I'm using the illustration from this question:

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Suppose A and B are d light years away, and at rest. Then they symmetrically start to travel toward each other (symmetric acceleration process in negligible time) at a high speed v. How each one sees the other one's clock working?

"Always faster" and "always slower" are both logically impossible, since due to the symmetry, they must agree that same time has elapsed on both clocks when they meet. So it's either "always normal" or probably "faster for some time and slower for some time". What's the mathematical expression for the seeming rate of clocks?

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    $\begingroup$ The logical flaw in your question is in "they symmetrically start to travel". By "symmetrically" you mean simultaneously. However, simultainety is relative. Each one will see the other starting to travel later. This question is essentially a duplicate of the one you asked yesterday. The Wiki page you link there explains everything in details. You should read it with attention and, if still confused, ask about the particular section or formula in the Wiki article that seems unclear. $\endgroup$ – safesphere May 26 '19 at 7:36
  • $\begingroup$ At the beginning they are at the same from of reference. Let's say there's a pulse generator also at rest, exactly midway the distance. Travelers agree to start moving once they received the pulse. Isn't this simultaneity? $\endgroup$ – Asmani May 26 '19 at 7:51
  • $\begingroup$ Furthermore, if A and B believe in the exactly same type/amount of asymmetry, then the situation is indeed symmetric, which means they're both (symmetrically!) wrong. Are you saying that it could never be the case that two physicist travelers meet each other and both their observations and calculations about their journey (written on paper and exchanged when they meet) be identical? $\endgroup$ – Asmani May 26 '19 at 8:01
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    $\begingroup$ Everything is fine. Yes, there is a frame in which they have started moving simultaneously. One may say that they started moving at different times in all other relatively moving frames, but that doesn't eliminate frame of "simultaneous start" .Do you ask what clock rate they will see measuring relativistic Doppler Shift? $\endgroup$ – Albert May 26 '19 at 8:03
  • $\begingroup$ "Isn't this simultaneity?" - No, not according to the travelers. It is a common mistake to think of "A relative to B in the frame of C". No such concept exists in relativity, no third parties, but only "A relative to B" and "B relative to A". In your scenario, each traveler would see the other departing later, so their notes would perfectly match. The Wiki page explains that time of a remote traveler moves slower, but the Doppler effect makes an illusion of it moving faster. Read the article. $\endgroup$ – safesphere May 26 '19 at 9:19

There is some difference between “seeing” and “calculation” or “interpretation”. At relativistic velocities you must include time dilation into Doppler Shift, either to A-clock or to B-clock or, as in your case - to the both. That does not affect measured frequency shift, but changes judgment regarding the speed of ticking of the "other clock".

Approaching each other the both will "see" or measure very high frequency, so they will "see" each other clock ticking very fast. However, that says nothing about “actual” clock rate of any of them; it is frame dependent and is a matter of interpretation.

Relativistic Doppler Effect is very simple – it is “ordinary” Doppler shift plus time dilation. If you think that B is moving in the frame of A, then you attach time dilation to B. If you think that A is moving in the frame of B, then you attach time dilation to A. If you think that they both move with equal but opposite velocities, then you can share full amount of time dilation – Lorentz factor in equal proportions. Measured blueshift of frequency will be the same, but contributions of time dilation will be frame dependent.

Let's B is emitting monochromatic light towards A. For example, one may think like that: A is at rest and B is moving towards A with velocity very, very close to c. B catches up all emitted by him wave fronts and emits new ones; hence the wavefronts gather straight in the front of him. These wavefronts will hit A almost at once, like fighter jet sonic boom. A will see very intense blueshift of frequency, but due to dilation of B clock it will be less intense than it could be in the classical case.

Or you may think that B is at rest and A is moving towards B. In this case maximum observed by A frequency (in classical case) should be $2f$ (wavefront is moving with velocity c and A is moving towards wavefronts with velocity close to c). But, due to dilation if A’s clock A will see this frequency as very intense (blueshift). A will explain it this way: my clock is ticking very slowly; I have turned into dawdler; that’s why I see all processes around me as very, very fast.

In your exact case every observer will explain or interpret that the other clock is ticking at the same rate as his own; B clock dilates and A clock dilates at the same magnitude and that time dilation cancel each other; so the measured Doppler Shift is no different from classic one.

So, since in your case they move with the same but opposite velocities, their clocks slow down at the same rate indeed.

Please look for relativistic Doppler Shift in Feynman Lectures (Relativistic Effects in Radiation).

Finally, if the both move at parallel lines, at certain moment (points of closest approach) they may see purely clock rate of each other clock – Transverse Doppler Effect. For "moving" source - light emitted at points of closest approach redshifs ; for "moving observer" - light received at points of closest approach blueshifts.

Sure, they will measure zero - Transverse Doppler Effect for the frame, in which they move with equal but opposite velocities,

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  • $\begingroup$ "In your exact case every observer will explain or interpret that the other clock is ticking at the same rate as his own; B clock dilates and A clock dilates at the same magnitude and that time dilation cancel each other; so the measured Doppler Shift is no different from classic one." - This doesn't sound right. In which frame both time dilations cancel? (And what does this even mean?) $\endgroup$ – safesphere May 26 '19 at 9:28
  • $\begingroup$ @safesphere. Could you please read carefully "transverse Doppler Effect" chapter, in particular fig. 4 and corresponding explanation, As well figures 2, 3 ,6. I have posted a link in my answer. Fig. 2 - an observer is moving. Fig. 3 - a source is moving. Fig 4 - the both are moving, there is no transverse doppler effect, i.e. relative time dilation. Again: en.wikipedia.org/wiki/Relativistic_Doppler_effect $\endgroup$ – Albert May 26 '19 at 9:37
  • $\begingroup$ It is very simple to introduce a frame, in which the both move with the same, but opposite velocities. It is the case, when two observers rotate on opposite ends of the rim of a centrifuge - this is Champeney and Moon experiment, it demonstrates absence of relative time dilation. Obviously, because their clock dilate at the same magnitude. If you have some doubts about their non - inertiality, simply assume momentarily co-moving and coinciding with them nertial frame. Nothing changes then, absence of relative time dilation. The article in Wikipedia explains everything very clearly. $\endgroup$ – Albert May 26 '19 at 9:58
  • $\begingroup$ I see now what you mean, but you did not clarify this in the part I quoted. You said, "each observer", and then, "dilation cancel", without stating that the cancelation is in the frame of a third party. The same applies to the Doppler shift. You should be more careful in these descriptions without assuming that others know what you think and imply without stating it clearly. $\endgroup$ – safesphere May 26 '19 at 10:28
  • $\begingroup$ I beleive there is a logical flaw by attaching "frame" to myself and see all others as moving ones. If I look through a window at another relatively moving train and see, that either the train is moving and his clock is ticking slower, or I move myself relatively to the train and my clock is ticking slower. Or: the both opposite train and my train move with the same but opposite velocities. I don't need any third party. Do you wish to insist, that according to some teories, I cannot keep myself moving in the frame of the other train? Shoud you wish to have some fun, I can drop you one task. $\endgroup$ – Albert May 26 '19 at 10:36

One important assumption that I've omitted is that the travelers have had been at rest for long enough. Additionally let's assume they have synchronized their clocks.

"Always faster" is logically possible, and is indeed the case. How? Suppose $d=10$ light years, and $v$ is so that the travelers meet after 2 years, according to their own clocks. At the starting point of the journey, each one observes the other one's clock showing $R1=-10$ years. When they meet, the second reading will be $R2=2$ years. Therefore, the average observed rate of clocks will be $12/2=6x$.

More specifically, each one observes a step function: A constant rate of $S1$ for $t$ years, which jumps to $S2$, and remains constant for the remaining $2-t$ years. I'm not skilled enough to calculate $t$, $S1$ and $S2$, but obviously, $S2>S1>1$ and $(t*S1+(2-t)*S2)/2=6$.

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