Definition for the term gauge symmetry

[alphaone]

Could somebody please give me a definition for the term gauge symmetry in contrast to any other symmetry? Is the decisive difference that a gauge symmetry is local i.e. a function of the coordinates in contrast to being constant? I would also appreciate it if it could be pointed out how the defining difference gives rise to a redundancy of description of the physical system.

 

[ZapperZ]

You might want to read this:

http://xxx.lanl.gov/abs/hep-ph/0012061

Zz.

 

[alphaone]

Thanks very much for the reply and the reference. It really is a nice article, however it discusses the topic in a way I would like to avoid: I am aware of the meaning of gauge invariance in quantum electrodynamics and where it comes from historically but I would like to know a compact mathematical definition of the term in general, which is valid not just in electrodynamics or Yang Mills. For example General Relativity is a gauge theory so that the definition given in the reference does not agree with how I am used to the term gauge theory.

 

[dextercioby]

Could somebody please give me a definition for the term gauge symmetry in contrast to any other symmetry? Is the decisive difference that a gauge symmetry is local i.e. a function of the coordinates in contrast to being constant? I would also appreciate it if it could be pointed out how the defining difference gives rise to a redundancy of description of the physical system.

The gauge symmetry appears in connection with the existence of first class constraints for a classical dynamical system. The equations of motion, in the presence of gauge transformations, do not have unique solutions corresponding to prescribed initial conditions, unlike the case when gauge symmetry is absent.

 

[alphaone]

Thanks for the reply. I am not used to the term first constraint but from what I understand about it, it is a system of functions which have vanishing poison bracket and which also poisson commute with the Hamiltonian. If this is indeed the correct notion could you please let me know if there is a way to see directly from a Lagrangian whether such a constraint exists and could you also point out the difference between a local symmetry and gauge invariance(meaning equations of motion which are not uniquely determined) as it somehow seems to me that having a local symmetry should be enough to get non-unique solutions, but maybe I am totally confusing things here.

 

[dextercioby]

Nope, the constraints are "invisible" at lagrangian level, they first occur when one's trying to do the Legendre transformation of the lagrangian. This transformation is not fully achieveble for systems with constraints.

The eqns of motion ARE uniquely determined, it's just that their solutions are "ill".

 

[Haelfix]

Alphaone you will want to google for Batalin-Vilkovisky formalism for the origins of the terminology first class constraints. I believe Weinberg in volume2 also spends some time with it (and where I first learned it)

 

[kharranger]

I believe a general definition of a gauge symmetry can be given as any transformation of an action whose conserved Noether current is trivial. A complete set of such transformations can then be grouped into the form of a symmetry transformation whose parameter is an arbitrary function of spacetime.

 

[alphaone]

Thanks for all the helpfull replies! Could somebody please show me an explicit example of a local syymetry which is not a gauge symmetry?

 

[AlphaNumeric]

The Lorentz or Poincare symmetries are not used as gauge groups in non-gravity theories.

 

[alphaone]

Yeah, but they are global symmetries, right?(meaning same transformation at every point in space) in contrast to being local symmetry(transformation dependent on point in space-time). So does anybody have another candidate for a local symmetry which is not a gauge symmetry?

 

[ap_nhp]

I'm from Viet Nam ,i've studied gauge theory, i want to make friends with some body to discuss about it.My yahoo nick is ap_nhp.

 

[kakarukeys]

Yeah, but they are global symmetries, right?(meaning same transformation at every point in space) in contrast to being local symmetry(transformation dependent on point in space-time). So does anybody have another candidate for a local symmetry which is not a gauge symmetry?

Consider the scalar field on curved space-time
$$S = \int (\frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2) \sqrt{-g}d^4x$$

(Active) Diffeomorphism that acts on the field (but not the metric)
$$\phi(x) \rightarrow \phi'(x) = \phi(y(x))$$
is a local symmetry of the theory, it's not a gauge symmetry. When you do a Hamiltonian Analysis, you won't find any first class constraint.

You can verify $$S[\phi] = S[\phi']$$, both $$\phi$$ and $$\phi'$$ satisfy the same field equation.

I will be interested to know how to tell a symmetry is gauge symmetry from the Action without going to Hamiltonian theory, some says it has to do with Noether's 2nd theorem. Someone pls show me.

 

[samalkhaiat]

I will be interested to know how to tell a symmetry is gauge symmetry from the Action without going to Hamiltonian theory, some says it has to do with Noether's 2nd theorem. Someone pls show me.

I will state without proof the 2nd Noether theorem and its connection with constraints.

The Lagrangian $$\mathcal{L}(\phi_{A},\partial_{a}\phi_{A})$$ with $$\phi_{A}$$ (A=1,2,...,N) being generic fields is invariant under the infinitesimal transformations;

$$\phi^{A} = G^{A}_{\alpha}(\phi) \Delta^{\alpha}(x) + T^{Aa}{}_{\alpha}\partial_{a}\Delta^{\alpha}(x)$$
$$\alpha = 1,2,..,r$$

with arbitrary x-dependent functions $$\Delta^{\alpha}$$ , i.e.,

$$\delta\mathcal{L}(x) = 0$$

if and only if the following relations hold identically (i.e. irrespective of whether or not $$\phi_{A}$$ are solutions of the field equations):

$$\partial_{a}(T^{Aa}{}_{\alpha}\frac{\delta}{\delta\phi^{A}}\mathcal{L}) = G^{A}{}_{\alpha}\frac{\delta}{\delta\phi^{A}}\mathcal{L} \ \ (1)$$

$$\partial_{a}K^{ab}{}_{\alpha} + J^{b}{}_{\alpha}=0 \ \ (2)$$

$$K^{ab}{}_{\alpha}+K^{ba}{}_{\alpha}=0$$

where
$$J^{a}_{\alpha}=G^{A}_{\alpha} \frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}} + T^{Aa}{}_{\alpha}\frac{\delta\mathcal{L}}{\delta\phi^{A}}$$

$$K^{ab}{}_{\alpha} = T^{Ab}{}_{\alpha}\frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}}$$

and

$$\frac{\delta\mathcal{L}}{\delta\phi^{A}}=\frac{\partial\mathcal{L}}{\partial\phi^{A}} - \partial_{a}(\frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}})$$

denotes Euler derivative.
Now, Eq(1) means that N-Euler derivatives are interrelated by r equations ( $$\alpha = 1,2,...r$$ ), and hence, that the number of independent quantities among Euler derivatives is (N-r) in general. This means that the field equations:

$$\frac{\delta\mathcal{L}}{\delta\phi^{A}} = 0$$

for N unknown quantities $$\phi_{A}$$ are not independent of each other, but that only N-r equations are independent owing to the constraints Eq(1).


regards

sam

 

[kakarukeys]

are you saying
if the r equations are all trivial relations, then there is no gauge symmetry?

for example $$S = \int \frac{1}{2}m(\dot{q}_1+\dot{q}_2)^2 dt$$
there is gauge symmetry in it, oweing to the fact that one of the variable is redundant.

from all symmetries of the system, including time translation, space translations, gauge symmetry, I expect Noether's 2nd theorem to give me only one non-trivial relation, that is (I guess)

$$\partial_0 (\frac{\delta}{\delta q_1}L - \frac{\delta}{\delta q_2}L) = 0$$

showing that there is only one gauge symmetry. As a result, given a set of initial conditions, there is no unique evolution. But it's hard to use your formulae, is there a more user-friendly version of the equations?

 

[kakarukeys]

Consider the scalar field on curved space-time
$$S = \int (\frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2) \sqrt{-g}d^4x$$

(Active) Diffeomorphism that acts on the field (but not the metric)
$$\phi(x) \rightarrow \phi'(x) = \phi(y(x))$$
is a local symmetry of the theory, it's not a gauge symmetry. When you do a Hamiltonian Analysis, you won't find any first class constraint.

You can verify $$S[\phi] = S[\phi']$$, both $$\phi$$ and $$\phi'$$ satisfy the same field equation.

I missed the point that the symmetry has to be generated by Killing vector fields, that is an isometry (metric-preserving).