Monday Math 130

Suppose we have a twice-differentiable vector field F(x) in three dimensions, which as r=|\mathbf{x}|\to\infty goes to zero faster than \frac{1}{r}. Then we can take the Fourier transform of F(x), its divergence, and its curl. With
\mathbf{\tilde{F}}(\mathbf{k})=\mathcal{F}\left[\mathbf{F}(\mathbf{x})\right](\mathbf{k}) the Fourier transform of F, then our results from last week tell us that
\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{F}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathbf{\tilde{F}}(\mathbf{k}).
and
\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{F}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathbf{\tilde{F}}(\mathbf{k}).

Now, the ability to decompose a vector into components parallel and perpendicular to a given vector means that an arbitrary vector field in k-space may be decomposed (uniquely) into components parallel and perpendicular to k:
\mathbf{\tilde{F}}_{||}(\mathbf{k})=(\mathbf{\hat{k}}\cdot\mathbf{\tilde{F}})\mathbf{\hat{k}}=\frac{\mathbf{k}\cdot\mathbf{\tilde{F}}}{\mathbf{k}\cdot\mathbf{k}}\mathbf{k}
is the component parallel to k. The perpendicular component is then \mathbf{\tilde{F}}_{\perp}(\mathbf{k})=\mathbf{\tilde{F}}(\mathbf{k})-\mathbf{\tilde{F}}_{||}(\mathbf{k}).
(See “vector projection” here.)
So we have
\mathbf{\tilde{F}}(\mathbf{k})=\mathbf{\tilde{F}}_{||}(\mathbf{k})+\mathbf{\tilde{F}}_{\perp}(\mathbf{k}) as our decomposition. Now, since \mathbf{\tilde{F}}_{||} is parallel to k, their cross product is zero; and since \mathbf{\tilde{F}}_{\perp} is perpendicular to k, their dot product is zero:
\mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k})=0
\mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k})=0.

Let \mathbf{F}_{||}(\mathbf{x}) and \mathbf{F}_{\perp}(\mathbf{x}) be the inverse Fourier transforms of \mathbf{\tilde{F}}_{||}(\mathbf{k}) and
\mathbf{\tilde{F}}_{\perp}(\mathbf{k}), respectively. Then, due to the linearity of the Fourier transform and its inverse,
\mathbf{F}(\mathbf{x})=\mathbf{F}_{||}(\mathbf{x})+\mathbf{F}_{\perp}(\mathbf{x}).
Then from our results from last week,
\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{F}_{||}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k}),
and
\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{F}_{\perp}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k}).
But \mathbf{k}\times\mathbf{\tilde{F}}_{||}(\mathbf{k})=0 and \mathbf{k}\cdot\mathbf{\tilde{F}}_{\perp}(\mathbf{k})=0, so these are both zero, and a field’s Fourier transform is zero if and only if it is zero;
thus
\mathbf{\nabla}\times\mathbf{F}_{||}(\mathbf{x})=0
and
\mathbf{\nabla}\cdot\mathbf{F}_{\perp}(\mathbf{x})=0.

This means that the vector field F(x) may be uniquely decomposed into the sum of two fields, \mathbf{F}_{||}(\mathbf{x}) and \mathbf{F}_{\perp}(\mathbf{x}), the first of which is irrotational (zero curl), and the other is solenoidal (zero divergence). This decomposition is known as the Helmholtz decomposition. The statement that any sufficiently smooth and decaying three-dimensional vector field may be decomposed in this way is Helmholtz’s theorem, also known as the “fundamental theorem of vector calculus.” In physics, the irrotational and solenoidal components are known as the longitudinal and transverse components, respectively.

Since \mathbf{F}_{||}(\mathbf{x}) is irrotational, it may be expressed as the gradient of a scalar field, and since \mathbf{F}_{\perp}(\mathbf{x}) is solenoidal, it may be expressed as the curl of a vector field. Thus, we can also express the Helmholtz decomposition as
\mathbf{F}=-\mathbf{\nabla}\phi+\mathbf{\nabla}\times\mathbf{A}
(where the minus sign arises from the usage in physics, particularly electrodynamics).

There is a related theorem which is also sometimes referred to as “Helmholtz’s theorem,” and which states that a vector field can be constructed with a specified divergence and a specified curl; and that if these divergence and curl are sufficiently smooth and vanish fast enough at infinity, the vector field is uniquely specified by them. This is crucial in electrodynamics, as Maxwell’s equations specify the divergence and curl of the electric and magnetic fields; this theorem tells us that (given physically appropriate boundary conditions at infinity) these are enough to determine the fields.

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One Response to “Monday Math 130”

  1. Monday Math 135 « Twisted One 151's Weblog Says:

    […] to say, can we construct a field with a specified divergence and curl? As noted in final remarks here, the answer is yes, we can, and given our boundary conditions, it is unique. More specifically, […]

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