Irodov 3.366: Galilean Fields as the Slow Limit – Amit & Deepali Sinha Foundation

Problem Irodov 3.366

An inertial frame $K'$ moves relative to an inertial frame $K$ with the constant velocity $\vect v_0$ . With $\beta = v_0/c$ , and with $\parallel$ and $\perp$ marking the components of a field along $\vect v_0$ and across it, the fields in the two frames are related by

\begin{gathered} \vect E'_\parallel = \vect E_\parallel, \qquad \vect B'_\parallel = \vect B_\parallel,\\ \vect E'_\perp = \frac{\vect E_\perp + \vect v_0\times\vect B}{\sqrt{1-\beta^2}}, \qquad \vect B'_\perp = \frac{\vect B_\perp - \vect v_0\times\vect E/c^2}{\sqrt{1-\beta^2}} \end{gathered}

Show that for $v_0 \ll c$ these reduce to $\vect E' = \vect E + \vect v_0\times\vect B$ and $\vect B' = \vect B - \vect v_0\times\vect E/c^2$ .

Given

the relativistic transformation of $\vect E$ and $\vect B$ for a frame moving with $\vect v_0$
$v_0 \ll c$

Find

the low-speed transformation

Solution

Step 1Split each field along and across the velocity

Each field is the sum of its part along $\vect v_0$ and its part across it, in $K$ and in $K'$ alike:

\vect E = \vect E_\parallel + \vect E_\perp, \qquad \vect B = \vect B_\parallel + \vect B_\perp \tag{1}

A field E resolved along the frame velocity v_0 and across it into E_∥ and E_⊥; B splits the same way. — A field $\vect E$ resolved along the frame velocity $\vect v_0$ and across it into $\vect E_\parallel$ and $\vect E_\perp$ ; $\vect B$ splits the same way.

Step 2The Lorentz factor at low speed

The relativistic relations differ from the low-speed ones only through the factor dividing the perpendicular parts. Expanding it in powers of $\beta$ :

\frac{1}{\sqrt{1-\beta^2}} = 1 + \frac{\beta^2}{2} + \dots \tag{2}

Its departure from 1 is of second order in $\beta$ . The transformation itself is of first order, since $\vect v_0\times\vect B$ and $\vect v_0\times\vect E/c^2$ carry one power of $v_0$ . To first order in $\beta$ the factor (2) is therefore 1, and the perpendicular parts become

\vect E'_\perp = \vect E_\perp + \vect v_0\times\vect B, \qquad \vect B'_\perp = \vect B_\perp - \frac{\vect v_0\times\vect E}{c^2} \tag{3}

Step 3Reassemble the fields

The parallel parts are unchanged in both theories. Adding them to (3) and using (1) in $K$ :

\begin{aligned} \vect E' &= \vect E'_\parallel + \vect E'_\perp = \vect E_\parallel + \vect E_\perp + \vect v_0\times\vect B\\ \vect B' &= \vect B'_\parallel + \vect B'_\perp = \vect B_\parallel + \vect B_\perp - \frac{\vect v_0\times\vect E}{c^2} \end{aligned}

The first two terms on each right side add back to the whole field, which is the low-speed transformation:

\begin{gathered}\vect E' = \vect E + \vect v_0\times\vect B\\ \vect B' = \vect B - \frac{\vect v_0\times\vect E}{c^2}\end{gathered}

(4)

Check

Consistency to first order: going from $K'$ back to $K$ is a boost with $-\vect v_0$ , so applying (4) twice should return the original field up to terms in $\beta^2$ . It does:

\vect E + \vect v_0\times\vect B - \vect v_0\times\Bigl(\vect B - \frac{\vect v_0\times\vect E}{c^2}\Bigr) = \vect E + \frac{\vect v_0\times(\vect v_0\times\vect E)}{c^2}

and the leftover term is of order $\beta^2 E$ . Units: $v_0 B$ is $(\text{m/s})\cdot\text{T} = \text{V/m}$ , a field strength, and $v_0E/c^2$ is $(\text{V/m})/(\text{m/s}) = \text{T}$ .

Remark

The dropped terms are $\beta^2/2$ times the perpendicular field that is kept, so the low-speed pair is accurate to that fraction whichever field dominates. The two relations are not equally familiar: $\vect E' = \vect E + \vect v_0\times\vect B$ is the Lorentz force per unit charge on a charge at rest in $K'$ , while $-\vect v_0\times\vect E/c^2$ is the magnetic field that charges at rest in $K$ produce once they are seen moving. The leftover of the round trip in the check shows that this pair cannot be an exact law at any speed; the relativistic factor $1/\sqrt{1-\beta^2}$ is what makes successive changes of frame agree.