Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 1 / iv

Table of contents
- Solution iv

Solution

At late time

f (k, τ_{0}) \to H / 2 k^{3}

and

a = - 1/ (H τ)

. Define the elementary symmetric polynomials

e_{2} = \sum_{a < b} k_{a} k_{b}, e_{3} = \sum_{a < b < c} k_{a} k_{b} k_{c}, e_{4} = k_{1} k_{2} k_{3} k_{4} .

Then part ii reduces to

⟨ O ⟩_{λ}^{'} = - \frac{iλ H ^{4} E}{8 \prod _{a} k _{a}^{3}} Im J (τ_{0}),

where a prime removes the momentum-conserving Dirac delta distribution and

J (τ_{0}) = \int_{- \infty (1 - i ϵ)}^{τ_{0}} \frac{d τ}{τ ^{4}} e^{- i k_{T} τ} \prod_{a = 1}^{4} (1 + i k_{a} τ) .

Expanding the product gives

\frac{1}{τ ^{4}} \prod_{a} (1 + i k_{a} τ) = \frac{1}{τ ^{4}} + \frac{i k _{T}}{τ ^{3}} - \frac{e _{2}}{τ ^{2}} - \frac{i e _{3}}{τ} + e_{4} .

Repeated integration by parts reduces every negative power to the supplied logarithmic integral. The power divergences are real and disappear when the imaginary part is taken. Writing

L_{T} = γ_{E} + lo g ∣ k_{T} τ_{0} ∣,

one obtains

Im J = (- \frac{k _{T}^{3}}{3} + k_{T} e_{2} - e_{3}) L_{T} + \frac{4 k _{T}^{3}}{9} - k_{T} e_{2} + \frac{e _{4}}{k _{T}} + o (1) .

Consequently the late-time parity-odd primordial trispectrum is

⟨ ϕ_{1} (k_{1}) ϕ_{2} (k_{2}) ϕ_{3} (k_{3}) ϕ_{4} (k_{4}) ⟩^{'} = - \frac{iλ H ^{4}}{8 \prod _{a} k _{a}^{3}} [k_{2} \cdot (k_{3} \times k_{4})] [(- \frac{k _{T}^{3}}{3} + k_{T} e_{2} - e_{3}) L_{T} + \frac{4 k _{T}^{3}}{9} - k_{T} e_{2} + \frac{e _{4}}{k _{T}}] .

Its factor of

i

is required by reality of a momentum-space scalar correlator: reversing all momenta complex-conjugates the correlator, while the scalar triple product changes sign.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version