Twin Paradox- a quick(ish) question

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  • #1
Smith987
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Hi all, I'm new here, I was on another forum asking this question but there was a lack of a response, so I hope you guys can help me out!

Ok, so after a few weeks of grappling with the twin paradox, I finally accept that the twin that travels on the rocket and back is the one that ages less.

(however)

If both twins had a stopwatch, the stay at home twin's watch would register a longer time when the twins meet again. My question is: what would the twins observe about the other twins stopwatch (imagine they have extremely good vision :P )?
The traveling twin would have to, at some point, see the stay at home twin's watch go faster, other wise how would it reach a larger value than his own?

Hope that makes sense, please help!
 

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  • #3
George Jones
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I wrote a(n unreadably) terse quantitative analysis of what atty posted,

https://www.physicsforums.com/showthread.php?p=1384776#post1384776.

A simpler version:

Two twins, Alfred and Betty, are together on the planet Omicron 7. After synchronizing their watches, Betty sets off on a return trip from Omicron 7 to Earth and back to Omicron 7, and Alfred remains the whole time on Omicron 7. The distance between Omicron 7 and Earth is 3.75 lightyears in the (approximately) inertial reference frame of Omicron 7 and the Earth. Betty takes the most direct route and moves at a constant speed of 3/5 lightspeed during both the outgoing and incoming segments of the trip. Both the outgoing an incoming legs of Betty's trip take 5 years according to her watch.

As Betty travels, Betty uses a telescope to watch Alfred's wristwatch. During Betty's outgoing leg, she sees Alfred's second hand spin slower than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5/2 = 2.5 years during the outgoing leg. During Betty's incoming leg, she sees Alfred's second hand spin faster than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5*2 = 10 years during the incoming leg.

During the whole trip:

Betty's watch advances by 5 + 5 = 10 years;

Alfred's watch advances by 2.5 + 10 = 12.5 years.
 
  • #4
Smith987
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Thanks! Makes sense now :)

That's cleared up a lot for me so thanks again!
 
  • #5
DrGreg
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Two twins, Alfred and Betty, are together on the planet Omicron 7. After synchronizing their watches, Betty sets off on a return trip from Omicron 7 to Earth and back to Omicron 7, and Alfred remains the whole time on Omicron 7. The distance between Omicron 7 and Earth is 3.75 lightyears in the (approximately) inertial reference frame of Omicron 7 and the Earth. Betty takes the most direct route and moves at a constant speed of 3/5 lightspeed during both the outgoing and incoming segments of the trip. Both the outgoing an incoming legs of Betty's trip take 5 years according to her watch.

As Betty travels, Betty uses a telescope to watch Alfred's wristwatch. During Betty's outgoing leg, she sees Alfred's second hand spin slower than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5/2 = 2.5 years during the outgoing leg. During Betty's incoming leg, she sees Alfred's second hand spin faster than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5*2 = 10 years during the incoming leg.

During the whole trip:

Betty's watch advances by 5 + 5 = 10 years;

Alfred's watch advances by 2.5 + 10 = 12.5 years.
George's argument is one the simplest explanations of the twins paradox and it follows directly from Einstein's postulates of relativity. The fact the same factor of 2 applies to the outward and inward journeys is no coincidence. Consider a third person Charlie who is on Earth all the time watching Alfred and Betty through a telescope. On the outward journey, of course he sees Alfred's clock hand spin at half the speed of Betty's hand, because that is what Betty sees (and it is the same light traveling from Alfred to Betty that continues to Charlie).(1) But Charlie and Alfred are stationary relative to each other, so they see each other's clock hands spin at the same rate. Therefore Charlie sees his own clock hand spin at half the speed of Betty's hand. Or in other words, as Betty travels towards Charlie, Charlie sees Betty's hand spin at twice the speed as his own.

On the return journey, Betty sees Alfred moving towards her at exactly the same speed that Charlie saw Betty moving towards him in the outward journey. By the principle of relativity,(2) both situations are equivalent, so Betty sees Alfred's hand spin at twice the speed as her own.

This argument is an example of k-calculus. (So called because you can use the same argument with any other number instead of 2, e.g. k.)


(1) This is an application of Einstein's 2nd postulate: light emitted by Alfred towards Betty and Charlie travels at the same speed as light emitted by Betty towards Charlie.

(2) i.e. Einstein's 1st postulate
 
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  • #6
NWH
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But Charlie and Alfred are stationary relative to each other, so they see each other's clock hands spin at the same rate.
You're assuming the gravity on Earth and Omicron 7 are the same though, right?
 
  • #7
George Jones
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You're assuming the gravity on Earth and Omicron 7 are the same though, right?

I assumed that gravity is negligible, i.e., special relativity.
 
  • #8
NWH
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Got you!
 
  • #9
otg
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As Betty travels, Betty uses a telescope to watch Alfred's wristwatch. During Betty's outgoing leg, she sees Alfred's second hand spin slower than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5/2 = 2.5 years during the outgoing leg. During Betty's incoming leg, she sees Alfred's second hand spin faster than hers by a factor of 2. Thus, Betty sees Alfred's watch advance by 5*2 = 10 years during the incoming leg.

During the whole trip:

Betty's watch advances by 5 + 5 = 10 years;

Alfred's watch advances by 2.5 + 10 = 12.5 years.

I wonder, how is it [that Alfred's watch is speeding from Betty's point of view] consistent with the fact (?) that a watch relative to which one is moving, slows down?

I see the argument with Doppler shift as clear as a day, but I don't find a "pure" time dilation argument to why the symmetry is broken. The gamma factor does not take into account whether one is approaching or moving away.

If one does not use Doppler shift as an argument, I don't find a way to prove that the situation is not possible to reverse [as in: Betty is at rest and Alfred is moving away and Alfred is younger at the reunion]

Any ideas?
 
  • #10
JesseM
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I wonder, how is it [that Alfred's watch is speeding from Betty's point of view] consistent with the fact (?) that a watch relative to which one is moving, slows down?

I see the argument with Doppler shift as clear as a day, but I don't find a "pure" time dilation argument to why the symmetry is broken. The gamma factor does not take into account whether one is approaching or moving away.

If one does not use Doppler shift as an argument, I don't find a way to prove that the situation is not possible to reverse [as in: Betty is at rest and Alfred is moving away and Alfred is younger at the reunion]

Any ideas?
The time dilation equation only works in inertial frames, it can't be applied in a non-inertial one where Betty is at rest throughout the trip. It is true that if you take the inertial frame where Betty is at rest during the outbound phase before the turnaround (assuming Betty moves inertially prior to turning around), then in this frame Alfred ages more slowly than Betty during the outbound phase; but after the turnaround, in this frame Betty has an even higher velocity than Alfred, and therefore ages more slowly than him. Likewise, in the inertial frame where Betty is at rest during the inbound phase after the turnaround, in this frame Alfred ages more slowly than Betty during the inbound phase, but in this frame Betty had a higher velocity than Alfred during the outbound phase and was aging slower than him during that phase. And you can't combine results from the two frames for the half of the trip where Betty was at rest in each frame, because the two frames disagree about simultaneity, and therefore have a large disagreement about Alfred's age at the moment of the turnaround. For example, suppose Betty moves at 0.6c for 25 years in Alfred's frame, then returns at 0.6c for another 25 years, so Alfred ages 50 years by the time she returns but Betty only ages 25*sqrt(1 - 0.6^2) = 25*0.8 = 20 years during each phase, so she is only 40 years older when she returns. In the first of Betty's two different rest frames, the frame where Betty was at rest during the outbound phase, at the moment of the turnaround Betty is 20 years older than when she departed Alfred, while Alfred is only 20*0.8 = 16 years older than when Betty departed. But then in the frame where Betty was at rest during the inbound phase, Betty is still 20 years older at the moment of the turnaround but Alfred is already 34 years older, And she'll age another 20 years from then until when she reunites with Alfred while he'll only age 16 more years, but because he was already 34 at the moment of the turnaround in this frame he'll be 34 + 16 = 50 years older when they reunite.
 
  • #11
otg
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thank you, I knew I had seen it somewhere, it's just that thing about "Alfred har aged 16 years and instantaneosly he has aged 34" that bothers me... but I guess I have to live with that :)
 
  • #12
JesseM
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thank you, I knew I had seen it somewhere, it's just that thing about "Alfred har aged 16 years and instantaneosly he has aged 34" that bothers me... but I guess I have to live with that :)
But it isn't "instantaneous" in that frame, it's only "instantaneous" if you try to imagine constructing a non-inertial frame where Betty is at rest throughout the trip, a frame whose measurements match those of the first inertial frame during the outbound phase and those of the second inertial frame during the inbound phase. In the inertial frame where Betty is at rest during the inbound phase, prior to the turnaround Betty had been traveling even faster than Alfred--using the velocity addition formula, we can see that in this frame Betty must have had a velocity of (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c, so in this frame her clock must have been slowed by sqrt(1 - 0.88235^2) = 0.4706. So if she aged 20 years during the outbound phase in this frame, the "actual" time in this frame before turning around must have been 20/0.4706 = 42.5 years. And in this frame Alfred is always moving at 0.6c so he always has a time dilation factor of sqrt(1 - 0.6^2) = 0.8, so during those 42.5 years he must have aged 42.5*0.8 = 34 years in this frame. So you see in this frame nothing special happened to Alfred at the moment of the turnaround itself, their relative ages at the moment of the turnaround are just a consequence of their velocities in this frame since Betty departed from Alfred.
 
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  • #13
Ich
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that thing about "Alfred har aged 16 years and instantaneosly he has aged 34" that bothers me...
Yes, it should bother you. Nobody claimed that.
See it as a change in perspective, like in this analogon: if you look at Rigel, 800 ly away, and then turn your head 90 degrees to the right, Rigel is suddenly 1100 ly further to the left. It would certainly bother you if someone claimed that Rigel moved 1100 ly in less than a second. You don't have to live with that.
 
  • #14
otg
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So you see in this frame nothing special happened to Alfred at the moment of the turnaround itself, their relative ages at the moment of the turnaround are just a consequence of their velocities in this frame since Betty departed from Alfred.

Yes, and I realized my error in formulation [that's why I'm back, but I see that I don't have to rephrase myself :)].

thank you very much for the comments
 
  • #15
otg
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In the inertial frame where Betty is at rest during the inbound phase, prior to the turnaround Betty had been traveling even faster than Alfred

hmm... gave it some more thought, and I can't really convince myself that it would be any different to make the same argument in the inertial frame where Alfred is at rest during the inbound leg, for instance... I find that Alfred is younger than Betty at the reunion...
Inbound Alfred has a velocity relative to the outbound Alfred which is higher than the velocity relative to Betty at all times...

Where's my glitch?

Darn my head is turning inside out :)
 
  • #16
otg
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Where's my glitch?

perhaps it was in that I forgot to shrink the distance earth-star in the spaceship-outbound-rest-frame... yikes this is not to be done late at night...
 
  • #17
otg
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perhaps it was in that I forgot to shrink the distance earth-star in the spaceship-outbound-rest-frame...

nope... didn't help me...
 
  • #18
JesseM
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hmm... gave it some more thought, and I can't really convince myself that it would be any different to make the same argument in the inertial frame where Alfred is at rest during the inbound leg, for instance... I find that Alfred is younger than Betty at the reunion...
Inbound Alfred has a velocity relative to the outbound Alfred which is higher than the velocity relative to Betty at all times...
I thought Alfred was the one moving inertially--if you pick the inertial frame where he's at rest during the inbound leg, then he's also at rest during the outbound leg, no?
 
  • #19
Mentz114
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Somewhere in this forum ( can't find it ) is a descrption of what a traveller sees as he accelerates away from the earth, turns round and comes back. On the outward trip the traveller sees the Earth apparently turning more and more slowly as he moves away, and on the return trip sees the Earth spinning faster and faster ( due to separation and light delay) until, when she touches down the number of rotations seen is equal to the number experienced by those left behind. Anything else would be a contradiction. But the travellers clock will still be behind those left on earth. So the sight of planet Earth turning more or less slowly or quickly means nothing since they'll agree at the end on the same number. But the travellers clock will still be showing less elapsed time than those left behind.

It doesn't matter a swat which frames are inertial or not, the elapsed times on clocks traveling a worldline ( according to SR and GR) is the sum of dt2-dx2. End of story. You don't need to know much else to resolve the twins non-paradox.
 
  • #20
sylas
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Somewhere in this forum ...

George does it in [post=2186296]msg #3[/post] of this thread.
 
  • #21
Mentz114
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George does it in [post=2186296]msg #3[/post] of this thread.
Yes indeed he has. But The OP doesn't get it even after that. It's a message that's going to be repeated a lot on this forum. An automatic answer would save everyone a lot of time.

[sylas, did you see the visitor message I left on your profile page ?]
 
  • #22
JesseM
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Yes indeed he has. But The OP doesn't get it even after that.
Why do you say that? The OP said:
Thanks! Makes sense now :)

That's cleared up a lot for me so thanks again!
The poster who's currently still confused, otg, isn't really asking a question about visual appearances as far as I can tell.
 
  • #23
sylas
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Yes indeed he has. But The OP doesn't get it even after that. It's a message that's going to be repeated a lot on this forum. An automatic answer would save everyone a lot of time.

[sylas, did you see the visitor message I left on your profile page ?]

Yes, thanks. I'm not sure how to reply... I left a visitor message to myself as a reply.

IMO, each new generation of students gets their minds blown by relativity. It's perfectly normal to take quite a while to wrap your head around it. It took me quite some time. There are a number of good canned responses around, and newcomers will still want to talk their way through it, and still find aspects of the description confusing.

Hence I expect threads talking about the twin paradox to continue, indefinitely. And that's a good thing, as a way for people to develop their understanding of how it works. It's also handy for amateurs to try their hand at giving the explanations... though there's always a risk of an amateur actually getting the explanations wrong. I'm in this category. I think I've figured out SR pretty well now, but I'd not be a reliable guide on GR yet.

Cheers -- sylas
 
  • #24
matheinste
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Hello all.

Regarding the twin paradox. It is a useful teaching device which raises many points for a beginner the resolution of which helps to gain a better understanding of SR. But as Mentz implies, once you see it from the point of view of spacetime intervals you wonder what all the fuss is about.

Matheinste.
 
  • #25
otg
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I thought Alfred was the one moving inertially--if you pick the inertial frame where he's at rest during the inbound leg, then he's also at rest during the outbound leg, no?

But, if you don't introduce concepts of acceleration or doppler shift or whatever, how is it possible to distinguish Alfred as the inertial one during the whole trip? It's just a matter of changing the names of the persons (and changing the word "space ship" to "earth")

If I place Betty in a rest frame in which she sees Alfred move away with the Earth and stars and everything, and then Alfred comes back again with the Earth and everything, how can I claim that this is any different from seeing Alfred at rest during the whole thing?

The relative velocity-argument from "12 JesseM" is, imho, reversible, since I just changed the names of the persons and their vessels, sort of. But I am also convinced that it isn't, so I need to find out what the difference is...
 
  • #26
But, if you don't introduce concepts of acceleration or doppler shift or whatever, how is it possible to distinguish Alfred as the inertial one during the whole trip?
Because it's the lack of acceleration that makes an observer "inertial" by definition. The concept of an inertial reference frame cannot exist without the concept of acceleration preceding it.
 
  • #27
otg
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Because it's the lack of acceleration that makes an observer "inertial" by definition. The concept of an inertial reference frame cannot exist without the concept of acceleration preceding it.

But I don't need the concept of acceleration to see that there is a change of frame for Betty at the turnaround. I can simply let her synchronize her clock with a third person who seems identical to Betty in Betty's outbound frame at the event of synchronization although she's moving in the opposite direction.
Why can't I change the name "Betty" to the name "Alfred" without changing her age at the reunion?
If I remove the earth, I can place Alfred in a spaceship where the Earth "used to be". How can Betty tell that it's not the Alfredian spaceship which is moving, changing frame and returning?
Why do we need the concept of acceleration to explain why her frame is non-inertial, if there is no such thing as an absolute space in relation to which one accelerates? That is, Betty needs to accelerate to return to Alfred, yes, but Alfred has to return to Betty in the other point of view, how can he do that without accelerating? Why is he returning without changing frame?

If there are three people involved as above, how can we tell that there are two Bettys and not two Alfreds?

I can calculate the dilations in the respective frames, but if I code the A, B and C on my gamma on my paper, I cannot tell which letter "actually" is which frame, so I cannot distinguish Alfred at rest all the time [where he moves away all the time and halfway meets an identical Alfred moving in the opposite direction] from the one where Betty is at rest all the time [where she moves away all the time and halfway meets an identical Betty moving in the opposite direction
 
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  • #28
sylas
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But I don't need the concept of acceleration to see that there is a change of frame for Betty at the turnaround. I can simply let her synchronize her clock with a third person who seems identical to Betty in Betty's outbound frame at the event of synchronization although she's moving in the opposite direction.
Why can't I change the name "Betty" to the name "Alfred" without changing her age at the reunion?

Her age is determined by the path she takes through space and time. For any path, there is an associated "proper time".

If someone is moving in the opposite direction, they WON'T be identical. They will have a large velocity relative to Betty.

If I remove the earth, I can place Alfred in a spaceship where the Earth "used to be". How can Betty tell that it's not the Alfredian spaceship which is moving, changing frame and returning?

If she is the one that turns around, then she will notice a significant reduction in the angular size of stay-at-home Alfred in the sky. Alfred, on the other hand, knows that it is Betty who is turned around because she's still the same angular size in the sky.

This is precisely because when you turn around, you have a new perspective, in which the other twin is much further away. Not "seems" further away... IS further away, by any means open to them for measuring distances. If the other twin is the one who turns, the distance to that twin is the same just before and just after the turn around.

This sounds impossible... but only because we are used to life with much smaller velocities where the effect is not noticeable. If we habitually moved near light speed, this change of perspective would be as natural as the fact that when WE turn around, the direction towards another person is altered... but when they turn around, the direction is not altered.

Cheers -- sylas
 
  • #29
matheinste
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Her age is determined by the path she takes through space and time. For any path, there is an associated "proper time".

If someone is moving in the opposite direction, they WON'T be identical. They will have a large velocity relative to Betty.



If she is the one that turns around, then she will notice a significant reduction in the angular size of stay-at-home Alfred in the sky. Alfred, on the other hand, knows that it is Betty who is turned around because she's still the same angular size in the sky.

This is precisely because when you turn around, you have a new perspective, in which the other twin is much further away. Not "seems" further away... IS further away, by any means open to them for measuring distances. If the other twin is the one who turns, the distance to that twin is the same just before and just after the turn around.

This sounds impossible... but only because we are used to life with much smaller velocities where the effect is not noticeable. If we habitually moved near light speed, this change of perspective would be as natural as the fact that when WE turn around, the direction towards another person is altered... but when they turn around, the direction is not altered.

Cheers -- sylas

Although I agree that otg is incorrect I think also that your statement about distances and angular size are also incorrect.The distance between the twins does not change immediately at turnaround.

It is easy to tell which twin is accelerating. It is the one that feels the force of acceleration.
But as has been said before, it is the different paths taken through spacetime by the twins that accounts for the differential ageing.

Matheinste.
 
  • #30
sylas
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Although I agree that otg is incorrect I think also that your statement about distances and angular size are also incorrect.The distance between the twins does not change immediately at turnaround.

Yes, it does change immediately at turn around. It changes immediately at turn around in precisely the same way as the direction to the other twin changes immediately at turn around.

If you switch from 60% light speed outbound, to 60% light speed inbound, the angular size of the objects in your direction of motion change instantly by a factor of 4, reflecting their new distance in the new reference frame. Recall that the size of an object in the sky depends not on the "current distance", but the distance that has been traveled by the photons now reaching you.

As you return back towards Sol, it will grow steadily in apparent size, consistent with Sol's velocity (in your frame) of 60% light speed.

Acceleration is not required. This is another common mistake. Acceleration is not what matters. It is the turn around, or the change of perspective, that matters. The same thing would happen if you use a teleporter to move from an outbound Enterprise to an inbound Klingon freighter, with no accelerations involved. You don't even need to transport... you can just communicate with a friendly passing Klingon. If you are headed out from Sol, at 60% light speed, and they are headed towards Earth at 60% light speed, then the solid angle subtended by Sol will be four times smaller for the inbound Klingons than for you, outbound. That's because it is further away in their reference frame than in yours.

Cheers -- sylas

Addendum: Oops. Actually the solid angle is 16 times smaller... each of the two transverse angles is 4 times smaller.
 
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  • #31
otg
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Yes, it does change immediately at turn around. It changes immediately at turn around in precisely the same way as the direction to the other twin changes immediately at turn around.

You can just communicate with a friendly passing Klingon. If you are headed out from Sol, at 60% light speed, and they are headed towards Earth at 60% light speed, then the solid angle subtended by Sol will be four times smaller for the inbound Klingons than for you, outbound. That's because it is further away in their reference frame than in yours.

But that doesn't say anything about why it is not the Earth that is communicating with a friendly passing Earth with a klingon on it. Then the far-away spaceship would be different in size depending on whether you are the "original" earthling or the klingon. You say no, I say why?

I know I'm wrong, but I don't feel where the difference is. And I don't know if you understand what I'm looking for either... :)

If I draw a pic of the situation, I can replace the triangles symbolizing the spaceship frames in relative motion with circles symbolizing earth-frames in relative motion, and the circle symbolizing the earth-frame at rest with a triangle symbolizing a spaceship frame at rest.
[What if the Earth were triangular in shape? ;)]

But no, I can't. I would like to avoid the spacetime diagrams. (also, why can't I let the time axis be the spaceship and the Earth move farther than the ship in a diagram?)

[And thank you very much for all your efforts, I've been looking around in the forum and tried to find the answer but I cannot find the answer anywhere, that's why I'm terrorizing you in yet-another-thread :)]
 
  • #32
sylas
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But that doesn't say anything about why it is not the Earth that is communicating with a friendly passing Earth with a klingon on it. Then the far-away spaceship would be different in size depending on whether you are the "original" earthling or the klingon. You say no, I say why?

The difference between communication with a passing Klingon and a Klingon at a great distance is that you can no longer treat communication as a single well defined event. It is the event of transmission, and then another event of reception, some time later. Further more, the distance traveled by the photon for a given communication will vary depending on who is measuring it.

Hence you can synchronize your watch with a passing Klingon, but not with a distant Klingon.

But no, I can't. I would like to avoid the spacetime diagrams. (also, why can't I let the time axis be the spaceship and the Earth move farther than the ship in a diagram?)

I don't know what's going to work best for you. Everyone is different. Personally, I have found space time diagrams very useful. You can do a diagram where Earth is stationary, or where the outbound ship is stationary, or where the inbound ship is stationary. Or you can use a passing Klingon with a totally different velocity of their own. All the diagrams can show the same events and calculate the same "proper time" elapsed between events for any of the travelers.

If you don't like diagrams, then you could use the Lorentz transformations.

What you can't do is hold on to your current intuitions. You are going to need to retrain your intuitions to match what is really happening at high velocities.

Cheers -- sylas
 
  • #33
otg
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Hence you can synchronize your watch with a passing Klingon, but not with a distant Klingon.

Yes, but I never mentioned any distant klingon. So far it seems that we have concluded two apparently identical situations which are different:

1.
Spaceship-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the Earth should see the distant Earth at a different angle than I do.

2.
Earth-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the spaceship should NOT see the distant spaceship at a different angle than I do.

The only thing that differs from the points of view is the placement of the words "spaceship" and "earth", which have changed places.
This redefinition of the words induces a "NOT" in the second situation. How is this possible to defend without calculations? And how is this possible to defend with calculations? Why do the calculations (or spacetime diagrams) take into account the definition of the words "earth" and "spaceship"?
 
  • #35
sylas
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Yes, but I never mentioned any distant klingon. So far it seems that we have concluded two apparently identical situations which are different:

1.
Spaceship-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the Earth should see the distant Earth at a different angle than I do.

This is what happens.

2.
Earth-klingon moving in the opposite direction from me with whom I sync my watch in the frame where I move away from the spaceship should NOT see the distant spaceship at a different angle than I do.

This doesn't happen. A Klingon passing by the Earth at high speed WILL be in a different frame to the Earth, and will be at a different distance from any other remote events by comparison with Earth-bound observer. Even though they are at the same place and time.

Distances and times to remote events depend on your velocity, as well as your location in spacetime.

The only thing that differs from the points of view is the placement of the words "spaceship" and "earth", which have changed places.
This redefinition of the words induces a "NOT" in the second situation. How is this possible to defend without calculations? And how is this possible to defend with calculations? Why do the calculations (or spacetime diagrams) take into account the definition of the words "earth" and "spaceship"?

All you've done wrong is presume that the Klingon passing by the Earth will see things at the same size, and distance as someone on Earth. They won't. Different frame, different distances and times for remote events.

Cheers -- sylas
 

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