It is true when a wave is traveling in a constant velocity, there is an inverse relationship between a wavelength and frequency. For example, if a wavelength is increasing, frequency should decrease because it is an inverse relationship.

However, what if the wave is not traveling constant?

Will the formula and the concepts change?


2 Answers 2


When a wave moves from one medium with different speeds, the frequency remains unchanged, and the wavelength changes to accommodate the new speed.


The wavelength cannot be changed by the motion of the observer, but the frequency and the speed of the waves (relative to the observer) do change:

http://physics.bu.edu/~redner/211-sp06/class19/class19_doppler.html "Let's say you, the observer, now move toward the source with velocity vO. You encounter more waves per unit time than you did before. Relative to you, the waves travel at a higher speed: v'=v+vO. The frequency of the waves you detect is higher, and is given by: f'=v'/λ=(v+vO)/λ."

Actually any interpretation of the Doppler effect (moving observer) proves, explicitly or implicitly, that the speed of light relative to the observer varies with the speed of the observer, in violation of Einstein's relativity:

http://www.hep.man.ac.uk/u/roger/PHYS10302/lecture18.pdf "The Doppler effect - changes in frequencies when sources or observers are in motion - is familiar to anyone who has stood at the roadside and watched (and listened) to the cars go by. It applies to all types of wave, not just sound. (...) Moving Observer. Now suppose the source is fixed but the observer is moving towards the source, with speed v. In time t, ct/λ waves pass a fixed point. A moving point adds another vt/λ. So f'=(c+v)/λ. (...) Relativistic Doppler Effect (...) If the source is regarded as fixed and the observer is moving, then the observer's clock runs slow. They will measure time intervals as being shorter than they are in the rest frame of the source, and so they will measure frequencies as being higher, again by a γ factor: f'=(1+v/c)γf..."

That is, according to the above interpretation,

f' = (c+v)/λ

when v is low (relativistic corrections are negligible), and

f' = (1+v/c)γf = γ(c+v)/λ

when v is high. Accordingly, the speed of the light relative to the moving observer is

c' = c+v

when v is low, and

c' = γ(c+v)

when v is high. Einstein's relativity is violated in either case.

  • 2
    $\begingroup$ Your claims about relativity are incorrect; the Doppler formula is altered for light, and shows no violation. In addition, this response does not address the OP. $\endgroup$ Apr 5, 2016 at 0:28

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