# Are possible gauge fields in a Lagrangian theory always determined by the structure of the charged degrees of freedom?

An elementary example to explain what I mean. Consider introducing a classical point particle with a Lagrangian $L(\mathbf{q} ,\dot{\mathbf{q}}, t)$. The most general gauge transformation is $L \mapsto L + \frac{d}{dt} \Lambda(\mathbf{q},t)$ which implies the usual transformations of the canonical momentum $p \to p+ \nabla_q \Lambda$. Generalizing this derivative as an extended one gives the connection of electromagnetism. Once the particle motion is quantized, we recognize this as a local $U(1)$ "internal" symmetry of the quantum-mechanical phase.

Is this a fundamental property of gauges symmetries - being implied by non-dynamical symmetries of the action? By "non-dynamical symmetries" I mean those coming from the structure of and the freedom of labeling for the degrees of freedom under consideration.

EDIT: After reflecting on the comments below, I'd re-formulate the question as:

Do non-dynamical symmetries of a local Hamiltonian exhaust all possible types of gauges fields that it can couple to in a gauge-invariant manner?

EDIT-2: The reason I'm asking is that it appears that the very possibility for a particle to couple to electromagnetism and gravity come from the applicability of the action formalism and is already built-in as the symmetry under addition of a total time derivative (which as I understand to be one of the possible general definitions of gauge symmetry).

Some comments suggest the answer is a trivial yes, presumably because non-dynamical symmetries are gauge symmetries by definition. A concise expert answer would be helpful to close the question.

• Maybe you are asking for the fully general definition of gauge symmetries of a given local action functional? The canonical reference that discusses this at great length is Henneaux-Teitelboim ncatlab.org/nlab/show/Quantization+of+Gauge+Systems . Sep 15, 2011 at 13:36
• @Urs: Thanks, this book indeed is a great in-depth source, will try to fish out an answer I'm looking for. Sep 15, 2011 at 13:48
• Still, I'd appreciate a concise answer from a live expert Sep 15, 2011 at 14:02
• I would have given a more direct answer if I understood what you are asking for. Can you maybe try to clarify? You seem to have asked "is every gauge symmetry induced by a non-dynamical symmetry"? By the only sense that I seem to be able to make of that this phrase it is trivially true. Can you be more specific, maybe? Sep 15, 2011 at 14:17
• Thanks for you input, I've edited the question in response. My feeling that it belongs more to Physics.SE seems to have been vindicated but let's wait for a definitive answer. Sep 15, 2011 at 16:31

I have been trying to understand what you may possibly mean by a "non-dynamical symmetry" (which is surely not a term that is normally used in papers from "mainstream" authors, to put it politely) and I became convinced that it cannot mean anything.

The problem arises in the third sentence when you write that the "most general gauge transformation" is $$L \to L+\frac{d\Lambda}{dt}.$$ But this is not a "transformation" in any sensible sense I can think of. This is a result telling you how the Lagrangian transforms under something – it transforms into itself up to a total derivative. But to define a transformation, you actually have to say how the fundamental fields $q,p$ actually transform, and not just how the Lagrangian transforms.

If a Lagrangian transforms to itself up to a total derivative, it means that the action $$S = \int dt\,L$$ may remain invariant given some favorable initial conditions at $\pm\infty$. So quite in general, it is allowed if symmetries transform the Lagrangian (or the Lagrangian density) up to itself plus a total derivative – or up to the divergence $\partial_\mu V^\mu$ in the field theory (multi-dimensional) case. In the component formalism (not superspace), this addition of total derivatives/divergences is inevitable e.g. for supersymmetry transformations.

But this result, how the Lagrangian itself transforms, is an extremely small part of the information that you need to actually define a transformation or a symmetry. So I don't think that you have defined any symmetry by saying how the Lagrangian transforms under it. There are infinitely many transformations that have this property.

The possibility to add a total derivative to the Lagrangian is completely general but specific gauge symmetries – such as Yang-Mills symmetry, diffeomorphisms, or local SUSY – are much more particular.

I think that the reason why you think that you're "deriving" a U(1) symmetry from the total derivative boils down to your confusing symbol $\Lambda$ whose total derivative is added to the Lagrangian. But the thing $\Theta$ whose derivative is added to the Lagrangian is a priori not the same thing as the parameter of a U(1) transformation. Instead, $\Theta$ may be an arbitrary complicated function of the fields (degrees of freedom) as well as the parameters of all the gauge transformations and perhaps derivatives of everything.

For a simple collection of classical particles and a U(1) electromagnetic symmetry, $\Theta$ may be a simple function of $\Lambda$ only (it's actually the sum of $\Lambda(\vec x_i)$ evaluated at the positions of all the particles, and summed over these particles, so the relationship is not as trivial as you suggest); for other symmetries, it's a more complicated function. But you actually need to study how the degrees of freedom transform under a would-be gauge symmetry to determine whether it's there or not; you can't just look at how the Lagrangian should transform. When you do so, you discover Yang-Mills symmetries, diffeomorphisms, local SUSY, and a few others as sensible local symmetries. But this work can't be done just by looking at total derivatives.

• Thank you, @Luboš, your answer is rich and useful (as usual). The gist of my observation is that adding total derivative to a Lagrangian adds a gradient to the canonical momentum, which is one of possible ways to couple to a gauge filed. I'll need to spell out the case of electromagnetism explicitly. Will either post it if part of my question remains, or accept your answer. Oct 2, 2011 at 13:32