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Homework Help: Directional Derivatives and Commutation

  1. Oct 13, 2013 #1


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    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:

    [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: Oct 13, 2013
  2. jcsd
  3. Oct 13, 2013 #2


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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: Oct 13, 2013
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