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Comparative Analysis of Perturbed $f(R)$ Gravity and Perturbed Rastall Gravity Models in Describing Cosmic Evolution from Early to Late Universe Relative to the $\Lambda$CDM Model
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abstract
This study conducts a meticulous examination of the cosmological implications inherent in Rastall gravity and $f(R)$ gravity models, assessing their efficacy across distinct cosmic epochs, from early universe structure formation to late-time acceleration. In the initial stages, both models exhibit commendable compatibility with observed features of structure formation, aligning with the established $\Lambda$CDM model. The derived Jeans' wavenumbers for each model support their viability. However, as the cosmic timeline progresses into the late universe, a discernible disparity surfaces. Utilizing the Markov Chain Monte Carlo method, we reconstruct the deceleration parameter $(q)$ and identify Deceleration - Acceleration redshift transition values. For $f(R)$ gravity, our results align closely with previous studies, emphasizing its superior ability to elucidate the recent cosmic acceleration. In contrast, Rastall gravity exhibits distinct redshift transition values. Our rigorous analysis underscores the prowess of $f(R)$ gravity in capturing the observed cosmic acceleration, positioning it as a compelling alternative to the conventional $\Lambda$CDM model. The discernible shifts observed in the peaks of the CMB power spectrum and evolution of deceleration parameter (q) for both $f(R)$ gravity and Rastall gravity models in the Early and Late universe, in relation to the $\Lambda $CDM model, provide compelling evidence supporting the proposition that these alternative gravity models can account for the anisotropy of the universe without invoking the need for dark energy.
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Works this paper leans on
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For ξ1: ϵ3 − ξ2 1 − ξ1 = 0 (30) This is a quadratic equation, so the solutions for ξ1 are: ξ1 = −1 ± √1 + 4ϵ3 2 (31)
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[2]
For ξ2: −ξ2ϵ1 = 0 (32) Since ϵ1 is not necessarily zero, we have ξ2 = 0. 5
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[3]
For ξ3: β − ξ2 3 − ϵ1ξ3 = 0 (33) This is another quadratic equation, so the solutions for ξ3 are: ξ3 = −ϵ1 ± p ϵ2 1 + 4β 2 (34)
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[4]
For ξ4: ξ4(1 − ξ1) = 0 (35) Therefore, ξ4 = 0 or ξ1 = 1
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For ξ5: −ξ5(1 + ξ3) = 0 (36) Thus, ξ5 = 0 or ξ3 = −1
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For ξ6: ϵ2 − ξ6(ξ1 + ϵ1) = 0 (37) Solving for ξ6: ξ6 = ϵ2 ξ1 + ϵ1 (38)
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For ξ7: ξ6 − ξ7ξ1 = 0 (39) Therefore, ξ7 = ξ6 ξ1 = ϵ2 ξ1(ξ1+ϵ1) . Step 2: Jacobian Matrix The Jacobian matrix J is formed by taking the partial derivatives of each differential equation with respect to each ξi: Jij = ∂ ∂ξj dξi dN (40) We will compute each partial derivative:
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For dξ1 dN = ϵ3 − ξ2 1 − ξ1: J11 = ∂ ∂ξ1 ϵ3 − ξ2 1 − ξ1 = −2ξ1 − 1 (41) All other partial derivatives J1j = 0 for j ̸= 1
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For dξ2 dN = −ξ2ϵ1: J22 = ∂ ∂ξ2 (−ξ2ϵ1) = −ϵ1 (42) All other partial derivatives J2j = 0 for j ̸= 2. 6
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For dξ3 dN = β − ξ2 3 − ϵ1ξ3: J33 = ∂ ∂ξ3 β − ξ2 3 − ϵ1ξ3 = −2ξ3 − ϵ1 (43) All other partial derivatives J3j = 0 for j ̸= 3
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For dξ4 dN = ξ4 − ξ4ξ1: J44 = ∂ ∂ξ4 (ξ4 − ξ4ξ1) = 1 − ξ1 (44) J41 = ∂ ∂ξ1 (ξ4 − ξ4ξ1) = −ξ4 (45) All other partial derivatives J4j = 0 for j ̸= 1, 4
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For dξ5 dN = −ξ5 − ξ5ξ3: J55 = ∂ ∂ξ5 (−ξ5 − ξ5ξ3) = −1 − ξ3 (46) J53 = ∂ ∂ξ3 (−ξ5 − ξ5ξ3) = −ξ5 (47) All other partial derivatives J5j = 0 for j ̸= 3, 5
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For dξ6 dN = ϵ2 − ξ6ξ1 − ϵ1ξ6: J66 = ∂ ∂ξ6 (ϵ2 − ξ6ξ1 − ϵ1ξ6) = −ξ1 − ϵ1 (48) J61 = ∂ ∂ξ1 (ϵ2 − ξ6ξ1 − ϵ1ξ6) = −ξ6 (49) All other partial derivatives J6j = 0 for j ̸= 1, 6
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For dξ7 dN = ξ6 − ξ7ξ1: J77 = ∂ ∂ξ7 (ξ6 − ξ7ξ1) = −ξ1 (50) J76 = ∂ ∂ξ6 (ξ6 − ξ7ξ1) = 1 (51) J71 = ∂ ∂ξ1 (ξ6 − ξ7ξ1) = −ξ7 (52) All other partial derivatives J7j = 0 for j ̸= 1, 6, 7. Jacobian Matrix J Putting all the elements together, the Jacobian matrix J is: 7 J = ...
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For ξ1 to be a critical point, ϵ3 must be exactly 6 .4581
ξ1 = −2.09 • Condition: ϵ3 = 6.4581 • Physical Interpretation: ξ1 typically represents a dimensionless parameter related to the scalar field or perturbations. For ξ1 to be a critical point, ϵ3 must be exactly 6 .4581. This indicates that the dynamics of the system at this poin...
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For ξ3 to be a critical point, β must satisfy the condition β = 1.0074
ξ3 = −0.69 • Condition: β = 0.4761 + 0.69 × 0.77 • Result: β = 0.4761 + 0.5313 = 1.0074 • Physical Interpretation: ξ3 could represent another dynamic variable, such as perturbations or field compo- nents. For ξ3 to be a critical point, β must satisfy the condition β = 1.0074. ...
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For ξ6 to be a critical point, ϵ2 must be related to ϵ1 by ϵ2 = −0.924
ξ6 = 0.7 • Condition: ϵ2 = 0.7 × 0.77 − 1.463 • Result: ϵ2 = 0.539 − 1.463 = −0.924 • Physical Interpretation: ξ6 could be related to field values, perturbations, or interaction terms. For ξ6 to be a critical point, ϵ2 must be related to ϵ1 by ϵ2 = −0.924. This indicates the s...
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[2023]
Joule-Thomson Expansion and Tidal Force Effects of AdS Black Holes Surrounded by Chaplygin Dark Fluid
“Joule-Thomson Expansion and Tidal Force Effects of AdS Black Holes Surrounded by Chaplygin Dark Fluid.” arXiv:2312.16224
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