Because the apparent recessional speeds of galaxies and quasars at great distances are close to the speed of light, the relativistic Doppler shift formula (Eq.37-31) must be used. The shift is reported as fractional red shift $\mathit{z}\mathbf{=}\frac{\mathbf{\Delta }\mathbf{\lambda }}{{\mathbf{\lambda }}_{\mathbf{0}}}$

(a)Show that, in termsof,$\mathit{z}$the recessional speedparameter$\mathit{\beta }\mathbf{=}\frac{\mathbf{v}}{\mathbf{c}}$isgiven by

$\mathit{\beta }\mathbf{=}\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{2}}$.

(b) A quasar in 1987 has $\mathit{z}\mathbf{=}\mathbf{4}\mathbf{.43}$. Calculate its speed parameter.

(c) Find the distance to the quasar, assuming that Hubble’s law is valid to these distances.

Consider the relativistic Doppler shift formula (Eq.37-31) as follows,

$\mathit{\lambda }\mathbf{=}{\mathit{\lambda }}_{\mathbf{0}}\sqrt{\frac{\mathbf{1}\mathbf{+}\mathbf{\beta }}{\mathbf{1}\mathbf{-}\mathbf{\beta }}}$…… (1)

(a)

Consider the Equation (1),

$\lambda ={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$.

From,$f=\frac{c}{\lambda }$, The Equation (1) can be written as,

${\lambda }_0=({\lambda }_0+\Delta \lambda )\sqrt{\frac{1-\beta }{1+\beta }}$,

Divide both the sides of the equation by ${\lambda }_0$,

$1=(1+z)\sqrt{\frac{1-\beta }{1+\beta }}$, where $z=\frac{\Delta \lambda }{{\lambda }_0}$

Solve the above equation for $\beta$ as follows,

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

$P=\frac{z^2+2z}{z^2+2+2}$…… (2)

Therefore, From the Equation (2) It has been shown that the recessional speed parameter$\beta =\frac{v}{c}$ can been expressed in terms of .data-custom-editor="chemistry" $\mathrm{z}$

b)

Consider the given value of quasar$z=4.43$ and substitute the value of $z$ in the Equation (3) from the solution of (a) as follows,

$P=\frac{z^2+2z}{z^2+2+2}$

$\begin{array}{l}\beta =\frac{{\left(4.43\right)}^2+2\left(4.43\right)}{{\left(4.43\right)}^2+2\left(4.43\right)+2}\\ \beta =0.934\end{array}$

Therefore, the speed parameter of the quasar is.$\beta =0.934$

c)

Consider the Hubble’s law,

$v=Hr$

The equation can be rewritten as,

$r=\frac{v}{H}$,

Substitute $\beta c$to $v$ into the equation as follows,

$r=\frac{\beta c}{H}$

$r=\frac{\left(0.934\right)(3.0\times {10}^8\text{m}/\text{s})}{0.0218\text{m}/\text{s}\cdot \text{ly}}$

$r=1.28\times {10}^{10}\text{ly}$

Therefore, The distance to the quasaris $r=1.28\times {10}^{10}\text{ly}$$r=1.28\times {10}^{10}\text{ly}$