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Let $\Phi: G \times M \rightarrow M$ be a group action on a symplectic manifold $M$ and $G$ be a Lie group. Furthermore, $x$ is a solution of the Hamilton equation $\dot{x}(t) = X_H(x(t))$ and for a any $t$ there is a $g(t) \in G_x:=\{g \in G; Ad^*_h(x)=x\}$ (here $x \in \mathfrak{g}^*$) such that $x(t) = \Phi(g(t),d(t)).$

Now this $d:I \rightarrow M$ is another path in $M$ that is not necessarily Hamiltonian, but can be transformed into one under the group action, where the group elements can be taken from $G_x.$

In this situation I found that $d$ already satisfies the Hamilton equation

$$\dot{d}(t) + Z_{\zeta(t)}(d(t))= X_H(d(t)),$$

where $Z_{\zeta(t)}(p) := \frac{d}{dt}|_{t=0} \Phi(e^{t \zeta(t)},p)$ and $\zeta(t):=dL_{g(t)^{-1}}\dot{g}(t).$

So actually $d$ satisfies a Hamiltonian equation with an additional term. Does anybody know how this result can be derived?

It should not be too hard, as we already have a strong relationship between $x$ and $d$, but I don't really see how we can get the equation modified Hamilton's equation shown above.

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I'm afraid I'm not entirely clear on what you are asking - you say you "found" a result, but then ask how it can be derived. Here's a comment that might be helpful.

In this situation I found that $d$ already satisfies the Hamiltonian equation $$\dot{d}(t)+Z_{\zeta(t)}(d(t)) = X_H(d(t)),$$

From your hypothesis, you should get $$ \dot{d}(t)+Z_{\zeta(t)}(d(t)) = d\Phi_{g(t)^{-1}}(X_H(x(t))). $$ Assuming $\Phi$ is a symplectic action, we have that $d\Phi_{g^{-1}}\cdot X_{H}(x) = X_{H\circ \Phi_g}(d)$. If $H$ is $G$-invariant, this gives the result quoted above.

Beyond that, maybe you could clarify your question (or give some context for it)?

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