Proof. Using the entry nabla acting on products, we first may write
In the brackets the first product is, according to Euler's theorem on homogeneous functions, equal to
. The second product can be written as
, which is
, i.e. . The third product is, due to the sodenoidalness, equal to
. The last product equals to (see the first formula for position vector). Thus we get the result
This is version 2 of example of vector potential, born on 2009-04-18, modified 2009-04-18.
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![$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \... ...}\cdot\nabla)\vec{r}-(\nabla\cdot\vec{U})\vec{r} +(\nabla\cdot\vec{r})\vec{U}].$](http://images.physicslibrary.org/cache/objects/654/l2h/img6.png "$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \... ...}\cdot\nabla)\vec{r}-(\nabla\cdot\vec{U})\vec{r} +(\nabla\cdot\vec{r})\vec{U}].$")
![$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \frac{1}{\lambda\!+\!2}[\lambda\vec{U}-\vec{U}-\vec{0}+3\vec{U}] = \vec{U}.$](http://images.physicslibrary.org/cache/objects/654/l2h/img13.png "$\displaystyle \nabla\times(\frac{1}{\lambda\!+\!2}\vec{U}\!\times\!\vec{r}) = \frac{1}{\lambda\!+\!2}[\lambda\vec{U}-\vec{U}-\vec{0}+3\vec{U}] = \vec{U}.$")
