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.

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    $\begingroup$ 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 . $\endgroup$ Sep 15, 2011 at 13:36
  • $\begingroup$ @Urs: Thanks, this book indeed is a great in-depth source, will try to fish out an answer I'm looking for. $\endgroup$
    – Slaviks
    Sep 15, 2011 at 13:48
  • $\begingroup$ Still, I'd appreciate a concise answer from a live expert $\endgroup$
    – Slaviks
    Sep 15, 2011 at 14:02
  • $\begingroup$ 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? $\endgroup$ Sep 15, 2011 at 14:17
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    $\begingroup$ 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. $\endgroup$
    – Slaviks
    Sep 15, 2011 at 16:31

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Slaviks
    Oct 2, 2011 at 13:32

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