- #1
CharlesB
- 6
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Given the following parametric form of the Friedmann Equation for an open, dust-filled (matter-dominated) universe:
[tex]a(x)={a_0 \Omega \over 2(1-\Omega)}(cosh(x)-1)[/tex]
[tex]t(x)={\Omega \over 2 H_0 (1- \Omega)^{3/2}}(sinh(x)-x)[/tex]
I am trying to calculate the Hubble Radius, R=c/H(t) where H(t)=(da/dt)/a and have come up with the following solution:
[tex]{da \over dt} = {da \over dx} {dx \over dt}[/tex]
[tex]{da \over dx}={a_0 \Omega \over 2(1-\Omega)}sinh(x)[/tex]
[tex]{dt \over dx}={\Omega \over 2H(1- \Omega)^{3/2}}(cosh(x)-1)[/tex]
[tex]{da \over dt}=Ha_0(1-\Omega)^{1/2}coth({x \over 2})[/tex]
Integrate to find a:
[tex]a=Ha_0(1-\Omega)^{1/2}coth({x \over 2})t[/tex]
[tex]{{da \over dt} \over a}={1 \over t}[/tex]
Therefore, the Hubble Radius:
[tex]R_H={c a \over ({da \over dt})}=ct[/tex]
I just can't seem to find any references to this solution, so I don't know if it is actually correct.
Can anyone confirm?
[tex]a(x)={a_0 \Omega \over 2(1-\Omega)}(cosh(x)-1)[/tex]
[tex]t(x)={\Omega \over 2 H_0 (1- \Omega)^{3/2}}(sinh(x)-x)[/tex]
I am trying to calculate the Hubble Radius, R=c/H(t) where H(t)=(da/dt)/a and have come up with the following solution:
[tex]{da \over dt} = {da \over dx} {dx \over dt}[/tex]
[tex]{da \over dx}={a_0 \Omega \over 2(1-\Omega)}sinh(x)[/tex]
[tex]{dt \over dx}={\Omega \over 2H(1- \Omega)^{3/2}}(cosh(x)-1)[/tex]
[tex]{da \over dt}=Ha_0(1-\Omega)^{1/2}coth({x \over 2})[/tex]
Integrate to find a:
[tex]a=Ha_0(1-\Omega)^{1/2}coth({x \over 2})t[/tex]
[tex]{{da \over dt} \over a}={1 \over t}[/tex]
Therefore, the Hubble Radius:
[tex]R_H={c a \over ({da \over dt})}=ct[/tex]
I just can't seem to find any references to this solution, so I don't know if it is actually correct.
Can anyone confirm?
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