Knowing that a body in motion experiences time dilation, "also" knowing when two objects travel at a great speed away from one an other, both observers experience the others clock as moving slower relative to themselves due to the Doppler effect "I presume". So is it safe to speculate that if observer A where stationary and observer B where in motion (disregarding time dilation for observer B) that they would mutually experience each other's time as slowing down due to the Doppler effect alone? However; taking time dilation into account for observer B, wouldn't the observed Doppler effect of observer A's slowing clock combine with the time dilation of B's to be an in-sync operation? Where if B looking back at A's clock would appear to B as time moving normal? (just to clarify: I realize this would have the reverse effect if B where to travel towards A rather then away, meaning B's time dilation combined with its more frequently emitted light pulse would then appear to A as B's time moving normally instead).
Special Relativity – How Does the Relativistic Doppler Effect Interact with Time Dilation?
doppler effectspacetimespecial-relativitytimetime-dilation
Related Solutions
Special relativity (let's leave aside GR for now) is notoriously unintuitive - generations of physics students have found this to their cost, so you are far from alone. So there is no simple intuitively clear explanation over what is going one. That said, I will attempt a quasi-intuitive explanation.
I think the mistake students make is to take time dilation in isolation. It's easy to think here's a phenomenon called time dilation: what causes it? What actually happens is that different observers will disagree about what constitutes space and what constitutes time and time dilation is just part of a bigger phenomenon.
As I sit here at my keyboard I'm not moving in space, but I am moving in time. So for some activity (e.g. from the start to the end of me typing this sentence) my $\Delta x = 0$ and my $\Delta t = T$ for some time $T$. However the bug eyed alien that has just zoomed past at $0.9c$ disagrees. The alien, seated at their own typewriter, sees me moving at $-0.9c$, so in between starting and finishing my sentence the alien sees that I have moved some distance $\Delta x = d$. But in SR space and time are linked, so if the alien measures a different $\Delta x$ they must also measure a different $\Delta t$. The two are linked by the relationship:
$$ c^2 \Delta t^2 - \Delta x^2 = c^2 \Delta t'^2 - \Delta x'^2 $$
where the unprimed $t$ and $x$ are what I measure and the primed $t'$ and $x'$ are what the alien measures. Even without doing the maths it should be obvious that because $\Delta x < \Delta x'$ it follows that $\Delta t < \Delta t'$. In other words when the alien times how long it takes me to type the sentence they measure a longer time than I do. For the alien my time has been dilated.
You've probably also heard of length contraction. Well this is the other side of the coin. Time dilation and length contraction always occur together because in effect some of the time is being converted into length and vice versa.
All this follows from a fundamntal symmetry called Lorentz invariance. This states that if we measure a property called the line interval and defined by:
$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$
then the quantity $ds^2$ is an invariant and all observers will measure the same value for it. To get the equation above linking my $(t, x)$ with the alien's $(t', x')$ I just exploited this invariance to require that $ds^2 = ds'^2$.
All very well, but I have really only pushed the non-intuitiveness one stage farther down, since my explanation assumes Lorentz invariance and this in turn is unintuitive. Still, hopefully this allows you to get some handle on what is going on.
So, if time going slower always causes a redshift, does that mean that if we see a blueshift it means that time appears to move faster?
Yes.
The machine that produces the wave-crests that appear to follow each other extra rapidly, appears to work extra rapidly.
For example if wave-crests appear to follow each other at one nano-seconds intervals, then the machine that produces those wave-crests appears to produce one wave-crest each nano-second. I mean, when looked very closely through a big telescope, then the machine can be seen to do that.
Those machines that appear to person X to work extra rapidly work extra slowly according to person X. I mean, person X subtracts the directional blue shift, and notes that there is a redshift.
Best Answer
You are right that there are two effects at play here.
Firstly, suppose that we turn relativity off --- suppose that we consider our universe to be Newtonian, with a finite speed of light propagation. It would indeed be the case that if an observer A were moving relative to another observer B, and emitting a light signal towards B, then the rate at which the signal arrived at B would be different to that at which it was emitted. This is the plain old Doppler effect, just the same as the effect experienced when listening to an ambulance moving towards or away from you. If the signal emitted by A was the light from her clock face, then indeed observer B would see the clock tick at a different rate.
But would she see it tick more quickly or more slowly? Now that depends on whether observer A were moving towards or away from observer B. If A were moving towards B, B would see the clock tick more quickly, just as you hear the ambulance siren at a higher pitch if it is moving towards you. Likewise, if A were moving away from B, B would see the clock tick more slowly, just as you hear the ambulance siren at a lower pitch if it is moving away from you. Of course, for speeds small compared to the speed of light, this effect would be very small.
Now suppose we turn relativity on. A new effect comes into play --- one much more profound and weird. The (relativistic) phenomenon of time dilation would entail that, according to B, the clock belonging to observer A runs intrinsically more slowly. It's not that she merely sees the the clock tick more slowly, it's that time itself has slowed down aboard A's spaceship, according to B. What this means is that if B were to measure the rate of arrival of signals from A and then do some maths to undo the Doppler effect, such that the resulting quantity obtained were the rate of emission of the signal, she would find the rate to be less than that at which A would say she emitted the signal. Which is pretty nutty.
So what is the combined effect? Well, we can see immediately that they won't cancel out in all cases by considering observer A to be moving away from observer B. Then time dilation and the Doppler effect combine, such that observer B sees the clock aboard A's ship running particularly slowly. The correct formula (I won't give a derivation here, but you can find many on the web, e.g. on Wikipedia) turns out, in fact, to be
$$ f_O = f_E \sqrt{\frac{1-\beta}{1+\beta}} \,,\qquad \beta = v/c\,,$$
where $f_E$ is the emitted frequency and $f_O$ the observed frequency. For all positive velocities, which in this case corresponds to observer A moving away from observer B, the top of the fraction is smaller than the bottom, and so the observed frequency is less than the emitted frequency. Likewise, for all negative velocities, which in this case corresponds to observer A moving towards observer B, the top of the fraction is larger than the bottom, and so the observed frequency is greater than the emitted frequency.
Qualitatively, then, the two effects combined are no different from the Doppler effect by itself. The only circumstance under which the two effects 'cancel out' is when the observers' relative velocity is zero, and in that case the two effects are individually zero! Whilst the Doppler effect results in an increased rate for $\beta < 0$ and a decreased rate for $\beta >0$, the phenomenon of time dilation always acts to slow the rate down. Then it just happens to be the case that for no choice of relative velocity can time dilation slow the rate down quite enough to counteract an increased rate due to the Doppler effect.
I've plotted the square root in the above formula below:
You can hopefully see that this function is always greater than 1 for $\beta$ less than zero, and always less than 1 for $\beta$ greater than zero. Hope this helps!