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In electromagnetism textbooks, the gauges most often talked about are the Lorenz gauge and Coulomb gauge. Sometimes it's convenient to work in a gauge in which there is only the vector potential $\vec{A}$ but no scalar potential $\phi$. The following gauge transformation transforms a general pair of potentials $(\vec{A},\phi)$ into $(\vec{A}',0)$, such that

$$\vec{A}'=\vec{A}+\int_0^t\nabla\phi\,dt,\quad \phi'=0.$$

Then one could work with only the vector potential $\vec{A}'$ to produce both the electric field

$$\vec{E}=-\frac{\partial\vec{A}'}{\partial t}=-\frac{\partial\vec{A}}{\partial t}-\nabla\phi,$$

and the magnetic field

$$\vec{B}=\nabla\times\vec{A}'=\nabla\times\vec{A}.$$

The above procedure seems to work generally without assuming there being no electric charge (which would produce the retarded scalar potential in Lorenz gauge). Is there a name for this $\,\phi=0\,$ gauge?

Qmechanic
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Zhuoran He
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2 Answers2

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The gauge $\phi = A_0 = 0$ is called Weyl gauge or temporal gauge.

This gauge is incomplete, as one can see from the definition of a gauge transformation, $$A_\mu \to A_\mu + \partial_\mu \alpha(x).$$ We can still perform any gauge transformation with gauge parameter $\alpha$ independent of $t$, as this keeps $A_0$ the same. To remove some of the residual gauge freedom we could, e.g. impose $$A_z|_{t = 0} = 0.$$ The proof this gauge can be reached is just the exact same as the proof that Weyl gauge can be reached, except with effectively one less dimension since nothing depends on $t$. At this point we are still not done, because we can still preserve both gauge conditions using any $\alpha$ independent of both $t$ and $z$. So we impose the further condition $$A_y|_{t = z = 0} = 0$$ leaving only $\alpha$ dependent on $x$, which are removed by imposing $$A_x|_{t = z = y = 0} = 0.$$ These four conditions together are a complete gauge fixing. It's quite a mouthful, which is why you won't see it written out in textbooks too often.

Whether or not you want to perform a complete gauge fixing is up to taste. For example, in the standard presentation of the QCD $\theta$-vacua, one takes the incomplete gauge fixing $A_0 = 0$ and then argues there are multiple vacua $|n \rangle$. But there is a completely equivalent presentation where one takes the complete gauge fixing I gave above (also mentioned here), and finds a unique vacuum but the exact same physical effects. This is related to whether one chooses to regard large gauge transformations as "do-nothing" transformations.

knzhou
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    Another (potential) downside of complete gauge fixing is that it necessarily breaks more symmetries: the incomplete temporal gauge breaks the 10d Poincare symmetry of EM down to the 7d (Euclidean group times time translation). But complete gauge fixing breaks it down to a finite group of some reflections. – tparker Nov 25 '18 at 15:43
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    This same idea can be described (and generalized) in lattice gauge theory like this: The gauge field consists of one group element per nearest-neighbor pair of sites (the group being U(1) for EM). A gauge transformation is any replacement of these group elements that leaves invariant all traces of products around closed loops. Given any "tree" that joins all lattice sites with no loops, we can fix the gauge completely by setting all group elements on the links of the tree equal to the identity element (zero $A_\mu$). @knzhou's answer uses the continuum version of one such tree. Any tree works. – Chiral Anomaly Nov 27 '18 at 04:37
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    Reference: Montvay and Münster (1994), Quantum Fields on a Lattice, section 3.2.5, page 108 in the paperback edition: "the temporal gauge... is not a complete gauge fixing, however... A complete gauge fixing is achieved by fixing the link variables on a maximal tree, which is a maximal set of links without closed loops." (I'm citing a book about quantum field theory because it's the only reference I could find, but the same conclusion also holds in the lattice version of classical electrodynamics.) – Chiral Anomaly Nov 27 '18 at 05:31
  • "These four conditions together are a complete gauge fixing. It's quite a mouthful, which is why you won't see it written out in textbooks too often." So Weyl gauge is as general as Coulomb or Lorentz gauge and can be applied in any electrodynamics problem? – Solidification Dec 01 '18 at 12:44
  • @mithusengupta123 Not sure what you mean. You can always use any gauge you want in any problem. – knzhou Dec 01 '18 at 12:45
  • But radiation gauge cannot be used in all problems. – Solidification Dec 01 '18 at 12:45
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    @mithusengupta123 Oh, sure. Yes, you can always set $\phi = 0$, even when matter is present. – knzhou Dec 01 '18 at 12:48
  • I have another pain! Gauge freedom allows me to choose only one gauge function $\alpha(x)$. Using that freedom I can fix only one condition, for example, $\nabla\cdot \vec{A}=0$ or $\partial_\mu A^\mu=0$. Here, we are apparently fixing two independent conditions using a single gauge function without any loss of generality. This usually do not happen. A gauge function which satisfies Coulomb gauge do not satisfy Lorentz gauge in general or vice versa. – Solidification Dec 01 '18 at 12:55
  • @mithusengupta123 The counting is simply more complicated than that. For example, the electric field $\mathbf{E}$ contains three components. But the condition $\nabla \times \mathbf{E} = 0$ contains three constraints. Since there are $3 - 3 = 0$ degrees of freedom left, does that mean any electrostatic field must be zero? No. Counting like this simply doesn't work. – knzhou Dec 01 '18 at 12:58
  • Can you suggest me an undergraduate/graduate level textbook to learn more about Weyl gauge and other gauges in general (except Coulomb and Lorentz)? – Solidification Dec 01 '18 at 13:22
  • @mithusengupta123 Besides "$\phi = 0$", what else do you want to know? – knzhou Dec 01 '18 at 13:24
  • applying this gauge to solve a specific problem, whether there are other gauges possible etc. – Solidification Dec 01 '18 at 13:28
  • @mithusengupta123 I would imagine any graduate-level E&M textbook, such as Jackson, would have plenty of information. But there aren't entire books on just gauges. – knzhou Dec 01 '18 at 14:02
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Your specified gauge is actually incomplete. In general, though, any gauge with $\phi = 0$ is a Weyl gauge.

Riley Scott Jacob
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    Wiki says it's incomplete, too. Does it mean it can satisfy more conditions than just $\phi=0$? – Zhuoran He Nov 11 '18 at 06:57
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    @Zhouran He, Yes. Gauges have the freedom to satisfy both the Weyl gauge ($\phi =0$) and the Coulomb gauge ($\nabla \cdot A = 0$), where $A$ is the vector potential. These are so commonly used together that people often mean both, when they say Coulomb gauge. – Craig Mar 28 '19 at 15:43