# Hamilton's equations in terms of initial conditions

I'm trying to understand the way that Hamilton's equations have been written in this paper. It looks very similar to the usual vector/matrix form of Hamilton's equations, but there is a difference.

$$\frac{{\bf dZ}(t)}{dt} ~=~ J \frac{\partial H({\bf Z(t)})}{\partial z}, \qquad\qquad{\bf Z(0)} ~=~ z,$$

where $J$ is the block matrix $((0,1),(-1,0))$. $\bf{Z}(t)$ represents the point in phase space (positions and momenta). The part I don't understand is that the derivative of $H$ is taken with respect to the initial value (lowercase $z$) of ${\bf Z(t)}$. How does this form follow from the usual way of writing Hamilton's equations in vector form?

Note that my question is not about the vector notation for Hamilton's equations, which can be found in any introductory textbook on classical mechanics. I am specifically asking about the derivatives with respect to $z$, the initial values of ${\bf Z(t)}$.

Edit: I could not find an arxiv version of the paper, but it seems to be available as the first publication on one of the authors' websites, under the section "Coarse-graining with proper dynamics." The equation in question is the first one in section II of the paper.

• Can you please link the arXiv version, if possible, so that we can see the equations in context? – DaniH Mar 19 '12 at 20:06

The choice of letters in the paper is sloppy, and by abuse of notation, the author uses $z$ rather than $z_0$ to denote the initial condition. (Thus you need to replace $z$ everywhere by $z_0$ where it denotes the initial condition to get a onsistent reading. Alternatively, replace the $z$ in $\partial z$ by a capital $Z$.) Primarily, $z$ is the generic argument of $H(z)$; then everything makes sense:

$\partial H/\partial z = [\partial H/\partial p, \partial H/\partial q]$ (as a column vector in the right order). The factor $J$ changes the sign of the $q$ part and thus gives the standard conditions.

Thus the derivatives are not to be taken with respect to the inital values.

This kind of intelligent interpretation of a formula to make sense of it is often needed when reading papers by others. You know that you have the right interpretation once you have one that makes sense. Misprint correction in formulas works by the same principle.

• Thanks, but my question is specifically about the issue of evaluating the derivative with respect to the initial conditions. In other words, how do I get the "usual" Hamilton's equations starting with the ones in the question. – Greg P Mar 19 '12 at 20:23
• If you plug my answer into your formula you get the result, as the more explicit answer by pcr shows. – Arnold Neumaier Mar 19 '12 at 20:49
• In the paper, $z$ represents the initial values, that is, $z = {\bf Z(0)}$. I still do not see how to get these equations involving $z = {\bf Z(0)}$ from the usual Hamilton's equations which involve positions and momenta evaluated at the same time $t$ as on the left hand side. – Greg P Mar 19 '12 at 21:01
• The choice of letters in the paper is sloppy, and by abuse of notation, the author uses $z$ rather than $z_0$ to denote the initial condition. Primarily, $z$ is the generic argument of $H(z)$; then everything makes sense. – Arnold Neumaier Mar 19 '12 at 21:06
• The paper explicitly defines $z = {\bf Z(0)}$, and both $z$ and ${\bf Z}$ are used elsewhere in the paper, so I'm still not sure. – Greg P Mar 19 '12 at 21:21

From $$\dot{Z} = J \frac{\partial H}{\partial z}$$

we can represent Z as (q,p) explicitly

$$\left(\begin{array}{c} \dot{q} \\ \dot{p} \end{array} \right) = \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \left( \begin{array}{c} \partial H /\partial q \\ \partial H/\partial p \end{array}\right) = \left( \begin{array}{c} \partial H /\partial p \\ -\partial H/\partial q \end{array}\right)$$

Edit: Here is an explicit example to support Arnold's argument

Consider a harmonic oscillator (mass = 1)

$$x = x_0 \cos wt + p_0 \sin wt$$ $$p = p_0 \cos wt + - wx_0\sin wt$$ with $$H = \frac12(p^2 + w^2x^2) = \frac12 (p_0^2+w^2x_0^2)$$ Now, you can try

$$\partial H(x,p) / \partial p_0 = p_0$$

which is clearly not $\dot{x}$

• This is the standard vector version of Hamilton's equations. How are these related to the ones in the paper (and the question) which include derivatives with respect to the initial values of p and q? That was my question. – Greg P Mar 19 '12 at 20:24
• @GregP "Note that my question is not about the vector notation for Hamilton's equations, which can be found in any introductory textbook on classical mechanics." Sorry I missed this sentence. – pcr Mar 19 '12 at 21:05
• No, you didn't miss it ;-) I added it after your answer to clarify the question. – Greg P Mar 19 '12 at 21:13