Special Relativity - speed of light question Just a basic question:
I know that if you are traveling at $x$ speed the time will pass for you slower than to an observer that is relatively stopped. That's all just because a photon released at the $x$ speed can't travel faster than the $c$ limit. 
I want to know what happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? 
And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same? 
I just want to know if I got it right...
 A: The speed of light will be the same, yes. This is the fundamental tenet of special relativity - that all inertial observers see the same laws of physics, including universal constants like the speed of light.
And yes, the wavelength $\lambda$ will change. The frequency $\nu$ will also change, since after all we still must have
$$ \lambda \nu = c. $$
A: Yes, every observer who makes a local measurement of the speed of light will get the same result, i.e. $c$, regardless of where the light came from.
I've used the word local because in general relativity the speed of light can differ from $c$ if it's not measured at your location. The most famous example of this is probably the fact that light slows down to a halt as it approaches the event horizon of a black hole. However even in general relativity a local measurement always returns the value $c$.
A: Yes. There are two things you gotta take care of. In order for the speed of light to be constant (SR's postulate) relative to the oppositely rushing observers $A$ and $B$, we can use the velocity addition formula. Another Yes. The wavelength will be contracted or elongated and the light seen by $B$ or $A$ would be blue-shifted or red-shifted according to the Relativistic Doppler effect, by how they're moving relatively (towards or away).
