(i) Starting from the field-strength tensor $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$, where $A^{\mu}=(\phi / c, \mathbf{A})$ is the 4-vector potential with components such that

$$\mathbf{E}=-\frac{\partial \mathbf{A}}{\partial t}-\boldsymbol{\nabla} \phi \quad \text { and } \quad \mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A}$$

derive the transformation laws for the components of the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ under the standard Lorentz boost $x^{\prime \mu}=\Lambda_{\nu}^{\mu} x^{\nu}$ with

$$\Lambda_{\nu}^{\mu}=\left(\begin{array}{cccc} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$

(ii) Two point charges, each with electric charge $q$, are at rest and separated by a distance $d$ in some inertial frame $S$. By transforming the fields from the rest frame $S$, calculate the magnitude and direction of the force between the two charges in an inertial frame in which the charges are moving with speed $\beta c$ in a direction perpendicular to their separation.

(iii) The 4-force for a particle with 4-momentum $p^{\mu}$ is $F^{\mu}=d p^{\mu} / d \tau$, where $\tau$ is proper time. Show that the components of $F^{\mu}$ in an inertial frame in which the particle has 3 -velocity $\mathbf{v}$ are

$$F^{\mu}=\gamma(\mathbf{F} \cdot \mathbf{v} / c, \mathbf{F})$$

where $\gamma=\left(1-\mathbf{v} \cdot \mathbf{v} / c^{2}\right)^{-1 / 2}$ and $\mathbf{F}$ is the 3-force acting on the particle. Hence verify that your result in (ii) above is consistent with Lorentz transforming the electromagnetic 3 -force from the rest frame $S$.