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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 3

Assume that $\mathbf v$ is constant. Set
$$
\tau=t_{\rm ret},\qquad
n=\frac{\mathbf R}{R},\qquad
\beta=\frac{\mathbf v}{c},\qquad
\kappa=1-n\cdot\beta,\qquad
D=R-\beta\cdot\mathbf R=R\kappa.
$$
<Implicit differentiation> of $t=\tau+R(\tau)/c$ gives
$$
\partial_t\tau=\frac1\kappa,
\qquad
\nabla\tau=-\frac{n}{c\kappa}.
$$
Since $d\mathbf R/d\tau=-\mathbf v$, direct differentiation of $D$ yields
$$
\partial_tD=\frac{c(\beta^2-n\cdot\beta)}{\kappa},
\qquad
\mathbf v\cdot\nabla D
=-\frac{c(\beta^2-n\cdot\beta)}{\kappa}.
$$
Therefore
$$
(\partial_t+\mathbf v\cdot\nabla)D=0.
$$
The <Liénard–Wiechert potentials> for uniform motion have $\phi=q/(4\pi\epsilon_0D)$ and $\mathbf A=\mathbf v\phi/c^2$. The preceding identity consequently implies
$$
\partial_t\phi+\mathbf v\cdot\nabla\phi=0.
$$
As $\mathbf v$ is constant,
$$
\boxed{
\frac1{c^2}\frac{\partial\phi}{\partial t}
+\nabla\cdot\mathbf A
=\frac1{c^2}
\left(\partial_t\phi+\mathbf v\cdot\nabla\phi\right)=0.}
$$
\b[Thus the explicit uniformly moving potentials satisfy the <Lorenz gauge>.]

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 36A / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 36A / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 36A / d / Solution

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