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A gauge symmetry is a symmetry (often local) which doesn't affect the physical quantities or the Lagrangian.

A translation of the whole Universe 3 meters to the left won't affect anything physically.

So does this mean translation is a gauge symmetry?

If so, how do we fix a gauge in this symmetry?

I imagine a scalar field under a translation in the direction $a^\mu$ has the transformation $\delta \phi(x) = a^\mu\partial_\mu \phi(x)$. Can we fix a gauge in this case?

But if we fixed a gauge does this mean the gauge-fixed theory is no-longer translation invariant?

  • "A gauge symmetry is a symmetry (often local) which doesn't affect the physical quantities or the Lagrangian." [citation needed] 1. (Pseudo-)symmetries can affect the Lagrangian by a total derivative. 2. It's not clear how this phrase is supposed to separate gauge symmetries from ordinary symmetries. – ACuriousMind Sep 09 '19 at 17:10
  • @ACuriousMind exactly. –  Sep 09 '19 at 17:12
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    My point is that this is not the correct definition of "gauge symmetry", and I'm asking you to provide a citation for where you got it from. If your actual question is "what is a gauge symmetry", please do ask that explicitly and make sure it is not already answered e.g. in https://physics.stackexchange.com/q/266992/50583 and its linked questions. – ACuriousMind Sep 09 '19 at 17:16
  • Gauge transformations correspond to redundancies in your description, they map physically equivalent configurations into each other. If your description has such redundancies, then nothing physical should depend on them and therefore every physical quantity must be invariant under such transformations. However, physical symmetries are transformations that map between physically DISTINCT configurations which leave your theory invariant. Such symmetries are physical, often lead to conservation laws, degeneracies in your spectrum etc and cannot be 'gauge fixed'. 1/2 – Heidar Sep 09 '19 at 17:20
  • Translation symmetry (in space-time) is a physical symmetry and responsible for energy and momentum conservation. 2/2 – Heidar Sep 09 '19 at 17:20
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    But what I don't understand is why is translating the whole Universe to the left a physical symmetry. When it wouldn't change anything physically? Every physical thing such as the distance between particles would remain the same. Don't gauge symmetries correspond to charge conservation? –  Sep 09 '19 at 17:25
  • If you move an object to a different position, you change the physical state of that object. If your theory is invariant under such translation, it means that your theory has this symmetry. A symmetry is not the same as a gauge invariance. Gauge invariance is due to the fact that you have redundancies in your description, it's not a physical thing per se (it can be quite subtle to understand). – Heidar Sep 10 '19 at 01:47
  • And no, gauge symmetries do not lead to charge conservation. Sometimes it is stated like that, but that's not correct. In a theory with local $U(1)$ symmetry (say, electromagnetism), local gauge transformations are maps of the form $f:M\rightarrow U(1)$, where $M$ is your space-time manifold. In particular functions of the form $f(x)=e^{i\epsilon(x)}$ (let's call the group of all these local transformations G). However, not all these transformations correspond to gauge redundancies because boundary conditions on $\partial M$ are physical and gauge redundancies are not allowed to change those. – Heidar Sep 10 '19 at 01:53
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    The subgroup $G_g\subset G$ corresponding to all transformations that do not change the boundary conditions are the genuine gauge transformations (redundancies). Since the rest of your transformations still leave your theory invariant, $G/G_g$, they are physical symmetries. For example in the case of $M=\mathbb R^4$, fields are often given 'boundary' conditions like $\phi(x)\rightarrow 0$, as $|x|\rightarrow \infty$. The set of gauge transformations that leave this invariant would be those that become trivial transformation at infinity, thus $f(x)\rightarrow 1$ as $|x|\rightarrow \infty$. – Heidar Sep 10 '19 at 01:58
  • Note that a GLOBAL $U(1)$ symmetry does not become identity at infinity and is thus not a gauge transformation, but a physical global symmetry. It is this symmetry which gives rise to charge conservation. Some books would call this a "global gauge transformation", but that is sloppy language. – Heidar Sep 10 '19 at 02:00
  • zooby: The concept of strict symmetry (gauge or otherwise) is meaningful only in a model, and trying to define what the words "physical symmetry" should mean quickly runs into some elusive subtleties of how we relate models to reality. Given a model, we can usefully distinguish between $G_g$ and $G/G_g$ as in @Heidar's comments, but you seem to be asking why we refer to $G/G_g$ as "physical." Is this a fair interpretation of your question? – Chiral Anomaly Sep 10 '19 at 03:52
  • I am still slightly confused. A particle at x at time 0 moves to y at time t. The distance measured is |x-y|. In a one particle theory, surely one could "fix" the coordinate of the particle always to be at the origin at time 0. How is this any different than fixing the gauge of a field? The one particle theory also has redundant degrees of freedom. Kind of makes me think ALL symmetries are unphysical. But it seems there is some answer relating to boundary conditions which is probably the right answer. –  Sep 10 '19 at 19:15

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