Exponential cosmological solutions with two factor spaces in EGB model with revisited

V. D. Ivashchuk; A. A. Kobtsev

doi:10.1140/epjc/s10052-019-7329-8

EPJ

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2023 Impact factor 4.2

Particles and Fields

Open Access

Eur. Phys. J. C (2019) 79: 824
https://doi.org/10.1140/epjc/s10052-019-7329-8

Regular Article - Theoretical Physics

Exponential cosmological solutions with two factor spaces in EGB model with $\Lambda = 0$ revisited

V. D. Ivashchuk¹^,2^* and A. A. Kobtsev³

¹ Institute of Gravitation and Cosmology, Peoples’ Friendship University of Russia (RUDN University), 6 Miklukho-Maklaya Street, Moscow, 117198, Russian Federation
² Center for Gravitation and Fundamental Metrology, VNIIMS, 46 Ozyornaya Street, Moscow, 119361, Russian Federation
³ Institute for Nuclear Research of the Russian Academy of Sciences, Moscow, Troitsk, 142190, Russian Federation

^* e-mail: ivashchuk@mail.ru

Received: 12 April 2019
Accepted: 20 September 2019
Published online: 8 October 2019

Abstract

We study exact cosmological solutions in D-dimensional Einstein–Gauss–Bonnet model (with zero cosmological term) governed by two non-zero constants: $\alpha _1$ and $\alpha _2$ . We deal with exponential dependence (in time) of two scale factors governed by Hubble-like parameters $$H >0$$ and h, which correspond to factor spaces of dimensions $$m >2$$ and $$l > 2$$ , respectively, and $$D = 1 + m + l$$ . We put $h \ne H$ and $mH + l h \ne 0$ . We show that for $\alpha = \alpha _2/\alpha _1 > 0$ there are two (real) solutions with two sets of Hubble-like parameters: $$(H_1, h_1)$$ and $$(H_2, h_2)$$ , which obey: $$ h_1/ H_1< - m/l< h_2/ H_2 < 0$$ , while for $\alpha < 0$ the (real) solutions are absent. We prove that the cosmological solution corresponding to is stable in a class of cosmological solutions with diagonal metrics, while the solution corresponding to $$(H_1, h_1)$$ is unstable. We present several examples of analytical solutions, e.g. stable ones with small enough variation of the effective gravitational constant G, for $$(m, l) = (9, l >2), (12, 11), (11,16), (15, 6)$$ .