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Leaving technical issues like Gribov copies and residual gauge freedom aside, how do gauge fixing conditions like the Coulomb condition \begin{equation} \partial_i A_i =0 \end{equation} or the axial condition \begin{equation} A_3 =0 \end{equation} help in getting rid of the gauge redundancy?

A first idea is that conditions like this help us to fix the gauge function $\eta(x_\mu)$. Before any gauge fixing takes place, we have the freedom to use different configurations $A_\mu$ to describe the same physical situation that are related by $$A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ).$$ If we plug $A'_\mu$ into a gauge condition, like the Coulomb condition, we can derive an for the gauge function $\eta(x_\mu$): \begin{align} 0 &=\partial_i A'_i \\ &= \partial_i A_i (x_\mu) + \partial_i \eta(x_\mu ) \\ \to \partial_i \eta(x_\mu ) &= -\partial_i A_i (x_\mu) \tag{1} \end{align} If we now use a specific solution of the equation of motion $A_\mu$, we can solve this equation to find a specific gauge function $\eta(x_\mu )$. But why is this useful at all if there is nothing that specifies what $A_\mu$, we should put on the right-hand side of Eq. 1?

In more concrete terms, let's say we have a solution of the equation of motion $A_\mu$ and another configuration $A'_\mu$ that is related to $A_\mu$ by a gauge transformation. My problem is that even if we choose one specific gauge function $\eta$, we still have no idea whether we should use $A_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ or $A'_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ and thus the gauge redundancy is still there. After all, there is nothing that tells us that $A'_\mu$ isn't the "original" solution of the equation of motion and $A_\mu$ just a gauge transformed version of it.

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    $\begingroup$ how is this different from your two previous questions physics.stackexchange.com/q/505029/84967 and physics.stackexchange.com/q/505049/84967 ? $\endgroup$ – AccidentalFourierTransform Sep 28 at 11:47
  • $\begingroup$ @AccidentalFourierTransform my previous question asked why, for example, the Coulomb gauge condition is sufficient to get rid of the gauge redundancy completely, which it isn't, as pointed out in the answers. In contrast, in this question I ask why a condition like the Coulomb gauge condition is helpful at all in getting rid of gauge redundancy (even if there remains a residual gauge symmetry). In other words, I'm here trying to understand how a condition like $\partial_i A_i =0$ reduces our gauge freedom. $\endgroup$ – jak Sep 28 at 12:16
  • $\begingroup$ Gauge fixing is a somewhat misleading term. Gauge choosing would be better. $\endgroup$ – my2cts Sep 28 at 12:42
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Leaving technical issues like Gribov copies and residual gauge freedom aside, how do gauge fixing conditions like the Coulomb condition $∂_𝑖 𝐴_𝑖=0$ or the axial condition $𝐴_3=0$ help in getting rid of the gauge redundancy?

Because while there are a huge number of fields $A_\mu$ that satisfy the defining equation $\partial_\mu A_\nu - \partial_\nu A_\mu = F_{\mu \nu}$ (where $F_{\mu \nu}$ is uniquely physically determined), there are a much smaller (though still infinite) number of fields $A_\mu$ that satisfy that equation and also solve the partial gauge fixing equations that you list above. Sometimes this narrowing down is useful, sometimes it isn't - it depends on the situation. For example, the covariant version of the two sourced Maxwell's equations takes on a simpler mathematical form in Lorentz gauge than in an arbitrary gauge. But these partial gauge fixing conditions don't fully pin down a unique $A_\mu$ field by themselves.

A first idea is that conditions like this help us to fix the gauge function $\eta(x_\mu)$. Before any gauge fixing takes place, we have the freedom to use different configurations $A_\mu$ to describe the same physical situation that are related by $$A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ).$$ If we plug $A'_\mu$ into a gauge condition, like the Coulomb condition, we can derive an for the gauge function $\eta(x_\mu$): \begin{align} 0 &=\partial_i A'_i \\ &= \partial_i A_i (x_\mu) + \partial_i \eta(x_\mu ) \\ \to \partial_i \eta(x_\mu ) &= -\partial_i A_i (x_\mu) \tag{1} \end{align} If we now use a specific solution of the equation of motion $A_\mu$, we can solve this equation to find a specific gauge function $\eta(x_\mu )$. But why is this useful at all if there is nothing that specifies what $A_\mu$, we should put on the right-hand side of Eq. 1?

You're right, there is no unique transition function $\eta$ that takes you into (say) Coulomb gauge - it depends on whichever $A_\mu$ you happened to start with. How you get to Coulomb gauge depends on where you're starting from, just as how to get to Chicago depends on whether you're starting from Boston or Seattle. In practice, you just start with a totally random choice of $A_\mu$, then solve the equation you give above to find an appropriate $\eta$, then add the gradient of that $\eta$ to your original $A_\mu$ to form $A_\mu'$, then you can completely forget about your original $A_\mu$ and $\eta$.

In more concrete terms, let's say we have a solution of the equation of motion $A_\mu$ and another configuration $A'_\mu$ that is related to $A_\mu$ by a gauge transformation. My problem is that even if we choose one specific gauge function $\eta$, we still have no idea whether we should use $A_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ or $A'_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ and thus the gauge redundancy is still there. After all, there is nothing that tells us that $A'_\mu$ isn't the "original" solution of the equation of motion and $A_\mu$ just a gauge transformed version of it.

You don't "choose one specific [transition] function $\eta$" ahead of time in order to gauge-fix. The choice of transition function inherently depends on your (arbitrary) starting field and so itself is pretty arbitrary.

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My problem is that even if we choose one specific gauge function $\eta$, we still have no idea whether we should use $A_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ or $A'_\mu (x_\mu) + \partial_\mu \eta(x_\mu )$ and thus the gauge redundancy is still there. After all, there is nothing that tells us that $A'_\mu$ isn't the "original" solution of the equation of motion and $A_\mu$ just a gauge transformed version of it.

Exactly, there is nothing that tells you to use a certain gauge fixing. In physics you can't delegate everything to math or some principle, sometimes you have to do the choice.

Like in the choice of a coordinate system, there isn't a mathematical principle that tells you to use spherical coordinates to compute the volume of a sphere, you may compute it in cartesian coordinates if you like and there would be nothing wrong with it.

The same thing applies to gauge fixing, you may compute anything you want in any gauge you want, but you decide what's the best gauge to do it, and it can depend from case to case and on your own personal taste.

The physics doesn't change no matter which gauge you choose, gauge fixing is just a tool useful to compute physical quantities, in other words, if you don't fix the gauge you can't really compute anything interesting.

To conclude, there is no original solution to the equation of motion, the equation of motion are gauge invariant, that means that the whole class of potentials related by gauge transformations satisfies them, they are all the same level, there isn't an original and a transformed one.

Now let's analyze your case more practically: The Coulomb gauge

$$ \partial_k A^k = 0 \, ; \quad A_0 = 0 $$

uniquely fixes the 4-vector potential, because the 0-th component is chosen to be null, and the other condition, as you can see fixes the 3-divergence of the spatial part: $ \nabla \cdot {\bf A} = 0$.

As you may know Maxwell equations fix the curl of ${\bf A}$ and Helmholtz Decomposition Theorem tells you that a 3-vector is completely described by its curl and its divergence, therefore fixing the divergence of the 3-vector potential you fixed it and uniquely chosen a potential from the infinite possibilities you had: you fixed the gauge.

As a side not, Helmholtz Decomposition Theorem is the reason why Maxwell's equations are given in terms of curl and divergence of Electric and Magnetic Fields, its because once you know their curl and their divergence, you know the fields, therefore you don't need more than two equations for each field.

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  • $\begingroup$ Thanks for your answer. My problem, however, isn't that I'm unsure about which gauge to use. Instead, I'm trying to understand why we fix the gauge (partially) by writing down a condition like $\partial_i A_i =0$? How does a condition like this (or any other condition) help in reducing the gauge redundancy? $\endgroup$ – jak Sep 28 at 12:19
  • $\begingroup$ Because you are putting constraints on the Vector potential if you add a condition. If the potential have to satisfy a certain relationship it is constrained, it has to satisfy that relationship, and not all the potentials related by gauge transformation can satisfy the constraint you're imposing. For example in the Lorentz gauge you're picking a class of potentials that satisfy the Lorenz condition, there is less redundancy because many potentials related by gauge transformation don't satisfy the Lorenz relation. I'll try be more explicit in the answer editing it later $\endgroup$ – AnOrAn Sep 28 at 12:32
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    $\begingroup$ @jak I edited the answer completing it with my comments, I'll delete the comments. $\endgroup$ – AnOrAn Sep 28 at 13:36

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