Monday Math 131

Let us consider a (real) vector field F(x) in an n-dimensional Euclidean space, and let it be a central field. That is to say, that for any point in the space (save possibly the origin, where it may be undefined), the vector field points in the radial direction, directly away from or directly toward to the origin; and the magnitude of the vector depends only on the radial coordinate r, the distance between the point an the origin. Alternately, this may be stated as the field being symmetric under all rotations, proper and improper, about the origin (it has symmetry group O(n)).

Specifically, there is a scalar function f(r) such that
, where is the radial coordinate, and is the unit radial vector. The magnitude of F(x) is thus |f(r)|, with positive f(r) indicating an outward vector and negative f(r) an inward vector.

Now, let us consider a scalar central field G(x); this means there is function g(r) such that G(x)=g(r), where is the radial coordinate. Due to the symmetry, the gradient of such a scalar field must be a central vector field. Denoting the cartesian coordinates by xi, i=1,2,…,n, and letting denote the unit vector in the direction of xi, then
Now, , then
so if g(r) is an antiderivative of f(r), so that g‘(r)=f(r), then
. Thus, as long as f(r) is well-behaved enough to be integrable, then the central vector field F(x) is the gradient of a scalar function.

Next, we note that due to the symmetry, the divergence of the vector field F(x) must be a scalar central field. Specifically, using our notation as before,

, where
Fi(x) is the ith component of F(x). Since , the ith component is thus
, and so
Now, noting that , we see then that


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One Response to “Monday Math 131”

  1. Monday Math 132 « Twisted One 151's Weblog Says:

    […] Math 132 By twistedone151 Continuing from last week, let us have a central vector field in n dimensions. Using, as with this post the […]

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