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What kind of tensor is the electromagnetic field tensor?

  1. Mar 14, 2012 #1
    The covariant form of the Lorentz force can be written as

    [tex]m \ddot x^\mu =q F^{\mu \nu} \dot x_\nu [/tex]

    and such a relation should prove by the quotient rule that F is indeed a tensor.
    But what kind of tensor is it? One can show that it transforms from an unprimed
    to a primed system like

    [tex] F'^{\mu \nu} = \Lambda^\mu_{\ \alpha} \Lambda^{\ \nu}_{\beta} F^{\alpha \beta} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} F^{\alpha \beta}[/tex],

    where [tex]\Lambda^{\ \nu}_{\beta}[/tex] is the inverse Lorentz transformation matrix. But what kind of tensors transforms like this? Does it have a name? I know about covariant, contravariant and mixed tensors. The closest I get is a mixed tensor but it would transform like

    [tex] T^\mu_\nu = \frac{\partial x'^\mu}{\partial x^\rho }\frac{\partial x^\sigma}{\partial x'^\nu} T^\rho_\sigma[/tex].
     
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 14, 2012 #2

    tiny-tim

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  4. Mar 14, 2012 #3
    Hi Tim!

    But then it should transform like

    [tex] F'^{\mu\nu} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x'^\nu}{\partial x^\beta} F^{\alpha \beta}[/tex]

    and not as

    [tex]F'^{\mu \nu} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} F^{\alpha \beta}[/tex]

    which arises from the fact that the inverse lorentz transformation gives the equality

    [tex] \frac{\partial x^\mu}{\partial x'^\nu} = \Lambda_\nu^{\ \ \mu}[/tex]

    and that I have shown that

    [tex]F'^{\mu \nu} = \Lambda^\mu_{\ \ \alpha} \Lambda^{\ \ \nu}_{\beta} F^{\alpha \beta}[/tex]

    from transforming the lorentz force law in covariant form.
     
    Last edited: Mar 14, 2012
  5. Mar 14, 2012 #4
    Maybe there is a flaw in my proof? It goes as follows: Ignoring the constants m and q

    [tex] \ddot x'^\mu = F'^{\mu \nu} \dot x'_\nu =\Lambda^\mu_{\ \alpha} (\dot x^\alpha) = \Lambda^\mu_{\ \alpha} (F^{\alpha \beta} \dot x_{\beta}) = \Lambda^\mu_{\ \alpha} F^{\alpha \beta} \Lambda_\beta^{\ \ \nu} \dot x'_\nu [/tex],
    from which it should follow that

    [tex]F'^{\mu \nu} = \Lambda^\mu_{\ \alpha} \Lambda_\beta^{\ \ \nu}
    F^{\alpha \beta} = \frac{\partial x'^\mu}{\partial x^\alpha} \frac{\partial x^\nu}{\partial x'^\beta} F^{\alpha \beta}[/tex].

    Here I have used the lorentz transformation and the inverse transformation
    [tex] x'^\mu = \Lambda^\mu_{\ \ \nu} x^\nu[/tex]

    [tex] x'_\mu = \Lambda_\mu^{\ \ \nu} x_\nu[/tex]

    from which it follows that

    [tex]\frac{\partial x^\mu}{\partial x'^\nu} = \Lambda_\nu^{\ \ \mu}[/tex]
    and

    [tex]\frac{\partial x'^\mu}{\partial x^\nu} = \Lambda^\nu_{\ \ \mu}[/tex].
     
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