Obtain the field (41.5), (41.6) of a uniformly moving charge \(q\left[=\left(4 \pi \varepsilon_{0}\right)^{-1} Q\right]\) by the following alternative method: Assume that the field in the rest frame \(S^{\prime}\) of the charge is given by $$ \mathbf{e}^{\prime}=\left(Q / r^{\prime 3}\right)\left(x^{\prime}, y^{\prime}, z^{\prime}\right), \quad \mathbf{b}^{\prime}=0, \quad r^{\prime 2}=x^{\prime 2}+y^{\prime 2}+z^{\prime 2} $$ then transform this field to the usual second frame \(\mathrm{S}\) at \(t=0\). [Hint: obtain \(\mathbf{b}=\mathbf{u} \times \mathbf{e} / c^{2}\) from \((39.2)\); from the inverse of \((39.1)\) obtain \(\mathbf{e}=\left(Q \gamma / r^{3}\right)(x, y, z) ;\) finally use (41.7). ]

First, we need to find the magnetic field \(\mathbf{b}\) using equation (39.2). Since the charge is moving uniformly, its velocity \(\mathbf{u}\) remains constant. Equation (39.2) relates the magnetic field \(\mathbf{b}\) with the electric field \(\mathbf{e}\) and velocity \(\mathbf{u}\) as: $$ \mathbf{b}=\frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$ Given that the magnetic field in the rest frame \(S^{\prime}\) is zero (\(\mathbf{b}^{\prime}=0\)), we can write the equation as: $$ \mathbf{b}=\frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$

Next, we need to obtain the electric field \(\mathbf{e}\) in the rest frame \(S^{\prime}\). From the inverse of equation (39.1), we can find the relationship between the electric field in both frames as: $$ \mathbf{e} = \left(\frac{Q\gamma}{r^3}\right)(x, y, z) $$ Where \(\gamma\) is the Lorentz factor, which is defined as: $$ \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} $$

Finally, we need to transform the obtained electric field \(\mathbf{e}\) and magnetic field \(\mathbf{b}\) to the second frame \(S\) at \(t=0\). Using equation (41.7), we can write the transformation as: $$ t=0 \implies x^{\prime}=\gamma x \text{ and } \mathbf{r}^{\prime}=\gamma \mathbf{r} $$ Substituting \(x^{\prime}=\gamma x\) and \(\mathbf{r}^{\prime}=\gamma \mathbf{r}\) into the expressions for \(\mathbf{e}\) and \(\mathbf{b}\) (from steps 1 and 2), we get the fields in the second frame \(S\) as: $$ \mathbf{e} = \left(\frac{Q}{\gamma^2 r^3}\right)(\gamma x, y, z) \text{ and } \mathbf{b} = \frac{\mathbf{u} \times \mathbf{e}}{c^{2}} $$ Now we've obtained the electric field \(\mathbf{e}\) and magnetic field \(\mathbf{b}\) of the uniformly moving charge \(q\) in the second frame \(S\) at \(t=0\).