Perform suitable gauge transformations

[LeoJakob]

Homework Statement

Let the electromagnetic potentials be given by
$$
\Phi(\vec{r}, t)=-(\vec{a} \cdot \vec{r}) t, \quad \vec{A}(\vec{r}, t)=-\frac{\vec{a} r^{2}}{4 c^{2}}, \quad \vec{a}=\text { const. }
$$
Perform suitable gauge transformations
$$
\begin{array}{l}
\vec{A}(\vec{r}, t) \longrightarrow \vec{A}^{\prime}(\vec{r}, t)=\vec{A}(\vec{r}, t)+\vec{\nabla} \chi(\vec{r}, t), \\
\Phi(\vec{r}, t) \longrightarrow \Phi^{\prime}(\vec{r}, t)=\Phi(\vec{r}, t)-\frac{\partial \chi(\vec{r}, t)}{\partial t}
\end{array}
$$
so that the potentials comply with the respective gauge condition:

$$(i) \Phi^{\prime}=0 $$

$$(ii) \text{Lorenz gauge: } \vec{\nabla} \cdot \vec{A}^{\prime}+\frac{1}{c^{2}} \frac{\partial \Phi^{\prime}}{\partial t}=0 $$

Hint: To solve the differential equations for the gauge function $\chi(\vec{r}, t)$, use that $\square\left[(\vec{a} \cdot \vec{r})(c t)^{2}\right]=-2(\vec{a} \cdot \vec{r})$.


Verify your solution.

Relevant Equations

$$\vec{B}(\vec{r}, t) = \vec{\nabla} \times \vec{A}(\vec{r}, t), \quad \vec{E}(\vec{r}, t) = -\dot{\vec{A}}(\vec{r}, t) - \vec{\nabla} \Phi(\vec{r}, t), \\
\Delta \Phi + \frac{\partial}{\partial t}(\vec{\nabla} \cdot \vec{A}) = -\frac{\rho}{\varepsilon_{0}}, \quad \left(\Delta - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}\right) \vec{A} - \vec{\nabla}\left(\vec{\nabla} \cdot \vec{A} + \frac{1}{c^{2}} \dot{\Phi}\right) = -\mu_{0} \vec{j}, \\
\square \equiv \Delta - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}
$$

Hello, here is my solution attempt:

(i)

$$ \begin{aligned} 0 & =\Phi^{\prime}=\Phi-\frac{\partial \chi}{\partial t} \Rightarrow \Phi=\frac{\partial \chi}{\partial t} \\ & \Rightarrow \int \Phi dt=\chi \\ & \Rightarrow \chi=\int \limits_{0}^{t}-(\vec{a} \cdot \vec{r}) t^{\prime} d t=-\frac{1}{2}(\vec{a} \cdot \vec{r}) t^{2}\end{aligned} $$

(ii)

$$ \begin{align*}
&\vec{\nabla} \cdot \vec{A}^{\prime}+\frac{1}{c^{2}} \frac{\partial \phi^{\prime}}{\partial t}=0 \\
&\Leftrightarrow \vec{\nabla} \cdot \vec{A}^{\prime}=-\frac{1}{c^{2}} \frac{\partial \Phi^{\prime}}{\partial t} \text{(I) }\\
&\vec{\nabla} \cdot \vec{A}^{\prime}=\vec{\nabla} \cdot(\vec{A}+\vec{\nabla} \chi)=\vec{\nabla} \cdot \vec{A}+\Delta \chi \text{(II) }\\
&\frac{\partial \Phi^{\prime}}{\partial t}=\dot{\Phi}-\ddot{\chi}=\frac{\partial}{\partial t}(\Phi-\dot{\chi}) \text{(III) }\\
&=-\vec{a} \cdot \vec{r}-\frac{\partial^2 \chi}{\partial t^2} \text{ with } \dot{\Phi}=\frac{\partial}{\partial t}(-\vec{a} \cdot \vec{r} t)=-\vec{a} \cdot \vec{r} \\
&\vec{\nabla} \cdot \vec{A}=-\frac{1}{4 c^{2}}\left(\begin{array}{c}
\partial_{x} \\
\partial_{y} \\
\partial_{z}
\end{array}\right) \cdot\left(\begin{array}{cc}
a_{1} r^{2} \\
a_{2} r^{2} \\
a_{3} r^{2}
\end{array}\right) \\
&=-\frac{1}{4 c^{2}} \cdot 2(a_{1} \chi+a_{2} y+a_{3} z) \\
&=-\frac{1}{2 c^{2}} \vec{a} \cdot \vec{r}=\frac{1}{2 c^{2}} \dot{\Phi} \\
&\text{(I) } \frac{1}{2 c^{2}}A\dot{\Phi}+\Delta \chi=-\frac{1}{c^{2}}(\dot{\Phi}-\ddot{\chi}) \\
&\stackrel{\cdot c^2}{\Leftrightarrow} \frac{3}{2} \dot{\Phi}+c^{2} \Delta \chi-\ddot{\chi}=0 \\
\end{align*} $$

Can somebody help me with the next step?

 

[TSny]

$$ \frac{3}{2} \dot{\Phi}+c^{2} \Delta \chi-\ddot{\chi}=0 $$

This may be written $$ \Box \chi = -\frac{3}{2c^2} \dot{\Phi}$$
Use $\dot{\Phi} = -\vec a \cdot \vec r$ and the hint given in the problem statement.

 

[LeoJakob]

This may be written $$ \Box \chi = -\frac{3}{2c^2} \dot{\Phi}$$
Use $\dot{\Phi} = -\vec a \cdot \vec r$ and the hint given in the problem statement.

Thank you :)

$$ \begin{align}
\Delta \chi - \frac{1}{c^{2}} \ddot{\chi} &= -\frac{3}{2 c^{2}} \dot{\Phi} = -\frac{3}{2 c^{2}}(-\vec{a} \cdot \vec{r}) = \frac{3}{2 c^{2}}(\vec{a} \cdot \vec{r}) \\
\Leftrightarrow \quad \Box \chi &= -\frac{1}{c^{2}} \frac{3}{2} \dot{\Phi} = \frac{3}{2 c^{2}}(\vec{a} \cdot \vec{r}) \\
&= -\frac{1}{c^{2}} \frac{3}{2} \cdot \frac{1}{2} \Box \left[ (\vec{a} \cdot \vec{r})(c t)^{2} \right] \\
&= \Box \left[ -\frac{3}{4}(\vec{a} \cdot \vec{r}) t^{2} \right] \\
\overset{\text{Is this implication correct?}}{\Rightarrow } \chi &= -\frac{3}{4}(\vec{a} \cdot \vec{r}) t^{2}
\end{align}
$$

I have thus found a Lorenz gauge $\chi$ for the potentials $\Phi$ and $\vec A$.

 

[TSny]

Thank you :)

$$ \begin{align}
\Delta \chi - \frac{1}{c^{2}} \ddot{\chi} &= -\frac{3}{2 c^{2}} \dot{\Phi} = -\frac{3}{2 c^{2}}(-\vec{a} \cdot \vec{r}) = \frac{3}{2 c^{2}}(\vec{a} \cdot \vec{r}) \\
\Leftrightarrow \quad \Box \chi &= -\frac{1}{c^{2}} \frac{3}{2} \dot{\Phi} = \frac{3}{2 c^{2}}(\vec{a} \cdot \vec{r}) \\
&= -\frac{1}{c^{2}} \frac{3}{2} \cdot \frac{1}{2} \Box \left[ (\vec{a} \cdot \vec{r})(c t)^{2} \right] \\
&= \Box \left[ -\frac{3}{4}(\vec{a} \cdot \vec{r}) t^{2} \right] \\
\overset{\text{Is this implication correct?}}{\Rightarrow } \chi &= -\frac{3}{4}(\vec{a} \cdot \vec{r}) t^{2}
\end{align}
$$

Your work looks good. I agree with your result for $\chi$. I think the problem statement wants you to write explicitly your results for $\vec A'$ and $\chi'$ for parts $(i)$ and $(ii)$. The results are not unique since $\chi$ is not unique.