Inertial Frames distinguished by proper times

[yogi]

A first spaceship S1 departs from Earth and quickly accelerates to a velocity V = c/2. S1 travels the shortest path (dead heading) toward a distant planet Alpha so that it arrives in 20 years as measured by a clock on S1. One year after S1 is launched from Earth as measured by a clock on the earth, a second spaceship S2 is launched along the same trajectory (we assume neither Earth nor alpha have moved during this experiment). S2 quickly accelerates to a greater velocity than c/2 and at some point S2 overtakes S1. When S2 arrives at Alpha, the clock aboard S2 will read a lapse time of 10 years since it departed from earth.

Both S1 and S2 move at uniform velocity, so each is a valid inertial system, but the proper clock rate of the S2 clock is 1/2 the proper clock rate of the S1 clock. Based upon the difference in proper rates - if observers on each spaceship take the measure of the other as S2 passes S1, will an observer in S1 arrive at the same conclusion about contraction and clock rate in the S2 frame as S1 determines about contraction and clock rate in the S2 frame?

 

[Ich]

The proper clock rate in S1 and S2 is 1s/s. Proper (German: "eigen") means belonging to itself, contrary to "with respect to". Try to remember this before posting.

 

[pervect]

S2 quickly accelerates to a greater velocity than c/2 and at some point S2 overtakes S1. When S2 arrives at Alpha, the clock aboard S2 will read a lapse time of 10 years since it departed from earth.

Both S1 and S2 move at uniform velocity, so each is a valid inertial system

S2 is in at least two different inertial systems, because it accelerates. It's unclear if S1 breaks to a stop or not - if S1 does break to a stop, it accelerates as well.

Accelearting clocks are not inertial clocks by definition.

 

[robphy]

Is this a fair rephrasing of the problem?

The Earth and alpha are inertial and at rest with respect to each other (so, they have parallel worldlines, vertical in their frame).

Ignoring the time when S1 was on the Earth awaiting launch... at event P, S1 travels inertially toward alpha at speed c/2 and arrives at alpha at event P', after 20 years on S1's clock.

Ignoring the time when S2 was on the Earth awaiting launch... at event Q, one year after P according to the Earth's clocks, S2 travels inertially at toward alpha some speed greater than c/2 so as to overtake S1. That is, their worldlines cross before alpha. S2 arrives on alpha at event Q', after 10 years on S2's clock.

Is this the situation?

If so, then

Both S1 and S2 move at uniform velocity, so each is a valid inertial system,

true

but the proper clock rate of the S2 clock is 1/2 the proper clock rate of the S1 clock.

false, as Ich says. The "proper clock rate of S1" is the clock rate of S1 measured by clock S1. Likewise, for S2.

Based upon the difference in proper rates - if observers on each spaceship take the measure of the other as S2 passes S1, will an observer in S1 arrive at the same conclusion about contraction and clock rate in the S2 frame as S1 determines about contraction and clock rate in the S2 frame?

What it all boils down to is that you have two intersecting inertial worldlines.

S1 and S2 are inertial observers, between P and P' and between Q and Q', respectively. So, identical experiments they perform on each other (e.g. comparing doppler effects or time dilation effects) will agree.

In fact, their relative-velocity can be computed from the data you've given. I get 0.41c. (S2's velocity must be 0.755c and alpha is 11.5 light years away.) I'm too lazy right now to calculate the intersection event... but according to the earth, S1 will arrive after 23 years, and S2 after 16 years (including the 1 year wait). I hope I didn't make any mistakes.

You can figure out when and where they intersect according to the Earth frame and what each clock reads at that meeting.

 

[yogi]

Robphy:

"Is this a fair rephrasing of the problem?

The Earth and alpha are inertial and at rest with respect to each other (so, they have parallel worldlines, vertical in their frame).

Ignoring the time when S1 was on the Earth awaiting launch... at event P, S1 travels inertially toward alpha at speed c/2 and arrives at alpha at event P', after 20 years on S1's clock.

Ignoring the time when S2 was on the Earth awaiting launch... at event Q, one year after P according to the Earth's clocks, S2 travels inertially at toward alpha some speed greater than c/2 so as to overtake S1. That is, their worldlines cross before alpha. S2 arrives on alpha at event Q', after 10 years on S2's clock.

Is this the situation?"

Yes - that is a correct. We have two intersecting world lines, and S1 and S2 are both inertial observers whose local clocks are running at different rates. And yes the relative velocity can be computed - but the actual numbers are immaterial, as is the location where S2 overtakes S1. The issue is whether there is any experiment that can be performed by the spaceships themselves to detect the difference between the intrinsic rate of S1 and S2. Between the two spatial points Earth and alpha the S1 clock records 20 years, the S2 clock records 10 years. The proper time interval is the time recorded by a clock attached to the observed body. While it is true that an observer in SI will always measure the S1 clock to be runiing at one minute per minute, and likewise an observer in S2 will measure the S2 rate as one minute per minute, there is nontheless a difference in how the proper rates get established ...the ratio of 20 years per 20 years and 10 years per 10 years. Is there a way to recover by experiment the difference between the numbers that formed the ratios in the two systems?

 

[RandallB]

Assuming “quickly accelerates” = instant acceleration to use straight SR no GR.
I get pretty much what robphy gets:

S1 v= 0.5c
S2 v= 0.756
Distance to travel 11.58 LY
S1 23.16 Earth time (20 S1 time)
S2 16.32 Earth time (10 S2 time 11 including wait)

S1 time .866 of earth
S2 time .655 of Earth

Relative S1-S2 v= 0.411
S2 time .912 of S1 (& S2 time .912 of S1 of course)

They meet at 1.475 LY from earth
2.95 Earth time
2.6 s1 time
2.27 S2 time (including the 1 y wait on earth)
1.27 S2 elapsed time
(note: S2 had two different speeds to build 2.27 time since being together with S1)

Classical SR math.

 

[robphy]

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Here's my calculation using rapidities... analogous to angles in Euclidean trigonometry... in Maple code.

restart;
theta:=arctanh(1/2.);
                        theta := 0.5493061443
d:=20*sinh(theta);
                           d := 11.54700538
phi:=arcsinh(d/10.);
                         phi := 0.9866469608
tanh(phi);
                             0.7559289459
tanh(phi-theta);
                             0.4114378276
20*cosh(theta);
                             23.09401076
1+10*cosh(phi);
                             16.27525231
cosh(phi-theta);
                             1.097167541

with a little more trigonometry, you can get everything else you would want. (beta=(v/c)=tanh(rapidity), gamma=cosh(rapidity))

One can use radar methods to operationally measure all of the kinematic quantities.

 

[yogi]

RandallB/robphy - More than I expected - good tutorial(s). Thanks. My interest in posing the question relates the information which is lost (or apparently lost) if we simply consider the two spaceships S! and S2 passing each other. Without the the benefit of knowing the initial conditions, SR predicts that the S1 and S2 frames are equivalent (equally valid inertial frames as they pass one another with relative velocity v) - and accordingly each would arrrive at the same conclusion(s) regarding contraction and clock rate in the other frame (actually Einstein never really said this in his 1905 paper, but SR is generally given this interpretation). So even though the S1 clock and the S2 clock are running at their own proper rate of one second per second, a second in S1 is different than a second in S2 when both are observed in the earth-Alpha frame. So when an observer in S2 uses his watch to make a measurement of the apparent length of S1, he will be using a different time base than the S1 observer uses his watch to measure the apparent length of S2.

There is a common Gamma factor, but we have different times T1 and T2 to use as a bases for dilation. So is it correct to conclude that the S1 observer figures the T2 time dilation based upon the value of a second in the T1 frame, and the S2 observer figures the T1 time dilation based upon the value of a second in the S2 frame? If they transmit their results to the earth, their conclusions will not be the same because each is arriving at the value of a dilated second in the other frame in terms of his own time.

 

[robphy]

I'm not sure what the times T1 and T2 mean precisely.

Are they the elapsed times for each observer to complete their trips?
If so, just realize that these a proper times between two different sets of events... (it just so happens that the launch events occurred on the Earth's worldline and the arrival events occurred on alpha's worldline). So, there is nothing to be found in comparing T1 and T2.

If not,...
Are they the observed time-dilations in studying the duration of one tick of the clock? That is, is T1 the duration of 1 tick of S1's clock , as determined by S2? More precisely, let A and B be events on S1's worldline corresponding to two consecutive ticks of his clock... so, S1 says the "time-difference between A and B" is 1 tick. Is "T1 ticks" what S2 says is the "time-difference between A and B"?
If so, then (by the relativistic symmetry of these inertial observers.. which can be borne out by a radar experiment) T1=T2, which is essentially gamma, which is a function of the relative-speed between these inertial observers.

In any case, consider the following experiment.
One tick after their meeting (when S1 and S2 meet, as S2 overtakes S1), each observer sends off a light signal to the other. Each observer will measure the same elapsed time on his clock when that signal arrives from the other. So, their identical procedures yields identical measurements (which they can write down in their log books for later comparison)... reflecting their SR-symmetry. (In fact, if each observer understands that the received signal was sent after one tick according to the emitter, the ratio of these proper time-intervals (1 tick for the emitter, T1 ticks for the receiver) gives the Doppler factor, k, which can be used to determine their relative-speed.)

 

[yogi]

Thanks robphy - i want to ponder your response before replying

Yogi

 

[yogi]

T1 is simply an increment measured in the S1 frame - for example one second as measured by the clock in S1. Likewsie T2 is an increment measured in the S2 frame by the S2 clock - e.g., also one second as measured by the clock S2. If S1 sends out a light pulse every second as measured by his own clock, and S2 does the same as measured by his clock, an observer on earth, knowing the two recessional velocities, would be able to determine that one second as measured by the S1 clock is different than one second as measured by the S2 clock. In other words one second of proper time in S1 is different than one second of proper time in S2.

As I read your post 9 you are saying that, in accordance with SR, if S1 and S2 transmit back and forth radio signals they will arrive at reciprocal results - that is, as between S1 and S2 there is no significance to the fact that T1 seconds are not the same as T2 seconds, or in any event it cannot be determined by signals sent between S1 and S2 that there is a difference between T1 and T2 seconds, but only the relative velocity will be revealed.

But if S1 transmits a signal every second according to the S1 clock and S2 transmits a signal every second according to the S2 clock, then, since the relative velocity v between S1 and S2 is known, each can determine the extra delay caused by their relative recession velocity v (after S2 overtakes S1). Once the relative velocity is determined, either observer can compare his own one second clock ticks with the rate at which pulses are arriving from the other spaceship...After compensating for the changing distance, either observer will see that the received pulses are being sent at a different rate than one per second as measured by the clock in the receiving spaceship, and each will agree upon which spaceship has the longer seconds!

 

[jtbell]

But if S1 transmits a signal every second according to the S1 clock and S2 transmits a signal every second according to the S2 clock, then, since the relative velocity v between S1 and S2 is known, each can determine the extra delay caused by their relative recession velocity v (after S2 overtakes S1).

This is the relativistic Doppler effect. Both S1 and S2 observe the other's signals arriving at the rate

[tex]f_{observed} = f_{source}\sqrt{\frac{1+v/c}{1-v/c}}[/itex]<br /> <br /> where v is the relative velocity of the source and observer, and is positive if the source is approaching the observer. <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> either observer will see that the received pulses are being sent at a different rate than one per second as measured by the clock in the receiving spaceship, and each will agree upon which spaceship has the longer seconds! </div> </div> </blockquote><br /> If S1 and S2 send signals at the same rate, they each receive the other's signals at the same rate. By the formula above, they each compute the same source frequency.[/tex]

 

[Ich]

each will agree upon which spaceship has the longer seconds!

Yes. Always the other one.

 

[robphy]

T1 is simply an increment measured in the S1 frame - for example one second as measured by the clock in S1. Likewsie T2 is an increment measured in the S2 frame by the S2 clock - e.g., also one second as measured by the clock S2. If S1 sends out a light pulse every second as measured by his own clock, and S2 does the same as measured by his clock, an observer on earth, knowing the two recessional velocities, would be able to determine that one second as measured by the S1 clock is different than one second as measured by the S2 clock. In other words one second of proper time in S1 is different than one second of proper time in S2.

Just to clarify...
Technically speaking, "proper-time" is akin to an arc-length... it is a number (as the result of an integral or sum)... and it is invariant (all observers will agree on it). So, one second of proper time in S1 is THE SAME AS one second of proper time in S2. What is different, however, is that each clock has a different spacetime displacement-vector between successive ticks of his clock [i.e., each has a different unit timelike vector tangent to his inertial clock worldline]. In general, an observer will say that the time-components of these vectors (i.e. the apparent durations between ticks for each clock) are unequal. This is time-dilation,

As I read your post 9 you are saying that, in accordance with SR, if S1 and S2 transmit back and forth radio signals they will arrive at reciprocal results - that is, as between S1 and S2 there is no significance to the fact that T1 seconds are not the same as T2 seconds, or in any event it cannot be determined by signals sent between S1 and S2 that there is a difference between T1 and T2 seconds, but only the relative velocity will be revealed.

yes... regarding T1 and T2 to be the vectors descrbed above.

Conerning that last part of your post, jtbell and Ich addressed that. (I was too slow working on the first part )

 

[jtbell]

Conerning that last part of your post, jtbell and Ich addressed that. (I was too slow working on the first part )

Call it parallel processing on a small scale.

 

[yogi]

Granted, if the S1 signals are being sent at the same frequency as the S2 signals, each will draw the same conclusion. But the pulse rate frequencies are not the same. Each is using his own clock to measure one second; the S1 clock is running at a faster rate than the S2 clock.

Let me analogize to a GPS situation. One clock is considered to be synchronized with a point at the center of the Earth (the non rotating Earth centered reference frame) we call it E clock. A second clock C2 is first synchronized with E and then put into circular Earth orbit.

C2 will run at a different rate because of two factors (the height and the velocity). If C2 is corrected for altitude, what is left is an orbiting clock C2 that runs a slower rate than E clock.
In this case the distance does not vary - so the only correction would be a transverse Doppler - What is observed is C2 running slower than E. Seconds take longer in the Satellite frame than in the ECRF.. If both C2 and E are used to transmit pulses at one second intervals, the pulses received from C2 will have a slower frequency than one pulse per second when received by E and the pulses received from E will have a higher frequency than one pulse per second when measured by C2. C2 must be compensated to make the seconds equal. There is no ambiguity as to which clock is running slower and which clock must be compensated.

 

[yogi]

One more example using a linear experiment rather than a satellite clock in free fall - a pion created in the lab and qucikly accelerated to near c velocity. Here the pion clock can be considered to have sent two pulses, one at the time of creation and one at the time of disintegration. If the lifetime of an at rest pion (approx 0.02 usec) is extended by a factor of 20 because of its high velocity wrt the lab, the lab detector will receive 2 pulses in say 20 x 0.02usec and the lab observer will conclude that the pion clock is running slower than the lab clock. But if a lab transmitter is sending signals every 0.02 usec, then during the extended lifetime of the high velocity pion, 20 pulses would have been transmitted. If there is a hypothetical observer P attached to the pion, how could P reach the conclusion that the Earth clock is running slower than the pion clock?

 

[robphy]

One more example using a linear experiment rather than a satellite clock in free fall - a pion created in the lab and qucikly accelerated to near c velocity. Here the pion clock can be considered to have sent two pulses, one at the time of creation and one at the time of disintegration. If the lifetime of an at rest pion (approx 0.02 usec) is extended by a factor of 20 because of its high velocity wrt the lab, the lab detector will receive 2 pulses in say 20 x 0.02usec and the lab observer will conclude that the pion clock is running slower than the lab clock. But if a lab transmitter is sending signals every 0.02 usec, then during the extended lifetime of the high velocity pion, 20 pulses would have been transmitted. If there is a hypothetical observer P attached to the pion, how could P reach the conclusion that the Earth clock is running slower than the pion clock?

Using your data...
T=pion lifetime (0.02 us)
gamma=20
k=Doppler factor= gamma+sqrt(gamma+1)*sqrt(gamma-1)=39.97
(another formula: k=exp(arccosh(gamma)))

Note that gamma*T="20 x 0.02usec" is the "lab's apparent -duration of the period of the pion's signal-emissions". It is NOT the "lab's reception-period of the pion's signal-emissions", which is kT="39.97T".

Note further that, while the lab may emit 20 regular signals (at 0.02 us intervals according to the lab) in the lab-time gamma*T, the pion receives only the signal at the meeting... it does not receive any other. (To reach the pion before it dies, the lab must have sent a signal at T/k=0.02usec/39.97. Waiting for 0.02us is too long.) If the pion lived long enough, the pion would have received the lab's signal [sent at 0.02 us after meeting] when the pion's clock read kT=39.97*0.02 us... just like the lab's result.Using a notational variant of the diagram from https://www.physicsforums.com/showthread.php?t=113915

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[yogi]

robphy - You are assuming the lab transmitter and receiver to be at the starting point. We can place the lab transmitter and receiver midway between the starting point of the pion and the point of disintegration (for v=0.99c the disintegration length is only about 6 meters). With the lab transmitter and recorder so located, the pion will receive more pulses during the experiment than the lab recorder.

 

[robphy]

robphy - You are assuming the lab transmitter and receiver to be at the starting point. We can place the lab transmitter and receiver midway between the starting point of the pion and the point of disintegration (for v=0.99c the disintegration length is only about 6 meters). With the lab transmitter and recorder so located, the pion will receive more pulses during the experiment than the lab recorder.

I'll try to produce a diagram of what I think you are describing.
However, after a sketch I made on paper, the following issue presented itself to me.

The pion's worldline is a segment with a finite proper-time of 0.02us. With the lab "midway between the starting point of the pion and the point of disintegration", the lab worldline bisects the pion's worldline-segment. In order to make their experiments "identical", we should only consider the analogously bisected 0.02us-segment of the lab's worldline. Without that concession, the symmetry between the inertial observers is broken... one is allowed to live a longer proper-time than the other.

 

[Ich]

Without that concession, the symmetry between the inertial observers is broken... one is allowed to live a longer proper-time than the other.

Yes, that´s the point. The pion will do nothing but collect a blue-shifted series of pulses that have been emitted "long" before its creation. No way to claim that the time between the emission of the first and the last collected pulse corresponds to 0.02 µs in the lab frame.

 

[Ich]

To add another point of view:
Yogi, it seems like you´re still mixing up observers and reference frames. I guess that this creates the paradox you can´t resolve.
Classical example:
A (very long, very fast, gamma=2) train drives past your window. They switch the cabin lights on and off in intervals of 1 s.
My guess is that you´d expext your room to be lit in 2 s intervals.

 

[yogi]

Yes, that´s the point. The pion will do nothing but collect a blue-shifted series of pulses that have been emitted "long" before its creation. No way to claim that the time between the emission of the first and the last collected pulse corresponds to 0.02 µs in the lab frame.

The pion gets synchronized at its creation with the Earth clock - there can be another clock at the X origin in sync with the pion clock and the lab transmitter clock- nothing is transmitted before the pion is created - the lab transmiter is located at some point x=3 along the +x axis... the pion will intercept all of the pulses that were transmitted during the time it travels from the origin until it reaches x = 3. Let it be assumed that during the time the pion travels from x=0 to x=3, 10 pulses were transmitted by the lab clock located at 3. To make things simplier we will assume the pion generates a signal as it passes x=3
--this corresponds to 0.01usec of time on the pion clock. The pion would have received 10 signals and the lab would have received 2. How can an observer attached to the pion conclude that the Earth clock is running slower than the pion clock?

 

[yogi]

Ich - your post #22 - With a distributed light source the problem is not the same - if you have a single light source point, then I would expect to see flashes at a different frequency than what I would be transmitting if in fact the two trains had been originally synchronized, and it was the other train that had been accelerated (i.e., the Gamma factor of 2 arose from the fact that I was not occupying the train that got the acceleration). In that case, yes, I would expect to receive less flashes per minute than I would be sending - but correction would be required for the changing distance however. To eliminate the changing distance, I proposed the orbiting satellite (post 16) version of the same problem. In actuality, this is the only example to date that is actually carried out. The result is unambiguous - the clock that received the acceleration runs slow relative to the Earth clock - and the Earth clock (ECRF) runs fast relative to the orbiting clock (ignoring height factors). We never do the linear experiment, but if we ever did, it should conform with Einstein's original development of actual time dilation - specifically, when one of two synchronized clocks is accelerated, it will run slower than the non-accelerated clock. Although every refernce frame is equivalent for the purpose of making internal experiments, and they all measure c as constant, they are not equivalent from the standpoint of making measurments in another frame that is in motion relative thereto.

 

[Ich]

The pion gets synchronized at its creation with the Earth clock - there can be another clock at the X origin in sync with the pion clock and the lab transmitter clock- nothing is transmitted before the pion is created - the lab transmiter is located at some point x=3 along the +x axis...

When you start the emission of pulses as you described (synchronized in the lab frame), the clock will tick for a proper time of 0.01 µs - quite longer than the lifetime of the pion. You should not wonder why it generates more ticks when you compare different timespans. It´s the same old synchronization problem as ever.

 

[Ich]

We never do the linear experiment, but if we ever did, it should conform with Einstein's original development of actual time dilation - specifically, when one of two synchronized clocks is accelerated, it will run slower than the non-accelerated clock. Although every refernce frame is equivalent for the purpose of making internal experiments, and they all measure c as constant, they are not equivalent from the standpoint of making measurments in another frame that is in motion relative thereto.

I´m not sure whether you´re talking about SR here. If you did, you´re wrong. SR says the following:
Acceleration has nothing to do with it. The result is the same when you think of your room being a cabin on a initially accelerated train, or when both you and the train were accelerated by the same amount. The result is symmetric.
You will see you room be lit in 0.5 s intervals. At your location, you will see the train time going twice as fast as yours. And the train observers will see your time going half as fast than theirs. Both agree.
But the same is of course valid for a fixed point on the train. Again both will agree, but this time your time (measured at a fixed point on the train) will run fast.
It´s exactly this effect that you mistake for "different proper time rates" in your example of an orbiting satellite: The satellites trajectory is curved such that the Earth center clock always remains at the same x-position (x being in the direction of relative movement). The Earth center of course sees the satellite move along its x-axis. That´s the assymetry: you compare times at a fixed point in the satellite system, not in the Earth center system. Acceleration in the direction of movement has nothing to do with it.

 

[yogi]

Ich: Which clock experienced acceleration (changed its speed after clocks were synchronized in the same frame) has everything to do with it. I am not talking about the (v^2)/R acceleration of the orbiting satellite - the satellite clock is in free fall inertial frame, and because of its velocity v it runs at a slower rate than the clock in the ECRF (Ignoring altitude). One second pulses sent by a transmitter in sync with the ECRF will be received at shorter than one second intervals by the satellite receiver - signals sent from the satellite at one second intervals as measured by the satellite clock will be received at longer than one second intervals in the ECRF. Based upon these signals, each will conclude that the satellite clock is running slower than the clock in the ECRF.

 

[yogi]

Ich - as to your post 26, If two trains are separated by some significant distance and both initially at rest in the Earth frame - and each is equally accelerated toward the other until they reach a velocity v wrt the Earth frame, then as they pass one another, things will be symmetrical - each will see the other clock to be APPARENTLY running slower by the Gamma factor that relates their relative velocity. In actuality however, both clocks on the trains will be running at the same rate - what is observed is an apparent slowing of the other guys clock. As each train passes the other and continues to the point where the other train started, the actual amount of slowing as measured by an Earth frame clock at that point will be the same for both trains. This is a symmetrical situation. But because actual clock rate differences depend upon which train has accelerated (or which has accelerated the most) as per Einstein's 1905 paper, it is improper to interpret the relativistic formula T1 = Gamma(T2) as being reciprocal to situations involving actual time dilation.

 

[Ich]

NO
From your previous posts I got the impression that you really try to understand SR, that´s why I joined the discussion. And now its the same old story: You promote your views of SR and don´t listen when people try to explain the real thing.
So, again, for the record:
Your understanding of SR in general and Einsteins paper in special is plain wrong. Its not a different point of view, its WRONG.
And your basic mistake is

to interpret the relativistic formula T1 = Gamma(T2)

There is no such formula in the lorentz equations, and you will never understand SR by interpreting time dilation. Even worse, your formula gives t2>t1, the contrary of time dilation.
The correct formula is gamma*(t1-vx). Only when you set x=v*t1 you get t2=(1/gamma)*t1, that is, time dilation.
You can derive all the results I discussed, and also what Einstein said in his paper, simply by applying the correct formula. No need for additional information like who has been accelerated and such things. Everything is unambiguous.
My advice: Sit down, try to understand what I posted about the train, calculate it yourself. Read my post about your satellites. Calculate the example yourself, using the correct formula. Derive why v²/r(*r) is important, not any initial accelerations. All the paradoxes and especially the real/apparent nonsense will vanish into thin air.
If you encounter any difficulties on your way, ask the people here.

 

[yogi]

No yourself - again you, like many, do not understand the difference between real time dilation (that shows up and is measurable in both one way and round trip experiments), and the apparent time dilation that results from the blind application of the LT transforms to a situation where you have failed to take into account the intrinsic asymmetry of the frames

The relationship between two spacetime events in the "at rest" system and the same two spacetime events in the moving system is directly determined by the invariance of the interval.

Your statement is:

"Only when you set x=v*t1 you get t2=(1/gamma)*t1, that is, time dilation.

I didn't specify which frame was T1 and T2 - its the same formula I gave - its the direct result of the invariance of the interval.

Read again what Einstein said in 1905 - when two clocks are synchronized in one frame and one of them is put in motion, the one put in motion will run at a lower rate than the one which is not put in motion. Ignoring oblateness - a clock at the equator will run faster than a clock at the North pole (directly from Einstein). Einstein never says the situation is reciprocal - it can't be where time dilation is actual - and that is the only situation to which I am referring -

Do some reading yourself - you will see that SR is explained in different ways by different authors - some of the explanations are mutually exclusive - but however analysed, the experimental results show that the clock put in motion runs slower - and that the clock which remains at rest runs faster relative to the clock put in motion - the situation is not reciprocal and Einstein clearly asserts this. Too bad you didn't read what he was saying before you took up preaching.