In our mechanics class, we began to discuss rigid movement. Let me begin with our definitions.
Definition 1: A map $f: D \subset \mathbb{E}^3 \to \mathbb{E}^3$ is rigid, if it preserves distance between points, i.e. for any two points $P_1, P_2 \in D$, we have $|P_1 - P_2| = |f(P_1) - f(P_2)|$.
Definition 2: Rigid movement is a one-parametric family of rigid maps $f_t$, $t \in I \subset \mathbb{R}$, which includes identity.
Then we said (very informally, since we don't know anything about measure theory really), that mass is some measure $dm$, which is absolutely continuous with regard to Lebesgue (volume) measure $dV$. By Radon-Nikodyn theorem, there exists a positive measure $\rho$, such that $dm = \rho dV$.
Question: We then stated that for rigid motion, the law of conservation of mass follows from the definition of mass. Precisely; for any part $U(t)$ of rigid body $B(t)$, the mass of $U(t)$ is constant. I've been thinking a lot about it, but have absolutely no idea how to deduce it. Is it something that follows from the definitions, but would require a lot of knowledge about measure theory and we just mentioned the result, or is it something that's so obvious that we just skipped the proof for that reason?
Perhaps we can just say that rigid transformations preserve volume, so $\frac{d}{dt} dV = 0$ and therefore $\frac{d}{dt} dm = \frac{d}{dt} \rho dV = 0$? Or is that not ok?
