Irodov 3.363: One Field Becomes Two, at Right Angles

Problem Irodov 3.363

Using the relativistic transformation formulas for the electric and magnetic fields, show that if an inertial frame $K$ contains only an electric field or only a magnetic field, then in any other inertial frame $K'$ there are both an electric and a magnetic field at the same time, with $\vect E' \perp \vect B'$ .

Given

in $K$ : only $\vect E$ , or only $\vect B$
$K'$ moves relative to $K$ with velocity $\vect v$

Find

show that

\vect E'

and

\vect B'

both exist and

\vect E' \perp \vect B'

Solution

Step 1The transformation formulas

Split each field into its part along $\vect v$ (subscript $\parallel$ ) and its part across it ( $\perp$ ), and let $\beta = v/c$ . The fields in $K'$ are

\begin{aligned} \vect E'_\parallel &= \vect E_\parallel, &\qquad \vect E'_\perp &= \frac{\vect E_\perp + \vect v\times\vect B}{\sqrt{1-\beta^2}},\\ \vect B'_\parallel &= \vect B_\parallel, &\qquad \vect B'_\perp &= \frac{\vect B_\perp - \vect v\times\vect E/c^2}{\sqrt{1-\beta^2}} \end{aligned} \tag{1}

Only the perpendicular part of a field enters a cross product with $\vect v$ : $\vect v\times\vect E = \vect v\times\vect E_\perp$ , and the same for $\vect B$ . Write $\gamma = 1/\sqrt{1-\beta^2}$ .

Step 2Only an electric field in

K

With $\vect B = 0$ , (1) gives

\vect E' = \vect E_\parallel + \gamma\,\vect E_\perp, \qquad \vect B' = -\frac{\gamma}{c^2}\,\vect v\times\vect E_\perp

Only E in K, at an angle to v. In K' its perpendicular part grows by γ, giving E' in the page, while B' appears normal to the page. — Only $\vect E$ in $K$ , at an angle to $\vect v$ . In $K'$ its perpendicular part grows by $\gamma$ , giving $\vect E'$ in the page, while $\vect B'$ appears normal to the page.

The vector $\vect v\times\vect E_\perp$ is perpendicular to $\vect v$ , hence to $\vect E_\parallel$ , and perpendicular to $\vect E_\perp$ , so both terms of $\vect E'$ are orthogonal to $\vect B'$ :

\vect E'\cdot\vect B' = -\frac{\gamma}{c^2}\left(\vect E_\parallel + \gamma\,\vect E_\perp\right)\cdot\left(\vect v\times\vect E_\perp\right) = 0

And $\vect B' \neq 0$ whenever $\vect E_\perp \neq 0$ : both fields are present in $K'$ .

Step 3Only a magnetic field in

K

With $\vect E = 0$ the roles swap:

\vect E' = \gamma\,\vect v\times\vect B_\perp, \qquad \vect B' = \vect B_\parallel + \gamma\,\vect B_\perp

Only B in K: in K' the field B' lies in the page and E' appears normal to it. — Only $\vect B$ in $K$ : in $K'$ the field $\vect B'$ lies in the page and $\vect E'$ appears normal to it.

By the same argument $\vect v\times\vect B_\perp$ is perpendicular to both $\vect B_\parallel$ and $\vect B_\perp$ :

\vect E'\cdot\vect B' = \gamma\left(\vect v\times\vect B_\perp\right)\cdot\left(\vect B_\parallel + \gamma\,\vect B_\perp\right) = 0

So in both cases

\vect E'\cdot\vect B' = 0

(2)

So $\vect E' \perp \vect B'$ , and both fields exist in $K'$ provided the field in $K$ has a part across $\vect v$ .

Check

The product $\vect E\cdot\vect B$ is a Lorentz invariant, and in $K$ it is zero because one of the two fields vanishes; so $\vect E'\cdot\vect B'$ must vanish in every frame, as found. The one exception to coexistence: if $\vect v$ is parallel to the field in $K$ , its perpendicular part is zero, and $K'$ again sees a single field equal to the original one.

Remark

The other invariant, $E^2 - c^2B^2$ , sorts such fields for good: positive for a pure electric field in $K$ , negative for a pure magnetic one. In $K'$ both fields are present, but the sign of $E'^2 - c^2B'^2$ still records which one was alone in $K$ ; a field with $\vect E\perp\vect B$ and $E < cB$ can always be transformed into a pure magnetic field, and one with $E > cB$ into a pure electric field.