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Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$ $$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$ where $N = 2 \,dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* ... 4 There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for$x_1$is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ ... 3 As the first question has received sufficient exposition, I would like to make a point with regard to the second one. First thing to understand is that integrability and non-linearity of a system are two different concepts. It is true though that all linear systems in classical mechanics (i.e those that are described by systems of linear equations, be them ... 2 By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous. This is just a formal way of restating ... 2 I would say pressure is better defined by$$ \vec{F} = P \vec{A}. $$Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ... 1 We can start at the relationship: W=-\Delta U, which is work done by a conservative force. The math A (conservative) force F will do this work on an object when doing a displacement \Delta x, and W=F \Delta x. In the general case, the force might be different at different points as the object is moved (the force of gravity is not constant along ... 1 The reasoning is the same as the two-particle system:$$ E_i = E_{1i} + \frac{p_0^2}{2m_1} + E_i' + \frac{p_0^2}{2(M-m_1)}, $$so that$$ E_i - E_{1i} - E_i' = \frac{p_0^2}{2}\left(\frac{1}{m_1} + \frac{1}{M-m_1}\right) = \frac{p_0^2}{2}\frac{M}{m_1(M-m_1)}. $$Therefore$$ \frac{p_0^2}{2m_1} = \frac{M-m_1}{M}(E_i - E_{1i} - E_i').$$1 REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space. The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final ... 1 Yes, area is a vector, which is the normal to the surface. ($\vec{A}=A\vec{n}$)$\vec{F} = -P\vec{A}\$ In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal ...