I was reading Fermi's review on Dirac's "Quantum Theory of Radiation", which he published in 1932. I was unable to know why he expressed electric field as the following:

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I understand that from Maxwell's equation, electric field can be expressed as:

$$\vec{E}=-\nabla V -\frac{1}{c}\frac{\partial \vec{U}}{\partial t} $$ and the magnetic field as: $$\vec{H} =\nabla \times \vec{U}$$

But in Fermi's expression there seems to be missing the gradient of a potential. I don't understand why is that the case? Or am I missing some concept from the paper? Please help!

I also found that the following;


So, does this mean the slides are in accurate?

  • 3
    $\begingroup$ It is likely that Fermi chose a gauge in which $V=0$. This is quite common. $\endgroup$
    – Meng Cheng
    Aug 7, 2023 at 15:07

2 Answers 2


Fermi is working in the gauge $V =0$.

Suppose you have some choice of potentials $\mathbf{A}(\mathbf{x},t), V(\mathbf{x},t)$. Then define $\tilde{\mathbf{A}}(\mathbf{x},t) = \mathbf{A}(\mathbf{x},t) + \int_{t_0}^t \nabla V(x,t') dt'$. It follows that (in natural units, $c=1$) $$ \mathbf{E} = -\nabla V-\partial_t\mathbf{A} = -\partial_t\left[\mathbf{A} + \int_{t_0}^tdt' \nabla V(\mathbf{x}, t')\right] = -\partial_t\tilde{\mathbf{A}}\\ \mathbf{B} = \nabla\times\mathbf{A} = \nabla\times\tilde{\mathbf{A}} $$

It follows that $(V,A)$ and $(0,\tilde{A})$ give the same physical fields, and so one is (at least classically) free to choose either one as a representation of the system.

Fermi's expression uses the tilde'd gauge field, which is often more convenient for doing QFT.

The slides in the question are not exactly wrong - it claims "often" used when no sources are present, not never. The price you pay when handling charges in Coulomb gauge is that the temporal boundary conditions on $\mathbf{A}$ are a bit awkward, since a system with any charges in it needs $A$ to increase from $-\infty$ at the beginning of time to $\infty$ at the end of time. There is nothing wrong with this over finite timescales, however.


You can find a good description of this issue in Jackson, section 6.2 and 6.3. In short, starting with $$\mathbf B =\nabla \times \mathbf A \tag{1}\label{1}$$ $$\nabla \times \left(\mathbf{E} + \frac{\partial \mathbf A}{\partial t} \right) = 0 \tag{2}\label{2}$$ you get $$\mathbf E + \frac{\partial \mathbf A}{\partial t} = -\nabla \phi \tag{3}\label{3}$$ and then two equations, $$\nabla^2 \phi + \frac{\partial}{\partial{t}}\left( \nabla \cdot \mathbf A\right)= -\frac{\rho}{\epsilon_0} \tag{4}\label{4}$$ and $$\nabla^2 \mathbf A -\frac{1}{c^2} \frac{\partial^2 \mathbf A}{\partial{t^2}} -\nabla \left( \nabla \cdot \mathbf A +\frac{1}{c^2} \frac{\partial \phi}{\partial t} \right)= -\mu_0 \mathbf J \tag{5}\label{5}$$ Now $\eqref{1}$, $\eqref{2}$ along with $\eqref{4}$ and $\eqref{5}$ are invariant to a substitution for an arbitrary (smooth) $\Lambda$ $$\mathbf A' = \mathbf A + \nabla \Lambda \tag{6}\label{6}$$ $$ \phi' = \phi -\frac{\partial \Lambda}{\partial t} \tag {7}\label{7}$$ so you can choose $\Lambda$ to suit your "needs".

One common choice, and this is what Fermi did, is to choose $\Lambda$ so that $\nabla \cdot \mathbf A = 0$, and it is called the Coulomb gauge because in this gauge, $$\nabla ^2 \phi = -\frac{\rho}{\epsilon_0}\tag{8}\label{8}$$

Now if you also set $\rho = 0$, the source free condition together with the Coulomb gauge, then $\phi =0$ and $$\mathbf E = -\frac{\partial \mathbf A}{\partial t} \tag{9}\label{9}$$ and $$\mathbf B = \nabla \times \mathbf A \tag{10} \label{10}$$


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