How do derive the following transformation rule (J.D. Jackson third Edition 11.10) for electric and magnetic field? $$\vec E' = \gamma \left( \vec E + \vec \beta \times \vec B\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec E \right ) \tag{1}$$ $$\vec B' = \gamma \left( \vec B - \vec \beta \times \vec E\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec B \right ) \tag{2}$$ I know if $\beta$ is along positive x axis, the transformation of fields are given by $$\begin{align*} E_1' &= E_1 \\ E_2' &= \gamma (E_2 - \beta B_3)\\ E_3' &= \gamma (E_3 + \beta B_2) \\ \end{align*}\tag{3}$$

and for magnetic fields by

$$\begin{align*} B_1' &= B_1 \\ B_2' &= \gamma (B_2 + \beta E_3)\\ B_3' &= \gamma (B_3 - \beta E_2) \\ \end{align*} \tag{4}$$ here is how I think of it so far, taking $\beta = (\beta_1, 0, 0)$ and taking $x$ component of $(1)$ should give me first of equation $(3)$ but I get $$E_1' = E_1 \left( \gamma - \frac{\gamma^2}{1+\gamma} \beta_1^2\right) = \gamma E_1 $$

  • $\begingroup$ Seems to me that you just need to add the components of the vectors together and then work with the definitions of the cross and dot products. $\endgroup$
    – Kyle Kanos
    Feb 11, 2014 at 14:22
  • $\begingroup$ @KyleKanos i don't get that $\gamma^2$ part $\endgroup$ Feb 11, 2014 at 15:20
  • $\begingroup$ I would do a lorentz transformation on the field tensor. I might do this later if I have time. $\endgroup$ Feb 11, 2014 at 18:13

1 Answer 1


$\newcommand{\B}{\vec{B}^\times} \newcommand{\e}{\vec{E}} \renewcommand{\b}{\vec{\beta}} \newcommand{\bv}{\vec{B}}$ The field tensor can be written $\begin{pmatrix} 0 & -\e \\ \e & \B \end{pmatrix}$, Where $\B$ is the dual tensor to $\vec{B}$ defined by $\B \vec{v} = \vec{B} \times \vec{v}$. Equivalently, $(\B)_{ik} = \epsilon_{ijk} B_j$. Note $\vec{v}^T \B = (\vec{v} \times \vec{B})^T$. It will also be important to note that $$(\vec{v} \times \vec{w})^\times \vec{u} = \vec{w} (\vec{u} \cdot \vec{v}) - \vec{v} (\vec{u} \cdot \vec{w})$$, so that $$ (\vec{v} \times \vec{w})^\times = \vec{w} \otimes \vec{v} - \vec{v} \otimes \vec{w}$$.

The action of a Lorentz transformation can be written $$\begin{pmatrix} \gamma & -\gamma \b \\ -\gamma \b& 1+\alpha \b \otimes \b \end{pmatrix}$$, where $$\alpha = \frac{\gamma^2}{1+\gamma}$$. It will be important to note that $$\gamma^2 - \gamma \alpha = \gamma(\gamma - \alpha) = \gamma (\frac{\gamma + \gamma^2 - \gamma^2}{1 + \gamma}) = \frac{\gamma^2}{1+\gamma} = \alpha$$. Also $$1+\alpha \beta^2 = 1 + \alpha (1-1/\gamma^2) =1+ \frac{\gamma^2 -1}{1 + \gamma} =1+ \gamma -1 = \gamma$$

Anyway, the transformed field is

$$\begin{pmatrix} \gamma & -\gamma \b \\ -\gamma \b& 1+\alpha \b \otimes \b \end{pmatrix} \begin{pmatrix} 0& -\e \\ \e& \B \end{pmatrix} \begin{pmatrix} \gamma & -\gamma \b \\ -\gamma \b& 1+\alpha \b \otimes \b \end{pmatrix}$$.

Since the field tensor is antisymmetric, and the Lorentz transformation tensor is symmetric, we know the result must be antisymmetric. We will use this fact later. Let's start by compute the first product

$$\begin{pmatrix} \gamma & -\gamma \b \\ -\gamma \b& 1+\alpha \b \otimes \b \end{pmatrix} \begin{pmatrix} 0& -\e \\ \e& \B \end{pmatrix} = \begin{pmatrix} -\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\ \e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv) \end{pmatrix} $$.

Next we compute the second product. Since we already know this product will be antisymmetric, we will only calculate the right column. $$ \begin{pmatrix} -\gamma \b \cdot \e& -\gamma \e-\gamma \b \times \bv \\ \e + \alpha \b (\b \cdot \e)& \gamma \b \otimes \e + \B + \alpha \b \otimes (\b \times \bv) \end{pmatrix} \begin{pmatrix} \gamma & -\gamma \b \\ -\gamma \b& 1+\alpha \b \otimes \b \end{pmatrix} $$ $$ = \begin{pmatrix} 0 & \gamma^2 \b (\b \cdot \e) -\gamma \e-\gamma \b \times \bv -\alpha \gamma \b (\b \cdot \e) \\ \cdots & -\gamma \e \otimes \b - \alpha \gamma (\b \cdot \e) \b \otimes \b + \gamma \b \otimes \e + \B \\ & + \alpha \b \otimes (\b \times \bv) + \alpha \gamma (\b \cdot \e) \b \otimes \b + \alpha (\bv \times \b) \otimes \b \end{pmatrix} $$ $$ = \begin{pmatrix} 0 & -(\gamma (\e + \b \times \bv) - (\gamma^2 - \alpha \gamma) \b (\b \cdot \e)) \\ \cdots & \B -\gamma( \e \otimes \b - \b \otimes \e) - \alpha((\b \times \bv) \otimes \b - \b \otimes (\b \times \bv)) \end{pmatrix} $$ $$ =\begin{pmatrix} 0 & -(\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)) \\ \cdots & \B -\gamma( \b \times \e)^\times - \alpha(\b \times (\b \times \bv))^\times \end{pmatrix} $$ By now we have found the expected expression for the new electric field from the upper right entry: $\tilde{\e} =\gamma (\e + \b \times \bv) - \alpha \b (\b \cdot \e)$. Let's now focus on the bottom right entry. $$ \tilde{\bv}^\times=\B -\gamma( \b \times \e)^\times - \alpha((\b \cdot \bv) \b^\times - \beta^2 \B)$$ $$ = ((1+\alpha \beta^2)\bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) )^\times$$. Thus $$\tilde{\bv} = \gamma \bv - \gamma \b \times \e - \alpha \b (\b \cdot \bv) $$ $$ = \gamma(\bv - \b \times \e) - \alpha \b (\b \cdot \bv)$$ as was desired.


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