Lorentz and Doppler and signal pick up

[Stephanus]

Dear PF Forum,
After so many questions about Twins Paradox and Universe Frame of Reference, I'd like to know about Lorentz and Doppler. First.
Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2.

Okay here's the question.
Two probes A and B.
Clocks are synchorized.
Distance is 100 Ly.
Each year they're sending digital signal, containing their respective time.
A will send A0, A1, A2, A3 for each respective year.
B also send B0, B1, B2,... each year.

Signal from B is always late for 100 years
So each year A will receive:

Signal   A Time      Interval
B0       100.00     
B1       101.00      1.00 year
B2       102.00      1.00 year
..
..
B98      198.00
B99      199.00      1.00 year
B100     200.00      1.00 year

Is this true?

When B clock (and A clock) shows 100 year, B ignites its rocket RIGHT AFTER B sents B100. Catapulted at 86% c then the rocket stops, leaving B moves steadily at 86% c to the 'west', heading toward A
B will reach A at 115.4701 years.
Because ##\gamma = 2##, then B will reach A according to B clock for 57.7350 years.
When B arrive, A clock will read 115.4701
Is this true?

How will B PICK UP signal from A?

From where B stays and A1 there's 1 ly
B travels at V_b
V_b.t + ct = 1 ly, if V_b is relative to c, then c = 1
V_b.t + t = 1
##t = \frac{1}{1+V_b}/\gamma##
t = 0.27
Is this true?

Signal B Time  Interval
A0     100.00  
A1     100.27  0.27
A2     100.54  0.27
..            
..             
A212   156.81  0.27
A213   157.07  0.27
A214   157.34  0.27
A215   157.61  0.27 B meets A

How will A receive signal from B?

B is still sending signal.
Signal B101 is not sent 100 lys from A because B is moving.
Signal B101 is sent at V_b.t distance
Signal B101 should be sent at (100-0.86) ly from A. Because ##\gamma = 2##, so B101 is sent at 98.27 ly from A. Is this true?
B01 will be picked by A after 102 years, not 101 years, because 1 year for B is 2 years for A
So B101 will be received by A when T_a is 102 + 98.27 = 200.27. Is this true?

Signal   A Time      Interval  
B0       100.00      when A received B0, A still doesn't know that B has
                     already traveled.      
B1       101.00      1.00 year A still doesn't know that B has moved
B2       102.00      1.00
..
..
B98      198.00
B99      199.00      1.00   Still doesn't know
B100     200.00      1.00   Stil doesn't know
B101     200.27      0.27   Suddenly A begins receiving signals at 0.27
                            years interval
B102     200.54      0.27   Is this true?
B103     200.80      0.27
..
..

B155     214.74      0.27 
B156     215.01      0.27
B157     215.27      0.27 B meets A

Is the table above is true?

The difference with A scenario is that the instance B ignites its rocket, the time interval for signal receiving is 0.27. Whlie A has to wait for 200 years to have 0.27 interval.
Is this true?

 

[Mentz114]

Dear PF Forum,
After so many questions about Twins Paradox and Universe Frame of Reference, I'd like to know about Lorentz and Doppler. First.
Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2.

Okay here's the question.
Two probes A and B.
Clocks are synchorized.
Distance is 100 Ly.
Each year they're sending digital signal, containing their respective time.
A will send A0, A1, A2, A3 for each respective year.
B also send B0, B1, B2,... each year.

You've certainly put some effort into that but I don't see the point.

Did you read this http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html ?

I attach a spacetime diagram showing that approaching receivers experience an increase in frequency, receding receivers experience a lower frequency.

The formula for frequency shift is ##\frac{f_1}{f_2}=\sqrt{\frac{1\pm v}{1\mp v}}=\gamma ( 1\pm v)## where ##v## is the relative velocity between the receiver and sender.
The distance between sender and receiver is not in the formula.

 

[Stephanus]

You've certainly put some effort into that but I don't see the point.

Did you read this http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html ?

I attach a spacetime diagram showing that approaching receivers experience an increase in frequency, receding receivers experience a lower frequency.

The formula for frequency shift is ##\frac{f_1}{f_2}=\sqrt{\frac{1\pm v}{1\mp v}}=\gamma ( 1\pm v)## where ##v## is the relative velocity between the receiver and sender.
The distance between sender and receiver is not in the formula.

Thanks for the link. I'll contemplate it

 

[Stephanus]

You've certainly put some effort into that but I don't see the point.

Did you read this http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html ?

I attach a spacetime diagram showing that approaching receivers experience an increase in frequency, receding receivers experience a lower frequency.

The formula for frequency shift is ##\frac{f_1}{f_2}=\sqrt{\frac{1\pm v}{1\mp v}}=\gamma ( 1\pm v)## where ##v## is the relative velocity between the receiver and sender.
The distance between sender and receiver is not in the formula.

Ah, yes the distance is not in the formula. But WHEN they see the doppler effect?
Does B experience doppler effect immediately? While A has to wait for another 100 years to experience Doppler effect?

 

[PeterDonis]

Does B experience doppler effect immediately? While A has to wait for another 100 years to experience Doppler effect?

You asked basically this same question in another thread. The answer is still the same: an observer (such as A) can't see any effects from an event (such as B firing his rocket) until the light from that event reaches the observer.

 

[Stephanus]

You asked basically this same question in another thread. The answer is still the same: an observer (such as A) can't see any effects from an event (such as B firing his rocket) until the light from that event reaches the observer.

Thanks PeterDonis,
I think I'm getting close to my answer.
What I find interesting is that, altough motion is relative and both observer receive signal in 0.27 years interval. B experience 0.27 years before A does?
Does it look like an asymmetry?
I'm still reading this: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

 

[PeterDonis]

motion is relative and both observer receive signal in 0.27 years interval. B experience 0.27 years before A does?

I'm not sure what you mean by this.

 

[Stephanus]

What I find interesting is that, altough motion is relative and both observer receive signal in 0.27 years interval. B experience 0.27 years before A does?

I'm not sure what you mean by this.

I mean A has to wait 100 years before receiving signal from B in 0.27 year interval.
B immediately receiving signal from A at 0.27 years interval once he starts his engine.

 

[Janus]

I mean A has to wait 100 years before receiving signal from B in 0.27 year interval.
B immediately receiving signal from A at 0.27 years interval once he starts his engine.

Yes, B sees a change in the Doppler shift in the signal from A the moment he changes his velocity, because he is changing his velocity. A has to wait 100 yrs to see the change in Doppler shift from B, because he did not change his velocity, B did.

Try looking at it this way: Add two more observers C and D. C stays alongside A while D initially stays along B. Now at some point, B fires it engines to slow down and reverse its relative motion with respect to A and C. D does not. Now B and D have a relative motion with respect to each other, so you would not expect them to see the same Doppler shift from A, and since D did nothing, it shouldn't see any change, so B should. In fact, you can even imagine a 5th observer E, who is moving relative to A at the same velocity as B is after he changes velocity. B and D meet E coming from the other direction and B changes velocity to match that of E when they meet so that aftarward the are flying side by side. Again, it is obvious Once B and E match speeds they should see the same Doppler shift from A, after all they are flying side by side. And since E did nothing, it should be B that sees a change in the Doppler shift.

During all of this A and C are doing nothing and neither see any change in the Doppler shift in B until they see B math speeds with E, and it takes 100 yrs after the fact for this information to reach them.

 

[Stephanus]

Thank you very much Janus for your help.
Somehow I get the idea.
Actually I want to know about how twins paradox occures while the universe has no frame of reference.
Should motion is relative in space. Acceleration is not relative, this is right, isn't it.
Shouldn't A and B see the other participant is moving while each of them see themselves respectively at rest.
And what medium does light travel in space.
But..

Yes, B sees a change in the Doppler shift in the signal from A the moment he changes his velocity, because he is changing his velocity. A has to wait 100 yrs to see the change in Doppler shift from B, because he did not change his velocity, B did.

Somehow there's asymmetry here! It might, it might not be the answer to Twins Paradox or the universe frame of reference. But there is a proof, at least for me, that even though motion is relative, both observer experience a different situation!Supposed A and C are in the "west".
B and D are in the "east"

Try looking at it this way: Add two more observers C and D. C stays alongside A ^A while D initially stays along B. ^B Now at some point, B fires it engines to slow down and reverse its relative motion with respect to A and C. D does not. Now B and D have a relative motion with respect to each other, so you would not expect them to see the same Doppler shift from A, and since D did nothing, it shouldn't see any change, so B should. ^C In fact, you can even imagine a 5th observer E, who is moving relative to A at the same velocity as B is after he changes velocity. ^D B and D meet E coming from the other direction and ^E B changes velocity to match that of E when they meet so that aftarward the are flying side by side. Again, it is obvious Once B and E match speeds they should see the same Doppler shift from A, after all they are flying side by side. And since E did nothing, it should be B that sees a change in the Doppler shift.

^F During all of this A and C are doing nothing and neither see any change in the Doppler shift in B until they see B math speeds with E, and it takes 100 yrs after the fact for this information to reach them.

A: Supposed B and D travels at 0.8c to the west.
B: Supposed now B travels at 0.5c while D still travels at 0.8c to the west, can I say that?
C: Does E travels from the east to the west at 0.8c? You said who is moving relative to A at the same velocity as B
D: Or does E travels from the west to the east? B and D meet E coming from the other direction
If E travels from the east to the west at 0.8c, how can D meet E?
F: Will A and C notice that the Doppler shift for B somehow changes for a while, because between "B" and "E", B travels at 0.5c?
Thanks for any help

 

[Mentz114]

Thank you very much Janus for your help.
Somehow I get the idea.
Actually I want to know about how twins paradox occures while the universe has no frame of reference.
Should motion is relative in space. Acceleration is not relative, this is right, isn't it.
Shouldn't A and B see the other participant is moving while each of them see themselves respectively at rest.
And what medium does light travel in space.
But..
Somehow there's asymmetry here! It might, it might not be the answer to Twins Paradox or the universe frame of reference. But there is a proof, at least for me, that even though motion is relative, both observer experience a different situation!

Your questions have been answered many times but you seem to believe there is some mystery here that you can solve with an absolute reference frame.

There is no mystery, the twins'paradox' is well understood.

 

[Stephanus]

Your questions have been answered many times but you seem to believe there is some mystery here that you can solve with an absolute reference frame.

There is no mystery, the twins'paradox' is well understood.

Come on...
"The theory is not good if it can't be explained it to a six year old",
Albert himself say that, or because I'm not six years old

 

[A.T.]

"The theory is not good if it can't be explained it to a six year old"

It doesn't have to be good, just right.

 

[Stephanus]

Your questions have been answered many times but you seem to believe there is some mystery here that you can solve with an absolute reference frame.

There is no mystery, the twins'paradox' is well understood.

It's not mystery, I think I type asymmetry.
But you could say that the twins paradox is a mystery, at least for me.
Still hard to figure it out. But, I'm getting closer to my answer with Doppler effect. (or not closer?)

 

[Mentz114]

Maybe I misinterpreted what you mean. I don't mean to discourage you.

You can calculate the twins ages using Doppler.

I do know of any physical reason why clocks ( time itself) behaves this way.

 

[Stephanus]

Maybe I misinterpreted what you mean. I don't mean to discourage you.

You can calculate the twins ages using Doppler.

I do know of any physical reason why clocks ( time itself) behaves this way.

Just Doppler, not Doppler and Lorentz?

 

[Mentz114]

Just Doppler, not Doppler and Lorentz?

The Doppler shift formula for the ratio of frequencies is ##\gamma (c-v)## where ##v## is the relative velocity of source and receiver. There is a ##\gamma## there which is not in the pre-relativistic expression.

In one of your many threads there is a spacetime diagram ( thank ghwells) showing the clock ticks ( which are also subject to Doppler shift) of the traveller received by the home base. You should find it and try to understand it.

 

[Stephanus]

The Doppler shift formula for the ratio of frequencies is ##\gamma (c-v)## where ##v## is the relative velocity of source and receiver. There is a ##\gamma## there which is not in the pre-relativistic expression.

In one of your many threads there is a spacetime diagram ( thank ghwells) showing the clock ticks ( which are also subject to Doppler shift) of the travellerreceived by the home base. You should find it and try to understand it.

Perhaps I should understand spacetime diagram first.
Thanks for giving me the right direction.

 

[Mentz114]

OK, here is something that could help. Look at the diagram and then play the movie. Can you work out how the movie was generated from that diagram ?

 

[Stephanus]

OK, here is something that could help. Look at the diagram and then play the movie. Can you work out how the movie was generated from that diagram ?

Wait...

 

[Stephanus]

Okay...
The vertical blue line at the left of the .png file is the left block from the movie...
The vertical brown line at the right of the .png file is the right block from the movie...
The diagonal line is the moving block from the movie...
Both vertical lines receive the same amount of signals, but at different receiving frequency (and the frequency of the signal, too?, Blue shifted for brown line?) It is doppler, isn't it.
Is the brown line lower than the blue line?
I'll contemplate it. Still trying to figure out how it's got to do with time dilation. I think I'm closing into the answer.

 

[Mentz114]

Okay...
The vertical blue line at the left of the .png file is the left block from the movie...
The vertical brown line at the right of the .png file is the right block from the movie...
The diagonal line is the moving block from the movie...
Both vertical lines receive the same amount of signals, but at different receiving frequency (and the frequency of the signal, too?, Blue shifted for brown line?) It is doppler, isn't it.
Is the brown line lower than the blue line?
I'll contemplate it. Still trying to figure out how it's got to do with time dilation. I think I'm closing into the answer.

Yes, that's about right.

If you lay a ruler along the bottom of the picture and slide it up ( increasing time) then the position of the thing on the worldline ( block) is the intersection of the ruler edge with the worldline. The edge of the ruler is 'now' and the intersections with 'now' gives the positions.

Just like a movie camera, the program moves the 'now' line up, takes a snap, moves and snaps and so on to make frames.

Because the moving worldline is tilted the time between the frames ( along the diagonal ) is less than in the stationary frame. That is time dilation.
Geometrically relative velocity is the tilt.

Here is the same thing from the emitter frame.

 

[Stephanus]

There are many terms here beside Lorentz to understand relativity. (Hell, I don't grasp Lorentz transformation 100%)
Worldline, Event of simultaneity, Spacetime diagram, etc...
I'd like to understand those first.
Gotta go now. It's very late here.
Thanks.

 

[harrylin]

[..]
Both vertical lines receive the same amount of signals, but at different receiving frequency (and the frequency of the signal, too?, Blue shifted for brown line?) It is doppler, isn't it.[..] Still trying to figure out how it's got to do with time dilation. I think I'm closing into the answer.

Not sure if it has been said already, but by going quickly over these threads I did not see it and you may find it useful.
Based on the light postulate you will find for the frequency shift between emission and reception, if you derive it, the classical Doppler equations for a system in rest. However, according to SR the frequencies of a moving system themselves are also shifted ("time dilation"). "Relativistic Doppler" is the total effect between what two independent reference systems measure; it is the combination (multiplication) of the Doppler effect and time dilation.

An interesting and useful variant is reflection from a moving mirror. Here time dilation doesn't play a role, and the signal that is received back is simply classical Doppler (for an emitter-receiver that can be held to be in rest). This has been used for experiments such as "Gravity probe A" (Vessot et al) in order to distinguish between Doppler and time dilation. Instead of light and a mirror, they used radio signals and a transponder, but the principle is the same.

 

[PAllen]

Not sure if it has been said already, but by going quickly over these threads I did not see it and you may find it useful.
Based on the light postulate you will find for the frequency shift between emission and reception, if you derive it, the classical Doppler equations for a system in rest. However, according to SR the frequencies of a moving system themselves are also shifted ("time dilation"). "Relativistic Doppler" is the total effect between what two independent reference systems measure; it is the combination (multiplication) of the Doppler effect and time dilation.

An interesting and useful variant is reflection from a moving mirror. Here time dilation doesn't play a role, and the signal that is received back is simply classical Doppler (for an emitter-receiver that can be held to be in rest). This has been used for experiments such as "Gravity probe A" (Vessot et al) in order to distinguish between Doppler and time dilation. Instead of light and a mirror, they used radio signals and a transponder, but the principle is the same.

Can you clarify what you mean about a moving mirror being strictly classical Doppler?

To provide a reference for discussion:

http://arxiv.org/abs/physics/0605100

you see that classical Doppler for a moving mirror is not at all the same as relativistic Doppler for a moving mirror. As for classical Doppler for emitter-receiver at rest, there is none - nor is there relativistic Doppler. I genuinely have no idea what you mean in these statements. You may well have something valid to say, but you have been so brief that I cannot make out your claim. Please expand further.

 

[harrylin]

Can you clarify what you mean about a moving mirror being strictly classical Doppler?
I am talking about

To provide a reference for discussion:

http://arxiv.org/abs/physics/0605100

you see that classical Doppler for a moving mirror is not at all the same as relativistic Doppler for a moving mirror. As for classical Doppler for emitter-receiver at rest, there is none - nor is there relativistic Doppler. I genuinely have no idea what you mean in these statements. You may well have something valid to say, but you have been so brief that I cannot make out your claim. Please expand further.

[emphasis mine]

Classical Doppler equations refer to either measurements on moving objects with systems in rest, or with systems in motion. The light postulate corresponds to measurements with a system that is assumed to be in rest.

Perhaps I should have clarified that my comment about mirrors is valid for cases like the one discussed in this thread, in which signals are sent and received back at the same angle; it is not valid for reflection from tilted mirrors. If reflection is perpendicular or parallel to the signals, length contraction of the mirror is irrelevant. That was the case with gravity probe A: emitter and receiver were at the same point. Vessot et al used Doppler reflection to distinguish time dilation from the Doppler effect, so that they could measure the time dilation of the clock in the moving rocket in real time.*

The point that I tried to make is that "relativistic Doppler" is simply classical Doppler for a reference system that is in rest, multiplied with the time dilation factor (and accounting for length contraction if the mirror is tilted). I think that this is well known and understood; usually this is explained in discussions of "transverse Doppler", which was used for the first direct time dilation experiment.

However, a quick glance at the paper to which you link (was it peer reviewed?) does not discuss the tilting of a moving mirror due to length contraction but something else; the argumentation looks overly complex, but I think that his purpose is different from what you suggest.
[EDIT:] In fact, I see now that he apparently claims to demonstrate exactly what I stated:
"In this paper we [..] give an entirely non-relativistic derivation of Eq.3 [the formula for the frequency of the reflected light]".

Classical Doppler from moving objects at zero angle for a system in rest is based on the same assumptions as the light postulate, and Vessot could measure time dilation because reflected signals are not time dilated - indeed, that would be quite impossible! Classical Doppler and relativistic Doppler are inevitably the same for that case.

*Vessot et al, PRL 26, 1980.

 

[PAllen]

Note that the paper claims and justifies that non-relatavitically you would get a different Doppler formula for a moving mirror (they describe this as resulting from different velocity addition). In fact, a very straightforward analysis in a frame in which the mirror is at rest and an emmitter/receiver is moving directly towards or away from the mirror, establishes that the Doppler for this case is exactly the square of the Doppler measured by a receiver replacing the mirror. Thus, the relativistic Doppler for a moving mirror is the square of the normal relativistic Doppler, and the same would be expected for the non-relativistic case. Similarly, the logic of analysis in the mirror fame establishes that time dilation must be involved - the source/emitter is time dilated in the relativistic case and not otherwise.

However, I now understand your point as well. In a frame in which the emmitter/receive doesn't move, its time dilation is not involved, and the mirror's is not involved either because we are not worrying about a receiver mounted on the mirror. To me, this analysis still makes no sense non-relativistically because it can't be sensibly squared with the analysis in any other frame. If pre-relativistic physics were correct, you could square Doppler and Galilean relativity only using a corpuscular Doppler analysis which would be different in both frames from the SR result; or assume formula (c+v)/(c-v) will be found true only in a frame in which the emitter/receive is motionless with respect to aether. To have (c+v)/(c-v) be sensible in both frames, you must include relativistic effects. That's what was bugging me.

[Note that by putting together a derivation of (c+v)/(c-v) using only constant light speed and motionless emitter/receiver with moving mirror; combined with the analysis (which can be done purely conceptually) in the mirror frame that this must be the square of ordinary Doppler, combined with frame independence, provides yet another route to the normal Doppler formula √((c+v)/(c-v)) ]

 

[harrylin]

[..] However, I now understand your point as well.

My point is basically the same as his point: a mirror has no own frequency and no time dilation. Standard wave theory applies, as he also shows.

[..] To me, this analysis still makes no sense non-relativistically because it can't be sensibly squared with the analysis in any other frame.

According to an observer in the rest system, frequencies in the moving system are shifted by the Lorentz factor. This kind of Doppler analysis was used by Ives as well as by Vessot to measure time dilation of a moving resonator, in two famous publications. It's very useful for understanding the physics.

If pre-relativistic physics were correct [..]

None of us thinks such a thing. The discussion here is about Doppler and time dilation...

you must include relativistic effects. That's what was bugging me.

I tried to clarify to Stephanus how time dilation conceptually as well as in practice can be distinguished from the Doppler effect. I hope that it is clearer now.

 

[pervect]

Dear PF Forum,
After so many questions about Twins Paradox and Universe Frame of Reference, I'd like to know about Lorentz and Doppler. First.
Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2.

Okay here's the question.
Two probes A and B.
Clocks are synchorized.
Distance is 100 Ly.

You didn't mention the frame. I assume it's the common frame shared by A and B. It's good to get into the habbit of specifying the frame, because it matters in SR. Typically when you don't mention something, it reflects an underling thought process that hasn't quite accepted the fact that the choice of frame DOES matter, which leads inevitably to confusion.

Each year they're sending digital signal, containing their respective time.
A will send A0, A1, A2, A3 for each respective year.
B also send B0, B1, B2,... each year.

Signal from B is always late for 100 years
So each year A will receive:

Signal   A Time      Interval
B0       100.00     
B1       101.00      1.00 year
B2       102.00      1.00 year
..
..
B98      198.00
B99      199.00      1.00 year
B100     200.00      1.00 year

Is this true?

Yes

When B clock (and A clock)

It's better to just say B's clock, due to the fact that simultaneity is relative

shows 100 year, B ignites its rocket RIGHT AFTER B sents B100. Catapulted at 86% c then the rocket stops, leaving B moves steadily at 86% c to the 'west', heading toward A
B will reach A at 115.4701 years.

I think you mean by this that A's clock will read 215.47years when B reaches A.

Because ##\gamma = 2##, then B will reach A according to B clock for 57.7350 years.
When B arrive, A clock will read 115.4701
Is this true?

As I said above, I think you meant to say that when A and B arrive, A's clock reads 215.47. B's clock will read 100 + 115.47/2 = 157.735

will B PICK UP signal from A?

B will receive a signal from A every ##\gamma(1-v/c)## years as per wiki. There are several alternate equivalent formulae. I get this as a period of .2679 years, close to your figure.

From where B stays and A1 there's 1 ly
B travels at V_b
V_b.t + ct = 1 ly, if V_b is relative to c, then c = 1
V_b.t + t = 1
##t = \frac{1}{1+V_b}/\gamma##
t = 0.27
Is this true?

I'm not following this, sorry. It's not what I'm getting, either. So it's probably not true.

A will send out a total of 216 plus a fraction signals that B will receive, the first signal A0 will be received by B when B starts the trip at B's time of 100. (Note that the set {0,1,...,215} contains 216 elements.) The first signal is numbered A0, the last A215. The spacing between signals is .2679 years. The last signal from A, A215, will be received at a time of 215*.2679 = 57.6 years according to B's clock, near the end of B's trip.

B will send out a total of 58 signals on the trip, the first signal B100, the last signal B157. A will receive the first signal, B100, at 200 years on A's clock due to propagation delays, the same time at which A will see B's rocket flare through a telescope (if A is looking).

A will receive signals from that point at the rate of 1 signal every .2679 years, so the last signal, B57, will arrive at 200+.269*57 = 215.27 on A's clock, shortly before B arrives.

 

[Stephanus]

Dear PF Forum,
After so many questions about Twins Paradox and Universe Frame of Reference, I'd like to know about Lorentz and Doppler. First.
Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2.

Okay here's the question.
Two probes A and B.
Clocks are synchorized.
Distance is 100 Ly.

You didn't mention the frame. I assume it's the common frame shared by A and B. It's good to get into the habbit of specifying the frame, because it matters in SR. Typically when you don't mention something, it reflects an underling thought process that hasn't quite accepted the fact that the choice of frame DOES matter, which leads inevitably to confusion.

They are both at rest.

...Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2...

This v is not their speed NOW, it's just an arbirary for later discussion, has nothing to do with the statement above.

 

[Stephanus]

B will reach A at 115.4701 years.

I think you mean by this that A's clock will read 215.47years when B reaches A.
Because γ=2, then B will reach A according to B clock for 57.7350 years...

Very smart and meticulous of you, very careless of me! Yes that's right.
When A's clock read 215.47 year or 115.47 year from A's clock read 100 years which is the first time A receive signal from B.

B will receive a signal from A every γ(1−v/c) years as per wiki. There are several alternate equivalent formulae. I get this as a period of .2679 years, close to your figure.

Yes, thanks 0.2679 actually, just copied only 2 dec points.

A will send out a total of 216 plus a fraction signals that B will receive, the first signal A0 will be received by B when B starts the trip at B's time of 100. (Note that the set {0,1,...,215} contains 216 elements.) The first signal is numbered A0, the last A215. The spacing between signals is .2679 years. The last signal from A, A215, will be received at a time of 215*.2679 = 57.6 years according to B's clock, near the end of B's trip.

B will send out a total of 58 signals on the trip, the first signal B100, the last signal B157. A will receive the first signal, B100, at 200 years on A's clock due to propagation delays, the same time at which A will see B's rocket flare through a telescope (if A is looking).

A will receive signals from that point at the rate of 1 signal every .2679 years, so the last signal, B57, will arrive at 200+.269*57 = 215.27 on A's clock, shortly before B arrives.

Thank you, thank you. I learn much from this.