Consider a flat expanding universe with no cosmological constant and no curvatur

Question

Consider a flat expanding universe with no cosmological constant and no curvature (k=0 in the Friedmann equation shown in class). Show that if the Universe is made of "dust", so the energy density scales like 1/R^3, then the scale factor, R(t), grows as t^(2/3). Show if it is made of radiation (so the energy density scales as 1/R^4 -- the extra factor of R comes from the redshift), then it grows as t^(1/2). In both cases, show that for early times, the scale factor grows faster than light. Is this a problem?

Explanation / Answer

Friedman equation=(a./a)^2= 8pi G* epsilon(t)/3c^2-kc^2/Ro^2*1/a^2(t)

Ro= curvature radius

k curvature index=+1,-1,0

If flat universe then k=0, q0=1/2

If universe is made up of dust then p=0, a^3p=constant

Friedman equation now becomes

a.^2/a^2= 8piG/3po (a0/a)^3

da/dt=square root(8piGpoa^3/3)*1/a/2

=intergratation a/2da=2/3a^3/2 + K

=square root(8piGpo a^3)* t/3

At big bang t=0, a=0 therefore k=0

using a0=1, since universe is flat therefore po=pcr

a=(6piGpo)^1/3 t^2/3=(6piGpcr)

=6pi3Ho^2/8piG)^1/3*t^2/3

=(9(Ho^2)/4))^1/3 t^2/3

=(3Ho/2)^2/3 t^2/3

if scale factor a=a0=1 then age of universe to

to=2/3Ho

where H= hubble constant

If universe is made up of radiation then p=1/3p, a^4p= constant

Now friedman equation

a.^2/a^2=8piG/3 po (ao/a)^4

da/dt=square root( 8piGpo ao^4/3)*1/a

=integration ada=2a^2+K

=square root((8piGpoao^4)/3)*t

At the big bang since t=0, a=0 then K=0, ao=1 , po=pcr

Now a=(2/3piGpo)^1/4=(2/3piGpcr)^1/4

=(2/3piGpo)^1/4 t^1/2

=((2/3 piG 3Ho^2/8piG)1/4*t^1/2

=(Ho/2)*t1/2

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