Question

A space probe has a powerful light beacon that emits 500 nm light in its own rest frame. Relative to Earth, the space probe is moving at 0.8c . An observer on Earth is viewing the light arriving from the distant beacon and detects a wavelength of 500nm. Is this possible? Explain.

Short answer

The angle is $120°$that is other than $0°$and $180°$. Therefore, we can see the light from the beacon.

Step-by-step solution

Write the given data from the question

The source wavelength,${\lambda }_{source}=500\mathrm{nm}$

The observer wavelength,${\lambda }_{obs}=500nm$

The velocity of the source, v = 0.8 c

Relationship between the source and observer wavelength

The relationship between the source and observer wavelength is given as follows.

${\mathit{\lambda }}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\mathbf{=}{\mathit{\lambda }}_{\mathbf{o}\mathbf{b}\mathbf{s}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{v}{c}}\right)}^{\mathbf{2}}}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$ …… (i)

Here, c is the speed of light, $\mathit{\theta }$is the angle between the source of light and observer and v is the velocity of the source.

Calculate that is it possible to see the light from the beacon

Calculate the angle between the source and observer light.

Substitute 500 nm for ${\lambda }_{obs}\mathrm{and}{\lambda }_{source}$into equation (i).

$$\begin{gathered} 500=500\frac{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}{1+{\displaystyle \frac{v}{c}}\cos \theta } \\ 1=\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos \theta } \\ 1+\frac{v}{c}\cos \theta =\sqrt{1-{\left(\frac{v}{c}\right)}^2} \\ \left(1+\frac{v}{c}\cos \theta \right)=1-{\left(\frac{v}{c}\right)}^2 \end{gathered}$$

Solve further as,

$$\begin{gathered} 1+{\left(\frac{v}{c}\right)}^2{\cos }^2\theta +2\frac{v}{c}\cos \theta =1-{\left(\frac{v}{c}\right)}^2 \\ {\left(\frac{v}{c}\right)}^2{\cos }^2\theta +2\frac{v}{c}\cos \theta +{\left(\frac{v}{c}\right)}^2=0 \end{gathered}$$

Solve the above quadratic equation,

$$\begin{gathered} \cos \theta =\frac{-2{\displaystyle \frac{v}{c}}\pm \sqrt{{\left(2{\displaystyle \frac{v}{c}}\right)}^2-4{\left(\frac{v}{c}\right)}^2{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos \theta =\frac{-2{\displaystyle \frac{v}{c}}\pm \sqrt{{\left(2{\displaystyle \frac{v}{c}}\right)}^2-4{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos \theta =\frac{-2{\displaystyle \frac{v}{c}}\pm 2{\left(\frac{v}{c}\right)}^2\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos \theta =\frac{-2{\displaystyle \frac{v}{c}}\pm \sqrt{1-{\left(\frac{v}{c}\right)}^2}}{v} \end{gathered}$$

Substitute 0.8c for v into above equation.

$$\begin{gathered} \cos \theta =\frac{-1\pm \sqrt{1-{\left({\displaystyle \frac{0.8c}{c}}\right)}^2}}{\frac{0.8c}{c}} \\ \cos \theta =\frac{-1\pm \sqrt{1-0.64}}{0.8} \\ \cos \theta =\frac{-1\pm 0.6}{0.8} \\ \cos \theta =-0.5\mathrm{or}-2 \end{gathered}$$

From the above solutions,$\cos \theta =-2$ is not valid solution. So, calculate the angle for -0.5 .

$$\begin{gathered} \cos \theta =-0.5 \\ \theta ={\cos }^{-1}(-0.5) \\ \theta ={120}^{°} \end{gathered}$$

Since the angle is $120°$that is other than $0°$and $180°$. Therefore, we can see the light from the beacon.