 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q12CQ

Expert-verified Found in: Page 1028 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Is the relativistic Doppler effect consistent with the classical Doppler effect in the respect that $${{\rm{\lambda }}_{{\rm{obs}}}}$$ is larger for motion away?

As, the measured frequency is less than the source frequency, the observed wavelength is greater than the emitted wavelength. As a result, the relativistic Doppler effect and the traditional Doppler effect are compatible.

See the step by step solution

## Step 1: Define Special Relativity

The special theory of relativity, sometimes known as special relativity, is a physical theory that describes how space and time interact. Theoretically, this is known as STR theory.

## Step 2: Explanation

The equation used is: $${{\rm{f}}_{{\rm{obs}}}}{\rm{ = }}\sqrt {\frac{{{\rm{1 - v/c}}}}{{{\rm{1 + v/c}}}}} {{\rm{f}}_{{\rm{source}}}}$$.

It gives the relativistic Doppler effect for a receding source. It is due to the measured frequency is less than the source frequency, the observed wavelength is greater than the emitted wavelength.

Therefore, the observed frequency is seen smaller than the source frequency. It means that the observed wavelength is larger than the emitted one. So, accordingly, the relativistic Doppler effect is said to be consistent with the classical Doppler effect. ### Want to see more solutions like these? 