The retarded scalar potential $\varphi(t, \mathbf{x})$ produced by a charge distribution $\rho(t, \mathbf{x})$ is given by

$$\varphi(t, \mathbf{x})=\frac{1}{4 \pi \epsilon_{0}} \int_{\Omega} d^{3} x^{\prime} \frac{\rho\left(t-\left|\mathbf{x}-\mathbf{x}^{\prime}\right|, \mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|}$$

where $\Omega$ denotes all 3 -space. Describe briefly and qualitatively the physics underlying this formula.

Write the integrand in the formula above as a 1-dimensional integral over a new time coordinate $\tau$. Next consider a special source, a point charge $q$ moving along a trajectory $\mathbf{x}=\mathbf{x}_{0}(t)$ so that

$$\rho(t, \mathbf{x})=q \delta^{(3)}\left(\mathbf{x}-\mathbf{x}_{0}(t)\right),$$

where $\delta^{(3)}(\mathbf{x})$ denotes the 3 -dimensional delta function. By reversing the order of integration, or otherwise, obtain the Liénard-Wiechert potential

$$\varphi(t, \mathbf{x})=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{R-\mathbf{v} \cdot \mathbf{R}},$$

where $\mathbf{v}$ and $\mathbf{R}$ are to be determined.

Write down the corresponding formula for the vector potential $\mathbf{A}(t, \mathbf{x})$.

$$\begin{aligned} & E_{x}^{\prime}=E_{x}, \quad E_{y}^{\prime}=\gamma\left(E_{y}-v B_{z}\right), \quad E_{z}^{\prime}=\gamma\left(E_{z}+v B_{y}\right), \\ & B_{x}^{\prime}=B_{x}, \quad B_{y}^{\prime}=\gamma\left(B_{y}+v E_{z}\right), \quad B_{z}^{\prime}=\gamma\left(B_{z}-v E_{y}\right), \end{aligned}$$