# Expanding manifold with constant boundary

1. ### Mike2

OK. Suppose you have a surface with a closed curve as a boundary. Then suppose that surface grows like a soap bubble but the boundary is stationary like the orifice through which air passes to make the bubble grow. It would seem that the 2D surface grows in both dimensions in the middle of the bubble, but the buble is not growing in at least one dimension along the boundary. What would be the equation for the metric both in the middle and on the boundary and as it approaches the boundary?

I wonder all this because in a different thread I explore the possibility that matter may be the boundary of an expanding universe. If so, I wonder what distinction there is in the metic of space as it approaches the boundary (particles). I'm kind of thinking that matter may be like a stationary boundary where the growing space must somehow bend and stretch to accommodate a fixed boundary. But the photon particles may be where the boundary grows right along with the surrounding space. I suppose you could have a boundary that has portions that expand and protions that are fixed.

2. ### Mike2

I'd have to think that this would give a similar metric as used in GR. For if the surrounding space has expanded while the boundary surface of mass particles remains fixed, then you would have in effect a contraction of space near mass, just as in GR, right?

3. ### Mike2

Now I wonder if a manifold with a boundary can have a metric just as easily as a manifold without boundary? If it can, then we can also have a metric on a manifold with a boundary that is distributed into smaller closed boundaries, right?

Thanks.

4. ### Mike2

I suppose there would be a discontinuity at the boundary points themselves, right? But if there is an undefined metric beyond the boundary, how would that be expressed, by an infinity there, by a zero? Or would we simply specify that the metric is a function defined only for the manifold and not to be extended past the boundary?

8,147
Staff Emeritus

I don't see why a singularity is necessary. You could have nice boundary conditions at the boundary, with good limiting behavior of the derivatives. Note that the AdS/CFT research area is about a manifold with an important boundary.

6. ### Mike2

Thanks. Would you be kind enough to tell me a little more about this AdS/CFT research?

I'm stuck here at the top trying to justify my way to the bottom, and I'm not sure how to proceed. So I ask question about how GR or QM can be justified by first principles. It is probably irritating to the rest of you. But those kinds of questions are inevitable. Please be patient.

8,147
Staff Emeritus
AdS means anti deSitter space, a Riemannian manifold with constant negative curvature. It is the boundary of a brane which supports Conformal Field Theory (CFT). The conjecture is that what happens on the boundary space completely determines the physics on the conformal brane. This is called the holographic conjecture because it is analogous to the way a 2-dimensional hologram cvan accurately capture 3-d shapes in the round. This is a very active research program.

8. ### Mike2

I read a breif intro to AdS-CFt correspondence at:
http://arxiv.org/PS_cache/hep-th/pdf/0003/0003120.pdf

Most of it over my head, of course. Did I read right that a Quantum Field theory in d+1 was not a quantum theory in the AdS of dimension d? Does a path integral in the d+1 of CFT translate to a different kind of path integral in the d dimensions in AdS? Or does the path integral of CFT not not translate to a path integral of AdS? Wouldn't it be great if we could justify QM in d+1 from a classical view in AdS of d dimensions? That's probably too much to hope for.

8,147
Staff Emeritus
This is a valuable paper that I wasn't aware of. Thanks for finding it. The paper is written in the mathematical physics tradition of C*-algebras acting on Hilbert space, where path integrals and other quantization techniques don't appear. It specifically proves that the conformal quantum field theory on d-dimensional Minkowski space determines the quantum field theory on d+1-dimensional space. So no, we don't get a free pass from classical theory to quantum theory.

(added after scanning the paper)

In fact he proves that the algebras of local observables are the same, so the quantum theories are identical, although the physical interpretation of the observables is radically different in the two spaces. An example, as he stresses, of the algebraic approach.

Last edited: Apr 14, 2004
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