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I 6-dimensional representation of Lorentz group

  1. Jul 1, 2017 #1
    Hello! I understand that the vector formed of the scalar and vector potential in classical EM behaves like a 4-vector (##A^\nu=\Lambda^\nu_\mu A^\mu##). Does this means that the if we make a vector with the 3 components of B field and 3 of E field, so a 6 components vector V, will it transform as ##V^\nu=M(\Lambda^\nu_\mu) V^\mu##, with M being a 6-dimensional representation of the Lorentz group? So can we build all the tensorial formalism of EM using this 6-dimensional representation, in the same way we did with a 4-vector representation?
     
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  3. Jul 1, 2017 #2

    Dale

    Staff: Mentor

    Not quite. The relativistic object which contains the components of both the non relativistic E and B field is called the Faraday tensor. It is a rank 2 tensor in 4D spacetime, so it has 16 components. But the Faraday tensor is antisymmetric, so it only has 6 independent components (the diagonal elements are 0 and the upper triangle is the opposite of the lower triangle)
     
  4. Jul 1, 2017 #3
    So the vector I just described, doesn't transform in a relativistic from, it must be through a tensor?
     
  5. Jul 1, 2017 #4

    Dale

    Staff: Mentor

    Well, technically what you described is not even a vector. A vector has 4 components in relativity, so something with 6 components cannot be a vector.
     
  6. Jul 1, 2017 #5
    Oh, so what is the point of a n-dimensional representation of the Lorentz group, if it makes sense only for n=4?
     
  7. Jul 1, 2017 #6

    vanhees71

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    There are two convenient ways to get the transformation of the electromagnetic-field components. First you can use the Faraday tensor components,
    $$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu},$$
    which transform as any 2nd-rank-tensor components
    $$\bar{F}^{\mu \nu}(\bar{x}) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x).$$
    The second posibility is to use the Riemann-Silberstein vector. In terms of the field components in the 1+3-formalism it reads
    $$\vec{F}=\vec{E}+\mathrm{i} \vec{B}.$$
    The two invariants one can form from the ##F_{\mu \nu}## are ##F_{\mu \nu} F^{\mu \nu}## and ##\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}##, which are proportional to ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. Now using the usual scalar product for the complex RS vector, shows that under Lorentz transformations the scalar product
    $$\vec{F} \cdot \vec{F} = (\vec{E}^2-\vec{B}^2) +2 \mathrm{i} \vec{E} \cdot \vec{B}$$
    stays constant. Thus the representation of the proper orthochronous Lorentz group for the RS vector is ##\mathrm{SO}(3,\mathbb{C})##. It turns out that the subgroup ##\mathrm{SO}(3,\mathbb{R})## is the representation of the rotations as it must be since ##\vec{E}## and ##\vec{B}## are real three-vectors. Rotations around a fixed direction with imaginary angles, represent the boosts in the corresponding direction. They mix electric and magnetic components.
     
  8. Jul 1, 2017 #7

    Orodruin

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    The set of anti-symmetric rank two tensors form a 6-dimensional vector space so: Yes, indeed the Faraday tensor transforms according to a 6-dimensional representation of the Lorentz group.

    It is not a vector in the typical nomenclature of relativity, but it is an element of a 6-dimensional vector space (the anti-symmetric rank two tensors) that allows a representation of the Lorentz group.

    I would not use the regular space-time indices to denote the components of this vector space though.

    The Faraday tensor indeed transforms under a 6-dimensional representation of the Lorentz group.
     
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