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In electromagnetism, it is often referred to gauges of the electromagnetic field, such as the radiation or Coulomb gauge. As far as I know, the definition of a gauge helps us to redefine the problem in terms of a vector potential and a scalar potential that, since we have some freedom in choosing them, can be chosen in cleverest way it is possible for the given problem.

Here comes my question: is the choice of the gauge a mere mathematical simplification of the given problem? Does this choice have a physical meaning?

My troubles are actually in understanding the physical meaning of this choice of the gauge and what will change if I choose a different gauge.

Qmechanic
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JackI
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  • It's important to note that when electrodynamics interfaces with hamiltonian classical mechanics and with quantum mechanics, gauge changes also affect the description on the mechanical side; in particular, this includes the relationship between canonical momentum and kinematic momentum. As in EM, this means that in those mechanical descriptions, there are gauge-dependent quantities that do not have direct physical significance. – Emilio Pisanty Dec 08 '17 at 18:04
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    Related/possible duplicates (from specific to general): https://physics.stackexchange.com/q/247261/50583, https://physics.stackexchange.com/q/257018/50583, https://physics.stackexchange.com/q/13870/50583 – ACuriousMind Dec 09 '17 at 14:29

3 Answers3

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In classical physics, and also quantum gauge field theory with an abelian gauge group (like QED), the choice of gauge has no physical significance whatsoever. It's basically just like choosing where to place the origin of your coordinate system. In nonabelian gauge quantum field theory the situation is a bit more subtle, because large gauge transformations take you between physically distinct states. But this is a rather technical detail, and for the most part you can safely think of gauge choices as completely physically irrelevant.

tparker
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Physical observables in a gauge theory$^1$ are independent of gauge-fixing choices$^1$. Conversely, gauge-fixing choices are unphysical.

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$^1$ Here we have applied a narrow definition of a gauge theory where gauge symmetry represents a redundant description of a physical system, cf. e.g. this Phys.SE question. In other words, we have ignored (large) gauge transformations that actually change the physical configuration, cf. answer by tparker.

$^2$ By a gauge-fixing condition, we assume a condition that intersect each gauge-orbit precisely once. Note that some conditions do not actually fulfill this, e.g. only partially fixes a gauge. Also there might be Gribov problems.

Qmechanic
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    Ok, so I can consider it just a mere mathematical artifact useful to make calculations? – JackI Dec 08 '17 at 17:34
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    I don't think this is correct in the case of nonabelian gauge quantum field theory; see my answer. – tparker Dec 08 '17 at 17:55
  • @tparker OP is asking explicitly about electromagnetism, which is immune from any nonabelian subtleties. – Emilio Pisanty Dec 08 '17 at 18:06
  • @EmilioPisanty Yes, but Qmechanic's original answer didn't have that caveat, and so ran the risk of being interpreted as applying to all gauge theories, including nonabelian. – tparker Dec 09 '17 at 16:01
  • @Qmechanic Doesn't the restriction that you state in your first footnote make your updated answer vacuous? You're now just saying that for the class of gauge theories for which gauge transformations are unphysical, gauge transformations are unphysical. This is true but tautological. – tparker Dec 09 '17 at 16:08
  • Well, the concept of restricted type of gauge symmetries is by itself quite powerful in QFT. It e.g. yields Noether identities. – Qmechanic Dec 09 '17 at 16:24
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The gauge choice or another has the same physical importance as choosing a inertial reference frame or another... the possibility of doing that gets you a lot of truly profound physical implications (by Noether theorem, for example), so both answers are yes, in some sense.

Iliado Odiseo
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