Condition of Helmholtz theorem to hold.

  • Thread starter yungman
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  • #1
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I want to verify this, according to Griffiths p 557:
[tex]\vec{F}(\vec r')=-\nabla U+\nabla\times\vec A[/tex]
ONLY if both ##\nabla U\; and \; \nabla\times\vec A\rightarrow\;0## faster than ##\frac 1 {r^2}## as ##r\rightarrow\;\infty##.

But the requirement of ##\vec{F}(\vec r')## is only ##|\vec{F}(\vec r')|\rightarrow\;0## as ##r\rightarrow\;\infty##.

If divergence or curl of ##\vec{F}(\vec r')## has to roll off faster than ##\frac 1 {r^2}##, then ##\vec{F}(\vec r')## must roll off at least as fast as ##\frac 1 {r^2}##. Why is it not a requirement?

I read quite a few article, it is very unclear in this condition. Only Griffiths spelled out clearly, all the other books just beat around this.

Thanks
 
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Answers and Replies

  • #2
king vitamin
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It appears that the constraints on [itex]\vec{F}[/itex] come from a different place than the constraints on its divergence and curl.

The reason why one needs [itex]\nabla \cdot \vec{F}[/itex] and [itex]\nabla \times \vec{F}[/itex] to decay like they do is because the solution is given explicitly by integrals such as

$$ \int d^3\vec{r}' \frac{\nabla \cdot \vec{F}(\vec{r}')}{|\vec{r}-\vec{r}'|} .$$

In order for this integral to converge, the integrand needs to decay faster than the measure, so like [itex]\frac{1}{r'^3}[/itex], so this gives your constraint on [itex]\nabla \cdot \vec{F}[/itex]. Furthermore, in the course of the proof, you integrate by parts, and you set the boundary term to zero. The boundary term from the "divergence" part looks like

$$ \oint_{r' = \infty} d^2\vec{a}' \frac{\vec{F}(\vec{r}') \cdot \hat{n}}{|\vec{r}-\vec{r}'|} $$

so by setting this to zero, you are merely assuming that the integrand is zero at [itex]r' = \infty[/itex].

Now in practice, in order for the conditions on the divergence and curl to hold, I suppose there are more stringent conditions on the function [itex]\vec{F}(\vec{r}')[/itex], so the second assumption might be superfluous in applications. However, most physicists are comfortable with setting everything to zero at infinity as soon as they see fit and don't really look twice at either of these conditions (you usually read "decays sufficiently quickly" without qualifications).
 

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