4 added 1608 characters in body edited Jan 2 '18 at 20:31 OON 1,50466 silver badges1212 bronze badges In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exactly on its cosmologial event horizon. In general case the cosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. UPD: By the way there's a very simple way to understand that for any FRLW cosmology with event horizon the sphere of no return will always have the radius equal to half of the horizon radius. To do so you need to use conformal time, $$\begin{equation} \tau=\int \frac{dt}{a(t)},\quad ds^2=a^2(\tau)\Big(c^2 d\tau^2-d\vec{x}^2\Big) \end{equation}$$ The nice thing about the conformal time is that lightcones look very simple, just like in the flat spacetime $$\Delta x=c \Delta \tau$$. But how cosmological horizons then appear, what prevents lightcones from intersecting with some wordline $$x=\mathrm{const}$$? This happens when $$a$$ grows sufficiently fast, then $$\tau$$ happens to be restricted by some finite $$\tau_{max}$$ where $$a\rightarrow+\infty$$. So for lightcones this looks just like a Minkowski spacetime where you cut off everything after $$\tau_{max}$$. If the intersection lies exactly at $$\tau_{max}$$ this is a horizon, if it happens after that then no signal can pass. From that it's obvious that the cosmological horizon for $$x=0$$ is the past lightcone originating at $$\tau_{max}$$ i.e. $$|\vec{x}_H|=c(\tau_{max}-\tau)$$ To find the point of no return we send lightsignal $$x=c(\tau-\tau_0)$$ and look where it intersects with the horizon lightcone. It's pretty obvious that the intersection will happen at $$x_{NR}=x_H/2=\frac{c}{2}(\tau_{max}-\tau_0)$$. For better understanding I add the illustration. Again I remind you that this is true for any $$a(t)$$ growing sufficiently fast in time. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exactly on its cosmologial event horizon. In general case the cosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exactly on its cosmologial event horizon. In general case the cosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. UPD: By the way there's a very simple way to understand that for any FRLW cosmology with event horizon the sphere of no return will always have the radius equal to half of the horizon radius. To do so you need to use conformal time, $$\begin{equation} \tau=\int \frac{dt}{a(t)},\quad ds^2=a^2(\tau)\Big(c^2 d\tau^2-d\vec{x}^2\Big) \end{equation}$$ The nice thing about the conformal time is that lightcones look very simple, just like in the flat spacetime $$\Delta x=c \Delta \tau$$. But how cosmological horizons then appear, what prevents lightcones from intersecting with some wordline $$x=\mathrm{const}$$? This happens when $$a$$ grows sufficiently fast, then $$\tau$$ happens to be restricted by some finite $$\tau_{max}$$ where $$a\rightarrow+\infty$$. So for lightcones this looks just like a Minkowski spacetime where you cut off everything after $$\tau_{max}$$. If the intersection lies exactly at $$\tau_{max}$$ this is a horizon, if it happens after that then no signal can pass. From that it's obvious that the cosmological horizon for $$x=0$$ is the past lightcone originating at $$\tau_{max}$$ i.e. $$|\vec{x}_H|=c(\tau_{max}-\tau)$$ To find the point of no return we send lightsignal $$x=c(\tau-\tau_0)$$ and look where it intersects with the horizon lightcone. It's pretty obvious that the intersection will happen at $$x_{NR}=x_H/2=\frac{c}{2}(\tau_{max}-\tau_0)$$. For better understanding I add the illustration. Again I remind you that this is true for any $$a(t)$$ growing sufficiently fast in time. 3 added 30 characters in body edited Jan 2 '18 at 15:56 OON 1,50466 silver badges1212 bronze badges In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the Hubblecosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exaclyexactly on its Hubblecosmologial event horizon. In general case the Hubblecosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the Hubble horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exacly on its Hubble horizon. In general case the Hubble horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmological event horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exactly on its cosmologial event horizon. In general case the cosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. 2 added 200 characters in body edited Jan 2 '18 at 15:43 OON 1,50466 silver badges1212 bronze badges In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmologicalHubble horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exacly on its cosmologicalHubble horizon. In general case the cosmologicalHubble horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$. Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the cosmological horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exacly on its cosmological horizon. In general case the cosmological horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ from us. In the expanding universe the interval is, $$\begin{equation} ds^2=c^2 dt^2-a^2(t) d\vec{x}^2 \end{equation}$$ The interval along lightlike trajectory is zero, so from that we get, $$\begin{equation} \frac{dx}{dt}=\frac{c}{a(t)},\quad \Delta x(t_0,t_1)=c\int_{t_0}^{t_1}\frac{dt}{a(t)} \end{equation}$$ If you as usually done choose that $$a$$ at present time ($$t=0$$) is equal to $$1$$. the coordinate $$x$$ will tell you how far this is at the present. To learn how far it is at $$t=t_1$$ you should multiply it on $$a(t_1)$$. Now the equation above require you to know how exactly scale factor $$a(t)$$ will evolve in the future which is determined by the matter. Luckily the problem is much simplified as we entered the dark energy dominated era with dark energy most likely being just a cosmological constant. We then assume that the Hubble parameter is approximately constant (actually it will gradually drop by about 20% as matter will disperse and its contribution will decrease), $$\begin{equation} H\equiv\frac{1}{t_H}\equiv\frac{\dot{a}}{a}\simeq \mathrm{const},\Rightarrow a(t)=e^{t/t_H} \end{equation}$$ With $$t_H\approx 14.4 Gy$$ (where Gy stands for gigayear i.e. billion years) Then, $$\begin{equation} \Delta x(t_0,t_1)=ct_H\Big(e^{-t_0/t_H}-e^{-t_1/t_H}\Big) \end{equation}$$ Note that if we take $$t_0=0$$ (i.e. shoot the light pulse now) the fartherst point it may get is $$c t_H$$. This is the Hubble horizon - anything further can't be reached because of the universe expansion. The point of no return for light is defined in such a way that when light reaches it the point of departure is exacly on its Hubble horizon. In general case the Hubble horizon depends on $$t_0$$. However when $$H=\mathrm{const}$$ we get highly symmetric de Sitter spacetime for which $$a(t+\Delta t)=a(t)a(\Delta t)$$ and $$a(t_0)\Delta x(t_0,t_1)=\Delta x(0,t_1-t_0)$$. So to find the point of no return we simply take, $$\begin{equation} a(t_{NR})\Delta x(0,t_{NR})=ct_H \end{equation}$$ This gives us $$t_{NR}=t_H\ln{2}\approx 10 Gy$$. At the present this point is at $$\Delta x=\frac{ct_H}{2}\approx 7.2 Gly$$ ($$7.2$$ billion light-years) from us. To put those numbers into perspective the age of the universe is 13.8 Gy, the radius of the observable universe is 45.7 Gly. 1 answered Jan 2 '18 at 14:58 OON 1,50466 silver badges1212 bronze badges