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

In the classical Doppler effect, the change in the observed frequency (and wavelength) is due to the relative motion of the source and the observer. The classical Doppler effect for light can be given by the following formula: \(f_{obs} = \frac{c \pm v_{obs}}{c \pm v_{src}} \cdot f_{src}\) Here, \(f_{src}\) is the emitted frequency, \(f_{obs}\) is the observed frequency, \(v_{obs}\) is the velocity of the observer, \(v_{src}\) is the velocity of the source, and \(c\) is the speed of light. The choice of signs (positive or negative) depends on the relative motion between the source and the observer (away or towards). Now, we need to find the relation between observed and emitted wavelength to figure out if the observed wavelength is larger for motion away. We know the relation between frequency and wavelength: \(f = \frac{c}{\lambda}\) Applying this relation to both the emitted and observed frequencies and solving for \(\lambda_{obs}\): \(\lambda_{obs} = \frac{c \pm v_{obs}}{c \pm v_{src}} \cdot \lambda_{src}\)

The relativistic Doppler effect takes into account the time dilation experienced due to relative motion. The relation for the relativistic Doppler effect is: \(f_{obs} = \frac{1}{\gamma (1 \pm \beta)} \cdot f_{src}\) Here, \(\beta = \frac{v}{c}\) (v is the relative velocity) and \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\) is the Lorentz factor. The choice of signs depends on the relative motion between the source and the observer (away or towards). Similarly, we can find the relation between observed and emitted wavelength for the relativistic Doppler effect: \(\lambda_{obs} = \gamma (1 \pm \beta) \cdot \lambda_{src}\)

Now, let's compare both the classical and relativistic formulas to see if the observed wavelength is larger for motion away: - Classical Doppler effect when motion is away: \(\lambda_{obs} = \frac{c + v}{c - v} \cdot \lambda_{src}\) Since \(\frac{c+v}{c-v}>1\) (as \(v>0\)), \(\lambda_{obs} > \lambda_{src}\). - Relativistic Doppler effect when motion is away: \(\lambda_{obs} = \gamma (1 + \beta) \cdot \lambda_{src}\) Since \(\gamma \geq 1\) and \(\beta > 0\), \(\lambda_{obs} > \lambda_{src}\). Therefore, both the classical and relativistic Doppler effects are consistent with the behavior that the observed wavelength is larger for motion away from the observer.