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(This page includes several similer ways to develop transformation set).
let us see how Electric and magnetic fields transform under different frames of reference.
\begin{tikzpicture}
% S frame
\draw (0,2.6) -- (0,0) -- (3.6,0);
\draw[->] (0,2.2) -- (0,2.6);
\node[above right] at (0,2.6) {$S$};
\fill (1.1,1.55) circle (1.4pt);
\node[above] at (1.1,1.62) {\small $q$};
\node[below] at (1.1,1.40) {\small Rest};
\node[anchor=west] at (1.35,2.05) {\small $(E_x\ E_y\ E_z)$};
\node[anchor=west] at (1.35,1.55) {\small $(B_x\ B_y\ B_z)$};
% S' frame
\draw (4.6,2.6) -- (4.6,1.15) -- (7.4,1.15);
\node[above] at (4.6,2.65) {$S'$};
\draw[->] (4.6,2.05) -- (5.15,2.05) node[right] {$u$};
\node[anchor=west] at (5.6,2.35) {\small $(E_x',\ E_y',\ E_z')$ \& $(B_x'\ B_y'\ B_z')$};
\end{tikzpicture}
All these quantities are tensors
\[
\begin{aligned}
F_x &= qE_x\\ F_y &= qE_y\\ F_z &= qE_z
\end{aligned}
\qquad
\begin{aligned}
F_x' &= q\!\left( E_x' + (v'\times B')_x \right)\\
F_y' &= q\!\left( E_y' + (v'\times B')_y \right)\\
F_z' &= q\!\left( E_z' + (v'\times B')_z \right)
\end{aligned}
\]
\[
\vec{v}\,'\times\vec{B}\,' \;=\;
\begin{vmatrix}
\hat{\imath}\,' & \hat{\jmath}\,' & \hat{k}' \\
-u & 0 & 0\\
B_x' & B_y' & B_z'
\end{vmatrix}
\qquad
\left( \vec{v}\,' = -\vec{u} \right)
\]
(here \(\vec{v}\,'\) = velocity of particle as seen from \(S'\) frame)
\[
(\vec{v}\,'\times\vec{B}\,') \;=\; \hat{\imath}\,'(0) - \hat{\jmath}\,'(-uB_z') + \hat{k}'(-uB_y')
\]
\[
(\vec{v}\,'\times\vec{B}\,')_x = 0, \qquad (\vec{v}\,'\times\vec{B}\,')_y = uB_z', \qquad (\vec{v}\,'\times\vec{B}\,')_z = -uB_y'
\]
Now we will use transformation to see how \(\vec{E}\) field & \(\vec{B}\) transform
\[
\text{We know that,} \qquad F_x' \;=\; \frac{dp_x'}{dt'} \;=\; \frac{\left( dp_x - \beta\,dp^0 \right)\gamma}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{\left( F_x - \dfrac{u}{c} \right)\dfrac{1}{c}\dfrac{dE}{dt}}{\left( 1 - \dfrac{u v_x}{c^2} \right)}
\]
\[
\text{here} \quad \frac{dx}{dt} = \text{speed of particle from `}S\text{' frame.} \qquad \left( v_x = 0 \right)
\]
\[
\Rightarrow \qquad F_x' \;=\; \frac{\left( F_x - \dfrac{u}{c^2}\left( \vec{F}\cdot\vec{v} \right) \right)}{\left( 1 - u v_x/c^2 \right)} \;=\; \frac{\left( F_x - \dfrac{u}{c^2}F_xv_x \right)}{\left( 1 - \dfrac{u v_x}{c^2} \right)}
\qquad
\left\{
\begin{aligned}
\vec{F} &= \left( F_x, F_y, F_z \right)\\
\vec{v} &= \left( v_x, 0, 0 \right)\\
\vec{F}\cdot\vec{v} &= F_xv_x
\end{aligned}
\right.
\]
\[
\text{here} \quad v \;=\; \text{speed of charge particle in } S \text{ frame} \;=\; \frac{dx}{dt} \;=\; v_x \;=\; 0
\]
\[
\Rightarrow \qquad \left( F_x' \;=\; \frac{F_x - \dfrac{u}{c^2}(0)}{1-0} \;=\; F_x \right)
\]
\[
\text{Similerly we have,} \qquad F_y' \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{1-0} \;=\; \frac{F_y}{\gamma}
\]
\[
F_z' \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{1-0} \;=\; \frac{F_z}{\gamma}
\]
Now using, (1)
\[
F_x' \;=\; F_x \quad \Rightarrow \quad q\!\left( E_x' + (v'\times B')_x \right) \;=\; q E_x
\]
\[
E_x' - 0 \;=\; E_x
\]
\[
\boxed{\; E_x' = E_x \;}
\]
\[
(2)\quad F_y' = \frac{F_y}{\gamma} \quad \Rightarrow \quad q\!\left( E_y' + (\vec{v}\times\vec{B})_y \right) \;=\; \frac{qE_y}{\gamma}
\]
\[
\Rightarrow \qquad E_y' + uB_z' \;=\; \frac{E_y}{\gamma}
\]
\[
\boxed{\; E_y \;=\; \left( E_y' + uB_z' \right)\gamma \;}
\]
\[
(3)\quad F_z' = \frac{F_z}{\gamma} \quad \Rightarrow \quad q\!\left( E_z' + (\vec{v}\times B')_z \right) \;=\; \frac{qE_z}{\gamma}
\]
\[
E_z' - uB_y' \;=\; \frac{E_z}{\gamma}
\]
\[
\Rightarrow \qquad \boxed{\; E_z \;=\; \left( E_z' - uB_y' \right)\gamma \;}
\]
But we want to find \(E_x' = f(\vec{E},\vec{B})\), \(E_y' = g(\vec{E},\vec{B})\) & \(E_z' = h(\vec{E},\vec{B})\)
To achieve it we need to just change sign of velocity of frame of reference.
(velocity of particle in \(S'\) is \(v' = -u\))
\[
F_x' \;=\; \frac{F_x - \dfrac{u}{c^2}\left( \vec{F}\cdot\vec{v} \right)}{1 - \dfrac{u v_x}{c^2}}
\ \Rightarrow \
F_x \;=\; \frac{F_x' + \dfrac{u}{c^2}\left( \vec{F}\,'\cdot v' \right)}{1 + \dfrac{u V_x'}{c^2}}
\ \Rightarrow \
F_x \;=\; \frac{F_x' - \dfrac{u^2}{c^2}F_x'}{1 - \dfrac{u^2}{c^2}} \;=\; F_x'
\]
\[
F_y' \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}}
\ \Rightarrow \
F_y \;=\; \frac{F_y'\sqrt{1-u^2/c^2}}{1 + \dfrac{u V_x'}{c^2}}
\ \Rightarrow \
F_y \;=\; \frac{F_y'\sqrt{1-u^2/c^2}}{1 - u^2/c^2} \;=\; \gamma F_y'
\]
\[
F_z' \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}}
\ \Rightarrow \
F_z \;=\; \frac{F_z'\sqrt{1-u^2/c^2}}{1 + \dfrac{u V_x'}{c^2}}
\ \Rightarrow \
F_z \;=\; \frac{F_z'\sqrt{1-u^2/c^2}}{1-u^2/c^2} \;=\; \gamma F_z'
\]
(R.H.S should be quantities measured in \(S'\) frame)
So from our eqn set of \(\vec{F}\) & \(\vec{F}\,'\) we have,
\[
F_x = F_x' \ , \qquad F_y = F_y'\,\gamma \ , \qquad F_z = \gamma F_z'
\qquad
\text{Where} \quad \gamma = \frac{1}{\sqrt{1-u^2/c^2}}
\]
\[
\begin{aligned}
qE_x &= q\!\left( E_x' + (\vec{v}\,'\times B')_x \right)\\
q(E_y) &= \gamma\,q\!\left( E_y' + (v'\times B')_y \right)\\
q(E_z) &= \gamma\left( E_z' + (\vec{v}\,'\times\vec{B}\,')_z \right)
\end{aligned}
\]
\[
\boxed{\;
E_x = E_x' + 0 \ , \qquad
E_y = \gamma\left( E_y' + uB_z' \right) \ , \qquad
E_z = \gamma\left( E_z' - uB_y' \right) \;}
\]
\(\left( E_x',\ E_y',\ E_z' \right)\) we reverse the direction of moving frame \(u \leftrightarrow (-u)\)
\[
\boxed{\;
E_x' = E_x \ , \qquad
E_y' = \gamma\left( E_y - uB_z \right) \ , \qquad
E_z' = \gamma\left( E_z + uB_y \right) \;}
\]
We can find these transformation from below problem
\begin{tikzpicture}
\draw (0,2.1) -- (0,0.35) -- (2.5,0.35);
\draw[->] (0,1.75) -- (0,2.1);
\node[above right] at (0,2.1) {$S$};
\fill (0.9,1.15) circle (1.4pt);
\node[above] at (0.9,1.22) {\small $q$};
\draw[->] (1.0,1.15) -- (1.55,1.15) node[right] {\small $u$};
\draw (3.3,2.1) -- (3.3,1.15) -- (5.5,1.15);
\node[above] at (3.3,2.15) {$S'$};
\draw[->] (3.3,1.65) -- (3.85,1.65) node[right] {\small $u$};
\end{tikzpicture}
\[
\begin{aligned}
V &= V_{\text{particle}}\Big)_S \;=\; u\,\hat{\imath}\\
V_{S'} &= u\,\hat{\imath}\\
V_{\text{particle}}\Big)_{S'} &= V' \;=\; 0
\end{aligned}
\]
\[
\begin{aligned}
F_x &= q\!\left( E_x + (\vec{v}\times\vec{B})_x \right)\\
F_y &= q\!\left( E_y + (\vec{v}\times\vec{B})_y \right)\\
F_z &= q\!\left( E_z + (\vec{v}\times\vec{B})_z \right)
\end{aligned}
\qquad
\left.
\begin{aligned}
F_x' &= q\!\left( E_x' \right)\\
F_y' &= q\,E_y'\\
F_z' &= q\,E_z'
\end{aligned}
\right\}
\quad
\begin{aligned}&\text{Since } V' = 0\\ &(V'\times B') = 0\end{aligned}
\]
using force transformation
\[
F_x' \;=\; \frac{F_x - \dfrac{u}{c^2}\left( \vec{F}\cdot\vec{v} \right)}{1 - \dfrac{u V_x}{c^2}} \;=\; \frac{F_x - \dfrac{u^2}{c^2}F_x}{1 - u^2/c^2} \;=\; F_x
\]
\[
F_y' \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}} \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{1-u^2/c^2} \;=\; \gamma F_y
\]
\[
F_z' \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}} \;=\; \frac{F_z}{\sqrt{1-u^2/c^2}} \;=\; \gamma F_z
\]
Above transformation implies.
\[
1)\quad F_x' = F_x \quad \Rightarrow \quad q E_x' \;=\; q\!\left( E_x + (\vec{v}\times\vec{B})_x \right) \;=\; q\!\left( E_x + V_yB_z - V_zB_y \right)
\]
(with \(V_y = V_z = 0\))
\[
\Rightarrow \qquad \boxed{\; E_x' = E_x \;}
\]
\[
2)\quad F_y' = \gamma F_y \quad \Rightarrow \quad q E_y' \;=\; \gamma q\!\left( E_y + (\vec{v}\times\vec{B})_y \right) \;=\; \gamma q\!\left( E_y + V_zB_x - V_xB_z \right)
\]
\[
\left\{\; E_y' \;=\; \gamma\left( E_y - uB_z \right) \;\right\}
\]
\[
3)\quad F_z' = \gamma F_z \quad \Rightarrow \quad q(E_z') \;=\; \gamma q\!\left( E_z + (\vec{v}\times\vec{B})_z \right) \;=\; \gamma q\!\left( E_z + V_xB_y - V_yB_x \right)
\]
\[
\left\{\; E_z' \;=\; \gamma\left( E_z + uB_y \right) \;\right\}
\]
\[
\boxed{\;
\left.
\begin{aligned}
E_x' &= E_x\\
E_y' &= \gamma\left( E_y - uB_z \right)\\
E_z' &= \gamma\left( E_z + uB_y \right)
\end{aligned}
\right\}
\quad (\text{And}) \quad
\left\{
\begin{aligned}
E_x &= E_x'\\
E_y &= \gamma\left( E_y' + uB_z' \right)\\
E_z &= \gamma\left( E_z' - uB_y' \right)
\end{aligned}
\right. \;}
\]
(the right-hand set follows from the left by \(u \to -u\))
Together represents transformation rules for \(\vec{E}\) from one frame to another.
If \(\left( E_x = E_y = E_z = 0 \right)\) we can see that \(\vec{E}\,' \neq 0\)
\[
E_x' = 0, \qquad E_y' = -\gamma\,uB_z, \qquad E_z' = \gamma\,uB_y
\]
So \(\vec{E}\) is relativistic menifestation of magnetic field. and vice versa.
We have similar transformation of \(\vec{B}\) as well.
\[
\boxed{\; B_x' = B_x \ , \qquad B_y' = \gamma\left( B_y + \frac{u}{c^2}E_z \right) \ , \qquad B_z' = \gamma\left( B_z - \frac{u}{c^2}E_y \right) \;}
\]
& inverse transformations
\[
\boxed{\; \left( B_x = B_x' \right) \ , \qquad B_y = \gamma\left( B_y' - \frac{u}{c^2}E_z' \right) \ , \qquad B_z = \gamma\left( B_z' + \frac{u}{c^2}E_y' \right) \;}
\]
\begin{tikzpicture}
\draw (0,2.1) -- (0,0.35) -- (2.7,0.35);
\draw[->] (0,1.75) -- (0,2.1);
\node[above left] at (0.05,2.1) {$y$};
\node[above right] at (0.15,2.1) {$S$};
\draw[->] (2.5,0.35) -- (2.9,0.35) node[right] {$x$};
\fill (0.75,1.05) circle (1.4pt);
\node[left] at (0.70,1.05) {\small $q$};
\draw[->] (0.75,1.15) -- (0.75,1.75) node[above] {\small $V$};
% S'
\draw (3.6,2.1) -- (3.6,0.35) -- (6.4,0.35);
\draw[->] (3.6,1.75) -- (3.6,2.1);
\node[above left] at (3.65,2.1) {$y'$};
\node[above right] at (3.75,2.1) {$S'$};
\draw[->] (6.2,0.35) -- (6.7,0.35) node[right] {$x'$};
\draw[->] (3.6,1.35) -- (4.15,1.35) node[right] {\small $u$};
\end{tikzpicture}
\[
\begin{aligned}
F_x &= q\!\left( E_x + (v\times B)_x \right)\\
F_y &= q\!\left( E_y + (\vec{v}\times\vec{B})_y \right)\\
F_z &= q\!\left( E_z + (\vec{v}\times\vec{B})_z \right)
\end{aligned}
\qquad
\begin{aligned}
F_x' &= q\!\left( E_x' + (\vec{v}\,'\times\vec{B}\,')_x \right)\\
F_y' &= q\!\left( E_y' + (\vec{v}\,'\times\vec{B}\,')_y \right)\\
F_z' &= q\!\left( E_z' + (\vec{v}\,'\times\vec{B}\,')_z \right)
\end{aligned}
\]
\[
\text{let us find } V' = ? \qquad V_x' \;=\; \frac{dx'}{dt'} \;=\; \frac{\left( dx - u\,dt \right)\gamma}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{\left( V_x - u \right)}{1 - \dfrac{u V_x}{c^2}} \;=\; -u \qquad (V_x = 0)
\]
\[
\text{similarly,} \quad V_y' \;=\; \frac{dy'}{dt'} \;=\; \frac{dy}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{V_y\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}} \;=\; V\sqrt{1-u^2/c^2}
\]
\[
\text{and,} \quad V_z' \;=\; \frac{dz'}{dt'} \;=\; \frac{dz}{\left( dt + \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; 0 \qquad (V_z = 0)
\]
\[
\vec{v}\times\vec{B} \;=\; \hat{\imath}\left( V_yB_z - V_zB_y \right) + \hat{\jmath}\left( V_zB_x - V_xB_z \right) + \hat{k}\left( V_xB_y - V_yB_x \right)
\]
\[
\vec{v}\times\vec{B} \;=\; \hat{\imath}\left( V_yB_z \right) + \hat{\jmath}(0) + \hat{k}\left( -V_yB_x \right)
\]
\[
\vec{v}\,'\times\vec{B}\,' \;=\; \hat{\imath}\,'\!\left( V_y'B_z' - V_z'B_y' \right) + \hat{\jmath}\,'\!\left( V_z'B_x' - V_x'B_z' \right) + \hat{k}'\!\left( V_x'B_y' - V_y'B_x' \right)
\]
\[
\vec{v}\,'\times\vec{B}\,' \;=\; \hat{\imath}\,'\!\left( \frac{VB_z'}{\gamma} \right) + \hat{\jmath}\,'\!\left( +uB_z' \right) + \hat{k}'\!\left( -uB_y' - \frac{V}{\gamma}B_x' \right)
\]
\[
\begin{aligned}
F_x &= q\!\left( E_x + VB_z \right)\\
F_y &= q\,E_y\\
F_z &= q\!\left( E_z - VB_x \right)
\end{aligned}
\qquad
\begin{aligned}
F_x' &= q\!\left( E_x' + VB_z'/\gamma \right)\\
F_y' &= q\!\left( E_y' + uB_z' \right)\\
F_z' &= q\!\left( E_z' - uB_y' - \frac{V}{\gamma}B_x' \right)
\end{aligned}
\]
use, force transformation to find relations -
\[
(1)\quad F_x' \;=\; \frac{dp_x'}{dt'} \;=\; \frac{\left( dp_x - \beta\,dp^0 \right)\gamma}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{\left( F_x - \dfrac{u}{c}\dfrac{1}{c}\dfrac{dE}{dt} \right)}{\left( 1 - \dfrac{u V_x}{c^2} \right)} \;=\; \frac{F_x - \dfrac{u}{c^2}\left( \vec{F}\cdot\vec{v} \right)}{\left( 1 - \dfrac{u V_x}{c^2} \right)} \;=\; F_x - \frac{u}{c^2}F_yV_y
\]
\[
F_x' \;=\; F_x - \frac{u}{c^2}F_yV
\]
\[
\Rightarrow \qquad \left( \frac{F_x'}{q} + \frac{VB_z'}{\gamma} \right) \;=\; F_x + VB_z - \frac{u}{c^2}E_yV
\]
\[
\text{We know,} \quad \left( E_x' = E_x \right) \quad \text{So,} \qquad \boxed{\; B_z' \;=\; \gamma\left( B_z - \frac{u\,E_y}{c^2} \right) \;}
\]
\[
(2)\quad F_y' \;=\; \frac{dp_y'}{dt'} \;=\; \frac{dp_y}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{1 - \dfrac{u V_x}{c^2}} \;=\; \frac{F_y}{\gamma} \qquad \left( V_x = 0 \right)
\]
\[
q\!\left( E_y' + uB_z' \right) \;=\; \frac{qE_y}{\gamma} \qquad \Rightarrow \qquad uB_z' \;=\; \frac{E_y}{\gamma} - E_y'
\]
\[
B_z' \;=\; \frac{1}{u}\left( \frac{E_y}{\gamma} - \gamma\left( E_y - uB_z \right) \right) \;=\; \frac{1}{\gamma u}\left( E_y - \gamma^2\left( E_y - uB_z \right) \right)
\]
\[
B_z' \;=\; \frac{1}{u\gamma}\left( E_y(1-\gamma^2) + uB_z\gamma^2 \right) \;=\; \frac{\gamma^2}{u\gamma}\left( E_y\left( \frac{1}{\gamma^2} - 1 \right) + uB_z \right)
\]
\[
B_z' \;=\; \frac{\gamma}{u}\left( E_y\left( 1 - \frac{u^2}{c^2} - 1 \right) + uB_z \right)
\]
\[
B_z' \;=\; \frac{u}{u}\,\gamma\left( B_z - \frac{u\,E_y}{c^2} \right)
\]
\[
\boxed{\; B_z' \;=\; \gamma\left( B_z - \frac{u\,E_y}{c^2} \right) \;}
\]
\[
(3)\quad F_z' \;=\; \frac{dp_z'}{dt'} \;=\; \frac{dp_z}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{F_z}{\left( 1 - \dfrac{u V_x}{c^2} \right)\gamma} \;=\; \frac{E_z}{\gamma} \qquad \because\ \left( V_x = 0 \right)
\]
\[
\text{So,} \qquad q\!\left( E_z' - uB_y' - \frac{V}{\gamma}B_x' \right) \;=\; \frac{q\left( E_z - VB_x \right)}{\gamma}
\]
\[
E_z' - uB_y' - \frac{V}{\gamma}B_x' \;=\; \frac{E_z - VB_x}{\gamma}
\]
\[
\gamma\left( E_z + uB_y \right) - uB_y' - \frac{V}{\gamma}B_x' \;=\; \frac{E_z - VB_x}{\gamma}
\]
\[
uB_y' + \frac{V}{\gamma}B_x' \;=\; \gamma E_z + \gamma uB_y - \frac{E_z}{\gamma} + \frac{VB_x}{\gamma}
\]
\[
= \frac{E_z\left( \gamma^2 - 1 \right)}{\gamma} + B_y u\gamma + \frac{VB_x}{\gamma}
\]
\[
= \gamma E_z\left( 1 - \left( 1 - \frac{u^2}{c^2} \right) \right) + \cdots
\]
\[
\left( uB_y' + \frac{V}{\gamma}B_x' \;=\; \gamma\,E_z\,\frac{u^2}{c^2} + B_y u\gamma + \frac{VB_x}{\gamma} \right)
\]