Solving Constant Magnetic Field Vector Problem: Annoying Vectors

[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?

 

[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.

 

[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

 

[dx]

The gradient of a vector (a,b,c) just (grad a, grad b, grad c).

 

[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}?

 

[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

 

[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?

 

[dx]

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