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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

[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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