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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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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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