2 edited body edited Jan 6 '17 at 5:01 Andrew 5,98511 gold badge1313 silver badges2525 bronze badges Indeed, the age is not just $$1/H$$. (In this answer I'll use units with $$c=1$$. The age $$T$$ can be calculated by making repeated use of the chain rule, as well as $$da/dt=aH$$ and $$a=1/(1+z)$$ $$\begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_\infty^0 dz \frac{da}{dz} \frac{a}{H} = \int_\infty^0 dz \frac{1}{(1+z)^3 H(z)} \end{equation}$$$$\begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_0^\infty dz \frac{da}{dz} \frac{a}{H} = \int_0^\infty dz \frac{1}{(1+z)^3 H(z)} \end{equation}$$ where I've assumed a flat universe so I can normalize the scale factor by $$a(today)=1$$, and I set the origin of time at the Big Bang a(t=0)=0. However, using the Friedman equation and assuming a universe with matter, radiation, and a cosmological constant we can write $$\begin{equation} T=\frac{1}{H_0} \int_\infty^0 dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation}$$$$\begin{equation} T=\frac{1}{H_0} \int_0^\infty dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation}$$ Ignoring inflation, and using the $$\Lambda$$CDM best fit values for $$\Omega_{I,0}$$ ($$I=R,M,\Lambda$$), the integral is $$O(1)$$ (as you can check), so $$1/H_0$$ is a good approximation to the age of the universe. Indeed, the age is not just $$1/H$$. (In this answer I'll use units with $$c=1$$. The age $$T$$ can be calculated by making repeated use of the chain rule, as well as $$da/dt=aH$$ and $$a=1/(1+z)$$ $$\begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_\infty^0 dz \frac{da}{dz} \frac{a}{H} = \int_\infty^0 dz \frac{1}{(1+z)^3 H(z)} \end{equation}$$ where I've assumed a flat universe so I can normalize the scale factor by $$a(today)=1$$, and I set the origin of time at the Big Bang a(t=0)=0. However, using the Friedman equation and assuming a universe with matter, radiation, and a cosmological constant we can write $$\begin{equation} T=\frac{1}{H_0} \int_\infty^0 dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation}$$ Ignoring inflation, and using the $$\Lambda$$CDM best fit values for $$\Omega_{I,0}$$ ($$I=R,M,\Lambda$$), the integral is $$O(1)$$ (as you can check), so $$1/H_0$$ is a good approximation to the age of the universe. Indeed, the age is not just $$1/H$$. (In this answer I'll use units with $$c=1$$. The age $$T$$ can be calculated by making repeated use of the chain rule, as well as $$da/dt=aH$$ and $$a=1/(1+z)$$ $$\begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_0^\infty dz \frac{da}{dz} \frac{a}{H} = \int_0^\infty dz \frac{1}{(1+z)^3 H(z)} \end{equation}$$ where I've assumed a flat universe so I can normalize the scale factor by $$a(today)=1$$, and I set the origin of time at the Big Bang a(t=0)=0. However, using the Friedman equation and assuming a universe with matter, radiation, and a cosmological constant we can write $$\begin{equation} T=\frac{1}{H_0} \int_0^\infty dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation}$$ Ignoring inflation, and using the $$\Lambda$$CDM best fit values for $$\Omega_{I,0}$$ ($$I=R,M,\Lambda$$), the integral is $$O(1)$$ (as you can check), so $$1/H_0$$ is a good approximation to the age of the universe. 1 answered Jan 6 '17 at 2:40 Andrew 5,98511 gold badge1313 silver badges2525 bronze badges Indeed, the age is not just $$1/H$$. (In this answer I'll use units with $$c=1$$. The age $$T$$ can be calculated by making repeated use of the chain rule, as well as $$da/dt=aH$$ and $$a=1/(1+z)$$ $$\begin{equation} T=\int_0^T dt = \int_0^1 da \frac{dt}{da} = -\int_\infty^0 dz \frac{da}{dz} \frac{a}{H} = \int_\infty^0 dz \frac{1}{(1+z)^3 H(z)} \end{equation}$$ where I've assumed a flat universe so I can normalize the scale factor by $$a(today)=1$$, and I set the origin of time at the Big Bang a(t=0)=0. However, using the Friedman equation and assuming a universe with matter, radiation, and a cosmological constant we can write $$\begin{equation} T=\frac{1}{H_0} \int_\infty^0 dz \frac{1}{(1+z)^3 \sqrt{\Omega_{R,0}(1+z)^4+\Omega_{M,0} (1+z)^3 + \Omega_\Lambda}} \end{equation}$$ Ignoring inflation, and using the $$\Lambda$$CDM best fit values for $$\Omega_{I,0}$$ ($$I=R,M,\Lambda$$), the integral is $$O(1)$$ (as you can check), so $$1/H_0$$ is a good approximation to the age of the universe.