Solving the Twin Paradox with Lorentz Transformation

[Dale]

I think the objectionable statement by Dolby and Gull is this one: "Barbara’s hypersurfaces of simultaneity, which define ‘when events happened’ according to her, have consistently been misrepresented". That is probably overstating it, but they are certainly correct that any coordinate system which assigns multiple times to a single event is asking for trouble.

 

[phyti]

Fredrick;
Ignore, my drawing isn't correct.

 

[Fredrik]

When B-clock reads 12, a signal (blue) from the A-clock would show 4.

It's true that a radio signal from Earth that reaches B at the event where B's clock reads 12 must have been sent at an event where A's clock reads 4, but you seem to have misunderstood what the diagram is about, or the meaning of the word "when". From B's point of view, A has aged 7.2 years "when" B has aged 12 years. You find this result, not by following the world line of a light ray, but by following a simultaneity line. This is because the word "when" refers to a given time, and a simultaneity line is a set of events that are all assigned the same time coordinate.

The blue (and red) lines in my diagram are simultaneity lines, not light rays. And the text in the boxes isn't about what the twins see with their eyes or video cameras, but about what coordinates their co-moving inertial frames assign to these events.

 

[Fredrik]

I think the objectionable statement by Dolby and Gull is this one: "Barbara’s hypersurfaces of simultaneity, which define ‘when events happened’ according to her, have consistently been misrepresented". That is probably overstating it, but they are certainly correct that any coordinate system which assigns multiple times to a single event is asking for trouble.

I agree that D & G are overstating it in that sentence. Regarding the coordinate system that assigns multiple times to a single event, there are two easy ways out of that: a) Don't bother trying to combine all the co-moving inertial frames into a single coordinate system, and b) if we really want to combine them into a single coordinate system, just say that it's only defined in the region where no simultaneity lines intersect. (We will have to do that anyway when we use D & G's convention).

 

[stevmg]

To anyone out there -

How do I get quotations from different sections of a post (without the whole post) or quotations from different posts into a reply? Fredrik did just that above.

I have tried the radio button "multiquote" which does nothing.

Sorry to clutter this topic with this mundane question. If there is another site for these housekeeping type questions please direct me to it.

 

[Fredrik]

How do I get quotations from different sections of a post (without the whole post)

Use the quote button. Then delete the text you don't need. (In this case "To anyone out there -"). Then type [noparse][/quote][/noparse] after the text you want to quote, and type your reply to the quoted text after that. Always preview before you post, to make sure it looks OK. (If you e.g. mistype quote as qoute or leave out the /, the result will be ugly).

If you want to quote more text, you copy and paste the starting quote tag that was created automatically when you clicked quote. In this post I'm copying [noparse]

[/noparse] from the top and pasting it below this line.

or quotations from different posts into a reply? Fredrik did just that above.

I have tried the radio button "multiquote" which does nothing.

And of course, I still have to type a closing quote tag before this line.

To use multiquote. Either click the multiquote buttons next to each of the posts you want to reply to and then click the "new reply" button at the bottom, or click the multiquote button next to all the posts you want to reply to except the last one and click the quote button next to the last one. Note that when you click multiquote, the button changes color. You can deselect that post by clicking the button again. The color will change back.

If there is another site for these housekeeping type questions please direct me to it.

I've seen questions like these in the feedback section.

 

[phyti]

It's easier for me to have a drawing of each frame in its own scale, rather than one with two scales.
I use light rays because there's an event at each end, and the events help simplify the transformation from one frame to another without using the hyperbolic scale.
I do use the axis of simultaneity.
The left drawing shows A's description in agreement Fredrick's post 3.

This is a good example to demonstrate how the instantaneous reversal produces misleading results. The middle drawing shows the A-clock running slower than the B-clock 14.4/24. After adding the missing 25.6 at reversal, the A-clock is running faster.
The conflict is lack of symmetry. When two clocks diverge from a common origin, the clock that returns to the other accumulates the least time. In this case that is the B-clock. The number of events 40:24, does not change with a transformation.

The right drawing shows the doppler effect as mentioned by DH, with each half of the A-events at frequencies of 1/3 and 3. It also shows why I select event A4, because it is the last event arriving at B prior to reversal, and the changeover for frequencies. The A-path extends past the vertex at B12 because A never changed course at A20. Just more distortion when trying to make frames equivalent when they are not.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=Ktj2Xn4gX81MnzXBpCtjs7vXFLXm1X7M8rx&id=124603&fn=twin-pardx2.gif

 

[Fredrik]

I don't see how any of this demonstrates that "the instantaneous reversal produces misleading results".

Note by the way that if we consider a scenario that takes much longer, say 40 million years of Earth time instead of 40 years. Then we can do the turnaround using a constant proper acceleration for a long time, and the diagram would still look more or less the same. If you make it fit on your screen, you probably won't even see the details that reveal that the turnaround took a long time, and yet it's the details you would see that lead to the description you've been arguing against.

 

[phyti]

We know that Fredrik, so what's new?

Since the B path is discontinuous at B12, when he switches frames, B's description can be done in two parts as shown. His distance is 10.6 instead of 9.6, because he never decelerates to 0. That's the difference.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=Mg79nCX3KJpBFgnnbxzbxtOXsLD7LXxxNMX&id=124665&fn=twin-pardx3.gif

 

[stevmg]

Fredrik -

Thanks for the input on multiquoting.

To both Fredrik and phyti, instead of thinking of this as twin B turning around, just visualize the first leg of the "moving" twin's journey as one rocket ship alread at Mach ten zillion from the Earth to the "turnaround point." This rocket ship just blasts on by.. Then have a second rocket ship at mach ten zillion which is going the opposite direction and it crosses the first rocket ship exactly at the same instant and place as the first rocket ship at what is called the "turnaround point" and goes to Earth, You will have two separate trips in exact coordination from start to finish with NO acceleration/deceleration. Adding the two times up gives you the elapsed age of the traveler.

This approach was posted by DaleSpam and JesseM to avoid the problems associated with acceleration/deceleration and any semantic questions that could arise. It is not precise but gives the general idea.

 

[Fredrik]

Phyti: I still have no idea what you're talking about. Sorry.

I'm also having a hard time trying to make sense of your B diagrams. In particular I don't see the point of representing light signals going from A to B with lines that have the same slope as the ones representing light signals going from B to A. I haven't got a clue what these 9.6, 10.6, 5.3 numbers are.

 

[phyti]

Fredrik -

Thanks for the input on multiquoting.

To both Fredrik and phyti, instead of thinking of this as twin B turning around, just visualize the first leg of the "moving" twin's journey as one rocket ship alread at Mach ten zillion from the Earth to the "turnaround point." This rocket ship just blasts on by.. Then have a second rocket ship at mach ten zillion which is going the opposite direction and it crosses the first rocket ship exactly at the same instant and place as the first rocket ship at what is called the "turnaround point" and goes to Earth, You will have two separate trips in exact coordination from start to finish with NO acceleration/deceleration. Adding the two times up gives you the elapsed age of the traveler.

This approach was posted by DaleSpam and JesseM to avoid the problems associated with acceleration/deceleration and any semantic questions that could arise. It is not precise but gives the general idea.

That was done in my post, but it doesn't include the 'turnaround point' because it never happens with switching frames on the fly. The essence of this case is that it's not symmetrical. The closed path decides one twin aged more than the other, and you can't straighten out B's path or bend A's, to make it reciprocal.

 

[phyti]

Phyti: I still have no idea what you're talking about. Sorry.

I'm also having a hard time trying to make sense of your B diagrams. In particular I don't see the point of representing light signals going from A to B with lines that have the same slope as the ones representing light signals going from B to A. I haven't got a clue what these 9.6, 10.6, 5.3 numbers are.

I agree, they are not clear enough, so here is a new and improved one.
When B switches frames at B12, event A4 gets shifted from 5.3 ly distant to 48 ly,
because of Einstein's simultaneity convention.
The light paths are cyan, and red denotes the portions of A-events observed by B.
I wanted to show the asymmetry of B's description vs. that of A.

[PLAIN]https://www.qdrive.net/en/home/images/?userid=2104&ky=N9XnI2MOt3NCB8C84XKpPxXO32x6M6mNXrr&id=125014&fn=twin-pardx4.gif

 

[stevmg]

Fredrik -

The reason why there are an "absurd" number of posts on this subject is that this is one of the most elementary areas of relativity and simpletons like me can actually delve into it. To wit, I know what Doppler is and how it works with sound, but the relativistic argument about it requires me to state "If you say so."

To me when DaleSpam used the 4-space depiction of the space-time coordinates this illustrated the inversion (for want of a better word) of the Pythagorean theorem in which the longer pathway is shorter in proper time. Below, AC > AB + BC

t
C
...\
...\
...\
...B (turnaround point)
.../
.../
.../
A...x

I works out that way when you use the norm of vectors but it is surely counter intuitive. The diagrams shown above make sense in a weird way if I remember that counter-intuitive relationship.

Kind of intereseting from basic arithmetic: Take 12 and divide it by 2. This equals 6 + 6
Take 12² and 6² + 6²
We get 144 and 36 +36
Taking the square root of both sides,
we get 12 and 8.45. That's kind of the way it works in this paradox problem.

Maybe God did make the universe follow mathematical rules.

 

[AJ Bentley]

Here's an explanation of the famous 'twin paradox' of relativity. WITHOUT ANY MATH.

Relativity says time passes more slowly for a moving body than a stationary one. At the same time it says you can't tell which of the two is moving.
The paradox revolves around a pair of twins. If you don't know which one was moving, how can you tell which finishes up the older if one of them takes a round trip?

Here you go.

Both twins agree that Joe (the 'moving' twin) passed Barnard's star at the age of 45 and his hair fell out the same day.

Both twins agree that at the age of 75 Moe lost his teeth.

However, they cannot agree that these two events were simultaneous - Moe says yes, they did happen at the same time and what's more, Moe is much younger than he is - a consequence of his motion.
Joe says, on the contrary, he was over a hundred by the time Moe reached 75 and when he passed Barnard's star Moe was celebrating his 30th - not his 75th - and anyway, Moe is the younger - because he is moving!.

That is the ONLY point of disagreement between them - the issue of which past events happened at the same time. The disagreement is quite profound. Each looking into his telescope can see the other twin and how old he is. And allowing for the distance can calculate how old the twin is 'now'. Each sees the other is moving and ageing very slowly and that even allowing for the time taken for the light image to arrive, the other twin must be far younger.

In order to bring simultaneity back into line, one twin or the other (or both) has to perform an action on his world view.
Joe can 'stop' thereby moving back into Moe's frame. Or Moe can start after Joe, entering his frame.

For the twin that performs this action (acceleration) the catalogue of simultaneous events completely changes.
As he changes his velocity, all the events he was yet to see but which have already happened for the other twin sweep past him in a great flood. He sees the other twin who he thought was much younger, suddenly grow old. Of course, this is the key to the 'paradox'. If you ignore this consequence of changing velocity, you get the paradox. Why does it happen? because it HAS to happen. For the two to ever meet over dinner again, one or the other has to go through this experience. In Treckie terms, they are on opposite sides of a 'simultaneity rift'. One or the other must cross to the other side.


This is because simultaneity as a concept has literally no meaning when applied between frames. Joe and Moe are unable to agree which events were simultaneous , neither is wrong.

 

[stevmg]

Somehow, I think the math explanation by DaleSpam or JesseM is easier to understand...

 

[MikeLizzi]

This is how to resolve the “Twins Paradox”. You must recognize the following.

1. To an inertial observer the clock of an inertially moving or accelerating body is always running SLOWER than his/her own.

2. To an accelerating observer the clock of an inertially moving body is always running FASTER than his/her own.

3. The earthbound twin is assumed to be an inertial observer throughout the course of the astronaut’s trip.

4. The astronaut is an inertial observer during some portions of the trip and is an accelerating observer during other portions.

So, the earthbound observer always calculates the astronaut’s clock as running slower. Therefore the earthbound observer must conclude that the astronaut returns younger.

But the astronaut sometimes calculates the earthbound observer’s clock to be running slower and sometimes faster. If the astronaut actually makes a calculation using numbers it turns out that the acceleration/faster clock rate portion of his trip overwhelms the inertial/slower clock rate portion and the astronaut concludes he/she will return younger.

 

[stevmg]

You're making it hard

Take Twin A...B

If twin B moves to the right then he has to move to the left to get back. On that portion his relative speed to twin A is much faster than his outbound speed from A. That's the asymmetry. Twin B never is the "center of the universe" as he has to change direction. Look up JesseM's solution for the math. Twin A never changes direction.

 

[Austin0]

Among those I personally like the Doppler shift explanation the best. My education is in physics but my career has been in engineering. I like explanations based on what one can see/measure. The acceleration explanation, e.g., A ages 25.6 years during B's turnaround event just doesn't jibe with the hardboiled engineer in me.

The Doppler shift explanation does jibe with that hardboiled engineer. I'll use Fredrik's example of space traveler B going at 0.8 c to/from a star that is 16 light years away. First, some assumptions:

• A and B continuously transmit a signal to and receive a signal from one another. Occasionally A and B will use this transmission to send messages, view family pictures, whatever.
• These continuous transmissions include a timing signal. For example, the mission elapsed time as measured on the sender's clock will be embedded once per second as measured by the sender's clock.
• B's spacecraft is equipped with sensors that can measure the distance (in B's local frame) to the target star (outbound leg) / Sun (return leg).
• B's spacecraft can similarly measure the relative velocity to the target star (outbound leg) / Sun (return leg). For example, the spacecraft might detect the frequency of the star's hydrogen alpha line and compute the velocity from the observed blueshift.

Just after B has accelerated to 0.8 c on the outbound leg, and for some amount of time thereafter, both A and B will see, by means of the communications link, the other as aging at 1/3 the rate at which they themselves are aging. Just before B returns to Earth, and for some amount of time before, both A and B will see, by means of the communications link, the other as aging at 3 times the rate at which they themselves are aging. Somewhere in between, each twin will see a transition from that 1/3 aging rate to a factor of 3 aging rate. The resolution of the paradox is that this transition occurs at distinctively different times.


What B sees
At the moment of departure (i.e., right before accelerating), B will see the target star as being 16 light years away. Assuming a rapid acceleration event, B will see the distance to the target star shrink to 9.6 light years upon reaching 0.8 c. B calculates the time needed to reach the target star as 9.6 light years / 0.8 c, or 12 years. B will see A doing things in slow motion during this 12 year long outbound leg. Suppose A gets married and has a child four years (A's clock) after B departs. When B arrives at the target star 12 years later (by her own clock), she will have just received a picture of A's newborn child. The signal from A will indicate a mission elapsed time of 4 years.

Now B stops, takes a few pictures, and turns around, doing all rather quickly. During the brief stop, B will sense that the Sun is 16 light years distance. Upon accelerating to 0.8 c, she will once again sense that distance has shrunk to 9.6 light years. Now B sees things happening to A and the child in fast motion (3x speed, to be precise). B will see A turn into an elderly gentleman and the baby zoom through life. By the time B reaches Earth 12 years later (her clock), she will see A as having aged by another 36 years. If the voyage started when A and B were both 20 years old, B will see herself as being 44 years old upon return, A as being 60, and the child as 38 -- only six years younger than B!


What A sees
From A's perspective it is B that is doing every in slow-mo on the outbound leg. A will get married, have a child, and the child will have just graduated from high school by the time A calculates that B has reached the target star. This is just a calculation, however. As far as what A can tell based on seeing, in the signal sent by B, B has a long ways to go to reach the star. A will continue to receive a slowed-down signal for another 16 years after the calculated turn around time. It will take 36 years before A sees the pictures from the target star and receives a congrats message on the birth of the child. At this point, when A is 56 years old, B will report to A that she is 32 years old. For the next 4 years, A will see B as working in fast motion and aging rapidly. By the time B returns four years later, A will be 60 years old, B will be 44, and the child will be 38.


Summary
Per this explanation, the paradox vanishes due to the disparity in the time at which the received signal switches from slow to fast. For B this happens right at the turnaround event. A has spent half of B's 24 year journey aging slowly and the other half aging quickly. From B's perspective, she has aged 24 years while A has aged 12/3 + 12*3 = 40 years. For A the transition occurs when the turnaround signal comes back to Earth. From A's perspective, A has aged 40 years while B has aged 36/3 + 4*3 = 24 years.

Hi DH With your engiuneering perspective you may be the one to answer a question I have had.

With a slight alteration of your scenario.

Earth and an inertial frame at some distance. COnsider instant accerelation for simplicity.

The ship makes a round trip and both the ship and Earth send out the timed signals.
At the end of the trip they will of course agree that the number of signal sent and received was proportianal to gamma factor.

During the trip both frames will of course have a count of received signals that is behind the current number by the time interval of the propagation time of the last received signal.
The rest being spread out through space.
Could not the ship [or earth] calculate that propagation time from D/c aand then look back through their log , subtract that dt and see how many signals they had sent at that point??
Compare that number with the number of the last received signal and get a number that was accurate at the time that last signal was sent?
Getting a comparison which would be out of date but meaningful.
That because it is based simply on their own measurement of proper time and distance ,this would still apply with acceleration and that Doppler or turn around etc. would not make any difference??

If this concept is not completely off the wall it would seem to mean that relative dilation would be apparent and quantifiable along the way albeit with the qualification that it would not be completely current.

SO what do you think?

Thanks

 

[AJ Bentley]

Doppler shift has nothing to do with it. That's a Newtonian concept that won't help you to understand SR.

The twin paradox is based on a naive view of simultaneity, which was the gist of Einstein's original point.

In order to understand relativity at all, you have to have a clear grasp of the twin paradox in such a way that you see it isn't a paradox at all.

Mathematics won't help, you can push symbols around all day and still have no idea what they ultimately mean. Especially when you elect to use mathematics that don't relate to your problem.

The ONLY thing that the twins disagree on is which events for one twin were simultaneous with those for the other. That's hardly a paradox - it's simply an interesting, mildly surprising fact.

 

[stevmg]

I think this is getting really complicated when it doesn't have to be so.

This post by JesseM explains the conceptual and simple mathematical approach to this problem

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FOR is proper time as he ain't movin'. Folks who are moving experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.

The supposed symmetrical situation is to look from B's FOR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does. When all the times are added up the elapsed time for B is the same this way as it was looking at it the first way in the above paragraph.

Guess what! This is NOT a symmetrical situation is it? In the first case one of the twins sits still (and gets older) while the other twin moves and gets older slower. The first twin never moves.

In the second case, BOTH twins move although one sits still for a while before mving.

These are NOT symmetrical approaches. There's no way to make them symmetrical. JesseM's calculations and working through the problem is self explanatory.

A symmetrical scenario would be to have both twin A and B depart the reference frame in opposite directions at the same speed for the same time and both turn around and come back to meet. In this case, they both would age at the same rate (though not as fast as their triplet brother who remained on Earth) and be the same age when they rejoined. Their triplet brother who remained on Earth would be older than both.

I am not going through the calculation looking at it from A's FOR or B's FOR but I guarantee you, it would work out. Who cares about Doppler, acceleration, deceleration and all that?

Don't make this more complicated than it is, which it isn't (really.)

 

[MikeLizzi]

I think this is getting really complicated when it doesn't have to be so.

This post by JesseM explains the conceptual and simple mathematical approach to this problem

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FOR is proper time as he ain't movin'. Folks who are moving experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.

The supposed symmetrical situation is to look from B's FOR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does.

"In this case A moves left - BUT never stops." ??

The Twins "paradox" involves observing and calculating aging from the point of view of each twin. That means when Twin A is the observer, all motion is described with respect to Twin A. (B goes out and back) And when twin B is the observer all motion is described with respect to twin B. (A goes out and back) That's the definition of the problem. Wether you are doing SR or Newtonian physics that's what happens from the point of view of each observer.

We are all free to define and solve any problems we want but we ought to give dfferent problems different names. The problem being solved in your reference is not the Twins Paradox.

 

[DrGreg]

I think this is getting really complicated when it doesn't have to be so.

This post by JesseM explains the conceptual and simple mathematical approach to this problem

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FOR is proper time as he ain't movin'. Folks who are moving experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.

The supposed symmetrical situation is to look from B's FOR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does.

"In this case A moves left - BUT never stops." ??

The Twins "paradox" involves observing and calculating aging from the point of view of each twin. That means when Twin A is the observer, all motion is described with respect to Twin A. (B goes out and back) And when twin B is the observer all motion is described with respect to twin B. (A goes out and back) That's the definition of the problem. Wether you are doing SR or Newtonian physics that's what happens from the point of view of each observer.

We are all free to define and solve any problems we want but we ought to give dfferent problems different names. The problem being solved in your reference is not the Twins Paradox.

If you read what stevemg read a little more carefully, it is the Twins Paradox. It's just that in the final paragraph quoted here, the phrase "B's FOR" is unclear. What he should have said was something like "the inertial frame of reference in which B is initially (i.e. for the "outward" journey) at rest", and then it all makes sense.

 

[Aaron_Shaw]

Look at this diagram:

The shows something that could be the two twins. Alice at home and Bob moving. It shows the situation of Alice's viewpoint. Her time and space axes are in white.
The diagram also shows Bob's time and space axes in bold blue. They are skewed, as expected, because he is moving relative to Alice.

The red dots are events of either light being transmitted, reflected, or recieved. You can tell from the arrows. What this diagram shows is that events that are simultaneous for bob (the four red dots along his space axis) are not simultaneous for Alice as they are at different time coordinates according to her time axis.Now I've edited this image:

I've added some thin red lines to emphasize the simultaneity lines according to Bob. I've also highlighted a section in green. Here you can see the simultaneity lines (thin red) which will show simultaneous events according to Bob with regard to Alice's clock time and his own clock time.

This represents Bob's outward journey. Now if we flip it over to represent the inbound journed we have this:

Note the change in orientation of the simultaneity lines. So now our diagram of simultaneity looks something like this:

Following these lines we can clearly see that Bob observes Alice's clock ticking lesson each leg of the round trip. That's where the apparent paradox comes into play because we would normally expect Bob to return younger, yet he sees Alice aging less the whole time.

The important part of the diagram is the lines of simultaneity around the turning point. Here you can see that the orientation of the lines shift dramatically at the turn around point. At this point Bob observes Alice's experiencing a short but dramatic increase in clock ticking rate. So much so that Alice ages enough to enable Bob to then continue observing her clock run more slowly for the remainder of the journey so that when the two are reunited Alice is infact older than Bob even though he saw her clock running more slowly.

 

[AJ Bentley]

Look at this diagram:
The important part of the diagram is the lines of simultaneity around the turning point.

At last! - absolutely spot on!

SR relates to inertial F.O.R. Whichever twin steps out of his inertial frame sees this sudden change in his perspective on simultaneity.

The Lorentz formulae allow you to put numbers to it - but this is the basic Physical understanding.

 

[Aaron_Shaw]

Following these lines we can clearly see that Bob observes Alice's clock ticking lesson each leg of the round trip.

Actually I'm not quite sure if I'm right here. I know it to be the case and i know that the changing orientation of the simultaneity lines "fixes" the "paradox". But I'm not sure if I'm correct in saying that the lines show the clock running slower. It doesn't matter. It's besides the point.

 

[AJ Bentley]

But I'm not sure if I'm correct in saying that the lines show the clock running slower. It doesn't matter. It's besides the point.

As you say, it doesn't matter, the point is the change of view when changing F.O.R.

I haven't ever bothered to think about the physical implication of details of the Minkowski diagram in this context - they have to accord with the reality they describe.
And the plain simple fact is that while the two are moving relative to each other, each sees the other as moving slower. (Of course the fact that light takes time to travel the intervening distance is a complication but again, that's irrelevant.)

Thanks for a very clear explanation.

 

[AJ Bentley]

so that when the two are reunited Alice is infact older than Bob even though he saw her clock running more slowly.

You'e missing out the point that Bob also has to stop at journey' end - creating a second realignment.
The trouble with this 'paradox' is that it creates a lot of complication by having two journeys involved.

I prefer to think of a single very short, very, very fast journey (say a foot or two!). Then you don't have the 'coming back' problem.

 

[stevmg]

If you read what stevemg read a little more carefully, it is the Twins Paradox. It's just that in the final paragraph quoted here, the phrase "B's FOR" is unclear. What he should have said was something like "the inertial frame of reference in which B is initially (i.e. for the "outward" journey) at rest", and then it all makes sense.

I couldn't have said it more eloquenlty, and I didn't. That's why you "tightened up" my phraseology.

Thank you, Dr. Greg.

Steve G

Melbourne, FL

 

[AJ Bentley]

There's nothing wrong with Stevemg's post. It too is correct, as far as it goes.

But the common misunderstanding in the paradox is not that the viewpoint is asymmetric - that's a given. It's the fact that the velocity difference causes a disparity in the view of what is and is not simultaneous, which is only resolved by matching velocities.

That resolution is what causes the age difference and most people have difficulty grasping how that comes about. The reason they have trouble is because they lose sight of the basic premise - that there is literally no meaning to the common word 'simultaneous'. (Or at most a very restricted one).

When I look at an explanation of the paradox, I simply look for the word - if it's not there, I know this answer doesn't help.

PS. Substitute 'Now' for Simultaneous' - possibly a bit clearer.