0
$\begingroup$

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 $$E_1' = E_1 \left( \gamma - \frac{\gamma^2}{1+\gamma} \beta_1^2\right) = \gamma E_1 $$

$\endgroup$
  • $\begingroup$ 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 '14 at 14:22
  • $\begingroup$ @KyleKanos i don't get that $\gamma^2$ part $\endgroup$ – Mula Ko Saag Feb 11 '14 at 15:20
  • $\begingroup$ I would do a lorentz transformation on the field tensor. I might do this later if I have time. $\endgroup$ – Brian Moths Feb 11 '14 at 18:13
1
$\begingroup$

$\newcommand{\B}{\vec{B}^\times} \newcommand{\e}{\vec{E}} \renewcommand{\b}{\vec{\beta}} \newcommand{\bv}{\vec{B}}$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.