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.

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