Irodov 3.367: Only the Cross Part Grows – Amit & Deepali Sinha Foundation

Problem Irodov 3.367

In an inertial frame $K$ there is no magnetic field, only a uniform electric field of strength $E = 8$ kV/m. A second inertial frame $K'$ moves relative to $K$ with a constant velocity $\vect v$ at the angle $\alpha = 45^\circ$ to $\vect E$ , with speed $v = \beta c$ , $\beta = 0.60$ . In $K'$ , find the magnitude and direction

of the electric field $\vect E'$ ;
of the magnetic field $\vect B'$ .

The uniform field E of the frame K at the angle α to the velocity v of the frame K'; the axes x and x' run along v. — The uniform field $\vect E$ of the frame $K$ at the angle $\alpha$ to the velocity $\vect v$ of the frame $K'$ ; the axes $x$ and $x'$ run along $\vect v$ .

Given

$E = 8$ kV/m, $\vect B = 0$ in $K$
$\alpha = 45^\circ$ between $\vect v$ and $\vect E$
$\beta = v/c = 0.60$

Find

(a)

E'

and its direction · (b)

B'

and its direction

Solution

Step 1Split the field along and across the velocity

The relativistic transformation treats the parts of a field along $\vect v$ and across it differently, so resolve $\vect E$ that way:

E_\parallel = E\cos\alpha, \qquad E_\perp = E\sin\alpha \tag{1}

With $\vect B = 0$ in $K$ and $\gamma = 1/\sqrt{1-\beta^2}$ , the transformation of the fields to $K'$ reads

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

Step 2The magnitude and direction of E’

Adding the squares of the two parts of $\vect E'$ from (2), with (1):

E'^2 = E^2\cos^2\alpha + \frac{E^2\sin^2\alpha}{1-\beta^2} = E^2\,\frac{(1-\beta^2)\cos^2\alpha + \sin^2\alpha}{1-\beta^2} = E^2\,\frac{1-\beta^2\cos^2\alpha}{1-\beta^2}

Numerically, with $\cos^2\alpha = 0.50$ :

E' = 8\sqrt{\frac{1 - 0.36\cdot 0.50}{1 - 0.36}}\ \text{kV/m}

E' = E\sqrt{\frac{1-\beta^2\cos^2\alpha}{1-\beta^2}} \approx 9.1\text{ kV/m}

(3)

The vector $\vect E'$ lies in the plane of $\vect v$ and $\vect E$ . Its angle $\alpha'$ with $\vect v$ is fixed by the ratio of its two parts:

\tan\alpha' = \frac{E'_\perp}{E'_\parallel} = \frac{E\sin\alpha}{\sqrt{1-\beta^2}\,E\cos\alpha}

\tan\alpha' = \frac{\tan\alpha}{\sqrt{1-\beta^2}}, \quad \alpha' \approx 51^\circ

(4)

So $\vect E'$ is turned away from $\vect v$ , from $45^\circ$ to about $51^\circ$ , on the same side as $\vect E$ .

The fields in K', drawn for β = 0.60: E' keeps the part of E along v and stretches the cross part by γ, so it leans away from v at α'; B' points into the page. — The fields in $K'$ , drawn for $\beta = 0.60$ : $\vect E'$ keeps the part of $\vect E$ along $\vect v$ and stretches the cross part by $\gamma$ , so it leans away from $\vect v$ at $\alpha'$ ; $\vect B'$ points into the page.

Step 3The magnetic field in K’

The magnitude of $\vect v\times\vect E$ is $vE\sin\alpha$ , so by (2), with $v = \beta c$ ,

B' = \gamma\,\frac{vE\sin\alpha}{c^2} = \frac{\gamma\beta E\sin\alpha}{c}

Numerically:

B' = \frac{0.60\cdot(8\times10^{3})\cdot 0.707}{(3.00\times10^{8})\cdot 0.80}\ \text{T}

B' = \frac{\beta E\sin\alpha}{c\sqrt{1-\beta^2}} \approx 14\ \mu\text{T}

(5)

Its direction is that of $-\vect v\times\vect E = \vect E\times\vect v$ : normal to the plane of $\vect v$ and $\vect E$ , and so perpendicular to $\vect E'$ as well.

Check

Limits: at $\beta \to 0$ both answers give $E' = E$ , $\alpha' = \alpha$ , $B' = 0$ ; at $\alpha = 0$ (motion along the field) $E' = E$ and $B' = 0$ . Invariant: from (3) and (5),

E'^2 - c^2B'^2 = E^2\,\frac{1-\beta^2\cos^2\alpha - \beta^2\sin^2\alpha}{1-\beta^2} = E^2

the same as in $K$ ; with the numbers, $9.055^2 - 4.243^2 = 64.0 = 8^2$ in $(\text{kV/m})^2$ , where $cB' = 4.243$ kV/m.

Remark

Because only the cross part is stretched, a moving observer always sees the electric field lean away from the observer’s own velocity, toward the plane normal to it: the same flattening that crowds the field of a fast charge into its equatorial plane. A field that is purely electric in one frame is purely electric in no other, unless the second frame moves along the field; and since $\vect B'$ comes out normal to $\vect E'$ , the two fields seen in $K'$ are always at right angles.