If there is any superluminal speed measured by the observer in the relativistic rocket, how does it affect the Doppler effect?

Question

Assume that a point rocket, as well as a diode, moves with a constant velocity $v$ towards a planet on which an observer $A$ is located. (Locate the diode, rocket and $A$ respectively from left to right along a specific line.) Although the planet's mass is much greater than the rocket's, it is assumed that the gravitational field of the planet is negligible. The diode is separated by a distance of $x$ from the rocket both approaching the planet at $v$ relative to $A$. It is evident that the mentioned distance is measured $\gamma_vx$ by an observer $B$ located in the rocket's rest frame where $\gamma_v=1/\sqrt{1-v^2/c^2}$.

If the rocket hits the planet undergoing, say, a very great constant acceleration of $a$ through an inelastic collision as measured by $A$, it is anticipated that the measurements of $B$ suddenly become similar to those of $A$ since $B$'s frame of reference, after rocket's deceleration, shifts into $A$'s. Therefore, $B$ claims that the diode's distance of $\gamma_vx$ shall abruptly collapse into $x$ the same as measured by $A$.

If the deceleration is great enough, does this mean that the diode's speed can exceed that of light from the viewpoint of $B$ during the tiny time-interval of deceleration? One may say that because $B$ is non-inertial, he allows to measure superluminal speeds, however, my problem is bout how this superluminal speed affects the relativistic Doppler effect? If $B$ observes that the diode sends a photon (with a frequency of $\nu_0$ measured in the diode's rest frame), how does this frequency change for the very moment at which the diode has a velocity greater than $c$? Can we say that the frequency does not change because, as well as the diode's velocity, the velocity of the photon exceeds $c$, or is there a considerable abrupt blueshift for the photon at that moment otherwise? What is the correct Doppler equation for this moment? (Forget about the additional frequency change due to the gravitational blueshift during deceleration.)

Answer 1

Unfortunately, the question is somewhat incomplete as stated. Since B is non-inertial there is no unique standard meaning to the term “B’s frame of reference”, and it is not necessarily correct that the coordinates of B become suddenly similar to A’s. Depending on the exact definition it may be sudden or gradual and depending on the exact definition the diode may have $v>c$ or not*. And critically to your question the Doppler shift depends on the exact definition of the reference frame.

One of my favorite methods for exactly defining coordinates in a non-inertial frame is Dolby and Gull’s radar coordinates. In those coordinates light always moves at $c$ and all objects have $v<c$ at all times. However, in those coordinates the transition to matching A’s coordinates is not sudden but instead propagates outward at $c$.

Although that is my favorite, there is no requirement to use it and you are free to choose your own preferred method. However, you do need to be explicit.

Without being explicit all that can be said is generic statements based on tensors. For example, we can say that the frequency received from the diode is $g_{\mu\nu}K^{\mu}U^{\nu}$ which would be the final measured frequency and is manifestly invariant. But we cannot say what is the timelike component of $K$, which would be the gravitational part of the Doppler shift.

*Note that the diode is never superluminal as “superluminal” means “faster than light” and not “v>c” and in non-inertial frames those are not the same thing