Questions about the Casimir energy equation

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The discussion centers on the Casimir effect and the associated electromagnetic quantum vacuum energy equation, specifically E= (i / 2τ) Tr ln Γ. The term "i" refers to the imaginary unit, the square root of -1. The variable τ denotes the infinite duration of the configuration, which pertains to the arrangement of two uncharged conducting parallel plates. The term Γ is indeed a matrix, and the trace of the natural logarithm of Γ is being calculated. Additionally, ω, μ, and ε represent angular frequency, permeability, and electric permittivity, respectively, with ε being a vector quantity.

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I've been studying the Casimir effect and zero point energy recently. In a scholarly article, I came across this formula for the electromagnetic quantum vacuum energy:

E= (i / 2τ) Tr ln Γ where Γ is Green's dyadic that satisfies the equation:

[(1/ω2 * ∇) × (1/μ * ∇) - ε ] Γ

Here is a link to the article where you can see the equation in the beginning of the article:
http://www.worldscientific.com/doi/pdf/10.1142/S2010194512007325

Now I just have a few questions about this equation:

1. First I just want to verify whether or not the term "i" in the equation refers to the imaginary number i or some other variable i. Is this i the square root of -1 or is it something else (and if so what is it)?

2. The article said that τ is the "infinite" time that the configuration exists. What is this "configuration exactly"? I assumed that it was the configuration of the two uncharged conducting parallel plates that is used to invoke the Casimir effect, but I wanted to make sure of this.

3. I assume that the term Γ is a matrix because from what I read in the formula, it seems that you are taking the trace of the natural logarithm of Γ. Is this correct? Are you taking the trace of the natural log of a matrix?

4. What are ω, μ, and ε? Is ε some electric permitivity vector? I assume ε is a vector quantity since it wouldn't make sense to subtract a scalar from a cross product (which is a vector).

Thank you.
 
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Sorry to bug people. It's just that it has been multiple days since I posted this and nobody has replied , so I thought I'd just bump up the thread a little to see if anyone may reply now.
 

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