5 ms·

So I've read that it's enough to know that "there is a U(1) gauge symmetry in the Lagrangian" to derive the Maxwell equations from. Basically U(1) => Maxwell eq

• danharaj 9y ago
  I don't know gauge theory and haven't read the link, but the Lagrangian also needs to be relativistically invariant. This is a global symmetry instead of a gauge symmetry. I don't know if there's such a thing as, say, Euclidean gauge theory.
  • zbyszek 9y ago
    There is. A field theory in Minkowski space can be formulated in Euclidean space (and vice versa) with a technique called Wick rotation.
• Y_Y 9y ago
  It's a great question, but slightly subtle. If you already know the Lagrangian (-1/4 F^{\mu\nu} F_{\mu\nu}) all you need is Noether's theorem. If you want to come up with the Lagrangian you can do so by trying to find an expression with local gauge symmetry (the A field is invariant under addition of the four-grad of a scalar field). I'm afraid I don't have a good reference off the top of my head, but it's not difficult.
  • danbruc 9y ago
    Is that why those symmetry groups are so often mentioned, because they single out unique theories? As far as I know, but I am not a physicist, gauge symmetries are more of a defect than a feature of field theories. Because we lack theories without redundancies we just declare all states in the mathematical model that are related by a gauge transformation to be equivalent to each other and to correspond to just one physical state. In consequence I always wondered why one would then stress those symmetries but it would of course make sense if they were some kind of unique labels.
    • Y_Y 9y ago
      Yes! The symmetries dictate the physics, gauge and otherwise. Unfortunately there isn't a nice way to see that without a ton of field theory background.
      • danbruc 9y ago
        Which seems quite strange to me. I assume we would be really happy to find theories equivalent to existing field theories but without the gauge redundancies so that we have a one-to-one mapping between physical states and mathematical states. Why would a part we want to get rid of determine the part we are really interested in?
        • Retra 9y ago
          The 'mathematical states' here are descriptions of nature. A symmetry in the description doesn't correspond to identified physical states, it corresponds to a constraint on the description. For instance, if you want to prove that something is unique, you have to show how all the different ways of describing it lead to the same conclusions.
          • danbruc 9y ago
            A symmetry in the description doesn't correspond to identified physical states [...] As I said before, I am not a physicist, but I am pretty sure that you are wrong here. Unless you are a physicist and you are sure about that. My understanding is that gauge symmetries are a misnomer and should better be called gauge redundancies because they essentially establish equivalence classes of mathematical states corresponding to identical physical states. This is not the same kind of symmetry as for example the Poincaré symmetry of special relativity. Let me illustrate my understanding with a silly example that would almost certainly not actually work out mathematically but it should get the point accross. Say all you have in your mathematical toolbox are complex numbers and you come up with a theory of thermodynamics where temperatures are described by complex numbers. Everything works out nicely besides that there is no discernable physical difference between all the temperatures differing only by their imaginary parts. So you declare that there is a gauge symmetry in your theory of thermodynamics and all temperatures only differing by their imaginary parts are actually the same physical temperature. This theory has translations along the imaginary axis as a gauge symmetry.