Why are E and gamma considered canonically conjugate in the Gauss law?

[naima]

Hi PF
I read a paper in which Lewandowski writes:
the Gauss law has the form
$\partial E^a / \partial x^a + c_{jk}E^{aj}\gamma ^k_a = 0$
wherec are the structure constants
he then writes that if we are in a semisimple algebra they are skew symmetric in the indices and it can be rewritten as
$\partial E^a / \partial x^a - c^j_{k}E^{a}_j\gamma ^k_a = 0$
And he writes (that is my question):
"where NOW E and $\gamma$ are canonically conjugate".
Can you explain why?

 

[dextercioby]

When reading a paper (presumably found online) it's a proof a courtesy to offer a link to that paper, so that any respondents could have the context of your question (the only Lewandowski I know plays for Bayern Muenchen :-))

 

[samalkhaiat]

Hi PF
I read a paper in which Lewandowski writes:
the Gauss law has the form
$\partial E^a / \partial x^a + c_{jk}E^{aj}\gamma ^k_a = 0$
wherec are the structure constants
he then writes that if we are in a semisimple algebra they are skew symmetric in the indices and it can be rewritten as
$\partial E^a / \partial x^a - c^j_{k}E^{a}_j\gamma ^k_a = 0$
And he writes (that is my question):
"where NOW E and $\gamma$ are canonically conjugate".
Can you explain why?

Further to what Dexter have said, it would be more helpful, if you tell us, for example, why the structure constant carries two indices instead of three, the meaning of different indices, symbols, etc.
Any way, in any field theory with (compact) non-Abelian symmetry $G$, the conjugate pair $(\varphi^{A} , \pi_{A})$, $A = 1,2, \cdots , \mbox{dim}\left(\rho_{\varphi}(G)\right)$, appear in (the Noether expression of) the generator $Q_{a}$ , $a = 1, \cdots , \mbox{dim}(G)$, of the infinitesimal symmetry transformation $$\delta_{a}\varphi^{A} = [Q_{a} , \varphi^{A}] = \left(T_{a}\right)^{A}{}_{B} \ \varphi^{B} ,$$ $$[T_{a} , T_{b}] = C_{abc}T_{c} .$$ In Noether theorem, $Q_{a}$ is given by the integral $$Q_{a} = \int d^{3}\vec{x} \ J^{0}_{a}(x) = \int d^{3}\vec{x} \ \pi_{A}(x) \left(T_{a}\right)^{A}{}_{B} \ \varphi^{B}(x) , \ \ \ (1)$$ where $J^{\mu}_{a}$ is the conserved 4-vector Noether current.
For pure Yang-Mills theory, $\mathcal{L} = - (1/4) F^{\mu\nu a}F_{\mu\nu}^{a}$, the Noether current (corresponding to the global symmetry) is given by $$J^{\mu}_{a} = - C_{abc} F^{\mu\nu b}A^{c}_{\nu} ,$$ where the vector potential $A^{a}_{\mu}$ transforms in the adjoint representation $C_{abc} = (T_{a})_{bc}$. Integrating the $J^{0}_{a}$ component, we obtain $$Q_{a} = - \int d^{3}\vec{x} \ F^{0j}_{b}\ \left(T_{a}\right)^{b}{}_{c} \ A^{c}_{j} .$$ Using, the definition of the non-Abelian electric field $E^{j}_{b} = - F^{0j}{}_{b}$, we get $$Q_{a} = \int d^{3}\vec{x} \ E^{j}_{b}(x) \ \left(T_{a}\right)^{b}{}_{c} \ A^{c}_{j}(x) . \ \ \ (2)$$ Comparing (2) with the generic form (1), we can identify the conjugate (Yang-Mills) pair with $( A^{a}_{j} , E_{a}^{j})$. You can also arrive at the same construction, using the generator of local gauge symmetry, i.e., the integral form of Gauss’ law.

 

[naima]

I did not give the link because it is not free
It is "
Variations of the parallel propagator and holonomy operator and the Gauss
law constraint"
I do not think that everything must be free but this paper is sold by scammers!
You can buy these 10 pages for 30 dollars or read it for 4 dollars. But when you click on read it, they say choose 5 articles for 20 dollars.
Yes i forgot the third indice, sorry

 

[DrClaude]

The proper citation is Jerzy Lewandowski, Ezra T. Newman, and Carlo Rovell, Variations of the parallel propagator and holonomy operator and the Gauss law constraint, J. Math. Phys. 34, 4646 (1993) http://dx.doi.org/10.1063/1.530362

I do not think that the American Institute of Physics can be scalled "scammers." Anyway, most of those who can help you will have access to this paper through an institutional subscription. Please always give full references.

 

[dextercioby]

Oh, then it's fair to say that http://scitation.aip.org/content/aip/journal/jmp/34/10/10.1063/1.530362 is the legal path to the article. Some of the readers of PF can access the paywall articles with their uni/research inst. logins, so they can read the excerpt you refer to and answer your question. Well, I guess samalkhaiat needn't read this, to offer an overwhelming answer.

 

[dextercioby]

[...]
I do not think that the American Institute of Physics can be called "scammers." [...]

Well, if they ask you to pay 20 bucks to read five articles (4 you probably don't need), when you are offered to pay only 4 to read the one you need, it's almost a scam.

 

[vanhees71]

The legal link is

http://dx.doi.org/10.1063/1.530362

 

[naima]

The i indice that i ommitted does not occur in a sum. there is a (4.2) constraint for each i.