- #1

- 46

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter alphaone
- Start date

- #1

- 46

- 0

- #2

- 35,847

- 4,676

- #3

- 46

- 0

- #4

- 13,033

- 588

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.

- #5

- 46

- 0

- #6

- 13,033

- 588

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

- #7

Haelfix

Science Advisor

- 1,955

- 222

- #8

- 17

- 0

- #9

- 46

- 0

- #10

- 289

- 0

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

- #11

- 46

- 0

- #12

- 7

- 0

- #13

- 190

- 0

Consider the scalar field on curved space-time

[tex]S = \int (\frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2) \sqrt{-g}d^4x[/tex]

(Active) Diffeomorphism that acts on the field (but not the metric)

[tex]\phi(x) \rightarrow \phi'(x) = \phi(y(x))[/tex]

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 [tex]S[\phi] = S[\phi'][/tex], both [tex]\phi[/tex] and [tex]\phi'[/tex] 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.

Last edited:

- #14

samalkhaiat

Science Advisor

- 1,736

- 1,045

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

[tex]\phi^{A} = G^{A}_{\alpha}(\phi) \Delta^{\alpha}(x) + T^{Aa}{}_{\alpha}\partial_{a}\Delta^{\alpha}(x)[/tex]

[tex]\alpha = 1,2,..,r[/tex]

with arbitrary x-dependent functions [tex]\Delta^{\alpha}[/tex] , i.e.,

[tex]\delta\mathcal{L}(x) = 0[/tex]

if and only if the following relations holdidentically(i.e. irrespective of whether or not [tex]\phi_{A}[/tex] are solutions of the field equations):

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

[tex]\partial_{a}K^{ab}{}_{\alpha} + J^{b}{}_{\alpha}=0 \ \ (2)[/tex]

[tex]K^{ab}{}_{\alpha}+K^{ba}{}_{\alpha}=0[/tex]

where

[tex]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}}[/tex]

[tex]K^{ab}{}_{\alpha} = T^{Ab}{}_{\alpha}\frac{\partial\mathcal{L}}{\partial\partial_{a}\phi^{A}}[/tex]

and

[tex]\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}})[/tex]

denotes Euler derivative.

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

[tex]\frac{\delta\mathcal{L}}{\delta\phi^{A}} = 0[/tex]

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

regards

sam

Last edited:

- #15

- 190

- 0

are you saying

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

for example [tex]S = \int \frac{1}{2}m(\dot{q}_1+\dot{q}_2)^2 dt[/tex]

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)

[tex]\partial_0 (\frac{\delta}{\delta q_1}L - \frac{\delta}{\delta q_2}L) = 0[/tex]

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?

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

for example [tex]S = \int \frac{1}{2}m(\dot{q}_1+\dot{q}_2)^2 dt[/tex]

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)

[tex]\partial_0 (\frac{\delta}{\delta q_1}L - \frac{\delta}{\delta q_2}L) = 0[/tex]

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?

Last edited:

- #16

- 190

- 0

Consider the scalar field on curved space-time

[tex]S = \int (\frac{1}{2}\partial^\mu\phi\partial_\mu\phi - \frac{1}{2}m^2\phi^2) \sqrt{-g}d^4x[/tex]

(Active) Diffeomorphism that acts on the field (but not the metric)

[tex]\phi(x) \rightarrow \phi'(x) = \phi(y(x))[/tex]

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 [tex]S[\phi] = S[\phi'][/tex], both [tex]\phi[/tex] and [tex]\phi'[/tex] 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).

Share: