Appendix

Derivation of learning equations

The equations for the system without temperature coupling described by the variables \(\mathbf{Y}\) are:

\[ -h \frac{dw^{i}}{dt} = k\{w^{i}, H\} = k\frac{\partial H}{\partial Y_{i}} \qquad -\frac{dY_{i}}{dt} = k\{Y_{i}, H \} = -k\frac{\partial H}{\partial w^{i}} \]

\[ -\frac{dU}{dt} = k\frac{\partial H}{\partial \beta} \qquad -h\frac{d\beta}{dt} = -k\frac{\partial H}{\partial U} \]

Let us now see how to derive the equations with the coupling \(\mathbf{Y}=\beta \mathbf{X}\). From the Leibniz rule for the bracket containing the coordinates of \(\mathbf{Y}\), given a Hamiltonian \(H\):

\[ \{Y_{i}, H\} = \{\beta X_{i}, H\} = \beta\{X_{i}, H\} + X_{i}\{\beta, H\} \]

Given that we know:

\[ \{Y_{i}, H \} = -\frac{\partial H}{\partial w^{i}} \qquad \{\beta, H\} = -\frac{\partial H}{\partial U} \]

We substitute these values to obtain the Poisson brackets of the system in terms of the momenta \(\mathbf{X}\):

\[ -\frac{\partial H}{\partial w^{i}} = \beta \{X_{i}, H\} - X_{i} \frac{\partial H}{\partial U} \Rightarrow \{X_{i}, H\} = -\frac{1}{\beta}\frac{\partial H}{\partial w^{i}} + \frac{X_{i}}{\beta} \frac{\partial H}{\partial U} \]

And we recover the proposed H-equation:

\[ -h\frac{dX_{i}}{dt} = k\{X_{i}, H\} = -\frac{k}{\beta}\frac{\partial H}{\partial w^{i}} + \frac{kX_{i}}{\beta} \frac{\partial H}{\partial U} \]

In order to reconcile the evolution of the internal energy with the proposed evolution, we recall the following thermodynamic identity:

\[ \left(\frac{\partial H}{\partial \beta} \right)_{\mathbf{Y}} = \left(\frac{\partial H}{\partial \beta} \right)_{\mathbf{X}} + \sum_{j} \left(\frac{\partial H}{\partial X_{j}} \right)_{\beta, X_{i \neq j}} \left(\frac{\partial X_{j}}{\partial \beta} \right)_{\mathbf{Y}} \]

The subscripts indicate which parameters are considered constant. This notation is common in thermodynamics and was omitted from the formalism, as we specified at each step the symplectic form upon which we are working.

Let us now consider how to find the partial derivative of the momentum with respect to the parameter \(\beta\). From the coupling relation:

\[ X_{i} = \frac{Y_{i}}{\beta} \Rightarrow \frac{\partial X_{i}}{\partial \beta} = -\frac{1}{\beta^2} Y_{i} = -\frac{1}{\beta} X_{i} \]

We can then obtain the evolution of the internal energy:

\[ -h\frac{dU}{dt} = k\frac{\partial H}{\partial \beta} - \frac{1}{\beta} \sum_{j} \frac{\partial H}{\partial X_{j}}X_{j} \]

For the weight-momentum brackets let us note that:

\[ \frac{\partial H}{\partial Y_{i}} = \frac{\partial H}{\partial X_{i}} \frac{\partial X_{i}}{\partial Y_{i}} = \frac{1}{\beta} \frac{\partial H}{\partial X_{i}} \]

Then:

\[ -h\frac{dw^{i}}{dt} = \frac{k}{\beta} \frac{\partial H}{\partial X_{i}} \]

Finally, the equation containing the changes of the \(H\) in the direction \(\partial/\partial U\) is identical, given that the variable U is not coupled in the same way as \(\beta\)."