http://www.phys.uconn.edu/~gibson/No...6_3/Sec6_3.htm
Professor George N. Gibson, University of Connecticut: "To understand the moving observer, imagine you are in a motorboat on the ocean. If you are not moving, the boat will bob up and down with a certain frequency determined by the ocean waves coming in. However, imagine that you are moving into the waves fairly quickly. You will find that you bob up and down more rapidly, because you hit the crests of the waves sooner than if you were not moving. So, the frequency of the waves appears to be higher to you than if you were not moving. Notice, the waves themselves have not changed, only your experience of them. Nevertheless, you would say that the frequency has increased. Now imagine that you are returning to shore, and so you are traveling in the same direction as the waves. In this case, the waves may still overtake you, but at a much slower rate - you will bob up and down more slowly. In fact, if you travel with exactly the same speed as the waves, you will not bob up and down at all. The same thing is true for sound waves, or ANY OTHER WAVES. (...) The formula for the frequency that the observer will detect depends on the speed of the observer; the larger the speed the greater the effect. If we call the speed of the observer, Vo, the frequency the observer detects will be: f'=f(1+Vo/Vwave). Here, f is the original frequency and Vwave is the speed of the wave."

For light waves we have

f'= f(1+Vo/Vwave) = f(1+Vo/c) = (c+Vo)/λ

where λ is the wavelength and c'=c+Vo is the speed of the waves relative to the moving observer.

Clearly the speed of light relative to the observer varies with the speed of the observer, in violation of Einstein's relativity.

Taking into account the relativistic corrections (time dilation) will change the above formula into

f' = γ(c+Vo)/λ

where γ is the Lorentz factor. In this case the speed of the light relative to the moving observer is

c' = γ(c+Vo)

and Einstein's relativity is violated again.

Pentcho Valev