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Grad x (grad x B)

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that:

    ∇x(∇xB) = (B∇)B - ∇ (1/2B2)

    2. Relevant equations

    r = (x,y,z) = xiei

    ∂xi/∂xj = δij

    r2 = xkxk

    δij = 1 if i=j, 0 otherwise (kronecker delta)
    εijk is the alternating stress tensor and summn convn is assumed.

    3. The attempt at a solution

    On the LHS I simplified to get:

    εijk2/∂xj∂xk

    but was unsure what to do next because the RHS contains only first order derivatives

    On the RHS I was able to get to:

    (B∇)B - ∇ (1/2B2) = B(∂Bi/∂i)-B
    = B(∂Bi/∂i-1)

    I feel like I'm just not seeing some simple trick, or there is a rule that I don't remember/haven't learned. This is for my Classical Mechanics class BTW.
     
  2. jcsd
  3. Mar 12, 2014 #2

    vanhees71

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    2016 Award

    There must be something wrong in your problem statement or how can you get an expression which is quadratic in [itex]\vec{B}[/itex] taking derivatives of an expression that contains only one [itex]\vec{B}[/itex]? The correct equation to prove is
    [tex]\vec{\nabla} \times (\vec{\nabla} \times \vec{B})=\vec{\nabla} (\vec{\nabla} \cdot \vec{B}) - \vec{\nabla}^2 \vec{B},[/tex]
    which holds, however, only in Cartesian coordinates!
     
  4. Mar 12, 2014 #3
    that's what I was thinking, but the assignment is what I wrote above
     
  5. Mar 12, 2014 #4

    maajdl

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    Gold Member

    (∇xB)xB = (B∇)B - ∇ (B²/2)

    is famous in MHD
     
  6. Mar 12, 2014 #5
    that's still not what i'm asking. but maybe showing a proof might help me out a bit
     
  7. Mar 13, 2014 #6

    maajdl

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    Gold Member

    Showing a proof of what?
    What are you asking, actually?
    The original identity is obviously wrong (∇x(∇xB) = (B∇)B - ∇ (1/2B2) is wrong).
    Shall we advise you to complain to your teacher?

    The proof of the second identity, (∇xB)xB = (B∇)B - ∇ (B²/2), is straightforward by using components representation.

    Using the notation "eik" for the Levi-Civita tensor,
    using 'F,l" to denote the derivative of F with respect to xl,
    (∇xB)xB can be developed as follows:

    ((∇xB)xB)i
    = eijk (ejlm Bm,l) Bk
    = - eikj elmj Bm,l Bk
    = -(eil ekm - eim ekl) Bm,l Bk
    = - Bk,i Bk + Bi,k Bk
    = - (Bk²/2),i + Bi,k Bk

    which ends the proof.

    Reading you initial post:

    "εijk is the alternating stress tensor ..."
    "On the LHS I simplified to get: εijk∂2/∂xj∂xk"

    I have the feeling you lack some basic understanding, since it makes almost no meaning.
    I don't know if your question is part of a math course or a physics course (electrodynamics).
    In any case, you need to go back to the basics.
    The strange thing is that the identity "(∇xB)xB = (B∇)B - ∇ (B²/2)" is indeed related to the Maxwell stress tensor in electrodynamics (if B is the magnetic field). The second term is then called the magnetic pressure.
    As you posted in the "Calculus & Beyond Homework" section I wonder how you could have mixed that "math exercise" with electrodynamics. Is Google the reason?
     
    Last edited: Mar 13, 2014
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