#### The Field Tensor
red.rem.Fmunu

We have seen that a four-vector transforms according to \[ \bar{a}^\mu = \Lambda^\mu{}_\nu a^\nu \] in which \(\Lambda\) is a matrix representing the Lorentz transformation. The form this matrix takes depends on the actual transformation: for the specific case of motion in the \(x\) direction with velocity \(v\), \[ \Lambda^\mu{}_\nu = \left( \begin{array}{cccc} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \] (in which \(\mu\) is the "row" index and \(\nu\) the "column" one).

A four-vector is synonymous to a rank-one tensor. Higher-rank tensors are simply objects carrying more indices. For example, a rank-two tensor transforms as \[ \bar{t}^{\mu \nu} = \Lambda^\mu{}_\lambda \Lambda^\nu{}_\sigma t^{\lambda \sigma}. \] Such a rank-two tensor can be represented similarly to a matrix: \[ t^{\mu \nu} = \left( \begin{array}{cccc} t^{00} & t^{01} & t^{02} & t^{03} \\ t^{10} & t^{11} & t^{12} & t^{13} \\ t^{20} & t^{21} & t^{22} & t^{23} \\ t^{30} & t^{31} & t^{32} & t^{33} \end{array} \right). \] Special cases include symmetric \(t_s^{\mu \nu} = t_s^{\nu \mu}\) and antisymmetric \(t_a^{\mu \nu} = -t_a^{\nu \mu}\) tensors. The latter contains six independent elements: \[ t_a^{\mu \nu} = \left( \begin{array}{cccc} 0 & t_a^{01} & t_a^{02} & t_a^{03} \\ -t_a^{01} & 0 & t_a^{12} & t_a^{13} \\ -t_a^{02} & -t_a^{12} & 0 & t_a^{23} \\ -t_a^{03} & -t_a^{13} & -t_a^{23} & 0 \end{array} \right) \] Under the Lorentz transformation along \(x\) defined above, we can work out how the nonvanishing elements of an antisymmetric tensor transform:

\begin{align} \bar{t}_a^{01} &= \Lambda^0{}_\mu \Lambda^1{}_\nu t_a^{\mu \nu} = \Lambda^0{}_0 \Lambda^1{}_0 t_a^{00} + \Lambda^0{}_1 \Lambda^1{}_0 t_a^{10} + \Lambda^0{}_0 \Lambda^1{}_1 t_a^{01} + \Lambda^0{}_1 \Lambda^1{}_1 t_a^{11} \\ & = (\Lambda^0{}_0 \Lambda^1{}_1 - \Lambda^0{}_1 \Lambda^1{}_0) t_a^{01} = \gamma^2 (1 - \beta^2) t_a^{01} = t_a^{01}, \\ \bar{t}_a^{02} &= \Lambda^0{}_\mu \Lambda^2{}_\nu t_a^{\mu \nu} = \Lambda^0{}_0 \Lambda^2{}_2 t_a^{02} + \Lambda^0{}_1 \Lambda^2{}_2 t_a^{12} = \gamma (t_a^{02} - \beta t_a^{12}), \\ \bar{t}_a^{03} &= \Lambda^0{}_\mu \Lambda^3{}_\nu t_a^{\mu \nu} = \Lambda^0{}_0 \Lambda^3{}_3 t_a^{03} + \Lambda^0{}_1 \Lambda^3{}_3 t_a^{13} = \gamma (t_a^{03} - \beta t_a^{13}), \\ \bar{t}_a^{12} &= \Lambda^1{}_\mu \Lambda^2{}_\nu t_a^{\mu \nu} = \Lambda^1{}_0 \Lambda^2{}_2 t_a^{02} + \Lambda^1{}_1 \Lambda^2{}_2 t_a^{12} = \gamma (t_a^{12} - \beta t_a^{02}), \\ \bar{t}_a^{13} &= \Lambda^1{}_\mu \Lambda^3{}_\nu t_a^{\mu \nu} = \Lambda^1{}_0 \Lambda^3{}_3 t_a^{03} + \Lambda^1{}_1 \Lambda^3{}_3 t_a^{13} = \gamma (t_a^{13} - \beta t_a^{03}), \\ \bar{t}_a^{23} &= \Lambda^2{}_\mu \Lambda^3{}_\nu t_a^{\mu \nu} = \Lambda^2{}_2 \Lambda^3{}_3 t_a^{23} = t_a^{23}. \end{align}

Comparing with the transformation rules for the electromagnetic field which we obtained in EMtr, we can define the

Electromagnetic Field Tensor

\[ F^{\mu \nu} = \left( \begin{array}{cccc} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & B_z & -B_y \\ -E_y/c & -B_z & 0 & B_x \\ -E_z/c & B_y & -B_x & 0 \end{array} \right) \tag{Fmunu}\label{Fmunu} \]

together with the handy

Dual Field Tensor

\[ G^{\mu \nu} = \left( \begin{array}{cccc} 0 & B_x & B_y & B_z \\ -B_x & 0 & -E_z/c & E_y/c \\ -B_y & E_z/c & 0 & -E_x/c \\ -B_z & -E_y/c & E_x/c & 0 \end{array} \right) \tag{Gmunu}\label{Gmunu} \]

obtained from the field tensor by the substitution \({\boldsymbol E}/c \rightarrow {\boldsymbol B}\), \({\boldsymbol B} \rightarrow -{\boldsymbol E}/c\).

Our electromagnetic field transformation laws then become the simple

Lorentz Transformation Rules for EM Fields

\[ \bar{F}^{\mu \nu} = \Lambda^\mu{}_\lambda \Lambda^\nu{}_\sigma F^{\lambda \sigma} \tag{LorF}\label{LorF} \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42