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Since frequency of a wave is a function of time, then for a particular ray of the wave, why would the frequency remain the same when observed from a moving reference frame? The frequency should change according to time dilation. No?

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Who says it doesn't? –  Emilio Pisanty Mar 1 '13 at 13:42

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Yes, the frequency does change. This is a component of the relativistic Doppler effect.

The effect comes from two sources: first, there is the classical (nonrelativistic) Doppler effect, which comes from the fact that successive crests of the wave are emitted from closer and closer to the receiver (I use the case where the emitter is moving toward the receiver). This changes the frequency as

$$f_\text{received} = \biggl(1 + \frac{v}{c}\biggr)f_\text{emitted}$$

That part has nothing to do with relativity; it's responsible for the Doppler effect we hear in sound waves, for example.

Then on top of that, for light, the frequency is subject to the Lorentz transformation. Since frequency is the reciprocal of a time, it has the inverse of the transformation for time, namely $f\to \gamma f$. So the combined result from the two effects is

$$f_\text{received} = \gamma\biggl(1 + \frac{v}{c}\biggr)f_\text{emitted} = \sqrt{\frac{c + v}{c - v}}f_\text{emitted}$$

for relative approach.

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Being a BOF of many years training I would have preferred to see this proved directly using the Lorentz transforms. If I get time I'll update my answer ... –  John Rennie Mar 1 '13 at 9:00
Yeah, it is nice to see a proof, but I didn't have much time to put into this. It's just what I could type up in a couple minutes or so. –  David Z Mar 1 '13 at 9:04
Is the relativistic Doppler effect applicable for rays reflected from an object. i.e. is it applicable for an incoherent light source.... –  Physics n00b Mar 1 '13 at 9:14
@Physicsman Yes. –  daaxix Mar 2 '13 at 0:38
@Physicsman why did you ask if you don't want to accept the answer? In any case daaxix is right (and you are wrong), the effect applies to all electromagnetic waves. –  David Z Mar 2 '13 at 23:50

Both the the frequency and wavelength change. Obviously so because $\nu\lambda = c$ and $c$ is a constant. If you have a specific example of where you think the frequency doesn't change update your question with the details and I'll update my answer to address it.

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