31.1.3 Almost Relativity

FIGURE 31.9a shows two positive charges moving side by side through frame A with velocity $\vec{v}_{CA}$ . Charge $q_{1}$ creates an electric field and a magnetic field at the position of charge $q_{2}$ . These are

\vec{E}_{A} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} ĵ and \vec{B}_{A} = \frac{μ_{0}}{4 π} \frac{q_{1} v_{CA}}{r^{2}} \hat{k}

where r is the distance between the charges, and we’ve used $\hat{r} = ĵ$ and $\vec{v} \times \hat{r} = v \hat{k}$ .

The figure consists of two parts labeled "a" and b that illustrate the charges placed on two frames in an x y coordinate plane.

FIGURE 31.9 Full Alternative Text

How are the fields seen in frame B, which moves with $\vec{v}_{BA} = \vec{v}_{CA}$ and in which the charges are at rest? From the field transformation equations,

\begin{array}{l} {\vec{B}}_{B} & = & {\vec{B}}_{A} - \frac{1}{c^{2}} {\vec{v}}_{BA} \times {\vec{E}}_{A} = \frac{μ_{0}}{4 π} \frac{q_{1} v_{CA}}{r^{2}} \hat{k} - \frac{1}{c^{2}} (v_{CA} \hat{ι} \times \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} \hat{j}) \\ = & \frac{μ_{0}}{4 π} \frac{q_{1} v_{CA}}{r^{2}} (1 - \frac{1}{ϵ_{0} μ_{0} c^{2}}) \hat{k} \end{array}

(31.11)

where we used $î \times ĵ = \hat{k}$ . But $ϵ_{0} μ_{0} = 1 / c^{2}$ , so the term in parentheses is zero and thus $\vec{B}_{B} = \vec{0}$ . This result was expected because $q_{1}$ is at rest in frame B and shouldn’t create a magnetic field.

The transformation of the electric field is similar:

\begin{array}{l} {\vec{E}}_{B} & = & {\vec{E}}_{A} + {\vec{v}}_{BA} \times {\vec{B}}_{A} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} \hat{j} + v_{BA} î \times \frac{μ_{0}}{4 π} \frac{q_{1} v_{CA}}{r^{2}} \hat{k} \\ = & \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} (1 - ϵ_{0} μ_{0} v_{BA}^{2}) \hat{j} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} (1 - \frac{v_{BA}^{2}}{c^{2}}) \hat{j} \end{array}

(31.12)

where we used $î \times \hat{k} = - ĵ, \vec{v}_{CA} = \vec{v}_{BA}$ , and $ϵ_{0} μ_{0} = 1 / c^{2}$ . FIGURE 31.9b shows the charges and fields in frame B.

But now we have a problem. In frame B where the two charges are at rest and separated by distance r, the electric field due to charge $q_{1}$ should be simply

\vec{E}_{B} = \frac{1}{4 π ϵ_{0}} \frac{q_{1}}{r^{2}} ĵ

The field transformation equations have given a “wrong” result for the electric field $\vec{E}_{B}$ .

It turns out that the field transformations of Equations 31.10, which are based on Galilean relativity, aren’t quite right. We would need Einstein’s relativity—a topic that we’ll take up in Chapter 36—to give the correct transformations. However, the Galilean field transformations in Equations 31.10 are equivalent to the relativistically correct transformations when $v ≪ c$ , in which case $v^{2} / c^{2} ≪ 1$ . You can see that the two expressions for $\vec{E}_{B}$ do, in fact, agree if $v_{BA}^{2} / c^{2}$ can be neglected.

Thus our use of the field transformation equations has an additional rule: Set $v^{2} / c^{2}$ to zero. This is an acceptable rule for speeds $v < 10^{7} m / s$ . Even with this limitation, our investigation has provided us with a deeper understanding of electric and magnetic fields.