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Zangwill, problem 10.19 - A Matching condition for the vector potential A

  1. May 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that the normal derivative of the coulomb gauge vector suffers a jump discontinuity at a surface endowed with a current density [tex] K(\vec r_s ) [/tex]

    2. Relevant equations
    The vector potential A is given by:
    [tex] A=\frac{\mu_0}{4\pi}\int{\frac{J(x')}{|x-x'|}d^3x} [/tex]

    The magnetic field has a jump in the parallel component given by:
    [tex]\mu_0K(\vec r_s )=\hat{n}\times(\vec{B_2}-\vec{B_1})[/tex]

    3. The attempt at a solution
    Starting with the expression for the jump discontinuity, one can substitute the relation:
    [tex]B=\nabla\times{A}[/tex]
    to get:
    [tex]\mu_0K(\vec r_s )=\hat{n}\times(\nabla\times{A_2}-\nabla\times{A_1})[/tex]

    From here I hoped I can distribute the n - hat to get:

    [tex]\mu_0K(\vec r_s )=\hat{n}\times{\nabla\times{A_2}}-\hat{n}\times{\nabla\times{A_1}}[/tex]

    However, I noticed most vector calculus identities refer to identities where the nabla is the first element, and trying to use the general identities for vectors lead me to have nablas that do not operate on anything. Am I in the right direction? I will appreciate advice on how to tackle this problem.

    Thanks!
     
  2. jcsd
  3. May 20, 2017 #2

    kuruman

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    I suppose you are referring to the "BAC - CAB" rule for the triple cross product, namely
    ##\vec{A}\times(\vec{B}\times \vec {C})=\vec{B}(\vec{A}\cdot \vec {C})-\vec{C}(\vec{A}\cdot \vec {B} ) ##
    Although order of operation matters on the left side, it does not on the right side. The vectors forming the dot products on the right side commute and so do the dot products themselves with the vectors that multiply them. So substitute
    ##\hat{n} = \vec{A}##, ##\vec{\nabla} = \vec{B}## and ##\vec{A}=\vec{C}##. Then you make a first pass and write
    ##\hat{n}\times(\vec{\nabla}\times \vec{A})=\vec{\nabla}(\hat{n}\cdot \vec{A})-\vec{A}(\hat{n}\cdot \vec{\nabla} ) ##
    Finally you make sure the "nabla" has something to operate on and rearrange the second term on the right to get
    ##\hat{n}\times(\vec{\nabla}\times \vec{A})=\vec{\nabla}(\hat{n}\cdot \vec{A})-(\hat{n}\cdot \vec{\nabla} )\vec{A} ##
     
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