Chapter 22, Problem 57P

To determine

The speed with which one has to drive to see a red light as green.

Answer to Problem 57P

The speed with which one has to drive to see a red light as green is $5\times {10}^7\text{ m/s}$ .

Explanation of Solution

To occur a Doppler shift of magnitude in which a red light is seen as green, a relativistic relative velocity is required.

The observer is approaching the source and the source of light is stationary in this case. This implies the relative velocity is greater than zero.

$v_{\text{rel}}>0$

Here, $v_{\text{rel}}$ is the relative velocity

Write the equation for the Doppler effect of EM waves.

$f_o=f_s\sqrt{\frac{1+v_{\text{rel}}/c}{1-v_{\text{rel}}/c}}$

Here, $f_o$ is the frequency of the wave as seen by the observer, $f_s$ is the frequency of the source and $c$ is the speed of EM waves in vacuum.

Divide both sides of the above equation by $f_s$ and square both sides of the equation.

$\begin{array}{c}{\left(\frac{f_o}{f_s}\right)}^2={\left(\frac{f_s}{f_s}\sqrt{\frac{1+v_{\text{rel}}/c}{1-v_{\text{rel}}/c}}\right)}^2\\ {\left(\frac{f_o}{f_s}\right)}^2=\frac{1+v_{\text{rel}}/c}{1-v_{\text{rel}}/c}\end{array}$

Rearrange the above equation.

$\begin{array}{c}{\left(\frac{f_o}{f_s}\right)}^2\left(1-v_{\text{rel}}/c\right)=1+v_{\text{rel}}/c\\ {\left(\frac{f_o}{f_s}\right)}^2-{\left(\frac{f_o}{f_s}\right)}^2\frac{v_{\text{rel}}}{c}=1+\frac{v_{\text{rel}}}{c}\\ {\left(\frac{f_o}{f_s}\right)}^2-1=\frac{v_{\text{rel}}}{c}+{\left(\frac{f_o}{f_s}\right)}^2\frac{v_{\text{rel}}}{c}\\ {\left(\frac{f_o}{f_s}\right)}^2-1=\frac{v_{\text{rel}}}{c}\left[1+{\left(\frac{f_o}{f_s}\right)}^2\right]\end{array}$

Rewrite the above equation for $v_{\text{rel}}$ .

$\begin{array}{c}{v}_{\text{rel}}=c\frac{{\left(\frac{f_o}{f_s}\right)}^2-1}{1+{\left(\frac{f_o}{f_s}\right)}^2}\\ =c\frac{{\left(f_o/f_s\right)}^2-1}{{\left(f_o/f_s\right)}^2+1}\end{array}$ (I)

Since frequency of the EM waves is inversely proportional to their wavelength, write .expression for $\frac{f_o}{f_s}$.

$\frac{f_o}{f_s}=\frac{{\lambda }_s}{{\lambda }_o}$

Here, ${\lambda }_s$ is the wavelength of the source and ${\lambda }_o$ is the wavelength observed by the observer

Put the above equation in equation (I).

$v_{\text{rel}}=c\frac{{\left({\lambda }_s/{\lambda }_o\right)}^2-1}{{\left({\lambda }_s/{\lambda }_o\right)}^2+1}$ (II)

Conclusion:

Given that the wavelength of red is $630\text{ nm}$ and that of green is $530\text{ nm}$ . The value of $c$ is $3.00\times {10}^8\text{ m/s}$ .

Substitute $3.00\times {10}^8\text{ m/s}$ for $c$ , $630\text{ nm}$ for ${\lambda }_s$ and $530\text{ nm}$ for ${\lambda }_o$ in equation (II) to find $v_{\text{rel}}$ .

$\begin{array}{c}{v}_{\text{rel}}=\left(3.00\times {10}^8\text{ m/s}\right)\frac{{\left(630\text{ nm}/530\text{ nm}\right)}^2-1}{{\left(630\text{ nm}/530\text{ nm}\right)}^2+1}\\ =5\times {10}^7\text{ m/s}\end{array}$

Therefore, the speed with which one has to drive to see a red light as green is $5\times {10}^7\text{ m/s}$ .