externo said:
This is how the observer will interpret it.
Not if the observer correctly understands relativity, and the difference between recession speed and relative velocity.
Also, you appear to be confusing the speed of the
light with the speed of the
light source.
Perhaps it might help to actually look at the math for this. As I understand your scenario, you have a stationary observer, O, a light source, S, and a light receiver, R. You have R moving at some speed ##v_R## relative to you (let's say in the positive ##x## direction), and S moving at some speed ##v_S## relative to you, where ##v_S > v_R##. Then S emits successive light pulses towards R at a constant rate according to S's clock. The question is at what rate R will receive the pulses according to R's clock.
First, let's dispose of the relative velocity issue. According to the relativistic velocity addition law, the speed of S relative to R will be:
$$
v = \frac{v_S - v_R}{1 - v_S v_R}
$$
Note the extra factor in the denominator: it is
not the case that ##v = v_S - v_R##. The quantity ##v_S - v_R## is what
@Ibix called the "recession speed", and is not the same as relative velocity, as the above shows. In fact, ##v## will be
larger than ##v_S - v_R##; the receiver R will see the source moving away
faster than the difference in their speeds as you see them.
You appear to be assuming that the light pulses move at speed ##-1## (I am using units in which ##c = 1##) relative to S, and then asserting that the pulses move
slower than ##- 1## according to you. But according to the relativistic velocity addition law, if ##v_L## is the velocity of the light pulses relative to you, we have
$$
v_L = \frac{v_S - 1}{1 - v_S} = -1
$$
So relativity says the light pulses move at speed ##-1## relative to you. (A similar calculation using ##v## as obtained above shows that the light pulses also move at speed ##-1## relative to R.)
Now, let's look at the redshift. Light pulses are emitted by S at a constant rate relative to S's clock. But relative to you, S's clock is time dilated by a factor ##\sqrt{1 - v_S^2}##. So if S emits light pulses at a rate ##\nu##, the rate at which
you see S's light pulses being emitted is ##\nu \sqrt{1 - v_S^2}##.
The receiver R, however, is
also time dilated relative to you, by a factor ##\sqrt{1 - v_R^2}##, so if S emits light pulses at a rate ##\nu##, the rate at which you calculate that R would receive S's light pulses,
not allowing for the change in light travel time, is ##\nu \sqrt{1 - v_S^2} / \sqrt{1 - v_R^2}##.
[Note: I have edited the previous paragraph and added the next one to correct the analysis; the final comparison has also been edited.]
However, since S is moving away from R, there is an additional effect due to the change in light travel time between pulses. This adds an extra factor which, from your point of view, can be expressed as the ratio of the "recession speed" of the light relative to S and R, i.e., a factor ##\left( 1 + v_R \right) / \left( 1 + v_S \right)##. So your final prediction for the rate at which R receives S's light pulses is ##\nu \sqrt{1 - v_S^2}\left( 1 + v_R \right) / \sqrt{1 - v_R^2} \left( 1 + v_S \right)##
What does the relativistic Doppler shift formula tell us? It tells us that, if S is moving at speed ##v## away from R (where ##v## is given by the formula we derived above), and S emits light pulses at the rate ##\nu##, then R receives light pulses at the rate
$$
\nu \sqrt{\frac{1 - v}{1 + v}}
$$
Now we just substitute and do the algebra:
$$
\nu \sqrt{\frac{1 - v}{1 + v}} = \nu \sqrt{\frac{1 - v_S v_R - v_S + v_R}{1 - v_S v_R + v_S - v_R}} = \nu \sqrt{\frac{\left( 1 - v_S \right) \left( 1 + v_R \right)}{\left( 1 + v_S \right) \left( 1 - v_R \right)}} = \nu \sqrt{\frac{\left( 1 - v_S^2 \right) \left( 1 + v_R \right)^2}{\left( 1 - v_R^2 \right) \left( 1 + v_S \right)^2}}
$$
which is the same as the formula that we derived above. So you calculate that R will receive S's light pulses at exactly the same rate that the relativistic Doppler shift formula predicts.
