Electromagnetic field tensor: Lorentz transformations

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 7; Problems 7.1.

Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.48.

To apply this to Griffiths problem 12.48, use

\displaystyle  t^{ij}=\left[\begin{array}{cccc} 0 & t^{01} & t^{02} & t^{03}\\ -t^{01} & 0 & t^{12} & t^{13}\\ -t^{02} & -t^{12} & 0 & t^{23}\\ -t^{03} & -t^{13} & -t^{23} & 0 \end{array}\right] \ \ \ \ \ (1)

in place of {F^{ij}} in what follows. The results are the same.

The electromagnetic field tensor is

\displaystyle  F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (2)

We can use the usual tensor transformation rules to see how the electric and magnetic fields transform under a Lorentz transformation. We get

\displaystyle   F^{\prime ij} \displaystyle  = \displaystyle  \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}}F^{kl}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \Lambda_{\;\; k}^{i}\Lambda_{\;\; l}^{j}F^{kl} \ \ \ \ \ (4)

where the Lorentz transformation matrix is

\displaystyle  \Lambda_{\;\; k}^{i}=\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] \ \ \ \ \ (5)

As we saw when discussing the inertia tensor, we can write this transformation as a matrix equation

\displaystyle  F^{\prime}=\Lambda F\Lambda^{T} \ \ \ \ \ (6)

The first product is

\displaystyle  \Lambda F=\left[\begin{array}{cccc} \gamma\beta E_{x} & \gamma E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -\gamma E_{x} & -\gamma\beta E_{x} & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (7)

The final product is

\displaystyle  F^{\prime}=\Lambda F\Lambda^{T}=\left[\begin{array}{cccc} 0 & \gamma^{2}\left(1-\beta^{2}\right)E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -\gamma^{2}\left(1-\beta^{2}\right)E_{x} & 0 & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -\gamma E_{y}+\gamma\beta B_{z} & \gamma\beta E_{y}-\gamma B_{z} & 0 & B_{x}\\ -\gamma E_{z}-\gamma\beta B_{y} & \gamma\beta E_{z}+\gamma B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (8)

Using {\gamma=1/\sqrt{1-\beta^{2}}} we get

\displaystyle  F^{\prime}=\left[\begin{array}{cccc} 0 & E_{x} & \gamma E_{y}-\gamma\beta B_{z} & \gamma E_{z}+\gamma\beta B_{y}\\ -E_{x} & 0 & -\gamma\beta E_{y}+\gamma B_{z} & -\gamma\beta E_{z}-\gamma B_{y}\\ -\gamma E_{y}+\gamma\beta B_{z} & \gamma\beta E_{y}-\gamma B_{z} & 0 & B_{x}\\ -\gamma E_{z}-\gamma\beta B_{y} & \gamma\beta E_{z}+\gamma B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (9)

From this, we see that

\displaystyle   E_{x}^{\prime} \displaystyle  = \displaystyle  E_{x}\ \ \ \ \ (10)
\displaystyle  E_{y}^{\prime} \displaystyle  = \displaystyle  \gamma E_{y}-\gamma\beta B_{z}\ \ \ \ \ (11)
\displaystyle  E_{z}^{\prime} \displaystyle  = \displaystyle  \gamma E_{z}+\gamma\beta B_{y}\ \ \ \ \ (12)
\displaystyle  B_{x}^{\prime} \displaystyle  = \displaystyle  B_{x}\ \ \ \ \ (13)
\displaystyle  B_{y}^{\prime} \displaystyle  = \displaystyle  \gamma\beta E_{z}+\gamma B_{y}\ \ \ \ \ (14)
\displaystyle  B_{z}^{\prime} \displaystyle  = \displaystyle  -\gamma\beta E_{y}+\gamma B_{z} \ \ \ \ \ (15)

Unlike lengths, the components of {E} and {B} in the direction of motion are unchanged, while those perpendicular to the motion are altered.

9 thoughts on “Electromagnetic field tensor: Lorentz transformations

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  2. Peter

    I’m not sure, but I think that the last expression of B’z should be = -γβEy + γBz and B’y = γβEz + γBy

    In any case the article was really useful, thank you for that!

    Reply
  3. Anonymous

    “The first product is:
    \Lambda F = [Matrix]”

    But the first element of the second line is $+\gamma E_x$, and not $-\gamma E_x$

    Reply
    1. growescience

      Element {left(Lambda Fright)_{21}} in equation 6 is calculated from the second row of {Lambda} multiplied into the first column of {F}, which gives {-gamma E_{x}}, so it’s correct as it is.

      Reply
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