Stupider Twins Question

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  • #1
Say we have the twins plus an observer at a space station(at rest with earth) 8 ly from earth. v = 0.8c. All clocks read 0 when the ship leaves Earth (already at 0.8c). When the ship's twin is observed to reach the space station, all observers stop their clocks. Ship's clock should read 6 yrs, Space station clock should read 10 years, Earth clock should read 18 yrs, but the Earth twin knows to subtract 8 yrs due to signal delay. Everyone agrees that when the ship reached the space station(a single event observed locally from each frame), the ship's twin was 4 yrs younger than the Earth twin.

Nobody accelerated.
Nobody changed frames.


Whether or not the twins ever meet again will not change the fact that the ship's twin aged less during this trip.

Of course this trip is not symmetrical, but acceleration and changing of reference frames are both irrelevant, because they never happened.

Comments? Please feel free to tell me how I botched this up. Besides the fact that I left out the novelty of the ship's twin getting to sneer at his brother in person for being older.

Thanks,
Al
 
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Answers and Replies

  • #2
Fredrik
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You didn't botch it. That's the way it is. While the twin on the ship was moving at 0.8c, his brother was aging slower than him in the ship's frame. The moment before he stopped he was simultaneous in his own rest frame with t=3.6 years on Earth, but as soon as he stopped, he became simultaneous in his own rest frame (which is now another frame) with t=10 years on Earth.

That's right, he can't stop the ship without having his brother age 6.4 years.
 
  • #3
You didn't botch it. That's the way it is. While the twin on the ship was moving at 0.8c, his brother was aging slower than him in the ship's frame. The moment before he stopped he was simultaneous in his own rest frame with t=3.6 years on Earth, but as soon as he stopped, he became simultaneous in his own rest frame (which is now another frame) with t=10 years on Earth.

That's right, he can't stop the ship without having his brother age 6.4 years.
In my example, the ship didn't stop. Just compared clocks. The ship's clock read 6 yrs, and the space station clock read 10 yrs. Although there is a lack of simultaneity, the ship's twin can see that the Earth twin ages more during the trip by just looking at the space station clock which is at rest with earth. He doesn't have to actually stop because he can compare clocks locally with the space station. This point was the reason for my post.

Yes, I realize that the Earth clock and space station clock were not started simultaneously in the ship's frame, but the ship's twin, being a smart guy, realizes that they started simultaneously in their own rest frame, and that his brother is at rest in that frame, and that the space station clock shows the elapsed time on earth, in Earth's frame, since the ship left earth.

Thanks,
Al
 
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  • #4
Fredrik
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In my example, the ship didn't stop. Just compared clocks. The ship's clock read 6 yrs, and the space station clock read 10 yrs. Although there is a lack of simultaneity, the ship's twin can see that the Earth twin ages more during the trip by just looking at the space station clock which is at rest with earth.
OK. In that case, this statement is wrong: "the ship's twin can see that the Earth twin ages more during the trip by just looking at the space station clock".

Yes, he can read the time on that clock, but the correct conclusion isn't that his brother is ten years older at that event. His brother isn't at that event. What he can do is say how old his brother is at events that are simultaneous with that event, and to do that he must specify what frame he's talking about. In his own rest frame, the event where he reads the clock is simultaneous with the event where the clock on Earth shows t=3.6 years. In the space station's frame, the clock-reading event is simultaneous with the event where the clock on Earth shows t=10 years.
 
  • #5
OK. In that case, this statement is wrong: "the ship's twin can see that the Earth twin ages more during the trip by just looking at the space station clock".

Yes, he can read the time on that clock, but the correct conclusion isn't that his brother is ten years older at that event. His brother isn't at that event. What he can do is say how old his brother is at events that are simultaneous with that event, and to do that he must specify what frame he's talking about. In his own rest frame, the event where he reads the clock is simultaneous with the event where the clock on Earth shows t=3.6 years. In the space station's frame, the clock-reading event is simultaneous with the event where the clock on Earth shows t=10 years.
I guess I should revise that statement to read:

"the ship's twin can see that the Earth twin ages more during the trip (as measured in his own rest frame) than the ship's twin does (as measured in his own rest frame) by just looking at the space station clock".

Why would the ship's twin judge how much his brother aged by calculating what Earth's clock should read in the ship's frame? The Earth twin is not at rest in the ship's frame.

The Earth twin aged 10 yrs between the events as measured by a clock at rest with him.
The ship's twin aged 6 yrs between events as measured by a clock at rest with him.
This is the asymmetry that was my point. And with no acceleration or change of frames by anyone.

In other words, if each twin took a picture of himself simultaneously with the ship reaching the space station (each in his own respective rest frame), the Earth twin will have more grey hair when and if the photos are compared.

Thanks,
Al
 
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  • #6
MeJennifer
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The Earth twin aged 10 yrs between the events as measured by a clock at rest with him.
The ship's twin aged 6 yrs between events as measured by a clock at rest with him.
This is the asymmetry that was my point. And with no acceleration or change of frames by anyone.
Which events? As far as I can see there are no events in your scenario. A spaceship "leaving" Earth which is already traveling at 0.8c and "arriving" at the spacestation by passing it with a whopping 0.8c are not events.
 
  • #7
Which events? As far as I can see there are no events in your scenario. A spaceship "leaving" Earth which is already traveling at 0.8c and "arriving" at the spacestation by passing it with a whopping 0.8c are not events.
Well, since I defined them as events, and they are easily observed and distinguished in time and space, I would call them events. But they are not acceleration events, which was my whole point.

If it would help, we can say that there were light flashes involved, but I wanted to keep it simple.

Thanks,
Al
 
  • #8
Fredrik
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"the ship's twin can see that the Earth twin ages more during the trip (as measured in his own rest frame) than the ship's twin does (as measured in his own rest frame) by just looking at the space station clock".
This statement doesn't make sense. A person's age can't be specified by specifying a frame. You have to specify an event. You can do that either by specifying that event explicitly (e.g. "at the space station, when its clock shows t=10 years"), or by specifying a time in that frame (e.g. "t=6 years in the ship's frame"), in which case the event is implicitly specified as the intersection of the person's world line with the line that has the specified time coordinate in the specified frame.

It seems that what you had in mind is that the Earth twin's age at the event where the Earth clock shows t=10 years is greater than the ship twin's age at the event where the ship's clock shows t=6 years. This statement is correct in both SR and prerelativistic physics. What you don't seem to realize is that this is a very different claim from "everyone agrees the Earth twin is older". What the ship twin really thinks when he passes the space station is this: "My brother will have aged 10 years, 10 years and 8 months from now". He's not thinking "My brother has aged 10 years now".

Why would the ship's twin judge how much his brother aged by calculating what Earth's clock should read in the ship's frame? The Earth twin is not at rest in the ship's frame.
I don't know how to explain that. It seems obvious to me that since he's trying to answer the question "how old is my brother now?", he must use his own "now" rather than someone else's. And in his own "now" (as he passes the space station) his brother has only aged 3.6 years.

If you don't see that, you probably haven't fully understood how simultaneity works in SR.
 
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  • #9
This statement doesn't make sense. A person's age can't be specified by specifying a frame. You have to specify an event. You can do that either by specifying that event explicitly (e.g. "at the space station, when its clock shows t=10 years"), or by specifying a time in that frame (e.g. "t=6 years in the ship's frame"), in which case the event is implicitly specified as the intersection of the person's world line with the line that has the specified time coordinate in the specified frame.

It seems that what you had in mind is that the Earth twin's age at the event where the Earth clock shows t=10 years is greater than the ship twin's age at the event where the ship's clock shows t=6 years. This statement is correct in both SR and prerelativistic physics. What you don't seem to realize is that this is a very different claim from "everyone agrees the Earth twin is older". What the ship twin really thinks when he passes the space station is this: "My brother will have aged 10 years, 10 years and 8 months from now". He's not thinking "My brother has aged 10 years now".


I don't know how to explain that. It seems obvious to me that since he's trying to answer the question "how old is my brother now?", he must use his own "now" rather than someone else's. And in his own "now" (as he passes the space station) his brother has only aged 3.6 years.

If you don't see that, you probably haven't fully understood how simultaneity works in SR.

But he's not asking "how old is my brother now?", he's asking "How much does my brother age between these two events?" This point is totally different. And I do understand that there is a loss of simultaneity between frames. I do understand that in the ship's frame, the space station clock and the Earth clock did not start simultaneously.

I'm referring to the time that passes in each respective frame between the two events. That's what I meant by "during the trip". Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.

Or, in other words, these two events are separated in time by 10 years in Earth's frame, the same two events are separated in time by 6 yrs in the ship's frame. Is this statement true?

I'm not asking if it's relevant, just if it's true.

Thank's
Al
 
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  • #10
Fredrik
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Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.
This is the part you keep getting wrong. There are three events:

Event A: Ship leaving Earth
Event B: Ship reaching space station
Event C: Earth at t=10 years

Or, in other words, these two events are separated in time by 10 years in Earth's frame, the same two events are separated in time by 6 yrs in the ship's frame. Is this statement true?
The statement doesn't make sense, since there are 3 events, not 2.
 
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  • #11
It seems that what you had in mind is that the Earth twin's age at the event where the Earth clock shows t=10 years is greater than the ship twin's age at the event where the ship's clock shows t=6 years. This statement is correct in both SR and prerelativistic physics.
True, but only in SR is the event of Earth's clock showing 10 yrs simultaneous (in Earth's frame) with an event (ship at space station) which is simultaneous (in the ship's frame) with the event of the ship's clock showing 6 yrs.

Thanks,
Al
 
  • #12
rahuldandekar
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Say we have the twins plus an observer at a space station(at rest with earth) 8 ly from earth. v = 0.8c. All clocks read 0 when the ship leaves Earth (already at 0.8c).

There is the problem. Space ship and Earth are separated by 8 ly. How are all clocks going to read zero simultaneously? Simultaneously in which frame?
 
  • #13
MeJennifer
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Well, since I defined them as events, and they are easily observed and distinguished in time and space, I would call them events. l
You may call them so but with all respect if you want to learn about relativity it would be better to understand why something zooming at 0.8c past something else is not the same event, their respective worldlines are are not even close in spacetime.
 
  • #14
Al68 said:
Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.
This is the part you keep getting wrong. There are three events:

Event A: Ship leaving Earth
Event B: Ship reaching space station
Event C: Earth at t=10 years
How about Event D: Ship at t=6 yrs.

Ah, but since events B and C are simultaneous in Earth's frame, 10 yrs passed between event A and event B/C in Earth's frame.
And since events B and D are simultaneous in the ship's frame, 6 yrs passed between event A and event B/D in the ship's frame.

Obviously more than two events exist, but my statement only referred to the two.

Thanks,
Al
 
  • #15
yuiop
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... Ship's clock should read 6 yrs, Space station clock should read 10 years, Earth clock should read 18 yrs, but the Earth twin knows to subtract 8 yrs due to signal delay. ...

This part is wrong. When the traveling twin paases the spacestation and looks back towards the clock on the Earth he sees the other twin as 2 years old. This is because light leaving the Earth twin when he is 2 years old takes 8 years to travel the distance and arrives at the spacestation after 10 years Earth time just as the traveling twin is passing.

The traveling twin might argue that that since the spacestation clock is synchronised with the Earth clock, then if the spacestation clock reads 10 years then the Earth twin must be 10 years old too. Open and shut case? No! If the traveling twin was at the front of a rocket that happened to have a proper length of 13.333 light years long, then due to length contraction the tail of the ship will be level with the Earth when the traveling twin arrives at the spacestation. Now if the clocks are synchronised on the ship as they probably would be if it was always traveling at 0.8c then the clock at the tail of the ship passing Earth will be reading 16.666 years. The Earth twin then argues that since the clocks on the rocket are synchronised, the traveling twin MUST also be be 16.666 years old while he is only 10 years old.

Each has a valid argument that they are 60% of the age of the other twin.
 
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  • #16
There is the problem. Space ship and Earth are separated by 8 ly. How are all clocks going to read zero simultaneously? Simultaneously in which frame?
Issue already addressed:
Al68 said:
Yes, I realize that the Earth clock and space station clock were not started simultaneously in the ship's frame

They don't all read zero simultaneously. The space station clock is synched with Earth's clock in their rest frame.

Thanks,
Al
 
  • #17
Fredrik
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B and D are not just simultaneous in the ship's frame. They are by definition the same event.

In Earth's frame, B and C are both 10 years later than A.
In the ship's frame, B is 6 years later than A.
 
  • #18
Fredrik
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In other words, this statement is correct if the two events are A and B:
Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.


But your original statement isn't:
Everyone agrees that when the ship reached the space station(a single event observed locally from each frame), the ship's twin was 4 yrs younger than the Earth twin.
If everyone agrees, it includes the twin on the ship, and he was 2.4 years older than his brother at this time in his rest frame.
 
  • #19
In other words, this statement is correct if the two events are A and B:
Al68 said:
Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.


But your original statement isn't:
Al68 said:
Everyone agrees that when the ship reached the space station(a single event observed locally from each frame), the ship's twin was 4 yrs younger than the Earth twin.
If everyone agrees, it includes the twin on the ship, and he was 2.4 years older than his brother at this time in his rest frame.
Yeah, you're right, that second statement isn't worded right. I hereby retract it. :smile:

This one:
Al68 said:
Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.
is my point. Isn't this asymmetrical?

Thanks,
Al
 
  • #20
yuiop
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This is the part you keep getting wrong. There are three events:

Event A: Ship leaving Earth
Event B: Ship reaching space station
Event C: Earth at t=10 years


The statement doesn't make sense, since there are 3 events, not 2.

I agree with Fredrik,

The interval between the events (Ship leaving Earth) and (Ship reaching space station) is measured as 10 years by the Earth frame observers and as 6 years by the traveling twin. That conclusion in no way implies that the event (travelling twin celebrates 6th birthaday) is universally simultaneous with (Earth twin celebrates 10th birthday) according to any observer. It is true to observers in the Earth frame but to observers in the traveling twins frame the events (travelling twin celebrates 6th birthaday) is simultaneous with (Earth twin is 3.6 years old).
 
  • #21
You may call them so but with all respect if you want to learn about relativity it would be better to understand why something zooming at 0.8c past something else is not the same event, their respective worldlines are are not even close in spacetime.

I'd say they're pretty close, since they're local. For example, the ship is at Earth at t=t'=0. That's Event A if I call it Event A. With all respect.

Thanks,
Al
 
  • #22
Fredrik
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This one:is my point. Isn't this asymmetrical?
Not really. Different observers will always agree about what happens at a specific event, but not about how much a specific coordinate changes from one event to another.
 
  • #23
rahuldandekar
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Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.

is my point. Isn't this asymmetrical?

Thanks,
Al

Yes, the clock readings are asymmetrical in the two frames. But that doesn't mean there's a paradox. The situation itself is asymmetrical in the two frames. Looking from the Earth's frame, one clock is moving, two clocks are stationary. In the spaceship frame, one clock is stationary, two clocks, at different locations, are moving. The asymmetry in the situation results in an asymmetry in the clocks.
 
  • #24
MeJennifer
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I'd say they're pretty close, since they're local. For example, the ship is at Earth at t=t'=0. That's Event A if I call it Event A. With all respect.

Thanks,
Al
Remember you wrote there was no acceleration involved. So if the initial speed is already 0.8c and the speed when the traveler is spatially close to the spacestation is also 0.8c there are no common spacetime events and the respective wordlines in spacetime are not even close to each other.
 
  • #25
Fredrik
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Not really.
Yes,...
It may sound like we disagree, but we just interpreted "asymmetrical" differently. My interpretation: "weird, paradoxial". His interpretation: "different".
 
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  • #26
Fredrik
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...and the respective wordlines in spacetime are not even close to each other.
The world lines of the twins intersect at Earth, t=t'=0. This is event A. After that the world lines don't come near each other. Event B is on the world line of the ship, at t'=6.

He has changed his original claim to just saying that it seems weird that t=10 at event B.
 
  • #27
I agree with Fredrik,

The interval between the events (Ship leaving Earth) and (Ship reaching space station) is measured as 10 years by the Earth frame observers and as 6 years by the traveling twin. That conclusion in no way implies that the event (travelling twin celebrates 6th birthaday) is universally simultaneous with (Earth twin celebrates 10th birthday) according to any observer. It is true to observers in the Earth frame but to observers in the traveling twins frame the events (travelling twin celebrates 6th birthaday) is simultaneous with (Earth twin is 3.6 years old).
Of course. I, too agree with Fredrik on that.

The asymmetry I'm trying to point out is the interval between the events in each frame.

I've never seen this asymmetry mentioned anywhere. And it has nothing to do with acceleration or observers changing frames.

Thanks,
Al
 
  • #28
MeJennifer
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The world lines of the twins intersect at Earth, t=t'=0. This is event A. After that the world lines don't come near each other. Event B is on the world line of the ship, at t'=6.
He wrote (bold text by me):
Say we have the twins plus an observer at a space station(at rest with earth) 8 ly from earth. v = 0.8c. All clocks read 0 when the ship leaves Earth (already at 0.8c). ...

Nobody accelerated.
Nobody changed frames.
In flat spacetime if there are two events and two travelers between them and they claim to have taken a different path in spacetime, which means their clocks are no longer in sync, then it must be true that they accelerated with respect to each other at least twice.
 
  • #29
Yes, the clock readings are asymmetrical in the two frames. But that doesn't mean there's a paradox.
I didn't say there was a paradox. Just an asymmetry.
The situation itself is asymmetrical in the two frames. Looking from the Earth's frame, one clock is moving, two clocks are stationary. In the spaceship frame, one clock is stationary, two clocks, at different locations, are moving. The asymmetry in the situation results in an asymmetry in the clocks.
Would you agree that the "twins paradox" contains this same asymmetry?

Thanks,
Al
 
  • #30
He wrote (bold text by me):

In flat spacetime if there are two events and two travelers between them and they claim to have taken a different path in spacetime, which means their clocks are no longer in sync, then it must be true that they accelerated with respect to each other at least twice.

Uh, only one traveler was at both events.

Thanks,
Al
 
  • #31
MeJennifer
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Uh, only one traveler was at both events.
All objects travel from one event to the other in spacetime.

Remember this is relativity, if there is no acceleration you cannot be sure who traveled. Without acceleration it is equally valid to consider that the Earth and spacestations moved at 0.8c and the traveler was standing still.
 
  • #32
The world lines of the twins intersect at Earth, t=t'=0. This is event A. After that the world lines don't come near each other. Event B is on the world line of the ship, at t'=6.

He has changed his original claim to just saying that it seems weird that t=10 at event B.
I didn't say it seems weird. I don't think it seems weird, or that there is any paradox. My original and current claim is that the situation is asymmetrical. Asymmetrical without acceleration.

And now that this same (acceleration independent) asymmetry is also present in the "twins paradox".

Thanks,
Al
 
  • #33
All objects travel from one event to the other in spacetime.

Remember this is relativity, if there is no acceleration you cannot be sure who traveled. Without acceleration it is equally valid to consider that the Earth and spacestations moved at 0.8c and the traveler was standing still.
In which case there was still only one observer present at both events. The Earth twin wasn't present at both events, neither was the space station observer. Only the ship's twin was at both events. Whether he "traveled" between the events, or not.

Thanks,
Al
 
  • #34
rahuldandekar
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I didn't say there was a paradox. Just an asymmetry.
Would you agree that the "twins paradox" contains this same asymmetry?

Thanks,
Al

No, the twins paradox contains an asymmetry of accelerating vs. non accelerating frames. We cannot make that situation symmetrical whatever we do.

There is a way to make this situation symmetrical. Forget the space station clock, just compare the Earth's clock with the spaceship clock.

Permit me to analyse the whole situation again.

Call event "spaceship leaves earth" as A and "spaceship meets space station" as B. Assume the space station and Earth clocks are synchronized in the Earth's frame. (Then Earth's clock lags in the spaceship frame.) Proper distance between Earth and spaceship is 8 light years.

In the Earth frame, spaceship covers 8 light years at 0.8c in 10 years. Due to time dilation, it sees 6 years pass in the spaceship's clock.

In spaceship frame, Earth's clock is initially at zero, and the distance between Earth and station is contracted to 4.8 ly. The space station covers this in 6.4 ly / 0.8c = 6 years. The space station clock, however, is seen running slower, and the spaceship sees 6/gamma = 3.6 years pass in the space station frame.

However, the space station frame reads different. Remember the spaceship's clock initially agreed with the Earth's clock. The space station clock, in the spaceship frame, was ahead of Earth's clock by L*v/c2 = 8 ly * 0.8c/ (c*c) = 6.4 years. (L is proper length.) So, the space station clock reads 6.4 + 3.6 = 10 years in the spaceship frame.

Hence the asymmetry. Spaceship's clock reads 6 years in both frames, and space station's clock reads 10 years in both frames.

However, forget the spaceship clock and consider the Earth's clock. It reads 3.6 years in the spaceship's frame between events A and B. The symmetry is now obvious, since we're comparing one moving clock with one stationary clock in each frame:

time on Earth's clock in spaceship frame = (time on spaceship's clock in spaceship frame)/gamma
time on spaceship's clock in Earth's frame = (time on Earth's clock in Earth's frame)/gamma.

Symmetry: interchange "earth" and "spaceship", and get the same equation.

Wait, there's still one asymmetry: time on Earth's clock in Earth frame is longer than time on spaceship clock in spaceship frame. But that creeps in because we say that spaceship is at rest with respect to earth, so, proper length is 8 ly. If we considered, say, another spaceship at 0.8c, 8 light years behind our spaceship, and considered the time difference between events C and D which are "spaceship one passes earth" and "spaceship two passes earth" respectively, we'd get exactly the reverse results.
 
  • #35
Fredrik
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I didn't say it seems weird. I don't think it seems weird, or that there is any paradox. My original and current claim is that the situation is asymmetrical. Asymmetrical without acceleration.
I have no idea what you mean by "asymmetrical". I certainly wouldn't describe the fact that two different coordinate systems assign different coordinates to the same event as "asymmetrical". This is what coordinate systems do. It's how they are defined. If they would assign the same coordinates to every event they'd be the same coordinate system.

It also shouldn't come as any surprise to someone who understands simultaneity that event B has different time coordinates in the two frames. Relativity of simultaneity means that the two twins have completely different ideas about which slices of space-time are "space", but they would have to agree on simultaneity to be able to assign the same time coordinate to event B.
 

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