## Conservative Vector Fields.

### May 28, 2010

Just a little post.  Let’s suppose that we have some vector field $g$, and let’s say that, because we’re so clever, we find that there is some function $f$ such that $\nabla f = g$.  How lucky!  Then we can do something really cute.

Let’s let $C$ be a loop.  That is, $C: [0,1] \rightarrow {\mathbb R}^{n}$ is a curve such that $C(0) = C(1)$.  In other words, $C$ starts and ends in the same place.  Then, what happens if we wanted

$\displaystyle \int_{C} g\cdot dS$

which is the line integral that starts somewhere, travels on the vector field $g$, and then ultimately goes back to where it started.  Well, since we know something about $g$ we might as well use it:

$\displaystyle \int_{C} g\cdot dS = \int_{C} \nabla f\cdot dS$

and in a previous post, we note that this is very similar to the fundamental theorem of calculus: the $\nabla f$ acts as a kind of derivative of $f$.  This means:

$\displaystyle \int_{C} \nabla f\cdot dS = f(C(1)) - f(C(0))$

But, hold on a second.  We said that the curve $C$ started and ended at the same point; let’s say for the sake of ease, that they end at the origin.  Then $f(C(1)) - f(C(0)) = f(0) - f(0) = 0$.  This means that, ultimately,

$\displaystyle \int_{C} g\cdot dS = 0$

This means that, in particular, vector fields that are the gradient of some sufficiently nice function are really sweet: any line integral around a loop is just zero!  We call a vector field which is the gradient of some sufficiently nice function a conservative vector field.

As a corollary to this loop integrating stuff, we have path independence; that is, in a conservative vector field, the line integral from a point to another point is the same regardless of the path we choose to take.  Let’s think about why for a second…

Given some path from point $a$ to point $b$ (let’s call this path $D$), let’s make another path, different from $D$, that goes from point $b$ to $a$ and let’s call this $E$.  Then we can say that around $D\cup E$, where the union here simply means the concatenation of the two paths to make a loop, the integral is 0 (since this is a loop in a conservative vector field).  This means that:

$\displaystyle \int_{D\cup E}g\cdot dS = \int_{D}g\cdot dS + \int_{E}g\cdot dS = 0$

which implies immediately that

$\displaystyle \int_{D}g\cdot dS = -\int_{E}g\cdot dS$

and we have not proven this yet, but it should stand to reason that if we were to travel down path $E$ backwards  (that is, from point $a$ to point $b$, instead of the way it was stated before) we’d get our original integral but the opposite sign.  Hence the negative sign on the right-hand-side.  Because both of these curves were arbitrary, every path from $a$ to $b$ has the same line integral value.  Cool.