# Cosmology - Age of Universe

As discussed in the earlier chapters, the time evolution of the Hubble parameter is given by −

$$H(z) = H_0E(z)^{\frac{1}{2}}$$

where **z** is the red shift and **E(Z)** is −

$$E(z) \equiv \Omega_{m,0}(1+z)^3 + \Omega(1+z)^4 +\Omega_{k,0}(1+z)^2 + \Omega^{\wedge,0}$$

If the expansion of the universe is constant, then the true age of the universe is given as follows −

$$t_H = \frac{1}{H_0}$$

If it is the matter dominated universe, i.e., Einstein Desitter universe, then the true age of universe is given by −

$$t_H = \frac{2}{3H_0}$$

Scale and Redshift is defined by −

$$a=\frac{a_0}{1+z}$$

Age of the universe in terms of the cosmological parameter is derived as follows.

The Hubble Parameter is given by −

$$H = \frac{\frac{da}{dt}}{a}$$

Differentiating, we get −

$$da = \frac{-dz}{(1+z)^2}$$

where **a _{0} = 1** (present value of the scale factor)

$$\frac{\mathrm{d} a}{\mathrm{d} t} = \frac{-1}{(1+z)^2}$$

$$\frac{\mathrm{d} a}{\mathrm{d} t} = \frac{\mathrm{d} a}{\mathrm{d} t}\frac{\mathrm{d} z}{\mathrm{d} t}$$

$$H = \frac{\dot{a}}{a} = \frac{\mathrm{d} a}{\mathrm{d} t}\frac{\mathrm{d} z}{\mathrm{d} t} \frac{1+z}{1}$$

$$\frac{\dot{a}}{a} = \frac{-1}{1+z}\frac{\mathrm{d} z}{\mathrm{d} t}\frac{1}{1}$$

$$H(z) = H_0E(z)^{\frac{1}{2}}$$

$$dt = \frac{-dz}{H_0E(z)^{\frac{1}{2}}(1+z)}$$

If we want to find the age of the universe at any given redshift **āzā** then −

$$t(z) = \frac{1}{H_0}\int_{\infty}^{z_1} \frac{-1}{E(z)^{\frac{1}{2}}(1+z)}dz$$

where **k** is the curvature density parameter and −

$$E(z) \equiv \Omega_{m,0}(1+z)^3 + \Omega_{rad,0}(1+z)^4 + \Omega_{k,0}(1+z)^2 + \Omega_{\wedge,0}$$

To calculate the present age of the universe, take **z _{1} = 0**.

$$t(z=0) = t_{age} = t_0 = \frac{1}{H_0}\int_{\infty}^{z_1} \frac{-1}{E(z)^{\frac{1}{2}}(1+z)}dz$$

For the Einstein Desitter Model, i.e, $\Omega_m = 1$, $\Omega_{rad} = 0$, $\Omega_k = 0$, $\Omega_\wedge = 0$, the equation for the age of the universe becomes −

$$t_{age} = \frac{1}{H_0}\int_{0}^{\infty} \frac{1}{(1+z)^{\frac{5}{2}}}dz$$

After solving the integral, we get −

$$t_H = \frac{2}{3H_0}$$

The night sky is like a **Cosmic Time Machine.** Whenever we observe a distant planet, star or galaxy, we are seeing it as it was hours, centuries or even millennia ago. This is because light travels at a finite speed (the speed of light) and given the large distances in the Universe, we do not see objects as they are now, but as they were when the light was emitted. The time elapsed between ā when we detect the light here on Earth and when it was originally emitted by the source, is known as the **Lookback Time (t _{L}(z_{1}))**.

So, the lookback time is given by −

$$t_1(z_1) = t_0-t(z_1)$$

The lookback time for the Einstein Desitter Universe is −

$$t_L(z) = \frac{2}{3H_0}\left [ 1- \frac{1}{(1+z)^{\frac{3}{2}}} \right ]$$

## Points to Remember

Whenever we observe a distant planet, star or galaxy, we are seeing it as it was hours, centuries or even millennia ago.

The time elapsed between ā when we detect the light here on Earth and when it was originally emitted by the source, is known as the lookback time.