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In special relativity, one knows the Euclidean and Minkowski metric to raise or lower the index of a covector or vector. The Euclidean metric is a (1,1) tensor and is represented by a 3x3 identity matrix while the Minkowski metric in matrix form is a 4x4 diagonal matrix with all of its diagonal element unity except for the last one which is -1. But in the case of other field of physics which involves tensors like dielectric permitivity tensor (consider a rank 2 dielectric tensor) in nonlinear optics, what metric do I use to raise/lower the indexes?

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You use the metric of your coordinate system; if that's Cartesian, then it's a Kronecker delta. Since permitivity is measured in Farads per meter- emphasis "per meter"-- it should transform accordingly.

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  • $\begingroup$ If the metric is the Kronecker delta, then to change from E field vector into E field covector, I simply transpose it without complex conjugation? In other words the E field in vector and covector form are basically identical? Moreover, would you please describe what a contravariant tensor is? I only know contravariant coordinate. $\endgroup$
    – nougako
    May 15 '16 at 4:40
  • $\begingroup$ By the way, I think electric permittivity tensor does not act on a position vector, it acts on electric field vector. $\endgroup$
    – nougako
    May 15 '16 at 4:43
  • $\begingroup$ I think I meant covariant. To get a 1st understanding of the 2 types of transformations (of a vector)--consider a simple coordinate transform x --> 2x, that is double the length of the basis vector. Positions are cut in 1/2--which is "contra" to the variance of the basis vector--hence contravariance. Meanwhile, a gradient, such as Volts per meter would double--since the "volts" in space don't change, but the length of "delta x" doubles. So gradients vary with the change in basis: "covariance". $\endgroup$
    – JEB
    May 15 '16 at 4:47
  • $\begingroup$ Yes: it acts on the electric field (V/m) to give the displacement field (C/m**2)--hence its units are F/m. $\endgroup$
    – JEB
    May 15 '16 at 4:51
  • $\begingroup$ Then which metric do I use to change from E field vector to E field covector? If it's the kronecker delta like you said, then the covector is the same as the vector. I asked this because I am used to conjugate transpose which one uses in an inner product. I have got the impression that a tensor of rank (1,1) is similar to a inner product. $\endgroup$
    – nougako
    May 15 '16 at 4:56

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