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I am trying to understand symplectic transformations. Assume that $H(q,p)$ is a Hamiltonian and the corresponding Hamiltonian equations are given as,

\begin{cases} \dot q = \dfrac{\partial H}{\partial q}, \\[2ex] \dot p = -\dfrac{\partial H}{\partial p}. \end{cases}

Now lets assume that $z = [p;q]$ then

\begin{split} \dot z = \mathbb{J}\frac{\partial H}{\partial z}, \qquad \mathbb{J}= \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}. \end{split}

Now let $\phi$ be a transformation such that $\tilde z = \phi(z)$. Then \begin{split} \dot{\tilde z} = \frac{\partial \phi }{\partial z} \dot z = \frac{\partial \phi }{\partial z} \mathbb{J}\frac{\partial H}{\partial z}\stackrel{?}{=} \frac{\partial \phi }{\partial z} \mathbb{J} \frac{\partial \phi }{\partial z}^T \frac{\partial H}{\partial \tilde z}. \end{split}

I do not fully understand the final step in the previous equation. If that step is correct then obviously $\phi$ is symplectic if and only if \begin{split} \frac{\partial \phi }{\partial z} \mathbb{J} \frac{\partial \phi }{\partial z}^T = \mathbb{J}. \end{split}

How can we know that $\frac{\partial}{\partial z} = \frac{\partial \phi }{\partial z}^T \frac{\partial}{\partial \tilde z}$ when only thing we know is $\tilde z = \phi(z)$?

Xander Henderson
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user139698
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1 Answers1

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It's just the chain rule. To avoid confusion, write $H(z) = G(\tilde{z}) = G(\phi(z))$. Then $$ \frac{\partial H}{\partial z_k}(z) = \sum_i \frac{\partial G}{\partial \tilde{z}_i}(\phi(z)) \, \frac{\partial \phi_i}{\partial z_k}(z) . $$

Hans Lundmark
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  • Thank you very much, it seems so simple now. – user139698 Mar 03 '15 at 16:23
  • can you help me with an answer regarding the transpose. I don't know from where appears that "T" in the equation: \begin{split} \frac{\partial \phi }{\partial z} \mathbb{J} \frac{\partial \phi }{\partial z}^T = \mathbb{J} \end{split} thanks! – Iuli Sep 10 '15 at 13:56
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    @luli: It's because the last factor is $\partial \phi_i/\partial z_k$ and not $\partial \phi_k/\partial z_i$ (which is what you would get without transposing). – Hans Lundmark Sep 11 '15 at 05:47
  • @HansLundmark I have one more question. I think I don't understand very well what is about the transpose of the jacobian. I want to ask you if you can follow the following link: http://www.damtp.cam.ac.uk/user/tong/dynamics/clas.pdf (page 100). Why the relation (4.93) and (4.94) are equivalent? In equation (4.93) $\frac{\partial H}{\partial y_l}$ has't same range with the others terms in that equation and the matrix product shouldn't be possible. In equation (4.94) it is a rearrangement. Why this is possible? Because the product of matrix is not commutative.. How does the transpose appear? – Iuli Sep 12 '15 at 11:43
  • @luli: In (4.93), you can rearrange, because there it's just multiplication of numbers. And if you swap the last two factors there, you get something which is clearly equal to the matrix product (4.94) (just use the definition of matrix product, $(AB){ij} = \sum_k A{ik} B_{kj}$). – Hans Lundmark Sep 12 '15 at 13:56