I am currently reading Griffiths, 'Introduction to Electrodynamics', 3rd ed, Chapter 10.1.3, the section on Gauge Invariance, and was reached a point of confusion. In particular, the differential equations that arose from choosing the Coulomb gauge $\nabla \cdot \vec{A}=0$:

$$\nabla^2 V=-\frac{1}{\epsilon_0}\rho \tag{10.9}$$

$$\nabla^2\vec{A}-\mu_0\epsilon_0\frac{\partial^2\vec{A}}{\partial t^2}=-\mu_0\vec{J} + \mu_0\epsilon_0\nabla(\frac{\partial V}{\partial t}) \tag{10.11}$$

I am confused about the structure of $\vec A$ and $V$, shouldn't they have some sort of $\nabla \lambda$ and $-\frac{\partial \lambda}{\partial t}$ to account for the gauge choice?

Here is my logical progression:

  1. We have differential equations directly from Maxwell's equations (no choice of gauge yet):

$$\nabla^2 V+\frac{\partial}{\partial t}(\nabla \cdot \vec{A})=- \frac{1}{\epsilon_0}\rho\tag{10.4}$$

$$(\nabla^2 \vec{A} - \mu_0 \epsilon_0 \frac{\partial^2 \vec{A}}{\partial t^2})-\nabla(\nabla \cdot \vec{A}+\mu_0 \epsilon_0 \frac{\partial V}{\partial t})=-\mu_0\vec{J}\tag{10.5}$$

  1. We choose the Coulomb gauge of $\nabla \cdot \vec{A_C}=0$ and therefore choose a scalar function, $\lambda$, which has the following properties:

\begin{align*} \nabla \cdot \vec{A_C}& =\nabla \cdot (\vec{A_0}+\nabla\lambda_C)\\ & =\nabla \cdot \vec{A_0} + \nabla^2\lambda_C\\ &=0 \end{align*}

$$ \nabla \cdot \vec{A_0} =- \nabla^2\lambda_C$$ So, in order for the Coulomb Gauge to be satisfied, we must add a scalar field, $\lambda_C$, that will satisfy the last expression above. This should have some consequences on the $V$ as well since we need to subtract $\frac{\partial \lambda_C}{\partial t}$ from it.

  1. Now, we try simplifying the scalar and vector potential differential equations 10.4 and 10.5 with our new gauge choice (terms with $\nabla \cdot \vec{A}$ will be zero):

$$\nabla^2 (V-\frac{\partial \lambda_C}{\partial t})+\frac{d}{\partial t}(0)=- \frac{1}{\epsilon_0}\rho\tag{New 10.4}$$

$$(\nabla^2 (\vec{A}+\nabla\lambda_C) - \mu_0 \epsilon_0 \frac{\partial^2 (\vec{A}+\nabla\lambda_C)}{\partial t^2})-\nabla(0+\mu_0 \epsilon_0 \frac{\partial (V-\frac{\partial \lambda_C}{\partial t})}{\partial t})=-\mu_0\vec{J}\tag{New 10.5}$$

Now, this is where my confusion starts. How does Griffiths go from my "New 10.4 and New 10.5" to his "10.4 and 10.5? Seems like he just ignored the Coulomb scalar function relations $\lambda_C$, but I doubt that can be it.


1 Answer 1


You have done all the work. Now just set $$\vec{A_c} = \vec{A} + \nabla\lambda_C$$ and $$V_c = V - \frac{\partial\lambda_C}{\partial t}\,,$$ and your "New" equations reduce to the Coulomb gauge equations.


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