Two norms $||·||_1$ and $||·||_2$ over a space $X$ are equivalent iff there exist positive $c, C\in \mathbb{R}$ such that for all $x\in X$ $c||x||_1\leq ||x||_2\leq C||x||_1$.
Two metrics $d_1$ and $d_2$ over a space $X$ are strongly equivalent iff there exist positive $c, C\in \mathbb{R}$ such that for all $x,y\in X$ $cd_1(x,y)\leq d_2(x,y)\leq Cd_1(x,y)$.
Is there a similar notion for total (pre)orders over rings? In other words, is there a definition along the lines of ''two orders $\leq_1$ and $\leq_2$ are equivalent (or compatible or friends or something else) iff there exist $c, C\in R$, $0<_1 c$ and $0<_2 C$, such that for all $x\in R$ $c x \leq_1 x \leq_2 Cx$?''.
I am interested in comparing two partial preorders $\leq_1$ and $\leq_2$ on a ring $R$ induced by two order-preserving homomorphisms $f, g: S \rightarrow R$, where $S$ is a totally ordered ring. On $Im(f)\subseteq R$ the relation $\leq_1$ is defined as $f(x)\leq_1f(y)$ iff $x\leq y$; similarly on $Im(g)$ one has $g(x)\leq_2g(y)$ iff $x\leq y$. In this case the two preorders are not equivalent if for instance $f(x)=0$ but $g(x)\ne0$.
This question has come up in an original research.