Sure - you get doppler shift. But for redshift you have to change the frequency enough to change visible spectrum to predominantly red. I mean, I've tried - but I've never been able to see a car moving away from me change its colour.

Redshift or blueshift means decrease or increase of frequency, that is, going towards the red (less energy) or going towards the blue (more energy). Is has nothing to do with being red or blue. A redshifted very energetic light emitter doesn't look red to us, just a bit more red

I guess what you meant by your first comment on this matter that at the moment I didn't get is that there are mainly two kinds of redshifts on the universe: Doppler and cosmological. The Doppler redshift (or blueshift sometimes) is due to the relative velocity of the body, like what the traffic radar does. This happens with close objects. Cosmological redshift, which is the most famous (Hubble constant etc etc) is due to the expansion of universe, a property of spacetime that is expanding and that make photons to loose energy as they travel. The farthest they are, the more energy they loose until they arrive at us, or the more redshifted we observe them.

Doppler effect on space needs a relativistic treatment indeed, which complicates a bit the formulas, but not too much.

Actually I'd think you have to on principle. It might be hard to measure, but there is no question the wavelengths have to be shifted longer by the speed of a car departing.

Now, with respect to Doppler effect here on earth, if the speed of the wave is much greater than the speed of the moving body, and the direction of the body is the same as the direction of the wave, the formula that relates the initial and the received frequency (

f) or wavelength (

l) is quite simple (

I make use of

c = f*l):

f' = (1 + v/c)*fº l' = (1 + v/c)^-1*lº = lº/(1 + v/c)
Where

c is the speed of the wave (EM wave in this case, so ~

3e8 m/s) and v is the negative of the relative velocity and hence the coefficient (

1 + v/c) is greater than 1 if the moving body is approaching the emitter (

v is positive) and less than 1 if it's moving away from it, which implies that in the first case the frequency increases and in the second case decreases, and the opposite with the wavelength. Which is what we know intuitively.

At a given velocity of the moving body, the ratio change of frequency or wavelength is the same, independently of the frequency itself, meaning that a microwave or visible light will be subjected to the same change, proportionally. If the relative velocity is

30 m/s, then

1 + v/c = 1 + 3e1/3e8 = 1 + 1e-7 = 1.0000001, which is is the same coeficient in both cases, so the change is tiny in both cases in relative terms. That explains why

@tonycollinet doesn't appreciate a change in color when looking at a moving red car, no matter how hard he tries. However, with eyes sensible to microwaves, I don't think he could do any better either.

In absolute terms, just for curiosity, we can calculate what is called the doppler frequency (

fd) or the difference in frequency:

fd = 2v*fº/(c - v) ~ 2v*fº/c
And if we use a microwave of lets say

10 GHz = 1e10 Hz and a visible light of

400 THz = 4e14 Hz, that is, one

4e4 greater than the other, the difference in frequency in both cases is:

fd_microwave = 2*30*1e10/3e8 = 6e11/3e8 = 2e3 = 2000 Hz

fd_visible = 2*30*4e14/3e8 = 2.4e16/3e8 = 8e7 = 80000000 Hz
exactly,

4e4 times greater than the other xD

So in absolute terms of course the difference in frequency will be quite different.

Please correct me if I'm missing something or (specially) if not getting it at all.