TITLE: Why is a gradient field a special case of a vector field?
QUESTION [8 upvotes]: My calculus manual suggests a gradient field is just a special case of a vector field. That implies that there are vector fields that there are not gradient fields. The gradient field is composted of a vector and each $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ component (using 3 dimensions) is multiplied by a scalar that is a partial derivative. Is this because the scalar may be of the form that there is no antiderivative of? Perhaps of a function that can not be differentiated? Maybe a vector field that has sharp turns but I can't come up with any that I can't find a derivative for. Does anybody have any so I can get some intuition of the problem?

REPLY [2 votes]: General vector fields can generate flows and circulations. A gradient fields and only gradient fields (under some additional regularities) always generate circulations that amount to zero. Try to compute the circulation of a gradient field in a plane around an elementary rectangle centered at any point whose sides are parallel to the coordinate axes. You will get an elementary circulation that is zero simply from the fact that mixed second order partial derivatives of the scalar field does not depend on the order of derivation. Notice that the density of such elementary circulations (that is, the curl of the gradient field in every point) is also null. 
So gradient fields and only gradient fields (under additional regularities) have curl identically equals to zero.
You can also see that there are fields whose flows (and elementary flow density in every point, that is their divergence) always amount to zero.