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Confusing arken/weber problem (del in triple vector product)

  1. Sep 25, 2005 #1
    the following problem is in the arfken/weber textbook and was also on a practice exam for my mathematical methods course:

    Verify that
    [tex]
    \mathbf{A} \times (\nabla \times \mathbf{A}) = \frac{1}{2} \nabla(A^2) - (\mathbf{A} \cdot \nabla)\mathbf{A}.
    [/tex]

    i used the BAC-CAB rule, but i don't get the factor of 1/2.

    the solutions booklet that came with the textbook very tersely explains, "the factor of 1/2 occurs because the del only operates on one of the A's."

    i would very much appreciate an explanation that is perhaps more informative than this one sentence blurb! :tongue:

    thank you.
     
    Last edited: Sep 25, 2005
  2. jcsd
  3. Sep 25, 2005 #2

    robphy

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    I think the "product rule" was used.
    Have you tried an index-based calculation...
    [tex]\epsilon_{ijk}A^j\epsilon^{klm}\nabla_l A_m
    =\left( \delta_i^l\delta_j^m- \delta_i^m\delta_j^l)A^j\nabla_l A_m[/tex]
     
  4. Sep 26, 2005 #3

    lurflurf

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    You can't just go around baccabing things. In particular as the one sentence you dislike states the del only operates on one of the A's. That other half is busy over here
    [tex](\mathbf{A} \times \nabla) \times \mathbf{A}= \frac{1}{2} \nabla(A^2) - \mathbf{A}(\nabla\cdot\mathbf{A})[/tex]
    since
    [tex](a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b[/tex]
     
  5. Sep 26, 2005 #4

    lurflurf

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    what is
    [tex]\nabla(A^2)[/tex]
    If one A is constant?
    say A(x) then find
    [tex]\nabla(A(x).A(0))|_{x=0}[/tex]
    Compare the 1-var analog
    D[y(x)*y(x)]|{x=0}=2y(0)*y'(0)
    D[y(0)*y(x)]|{x=0}=y(0)*y'(0)
     
    Last edited: Sep 26, 2005
  6. Sep 26, 2005 #5

    i still don't understand. :grumpy:
     
  7. Sep 26, 2005 #6

    lurflurf

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    It is like asking where the 1/2 comes from in
    [tex]y \ \frac{d}{dx}y=\frac{1}{2} \ \frac{d}{dx}y^2[/tex]
    You should avoid trying to use vector identities on del that do not apply.
    [tex]\mathbf{A_1} \times (\nabla_2 \times \mathbf{A_2}) =\nabla_2(\mathbf{A_1}\cdot\mathbf{A_2}) - (\mathbf{A_1} \cdot \nabla_2)\mathbf{A_2}[/tex]
    is true and makes what is happening clear. A1 and A2 are equal, but nabla2 acts only on A2. The grad(A^2) part is the place where the distinction matters. In order to switch back to more standard notation we note that
    [tex]\nabla_1(\mathbf{A_1}\cdot\mathbf{A_2})=\nabla_2(\mathbf{A_1}\cdot\mathbf{A_2})[/tex]
    and
    [tex]\nabla(\mathbf{A}\cdot\mathbf{A})=\nabla_1(\mathbf{A_1}\cdot\mathbf{A_2})+\nabla_2(\mathbf{A_1}\cdot\mathbf{A_2})[/tex]

    You infact did not use B(A.C)-C(A.B), you used B(A.C)-(A.B)C
    so you saw that not all vector idenetities hold if a vector is replaced by del, but you still used an identity that was not true.

    for comparison
    [tex](\mathbf{A_1} \times \nabla_2) \times \mathbf{A_2} =\nabla_2(\mathbf{A_1}\cdot\mathbf{A_2}) - \mathbf{A_1}(\nabla_2\cdot\mathbf{A_2})[/tex]
    thus
    [tex](\mathbf{A} \times \nabla) \times \mathbf{A}= \frac{1}{2} \nabla(A^2) - \mathbf{A}(\nabla\cdot\mathbf{A})[/tex]
     
    Last edited: Sep 26, 2005
  8. Sep 26, 2005 #7

    ah!

    thank you very much. that is exactly what i needed.

    thanks a lot.
     
  9. Oct 2, 2005 #8
    found another way to do the problem (without invoking the scary kronecker deltas and epsilon tensors!). this is the proof suggested by david griffiths' intro to electrodynamics:

    [tex]
    \nabla (\mathbf{A} \cdot \mathbf{B}) = \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A}) + (\mathbf{A} \cdot \nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A}.
    [/tex]

    when [tex]\mathbf{A} = \mathbf{B},[/tex]

    [tex]\nabla (\mathbf{A} \cdot \mathbf{A}) = 2\mathbf{A} \times (\nabla \times \mathbf{A}) + 2(\mathbf{A} \cdot \nabla)\mathbf{A}.[/tex]

    note: [tex]\mathbf{A} \cdot \mathbf{A} = A^2.[/tex]

    so

    [tex]\mathbf{A} \times (\nabla \times \mathbf{A}) = \frac{1}{2} \nabla(A^2) - (\mathbf{A} \cdot \nabla)\mathbf{A}.
    [/tex]
     
    Last edited: Oct 2, 2005
  10. Oct 2, 2005 #9

    robphy

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    Of course, someone had to derive [or give you] this first identity of the gradient of a dot-product in terms of the cross-products of curls... or is this identity just so intuitively obvious?

    Believe it or not, the epsilon and delta tensors make these "Vector Calculus identities" much simpler... with some practice. It's not unlike learning Euler's identity in complex algebra to make "Trigonometric identities" much simpler.
     
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