Velocity of approach of a beam of light and a moving object I came across a problem on Brian Greene's course on Special Relativity.
(http://www.worldscienceu.com/courses/6/elements/TnRUSU)

From George's perspective, how fast does a beam of light approach Gracie when she races away at speed $v$ relative to George? That is, from George's perspective, how quickly does the distance between the light beam and Gracie decrease?

I think the answer should be c, according to the relativistic velocity addition rule. However, the correct answer is (according to the course), $c-v$. I have no idea how that can be possible. With that sort of reasoning, if Gracie stopped and ran in the opposite direction, George would see her and the light's seperation decreasing at a rate more than $c$. Isn't that a violation?
The question comes with an ominous note:

Note: This is a simple, but slightly tricky, exercise. The question is not asking about the speed of light itself. Rather, the question is asking about the speed at which the beam of light approaches Gracie, from George's perspective. While this might sound like an irrelevant distinction, it's not. So ponder this for a moment before selecting your answer.

 A: Let me quote the relativistic velocity addition formula for easy reference:
$$v_{AB} = \frac{v_A - v_B}{1 + \frac{v_Av_B}{c^2}}\tag{SR}$$
I'm guessing you interpreted these quantities as follows:


*

*$v_A$ is the speed of the light beam relative to George

*$v_B$ is the speed of Gracie relative to George

*$v_{AB}$ is the speed of the light beam relative to Gracie


If so, your mistake is in the third interpretation. $v_{AB}$ is the speed of the light beam as measured by Gracie - that is, in Gracie's reference frame. That is not the same as the difference between the light beam's speed in George's reference frame ($v_A$) and Gracie's speed in George's reference frame ($v_B$). This is exactly what people mean when they say velocities don't add linearly in special relativity: the fact that the velocity of A in B's reference frame is not the same as the velocity of A minus the velocity of B.
Note that if $v_A,v_B\ll c$ in the equation above, the denominator is very close to $1$ and the formula becomes approximately
$$v_{AB} = v_A - v_B\tag{Galilean}$$
Galilean relativity is a theory which uses this equation instead of the one from the top of the post. It works fine for slow-moving objects.
A: The key to understanding this somewhat surprising result is that the relativistic velocity addition formula is not applicable to this calculation.
As an example of when to apply the velocity addition formula, sssume there is an object with (1D) velocity $\mathbf u$ in some inertial coordinate system.
Now, what is the velocity $\mathbf u'$ of that object in another inertial coordinate system moving with (uniform) velocity $\mathbf v$ with respect to the other system?
According to the relativistic velocity addition formula, the object has velocity
$$\mathbf u' =  \frac{\mathbf u - \mathbf v}{1 - \frac{\mathbf u \mathbf v}{c^2}}$$
in the relatively moving coordinate system.  This guarantees that $u' \le c$ as you expect.

However, in this problem, you're not asked to find the speed of an object,


*

*you're asked to find the rate at which the distance between two objects, moving relative to an inertial coordinate system, change.
While this quantity has units of speed, it is not the speed of an object and thus can be greater than $c$.
For example, consider two objects $A$ and $B$.  The position of object $A$ is, in some inertial coordinate system
$$x_A = vt$$
while the position of object $B$ is
$$x_B = -vt$$
where
$$v \lt c$$
The distance between the objects in this coordinate system is simply
$$d_{AB} = |x_A - x_B| = |vt - (-vt)| = 2vt$$
Then, the time rate of change of this distance is just
$$\frac{d}{dt}(d_{AB}) = 2v$$
Thus, the rate at which the distance changes can be arbitrarily close to $2c$.

In the particular case of the problem at hand, stipulate that in the inertial coordinate system in which George is at rest, the coordinates of Gracie are
$$x_{Gracie} = d_0 + vt$$
and the coordinates of the front of the light beam are
$$x_\gamma = ct$$
The distance between them is then
$$d = d_0 + vt - ct$$
and the distance changes at a rate of
$$\dot d = v - c < 0$$
which is to say the distance decreases at a rate of $c - v$.
