Question

Every years, more or less, The New York Times publishes an article in which some astronomer claims to have found an object traveling faster than the speed of light. Many of these reports result from a failure to distinguish what is seen from what is observed—that is, from a failure to account for light travel time. Here’s an example: A star is traveling with speed v at an angle$\mathit{\theta }$to the line of sight (Fig. 12.6). What is its apparent speed across the sky? (Suppose the light signal fromb reaches the earth at a timelocalid="1656138453956" $\mathbf{∆}\mathit{t}$after the signal from a, and the star has meanwhile advanced a distancelocalid="1656138461523" $\mathbf{∆}\mathit{s}$across the celestial sphere; by “apparent speed,” I meanlocalid="1656138468709" $(∆s/∆t)$. What anglelocalid="1656140989446" $\mathit{\theta }$gives the maximum apparent speed? Show that the apparent speed can be much greater than c, even if v itself is less than c.

Short answer

The apparent speed across the sky is $u=\frac{v\sin \theta }{\left(1-{\displaystyle \frac{v\cos \theta }{c}}\right)}$, the angle which gives the maximum apparent speed is ${\theta }_{\max }={\cos }^{-1}\left(\frac{v}{c}\right)$, and the apparent speed can be much greater than c, even if v itself less than c.

Step-by-step solution

Determine the apparent speed across the sky:

Let $t_a$be the time at which light signal leaves the point a and $t_b$be time at which the light signal leaves the point b.

Write the expression for the time when the light signal leaves from point a.

${\mathit{t}}_{\mathbf{a}}\mathbf{=}{\mathit{t}}_{\mathbf{a}}^{\mathbf{,}}\mathbf{+}\frac{{\mathbf{d}}_{\mathbf{a}}}{\mathbf{c}}$ …… (1)

Here,$d_a$is the distance from a to earth.

Write the expression for the time when the light signal leaves from point b.

${\mathit{t}}_{\mathbf{b}}\mathbf{=}{\mathit{t}}_{\mathbf{b}}^{\mathbf{,}}\mathbf{+}\frac{{\mathbf{d}}_{\mathbf{b}}}{\mathbf{c}}$ …… (2)

Here,$d_b$is the distance from b to earth.

Subtract equation (1) from (2) and calculate the time difference.

$$\begin{gathered} ∆t=\left(t_b^{,}+\frac{d_b}{c}\right)-\left(t_a^{,}+\frac{d_a}{c}\right) \\ ∆t=\left(t_a^{,}-t_b^{,}\right)+\frac{\left(d_a-d_a\right)}{c} \\ ∆t=∆t^{\text{'}}+\frac{\left(d_a-d_a\right)}{c} \end{gathered}$$

Here, $d_b-d_b=-v∆t^{\text{'}}\cos \theta$.

Hence, the equation becomes,

$$\begin{gathered} ∆t=∆t^{\text{'}}+\frac{\left(-v∆t^{\text{'}}\cos \theta \right)}{c} \\ ∆t=∆t^{\text{'}}+\left(1-\frac{v\cos \theta }{c}\right) \end{gathered}$$

Write the expression for the advanced distance by starting across the celestial spheres.

$∆s=v∆t^{\text{'}}\sin \theta$

Write the expression for the apparent speed.

role="math" localid="1656139838204" $$\begin{gathered} u=\frac{∆s}{∆t} \\ u=\frac{\left(v∆t^{\text{'}}\sin \theta \right)}{∆t^{\text{'}}+\left(1-{\displaystyle \frac{v\cos \theta }{c}}\right)} \\ u=\frac{v\sin \theta }{\left(1-{\displaystyle \frac{v\cos \theta }{c}}\right)}......\left(3\right) \end{gathered}$$

Determine the angle which gives the maximum apparent speed:

For the apparent speed to be maximum,

$\frac{du}{d\theta }=0$

Hence, substitute the value of u in the above expression.

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

On further solving,

$$\begin{gathered} \frac{du}{d\theta }=\frac{v\cos \theta -{\displaystyle \frac{v^2}{c}}\left(1\right)}{\left(1-\frac{v}{c}\cos \theta \right)}=0 \\ \cos {\theta }_{\max }=\frac{v}{c} \\ {\theta }_{\max }={\cos }^{-1}\left(\frac{v}{c}\right) \end{gathered}$$

Substitute the value of ${\theta }_{\max }$in equation (3) for the maximum angle.

$$\begin{gathered} u=\frac{\sqrt[v]{1-{\displaystyle \frac{v^2}{c^2}}}}{\left(1-\frac{v^2}{c^2}\right)} \\ u=\frac{v}{\sqrt[]{1-\frac{v^2}{c^2}}} \end{gathered}$$

Show that the apparent speed is greater than c even if v<c:

As$v\to c$, the apparent speed will be,

$u\to \infty$

Therefore, the apparent speed across the sky is $u=\frac{v\sin \theta }{\left(1-{\displaystyle \frac{v\cos \theta }{c}}\right)}$, the angle which gives the maximum apparent speed is ${\theta }_{\max }={\cos }^{-1}\left(\frac{v}{c}\right)$, and the apparent speed can be much greater than c, even if v itself less than c.