Buckethead said:
Considering that there is no way for the light pulse to have encoded any speed information from the emitter (since it will always be emitted at c) why is the pulse now blue when it was emitted with a red color.
Relativistic Doppler shift is very simple. For example, if a source and an observer move towards each other, proper frequency of the source (for example it was red) the observer will see blueshifted. Cause of this blueshift he can explain doubly:
a) An observer is "at rest" and the source moves towards the observer - the source’s clock dilates, so the source oscillates slower.
$$\omega = \frac {\omega_0 \sqrt {1-v^2/c2}}{1-v/c}$$
If velocity of the source ##v## was close to ##c##, according to this formula “classic” frequency would be “very infinitely intense” and all the waves would have gathered in a heap straight in front of the source. But Lorentz – factor (dilation of the source’s clock) “reduces” that frequency to ##\omega##.
b) The source is "at rest" and the observer moves toward it,
so the observer’s clock (and observer himself) slows down. Proper frequency was “red”, but it turns into “blue” due to dilation of observer’s clock.
$$\omega = \frac {\omega_0 (1+v/c)} {\sqrt {1-v^2/c2}}$$
If velocity of an observer ##v## was close to ##c##, according to this formula “classic” frequency would be ##2\omega_0##, but Lorentz – factor (dilation of observer’s clock) “increases” it to ##\omega##. That means the observer turns into a "dawdler" himself.
Outcome will be the same - measured frequency will be ##\omega##, but the reasons are frame - dependent. Feynman considers the both cases in his lectures – Relativistic Doppler Effect
http://www.feynmanlectures.caltech.edu/I_34.html
Worth to note, that Einstein himself described exactly the “moving observer” case in his 1905 article:
https://www.fourmilab.ch/etexts/einstein/specrel/www/ - § 7. Theory of Doppler's Principle and of Aberration
From the equation for ##\omega'## it follows that if an observer is moving with velocity ##v## relatively to an infinitely distant source of light of frequency ##\nu##, in such a way that the connecting line “source-observer” makes the angle ##\varphi## with the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light, the frequency ##\nu '## of the light perceived by the observer is given by the equation
$$\nu' = \nu \frac {1-\cos \varphi v/c}{\sqrt {1-v^2/c^2}}$$
We see that, in contrast with the customary view, when ##v= -c, \nu' = \infty##
It follows from these results that to an observer approaching a source of light with the velocity ##c##, this source of light must appear of infinite intensity.
Again, the formula shows, that in classical case (if ##v = -c##) measured by a moving observer frequency would be ##2\nu##. Wavefronts and the observer approach each other with equal velocities. According to Einstein, if an observer moves towards a source, we must
divide this frequency by ##\sqrt {1-v^2/c^2} ##, i.e. “actual” frequency ##\nu## appears ##\sqrt {1-v^2/c^2} ## times more intense due to dilation of the observer’s clock. That means, the observer sees that processes around him happen not slower, but faster.
Quite strange, that Einstein himself looked at Doppler Shift “by eyes of moving observer”, because this case graphically demonstrates, that the source's clock is running not slower, but ##\sqrt {1-v^2/c^2} ## times faster than observer's own clock. In Special Relativity an observer never moves himself, but Einstein has made some exception.
It is exactly as in the Lorentz theory.
Page 59 -
http://www.mpiwg-berlin.mpg.de/litserv/diss/janssen_diss/Chapter3.pdf