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How to determin the momentum eA of electromagnetic field?

  1. Jun 6, 2009 #1
    I got a question when I was reading a paper, it describe a periodic structure composed of many infinite metal wire, and then try to calculate the momentum per unit length of the wire like these when there is current through wire:


    electrons in a magnetic field have an additional contribution to theri momentum of eA,where e is charge value and A is vector potential of magnetic field B, and therefore the momentum per unit length of the wire is.
    [tex]e\pir^2nA(r)=\frac{\mu_0 e^2(\pi r^2n)^2}{2\pi}ln(a/r)[/tex]

    because
    [tex]H(R)=\frac{I}{2\pi R}=\frac{\pi r^2nve}{2\pi R}[/tex]
    where I is the current flowing in the wire, R is the distance from the wire and v is the mean electron velocity. we can express this field as
    [tex]\vec{H(R)}={\mu_o}^{-1}\nabla\times\vec{A}[/tex]
    where A(R)=\frac{\mu_0\pi r^2nve}{2\pi}ln(a/R)[/tex]
    and a is the lattice constant as listed in the picture.

    I know that from classical mechanics we could write momentum of a particle in EM field as P=mv+eA, and here when talking about the momentum of wire, we ignore "mv" because it is from electrons in the wire.but we should calculate moment of the EM field.

    but when calculate A in classical electro-magnetic theory, we use
    [tex]A_z=\frac{\mu I}{4\pi}\int_{-\infty}^{+\infty}\frac{dz}{\sqrt{z^2+a^2}}
    and if we want to get the result like
    A(R)=\frac{\mu_0\pi r^2nve}{2\pi}ln(a/R)[/tex]
    we should choose a point as zero vector potential, here we should choose it as R.
    but since A is arbitrary if we another point, so my question is that here why we choose R to get the momentum of wire.

    thanks for your patience.
     

    Attached Files:

  2. jcsd
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