Question

Some of the familiar hydrogen lines appear in the spectrum of quasar 3C9, but they are shifted so far toward the red that their wavelengths are observed to be 3.0 times as long as those observed for hydrogen atoms at rest in the laboratory. (a) Show that the classical Doppler equation gives a relative velocity of recession greater than c for this situation. (b) Assuming that the relative motion of 3C9 and Earth is due entirely to the cosmological expansion of the universe, find the recession speed that is predicted by the relativistic Doppler equation.

Short answer

(a) The relative velocity of recession is greater than speed of light.

(b) The recession speed of quasar by relativistic Doppler equation is $0.8c.$ .

Step-by-step solution

Identification of given data

The observed wavelengths of lines in quasar spectrum is $\lambda \text{'}=3\lambda$

The wavelength of light for observer on Earth is calculated by using the formula for Doppler effect in light and color of light is decided by wavelength of light.

Proof for classical Doppler equation relative velocity of recession greater than speed of light

(a)

The relative velocity of recession is given as

$v_s=c\left(\frac{\lambda \text{'}}{\lambda }-1\right)$

Here, $c$ is the speed of light and its value is $3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .

Substitute all the values in the above equation.

$$\begin{gathered} v_s=c\left(\frac{3\lambda }{\lambda }-1\right) \\ {v}_s=c\left(3-1\right) \\ {v}_s=2c \\ {v}_s\left(2c\right)>c \end{gathered}$$

Therefore, the relative velocity of recession is greater than speed of light.

Determination of recession speed by relativistic Doppler equation

(b)

The recession speed of quasar by relativistic Doppler equation is given as:

$u_s=c\left[\frac{1-{\left({\displaystyle \frac{\lambda }{\lambda \text{'}}}\right)}^2}{1+{\left({\displaystyle \frac{\lambda }{\lambda \text{'}}}\right)}^2}\right]$

Substitute all the values in the above equation.

$$\begin{gathered} u_s=c\left[\frac{1-{\left({\displaystyle \frac{\lambda }{3\lambda }}\right)}^2}{1+{\left({\displaystyle \frac{\lambda }{3\lambda }}\right)}^2}\right] \\ {u}_s=0.8c \end{gathered}$$

Therefore, the recession speed of quasar by relativistic Doppler equation is$0.8c.$ .