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

[Apashanka]

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=nk_BN_AT/V
P=(nN_A)k_BT/V
P=(nN_Am₀)k_BT/v
P=m_totalk_BT/vm₀
P=(m_total/V)k_BT/m₀
P=ρk_BT/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

3/2k_BT=m₀c²(total energy of each molecule)
k_BT/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??

 

[Orodruin]

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 $w \simeq 0$.

 

[Apashanka]

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.

 

[kimbyd]

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.

 

[Apashanka]

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