Pre-Quantum Electrodynamics

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

together with the handy

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