Poisson brackets for Hamiltonian descriptions

haushofer
Science Advisor
Insights Author
Messages
3,045
Reaction score
1,579
Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

If I generate my symmetry with a vector field \xi or \alpha, the statement which is often made is that

<br /> \{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]<br />

In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

Thanks in forward :)
 
Physics news on Phys.org
haushofer said:
Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

If I generate my symmetry with a vector field \xi or \alpha, the statement which is often made is that

<br /> \{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]<br />

In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

Thanks in forward :)

I hope, I have uderstood your question correctly!

In the real life, we use Noether procedure to write H_{\alpha} in terms of the dynamical variables on PS:

H_{\alpha} = \int dx \pi \mathcal{L}_{\alpha} \phi \ \ \ (R)

where \mathcal{L}_{\alpha} is Lie derivative along the vector field \alpha.
It follows immediately from eq(R) that

1) canonical transformations generated by the charge/constraint correspond precisely to the diffeomorphisms generated by the vector field \alpha ,i.e.,

<br /> \{ H_{\alpha} , \phi \} = \mathcal{L}_{\alpha} \phi, \ \ \mbox{etc.} \ (1)<br />

2) charges/constraints satisfy the algebra

\{ H_{\alpha}, H_{\xi} \} = H_{[\alpha , \xi ]} \ \ \ (2)

To confirm the consistency between eq{1) and eq(2), we use the Jacobi identity

\{ \{ H_{\alpha} , H_{\xi} \} , \phi \} = \{\{H_{\alpha}, \phi\} , H_{\xi} \} - \{\{H_{\xi} , \phi \} , H_{\alpha} \}

This gives

\{\{H_{\alpha} , H_{\xi} \} , \phi \} = \mathcal{L}_{\alpha} \mathcal{L}_{\xi} \phi - \mathcal{L}_{\xi}\mathcal{L}_{\alpha} \phi = \mathcal{L}_{[\alpha , \xi ]} \phi

However, the algebra of constraints is not, in general, a proper Lie algebra: the RHS of eq(2) may contain structure functions rather than structure constants. This means, in the BRST terminology, that we may have an open algebra. Indeed, it has been shown [Bargmann and Komar] that the algebra of constraints is not isomorphic to the Lie algebra of the "gauge group" of tetrad-gravity.

*****

Ok, here is a math question for you:

Given two function f,g : PS \rightarrow \mathbb{R}, their Poisson bracket is defined by

\{ f , g \} = \mathcal{L}_{X_{f}}g = - \mathcal{L}_{X_{g}}f

this means that PB turns the vector space of functions on PS into Lie algebra. Use this Lie-bracket to show that the linear map f \rightarrow X_{f} (which associates to f its (Hamiltonian) vector field) takes Poisson brackets of functions into commutators of vector fields:

[X_{f} , X_{g}] = X_{\{f,g\}} \ \ (3)

compair this with eq(2).

regards

sam
 
A, nice! I will look at this tomorrow! Sam to the rescue, again :D
 
haushofer said:
A, nice! I will look at this tomorrow! Sam to the rescue, again :D

It is a pleasure, very time.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top