Apparent speeds greater than the speed of light

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 6.

Although no object can travel faster than light, it is possible for the apparent speed of an object to be greater than ${c}$. A common example is that of the apparent motion of a star across the sky. Suppose a star is a distance ${a=L}$ from Earth at time ${t_{a}}$ and that its velocity ${\mathbf{v}}$ is towards Earth at an angle ${\theta}$ to the line of sight. At time ${t_{b}}$ it arrives at a point ${b}$ whose distance from Earth is

$\displaystyle b=L-v\cos\theta\left(t_{b}-t_{a}\right) \ \ \ \ \ (1)$

The times of arrival at Earth of the light emitted at distances ${a}$ and ${b}$ are

 $\displaystyle T_{a}$ $\displaystyle =$ $\displaystyle t_{a}+\frac{L}{c}\ \ \ \ \ (2)$ $\displaystyle T_{b}$ $\displaystyle =$ $\displaystyle t_{b}+\frac{1}{c}\left[L-v\cos\theta\left(t_{b}-t_{a}\right)\right] \ \ \ \ \ (3)$

During this time, the star moves a distance perpendicular to the line of sight of ${v\sin\theta\left(t_{b}-t_{a}\right)}$, so the apparent speed as seen from Earth is

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \frac{v\sin\theta\left(t_{b}-t_{a}\right)}{T_{b}-T_{a}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v\sin\theta\left(t_{b}-t_{a}\right)}{\left(t_{b}-t_{a}\right)\left(1-\frac{v}{c}\cos\theta\right)}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v\sin\theta}{1-\frac{v}{c}\cos\theta} \ \ \ \ \ (6)$

This speed has a maximum at an angle ${\theta}$ which can be found by setting the derivative to zero:

 $\displaystyle \frac{du}{d\theta}$ $\displaystyle =$ $\displaystyle {v\cos\left(\theta\right)\left(1-{\frac{v\cos\left(\theta\right)}{c}}\right)^{-1}}-{\frac{{v}^{2}\left(\sin\left(\theta\right)\right)^{2}}{c}\left(1-{\frac{v\cos\left(\theta\right)}{c}}\right)^{-2}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle {\frac{\left(\cos\left(\theta\right)c-v\right)cv}{\left(\cos\left(\theta\right)\right)^{2}{v}^{2}-2\, c\cos\left(\theta\right)v+{c}^{2}}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle \frac{v}{c} \ \ \ \ \ (10)$

At this angle, the apparent speed is

 $\displaystyle u_{max}$ $\displaystyle =$ $\displaystyle \frac{v\sqrt{1-v^{2}/c^{2}}}{1-v^{2}/c^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma v \ \ \ \ \ (12)$

Since ${\gamma\rightarrow\infty}$ as ${v\rightarrow c}$, ${u_{max}}$ can be much larger than ${c}$ even though the actual speed of the star is less than ${c}$. This again illustrates the importance of correctly interpreting the raw data that we see, and of allowing for the travel time of the light.