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

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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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