I Why ##\omega=0## for a matter dominated universe?

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In a matter-dominated universe, the equation of state parameter ω is set to 0, indicating that pressure (P) is negligible compared to energy density (ρ). This is derived from the kinetic energy of non-relativistic gas molecules, where thermal motion contributes minimally to energy density, leading to P being much less than ρ. The discussion highlights that for non-relativistic conditions, the pressure approaches zero as the velocity of particles is significantly lower than the speed of light. Additionally, it notes that in the early universe, matter behaved like radiation, complicating the expansion history until it transitioned to a non-relativistic state. Overall, the relationship between pressure and energy density is crucial in understanding the dynamics of the universe's expansion.
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From the energy equation E=m0c2/√(1-v2/c2) for non-relativistic gas molecules (v<<c) ,E reduces to m0c2...(1)
From ideal gas law PV=nRT
P=nRT/V
P=nkBNAT/V
P=(nNA)kBT/V
P=(nNAm0)kBT/v
P=mtotalkBT/vm0
P=(mtotal/V)kBT/m0
P=ρkBT/m0
(If n moles of a gas is taken in volume V at temp T and volume V,m0 being the mass of each molecule at equipibrium)
From equipartition theorem

3/2kBT=m0c2(total energy of each molecule)
kBT/m0=2/3c2
Putting this in P becomes
P=2/3ρc2
P=2/3ε(energy density)
Therefore (P=ωε) ω=2/3 (Nonrelativistic gas)
But for matter dominated universe we take ω=0
Why is it so??
 
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Apashanka said:
3/2kBT=m0c2(total energy of each molecule)
This is wrong. The kinetic energy of each particle is ##k_BT/2## per spatial dimension. For a gas to be non-relativistic, you need ##T \ll m_0## and so very little of the energy density is due to the kinetic energy due to thermal motion. In fact, it is exactly the reason why ##P \ll \rho## and therefore ##w \simeq 0##.
 
Orodruin said:
This is wrong. The kinetic energy of each particle is ##k_BT/2## per spatial dimension. For a gas to be non-relativistic, you need ##T \ll m_0## and so very little of the energy density is due to the kinetic energy due to thermal motion. In fact, it is exactly the reason why ##P \ll \rho## and therefore ##w \simeq 0##.
Ok it will be then P=εv2/3c2 and for NR v<<c and therefore P~0.
Thanks I got it now.
 
Apashanka said:
Ok it will be then P=εv2/3c2 and for NR v<<c and therefore P~0.
Thanks I got it now.
Yup! Which means that in the very early universe, matter behaved like radiation. This fact complicates the expansion history early-on, as matter became non-relativistic over time.
 
kimbyd said:
Yup! Which means that in the very early universe, matter behaved like radiation. This fact complicates the expansion history early-on, as matter became non-relativistic over time.
For relativistic particles P=β2/(3+3β2/2)ε , isn't it?? where β=v/c and for non relativistic it is only (β2/3)ε
 
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