latentcorpse
- 1,411
- 0
This must be a pretty standard proof but I'm having difficulty with part of it.
So we have from Biot Savart law that \vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int_V dV' \vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})
we take the curl of this and show the second term vanishes to leave us with \nabla \times \vec{B}(\vec{r})=\mu_0 \vec{J}(\vec{r}) 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
\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})
but in the book they have the scond term as (\vec{J}(\vec{r'}) \cdot \nabla) \nabla(\frac{1}{r}). why is this allowed?
the dot product doesn't commute when a grad is involved does it?
So we have from Biot Savart law that \vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int_V dV' \vec{J}(\vec{r'}) \times \nabla(\frac{1}{r})
we take the curl of this and show the second term vanishes to leave us with \nabla \times \vec{B}(\vec{r})=\mu_0 \vec{J}(\vec{r}) 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
\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})
but in the book they have the scond term as (\vec{J}(\vec{r'}) \cdot \nabla) \nabla(\frac{1}{r}). why is this allowed?
the dot product doesn't commute when a grad is involved does it?