Given a set of equations, show if it is a Hamiltonian system

[peguerosdc]

Homework Statement

Consider the system $\dot q = p$ and $\dot p = -q-\gamma p$.

a) Show that the system of equations CAN'T be expressed in a Hamiltonian form if (q,p) are considered canonical coordinates.
b) Is it possible to write the system in a Hamiltonian form? If so, do it; if not, prove it.

Relevant Equations

$\dot q = p$
$\dot p = -q-\gamma p$

Hi!

So this is my first homework ever of Hamiltonian dynamics and I am struggling with the understanding of the most basic concepts. My lecturer is following Saletan's and Deriglazov's and from what I have read and from my lectures, this is what I think I know. Please let me know if this is correct or not:

A system is called "Hamiltonian" if there exists a function $H(q,p,t)$ such that:

$\dot q = \frac {\partial H} {\partial p}$
$\dot p = -\frac {\partial H} {\partial q}$

Written in the symplectic form:

$\dot {X}^{\mu} = W^{\mu\nu} (x) \frac {\partial H} {\partial X^{\nu}}$

And (this statement I am not sure if it is correct) it takes the following form when it is canonical (is this the matrix only for when we are using canonical coordinates?):

$\begin{pmatrix} \dot {q_i} \\ \dot {p_i} \\ \end{pmatrix} = \begin{pmatrix} 0_{nxn} & 1_{nxn} \\ -1_{nxn} & 0_{nxn} \\ \end{pmatrix} \begin{pmatrix} \frac {\partial H} {\partial q_i} \\ \frac {\partial H} {\partial p_i} \\ \end{pmatrix}$

So, for part "a" I think I need to prove somehow that the given set of equations does not meet the definitions of $\dot {q}$ and $\dot {p}$, but I am not sure how to proceed as I am not given $H$. Also, I don't know how to use the requirement that the set of coordinates is canonical.

Also, from what I understood from the lecture:

For a transformation $(q,p) \rightarrow (Q_i,P_i)$ to be canonical, it has to meet the requirement that: $\{Q_i,P_j\}=\delta_{ij}$ and $\{Q_i,Q_j\}=\{P_i,P_j\}=0$.

Where $\{.,.\}$ are the Poisson brackets:

$\{ f(q_i,p_i), g(q_i,p_i) \} = \frac {\partial f} {\partial q_i} \frac {\partial g} {\partial p_i} - \frac {\partial f} {\partial p_i} \frac {\partial g} {\partial q_i}$

Is this also a requirement for the set of coordinates to be canonical? Then, how can I use it in my conditions for the system to be Hamiltonian?

Thank you!

 

[strangerep]

Consider the system $\dot q = p$ and $\dot p = -q-\gamma p$.

a) Show that the system of equations CAN'T be expressed in a Hamiltonian form if (q,p) are considered canonical coordinates.

A system is called "Hamiltonian" if there exists a function $H(q,p,t)$ such that:

$\dot q = \frac {\partial H} {\partial p}$
$\dot p = -\frac {\partial H} {\partial q}$

I'd approach part (a) on the basis that the variables $q,p$ are independent, i.e., $\partial_q p = 0 = \partial_p q$.

Not sure if this is too big a hint, but... integrate $$ \partial_p H ~=~ \dot q ~=~ p ~,$$ which gives $$H = \frac12 \, p^2 ~+~ f(q) ~,$$ where $f$ is an arbitrary function of $q$. To determine $f$, plug this $H$ into the 2nd Hamilton equation. What do you find?

For part(b), consider the relationship between the PB formula, the Jacobian of the transformation $(q,p) \leftrightarrow (Q,P)$, and the determinant of that Jacobian.

 

[peguerosdc]

To determine $f$, plug this $H$ into the 2nd Hamilton equation. What do you find?

Thanks for your reply! So, after plugging H into the second equation, I got:

$\dot p = - \frac {\partial f} {\partial q}$

But before proceeding with my interpretation of this, I would like to improve my understanding of the theory as that's the main reason I am not sure how to approach this problem if you don't mind

• Does the fact that $(p,q)$ are considered canonical implies that $p$ and $q$ are independent? Then, for non-canonical coordinates, can we have $p = p(q)$, $q = q(p)$?
• What is the sufficient and/or necessary condition to say that a system is "Hamiltonian"?

 

[strangerep]

Thanks for your reply! So, after plugging H into the second equation, I got:
$\dot p = - \frac {\partial f} {\partial q}$

Keep going -- you haven't finished this part. You must either find a suitable $H$, or prove that such an $H$ does not exist.

Does the fact that $(p,q)$ are considered canonical implies that $p$ and $q$ are independent?

Let's take a step back. In the Hamiltonian formulation of mechanics, the primary area is a phase space, with coordinates denoting position $q$ and momentum $p$. All dynamical quantities are expressed as functions of $q$ and $p$.

A priori, $q, p$ are independent -- they only gain an apparent interdependency when we restrict to solutions of the equations of motion (EoM), since the set of such solution curves picks out only a subset of phase space, in general. At this stage, the concept of "canonical" is empty -- the $q,p$ are simply independent variables in phase space.

Choosing a Hamiltonian function $H(q,p)$ defines a (class of) physical system. The solution curves, $\Big(q(t), p(t) \Big)$ of Hamilton's EoM express all possible behaviours of that (class of) system.

A canonical transformation (CT) in phase space is one which maps the solution curves among themselves. IOW, CT's preserve the form of Hamilton's EoM.

When we say that some other variables $Q(q,p)$ and $P(q,p)$ are "canonical", it means the dynamical quantities associated with the system can just as well be expressed in terms of $Q,P$ as $q,p$ (though possibly more conveniently).

What is the sufficient and/or necessary condition to say that a system is "Hamiltonian"?

That there exists an $H(q,p)$ such that the Hamilton EoM are indeed the EoM of the system being considered.

for non-canonical coordinates, can we have $p = p(q)$, $q = q(p)$?

I wouldn't think of it that way. Say rather that new coordinates $Q(q,p)$ and $P(q,p)$ are non-canonical if the transformation from $(q,p)$ to $(Q,P)$ fails to preserve Hamilton's EoM, which is equivalent to saying that the transformation does not, in general, map solutions of the EoM among themselves.

 

[peguerosdc]

Keep going -- you haven't finished this part. You must either find a suitable $H$, or prove that such an $H$ does not exist.

So, $\dot p = - \frac {\partial f} {\partial q}$ but we also know that $\dot p = - q - \gamma p$. Then, by comparison $\frac {\partial f} {\partial q} = q + \gamma p$ but we stated that $f=f(q)$, so we will not be able to find a suitable $f$ as we would need $f=f(q,p)$.

If this reasoning is correct, I would be left only with part b. As for it, I haven't came across a relation between the Jacobian and the PBs so I would need to read some more to avoid commenting something non-sense in here, but then are you suggesting to try to derive (if possible) a transformation that would make the given equations satisfy the Hamilton's equations?
If that's the case, I think that $\dot p \rightarrow \dot q$ and $\dot q \rightarrow \dot p$ would work, but I fear this is exactly what I was talking about when saying "avoid commenting non-sense" as I suspect that transformations can't include $\dot p$ and $\dot q$.

 

[strangerep]

So, $\dot p = - \frac {\partial f} {\partial q}$ but we also know that $\dot p = - q - \gamma p$. Then, by comparison $\frac {\partial f} {\partial q} = q + \gamma p$ but we stated that $f=f(q)$, so we will not be able to find a suitable $f$ as we would need $f=f(q,p)$.

Yes. So you have shown... what?

If this reasoning is correct, I would be left only with part b. As for [part (b)], I haven't came across a relation between the Jacobian and the PBs so I would need to read some more [...]

It's reasonably easy. Do you know what a Jacobian matrix is? If not, try Wikipedia. Then write out the relevant Jacobian matrix here for a generic transformation $q,p \to Q(q,p), P(q,p)$ and calculate its determinant. Notice anything?

are you suggesting to try to derive (if possible) a transformation that would make the given equations satisfy the Hamilton's equations?

Or, as part(b) said, prove that such a transformation does not exist.

If that's the case, I think that $\dot p \rightarrow \dot q$ and $\dot q \rightarrow \dot p$ would work,

That's equivalent to $P = q + \mbox{const},~~ Q = p + \mbox{const}$. Do you still think it would work? Try repeating part(a) for the case when this transformation has been performed on the original EoM. Can you now find a suitable $H$?

 

[peguerosdc]

Yes. So you have shown... what?

That we can't find a function $f$ such that $f = f(q)$ that makes my system satisfy Hamilton's canonical equations because we reached a contradiction defining $f = f(q)$ and then finding that we needed $f = f(q,p)$. So, if this is correct, then there is no $H$ for these equations of motions and thus it is not a Hamiltonian system.

It's reasonably easy. Do you know what a Jacobian matrix is? If not, try Wikipedia. Then write out the relevant Jacobian matrix here for a generic transformation $q,p \to Q(q,p), P(q,p)$ and calculate its determinant. Notice anything?

For a generic transformation, the Jacobian $J$ and its determinant would be:

$J = \begin{pmatrix} \partial _q Q & \partial _p Q \\ \partial _q P & \partial _p P \end{pmatrix} \rightarrow det(J) = \partial_q Q \partial_p P - \partial_p Q \partial_q P = \{Q,P\}$
So the determinant of the Jacobian is the same as the Poisson brackets {Q,P}! Though I don't understand why they are the same and how this is related to the fact of $(Q,P)$ being canonical or not.

Also, I was thinking of another way to approach part "b", though I don't know if it is correct:

So I need to find an H so that I can write:

$\begin{pmatrix} \dot q \\ \dot p \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \partial_q H \\ \partial_p H \\ \end{pmatrix}$

Let's define $W = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . I know W must be antisymmetric and invertible, so I can write it as:

$W = \begin{pmatrix} 0 & f \\ -f & 0 \end{pmatrix}$

Where $f$ could be $f=f(q,p)$. Then I can see that:

$\begin{pmatrix} 0 & -f \\ f & 0 \end{pmatrix} \begin{pmatrix} \dot q \\ \dot p \end{pmatrix} = \begin{pmatrix} \partial_q H \\ \partial_p H \\ \end{pmatrix}$

Where I have just inverted $W$. This means that:

$\partial_q H = f(q,p) (q + \gamma p) \\ \partial_p H = pf(q,p)$

So if I can prove that there is no $f$ that satisfies those relations, then the system can't be put in Hamiltonian form. On the contrary, if I can find such $f$, then I can write the equations of motion in the desired Hamiltonian form and the system would be Hamiltonian.

Is my approach correct? Because if it is, I am stuck at this last step.

 

[strangerep]

[...] there is no $H$ for these equations of motions and thus it is not a Hamiltonian system.

Yes, that's the same conclusion I reached before I responded to your opening post.

In the next bit, I've renamed the Jacobian matrix from "$J$" to "$M$", for reasons that will hopefully become apparent soon.

For a generic transformation, the Jacobian $M$ and its determinant would be:$$
M ~=~
\begin{pmatrix}
\partial _q Q & \partial _p Q \\
\partial _q P & \partial _p P
\end{pmatrix}
\rightarrow
det(M) = \partial_q Q \partial_p P - \partial_p Q \partial_q P = \{Q,P\} ~.
$$
So the determinant of the Jacobian is the same as the Poisson brackets {Q,P}!

Yes.

Though I don't understand why they are the same and how this is related to the fact of $(Q,P)$ being canonical or not.

Your "another way" from your previous post is (sort-of) struggling in the right direction (i.e., trying to find a transformation), but there's a more direct and powerful technique to pursue this.

Recall the matrix version of Hamilton's EoM in your opening post, i.e., $$
\begin{pmatrix} \dot {q_i} \\ \dot {p_i} \\ \end{pmatrix}
~=~
\begin{pmatrix}
0_{nxn} & 1_{nxn} \\
-1_{nxn} & 0_{nxn} \\
\end{pmatrix}
\begin{pmatrix} \frac {\partial H} {\partial q_i} \\ \frac {\partial H} {\partial p_i} \\ \end{pmatrix}
~\equiv~ J \, \begin{pmatrix} \frac {\partial H} {\partial q_i} \\ \frac {\partial H} {\partial p_i} \\ \end{pmatrix}
~, $$where I've named the antisymmetric matrix on the RHS as "$J$" (which is a common choice of name for that matrix -- that's why I renamed your earlier transformation matrix to be "$M$").

Now, can you prove that the transformation represented by $M$ is canonical (i.e., preserves the Hamilton EoM) iff $$M^{-1} J M ~=~ J ~~~~~~~~ (**)$$? (If you have difficulty with this, it should be findable in every textbook on Hamiltonian mechanics.)

When you have proven eqn (**), consider the simplest case that we're dealing with here ($2\times 2$ matrices), and determine what specific condition(s) a $2\times 2$ matrix $M$ must satisfy so that eqn(**) is indeed fulfilled. Then try to arrange the jigsaw pieces I've given you to construct a proof of part(b) by contradiction.

 

[peguerosdc]

Now, can you prove that the transformation represented by $M$ is canonical (i.e., preserves the Hamilton EoM) iff $$M^{-1} J M ~=~ J ~~~~~~~~ (**)$$? (If you have difficulty with this, it should be findable in every textbook on Hamiltonian mechanics.)

When you have proven eqn (**), consider the simplest case that we're dealing with here ($2\times 2$ matrices), and determine what specific condition(s) a $2\times 2$ matrix $M$ must satisfy so that eqn(**) is indeed fulfilled. Then try to arrange the jigsaw pieces I've given you to construct a proof of part(b) by contradiction.

So, I did some research on this and I found equation (**) slightly different on the textbooks as they state that the transformation is canonical if:

$M^T J M = J$

Where $T$ stands for "transpose". If this is the case, this would lead to the following relations for the Poisson brackets for canonical transformations:

$\{Q_i,P_j\} = \delta_{ij} , \{P_i,P_j\} = \{Q_i,Q_j\} = 0$

Which, for the particular case of the 2x2 M, it would require its determinant to equal $1$.

So let me ask two more questions as I would like to understand why this is the way to go (sorry if I am being repetitive, but I think I have some misconceptions from my lectures and what I have read from multiple sources and I am trying to clear those out).

It seems that we are trying to find (or prove it doesn't exist) a canonical transformation such that we can write the following system of equations for a new hamiltonian $K=K(Q,P)$:

$$
\begin{pmatrix} \dot {Q} \\ \dot {P} \\ \end{pmatrix}
~=~
\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}
\begin{pmatrix} \frac {\partial K} {\partial Q} \\ \frac {\partial K} {\partial P} \\ \end{pmatrix}
~\equiv~ J \, \begin{pmatrix} \frac {\partial K} {\partial Q} \\ \frac {\partial K} {\partial P} \\ \end{pmatrix}
~, $$
But:

1. Why are we restricting ourselves to trying to find canonical transformations?
2. (Kind of related to the previous question) Why are we restricting ourselves to the case where $J =\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$ ?

From what I recall, the only requirement for $J$ was that it should be invertible and antisymmetric, so shouldn't we consider the cases when $J_{1,2}$ and $J_{2,1}$ are not $1$? Or when you say:

[...] determine what specific condition(s) a $2\times 2$ matrix $M$ must satisfy so that eqn(**) is indeed fulfilled. Then try to arrange the jigsaw pieces I've given you to construct a proof of part(b) by contradiction.

Do you mean for any arbitrary $J$?

 

[strangerep]

So, I did some research on this and I found equation (**) slightly different on the textbooks as they state that the transformation is canonical if:
$M^T J M = J$
Where $T$ stands for "transpose".

Ah, good. You're awake and have detected my intentional mistake.

[...] It seems that we are trying to find (or prove it doesn't exist) a canonical transformation such that we can write the following system of equations for a new hamiltonian $K=K(Q,P)$:
$$
\begin{pmatrix} \dot {Q} \\ \dot {P} \\ \end{pmatrix}
~=~
\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}
\begin{pmatrix} \frac {\partial K} {\partial Q} \\ \frac {\partial K} {\partial P} \\ \end{pmatrix}
~\equiv~ J \, \begin{pmatrix} \frac {\partial K} {\partial Q} \\ \frac {\partial K} {\partial P} \\ \end{pmatrix}
~, $$
But:
1. Why are we restricting ourselves to trying to find canonical transformations?

Re-read my post #4. Canonical transformations are precisely those which map solutions of the EoM among themselves. I.e., restricting to that type of transformation ensures we're still dealing with the original (class of) physical system, not some other totally unrelated system.

(Kind of related to the previous question) Why are we restricting ourselves to the case where $J =\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$ ?

Because that's the crucial matrix that appears in Hamilton's EoM, (when the latter are expressed in matrix language).

From what I recall, the only requirement for $J$ was that it should be invertible and antisymmetric, so shouldn't we consider the cases when $J_{1,2}$ and $J_{2,1}$ are not $1$? Or when you say:

determine what specific condition(s) a 2×2 matrix M must satisfy so that $$M^T J M ~=~ J$$ is indeed fulfilled.

Do you mean for any arbitrary $J$?

No. We are trying to find transformations that specifically preserve $J$ (i.e., which preserve the Hamilton EoM).

I think maybe your next task should be to figure out why $$M^T J M ~=~ J$$ is the correct equation for preserving the Hamilton EoM. Hint: figure out how the derivative operators $\partial_q$ and $\partial_p$ transform, and also how the differentials $dq$ and $dp$ transform, under a change of variables $(q,p) \to \Big( Q(q,p) , P(q,p)\Big)$.