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So we have from Biot Savart law that [itex]\vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int_V dV' \vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})[/itex]

we take the curl of this and show the second term vanishes to leave us with [itex]\nabla \times \vec{B}(\vec{r})=\mu_0 \vec{J}(\vec{r})[/itex] which is Ampere's law in differential form.

however we take the curl of this using the identity a x (b x c) = b(a.c)-c(a.b). here this gives

[itex]\nabla \times \left(\vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})\right) = \nabla^2(\frac{1}{r}) \vec{J}(\vec{r'}) - (\nabla \cdot \vec{J}(\vec{r'})) \nabla(\frac{1}{r})[/itex]

but in the book they have the scond term as [itex](\vec{J}(\vec{r'}) \cdot \nabla) \nabla(\frac{1}{r})[/itex]. why is this allowed?

the dot product doesn't commute when a grad is involved does it?