Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.12

Statement

$14.3.12.$

Determine the distribution of the electric intensity and magnetic induction ofa charge q moving with

Solution

Rest frame (S')
The charge is at rest at the origin:

$\mathbf{E}' = \frac{q}{r'^3}\, \mathbf{r}' \quad (\text{CGS}),\qquad \mathbf{B}' = 0$

Transformation to the laboratory (S)

The system S' moves with velocity $\mathbf{v} = \beta c\, \hat{\mathbf{z}}$
relative to S. For a boost along z, the parallel and perpendicular components transform as:

$E_z = E'_z, \quad \mathbf{E}_\perp = \gamma,\mathbf{E}'_\perp, \qquad
B_z = 0, \quad \mathbf{B}_\perp = \gamma,\frac{\mathbf{v}}{c}\times\mathbf{E}'$

Change of coordinates

The positions are related by$ x' = x,\; y' = y,\; z' = \gamma(z - vt)$
The distance in S' as a function of the coordinates in S is:

$r' = r\, \frac{\sqrt{1-\beta^2\sin^2\alpha}}{\sqrt{1-\beta^2}}$

where$ \alpha$ is the angle between $\mathbf{v}$ and$ \mathbf{r} in S$

Electric field in S

Substituting the expression for r' into $\mathbf{E}'$ and applying the transformation of components, one obtains a radial field from the instantaneous position of the charge:

$\boxed{\mathbf{E} = \frac{q}{r^3}\, \frac{1-\beta^2}{\bigl(1-\beta^2\sin^2\alpha\bigr)^{3/2}}\; \mathbf{r}} \quad (\text{CGS})$

In SI, q is replaced by $\dfrac{q}{4\pi\varepsilon_0}$

Magnetic field in S

$\boxed{\mathbf{B} = \frac{\mathbf{v}}{c}\times\mathbf{E}} \quad (\text{CGS}),\qquad
\boxed{\mathbf{B} = \frac{1}{c^2},\mathbf{v}\times\mathbf{E}} \quad (\text{SI})$

Answer

$\boxed{\mathbf{B} = \frac{\mathbf{v}}{c}\times\mathbf{E}} \quad (\text{CGS}),\qquad
\boxed{\mathbf{B} = \frac{1}{c^2},\mathbf{v}\times\mathbf{E}} \quad (\text{SI})$