 Solution

Let the source's proper emission period be $T_0$, and let it approach a stationary observer directly with speed $v=\beta c>0$. In the observer's inertial frame, consecutive emissions are separated by $\gamma T_0$ through time dilation. The second crest starts $v\gamma T_0$ nearer the observer, reducing the arrival interval toSince light travels at $c$ in each inertial frame, the longitudinal relativistic Doppler effect isRecession reverses the sign of $\beta$. More generally, if $\Theta$ is the angle between the source velocity and the photon propagation direction in the observer's frame, a Lorentz transformation of the photon four-momentum gives ${\nu }_0=\gamma {\nu }_{\mathrm{obs}}(1-\beta \cos \Theta )$.

The common Doppler scaling of spectral-line separations follows because every line receives the same wavelength factor $D$. The supplied wavelengths have rest separation $50\mathrm{nm}$; the measured separation gives $D=51/50=1.02$. Defining ${\beta }_{\mathrm{rec}}>0$ for recession, inversion givesSource contradiction: a strictly approaching source has $D<1$ and must reduce the wavelength separation below $50\mathrm{nm}$, so no positive approach speed satisfies the stated $51\mathrm{nm}$. For example, approach at $0.02c$ gives a separation of $49.0098\mathrm{nm}$. The speed inferred from the numerical data is therefore about $0.0198c$ receding; retaining the word “approaching” would require a negative signed approach velocity. The two observed wavelengths consistent with the given separation are $357$ and $408\mathrm{nm}$.