# Directional Derivatives and Commutation

1. Oct 13, 2013

### tnb

1. The problem statement, all variables and given/known data

I need to prove that directional derivatives do not commute.

2. Relevant equations

Thus, I need to show that:
$$(\vec{A} \cdot \nabla)(\vec{B} \cdot \nabla f) - (\vec{B} \cdot \nabla)(\vec{A} \cdot \nabla f) = (\vec{A} \cdot \nabla \vec{B} - \vec{B} \cdot \nabla \vec{A}) \cdot \nabla f$$

3. The attempt at a solution

I used the following vector identity:

$$\nabla (\vec{C} \cdot \vec{D}) = (\vec{C} \cdot \nabla) \vec{D} + (\vec{D} \cdot \nabla) \vec{C} + \vec{C} \times (\nabla \times \vec{D}) + \vec{D} \times (\nabla \times \vec{C})$$

And got:

$$\vec{A} \cdot \left[ \vec{B} \times (\nabla \times \nabla f) + (\vec{B} \cdot \nabla)\nabla f + \nabla f \times (\nabla \times \vec{B}) + (\nabla f \cdot \nabla) \vec{B} \right] - \vec{B} \cdot \left[ \vec{A} \times (\nabla \times \nabla f) + (\vec{A} \cdot \nabla)\nabla f + \nabla f \times (\nabla \times \vec{A}) + (\nabla f \cdot \nabla) \vec{A} \right]$$

Then I reduced this to:

$$\vec{A} \cdot \left[ \nabla f \times (\nabla \times \vec{B}) + (\nabla f \cdot \nabla) \vec{B} \right] - \vec{B} \cdot \left[ \nabla f \times (\nabla \times \vec{A}) + (\nabla f \cdot \nabla) \vec{A} \right]$$

I am not sure how to proceed from here or if I even am on the right track. Any help is much appreciated. Thanks.

Last edited: Oct 13, 2013
2. Oct 13, 2013

### tnb

So I found a solution but would still find it useful if someone could explain the vector identity used:

(A⃗ ⋅∇)(B⃗ ⋅∇f)−(B⃗ ⋅∇)(A⃗ ⋅∇f) =
$$\vec{B} \cdot \left[ (\vec{A} \cdot \nabla ) \nabla f \right] + (\vec{A} \cdot \nabla \vec{B}) \cdot \nabla f - \vec{A} \cdot \left[ (\vec{B} \cdot \nabla ) \nabla f \right] + (\vec{B} \cdot \nabla \vec{A}) \cdot \nabla f$$

The second and third terms cancel and yield the given answer.

Last edited: Oct 13, 2013