In the case where the universe does not recollapse, the proper distance to the is...<math>d_{MAX}(t) = a(t) \int_{0}^{r_{MAX}(t)} \frac{dr}{\sqrt{1-K… - Steven Weinberg

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In the case where the universe does not recollapse, the proper distance to the is...<math>d_{MAX}(t) = a(t) \int_{0}^{r_{MAX}(t)} \frac{dr}{\sqrt{1-Kr^2}} = a(t)\int_{0}^{\infty} \frac{dt'}{a(t')}</math>... In the absence of a cosmological constant, <math>a(t)</math> grows like <math>t^{\frac{2}{3}}</math>, and the integral diverges, so there is no event horizon. But with a cosmological constant, <math>a(t)</math> will eventually grow as exp(<math>Ht</math>) with <math>H = H_0 \Omega^{1/2}_\Lambda</math> constant and... an event horizon... approaches... <math>d_{MAX}(\infty) = 1/H</math>. As time passes all sources of light outside our gravitationally bound will move beyond this... and become unobservable. The same is true for the quintessence theory... In that case <math>a(t)</math> eventually grows as exp(const <math>\times\, t^{2/{(2+\frac{\alpha}{2})}}</math>), so for any <math>\alpha \ge 0</math> the integral... [<math>d_{MAX}(t)</math>] converges.