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?
Light travels at the fastest (known) possible speed in the universe. Thus, the distance that light has traveled determines limit of the observable part of the universe.
Light travels at the fastest (known) possible speed in the universe. Thus, the distance that light has traveled determines limit of the observable part of the universe.
(b) Light satisfies the statement that \(ds^2 = 0\). 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 \(d\theta = d\phi = 0\), find the differential equation in terms of the coordinates t and r only.
Setting ds = 0, we know that \[0 = -c^2dt^2 + a^2(t)[\frac{dr^2}{1-kr^2} + r^2(d\theta^2 + \text{sin}^2\theta d\phi^2)]\] After we make the given assumptions, we are left with: \[c^2dt^2 = a^2(t)\frac{dr^2}{1-kr^2}\] Now we rearrange the equation to yield our differential equation: \[\frac{c^2dt^2}{a^2(t)} = \frac{dr^2}{1 - kr^2}\]
Setting ds = 0, we know that \[0 = -c^2dt^2 + a^2(t)[\frac{dr^2}{1-kr^2} + r^2(d\theta^2 + \text{sin}^2\theta d\phi^2)]\] After we make the given assumptions, we are left with: \[c^2dt^2 = a^2(t)\frac{dr^2}{1-kr^2}\] Now we rearrange the equation to yield our differential equation: \[\frac{c^2dt^2}{a^2(t)} = \frac{dr^2}{1 - kr^2}\]
(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 \(t = t_0\).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to \(r_{\text{horizon}}\) and t from 0 to \(t_0\)).
In the case of a flat universe, k = 0, so the RHS of our above differential equation becomes just \(dr^2\). From our previous worksheet, we know that: \[a(t) = a_0\big(\frac{t}{t_0}\big)^{2/3}\] After we take the square root of both sides of our differential equations, we can sub in the above for a(t) to get: \[dr = \frac{cdt}{a(t)} = \frac{cdt}{a_0\big(\frac{t}{t_0}\big)^{2/3}}\] Now we integrate the LHS from r to \(r_{\text{horizon}}\) and the RHS from 0 to \(t_0\): \[\int_0^{r_{\text{horizon}}}dr = \int_0^{t_0} \frac{cdt}{a_0\big(\frac{t}{t_0}\big)^{2/3}}\] This gives us: \[r_{\text{horizon}} = \frac{3c}{a_0}t_0\]
In the case of a flat universe, k = 0, so the RHS of our above differential equation becomes just \(dr^2\). From our previous worksheet, we know that: \[a(t) = a_0\big(\frac{t}{t_0}\big)^{2/3}\] After we take the square root of both sides of our differential equations, we can sub in the above for a(t) to get: \[dr = \frac{cdt}{a(t)} = \frac{cdt}{a_0\big(\frac{t}{t_0}\big)^{2/3}}\] Now we integrate the LHS from r to \(r_{\text{horizon}}\) and the RHS from 0 to \(t_0\): \[\int_0^{r_{\text{horizon}}}dr = \int_0^{t_0} \frac{cdt}{a_0\big(\frac{t}{t_0}\big)^{2/3}}\] This gives us: \[r_{\text{horizon}} = \frac{3c}{a_0}t_0\]
So elegantly presented!
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