Q12CQ - College Physics (Urone)

Question

Is the relativistic Doppler effect consistent with the classical Doppler effect in the respect that \({{\rm{\lambda }}_{{\rm{obs}}}}\) is larger for motion away?

 

Step-by-Step Solution

Answer

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.

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.