# Mass matrix in underdamped Langevin dynamics and convergence

I have two doubts regarding the underdamped Langevin dynamic.

1. The first one is that I noticed that the underdamped Langevin dynamic can be written in the following two ways:

\begin{align}\label{eq:ct_underdamped_langevin} \text{d}\mathbf{v}_t &= -\nabla U(\mathbf{x}_t) - \gamma \mathbf{v}_t\text{d}t + \sqrt{2 \gamma \tau} \mathbf{M}^{\frac{1}{2}}\text{d}\mathbf{W},\\ \text{d}\mathbf{x}_t &= \mathbf{M}^{-1}\mathbf{v}_t \text{d}t,\nonumber\end{align}

and

\begin{align} \text{d}\mathbf{v}_t &= -\nabla U(\mathbf{x}_t) - \mathbf{M}^{-1}\gamma \mathbf{v}_t\text{d}t + \sqrt{2 \gamma \tau}\text{d}\mathbf{W},\\ \text{d}\mathbf{x}_t &= \mathbf{M}^{-1}\mathbf{v}_t \text{d}t,\nonumber\end{align}

The difference between both cases is where the mass matrix $$\mathbf{M}$$ is written: in the first one is considered in the random term, while in the second one in the drift part Ornstein–Uhlenbeck part.

I think that the second case can be obtained easily from the second-order differential equation, by defining the first order derivative proportional to $$\mathbf{M}^{-1}\mathbf{v}_t$$. However, I can't see how to obtain the first case. My guess is that the Fokker-Planck operator is invariant under this change, so everything works in the same way. But I couldn't find the right answer.

1. The second one is how to prove the convergence to the stationary distribution. My understand is that one need to show that the gibbs distribution lies in the kernel of the Fokker-Planck operator. Of course this will not give a rate of convergence to the stationary regime, but does this is sufficient to say that the SDE will converge to the gibbs distribution?