For this metric,

- g = a^{2}

, while

g^{ηη} = - a^{- 2}

g^{zz} = a^{- 2}

and

g^{ij} = δ^{ij}

. The massless Klein-Gordon equation is consequently

□ ϕ = \frac{1}{a ^{2}} (- \partial_{η}^{2} + \partial_{z}^{2}) ϕ + δ^{ij} \partial_{i} \partial_{j} ϕ = 0.

Insert the spatial Fourier transform given in the question. Each wavenumber mode then satisfies

Φ_{k_{z} k}^{''} + ω_{k_{z} k}^{2} (η) Φ_{k_{z} k} = 0, ω_{k_{z} k}^{2} (η) = k_{z}^{2} + a (η)^{2} ∣ k ∣^{2} .

ii

Solution

In the two constant regions define

ω_{\pm} = k_{z}^{2} + a_{\pm}^{2} ∣ k ∣^{2} .

u_{\pm} (η) = \frac{e ^{- i ω_{\pm} η}}{2 ω _{\pm}},

up to the common spatial Fourier normalization. They satisfy the unit Wronskian condition

i (u_{\pm}^{*} u_{\pm}^{'} - u_{\pm} u_{\pm}^{*'}) = 1,

which is the mode form of the Klein-Gordon inner product. Because the mode equation contains no delta function at

η = 0

, both

Φ

and

Φ^{'}

are continuous there.

Continue the incoming positive-frequency mode through

η = 0

u_{-} (η) = α u_{+} (η) + β u_{+}^{*} (η) (η > 0) .

α = \frac{ω _{+} + ω _{-}}{2 ω _{+} ω _{-}}, β = \frac{ω _{+} - ω _{-}}{2 ω _{+} ω _{-}} .

The associated Bogoliubov transformation says that the incoming vacuum has expected outgoing occupation number

⟨ N_{k_{z} k}^{(+)} ⟩ = ∣ β ∣^{2} = \frac{( ω _{+} - ω _{-} ) ^{2}}{4 ω _{+} ω _{-}} .

It vanishes when

a_{+} = a_{-}

, as required when there is no change of geometry.

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