I have been asked to show that the Lorenz gauge condition, written as $$\nabla_T \bullet \vec{A} + \dfrac{1}{c^2}\dfrac{\partial}{\partial t}\Phi = 0$$ is mathematically consistent with the vector and scalar potentials written as $$\vec{A} \equiv \dfrac{1}{4 \pi \epsilon_0 c^2}\iiint \dfrac{\vec{J}}{r_{TS}}dVol$$ with $\vec{J} \equiv \rho_S \vec{v}_S$ and $$\Phi \equiv \dfrac{1}{4 \pi \epsilon_0}\iiint \dfrac{\rho_s}{r_{TS}}dVol.$$ Here, $r_{TS} \equiv \vec{r}_T - \vec{r}_S,$ $\vec{r}_T \equiv x_T \hat{x} + y_T \hat{y} + z_T \hat{z}$ and $$\nabla_T \equiv \dfrac{\partial}{\partial x_T}\hat{x} + \dfrac{\partial}{\partial y_T}\hat{y} + \dfrac{\partial}{\partial z_T}\hat{z}.$$

I use the relations $\nabla_T \dfrac{1}{r_{TS}}=\dfrac{-1}{r^2_{TS}}\hat{r}_{TS}$ and the facts that $\nabla_T \bullet \vec{v}_S = 0$ and that $\rho_S$ does not depend on $\vec{r}_T$, but does, in general, depend on time $t$.

Using all these definitions, I can not get the Lorenz gauge condition to equal zero unless $\nabla_T \bullet \vec{A} = 0$ and $\dfrac{\partial}{\partial t}\Phi = 0$.

It seems like a straight forward calculation. But for $\nabla_T \bullet \vec{A}$ I get

$\nabla_T \bullet \vec{A} = \dfrac{-1}{4 \pi \epsilon_0 c^2}\iiint \dfrac{\rho_S}{r^2_{TS}}\hat{r}_{TS} \bullet \vec{v}_SdVol$.

And for $\dfrac{1}{c^2}\dfrac{\partial}{\partial t}\Phi$ I get

$\dfrac{1}{c^2}\dfrac{\partial}{\partial t}\Phi = \dfrac{1}{4\pi \epsilon_0 c^2} \iiint \dfrac{1}{r_{TS}}\dfrac{\partial}{\partial t}\rho_S - \rho_S \dfrac{1}{r^2_{TS}}\dot{r}_{TS} dVol$.

Clearly, these do not, in general, sum to zero.

Can someone help with this calculation?

  • $\begingroup$ You should not be getting that $\dot {\mathbf r}_\text{TS}$ easily from $\nabla_\text{T}\cdot\mathbf A$ as neither the derivative nor integral has a time component to it. Indeed $\dot {\mathbf r}_\text{TS}=0$ because we are not using some non-inertial coördinate system; when we move from talking about particles to talking about fields, we no longer have position vectors as a function of time, they are just other inputs to the field function: it maps space and time to a field value. $\endgroup$ – CR Drost Apr 28 at 14:36
  • $\begingroup$ Thanks. I see that I should not get $\dot{r}_{TS}$ from the $\nabla_T \bullet \vec{A}$. I have corrected it. It should have been $\hat{r}_{TS} \bullet \vec{v}_S$. Still can't get divergence of $\vec{A}$ to cancel time derivative of $\Phi$. $\endgroup$ – user458534 Apr 28 at 16:11

It follows from the continuity equation.

Using the quotient rule of vector calculus $$\begin{align}\nabla \cdot \left(\frac{\mathbf{J}}{r_{TS}}\right) &= \frac{r_{TS} \nabla \cdot \mathbf{J} - (\nabla r_{TS}) \cdot \mathbf{J}}{r_{TS}^2} \\ & =\frac{1}{r_{TS}}\nabla \cdot \mathbf{J} - \dfrac{1}{r^2_{TS}}\hat{r}_{TS} \cdot \mathbf{J} \\ & =\frac{1}{r_{TS}}\nabla \cdot \mathbf{J} - \dfrac{1}{r^2_{TS}}\hat{r}_{TS} \cdot \mathbf{v_s}\rho_s \end{align}$$

Also $$\dfrac{\partial }{\partial t}\left(\frac{\rho_s}{r_{TS}}\right) = \dfrac{1}{r_{TS}}\dfrac{\partial \rho_s }{\partial t}- \dfrac{1}{r^2_{TS}}\dot{r}_{TS}\rho_s$$

By adding these expressions you will find the desired result due to the continuity equation given by $$\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0$$

You can find a related discussion here.

References: Andrew Zangwill, Modern Electrodynamics, 2012.

  • $\begingroup$ I see your approach, but no, this does not work. Adding the 2 expressions gives $$ \dfrac{1}{r_{TS}} \dfrac{\partial}{\partial t}\rho_S - \dfrac{1}{r^2_{TS}}\dot{r}_{TS} \rho_S + \dfrac{1}{r_{TS}}\nabla \bullet \vec{J} - \dfrac{1}{r^2_{TS}}\rho_S \hat{r}_{TS} \bullet \vec{v}_S$$ Then using the continuity relation $$ \dfrac{1}{r_{TS}} \dfrac{\partial}{\partial t}\rho_S - \dfrac{1}{r^2_{TS}}\dot{r}_{TS} \rho_S + \dfrac{1}{r_{TS}}(-\dfrac{\partial}{\partial t}\rho_S) - \dfrac{1}{r^2_{TS}}\rho_S \hat{r}_{TS} \bullet \vec{v}_S$$ This does not equal zero. $\endgroup$ – user458534 Apr 29 at 23:08
  • $\begingroup$ It does not equal zero because $$\dot{r}_{TS} = \hat{r}_{TS}\bullet (\vec{v}_T - \vec{v}_S) $$ so there is a $$\hat{r}_{TS}\bullet \vec{v}_T$$ left over. $\endgroup$ – user458534 Apr 29 at 23:14

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