Let $(K, <)$ be an order field, can I define the order "<" in $K$ ?
I know that $K \models 0<a \;$ if and only if there is $b$ in the real closure of $K$ such that $b*b = a$. Can I "interpret" the real closure of $K$ in $K$?
A slightly more convoluted solution than Hagen's:
Consider the field $K=\mathbb Q(t)$, any interpretation of $t$ as a transcendental real number will define an embedding of $\Bbb Q(t)$ into $\Bbb R$, and by that an ordering and real-closure. If we map $t=\pi$ and $t=-e$ we get that $t>0$ and $t<0$ respectively.
Hence, definability of the real-closure or the order is impossible.
Consider $K=\mathbb Q[X]/(X^2-2)$. There are two orders on $K$, depending on if we map $X\mapsto\sqrt 2 $ or $X\mapsto-\sqrt 2$. Hence $<$ cannot be recovered from the field $K$ alone.