To compute the size of the horizon, let us compute how far the light can travel since the Big Bang.
(a) First of all - Why do we use the light to figure out the horizon size?
(b) Light satisfies the statement that ds2 = 0. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that aptq as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
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| https://upload.wikimedia.org/wikipedia/commons/9/98/Observable_Universe_with_Measurements_01.png |
Since nothing travels faster than the speed of light, the observable universe is limited by the point farthest away whose emitted light has reached us.
(b) Light satisfies the statement that ds2 = 0. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
The metric is:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2(d\theta^2 + sin^2(\theta)d\phi)\right)\]
We are told to set dθ = dφ = 0, so:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}\right)\]
Since we are in a flat universe, k=0.
\[0=-c^2dt^2+a^2(t)dr^2\]
The metric is:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2(d\theta^2 + sin^2(\theta)d\phi)\right)\]
We are told to set dθ = dφ = 0, so:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}\right)\]
Since we are in a flat universe, k=0.
\[0=-c^2dt^2+a^2(t)dr^2\]
\[c^2dt^2=a^2(t)dr^2\]
Simplifying gives us our differential result:
Simplifying gives us our differential result:
\[\frac{dr}{dt}=\frac{c}{a(t)}\]
(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that a(t) as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
\[\frac{dr}{dt}=\frac{c}{a(t)}\]
Last worksheet, we proved that \(a(t)\propto t^{2/3}\). Let's plug this in.
\[\frac{dr}{dt}=\frac{c}{t^{2/3}}\]
Let's separate variables and integrate.
\[\frac{dr}{c}=\frac{dt}{t^{2/3}}\]
\[\int_0^{r_{horizon}} \frac{dr}{c}=\int_0^{t_0} \frac{dt}{t^{2/3}}\]
Last worksheet, we proved that \(a(t)\propto t^{2/3}\). Let's plug this in.
\[\frac{dr}{dt}=\frac{c}{t^{2/3}}\]
Let's separate variables and integrate.
\[\frac{dr}{c}=\frac{dt}{t^{2/3}}\]
\[\int_0^{r_{horizon}} \frac{dr}{c}=\int_0^{t_0} \frac{dt}{t^{2/3}}\]
\[\frac{r_{horizon}}{c}=3t_0^{1/3}\]
\[\boxed{r_{horizon}=3ct_0^{1/3}}\]
Worked with B. Brzycki, G. Grell, and N. James on this problem.
you need to use \( a(t) = a_0 (t/t_0)^{2/3} \) here. to get \( r_H = 3 c t_0 \)
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