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| Youngsub... |
 Posted: Mon Sep 07, 2009 11:21 pm |
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Guest
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Could someone explain to me why a Lagrangian with Chern-Simons term is
needed besides usual Einstein-Hilbert action when we consider the
boundary of manifold?
I came across with this fact from some papers that I am reading these
days, but I am not satisfied. Could someone point me to a good
reference?
Thanks in advance!
Youngsub |
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| Posted: Tue Sep 08, 2009 11:09 pm |
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Guest
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Youngsub <yyoon at (no spam) fas.harvard.edu> wrote:
[quote:58b581efad]Could someone explain to me why a Lagrangian with Chern-Simons term
is needed besides usual Einstein-Hilbert action when we consider the
boundary of manifold?
[/quote:58b581efad]
It would help if you provided the context. Sometimes a Chern-Simons
term is needed, sometimes not -- it depends on your choice of variables,
your action, and your boundary conditions.
The basic idea of an action principle is that the equations of motion
determine the extrema of the action, that is, that the variation of
the action is zero when evaluated at solutions of the equations of
motion.
Now, it is often the case that when you vary an action to get the
equations of motion, you have to integrate by parts. If your action is
an integral over a manifold with boundary, this partial integration
leads to boundary terms, which will depend on the fields and their
variations at the boundary. If these boundary terms are not zero,
the equations of motion *don't* determine an extremum -- the total
variation of the action isn't zero. To fix this, you need to add new
boundary terms to the original action, whose variation cancels the
contribution from integration by parts.
For example, the action for a scalar field has a term of the form
I = \int d^4x \partial_\mu\phi \partial^\mu\phi
Varying this action gives you a d'Alembertian, and the Klein-Gordon
equation, but only after you do an integration by parts. The boundary
term looks something like
\int d^3x (\delta\phi)\partial_n\phi
where \partial_n is the normal derivative. If you use Dirichlet boundary
conditions, fixing \phi at the boundary, then \delta\phi is zero in the
boundary term, and you have a real extremum. If you use Neumann
boundary conditions, though, fixing \partial_n\phi at the boundary, the
boundary term is not zero. Instead, you have to eliminate it, by adding
a boundary term of the form
-\int d^3x \phi\partial_n\phi
Just as in this simple case, the boundary term you need for general
relativity depends on the boundary conditions you choose, and also
on the fields you choose to write the action (metric, or tetrads, or
Ashtekar variables,...). For the standard metric formalism, if you fix
the metric at the boundary, Gibbons and Hawking showed that the term
you need is an integral over the extrinsic curvature at the boundary.
You can find a derivation in, for instance, Appendix E of Wald's textbook
or Poisson's _A Relativist's Toolkit_. No Chern-Simons term is needed.
If, on the other hand, you write the action in Ashtekar variables, you
need different boundary conditions (because the variables are not the
same), and therefore a different boundary term. In particular, if your
boundary is a horizon of a black hole, with "isolated horizon" boundary
conditions, you find that the term you need is a Chern-Simons term. I
don't know of any simple, intuitive derivation; it's just a calculation.
You can find it in http://arxiv.org/abs/gr-qc/9905089.
Steve Carlip |
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