What does the expression (A dot DEL)B mean where A,B are vectors, and DEL is the gradient operator?

Question

Physical significance would be nice, and cartesian components if possible. Perhaps there's an identity I'm missing that would shed some light on the situation.

Answer 1

The other answerers aren't considering something.

In the following I will use a '.' to represent the dot product.

So if:
(A . Del)B = A . Del(B) as he suggests, you would be taking the gradiant of a vector, which is not a vector (it's not well defined in strictly vector algebra - but can be in more advanced stuff). So that's not a well-defined equation. You either need to get into carternians or tensors, or just tensor notation. I'm sure that's not what you want to do. But sometimes you do in a physical application, which is why you really need parenthesis uses right.

What it really means is (A . Del) operating on vector B. (notice A . Del does not equal Del . A, so this is not a commutative process). (A . Del) gives you a scalar, not a vector. So when you operate it on B you're going from a vector to a vector since each component of B gets operated on by (A . Del) .

What that means physically depends on your application. I'm not sure if I can come up with a conceptual explanation. Maybe someone else can? I just know it shows up a lot in Physics.

After thinking about it, I'd say it tells you how each component of B changes with position in the A direction.

Answer 2

I think, A (dot DEL) B is the dot product of vector A and (del B).

Which means the component of A in the direction of (del B).

Hey, if it is not the correct answer pls let me know the correct answer when you get it. Good Luck.

Answer 3

One place I've seen this used is in the convective time derivative of hydrodynamics. If A is fluid velocity, it gives the total time derivative of B when added to the partial time derivative of B.