# Discovering vector potential from moving field analysis

First I would like to remind Lorentz transformation of length and time as a matrix: $$\begin{pmatrix} ct'\\ x'\\ y'\\ z' \end{pmatrix} = \begin{pmatrix} \gamma & - \frac{v}{c} \gamma & 0 & 0 \\ - \frac{v}{c} \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}$$ The Transform matrix 4x4 should be the same for all four vectors in Special Relativity. Look at the picture below: We know only scalar potential in both cases. My goal is to find vector potentials and to prove that $$\vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t}$$

Let me use Lorentz transformation for some imaginery four electric vector: $$\begin{pmatrix} \lambda \phi'\\ A_x'\\ A_y'\\ A_z' \end{pmatrix} = \begin{pmatrix} \gamma & - \frac{v}{c} \gamma & 0 & 0 \\ - \frac{v}{c} \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda \phi \\ A_x \\ A_y \\ A_z \end{pmatrix}$$

$$\phi '= - \frac{Q}{4 \pi \epsilon_0 r'}$$ and $$\phi = - \frac{1}{4 \pi \epsilon_0 } \cdot \frac{Q}{r' - \vec{r'} \cdot \frac{\vec{v}}{c} }$$

$$\lambda$$ is some constant for a while

It is looking even good until I begin manipulate the algebra to get $$A_x'$$ What do You think about this approach? Maybe Am I totally wrong? Please help

• I'm not sure I understand what you're doing here. Why are you introducing $\lambda$? The four-potential does transform like a four-vector. In your "rest" frame, the four potential is just $$A^\mu \equiv \begin{pmatrix}\phi\\0\\0\\0\end{pmatrix},$$ as there's no magnetic field. That should greatly simply things, no? Commented Aug 12, 2020 at 15:39
• What are you taking as your ground truth? Do you know the field in the $v>0$ and you are trying to confirm that you can get this using the potentials? Commented Aug 12, 2020 at 16:41

If I understood well you are trying to deduce an expression for $$\boldsymbol{A}$$. I'll try to give it just by a retarded-time discussion without explicity passing from Lorentz transformations. You should consider a particle with charge $$q_i$$ moving along a trajectory $$\boldsymbol{r}_i(t)$$ with a velocity $$\dot{\boldsymbol{r}}_i(\tau)\doteq\text{d}\boldsymbol{r}_i(\tau)/\text{d}\tau$$ where $$\tau$$ will be the retarded time, defined at the end of the argumentation. Let's start from the equations of the Lorentz gauge $$\begin{gather*} \frac{1}{\mu}\nabla\cdot\boldsymbol{A} +\varepsilon\frac{\partial\phi}{\partial t}=0 \\ \nabla^2\phi -\mu\varepsilon\frac{\partial^2\phi}{\partial t^2} = -\frac{\rho}{\varepsilon} \\ \nabla^2\boldsymbol{A} -\mu\varepsilon\frac{\partial^2 \boldsymbol{A}}{\partial t^2} = -\mu\boldsymbol{j} \end{gather*}$$ which have solutions $$\begin{gather*} \phi(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \int\limits_V \frac{ \rho\left(\boldsymbol{r}^\prime,t -\displaystyle{ \frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c} }\right) }{ |\boldsymbol{r}-\boldsymbol{r}^\prime| } \text{d}^3{\boldsymbol{r}^\prime} \\ \boldsymbol{A}(\boldsymbol{r},t) = \frac{\mu}{4\pi} \int\limits_V \frac{ \boldsymbol{j}\left(\boldsymbol{r}^\prime,t -\displaystyle{ \frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c} }\right) }{ |\boldsymbol{r}-\boldsymbol{r}^\prime| } \text{d}^3{\boldsymbol{r}^\prime} \end{gather*}$$ but you know what $$\rho,\boldsymbol{j}$$ are for a moving particle $$\begin{gather*} \rho_i(\boldsymbol{r},t) = q_i\delta(\boldsymbol{r}-\boldsymbol{r}_i(t)) \\ \boldsymbol{j}_i(\boldsymbol{r},t) = q_i\boldsymbol{u}_i(t)\delta(\boldsymbol{r}-\boldsymbol{r}_i(t)) \\ \phi_i(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \int\limits_V \frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t -\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right) }{ |\boldsymbol{r}-\boldsymbol{r}^\prime| } \text{d}^3{\boldsymbol{r}^\prime} \\ \boldsymbol{A}_i(\boldsymbol{r},t) = \frac{\mu}{4\pi} \int\limits_V \frac{q_i\dot{\boldsymbol{r}}_i\left(t-\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i\left(t -\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}}\right)\right) } {|\boldsymbol{r}-\boldsymbol{r}^\prime| } \text{d}^3{\boldsymbol{r}^\prime} \end{gather*}$$ But you don't like very much this integral so you operate a substitution $$\begin{gather*} t^\prime \doteq t -\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}^\prime|}{c}} \\ \boldsymbol{d}_i(t) \doteq \boldsymbol{r}-\boldsymbol{r}_i(t) \end{gather*}$$ such that the condition posed by the $$\delta$$ becomes $$\boldsymbol{r}^\prime=\boldsymbol{r}-\boldsymbol{d}_i(t^\prime)$$. So now you have $$\begin{gather*} \phi_i(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \int\limits_V \frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|} \text{d}^3{\boldsymbol{r}^\prime} \equiv \frac{1}{4\pi\epsilon} \int\limits_{t^{\prime\prime}} \int\limits_V \frac{q_i\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|} \delta(t^{\prime\prime}-t^\prime) \text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}} \\ \boldsymbol{A}_i(\boldsymbol{r},t) = \frac{\mu}{4\pi} \int\limits_V \frac{q_i\dot{\boldsymbol{r}}_i(t^\prime)\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^\prime)\right)}{|\boldsymbol{d}_i(t^\prime)|} \text{d}^3{\boldsymbol{r}^\prime} \equiv \frac{\mu}{4\pi} \int\limits_{t^{\prime\prime}}\int\limits_V \frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(\boldsymbol{r}^\prime-\boldsymbol{r}_i(t^{\prime\prime})\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|} \delta(t^{\prime\prime}-t^\prime) \text{d}^3{\boldsymbol{r}^\prime}\text{d}{t^{\prime\prime}} \\ \phi_i(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \int\limits_{t^{\prime\prime}} \frac{q_i\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|} \text{d}{t^{\prime\prime}} \\ \boldsymbol{A}_i(\boldsymbol{r},t) = \frac{\mu}{4\pi} \int\limits_{t^{\prime\prime}} \frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime}-t+\displaystyle{\frac{|\boldsymbol{r}-\boldsymbol{r}_i(t^{\prime\prime})|}{c}}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|} \text{d}{t^{\prime\prime}} \end{gather*}$$ You see? Now the integral is on the time domain! We are near the end. Now a new variable substitution $$\begin{equation*} t^{\prime\prime\prime} = t^{\prime\prime} -t +\frac{|\boldsymbol{d}_i(t^{\prime\prime})|}{c} \Longrightarrow \text{d}{t^{\prime\prime\prime}} = \text{d}{t^{\prime\prime}} +\frac{1}{c}\frac{\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}\text{d}{t^{\prime\prime}} \end{equation*}$$ but defining $$\begin{equation*} \boldsymbol{n}_i(t^{\prime\prime}) \doteq \frac{\boldsymbol{d}_i(t^{\prime\prime})}{|\boldsymbol{d}_i(t^{\prime\prime})|} \end{equation*}$$ You will see that $$\begin{equation*} {\text{d}|\boldsymbol{d}_i(t^{\prime\prime})|}{\text{d}t^{\prime\prime}}=-\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime}) \end{equation*}$$ and define $$\begin{equation*} \kappa_i(t^{\prime\prime}) \doteq 1-\frac{1}{c}\boldsymbol{n}_i(t^{\prime\prime})\cdot\dot{\boldsymbol{r}}_i(t^{\prime\prime}) \end{equation*}$$ such that $$\begin{gather*} \phi_i(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \int\limits_{t^{\prime\prime\prime}} \frac{q_i\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})} \text{d}{t^{\prime\prime\prime}} \\ \boldsymbol{A}_i(\boldsymbol{r},t) = \frac{\mu}{4\pi} \int\limits_{t^{\prime\prime\prime}} \frac{q_i\dot{\boldsymbol{r}}_i(t^{\prime\prime})\delta\left(t^{\prime\prime\prime}\right)}{|\boldsymbol{d}_i(t^{\prime\prime})|\kappa_i(t^{\prime\prime})} \text{d}{t^{\prime\prime\prime}} \end{gather*}$$ and finally the last definition $$\begin{gather*} \tau +\frac{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|}{c} \doteq t \\ \phi_i(\boldsymbol{r},t) = \frac{1}{4\pi\epsilon} \frac{q_i}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)} \\ \boldsymbol{A}_i(\boldsymbol{r},t) = \frac{\mu}{4\pi} \frac{q_i\dot{\boldsymbol{r}}_i(\tau)}{|\boldsymbol{r}-\boldsymbol{r}_i(\tau)|\kappa_i(\tau)} \end{gather*}$$ That are the potential of Liénard-Wiechart. The expression of the electric field that you cited is just a consequence of Maxwell equations and so electromagnetic field can be obtained by the potentials, being careful with the gradient and the time derivative. Hope this helps