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I am very new to the subject, so please forgive my very naïf question. I learned that there are some non-hamiltonian systems which can become hamiltonian, just by a change of coordinates. I was given the Susceptible-Infected-Removed (SIR) model as an example: \begin{cases} \frac{dS}{dt} = - \alpha SI \\ \frac{dI}{dt} = \alpha SI - \beta I \\ \frac{dR}{dt} = \beta I \end{cases}

with $\alpha$ and $\beta$ being real parameters.

This is clearly non-hamiltonian because it is associated with a vector field which has non zero divergence. However, by choosing $x= log(S)$ and $y= log(I)$, where $S$ and $I$ are the susceptible and infected respectively, the system becomes hamiltonian.

What I found really strange about this result is that I'm used at seeing quantities being preserved by a change of coordinates, while here the property of the system being hamiltonian clearly isn't intrisinc: is there something deeply conceptual that I'm missing about hamiltonian mechanics?

Qmechanic
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Nope
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    Since you mention divergence and invariant quantities, keep in mind how divergence transforms under change of coordinates; there is a volume element involved. So if you work backwards, the associated vector field is divergence free, you just have to calculate the divergence with respect to right volume element (the one for which x,y are euclidean coordinates, which is (SI)^-1). – mlk May 12 '20 at 13:14

1 Answers1

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  1. Note that the divergence ${\rm div}_{\rho}X=\rho^{-1}\partial_i(\rho X^i)$ of a vector field $X=X^i\partial_i$ in general depends on a density $\rho$, cf. above comment by user mlk. The possibility of a non-trivial $\rho$ makes it more difficult to identify which 1st-order systems are potentially Hamiltonian and which are not.

  2. A 3D phase space can never have a non-degenerate symplectic structure, but if we ignore the last coordinate $R(t)=\beta\int^t\!dt^{\prime} I(t^{\prime})$, then we have a 2D phase space, which always has a (local) Hamiltonian formulation, cf. this Phys.SE post.

  3. Concretely, the SIR-model $$\begin{align} \dot{S}~=~& - \alpha SI~=~\{S,H\}, \cr \dot{I}~=~& \alpha SI - \beta I~=~\{I,H\},\cr \dot{R}~=~& \beta I~=~\{R,H\},\end{align} \tag{1}$$ has non-canonical, degenerate Poisson structure $$\begin{align} \{S,I\}~=~&SI, \cr \{I,R\}~=~&\frac{\beta}{\alpha}I, \cr \{S,R\}~=~&0,\end{align} \tag{2} $$ and Hamiltonian $$ H~=~\beta \ln S -\alpha (S+I) .\tag{3} $$

  4. If we follow OP's suggestion to change coordinates $$\begin{align} s~:=~&\ln S, \cr i~:=~&\ln I , \cr r~:=~&R, \end{align}\tag{1'}$$ then the fundamental Poisson brackets (2) become constant $$ \begin{align} \{s,i\}~=~&1, \cr \{i,r\}~=~&\frac{\beta}{\alpha}, \cr \{s,r\}~=~&0, \end{align}\tag{2'} $$ and it becomes obvious that the Jacobi identity is satisfied as it should. The Hamiltonian reads $$ H~=~s -\alpha (e^s+e^i) \tag{3'} $$ in the new coordinates.

Qmechanic
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