Twin Paradox- a quick(ish) question

[Smith987]

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!

 

[atyy]

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

 

[George Jones]

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.

 

[Smith987]

Thanks! Makes sense now :)

That's cleared up a lot for me so thanks again!

 

[DrGreg]

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

 

[NWH]

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?

 

[George Jones]

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.

 

[NWH]

Got you!

 

[otg]

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?

 

[JesseM]

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.

 

[otg]

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 :)

 

[JesseM]

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.

 

[Ich]

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.

 

[otg]

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

 

[otg]

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 :)

 

[otg]

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...

 

[otg]

perhaps it was in that I forgot to shrink the distance earth-star in the spaceship-outbound-rest-frame...

nope... didn't help me...

 

[JesseM]

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?

 

[Mentz114]

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.

 

[sylas]

Somewhere in this forum ...

George does it in [post=2186296]msg #3[/post] of this thread.

 

[Mentz114]

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 ?]

 

[JesseM]

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.

 

[sylas]

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

 

[matheinste]

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.

 

[otg]

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...

 

[Al68]

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.

 

[otg]

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

 

[sylas]

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

 

[matheinste]

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.

 

[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.

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.

 

[otg]

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 :)]

 

[sylas]

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

 

[otg]

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"?

 

[Demystifier]

This may help:
http://xxx.lanl.gov/abs/physics/0004024 [Found.Phys.Lett. 13 (2000) 595-601]

 

[sylas]

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

 

[otg]

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.

Ah, but I didn't presume this. I never said that the Klingon passing the Earth would be in the same frame as me on the earth. This was only a result of the previous claim that the earth-at-rest-frame-point-of-view is different from the spaceship-at-rest-frame-point-of-view. I summarized the situations in two sentences in which I used the exact same words, except for the NOT, which made you say there is a difference between 1 and 2.

If what you write is true, that 1 will occur, but also that 2 will occur (if I remove the NOT, since it won't if I keep it), then 1 and 2 are indistinguishable. [if not, why? [my original [and so far only] question]

[I sometimes get a feeling that I think I ask something different than what you think you're answering... is this observation in my rest frame different in your rest frame? :)
You do have a good way of replying. Thank you for your patience]

 

[sylas]

Ah, but I didn't presume this. I never said that the Klingon passing the Earth would be in the same frame as me on the earth.

You effectively presumed this when you said that the distances are not changed, and when you say passing. If it is passing, then it is moving, and therefore it is NOT in the same frame. (text in blue added in edit)

If what you write is true, that 1 will occur, but also that 2 will occur (if I remove the NOT, since it won't if I keep it), then 1 and 2 are indistinguishable. [if not, why? [my original [and so far only] question]

I do not think you are defining the situation sufficiently carefully. This is why you badly need to come up with maths, or a diagram, or SOMETHING that will let you be precise. Without that, you will almost certainly make invalid assumptions without meaning to, and without even realizing it.

In your original statement for example, you say this:

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.

Note that you speak of TWO Klingons. One at Earth, one at a spaceship. And you also speak of synchronizing your watch to both of them.

That's impossible. You cannot synchronize with two events that are separated in space, because you and the Klingon don't have a common reference frame for times. I explained this previously, and you objected. This is almost certainly because you have not yet got your head around the idea that simultaneity is relative. Nearly always, the problems come down to this.

You can have an Earthbound twin, who synchronizes with a Klingon passing by the Earth, and a traveling twin, who synchronizes with another Klingon as they both pass by a distant star. You can only synchronize when you have observers at the same location in the same instant.

Try again. Spell out exactly where and when you have any synchronizations.

Here's an account of different events.

I am assuming we have Earth, and another star six light years away, which is at rest with respect to the Earth. The six light years distance is as measured from Earth, or from the star. Same thing, as they are at rest with respect to each other.

I am assuming we have a traveler, who travels from Earth to the star at 60% of the speed of light, and then returns at the same speed. It takes 20-Earth years to make the round trip. During the trip, they age 16 years.

Now. Where are the Klingons you speak of? If passing Earth, WHEN do they pass Earth. If passing the other star, WHEN do they pass the other star? At what speed, and what direction?

I'm proposing Klingon-A, who moves always in a straight line, at 60% light speed wrt Earth, and in a direction from the star towards Earth. This Klingon passes by the star in the same instant that the traveler from Earth arrives at the star.

I did this, because THIS Klingon can pick up the traveler by a short range teleportation device, and bring them back to Earth. It makes it convenient to describe where Earth is from the perspective of the inbound Klingon, or the outbound Enterprise, as the both pass by each other and by the star.

Clear enough? If you have another Klingon in mind, see how carefully you can identify where they are. You speak of them passing Earth. When? In what direction? At what speed?

Klingon-A, by the way, observes that Earth is approaching them at 60% light speed. As the star flies past the Klingon, they are observing light from Earth that was emitted twelve years previously, when Earth was at a distance of 12 light years.

The Enterprise, traveling in the other direction, observes that Earth is receding into the distance at 60% light speed. As the star flies past the Enterprise, they are observing light from Earth that was emitted three years previously, when Earth was at a distance of 3 light years.

For aliens living on the star, they are observing Earth with light that was emitted six years previously. Earth is not moving as far as these aliens are concerned, and is always 6 light years distant.

All three observations are looking at the same photons. If a Bomb is let off on Earth at the right time, all three observers -- Enterprise, Klingon-A, and the aliens at the star, will see the bomb in the same instant as Enterprise and the Klingon pass by the star. The distance to the explosion is different for all observers. It depends not only on time and location, but on velocity as well.

The bomb will need to be detonated 4 Earth-years after the Enterprise left.

Cheers -- sylas

 

[otg]

Note that you speak of TWO Klingons. One at Earth, one at a spaceship. And you also speak of synchronizing your watch to both of them.

Still haven't read the reply thoroughly, just wanted to make clear that I did not speak of two klingons to both of which I sync my watch. I simply [think that I] had the situation described in two different ways.

Either I see it as

person being at rest in the spaceship frame in the outbound leg [spaceship moving out away from the earth] meeting a klingon going the other direction and syncing with him

OR

person being at rest in the Earth frame at the outbound leg [earth moving out away from the spaceship] meeting a klingon going the other direction and syncing with him

I still have only renamed the spaceship "earth", but there is supposed to be a difference. I'll read your last reply and hopefully never come back to this forum in this issue :)

 

[sylas]

Still haven't read the reply thoroughly, just wanted to make clear that I did not speak of two klingons to both of which I sync my watch. I simply [think that I] had the situation described in two different ways.

Either I see it as

person being at rest in the spaceship frame in the outbound leg [spaceship moving out away from the earth] meeting a klingon going the other direction and syncing with him

OR

person being at rest in the Earth frame at the outbound leg [earth moving out away from the spaceship] meeting a klingon going the other direction and syncing with him

One difference is the relative velocity of your passing Klingon. Assuming that all Klingons are moving in the direction from star towards Earth, at 60% light speed wrt to Earth. Then the Klingons are moving at over 88% light speed wrt to the outbound traveller. (Actually 15/17 of light speed.)

The Klingon passing Earth and looking back at the receding star sees light that left the star 3 years previously. The Earth sees the same light, and that light left the star six years previously. Both distances are correct, for the different observers.

The Klingon passing the star looks forward to see the approaching Earth, and sees light that left Earth 12 years previously. The Enterprise passing the star looks back to see the receding Earth, and sees light that left Earth 3 years previously. (And this IS symmetrical with the Klingon passing the Earth and leaving the star behind.)

You can calculate these numbers by using the Lorentz transformations.

I still have only renamed the spaceship "earth", but there is supposed to be a difference. I'll read your last reply and hopefully never come back to this forum in this issue :)

Don't be shy... very few people get this straight away; I certainly didn't.

Cheers -- sylas

 

[JesseM]

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.

The fact remains that you are comparing a path made up of two segments that have different velocities as seen in every inertial frame with Alfred's clock which has a constant velocity in every inertial frame. What's more, in Betty's frame this third person's clock does not read the same time as her clock at the moment the third person passes Alfred, because in Betty's frame the third person's clock is slowed down.

Why can't I change the name "Betty" to the name "Alfred" without changing her age at the reunion?

What do you mean "change the name"? Do you mean having Alfred be the one that either accelerates to return to Betty, or just synchronizes clocks with a third person and never reunites with Betty himself?

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?

Because if an accelerometer is placed on Alfred's ship it will read zero throughout, showing that it was moving at constant velocity in every inertial frame. It's implied by the definition of an inertial frame that any object at rest or moving at constant velocity in an inertial frame must experience zero G-forces as measured by accelerometers.

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?

Because acceleration is absolute in SR--you can tell objectively whether you're accelerating or not by looking at an accelerometer.

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

I don't understand this question--obviously there must be an objective fact of the matter as to whether the third person passes Alfred or Betty first (and synchronizes clocks with whoever he passes first at the moment they pass). Relativity doesn't involve uncertainty about the identity of different worldlines, after all.

 

[matheinste]

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.

Cheers -- sylas

Addendum: Oops. Actually the solid angle is 16 times smaller... each of the two transverse angles is 4 times smaller.

One thing I cannot grasp is that if we consider a position just before turnaround and a position just after turnaround where the relative speeds are equal and opposite in direction
how can the "new" distance between traveller and Earth be different on these two occasions. Surely the distance depends only on relative speed and not on direction. As nobody else has any objection to this description I must assume that there is something lacking in my understanding of the situation. Perhaps you could throw in some mathematics to make it clearer to me.

I have never had a problem with the resolution of the twin paradox before but have never come across this aspect of it. I know that we can also resolve the paradox using the Lorentz contracted distance for the traveller which the earthbound twin does not experience, but as far as the traveller is concerned his inbound and outbound legs are symmetrical with regards to distance, although of course his line of simultaneity is different in and outbound.

This point is really bothering me and I would be glad of some help in its resolution.

To recap, my query is, how is the distance between the traveller and Earth different immediately before turnaround from immediately after turnaround.

Matheinste

 

[sylas]

One thing I cannot grasp is that if we consider a position just before turnaround and a position just after turnaround where the relative speeds are equal and opposite in direction
how can the "new" distance between traveller and Earth be different on these two occasions. Surely the distance depends only on relative speed and not on direction. As nobody else has any objection to this description I must assume that there is something lacking in my understanding of the situation. Perhaps you could throw in some mathematics to make it clearer to me.

Use the Lorentz transformations. Remember… velocity is signed. The reference frame for a ship passing the star with v = 0.6 is different from the one in which the ship passes the star with v = -0.6. I am using units where c=1.

Here is the general form of the Lorentz transform. Let (t,x) be a spacetime event in the frame of observer A. Assume that observer B has a velocity of v with respect to A.

To map (t,x) into the frame of B, use the following:

\begin{equation*}\begin{split}<br /> t&#039; &amp; = \gamma ( t - vx ) \\<br /> x&#039; &amp; = \gamma ( x - vt ) \\<br /> \gamma &amp; = \frac{1}{\sqrt{1 - v^2}}<br /> \end{split}\end{equation*}​

To map back into the frame of A, note that A has velocity of -v with respect to B. Hence we use

\begin{equation*}\begin{split}<br /> t&#039;&#039; &amp; = \gamma ( t&#039; + vx&#039; ) \\<br /> &amp; = \gamma^2 ( t - vx + v(x- vt) ) \\<br /> &amp; = \gamma^2 ( 1 - v^2 ) t \\<br /> &amp; = t<br /> \end{split}\end{equation*}​

A similar proof applies for x'' = x. In other words, it's all consistent. You map from the frame of A to the frame of B and back again with the same transformations, but remember that the sign of velocity is reversed.

OK. Make it concrete. Let (0,0) be the turn around event. In the frame of the star, the Earth is 6 light years away. Hence, in the frame of the star, the light that they are now seeing from Earth was emitted six years ago, and at a distance of six light years. Assume that the positive direction is from Earth towards the star. Thus we have the light from Earth being emitted at the spacetime event (-6,-6), in the frame of the star.

We can transform that event at (-6,-6) for a passing spaceship moving at speed 0.6. We get different results depending as the velocity is positive, or negative. (We are keeping to one dimension for simplicity throughout.)

Here's what you have for v=0.6, which is a spaceship outbound from Earth:

\begin{equation*}\begin{split}<br /> t &amp; = x = -6 \\<br /> v &amp; = 0.6 \\<br /> \gamma &amp; = 1.25 \\<br /> t&#039; &amp; = \gamma ( t - vx ) \\<br /> &amp; = 1.25 \times -6 \times (1-v) \\<br /> &amp; = 1.25 \times -6 \times 0.4 \\<br /> &amp; = -3 \\<br /> x&#039; &amp; = \gamma ( x - vt ) \\<br /> &amp; = 1.25 \times -6 \times (1-v) \\<br /> &amp; = -3<br /> \end{split}\end{equation*}​

They are seeing Earth from a distance of 3 light years, with light that left 3 years previously.

Here's what you have for v=-0.6, which is a spaceship inbound for Earth:

\begin{equation*}\begin{split}<br /> t &amp; = x = -6 \\<br /> v &amp; = -0.6 \\<br /> \gamma &amp; = 1.25 \\<br /> t&#039; &amp; = \gamma ( t - vx ) \\<br /> &amp; = 1.25 \times -6 \times (1-v) \\<br /> &amp; = 1.25 \times -6 \times 1.6 \\<br /> &amp; = -12 \\<br /> x&#039; &amp; = \gamma ( x - vt ) \\<br /> &amp; = 1.25 \times -6 \times (1-v) \\<br /> &amp; = -12<br /> \end{split}\end{equation*}​

They are seeing Earth from a distance of 12 light years, with light that left 12 years previously.

You can also use these transformations for a pinhole camera on board the spaceships, and use that to show that the angle subtended by light moving in the direction of motion has an angle that scales with the redshift.

Cheers -- sylas

 

[matheinste]

Hello sylas.

Thanks for a quick reply. I see what you are saying but I look at it as distance being the same but the line of simultaneity being different. That is, we are seeing the Earth as it is/was at different times, not from different distances.

I wil take a more detailed look later. Thanks.

Matheinste.

 

[sylas]

Hello sylas.

Thanks for a quick reply. I see what you are saying but I look at it as distance being the same but the line of simultaneity being different. That is, we are seeing the Earth as it is/was at different times, not from different distances.

I wil take a more detailed look later. Thanks.

Matheinste.

If you are seeing it at different times, it MUST also be seen at different distances. You are seeing it with light, that travels at the same speed for everyone. And note here that "different time" means the SAME instant, the SAME spacetime event -- which occurred a different amount of time previously depending on your perspective.

Just remember that what you see is not simultaneous with the observation. Hence simultaneity is not what you want for the angular size of something in the sky. What you see in the sky is events that have already occurred in the past.

For both the inbound and outbound ship, the distance to the Earth "now" is 4.8 light years. The problem is that they don't have the same idea of "now", and they aren't seeing the Earth "now".

They CAN identify a common event along a photon light path. They are both seeing light from the Earth that was emitted at an unambiguously identified event in the past. They just locate it at a different time and distance. The sudden change in angular size is another way to recognize that you have turned around, even if you fail to notice an acceleration for some reason.

Cheers -- sylas

 

[matheinste]

Hello sylvas.

In a standard spacetime diagram using the inertial frame of Earth there is no distance change immediately before/after the traveller's turnaround. Is this change represented in the spacetime diagram using the successive frames of the traveller?

Matheinste

 

[sylas]

Hello sylvas.

In a standard spacetime diagram using the inertial frame of Earth there is no distance change immediately before/after the traveller's turnaround. Is this change represented in the spacetime diagram using the successive frames of the traveller?

Matheinste

Yes, there is a distance change.

Recall, we are talking of the distance traveled by light from Earth, as seen at the turn around; not the distance between two events along anyone's plane of simultaneity.

Also, your standard spacetime diagram shows the perspective of one reference frame; and so you would need TWO diagrams to show what is seen before, and after, the turn around. Or you can use a non-standard diagram.

Here I show FOUR spacetime diagrams, all on the one grid. Scale is years (vertical) and lightyears (horizontal). There are four world lines shown.

The diagrams show four ways to look at the same history of events. This involves Capt. Kirk traveling outbound from Earth to another star, 6 light years from Earth as measured on Earth, at which point Kirk teleports across to a passing inbound Klingon freighter. A bomb is set off on Earth just at the right time to be seen by Kirk and the Klingons as teleportation occurs.

World lines:

• Green is the Earth, and a light green circle for the bomb event.
• Red is light from the bomb, received at the turn around; and also light from the turnaround going back to Earth.
• Brown is Capt. Kirk, outbound.
• Purple is the Klingon freighter, headed inbound for Earth at constant velocity.

Diagrams:

• Left most diagram is from Earth's perspective. Turn around is six light years away. The bomb is set off four years after Kirk's departurn. The turn around is seen 16 years after Kirk's departure. The Klingon freighter drops Kirk back home 20 years after departure.
• Next diagram across is from Kirk's outbound perspective. Note the distance to the bomb.
• Right most diagram is from the perspective of the Klingon freighter. Note also that they see Kirk approaching at 15/17 of the speed of light.
• The unconventional diagram is in the upper right. This is what Kirk sees during his round trip. All events are located according to Kirk's frame at the time he receives photons from that event. It is obtained by combining views of Kirk outbound with the view of the freighter.

All locations in these spacetime diagrams can be calculated using Lorentz transformations.

Cheers -- sylas

 

[otg]

I get a feeling that my original question, which I think was quite straightforward, got lost somewhere in all the lengthy discussions about Klingons... I have no issue with simultaneity being relative, and no problem accepting that different frames give different results, nor the relativity theory itself.
The issue here is just that it's not always easy to get the head right on what happens in which frame when one tries to see it in different ways, and where those ways are different even though they seem not to be.

My original question is this attached pdf. Why is the away-traveller older than the other in one view, and younger in the other? I only changed symbols and letters.
Correction: I think I only changed symbols and letters.

 

[sylas]

I get a feeling that my original question, which I think was quite straightforward, got lost somewhere in all the lengthy discussions about Klingons... I have no issue with simultaneity being relative, and no problem accepting that different frames give different results, nor the relativity theory itself.
The issue here is just that it's not always easy to get the head right on what happens in which frame when one tries to see it in different ways, and where those ways are different even though they seem not to be.

My original question is this attached pdf. Why is the away-traveller older than the other in one view, and younger in the other? I only changed symbols and letters.
Correction: I think I only changed symbols and letters.

Here is your diagram:

You don't say "older" with respect to which events.

Let me see if I understand correctly. In the upper part of the diagram, you have "A" being the Earth, at rest. "B" travels outbound from Earth, and passes "C", who travels back to Earth again. No problem there.

What is the next diagram trying to show? Is it the same events from the perspective of "C"? If so, you have "B" passing by "A" on the right, then synchronizing with "C", and then "C" ariving at "A"... except that this time, "A" is moving towards "C". So you really need to indicate "A" moving towards "C", but more slowly than "B".

When you say "older", what are you comparing? You have only one reference point where ages of "A" and "C" can be compared. If you try to get their ages from some other point, then you have the same problem that ALWAYS shows up in these discussions. You are (effectively) assuming simultaneity.

We can relate your diagrams to my diagrams as follows. "A" is the Earth (green world line in my diagrams) "B" is Capt. Kirk, who travels outbound from Earth (brown world line). "C" is the Klingon (purple world line) who travels back towards Earth again, having passed "B" (Capt. Kirk) and synchronized, at the turn around point.

Hence the upper part of your diagram corresponds to my left-most spacetime diagram, and the lower part of your diagram is... what? "C" being at rest? If so, this is the rightmost of my spacetime diagrams, for the Klingon's reference frame, and there's no basis to compare ages of "A" and "C", because the Klingon only has one point at which they can synchronize with Earth.

You have three events. "B" leaves from "A". Then "B" synchronizes with "C" as they pass by each other. Then "C" arrives at "A".

The age of "A" in this series of events is unambiguous, because "A" is there at the start, and at the finish. But what age of "C" can you compare? You have to identify a point in time when you start counting "C"'s age. If you make it simultaneous with "B" leaving "A", then you are not actually identifying a starting point at all, because you cannot synchronize "C" with the event of "B" leaving "A". They weren't there. I pointed this out previously that you can't synchronize with remote events.

If "C" and "B" are the same individual, however, (with a turn around) then you have what I have described for Capt. Kirk, who teleports into the Klingon ship for the return part of his journey.

Cheers -- sylas

 

[matheinste]

Hello sylvas.

I make no mention of Klingons. I think we are at cross purposes here. It's my fault for interupting someone elses thread so causing confusion. I will back out now and perhaps start another thread on this changeing distance point.

Thanks for your replies.

Matheinste.

 

[sylas]

Hello sylvas.

I make no mention of Klingons. I think we are at cross purposes here. It's my fault for interupting someone elses thread so causing confusion. I will back out now and perhaps start another thread on this changeing distance point.

Thanks for your replies.

Matheinste.

No problem. I don't think you caused any confusion. Here again are the three stock standard space time diagrams for the traveling twins, one for each perspective, but with all mention of Klingons removed. There is also a mixed diagram, in which the traveler is always at rest, and all other events are at spacetime co-ordinates for the frame of the traveler at the time the photons arrive to let the traveler see that event.

The traveler goes to another star at 60% light speed and then returns. At the point when they turn around, the light coming from Earth is coming from a point at 3 years previously (just before the jump) and from a point at 12 years previously (just after the jump). The Earth is being seen at the same instant in spacetime, but the distance and time to that event changes with the new inbound perspective of the traveler.

The only reason for introducing Klingons was to emphasize that acceleration is not what matters... only the new perspective. The same diagrams apply whether the traveler turns using an almost infinite acceleration, or a spacetime warp, or by teleporting into a passing spaceship headed back the way they have just come.

Cheers -- sylas