- #1
robousy
- 334
- 1
...on the off chance anyone knows this, I'm trying to get from:
[tex]V=\frac{1}{2A}Tr Log(\frac{-\Box}{\mu^2})[/tex]
to
[tex]V=\frac{(-1)^{\eta-1}}{4\pi^\eta\eta!}\frac{\pi}{L}^{D-1}\zeta'(1-D)[/tex]
I know this is a shot in the dark, but in case anyone has experience.
The paper I'm reading explains 'it is easy to show that' to get to the seconds equation ! I hate that. The paper also has a reference...the reference is Birrel and Davies, great I thought, I have that book, the reference is for p340, which takes me to the Index ! lol.
Anyway, I guess the key is figuring how the trace of the log of the box operator gives me the derivative of the zeta function.
Anyone??
:)
[tex]V=\frac{1}{2A}Tr Log(\frac{-\Box}{\mu^2})[/tex]
to
[tex]V=\frac{(-1)^{\eta-1}}{4\pi^\eta\eta!}\frac{\pi}{L}^{D-1}\zeta'(1-D)[/tex]
I know this is a shot in the dark, but in case anyone has experience.
The paper I'm reading explains 'it is easy to show that' to get to the seconds equation ! I hate that. The paper also has a reference...the reference is Birrel and Davies, great I thought, I have that book, the reference is for p340, which takes me to the Index ! lol.
Anyway, I guess the key is figuring how the trace of the log of the box operator gives me the derivative of the zeta function.
Anyone??
:)
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