Special theory of relativity - relative movement If Mr. Einstein traveled at 99.99999999% speed of light in one direction and shined light backward, would the photon hang static in the air? Traveling speed of light away from its source that travels speed of light the other direction, thus not moving anywhere. 
 A: If Einstein traveled 99.99999999% the speed of light, and shined a flashlight backwards, he would see photons moving at the speed of light. A stationary observer would also see photons moving at the speed of light. This is one of the fundamental principles of Special Relativity:
The speed of light is the same in all reference frames.
This principle isn't just plucked out of thin air. It's supported by various experiments done at the end of the 19th century, chief among them being the Michelson-Morley experiment, that failed to show any variation in the speed of light with direction, even though we should by all accounts be moving at fantastic speeds relative to, for example, the center of the galaxy.
In order for this principle to make any sense, we have to relax the "common sense" notion that time and space are absolute. In reality, the speed of time's passage and the length of objects are both dependent on the speed of the observer relative to the object. In particular, the Lorentz factor $\gamma$ defines the magnitude of this shift:
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
For the speed you quoted (.9999999999c), the Lorentz factor is $\gamma=70710$. This has four implications:


*

*Einstein will observe that clocks in the rest of the world tick 70710 times slower than the watch in his pocket;

*Einstein will observe that pens in the rest of the world are 70710 times shorter (in his direction of motion) than the pen in his pocket;

*The stationary observer will observe that Einstein's watch ticks 70710 times slower than his watch; and

*The stationary observer will observe that Einstein's pen is 70710 times shorter (in the direction of motion) than his pen.
If the first two seem to contradict the last two ("Who's right?" you may ask), remember that you can't be both Einstein and the stationary observer at the same time, just like you can't travel at two speeds at once. You have to pick a speed to travel at, and that fixes your notion of space and time in a way that is consistent with the laws of physics in your frame of reference.
So, what will the stationary observer think of the light beam that propagates backwards from Einstein's flashlight? He will observe that time is moving much, much more slowly, which, in isolation, might imply that the light beam would move slower. BUT he also observes that lengths are much, much shorter, which, in isolation, would mean that the light beam can cover a much greater distance in each timestep (which would make it move faster). It turns out that, for objects traveling at the speed of light, these two effects exactly cancel out! This means that, even though time is slowed down, the distance is also contracted in just such a way that light travels the same distance in the same amount of time. Hence, light has the same speed for both observers.
A: Let Mr. Einstein have a velocity $v$ and let him shine a light backward. For him, the light will travel at $-c$. Now we invoke something called the Einstein velocity addition formula which states:
$$v_{rel}=(v_a+v_b)/(1+v_av_b/c^2)$$
Putting everything in, we find :
$$v_{rel}=(v-c)/(1-vc/c^2)=-c$$
So even for us, it moves at the speed of light. And also notice that it doesn't depend upon the speed of Mr. Einstein, as long as its finite. But upon physical grounds, we can say $v<c$, but whatever it be, light is always traveling at the speed of light! 
