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Hello, I don't know if this is a good place for exposing my problem but I'll try...

I have a gauge theory with action:

$S=\int\;dt L=\int d^4 x \;\epsilon^{\mu\nu\rho\sigma} B_{\mu\nu\;IJ} F_{\mu\nu}^{\;\;IJ} $

Where $B$ is an antisymmetric tensor of rank two and $F$ is the curvature of a connection $A$ i.e: $F=dA+A\wedge A$, $\mu,\nu...$ are space-time indices and $I,J...$ are Lie Algebra indices (internal indices) I would like to find its symmetries. So I rewrite the Lagrangian by splitting time and space indices $\{\mu,\nu...=0..3\}\equiv \{O; i,j,...=1..3\}$ I find:

$L = \int d^3 x\;(P^i_{\;IJ}\dot{A}_i+B_i^{\,IJ}\Pi^i_{\,IJ}+A_0^{\;IJ}\Pi_{IJ})$

Where $\dot{A}_i = \partial_0 A_i$, $P^i_{\;IJ} = 2\epsilon^{ijk}B_{jk\,IJ}$ is hence the conjugate momentum of $A_i^{\,IJ}$

$B_i^{\,IJ}$ and $A_0^{\;IJ}$ being Lagrange multipliers we obtain respectively two primary and two secondary constraints:

$\Phi_{IJ} = P^0_{\;IJ} \approx0$

$\Phi_{\;\;IJ}^{\mu\nu} = P^{\mu\nu}_{\;\;IJ} \approx0$

$\Pi^i_{\,IJ} = 2\epsilon^{ijk}F_{jk\,IJ} \approx0$

$\Pi_{IJ}=(D_i P^i)_{IJ} \approx0$

Where $P^0_{\;IJ}$ are the conjugate momentums of $A_0^{\,IJ}$ and $P^{\mu\nu}_{\;\;IJ}$ those of $B_{\mu\nu}^{\;\;IJ}$. Making these constraints constant in time produces no further constraints.

Whiche gives us a general constraint:

$\Phi = \int d^3 x \;(\epsilon^{IJ}P^0_{\,IJ}+\epsilon_{\mu\nu}^{IJ}\;P^{\mu\nu}_{\;\;IJ}+\eta^{IJ}\Pi_{IJ}+\eta_i^{IJ}\Pi^i_{\;IJ})$

Each quantity $F$ have thus a Gauge transformation $\delta F = \{F,\Phi\}$ where $\{...\}$ denotes the Poisson bracket.

Knowing that this theory have the following Gauge symmetry:

$\delta A = D\omega$

$\delta B = [B,\omega]$

Where $\omega$ is a 0-form, I would like to retrieve these transformations using the relation below. (where $\Phi$ is considered as the generator of the Gauge symmetry) but my problem is that I don't know how to proceed, I already did this with a Yang-Mills theory and it worked... but for this theory it seems to le intractable! Someone to guide me?

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2 Answers 2

Unless for some reason you absolutely must work within the Hamiltonian approach, you can just directly look for the complete set of (infinitesimal Lie point) symmetries of the Euler--Lagrange equations or of the action itself. The procedure is standard and described in many good books. For instance, you can look into those by Olver (more math-y) or Stephani (somewhat closer to physics). Using the theory from these books you can also verify whether the transformation at the end of your question is indeed a symmetry.

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Although the second book looks very interesting, I hesitate to start reading for the following reason: - Symmetries in physics are related to the indetermination generated by the fact that the Hamiltonian contains constraints times arbitrary parameters, if one tries to write down two possible future states corresponding to two different parameters (using the Poisson bracket with the Hamiltonian), these two states will be physically equivalent whereas different by a term consisting of constraints times arbitrary parameters. Suggesting that constraints are the generators of these Gauge transf. –  Pedro Mar 14 '10 at 11:19
    
So I don't thing that a pure algebraic method will exhibit these "Gauge transformations"!! On the other hand, I confess I misunderstand the fact that a Lagrangian which did not ask anyone find itself invariant under symmetries exhibited by the passage to the Hamiltonian formalism!! I mean, a priori the form of a Lagrangian has no relation with will be done in the Hamiltonian framework. –  Pedro Mar 14 '10 at 11:28
    
PS: I don't want to verify that this is a symmetry, it's already done. I want to derive this symmetry, actually I have a little more complicated action which I need to analyse... –  Pedro Mar 14 '10 at 12:26
    
First of all, the method outlined above is systematic and it should yield, inter alia, the gauge symmetries (and also the obvious symmetries like, say, Poincare or conformal ones). Moreover, while the notion of symmetry can be considered within the Hamiltonian framework, the notion of symmetry (including but not restricted to the gauge symmetry) does make perfect sense without reference to this framework, see e.g. Olver's book I linked above. I hope that perhaps some other people can better understand the menaing of your comments and be of more help. –  mathphysicist Mar 15 '10 at 23:29

Although this question is over a year old, some readers may be interested in the following comments.

  1. The symmetries of the Lagrangian and the action are not identical.

  2. I don't know why you cannot recover the gauge transformations from the constraints, perhaps it has to do with not including the $B_{0i}$ fields in the second expression for the Lagrangian?

  3. The constraints give infinitesimal transformations leading to the usual gauge transformations, but there seem to be symmetries which are non-trivial extensions of these.

  4. For example, for a gauge group SU(n), the transformations $$ A' = UAU^\dagger + \phi $$

$$ B' = UBU^\dagger + UAU^\dagger\wedge\phi + \phi\wedge UAU^\dagger + \phi\wedge\phi $$

where $U(x) \in$ SU(n), and $\phi = -dUU^\dagger$, leaves the action invariant.

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Thank you for these notes, I just want to say that I finally found how to derive these Gauge transformations (more or less because I arrived to derive them for spatial components only), the demonstration s here docs.google.com/… until page 10 (in french) –  Pedro Apr 30 '11 at 8:38
    
You are welcome. Getting the transformations for the spatial components should be sufficient for the dynamics, and you can extend them to all the components. –  Amitabha Lahiri May 2 '11 at 7:32

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