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Equation of motion for cyclic model of the universe

Posted on 12/3/2013
UQ ID:375
Science
Physics
Original

I recently started to study the cyclic universe. I came across this article [1]. My question is about the action used for describing the cyclic model: where is Ricci scalar and is the metric. I solve the Euler-Lagrange equation for this action and find the equation of motion for : The radiation term is independent of so only enters the equation of motion. The zero component: where is the Hubble parameter. But this result is different from what was written in the article. The difference is the coefficient of the last term. In addition, I have a problem finding Friedmann equations for this action (again in finding coefficients). Can anybody elaborate on the reason?

Reference:

[1] P.J. Steinhardt and N. Turok, " A Cyclic Model of the Universe," Science 296 (2002), available at here

Keywords & Tags
cosmology
lagrangian-formalism
cosmological-inflation
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K2-Think

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Qwen2.5-32B
Sep 18, 2025, 01:50 PM

The discrepancy in your derived equation of motion for the scalar field (\phi) and the one presented in the article arises from two key issues in your derivation:

1. Incorrect Handling of the Kinetic Term's Derivative

The kinetic term in the action is (-\frac{1}{2} \sqrt{-g} (\partial\phi)^2), where ((\partial\phi)^2 = g^{\mu\nu} \partial_\mu\phi \partial_\nu\phi). When computing (...

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GPT-5

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Sep 2, 2025, 06:56 AM

Short answer: The coefficient 4 β^3 β,φ in your φ–equation is an artifact of treating the matter/radiation piece as S_m ∼ ∫√−g β^4(φ) ρ and varying it as if ρ were an external constant. That is not correct in GR. The matter action depends on the metric seen by matter, ĝμν = β^2(φ) gμν, and when you vary the action properly you find that φ couples to the trace of the matter stress–energy, not to ρ ...

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Claude 3.7 Sonnet

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Anthropic
Aug 1, 2025, 12:00 AM

Equation of Motion for Cyclic Model of the Universe

The question concerns the derivation of equations of motion from the action for Steinhardt and Turok's cyclic model of the universe. I'll analyze the issue step by step.

The Action

The action is given as:

Equati...

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DeepSeek R1

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DeepSeek
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Aug 1, 2025, 12:00 AM
<think> We are given a specific action for the cyclic model of the universe and are to derive the equation of motion for the scalar field φ. The action is:

[ S = \int d^{4}x \sqrt{-g} \left( \frac{1}{16\pi G} R - \frac{1}{2} (\partial \phi)^{2} - V(\phi) + \beta^{4}(\phi) (\rho_M + \rho_R) \right) ]

We note that ρ_M and ρ_R are the energy densities of matter and radiation, respectively. Howeve...

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Gemini 2.5 Pro

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Aug 1, 2025, 12:00 AM

Excellent question. This is a very common point of confusion when working with models that have non-minimal couplings, like the cyclic or ekpyrotic universe. The discrepancy you've found arises from a subtle but crucial aspect of how the matter fields are coupled to the scalar field and gravity.

Let's break down the problem into two parts: first, analyzing the action you've written, and second, s...

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o3

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Aug 1, 2025, 12:00 AM

Write the action in the form that we shall actually vary:

S[ g,φ ,ψ ] = ∫ d^4x √–g
[ 1 16πG R − 1 2 g^{μν}∂_μφ∂_νφ − V(φ) ]

  • ∫ d^4x √–g β^4(φ) ℒ_m ( ψ ,g) . (1)

The matter fields are collectively denoted by ψ.
In the Einstein frame the only place where the scalar field talks to the matter sector is through the conformal factor β(φ). For a perfect fluid one may write...

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Claude Opus 4

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Anthropic
Aug 1, 2025, 12:00 AM

I'll help you work through the equation of motion derivation for the cyclic model and identify where the discrepancy might arise.

The Action and Field Equations

The action you've written is:

Issue with the Matter Coupling

The key issue here is the interp...

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o4 mini

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Aug 1, 2025, 12:00 AM

When you go through Steinhardt‐Turok’s paper carefully what you discover is that the extra “4” in your variation simply comes from differentiating β^4 rather than from the genuine coupling of φ to the trace of the matter stress–tensor. In fact one shows most cleanly that

  1. all of the matter in the cyclic model lives in a “Jordan–frame’’ metric g^J_{μν} = β^2(φ),g_{μν},,

  2. one...

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o3 Pro

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Aug 1, 2025, 12:00 AM

The point where your derivation parts company with the one that appears in the cyclic/ekpyrotic papers is the way the matter part of the action is varied.
In the action that Steinhardt & Turok write

S = ∫ d⁴x √-g [ 1/(16πG) R − ½ (∂φ)² − V(φ) ] + S_matter [ ψ , ĝ_{μν}=β²(φ) g_{μν} ] (1)

the whole matter sector (dust, radiation, baryons, …) is assumed to couple not to the Einstein...

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