# Question regarding a twin paradox related example

Tags:
1. Aug 20, 2015

### Yichao Liu

Since I am new to the forum here, I apologize in advance that in case similar example has been explained in another thread. If so, please kindly refer me to the appropriate place to read further.

In my example, there will be four reference objects, earth, E, distant star, S, spaceship A, and spaceship B. Initially, all four objects are at rest arrange in the following way, A is right next to E; S is ahead of A and E with a distance of 0.866 light year, and B is behind A and E also at a distance of 0.866 light year. (I chose this number just so that the time dilation fact will end up being roughly 0.5 later)

S--------------------A/E--------------------B

Now, let’s assume A starts to accelerate at a very high rate with respect to E, so that A reaches a speed of 0.866c almost instantly. B will also accelerate in a way such that the proper distance between A and B remain constant at 0.866 light year all the time. And right before A reaches S, A decelerates almost instantly again and stop at S. Again, B will also decelerates to keep the distant between A and B constant. Before A starts to move, observers on S,E,A and B all synchronize their clocks.

I would like to claim the event (EAS)where spaceship A reaches distant star S, and the event(EBE) where spaceship B reaches Earth E happens simultaneously for both A/B and E/S. The reason I think is that the very moment A stops at S, A joins the reference frame of S/E. Since B moves in a way such that proper distant between A and B observed by A and B is always constant, the moment A stops, A should see B stops at E as well. Conversely, for observer on E/S, he should see EAS and EBE happens at same time as well.

Now, I will try to figure out what does each observer see during the event.

1 from an observer on Earth E

a. S is always at same distance away from E and the two clocks run the same.

b. A is at first rest next to E, then accelerate instantly to a speed of 0.866c, then move at that speed towards S for most of the distance, then A decelerates and stops at S. Assume the acceleration and deceleration are almost instant, the clock on E should show roughly 1 year has passed. And E should see that the clock on A shows roughly 0.5 year passed.

c. Due to the length contraction, E should B “jump” 0.433 light year towards E almost instantly, and the clock on B should also jump 0.5 year almost instantly. Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E should show 0.5 year. (This is the part I think I might I have made some mistakes)

2 from an observer on spaceship A

a. B and A are always on the same inertial frame, as defined, so B is always at same distance away from A and the two clocks run the same. (Not entirely sure about this either)

b. Immediately after the acceleration, the distance between S and E contract by half, A should see S “jump” 0.433 light year towards A, and again the clock on S instantly advances by 0.5 year. Then A continues to move at a constant speed of 0.866c. When A comes to a rest right next to S, the clock on A should show 0.5 year passed and clock on S shows 1 year.

c. A should see E accelerate away from A and continue move towards B at a constant speed. Right before A decelerates to stop, while A should almost reach S, A should see E almost reach halfway between A and B, and the time on both clocks should be 0.5 year. (distance between A and B is 0.866 light year, while distance between S and E as seen by A is 0.433 light year). Again, when A decelerates almost instantly, A should see E “jump” instantly towards B and the time on E should “jump” forward 0.5 year. Thus, when A stops at S, clock on A shows 0.5 year and clock on E shows 1 year.

3 from an observer on spaceship B

a. B should see E first accelerate then move at constant speed until E reaches almost half way between A and B. Then E will appears to “jump” right in front of B. When E stops at B, clock on E shows 1 year and clock on B shows 0.5 year.

So far, I think most of observations about the events agree with each other, and indicates that when A comes to a stop at S, time on A/B has passed roughly 0.5 year and time on E/S has passed 1 year. However, I am confused about the events I stated in 1c and 3a. Somehow they shows contradicting results.

I think the contradiction could arise from two main issues, 1st, I did not handle the “time jump” and “distance jump” during length contraction correctly, 2nd, I assumed that the event of A stops at S and the event of B stops at E happens simultaneously in both reference frames

So my ultimate question would be where I was wrong and what time the clocks on each object should show, when A first stops at S and also when B first stops at E (if the two events are not simultaneous)

2. Aug 20, 2015

### Staff: Mentor

How do you define this distance? Measured in which frame at which time?
A proper distance is well-defined between two points in spacetime, but not between two spaceships without as specific time reference.
If the distance stays 0.866 light years for E and S, it will be different for A and B while they are moving, and B will get the impression that A stops before B reaches E (B doesn't "see" this as the light needs more time to arrive).

3. Aug 20, 2015

### Stephanus

Dear Yichao Liu, welcome to PF Forum.
I'm new too in SR, but I see that you understand SR better than I. Since you ask the correct and better quesion than I did. If you allow me to explain to you...

In my example, there will be four reference objects, earth, E, distant star, S, spaceship A, and spaceship B. Initially, all four objects are at rest arrange in the following way, A is right next to E; S is ahead of A and E with a distance of 0.866 light year, and B is behind A and E also at a distance of 0.866 light year. (I chose this number just so that the time dilation fact will end up being roughly 0.5 later)
Yes you're right.

S--------------------A/E--------------------B

Now, let’s assume A starts to accelerate at a very high rate with respect to E, so that A reaches a speed of 0.866c almost instantly.

Again, you're right.
Instantly, so it's easier to study SR.
Almost. Right, Nothing can accelerate instantly.

B will also accelerate in a way such that the proper distance between A and B remain constant at 0.866 light year all the time. And right before A reaches S, A decelerates almost instantly again and stop at S. Again, B will also decelerates to keep the distant between A and B constant.

Before A starts to move, observers on S,E,A and B all synchronize their clocks.

Yes, synchronize their clock

I would like to claim the event (EAS)where spaceship A reaches distant star S, and the event(EBE) where spaceship B reaches Earth E happens simultaneously for both A/B and E/S.
Simultaneously with respect to (wrt) whom? We shall see later.

The reason I think is that the very moment A stops at S, A joins the reference frame of S/E.
Okay...

Since B moves in a way such that proper distant between A and B observed by A and B is always constant,
How can A observe B distance? (and how can B observe A distance)
The only way that A and B can (mutually) observe their distance is that A and B communicate each other.
If B reports its coordinate (send a digitan signal) to A, this will deceive A. Because A is changing reference frame. Twice. So does B.

the moment A stops, A should see B stops at E as well.
No, the moment A stops, (if B keeps sending signal to A), A still see that B is moving. But in this case, the moment A stops, A will see that B haven't moved yet!. See Pic 1

Conversely, for observer on E/S, he should see EAS and EBE happens at same time as well.
May be, but after EBE send signals to S and EAS sends signal to E, see Pic 1

Now, I will try to figure out what does each observer see during the event.

1 from an observer on Earth E

a. S is always at same distance away from E and the two clocks run the same.
Yes

b. A is at first rest next to E, then accelerate instantly to a speed of 0.866c, then move at that speed towards S for most of the distance, then A decelerates and stops at S. Assume the acceleration and deceleration are almost instant, the clock on E should show roughly 1 year has passed. And E should see that the clock on A shows roughly 0.5 year passed. Yes [ADD: after the light from EAS reaches E]

c. Due to the length contraction, E should [add: see] B “jump” 0.433 light year towards E almost instantly,
No, there's no "jump" here. I have thread asking about it before, see my profile. One - two months ago I have created a lot of thread about basic SR. See if you can find them.

In A moving frame. A moves first. A will deduct that A-B distance is actually expanding.

and the clock on B should also jump 0.5 year almost instantly.
No jump

Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E should show 0.5 year.
You mean B's clock shows 0.5 year and E clock's shows 1 year?
(This is the part I think I might I have made some mistakes) May be, but I'm not sure about it either

2 from an observer on spaceship A

a. B and A are always on the same inertial frame, as defined,

I think no. The moment A moves, the relative simultaneity of event will dictat that A moves first than B. See Pic 2

so B is always at same distance away from A and the two clocks run the same. (Not entirely sure about this either)
I'm not sure about this either, but because A moves first as I said. A-B distance is expanding wrt A. Dopler effect will prove it.

b. Immediately after the acceleration, the distance between S and E contract by half,
A should see S “jump” 0.433 light year towards A, and again the clock on S instantly advances by 0.5 year.
Again, no "jump" And how could A "see" where is S and what does S clock show? The only thing that A can know these informations is that S keeps sending signal to A, containing digital data that shows S' clock. And the moment A moves, A still see S as it was 0.866 years ago.

Then A continues to move at a constant speed of 0.866c. When A comes to a rest right next to S, the clock on A should show 0.5 year passed and clock on S shows 1 year.
Yes this is true. This I'm absolutely positive, but...
It's just number, there's no proof for A that S clock has advanced 1 year. The only thing that A can see from S is S'proper time. I you want to know about proper time, I can tell you. I didn't know then, but now I do.

c. A should see E accelerate away from A and continue move towards B at a constant speed. Yes
Right before A decelerates to stop, while A should almost reach S, A should see E almost reach halfway between A and B,
No. A would see E as it was 0.866 years ago.
I will calculate what A see E's clock...
A will see that E's clock still shows (1-0.866) = 0.134 year.
A will see that B's clock shows (1-0.866-0.866) = -0.7321 year
And "if" A "knows" that E's speed is 0.866c then A will deduct that EB distance is..., well 0 light year
See pic 1
Which in fact B is already at EBE, but at EAS there's no way for A to know where B is, only after 0.866 years later.

and the time on both clocks should be 0.5 year.
Which clock? A and B?
At EAS, A will see that A's clock is 0.5 year. But still see B clocks as it was 0.866 years later, and because A is changing frame, at EAS, A sees B's clock at -0.7321 year.
See Pic 1, the light ray line (yellow)

(distance between A and B is 0.866 light year, while distance between S and E as seen by A is 0.433 light year).
Again, measuring distance should involved doppler effect. The only informations that A is receiving is if E/S/B keep sending signal to A.
Perhaps I would calculate what A sees in each clock at EAS.
A clock: 0.5 year
S clock: 1 year
E: clock: (1-0.866) = 0.134 year
B: clock: (1-0.866-0.866) = -0.7321 year
Those are the only things that A can see.
About distance? Well, I don't know how to answer that. Perhaps some mentor/advisors can help us. But measuring distance can be deceptive.

Again, when A decelerates almost instantly, A should see E “jump” instantly towards B and the time on E should “jump” forward 0.5 year.

Thus, when A stops at S, clock on A shows 0.5 year and clock on E shows 1 year.
Yes, this is true. But A will see E's clock shows 1 year after A's clock shows 1.366 year.
See Pic 1, yellow light ray line from EBE to S world line

3 from an observer on spaceship B

a. B should see E first accelerate

B only see the doppler effect of E. If E keeps sending signal each 1 second, B will receive the signal every
$\sqrt{\frac{1+v}{1-v}} = 3.731 \text{ signals per second, each with E time}$ This is doppler factor adjusted by gamma factor. So I was told in this forum. And it makes sense for me.

then move at constant speed until E reaches almost half way between A and B.
How do B knows E is almost halfway? B can only deduct that from dopper effect which B realizes 0.866 years later.

Then E will appears to “jump” right in front of B. When E stops at B, clock on E shows 1 year and clock on B shows 0.5 year.
Hmm, to many "jump"s here.

So far, I think most of observations about the events agree with each other, and indicates that when A comes to a stop at S, time on A/B has passed roughly 0.5 year and time on E/S has passed 1 year. However, I am confused about the events I stated in 1c and 3a. Somehow they shows contradicting results.

---------------------------------------------------------------------------------------------------------------------------------
1c from an observer on Earth E

Due to the length contraction, E should [add: see] B “jump” 0.433 light year towards E almost instantly,

No, there's no "jump" here. I have thread asking about it before, see my profile. One - two months ago I have created a lot of thread about basic SR. See if you can find them.

In A moving frame. A moves first. A will deduct that A-B distance is actually expanding.

and the clock on B should also jump 0.5 year almost instantly.
No jump

Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E should show 0.5 year.
You mean B's clock shows 0.5 year and E clock's shows 1 year?
(This is the part I think I might I have made some mistakes) May be, but I'm not sure about it either
---------------------------------------------------------------------------------------------------------------------------------

3a from an observer on spaceship B

B should see E first accelerate

B only see the doppler effect of E. If E keeps sending signal each 1 second, B will receive the signal every
$\sqrt{\frac{1+v}{1-v}} = 3.731 \text{ signals per second, each with E time}$ This is doppler factor adjusted by gamma factor. So I was told in this forum. And it makes sense for me.

then move at constant speed until E reaches almost half way between A and B.
How do B knows E is almost halfway? B can only deduct that from dopper effect which B realizes 0.866 years later.

Then E will appears to “jump” right in front of B. When E stops at B, clock on E shows 1 year and clock on B shows 0.5 year.
Hmm, to many "jump"s here.
----------------------------------------------------------------------------------------------------------------

I think the contradiction could arise from two main issues, 1st, I did not handle the “time jump” and “distance jump” during length contraction correctly, 2nd, I assumed that the event of A stops at S and the event of B stops at E happens simultaneously in both reference frames.
Perhaps the space time diagram can explain it better to you? See pic 1 and Pic 2

So my ultimate question would be where I was wrong and what time the clocks on each object should show, when A first stops at S and also when B first stops at E (if the two events are not simultaneous)
From A and B inertia frame, event EAS and EBE are not simultaneous.
The clocks?
Ok, I'll try to calculate it for you. See the pictures for references. Focus on light ray yellow line.
EAS
A's clock: 0.5 year
S' clock: 1 year
E's clock: 1-0.866 = 0.134 year
B's clock: 1-0.866-0.866 = -0.7321 year
Actual clock:
A's clock: 0.5 year
S' clock: 1 year
E's clock: 1 year
B's clock: 0.5 year

EBE, oops I forgot to draw the light ray TO EBE, ok never mind
A's clock: Now, this is tricky. Has to use cartesian elimination: 0.267 year. Warning: I'm not sure about my calculations perhaps some mentor can step in?
S' clock: 0.134 year: Has tu use cartesian elimination Warning: I'm not sure about my calculations perhaps some mentor can step in?
E's clock: 1 year
B's clock: 0.5 year
Actual clock:
A's clock: 0.5 year
S' clock: 1 year
E's clock: 1 year
B's clock: 0.5 year

If there are some mistakes in my calculations I do apologize. It's not my itention to mislead the OP.
[ADD: mfb beats me while I typed]

Last edited: Aug 20, 2015
4. Aug 20, 2015

### Yichao Liu

In this case the distance is defined from either spaceship A or B, since they both accelerate, but keep the same distance from spaceships' point of view. So through out the whole journey, spaceship A will always see spaceship B at 0.866 light years behind it. Its kinda like reversed bell's spaceship paradox. In Bell's example, the two spaceships keep same distance from another inertia frame's view, so the distance from spaceship's view is increasing. However in my example, the frame of E/S will see two spaceship accelerate at different rates, but the spaceships see themselves always in the same frame. Or in other words, its almost like Spaceship A and B is just a super long spaceship with its head at A and tail at B. And if they accelerate at a slower speed, E/S will basically see spaceship B accelerates at a faster rate then A, and when they reach 0.866c, the distance between A and B contract to 0.433 light year from E/S.

yeah, i understand that if A/B is just moving at constant speed passing E/S then those events wont happen at the same time. However, in my case, both spaceship A and B decelerate to a stop, so the very moment spaceships stop, they join the frame of E/S, thus should see two events happen at same time.

5. Aug 20, 2015

### Yichao Liu

Thanks Stephanus, while I am still trying to process your post (It might take a while). I just have a few quick questions first.

"c. Due to the length contraction, E should [add: see] B “jump” 0.433 light year towards E almost instantly,
No, there's no "jump" here. I have thread asking about it before, see my profile. One - two months ago I have created a lot of thread about basic SR. See if you can find them.

In A moving frame. A moves first. A will deduct that A-B distance is actually expanding.

a. B and A are always on the same inertial frame, as defined,

I think no. The moment A moves, the relative simultaneity of event will dictate that A moves first than B. See Pic 2"

Are you saying there is no way for A and B to move in a way such that A and B keep at same distance all the time, and from another inertial frame (E/S), they will see both A and B accelerating?

I was trying to use the word "jump" to describe the results of almost instant acceleration/deceleration of the spaceships. Of course, if spaceship A and B are accelerating at a much slower speed, E/S will just see spaceship B accelerate at a faster rate then A.

I got the idea of "jump" from this paper http://studenci.fuw.edu.pl/~skfiz/stara/materialy/space_ships.pdf
It is basically trying to explain the bell's spaceship case. Under section 5 it says
"In other words, the spaceship A seems to warp from A’ to A’’. This is because an iso-time line jumps suddenly. For the pilot of B, A’ is identical to A’’."

Of course that is trying to explain case when two spaceships are observed from another inertial frame to be accelerating very fast but identical rate. while I am talking about a case similar to whats described in section 6, i.e. "WORLD LINE OF A SPACESHIP THAT PRESERVES THE PROPER DISTANCE"

"1 from an observer on Earth E

c. Due to the length contraction, E should B “jump” 0.433 light year towards E almost instantly, and the clock on B should also jump 0.5 year almost instantly. Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E should show 0.5 year. (This is the part I think I might I have made some mistakes)
"

Actually, here I do mean that the clock on B should show roughly 1 year has passed and the clock on E should show 0.5 year, but not the other way around, hence the contradiction I am facing. But as I said, I am not sure what is the proper way to describe what E sees the movement of B during the whole journey.

"harrylin said:
A little elaboration: suppose that at CERN in Geneva a particle is accelerated to 0.99999c in the direction of Lyon. Does that contract the distance between Geneva and Lyon?"

I think we are describing different things here. The distance between Geneva and Lyon never changes from either Geneva or Lyon's point of view. However, when the particle is accelerated to 0.99999c, the particle itself will see the distance between Geneva and Lyon subject to length contraction. So thats two different frames, so the distance is also different. Again, if the particle is accelerated almost instantly to 0.99999c, then the only way I describe what the particle sees is that Lyon all sudden almost instantly "jump" towards Geneva, so that the distance viewed by the particle is shorter. (The way i see it is that when the particle is at rest at Geneva, it will measure a distance between Geneva and Lyon as D. Now, lets suppose the particle is moving at a constant speed of 0.99999c, then it should measure the distance between Geneva and Lyon as D/223, with 1/223 being the contraction factor. However, if the acceleration to reach 0.99999c is almost instant, then the particle will effectively see Lyon "jump" 222/223 D towards Geneva almost instantly)

Last edited: Aug 20, 2015
6. Aug 20, 2015

### Staff: Mentor

That's not right because you don't change frames just because you change your state of motion. A frame is a convention for describing the motion of all the objects in the problem, and once you choose one you have to stick with it to get consistent results. If an object is changing its speed, then it may be at rest in the chosen frame at some times but not other times, but it's not leaving and then "rejoining" the frame - everything is always in all frames all the time, just not necessarily at rest.

7. Aug 20, 2015

### Yichao Liu

Ok, i see what you are saying. Now I have one more follow-up question. So for two inertial frames that are at rest with respect to each other, should any event appear to be identical to observers on both frames?

Also, I guess ultimately, I am really curious about, in my example, what is the proper way to describe what does observers on Earth E see the motion and timing for spaceship B for the duration of the journey(more specifically, when the observer on Earth sees that spaceship A just accelerates to 0.866c, whats the clock on Earth showing and what is the position/time/speed of spaceship B according Earth? Similarly, what's the position/time/speed of spaceship B according Earth, when Earth sees spaceship A right before its deceleration, and finally when the moment spaceship A stops at distant star S), and also what does spaceship B see Earth E during the journey.

8. Aug 20, 2015

### Staff: Mentor

"Two inertial frames that are at rest with respect to each other" is a contradiction in terms. If they're inertial and (their origins are) at rest relative to one another, they're the same frame. The coordinates of any event will be the same to within a translation, which makes them the same frame.

9. Aug 21, 2015

### Yichao Liu

While what you said make sense, I am more confused about the implications of your statement. Could you please answer a few simple questions regarding my example, hopefully that will clear things up for me.

The moment when spaceship A observes that distant start, S stops at A, what does A see Earth, E? Is E moving or stationary? Where is the position of E, somewhere in between A and B, right at B or beyond A and B?

Similarly, the moment, distant star S sees spaceship A stops at S, what does S see spaceship B? Is B moving or stationary? Where is the position of B, somewhere in between S and E, right at E or beyond S and E?

10. Aug 21, 2015

### Stephanus

Are you saying there is no way for A and B to move in a way such that A and B keep at same distance [ADD: wrt rest frame] all the time, and from another inertial frame (E/S), they will see both A and B accelerating?
If it is than for the spaceships, their distance must have expanded.
This is rather difficult to me. My understanding of SR is limited.
1. The second postulate of Relativity: Light always travel at c. And no object can travel faster than light.
2. To compensate that, there are length contraction. This is irrefutable. See this animation:
Click the up arrow for full description of this post. But I think you're familiar with this animation.
3. For the spaceships to remain at the same distance wrt rest frame. So their distance wrt spaceships must have expanded.
And this is where I can't think any further.

I was trying to use the word "jump" to describe the results of almost instant acceleration/deceleration of the spaceships. Of course, if spaceship A and B are accelerating at a much slower speed, E/S will just see spaceship B accelerate at a faster rate then A.

I got the idea of "jump" from this paper http://studenci.fuw.edu.pl/~skfiz/stara/materialy/space_ships.pdf
It is basically trying to explain the bell's spaceship case. Under section 5 it says
"In other words, the spaceship A seems to warp from A’ to A’’. This is because an iso-time line jumps suddenly. For the pilot of B, A’ is identical to A’’."

Of course that is trying to explain case when two spaceships are observed from another inertial frame to be accelerating very fast but identical rate. while I am talking about a case similar to whats described in section 6, i.e. "WORLD LINE OF A SPACESHIP THAT PRESERVES THE PROPER DISTANCE"

For this, I have had the same problem. But it's about train. If the length is contracted, it will threaten the train to be derailed. You use rocket.
Click the up arrow for further reading of that post.

"1 from an observer on Earth E

c. Due to the length contraction, E should B “jump” 0.433 light year towards E almost instantly, and the clock on B should also jump 0.5 year almost instantly. Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E should show 0.5 year. (This is the part I think I might I have made some mistakes)
"

Actually, here I do mean that the clock on B should show roughly 1 year has passed and the clock on E should show 0.5 year, but not the other way around, hence the contradiction I am facing. But as I said, I am not sure what is the proper way to describe what E sees the movement of B during the whole journey.
But, the clock on B can't jump right. I think you already know that. If B accelerates, then B's clock always shows 0.5 year.
Consider this little doppler experiment here.
I change the parameter so that the number is pretty round.
V = 0.6; Gamma = 1.25; Distance 15000 light sec;
B travels to E and sends signal every 200 sec:
B sends digital signal to E containing its clock and E sends back the signal to B adds E's clock.
At time 32,000 B starts travel at 0.6c;
Code (Text):

B Time    Data from E
30,000    B:   0; E: 15,000
30,200    B: 200; E: 15,200
30,400    B: 400; E: 15,400

Here we see that B receives its signal back every 200 sec. B reads E proper time; 200 secs. So 200 secs for B is 200 secs for E
Distance? B can't rely on E's clock. B should have calculated:
"I receive my own signal "B: 200; E:...." at 30,200. So the distance is 15,000 ls. B is at rest

Code (Text):

31,800    B: 1,800; E: 16,800
32,000<-  B: 2,000; E: 17,000;  B interval 100 secs
32,100<-  B: 2,200; E: 17,200;  B interval 100 secs
32,200    B: 2,400; E: 17,400

At 32,100 now B is receiving signal every 100 sec. B should have received signal every 125 sec if B travels at 0.6c, but B is receiving signal every 100 sec. This is already adjusted to Lorentz time dilation. B is receiving B's signal before B started to move.
E's clock interval is still 200

Code (Text):

46,900    B: 31,800; E: 46,800
47,000    B: 32,000; E: 47,000
47,050<-  B: 32,200; E: 47,100<- B interval 50 secs. E time interval 100 secs
47,100<-  B: 32,400; E: 47,200<- B interval 50 secs. E time interval 100 secs

At 47,000 now B is receiving signal every 50 sec. This is the signal when B started to move bounced back to B. E time interval is 100 secs.
Perhaps the "jump" effect is between 31999 to 32000, when B starts to move.
From 200 secs to 100 secs. Perhaps if B sends signal every micro seconds the interval is not 200 to 100, but 200, 199, 198, 197,... 102, 101, 100.
Perhaps this is the jump effect.
But there's no way that B can see E "jump".
Supposed, as I asked in the previous thread, E is not a Type 1a supernova, which is the standard candle. E is just a black small probe with a very strong power.
There's no way that B can see E. All B can see is the signal sent by E.
And one more thing. At earth clock 47,000 that the earth "realizes" that B has moved. When B moves, E can't see that. E can only know that B moves when B light cone is entering E world line. That's why at earth clock 47,000, E's clock interval is 100 sec, not 200 sec.

My understanding of SR is very limited but I think this statement is wrong.
c. Due to the length contraction, E should B “jump” 0.433 light year towards E almost instantly, and the clock on B should also jump 0.5 year almost instantly. Then B moves at constant speed then decelerates to then stop and both clocks on B should show roughly 1 year has passed and the clock on E
I have trouble understanding how "the clock on B should also jump 0.5 year almost instantly"

Yes, I agree with you 100%. It's two different frames. But I'm having trouble the particle can see Lyon distance. All it receives are radar signal,
Like Nugatory said.

11. Aug 21, 2015

### Yichao Liu

Sorry, I wasnt clear. what I meant is that spaceship A and B accelerate in a way such that the distance between then as measured by A or B is always the same, i.e. the spaceship A and B are behaving like the train example you talked about later. Basically, consider A as the head and B as the tail and the super long spaceship starts to accelerate, move at constant speed then decelerate etc. Or in other words, the distance between A and B is not constant from an observer on E/S

I am not saying time "jump" according to spaceship B, I am saying Earth, E should see the clock on B appears to "jump" (or rather moves very very fast according to the clock on earth) due to the almost instant acceleration.

12. Aug 21, 2015

### Stephanus

Then, what's the problem? It's the "usual" relativity, right.
Because A and B are comoving then, A and B will always measure their distance is the same. And E/S will measure A and B distance different.
The "problem" is when A-B distance as measured by E/S is constant, then there is "relativistic stress" as I read in Google. But there's no string in your spaceships right.

What do you mean by that?
Code (Text):
As seen by E
A               B               C
E     B         E     B         E      B
0     0         0     0         0    0.0
1    11         1     2         1    0.5
2    12         2     4         2    1.0
3    13         3     6         3    1.5
4    14         4     8         4    2.0
5    15         5    10         5    2.5
6    16         6    12         6    3.0
7    17         7    14         7    3.5
8    18         8    16         8    4.0
9    19         9    18         9    4.5
10    20        10    20        10    5.0
It's A or B or C? Spaceships B starts move at 0 second.

If their distance is 5, then E will see something like this.
Code (Text):
from E point of view                                      from B point ov view
E       dE        B       dB    dE/dB                   E       dE    B           dB    dB/dE

0.000            -5.000                              |  -5.000             0.000
0.577    0.577   -4.423    0.577    1.000 same rate  |  -3.923    1.077    0.289    0.289    3.732 diff rate
1.155    0.577   -3.845    0.577    1.000 same rate  |  -2.845    1.077    0.577    0.289    3.732 diff rate
1.732    0.577   -3.268    0.577    1.000            |  -1.768    1.077    0.866    0.289    3.732
5.000    3.268    0.000    3.268    1.000            |  -0.691    1.077    1.155    0.289    3.732
5.129    0.129    0.481    0.481    3.732 diff rate  |   0.387    1.077    1.443    0.289    3.732
5.258    0.129    0.962    0.481    3.732 diff rate  |   1.464    1.077    1.732    0.289    3.732
5.387    0.129    1.443    0.481    3.732            |   2.541    1.077    2.021    0.289    3.732
5.516    0.129    1.925    0.481    3.732            |   3.619    1.077    2.309    0.289    3.732
5.645    0.129    2.406    0.481    3.732            |   4.696    1.077    2.598    0.289    3.732
5.774    0.129    2.887    0.481    3.732            |   5.774    1.077    2.887    0.289    3.732

B will see the right side numbers.
Both of them will see the other counter part tick faster. But, B sees E ticks faster the moment B starts moving. E will see B ticks faster after B lightcone enters E world line.

13. Aug 22, 2015

### Mentz114

If the spaceships timed their launch to be $1/x$ secs apart. where $x$ is the position of the launch, they will appear to each other to be keeping the same distance apart.

Each ship will be following an integral curve of the congruence $u_a=-\frac{x}{\sqrt{x^2-t^2}}\ \partial_t +\frac{t}{\sqrt{x^2-t^2}}\partial_x$ which has expansion coefficient of zero.

Last edited: Aug 22, 2015
14. Aug 25, 2015

### harrylin

The point I made earlier is that it is not a valid point of view that, to give perhaps a clearer example, the distance between two towns changes when you measure it first in nautical miles and next in London miles - the distance cannot "jump" due to your change of measuring that same distance! A different measure of a fixed distance can only affect the measure of it and not the distance itself. Similarly when people say that they "bring objects closer" by looking at them through binoculars, that is only figurative speech and not meant to be taken seriously. Probably you understood that and so you may regard this as nitpicking over words, but sloppy phrasings can be detrimental for correct physical understanding.

PS. Note also that everyone can all the time choose which reference system to use, there is no need to switch reference system ("inertial frame") when accelerating. And "what a particle sees" depends on that choice of reference system.

Last edited: Aug 25, 2015