Directional Derivatives and Commutation

  • Thread starter tnb
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tnb
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Homework Statement



I need to prove that directional derivatives do not commute.

Homework Equations



Thus, I need to show that:
[tex]
(\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
[/tex]

The Attempt at a Solution



I used the following vector identity:

[tex] \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}) [/tex]

And got:

[tex] \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] [/tex]

Then I reduced this to:

[tex] \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] [/tex]

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:

Answers and Replies

  • #2
tnb
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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) =
[tex] \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 [/tex]

The second and third terms cancel and yield the given answer.
 
Last edited:

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