Can Emission Theory Produce Doppler-Shift Formula?

  • #1
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Emission theories propose that the velocity of light depends on the velocity of the light source. But the ordinary Doppler effect assumes the velocity of light remains as ##c## with respect to the ether medium, even when the light source is moving at speed ##v## with respect to the ether. They are saying different things, so how could they be consistent?

Explicitly, the original-source emission theory suggests that the velocity of light measured in the ether frame is ##c+v## when the source moves ahead at speed ##v## with respect to the ether, while the Doppler effect says the velocity of light is still ##c##.

Further, for a mirror moving at speed ##v## (with respect to the ether) towards a light source (stationary with respect to the ether), the Doppler effect gives the frequency of the reflected light as ##f=\frac{c+v}{c-v}f_o##, where ##f_o## is the frequency of the incident light. I don't understand how the original-source emission theory could come up with the same formula.

Explanations of the various emission theories are in the text below.

Source: Introduction to Special Relativity by Robert Resnick
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  • #2
Just compute the travel of two wave crests under the assumptions of the original emission theory as you quoted it. You readily arrive at the SR correct formula. Note that this analysis is the same as for SR in the emitter frame.
 

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