# Cross product of curl.

1. Mar 1, 2005

### Galileo

Is there any neat way/rule to write:

$$\vec B \times (\vec \nabla \times \vec A)$$
?

I've tried it myself and found (e.g) for the x-component:

$$\left(B_x\frac{\partial A_x}{\partial x}+B_y\frac{\partial A_y}{\partial x}+B_z\frac{\partial A_z}{\partial x}\right)-\left(B_x\frac{\partial A_x}{\partial x}+B_y\frac{\partial A_x}{\partial y}+B_x\frac{\partial A_x}{\partial z}\right)$$

I can write the last terms with the minus sign as: $\vec B \cdot \nabla A_x$, but I can't find a way to do something nice to the first term, except maybe:

$$\left(\vec B \cdot \frac{\partial}{\partial x}\vec A\right)$$
I've never seen such an expression before though.
The other 2 components are similar:
$$\left[\vec B \times (\vec \nabla \times \vec A)\right]_y=\left(\vec B \cdot \frac{\partial}{\partial y}\vec A\right)-\left(\vec B \cdot \nabla A_y\right)$$
$$\left[\vec B \times (\vec \nabla \times \vec A)\right]_z=\left(\vec B \cdot \frac{\partial}{\partial z}\vec A\right)-\left(\vec B \cdot \nabla A_z\right)$$

I figured I may see something if I combined them all into the general expression:

$$\left(B_x\frac{\partial A_x}{\partial x}+B_y\frac{\partial A_y}{\partial x}+B_z\frac{\partial A_z}{\partial z}\right)\hat x +\left(B_x\frac{\partial A_x}{\partial y}+B_y\frac{\partial A_y}{\partial y}+B_z\frac{\partial A_z}{\partial y}\right)\hat y+\left(B_x\frac{\partial A_x}{\partial z}+B_y\frac{\partial A_y}{\partial z}+B_z\frac{\partial A_z}{\partial z}\right)\hat z-(\vec B \cdot \vec \nabla)\vec A$$
There's definately a pattern in the first 3 terms, but the best I could come up with is writing these terms as:
$$B_x\nabla A_x+B_y\nabla A_y+B_z\nabla A_z$$
That has condensed it a lot. Looks like a dot product with B, but....

2. Mar 1, 2005

### robphy

identity: (from Griffiths, Introduction to EM)

grad(A dot B)=A cross (curl B) + B cross (curl A) + (A dot grad)B + (B dot grad)A

3. Mar 1, 2005

### robphy

Playing around with it more:

$$\vec B \times(\nabla \times \vec A) =\epsilon_{ijk}B_j ( \epsilon_{klm}\nabla_l A_m) =B_m \nabla_i A_m - B_l \nabla_l A_i$$
where I used
$$\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$

So, you've essentially got it.

4. Mar 1, 2005

### dextercioby

HINT:ALWAYS use cartesian tensors when proving vector identities...With objects from R^{n},of course.

Daniel.

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