\right)\hfill & =H\left(t_1,q_1,p_1\right)\hfill & \hfill \\ \hfill & =H\left(t_1,q_1,-{\partial }_1\overline{F}\left(t_1,q_1,t_2,q_2\right)\right),\hfill & (6.62)\hfill \end{array}$$

and similarly

$$\begin{array}{lll}{\partial }_2\overline{F}\left(t_1,q_1,t_2,q_2\right)\hfill & =-H\left(t_2,q_2,p_2\right)\hfill & \hfill \\ \hfill & =-H\left(t_2,q_2,{\partial }_3\overline{F}\left(t_1,q_1,t_2,q_2\right)\right).\hfill & (6.63)\hfill \end{array}$$

These are a pair of Hamilton–Jacobi equations, computed at the endpoints of the path.

The function $\overline{F}$ can be written in terms of an F₁ that satisfies a Hamilton–Jacobi equation for H. We can compute time evolution by using the F₁ solution of the Hamilton–Jacobi equation to express the state (t₁, q₁, p₁) in terms of the constants q′ and p′. Using the same solution we can then perform a subsequent transformation back from q′ p′ to the original state variables at a different time t₂, giving the state (t₂, q₂, p₂). The composition of these two canonical transformations is canonical (see exercise 5.12).

The generating function for the composition is the difference of the generating functions for each step:

$$\begin{array}{cc}{\overline{F}}_x\left(t_1,q_1,q\prime ,t_2,q_2\right)=F_1\left(t_2,q_2,q\prime \right)-F_1\left(t_1,q_1,q\prime \right),& \left(6.64\right)\end{array}$$

with the condition

$$\begin{array}{cc}{\partial }_2{F}_1\left(t_2,q_2,q\prime \right)-{\partial }_2{F}_1\left(t_1,q_1,q\prime \right)=0,& \left(6.65\right)\end{array}$$

which allows us to eliminate q′ in terms of t₁, q₁, t₂, and q₂. So we can write

$$\begin{array}{cc}\overline{F}\left(t_1,q_1,t_2,q_2\right)=F_1\left(t_2,q_2,q\prime \right)-F_1\left(t_1,q_1,q\prime \right).& \left(6.66\right)\end{array}$$

Exercise 6.4: Uniform acceleration

a. Compute the Lagrangian action, as a function of the endpoints and times, for a uniformly accelerated particle. Use this to construct the canonical transformation for time evolution from a given initial state.

b. Solve the Hamilton–Jacobi equation for the uniformly accelerated particle, obtaining the F₁ that makes the transformed Hamiltonian zero. Show that the Lagrangian action can be expressed as a difference of two applications of this F₁.

6.2   Time Evolution is Canonical

We use time evolution to generate a transformation

$$\begin{array}{cc}\left(t,q,p\right)={\mathcal{C}}_{\Delta }\left(t\prime ,q\prime ,p\prime \right)& \left(6.67\right)\end{array}$$

that is obtained in the following way. Let $\sigma \left(t\right)=\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right)$ be a solution of Hamilton's equations. The transformation ${\mathcal{C}}_{\Delta }$ satisfies

$$\begin{array}{cc}{\mathcal{C}}_{\Delta }\left(\sigma \left(t\right)\right)=\sigma \left(t+\Delta \right),& \left(6.68\right)\end{array}$$

or, equivalently,

$$\begin{array}{cc}{\mathcal{C}}_{\Delta }\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right)=(t+\Delta ,\overline{q}\left(t+\Delta \right),\overline{p}\left(t+\Delta \right)).& \left(6.69\right)\end{array}$$

Notice that ${\mathcal{C}}_{\Delta }$ changes the time component. This is the first transformation of this kind that we have considered.²

Given a state (t′, q′, p′), we find the phase-space path σ emanating from this state as an initial condition, satisfying

$$\begin{array}{ll}q\prime =\overline{q}(t\prime )\hfill & \hfill \\ p\prime =\overline{p}(t\prime ).\hfill & \left(6.70\right)\hfill \end{array}$$

The value (t, q, p) of ${\mathcal{C}}_{\Delta }\left(t\prime ,q\prime ,p\prime \right)$ is then $\left(t\prime +\Delta ,\overline{q}\left(t\prime +\Delta \right),\overline{p}\left(t\prime +\Delta \right)\right)$.

Time evolution is canonical if the transformation ${\mathcal{C}}_{\Delta }$ is symplectic and if the Hamiltonian transforms in an appropriate manner. The transformation ${\mathcal{C}}_{\Delta }$ is symplectic if the bilinear antisymmetric form ω is invariant (see equation 5.73) for a general pair of linearized state variations with zero time component.

Let ζ′ be an increment with zero time component of the state (t′, q′, p′). The linearized increment in the value of ${\mathcal{C}}_{\Delta }\left(t\prime ,q\prime ,p\prime \right)$ is $\zeta =D{\mathcal{C}}_{\Delta }\left(t\prime ,q\prime ,p\prime \right)\zeta \prime$: The image of the increment is obtained by multiplying the increment by the derivative of the transformation. On the other hand, the transformation is obtained by time evolution, so the image of the increment can also be found by the time evolution of the linearized variational system. Let

$$\begin{array}{ll}\overline{\zeta }\left(t\right)=\left(0,{\overline{\zeta }}_q\left(t\right),{\overline{\zeta }}_p\left(t\right)\right)\hfill & \hfill \\ \overline{\zeta }\prime \left(t\right)=\left(0,{\overline{\zeta }}_p^{\prime }\left(t\right),{\overline{\zeta }}_p^{\prime }\left(t\right)\right)\hfill & \left(6.71\right)\hfill \end{array}$$

be variations of the state path $\sigma \left(t\right)=\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right)$; then

$$\begin{array}{ll}\overline{\zeta }\left(t+\Delta \right)=D{\mathcal{C}}_{\Delta }\left(t,q\left(t\right),p\left(t\right)\right)\overline{\zeta }\left(t\right)\hfill & \hfill \\ \overline{\zeta }\prime \left(t+\Delta \right)=D{\mathcal{C}}_{\Delta }\left(t,q\left(t\right),p\left(t\right)\right)\overline{\zeta }\prime \left(t\right).\hfill & (6.72)\hfill \end{array}$$

The symplectic requirement is

$$\begin{array}{cc}\omega \left(\overline{\zeta }\left(t\right),\overline{\zeta }\prime \left(t\right)\right)=\omega \left(\overline{\zeta }\left(t+\Delta \right),\overline{\zeta }\prime \left(t+\Delta \right)\right).& (6.73)\end{array}$$

This must be true for arbitrary Δ, so it is satisfied if the following quantity is constant:

$$\begin{array}{lll}A\left(t\right)\hfill & =\omega \left(\overline{\zeta }\left(t\right),\overline{\zeta }\prime \left(t\right)\right)\hfill & \hfill \\ \hfill & =P\left(\overline{\zeta }\prime \left(t\right)\right)Q\left(\overline{\zeta }\left(t\right)\right)-P\left(\overline{\zeta }\left(t\right)\right)Q\left(\overline{\zeta }\prime \left(t\right)\right)\hfill & \hfill \\ \hfill & ={\overline{\zeta }}_p^{\prime }\left(t\right){\overline{\zeta }}_q\left(t\right)-{\overline{\zeta }}_p\left(t\right){\overline{\zeta }}_q^{\prime }\left(t\right).\hfill & \left(6.74\right)\hfill \end{array}$$

We compute the derivative:

$$\begin{array}{lll}DA\left(t\right)\hfill & =D{\overline{\zeta }}_p^{\prime }\left(t\right){\overline{\zeta }}_q\left(t\right)+{\overline{\zeta }}_p^{\prime }\left(t\right)D{\overline{\zeta }}_q\left(t\right)\hfill & \hfill \\ \hfill & -D{\overline{\zeta }}_p\left(t\right){\overline{\zeta }}_q^{\prime }\left(t\right)-{\overline{\zeta }}_p\left(t\right)D{\overline{\zeta }}_q^{\prime }\left(t\right).\hfill & 6.75\hfill \end{array}$$

With Hamilton's equations, the variations satisfy

$$\begin{array}{lll}D{\overline{\zeta }}_q\left(t\right)\hfill & ={\partial }_1{\partial }_{2}H\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right){\overline{\zeta }}_q\left(t\right)\hfill & \hfill \\ \hfill & +{\partial }_2{\partial }_{2}H\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right){\overline{\zeta }}_p\left(t\right),\hfill & \hfill \\ D{\overline{\zeta }}_p\left(t\right)\hfill & =-{\partial }_1{\partial }_{1}H\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right){\overline{\zeta }}_q\left(t\right)\hfill & \hfill \\ \hfill & -{\partial }_2{\partial }_{1}H\left(t,\overline{q}\left(t\right),\overline{p}\left(t\right)\right){\overline{\zeta }}_p\left(t\right).\hfill & \left(6.76\right)\hfill \end{array}$$