# Annoying vectors

1. Apr 16, 2009

### latentcorpse

for a constant magnetic field $\vec{B}$ everywhere, $\vec{A}=\frac{1}{2} \vec{B} \times \vec{r}$

because (im not going to use vector notation to save time)

$\nabla \times (B \times r)= (\nabla \cdot r)B + (r \cdot \nabla)B - (\nabla \cdot B)r - (B \cdot \nabla)r$

the first term gives 3B
the third term vanishes
the fourht term gives -B
so to get the answer i need the second term to vanish but i can't get it to go away - how do i do this?

2. Apr 16, 2009

### dx

The second term vanishes because the gradient of a constant vector field is zero. None of the components are changing, so their space derivatives and hence the gradient is zero.

Last edited: Apr 16, 2009
3. Apr 16, 2009

### latentcorpse

how can you take the gradient of a vector - you need index notation yes?

consider $(m \cdot \nabla)r$
is this just $m_j \partial_j r_i=m_i$

4. Apr 16, 2009

### dx

5. Apr 16, 2009

### latentcorpse

but $\nabla \phi=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z})$ so doesn't that give us a nine component vector - is that just a tensor?

e.g. $\partial_i r_j =\delta_{ij}$?

6. Apr 16, 2009

### dx

Sorry, I misread the equation. The second term is $$(r\cdot\nabla) B$$, which is

$$\sum_{i = 1}^3 (r_x \partial_x + r_y \partial_y + r_z \partial_z) B_i \mathbf{e}_i$$

7. Apr 16, 2009

### latentcorpse

how is that 0 though?

why can't this be done using eisntein summation convention i.e.
$r_j \partial_j B_i=...$ i can't get it to go any further?

8. Apr 16, 2009

### dx

The derivatives of $$B_i$$ are zero, so what you just wrote must be zero.