# Casimir Energy and AdS space

1. Oct 16, 2007

### robousy

Hey Folks,

Had a thought - if you work out the Cosmological constant in SI units it works out to be around $$10^{-10} J/m^3$$. I performed a quick calculation for the Casimir energy (EM modes) for a plate measuring $$1 m^2$$ for a plate separation of $$1 \mu m$$ and found it to be on the order of $$- 10^{-9} J$$ which works out to be around $$- 10^{-3} J/m^3$$.

Note this is NEGATIVE energy, and significantly more negative then the comsological constant is postive.

Would it be wrong (and if so why) to say that the space between the plates is AdS space, eg a space with a negative cosmological const?

2. Oct 16, 2007

### josh1

It would be wrong because AdS solutions have a global negative cosmological constant.

3. Oct 16, 2007

### BenTheMan

Also AdS is defined by negative curvature, and a warp factor, neither of which exist in your case. The space between the plates is still flat.

4. Oct 16, 2007

### josh1

Maybe I'm misunderstanding you. I know that warp factors appear in flux compactifications and that these solutions are AdS before lifting to De Sitter, but I've never seen the idea of warp factors presented as part of the definition of AdS.

5. Oct 16, 2007

### BenTheMan

Josh---

You are most likely correct. My experiences with AdS have al been from Randall-Sundrum. It was my impression that AdS was negatively curved, and the negative curvature in the radial direction was called the warp factor''.

Perhaps I missed the point---the cosmological constant IS negative curvature.

Last edited: Oct 16, 2007
6. Oct 16, 2007

### robousy

Aaah, ok, so would it be correct to say that there was a local NEGATIVE comsological constant between the plates?

Or how about the following: the Freidmann equations tell us:

$$\ddot{a}=-\frac{4}{3}G(\rho+3p)a$$

Let p go to zero, if you flip the sign of $$\rho$$ you would flip from inflation to deflation, which is what is happening between the plates. Comments??