# Expression with two vectors and del operator

1. Sep 21, 2012

### BOYLANATOR

1. The problem statement, all variables and given/known data
(A.∇)B

What does this mean and how do I go about trying to expand this (using cartesian components)?

2. Sep 21, 2012

### SammyS

Staff Emeritus
Well, $\displaystyle \vec{\text{A}}\cdot\vec{\nabla}=\text{A}_x\,\frac{ \partial}{\partial x}+\text{A}_y\,\frac{ \partial}{\partial y}+\text{A}_z\,\frac{ \partial}{\partial z}\ .$

So that $\displaystyle \left(\vec{\text{A}}\cdot\vec{\nabla}\right) \vec{\text{B}}=\text{A}_x\,\frac{ \partial\vec{\text{B}}}{\partial x}+\text{A}_y\,\frac{ \partial\vec{\text{B}}}{\partial y}+\text{A}_z\,\frac{ \partial\vec{\text{B}}}{\partial z}\ .$

3. Sep 21, 2012

### BOYLANATOR

Of course. It was the order (A.∇) rather than (∇.A) that confused me but now I realise that these are obviously equivalent. Thanks

4. Sep 21, 2012

### BOYLANATOR

Actually that's wrong isn't it?
If the order is reversed you find the derivatives of each component of A. In the problem case though the derivation is carried out on the components of B. I hope this is right as it makes sense in my head.

5. Sep 21, 2012

### dextercioby

Yes, it's correct. The whole expression is called the derivative of B along A.