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Found in: Page 1364

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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 ${\mathbit{z}}{\mathbf{=}}\frac{\mathbf{\Delta }\mathbf{\lambda }}{{\mathbf{\lambda }}_{\mathbf{0}}}$(a)Show that, in terms of,${\mathbit{z}}$ the recessional speed parameter${\mathbit{\beta }}{\mathbf{=}}\frac{\mathbf{v}}{\mathbf{c}}$ is given by ${\mathbit{\beta }}{\mathbf{=}}\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{2}}$.(b) A quasar in 1987 has ${\mathbit{z}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.43}}$ . Calculate its speed parameter.(c) Find the distance to the quasar, assuming that Hubble’s law is valid to these distances.

(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, $r=1.28×{10}^{10}\text{ly}$width="113" style="max-width: none; vertical-align: -4px;" $r=1.28×{10}^{10}\text{ly}$

See the step by step solution

Step 1:Explain given information

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

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

Step 2: 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,

${\lambda }_{0}=\left({\lambda }_{0}+\Delta \lambda \right)\sqrt{\frac{1-\beta }{1+\beta }}$ ,

Divide both the sides of the equation by ${\lambda }_{0}$ ,

$1=\left(1+z\right)\sqrt{\frac{1-\beta }{1+\beta }}$ , where $z=\frac{\Delta \lambda }{{\lambda }_{0}}$

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

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

$P=\frac{{z}^{2}+2z}{{z}^{2}+2+2}$…… (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" $\mathrm{z}$

Step 3: 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,

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

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

$r=1.28×{10}^{10}\text{ly}$

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