Irodov 3.371: A Fast Charge Flattens Its Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.371

A point charge $q$ moves in a straight line at the constant speed $v = \beta c$ , comparable with the speed of light. At some instant the vector from the charge to a field point is $\vect r$ , at the angle $\theta$ to the charge’s velocity. Find the electric field $\vect E$ of the charge at that point at that instant.

The charge q moving with velocity v (drawn for q > 0) and the vector r from its position to the field point P, at the angle θ to v. — The charge $q$ moving with velocity $\vect v$ (drawn for $q > 0$ ) and the vector $\vect r$ from its position to the field point $P$ , at the angle $\theta$ to $\vect v$ .

Given

$q$
uniform straight motion, $\beta = v/c$
$\vect r$ from the charge, at angle $\theta$ to $\vect v$

Find

\vect E

at the tip of

\vect r

Solution

Step 1The field in the charge’s rest frame

Let $K$ be the frame of the problem, with its $x$ axis along $\vect v$ , and $K'$ the frame moving with the charge, its origin on the charge and its $x'$ axis along $x$ . In $K'$ the charge is at rest, so its field is the Coulomb field and there is no magnetic field:

\vect E' = \frac{q\,\vect r'}{4\pi\varepsilon_0 r'^3}, \qquad \vect B' = 0 \tag{1}

Step 2Relating the positions

At the chosen instant of $K$ the charge is at the origin and the field point at $x = r\cos\theta$ , $y = r\sin\theta$ . The rest frame sees lengths along $\vect v$ longer by the Lorentz factor, $\gamma = 1/\sqrt{1-\beta^2}$ , and lengths across $\vect v$ unchanged:

x' = \gamma x, \qquad y' = y \tag{2}

Step 3The field back in the frame K

Relative to $K'$ the frame $K$ moves with $-\vect v$ . With $\vect B' = 0$ the transformation leaves the part of the field along $\vect v$ alone and multiplies the cross part by $\gamma$ :

E_x = E'_x, \qquad E_y = \gamma E'_y \tag{3}

Taking the components of (1) and using (2):

E_x = \frac{q\,x'}{4\pi\varepsilon_0 r'^3} = \frac{\gamma q\,x}{4\pi\varepsilon_0 r'^3}, \qquad E_y = \gamma\,\frac{q\,y'}{4\pi\varepsilon_0 r'^3} = \frac{\gamma q\,y}{4\pi\varepsilon_0 r'^3}

Both components carry the same factor, so the field is along $\vect r$ :

\vect E = \frac{\gamma q\,\vect r}{4\pi\varepsilon_0 r'^3} \tag{4}

Step 4The rest-frame distance

It remains to express $r'$ through $r$ and $\theta$ . By (2), with $\gamma^2(1-\beta^2) = 1$ :

r'^2 = \gamma^2x^2 + y^2 = \gamma^2r^2\bigl[\cos^2\theta + (1-\beta^2)\sin^2\theta\bigr] = \gamma^2r^2\,(1-\beta^2\sin^2\theta)

Putting $r'^3 = \gamma^3r^3(1-\beta^2\sin^2\theta)^{3/2}$ into (4), and $1/\gamma^2 = 1-\beta^2$ :

\vect E = \frac{q\,\vect r}{4\pi\varepsilon_0 r^3}\,\frac{1-\beta^2}{(1-\beta^2\sin^2\theta)^{3/2}}

(5)

The field points straight away from the charge’s present position (toward it for $q < 0$ ), but its strength now depends on the direction.

The field lines of the moving charge, drawn for β = 0.8 with equal flux between neighbouring lines in the drawn plane: they crowd toward the plane normal to v. E at the tip of r lies along r. — The field lines of the moving charge, drawn for $\beta = 0.8$ with equal flux between neighbouring lines in the drawn plane: they crowd toward the plane normal to $\vect v$ . $\vect E$ at the tip of $\vect r$ lies along $\vect r$ .

Check

At $\beta = 0$ the factor is 1 and (5) is Coulomb’s law. Ahead of the charge ( $\theta = 0$ ) the field is weaker by $1-\beta^2$ , abeam ( $\theta = 90^\circ$ ) stronger by $1/\sqrt{1-\beta^2}$ . Gauss: the flux through a sphere of radius $r$ about the charge, with $u = \cos\theta$ , so that $1-\beta^2\sin^2\theta = (1-\beta^2) + \beta^2u^2$ , is

\frac{q(1-\beta^2)}{4\pi\varepsilon_0}\cdot 2\pi\int_{-1}^{1}\frac{du}{\bigl[(1-\beta^2)+\beta^2u^2\bigr]^{3/2}} = \frac{q(1-\beta^2)}{2\varepsilon_0}\cdot\frac{2}{1-\beta^2} = \frac{q}{\varepsilon_0}

whatever $\beta$ , as it must be.

Remark

Although nothing travels faster than light, the field at $P$ points from where the charge is now, not from where it was when the field now arriving set out. For uniform motion the retardation and the motion conspire exactly to give this, and the field in $K$ is the rest-frame Coulomb field squeezed along $\vect v$ : as $\beta \to 1$ it collapses into a thin disc normal to the motion, the field a passing fast particle delivers as a short transverse pulse.