Can anyone give an intuitive example of what a gauge symmetry is? I am new to this concept and would like to understand it better!
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$\begingroup$ Wikipedia $\endgroup$– GhosterCommented Nov 25, 2023 at 7:03
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1$\begingroup$ A nice reference that ties together some of the answers below math.ucr.edu/home/baez/torsors.html $\endgroup$– Sachin ValeraCommented Nov 25, 2023 at 21:21
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1$\begingroup$ You might find Maldacena’s economics analogy intuitive arxiv.org/abs/1410.6753 $\endgroup$– innisfreeCommented Nov 26, 2023 at 12:21
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4 Answers
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Another example of of a Gauge Symmetry can be found in the basic $V=mgh$. Here you can have your "ground" anywhere you want. This freedom reflects the key idea in gauge theory.
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2$\begingroup$ Thank you. Can you think of this symmetry as an analogy with @MrDBrane’s answer? $\endgroup$ Commented Nov 25, 2023 at 7:27
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$\begingroup$ It gets a bit messy sometimes because the gauge groups have different dimensions. The "gauge fields", which reflect the gauge freedom can be of different forms, so you can always exchange interpretations. $\endgroup$ Commented Nov 25, 2023 at 7:31
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$\begingroup$ Yes. In the particular case of QED, the freedom to choose an electromagnetic gauge corresponds to the fact that you should come up with all the same object level predictions -- collisions of particles, for example -- no matter your path through spacetime, even if you accelerate. $\endgroup$ Commented Nov 28, 2023 at 12:22
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A Gauge Symmetry (not talking about large gauge transformations) refers to mathematical symmetries that are not physical but rather redundancies in our formulation.
It is a quite rich subject, but I will limit myself to intuitive answers.
Consider the magnetic field. We know that it is given by the curl of a vector potential
$$\mathbf{B = \nabla \times A}$$
Now notice that if you add a curl-less to A, the magnetic field stays the same, that is the magnetic field is insensitive to curl-less terms. This means that we have a symmetry where any transformation of A in the form of $\mathbf{A} \rightarrow \mathcal{A} + \nabla f$, where f is just a scalar function, leaves our physics invariant.
In physics it is efficient and effective to describe these symmetries by gauge groups, so groups that describe these redundancies, that is groups under whose action our model stays physically the same. The standard Model is described by $SU(3)\times SU(2) \times U(1) $, which all describe different aspects of the theory.
Gauge theory is a very rich subject that has shaped the way we approach physics.
Reading Recommendations:
Large Gauge Transformation (Gauge Transformations that cannot be thought of as redundancies.
Yang Mills (Wikipedia)
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$\begingroup$ To add to this point: note that we usually explicitly do experiments that "feel" the magnetic field, but we never seem to have direct access to the magnetic potential itself*; when we write things in terms of magnetic potential there is a redundancy in our description (this is the gauge symmetry). (*ignoring Aharanov-Bohm type subtelties) $\endgroup$ Commented Nov 26, 2023 at 22:54
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Take a potential that only depends on the distance of an object’s center from a center point. Then you can consider a rotation of your object. If this transformation acts locally, so the rotation for example is dependent o. You spacetime location, then you have a gauge symmetry and the rotations are gauge transformations.
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$\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$– Community BotCommented Nov 28, 2023 at 20:07
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A toy model of a gauge theory is $$ Z ~\propto~\int \! dx ~dy~ e^{iS(x)}, \tag{71.8}$$ cf. Ref. 1. The action $S(x)$ and the path integral $Z$ are invariant under a gauge transformation $$y\quad\longrightarrow\quad y^{\prime}~=~y+f(x).$$ The $y$ variable is redundant/unphysical. See Ref. 1 for further details on how to gauge-fix the theory.
In more realistic gauge theories, it is more difficult (if not impossible) to cleanly separate physical and unphysical DOF, but the underlying idea is the same.
References:
- M. Srednicki, QFT, 2007; chapter 71. A prepublication draft PDF file is available here.
answered Nov 25, 2023 at 7:41