Question

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.

Short answer

(a) It can be shown that,the recessional speed parameter$\beta =\frac{v}{c}$can been expressed in terms of $z$is given by .$\beta =\frac{z^2+2z}{z^2+2z+2}$

(b) Speed parameter, $\beta =0.934$

(c) The distance to the quasar, $r=1.28\times {10}^{10}\text{ly}$width="113">r=1.28×1010ly

Step-by-step solution

Step 1:Explain given information

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)

Show the recessional speed parameter in terms of .z

(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" $\mathrm{z}$

Calculate the speed parameter of the quasar.

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.$\beta =0.934$

Step 4:Find the distance to the quasar, assuming that Hubble’s law is valid to these distances.

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}$