Calcluating the Hubble Radius for an open universe?

[CharlesB]

Given the following parametric form of the Friedmann Equation for an open, dust-filled (matter-dominated) universe:
$$a(x)={a_0 \Omega \over 2(1-\Omega)}(cosh(x)-1)$$

$$t(x)={\Omega \over 2 H_0 (1- \Omega)^{3/2}}(sinh(x)-x)$$

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:

$${da \over dt} = {da \over dx} {dx \over dt}$$

$${da \over dx}={a_0 \Omega \over 2(1-\Omega)}sinh(x)$$

$${dt \over dx}={\Omega \over 2H(1- \Omega)^{3/2}}(cosh(x)-1)$$

$${da \over dt}=Ha_0(1-\Omega)^{1/2}coth({x \over 2})$$

Integrate to find a:

$$a=Ha_0(1-\Omega)^{1/2}coth({x \over 2})t$$

$${{da \over dt} \over a}={1 \over t}$$

Therefore, the Hubble Radius:

$$R_H={c a \over ({da \over dt})}=ct$$

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?

 

[Chalnoth]

Could you please enclose the latex in [noparse][tex][/tex][/noparse] tags?

e.g.

[noparse]$${dy \over dx} = x$$[/noparse]

 

[CharlesB]

Could you please enclose the latex in [noparse][tex][/tex][/noparse] tags?

e.g.

[noparse]$${dy \over dx} = x$$[/noparse]

Sorry, fixed now.

 

[Chalnoth]

Sorry, fixed now.

Thanks!

Unfortunately, I don't think it's correct. The step after "Integrate to find a:" appears to be wrong. Basically, x and t are not independent variables, so you can't integrate a function which includes x over t without first substituting x as a function of t.

Edit: Looks like a potential solution may be:

$$\int f(x)dt = \int {f(x) \over {dx \over dt} } {dx \over dt} dt = \int {f(x) \over {dx \over dt} } dx$$

 

[CharlesB]

Thanks, I thought that might have been the case. Any ideas how to get a solution for a then?

 

[Chalnoth]

Thanks, I thought that might have been the case. Any ideas how to get a solution for a then?

Heh, you replied too fast :) I think the edit in my above post may be a way to go.

 

[CharlesB]

Tricky. Thanks for your help!

 

[Chalnoth]

Tricky. Thanks for your help!

Heh, just realized that doesn't actually get you anywhere, because it just gives a(x), which you already have!

I also think I've noticed another mistake, where you say da/dt is proportional to the hyperbolic cotangent. I don't think this is the case, as it looks like you've dropped the -1 in dt/dx.

Anyway, you should be able to get the Hubble radius as a function of x no problem. Getting it as a function of t analytically may not be possible.

 

[CharlesB]

Yeah, I realized it was pretty stupid to integrate da/dt since I already had a(x). And I just used WolframAlpha to get coth(x/2), http://www.wolframalpha.com/input/?i=d/dx(cosh(x)-1)*(1/(d/dx(sinh(x)-x))). It's probably just as easy to use sinh(x)/{cosh(x) -1}.

Now to get the particle horizon!

 

[Chalnoth]

Yeah, I realized it was pretty stupid to integrate da/dt since I already had a(x). And I just used WolframAlpha to get coth(x/2), http://www.wolframalpha.com/input/?i=d/dx(cosh(x)-1)*(1/(d/dx(sinh(x)-x))). It's probably just as easy to use sinh(x)/{cosh(x) -1}.

Now to get the particle horizon!

Oh, I see, it's the hyperbolic cotangent of x/2, not x. That makes sense now. For some reason I missed the division by 2 before. Obviously they're using some interesting trigonometric identities to get that answer.