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Geometric product in electromagnetism

  1. Dec 22, 2009 #1
    Hi. I've been learning how to use geometric algebra and I've been stumbling when I apply it to E&M. I am hoping someone here can point out what I am doing wrong.

    The problem comes when trying to represent the field tensor in terms of the 4-potential. Here is the standard form:

    [tex] F^{\mu\nu} = \partial^\mu A^{\nu} - \partial^\nu A^{\mu}[/tex]

    [tex]\partial_\mu A^{\mu} = 0[/tex]

    Here is the form presented in "Geometric Algebra for Physicists":

    [tex] F = E + IB = \nabla A[/tex]

    Here it uses the space time vectors [tex]\nabla = \gamma^\mu \partial_\mu , A = \gamma_\nu A^\nu: \mu, \nu= 0,1,2,3[/tex], the space time bi-vectors [tex]E = E^i \gamma_i \gamma_0 ; B = B^i \gamma_i \gamma_0: i = 1,2,3[/tex], and spacetime pseudoscalar [tex] I = \gamma_0 \gamma_1 \gamma_2 \gamma_3[/tex]

    When I apply it to the 4-potential I get:

    [tex]\nabla A = \gamma^\mu \gamma_\nu \partial_\mu A^\nu = \gamma^0 \gamma_0 \partial_0 A^0 + ... + \gamma^3 \gamma_3 \partial_3 A^3[/tex]

    Now, at this point I make use of some identities: [tex]\gamma^0 = \gamma_0; \gamma_0 \gamma_0 = 1; \gamma^i = -\gamma_i; \gamma_i \gamma_i = -1: i = 1,2,3; \gamma_\mu \gamma_\nu = -\gamma_\nu \gamma_\mu[/tex]

    When I do this to collect terms I should get a scalar, which represents the divergence of A = 0, and a set of bi-vectors which represents the field. But, instead I get numerous wronge signs.

    [tex](\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3) + (\partial_0 A^1 + \partial_1 A^0)\gamma_0 \gamma_1 + (\partial_0 A^2 + \partial_2 A^0)\gamma_0 \gamma_2 + ...[/tex]

    The scalar is correct, but the bivectors should be, for example: [tex](\partial_0 A^1 - \partial_1 A^0)\gamma_0 \gamma_1[/tex]

    Thank you for anyone that can help point out my misunderstanding.
     
  2. jcsd
  3. Dec 23, 2009 #2

    tiny-tim

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    Hi Longstreet! :smile:

    (have a nabla: ∇ :wink:)

    Shouldn't F = Λ A ?
     
  4. Dec 23, 2009 #3
    Thank you for your reply. However, [tex]\nabla A = \nabla \cdot A + \nabla \wedge A = \nabla \wedge A[/tex], from the lorenz condition [tex]\nabla \cdot A = 0[/tex].

    I think I realized where the negative comes in. EG:

    [tex]
    (\partial_0 A^1 + \partial_1 A^0)\gamma_0 \gamma_1 = (\partial^0 A^1 - \partial^1 A^0)\gamma_0 \gamma_1
    [/tex]

    Because of the fact that [tex]x^\mu \gamma_\mu = x_\mu \gamma^\mu[/tex]. However, now I am a bit confused as to what difference it makes taking the derivative [tex]\partial_\mu[/tex] versus [tex]\partial^\mu[/tex] (besides a potential negative sign), and how to know when to use which one. I guess one interpretation is a unit like effect, as in bi-vectors have a area-like quantity associated with them.
     
    Last edited: Dec 23, 2009
  5. Dec 27, 2009 #4
    You can use either:

    [tex]\nabla = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu[/tex]

    (the sign change for raising [itex]\partial_k = - \partial^k[/itex] is cancelled by the corresponding sign change lowering [itex]\gamma^k = - \gamma_k[/itex])

    Use whichever is most convienent. For example, expanding the four-Laplacian in coordinates, use of both makes sense:

    [tex]\nabla^2 = \gamma^\mu \partial_\mu \cdot \gamma_\nu \partial^\nu = \partial_\mu \partial^\mu[/tex]
     
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