Normal bundle to fibers of a rational morphism

Question

Let $f:X\dashrightarrow C$ be a rational fibration from a 3-dimensional variety $X$ to a curve $C$ such that generic fiber is smooth and different fibers intrsect in smooth curves. Take $S$ to be a smooth fiber. Let $D$ be the sum of intersections of $S$ with all other fibers. It is an effective divisor on $S$.

Is it true that $N_{S/X}\cong \mathcal{O}_S(D)$?

If this is not true, maybe something weaker is true. For instance, is $H^0(N_{S/X})$ is always nonzero? Maybe it is even true that locally fibers are deformations of each other? Is it at least true that $c_1(N_{S/X})=D$?

Any other information about this normal bundle is also very interesting.

Edit(thanks to @potentially dense): Probably this was unclear in the first version of the question. By fiber of a rational morphism I mean the closure of a fiber of the regular morphism $f:U\to C$ where $U$ is the domain of definition of $f$.

Answer 1

If $C$ is not rational and $X$ is smooth, the map $f$ is a morphism. Indeed, you can view $f$ as a map into the Jacobian $J(C)$ and a map from a smooth variety to an abelian variety is always a morphism.

So if $C$ is not rational, the normal bundle to a fiber is trivial.

If $C={\mathbb P^1}$ then $f$ is given by a $1$-dimensional linear system $|S|$ and the normal bundle to an element $S\in |S|$ is ${\mathcal O}_S(S)$.

(I am assuming that $f$ has connected fibers; you can always reduce to this case by considering the Stein factorization)