Observed time period in distant clock when moving towards it

Question

Imagine, 2 persons ('A' & 'B') are 6 light years apart in space, stationary to each other and with no gravitation acting on anybody. Suppose 'B' starts his clock which also shows years, months and days. also lets say he flashes a signal at the same time. When this signal reaches at 'A' he starts his own clock which also shows year, month and time and also flashes a signal. But now another 6 year passes for B to see the clock of 'A' which is just started. now they can see each others clock and the ticks of the both clocks are in sync even though they show different timings.

from 'A's perspective his own clock shows same time as that of the clock of 'B'. and from 'B's perspective 'B' clock running 12 years ahead of 'A's clock. And They just remain in this setting for a brief time.

Now 'B' starts his journey towards 'A' with speed of 0.6c. lets say 'B' achieve his speed (0.6c) in 1 second (for the sake of simplicity). Lets say from 'B's perspective his clock showing 24 years and 'A's clock showing 12 years just before starting the journey.

from 'B' perspective the distance next to him will be contracted to 4.8 light years due to length contraction and hence he'll reach at 'A' after 8 years. so his final clock will show 32 years when he reach at 'A'.

from 'A' perspective the journey of 'B' will start when he see time 24 years in both the clocks but if he remove signal delay from his clock as the journey of 'B' starts, he can say that his actual clock was at 18 years when the journey started and also it'll take actual 10 years for 'B' to reach at him. So when 'B' reaches at the position of 'A' their respective clocks will show 32 years('B's clock) and 28 years ('A's clock).

first of all, IS MY MATH CORRECT ABOUT THE TIMINGS OF CLOCK 'A' AND 'B' WHEN 'B' REACHES AT THE POSITION OF 'A'?

Go further only If I am correct about the former question.

In reality, without the consideration of signal delays, both will perceive each others clocks as ticking faster.

From 'A's perspective the journey actually started when he see 24 years in each clock. And 4 more years pass for him When 'B' reaches at the position of 'A'.
So Does 'A' actually perceive 'B' as moving and the respective relativistic doppler shift only for 4 years? And what can he say about the velocity of the 'B'? As he moved 6 light years in just 4 years.

Also from 'B's perspective the clock of 'A' showing 12 years when he starts his journey. but similar as before if signal delay is removed, he can say that actual time is 18 years in the clock of 'A' just before departure. but then as he takes 8 years to reach at the position of 'A', 'A's clock should be time dilated and should pass only 6.4 years but that is not the case.

is it because of the acceleration of 'B' at the start of the journey there are 3.6 more years in 'A's clock?

Answer 1

first of all, IS MY MATH CORRECT ABOUT THE TIMINGS OF CLOCK 'A' AND 'B' WHEN 'B' REACHES AT THE POSITION OF 'A'?

Yes, up to this point your calculations appear to be correct. When they meet, $B$'s clock will show $32 \ \text{years}$ and $A$'s clock will show $28 \ \text{years}$.

But does 'A' actually perceive 'B' as moving and the respective relativistic doppler shift only for 4 years?

Yes, $A$ only sees a doppler shift for $4 \ \text{years}$ of his time, but he knows that due to light travel time, $B$ must have started moving $6 \ \text{years}$ prior to that making a total travel time of $10 \ \text{years}$ in $A$'s reference frame.

And what can he say about the velocity of the 'B'? As he moved 6 light years in just 4 years.

No one measures $B$ to travel $6 \ \text{lightyears}$ in $4 \ \text{years}$. $A$ measures $B$ to travel $6 \ \text{lightyears}$ in $10 \ \text{years}$ ($v=0.6c$) and $B$ measures $A$ to travel $4.8 \ \text{lightyears}$ in $8 \ \text{years}$ ($v=0.6c$).

Also from 'B's perspective the clock of 'A' showing 12 years when he starts his journey. but similar as before if removes the signal delay he can say that actual time is 18 years in the clock of 'A' just before departure. but then as he takes 8 years to reach at the position of 'A', 'A's clock should be time dilated and should pass only 6.4 years but that is not the case.

When $B$ accelerates, he has to re-synchronise his clocks and when he does that he will see that $A$'s clock is ahead by a factor of $3.6 \ \text{years}$ due to the relativity of simultaneity. Instead of reading $18 \ \text{years}$ at the time of acceleration, $A$'s clock now reads $21.6 \ \text{years}$ in $B$'s reference frame and $A$'s clock advances by $6.4 \ \text{years}$ in the $8 \ \text{years}$ measured by $B$ for $A$ to travel to $B$, giving the expected $0.8$ time dilation factor.

If an object of length $\ell$ is moving relative to us and there are clocks at either end of it, then the relativity of simultaneity formula $$ \Delta t = v \ell$$ tells us how much the clock at the rear is ahead of the clock at the front. In our case $A$ and $B$ are $6 \ \text{lightyears}$ apart and the relative velocity is $0.6c$, so the result is $v \ell = 0.6 \times 6 = 3.6 \ \text{years}$. Add this to $18$ and you get $21.6 \ \text{years}$. That is the missing $3.6 \ \text{years}$ you are looking for. $A$'s clock advances $28-21.6 = 6.4 \ \text{years}$ in $B$'s reference frame and this is equal to $8 \ \text{years}$ time dilated by a factor of $0.8$ as expected.

is it because of the acceleration of 'B' at the start of the journey there are 3.6 more years in 'A's clock?

The acceleration per se does not affect the time dilation. (See my $5$ counter examples to the claim that acceleration affects time dilation.)

It is the fact that $B$ changes to a different frame of reference that artificially creates a forward jump in $A$'s time from $B$'s point of view.

$B$ makes odd measurements because $B$ is not a valid inertial reference frame but accelerates in the middle. The best way to analyse the problem is to have a third observer $C$ that is always moving at $0.6c$ relative to $A$. $C$ is a valid inertial reference frame and things make sense if you draw the time space diagram from $C$'s point of view.

Also, if you want to make things simpler for yourself to understand and make it easier for people to understand you, you really should use the standard form of time synchronisation. That means when $B$ receives a timing signal from $A$ of $0 \ \text{years}$, $B$ should set his own clock to $+6 \ \text{years}$, to allow for the light travel time. Now the clocks are properly synchronised in $B$'s reference frame. When both clocks are stationary, $A$ sees signals from $B$ as $6 \ \text{years}$ behind and $B$ sees signals from $A$ as $6 \ \text{years}$ behind, and things are symmetrical. Both consider each others clocks to be running at the same rate and reading the same time.

In relativity, an inertial reference frame is not just one clock, but an array of rulers and clocks that potentially stretches to infinity.

When both are stationary with respect to each other, $A$ sets up his clocks and can lay them out all the way to $B$'s location and synchronises all his clocks to each other. $B$ does the same and sets clocks out at regular intervals all the way to $A$'s location, synchronising all his clocks. Now, for every clock belonging to $A$ there is a clock belonging to $B$ right next to it and there really is no good reason to set all of $B$'s clocks to read $6 \ \text{years}$ more than all of $A$'s clocks when they are right next to each other and stationary with respect to each other. It just adds a layer of confusion.

If you want to use the standard Lorentz transformations, it makes sense to use the standard clock synchronization method as that is assumed in the equations.