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Faraday Tensor Matrix Element

  1. Dec 3, 2014 #1
    The Faraday Tensor is given by:

    qwcQd.png

    Consider the following outer product with the 4-potential:

    Y1mIy.png

    The Faraday Tensor is related to the 4-potential:

    [tex]F^{mn} = \Box^{m} A^n - \Box^n A^m[/tex]

    For example, ## F^{01} = -\frac{1}{c} \frac{\partial A^x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x} ##

    How do I show that ## \frac{E_x}{c} = -\frac{1}{c} \frac{\partial A^x}{\partial t} - \frac{1}{c}\frac{\partial \phi}{\partial x} ##?
     
  2. jcsd
  3. Dec 3, 2014 #2

    Dale

    Staff: Mentor

    How are the scalar and vector potentials defined?
     
  4. Dec 3, 2014 #3
    [tex] E = -\nabla \phi - \frac{\partial A}{\partial t} [/tex]

    How do I prove this?
     
  5. Dec 3, 2014 #4

    Dale

    Staff: Mentor

    You don't prove it. All definitions are true by definition.
     
  6. Dec 3, 2014 #5
    Also, in einstein summation, why does the ##\mu## go to the bottom on the first term:

    [tex] X \dot X = X_{\mu} X^{\mu} [/tex]
     
  7. Dec 3, 2014 #6
    I thought usually electric field is simply the grad of potential? ## E = -\nabla \phi##
     
  8. Dec 3, 2014 #7

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    This is true in electro-statics, but not true in electro-dynamics. In electro-dynamics Faraday's law tells you that a changing magnetic field will produce an electric field. That is where the ##\frac{\partial \vec{A}}{\partial t}## term comes from.
     
  9. Dec 3, 2014 #8
    Is there a derivation for this?
     
  10. Dec 3, 2014 #9

    Matterwave

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    Science Advisor
    Gold Member

    It is basically a definition, not so much a derivation. By Gauss's law for magnetism we know ##\nabla\cdot\vec{B}=0## so that we can define ##\vec{B}\equiv\nabla\times\vec{A}## for some vector potential ##\vec{A}##, Faraday's law tells us ##\nabla\times\vec{E}=-\frac{1}{c}\frac{\partial \vec{B}}{\partial t}=-\frac{1}{c}\nabla\times\frac{\partial \vec{A}}{\partial t}## so we know that if we define ##\vec{E}\equiv -\nabla\phi-\frac{1}{c}\frac{\partial \vec{A}}{\partial t}## everything will work out (the first term will go away when you take the curl of it). Notice that because of Faraday's law, we can not simply define ##\vec{E}\equiv\nabla\phi## since this will imply ##\nabla\times\vec{E}=0## which is not true.
     
    Last edited: Dec 3, 2014
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