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I have a few questions on how to use Lie series as a canonical transformation, which are widely used in perturbation theory (celestial mechanics).

I know that these series are related to a Taylor expansion in $\varepsilon$ around the origin, but I struggle to reproduce the whole story up to the variable transformations given above. I'll give a summary of this process as given by Sections 5.2 and 5.3 of the book "Canonical Perturbation Theories: Degenerate Systems and Resonance" by Ferraz-Mello (2007).

Lie Series


Let's start by considering $(p,q)$ which are sets of $2N$ canonically conjugated variables which depend on a time-like parameter $\lambda$ and are related by the Lie generating function $W(p,q)$, that is

\begin{equation} \frac{dq_i}{d\lambda} = \frac{dW}{dp_i}\ \ ;\ \ \frac{dp_i}{d\lambda} = -\frac{dW}{dq_i},\ \ \ \ \ \ \ (i=1,...,N) \end{equation}

Now, let's consider the Taylor series of $(p,q)$ around $\lambda=0$

\begin{equation} q_i(\lambda) = \sum_{n=0}^\infty\frac{\lambda^n}{n!}\frac{d^nq_i}{d\lambda^n}\Big\rvert_{\lambda=0} \end{equation}

\begin{equation} p_i(\lambda) = \sum_{n=0}^\infty\frac{\lambda^n}{n!}\frac{d^np_i}{d\lambda^n}\Big\rvert_{\lambda=0} \end{equation}

Let's consider the Poisson bracket of any function $f(q(\lambda),p(\lambda))$ (which does not depend on $\lambda$ explicitly) with $W$

\begin{equation} \frac{df}{d\lambda}=\sum_{i=0}^{N}\frac{df}{dq_i}\frac{dq_i}{d\lambda} + \frac{df}{dp_i}\frac{dp_i}{d\lambda} = \sum_{i=0}^{N}\frac{df}{dq_i}\frac{dW}{dp_i} - \frac{df}{dp_i}\frac{dW}{dq_i} = \{f,W\} = D_W(f) \end{equation}

where $D_W(f)$ is defined as an operator known as the Lie derivative of f generated by W.

Inputing $\frac{df}{d\lambda}$ instead of $f$ in the previous equation gives us

\begin{equation} \frac{d^2f}{d\lambda^2} = \{\{f,W\},W\}=D_W(D_W(f))=D_W^2(f) \end{equation}

And so on

\begin{equation} \frac{d^nf}{d\lambda^n} = D_W^n(f) \end{equation}

If $f$ takes the very simple form of the variables $q_i,p_i$ we get an expression for its derivatives

\begin{equation} \frac{d^nq_i}{d\lambda^n} = D_W^n(q_i)\ \ ;\ \ \frac{d^np_i}{d\lambda^n} = D_W^n(p_i) \end{equation}

which we can use in their Taylor representations given above

\begin{equation} q_i(\lambda) = \sum_{n=0}^\infty\frac{\lambda^n}{n!}D_W^n(q_i)\Big\rvert_{\lambda=0} = E_Wq_i \end{equation}

\begin{equation} p_i(\lambda) = \sum_{n=0}^\infty\frac{\lambda^n}{n!}D_W^n(p_i)\Big\rvert_{\lambda=0} = E_Wp_i \end{equation}

where $E_Wq_i$ and $E_Wp_i$ are the Lie series representation of $(p,q)$ around $\lambda=0$.

Noting that the left-hand side of the previous equation depends on $(p,q)$ but the right hand side depends on $(p_0,q_0)=(p(0),q(0))$, we start to see where the canonical transformation to new variables is going.

In order to note that the right-hand side depends on these new variables, Ferraz-Mello rewrites the previous equation as follows:

\begin{equation} q_i = E_{W^*}q_i^* \end{equation}

\begin{equation} p_i = E_{W^*}p_i^* \end{equation}

where $(p^*,q^*)=(p_0,q_0)$ and $W^*=W(p^*,q^*)$.

It can be shown that this procedure can be extended for any function $f=f(p,q)$ (commutation theorem), giving

\begin{equation} f = E_{W^*}f(p^*,q^*) \end{equation}

Hori's method


Everything said is frequently applied in Hori's averaging method. Ferraz-Mello does aswell in Chapter 6, where we have a Hamiltonian $H=H(J,\theta)$ which depends on "old-variables" and we seek to transform it into $H^*=H^*(J^*,\theta^*)$.

The Lie series given above are known to be canonical and are proposed as transformations, that is

\begin{equation} \theta_i = E_{W^*}\theta_i^* \end{equation}

\begin{equation} J_i = E_{W^*}J_i^* \end{equation}

\begin{equation} H = E_{W^*}H(J^*,\theta^*) \end{equation}

which is actually a tranformation from new to old variables and not the other way around, but this way ends up being of practical use.

Questions


  1. Is the derivation given above (which uses Taylor series to arrive at Lie series) correct?

  2. With respect to which variables are the canonical equations with the generating function $W$ defined in Hori's method? Hori himself (and other sources) defines the generating function with respect to the new variables ($(J^*,\theta^*)$ in our case), although the explanation given above requires the relation with $W$ to be with the old variables.

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  1. Is the derivation given above (which uses Taylor series to arrive at Lie series) correct?

Yes, it is (unless there is some typo I am missing).

With respect to which variables are the canonical equations with the generating function W defined in Hori's method? Hori himself (and other sources) defines the generating function with respect to the new variables ($(J^*,θ^*)$ in our case), although the explanation given above requires the relation with $W$ to be with the old variables.

What you (or the text) calls Lie generating function $W(q,p)$ is actually a Hamiltonian function. The idea is that a Hamiltonian function $W(q,p)$ gives rise to a Hamiltonian vector field $J\nabla\, W_{(q,p)}$ whose phase flow $(q_{\lambda}, p_{\lambda}) = \Phi^{\lambda}_{W}(q_0,p_0)$ is a symplectomorphism, i.e. it is a canonical change of variables from $(q_0,p_0)$ to $(q_{\lambda}, p_{\lambda})$ and vise versa!

So basically, this one parameter family of canonical transformations $(q_{\lambda}, p_{\lambda}) = \Phi^{\lambda}_{W}(q_0, p_0)$ can be written in terms of series as follows (I have paraphrased a bit the expansion: one can think of $q$ and $p$ as $n-$vectors, and $D_W(q)$ and $D_W(q)$ are simply applied to these vectors component-wise):

\begin{align} q_{\lambda} \,=\, \sum_{i=0}^{\infty} \frac{\lambda^n}{n!} [D^n_{W}(q)]\big(q_0, p_0\big)\\ p_{\lambda} \,=\, \sum_{i=0}^{\infty} \frac{\lambda^n}{n!} [D^n_{W}(p)]\big(q_0, p_0\big)\\ \end{align}

Analogously, since $(q_{\lambda}, p_{\lambda}) = \Phi^{\lambda}_{W}(q_0, p_0)$ is the phase flow of the Hamiltonian vector field generated by $W(q, p)$ the inverse canonical transformation is then $(q_0,p_0) = \Phi^{-\lambda}_{W}(q_{\lambda}, p_{\lambda})$ and can be expanded as

\begin{align} q_0 \,=\, \sum_{i=0}^{\infty} \frac{(-\lambda)^n}{n!} [D^n_{W}(q)]\big(q_{\lambda}, p_{\lambda}\big)\\ p_0 \,=\, \sum_{i=0}^{\infty} \frac{(-\lambda)^n}{n!} [D^n_{W}(p)]\big(q_{\lambda}, p_{\lambda}\big)\\ \end{align}

This is simply because $\Phi^{\lambda}_{W} \circ \Phi^{-\lambda}_{W} = \text{id}$ and $\Phi^{-\lambda}_{W} \circ \Phi^{\lambda}_{W} =\text{id}$.

Since it is more natural to pull back functions than to push them forward, it is easier to formulate the change of variables of some Hamiltonian $H$ as a canonical transformation from the new coordinates to the old. But as you can see, there is not much difference in the underlying philosophy of how you expand the canonical transformation and its inverse. Both of them are embedded in the same phase flow of some Hamiltonian $W$. What I am trying to say is that it is more convenient to know $H(q, p)$ and to know the canonical transformation $(q, p) = \Phi(Q, P)$, which immediately allows you to write the Hamiltonian in new coordinates:

$$h(Q, P) = H\big(\Phi(Q,P)\big)$$

I guess the moral of the story is that $W$ encodes both directions of the generated canonical transformation.

As a side note, my impression is that in perturbation theory we know the Hamiltonian we are given in the original variables, we know the Hamiltonian we want to obtain in the new variables, and the goal is to find the Hamiltonian that generates the canonical change of variables that transforms the former Hamiltonian into the latter.

For examples something like this: we start with a Hamitlonian $$H = H_0 + \varepsilon H_1 + O(\varepsilon^2)$$ where $\varepsilon$ is a small parameter, and we want to kill the $\varepsilon H_1$ term by a canonical change of variables, close to identity, embedded in the phase flow of some Hamiltonain $W$. So we look for a $W$ that generates the floe $\Phi^{\varepsilon}_W$ such that $$H \circ \Phi^{\varepsilon}_W \,=\, H_0\,+\, O(\varepsilon^2) $$ or in more detail: $$ H_0 \circ \Phi^{\varepsilon}_W \,+\, \varepsilon H_1 \circ \Phi^{\varepsilon}_W \,+\, O(\varepsilon^2) \,=\, H_0\,+\, O(\varepsilon^2) $$

By the expansion property, \begin{align} H_0 \circ \Phi^{\varepsilon}_W \,&=\, H_0 \,+\, \varepsilon \{H_0, W\} \,+\, O(\varepsilon^2) \\ \varepsilon \, H_1 \circ \Phi^{\varepsilon}_W \,&=\, \varepsilon \,H_1 \,+\, \varepsilon^2 \{H_1, W\} \,+\, O(\varepsilon^3) \,=\, \varepsilon \,H_1 \,+\, O(\varepsilon^2) \\ \end{align}

so the equation above becomes $$H_0 \,+\, \varepsilon \{H_0, W\}\,+\, \varepsilon H_1 \,+\, O(\varepsilon^2) \,=\, H_0\,+\, O(\varepsilon^2) $$ which yields $$H_0 \,+\, \varepsilon \Big(\,\{H_0, W\}\,+\, H_1 \Big) \,+\, O(\varepsilon^2) \,=\, H_0\,+\, O(\varepsilon^2) $$ which leads us to the conlusion that $W$ must be chosen so that

$$\{H_0, W\}\,+\, H_1 = 0$$

This is a partial differential equation for the unknown function $W$, because we know $H_0$ and $H_1$. Often, $H_0$ and $H_1$ are some kind of functions polynomial in the $p$ variables and Fourier trigonometric terms in the $q$ variables, so one can solve the equation for $W$ in terms of polynomial times Fourier series expansion of $W$ by comparing Fourier coefficients. The canonical transformation that achieves the removal of the $\varepsilon H_1$ term can be written as $$\Phi^{\varepsilon}_W \,=\, \text{id} \,+\, \varepsilon\,\{\,\cdot\, , W\}$$ mapping the new variables to the old ones (we can drop the $O(\varepsilon^2)$ terms, as we drop them in the Hamiltonains too). This is convenient because, know, in the new variables, $\varepsilon^2$ is much smaller and can be ignored in $H_0 + O(\varepsilon^2)$, and since often we can solve the Hamiltonian system generated by $H_0$, we can transform the solutions of $H_0$, which are in the new coordinates, by mapping them back via $ \text{id} \,+\, \varepsilon\,\{\,\cdot\, , W\}$ into the old coordinates, and these are solutions of the hamiltonain system generated by $H_0 + \varepsilon H_1$ where the $\varepsilon^2$ terms have been ignored.

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  • $\begingroup$ Thank you for the detailed answer. I have one follow-up for now: When going back and forth between the phase flow and its inverse operator, does the generating function W stay the same? Because in the first case the Lie derivative involves derivatives with respect to the old variables, and W=W(p,q), but if we're going the other way around, are the derivatives involved now with respect to the new variables (q0,p0)? Is W now W=W(p0,q0)? In that case, is it a simple evaluation at (q,p)=(q0,p0)? Or is there a more complicated transformation of W? $\endgroup$ Aug 10, 2023 at 15:45
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    $\begingroup$ @MatíasCerioni The Hamiltonian function $W$ stays the same, because the Hamiltonian is conserved under its phase flow , i.e. $W\circ \Phi^{\lambda}_W = W$. Consequently, the series expansion is always the same, no matter what initial condition we choose, i..e what variables we start from. The only thing that changes are the labels of the variables. That is why I wrote both expansions with the variables written explicitly, to emphasize that. $\endgroup$ Aug 10, 2023 at 16:28
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    $\begingroup$ @MatíasCerioni When going back and forth between the phase flow and its inverse, the generating Hamiltonian function W stay the same. If we are going from the old variables $(q_0,p_0)$ to new variables $(q_{\lambda}, p_{\lambda})$, then $W=W(q_0,p_0)$ and the derivatives are with respect to $(q_0,p_0)$. If we're going from the new variables $(q_{\lambda}, p_{\lambda})$ to the old ones $(q_0,p_0)$ then $W=W(q_{\lambda}, p_{\lambda})$ and the derivatives are with respect to $(q_{\lambda}, p_{\lambda})$. It is a simple evaluation. No complicated transformation of W. $\endgroup$ Aug 11, 2023 at 13:43
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    $\begingroup$ @MatíasCerioni The same Hamiltonian $W(q,p)$ works for both directions: from $(q_{\lambda}, p_{\lambda})$ to $(q_{0}, p_{0})$ and from $(q_{0}, p_{0})$ to $(q_{\lambda}, p_{\lambda})$. $\endgroup$ Aug 24, 2023 at 23:15
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    $\begingroup$ @MatíasCerioni Yes, you are correct. This comes from the notion of a phase flow of a time-homogeneous vector field in general (and phase flow of a time-homogeneous Hamiltonian vector field in particular, as a special case). Maybe it is a good idea to introduce yourself to the notion of a phase flow of a vector field and even to go through some explicit examples. $\endgroup$ Sep 5, 2023 at 14:02

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