1. Jan 30, 2012

### AlonsoMcLaren

We know that A x (BxC)= (A·C)B-(A·B)C (*)

In the following example, we can treat ∇ as a vector and apply the formula (*) above to get the correct answer
∇x(∇xV)= ∇(∇·V)-∇^2 V

But in this example, the formula (*) seems to fail
∇x(UxV)≠U(∇·V)-V(∇·U)

Why?

2. Jan 30, 2012

### Char. Limit

Because ∇ is NOT a vector, no matter how much we want it to act like one. For one, ∇ isn't commutative. (Vectors are)

3. Jan 30, 2012

### AlonsoMcLaren

Then why the textbook simply uses the formula (*) when deriving ∇x(∇xV)= ∇(∇·V)-∇^2 V ? Is it just a coincidence that the formula (*) works for ∇x(∇xV)= ∇(∇·V)-∇^2 V ?

4. Jan 31, 2012