How Does the Twin Paradox Affect Aging in Relativity?

[JesseM]

Hey, Sports Fans...

A few days off and I'm back. Here is the simpleton's (that's me) approach to the twin paradox problem. We're going to use Simple Relativity and no references to anything else which may detract from this simple illustration which follows. Again, assume there are twins A and B - and believe me, that's how we would label them in the delivery room and nursery, even after Mom would give them their "real" names. Now, take twin B and put him/her on a spaceship at a velocity of 0.9949874371*c. Assume the jump to warp speed is instanenous. Start the two clocks (for A and for B) at that instant) t_A0 = t_B0 = 0.0.

Now send that little guy, B, off, say, to the right at that "warp"* speed: 0.9949874371*c while A remains here going nowhere. We will use the Earth as a frame of refrence F.O.R.. Keep him (B) going for ten (10) years Earth time t_A1.

Now, after 10 years Earth time or t_A1, turn him (B) around and return at the same speed [this time the velocity sign is reversed (from a plus (+) to a minus (-)], so make that -0.9949874371*c and he will get home in twenty years - Earth time t_A2. Now, that isn't too hard to wrap your brains around, is it?

We will proceed using Simple Relativity and will ignore the deceleration and subsequent acceleration back to warp speed for the return trip. Thus, we will not attempt to apply General Relativity. General Relativity would slow down the spaceship time even more because of the forces of acceleration/deceleration, so we can proceed with Simple Relativity and not lose the flavor of what we are trying to illustrate.

Using the time-dilation formula:

t = \gamma*t' or, in this case: t_A = \gamma*t_B

We have two F.O.R.s: The Earth (t_A and the spaceship t_B.)

What do you mean by t_B in this case? The spaceship does not remain at rest in any inertial frame, so there's no inertial frame whose time coordinate always keeps pace with the age of the twin on the spaceship (i.e. that twin's proper time). You could construct a non-inertial frame where the spaceship remains at rest and the coordinate time keeps pace with the ship's proper time, but then the normal formulas of SR (like the time dilation formula) would not apply to this frame.

Continuing on with the use of the time-dilation formula:

Remember, \gamma = 1/SQRT[(1 - v²/c²)

We get the ratio of t_A/t_B = 10

If t_A and t_B represented the time coordinates of two inertial frames this would not be correct. Remember, in relativity the situation between two inertial frames is always perfectly symmetric! If you and I are at rest in two different inertial frames, then if it is true in your frame that my clock is running ten times as slow as yours, it must be true in my frame that your clock is running ten times as slow as mine. This symmetry between frames is exactly where the idea for the twin paradox comes from--people think that you should be able to consider things from the perspective of the traveling twin's frame, and in this frame it would be the Earth twin's clock that's running slow, so that this frame would predict that the Earth twin is the younger one when they reunite. Fortunately that is not actually what relativity says, since the traveling twin does not remain at rest in a single inertial frame, and the time dilation formula only applies in inertial frames.

Here's when it would be valid to use the time dilation formula. Suppose you have two inertial frames A and B, and you look at two events that happen at the same position in the A frame, like two events on the worldline of a clock at rest in the A frame. If you want to know the time t_B between these events in the B frame, compared with the time t_A between them in the A frame (where they would just be equal to the proper time measured by a clock at rest in the A frame that's at the same position as both events), then the time dilation formula would say the answer is t_B = t_A*gamma, which means with a velocity of 0.9949874371 you have a ratio of t_B/t_A = 10. On the other hand, if you have two events that happen at the same position in the B frame, like events on the worldline of a clock at rest in this frame, then if the time between these events in the A frame is t_A and in the B frame it's t_B, then the time dilation formula would tell you t_A = t_B*gamma, so t_A/t_B = 10. In general, if you have a clock at rest in a given frame and you want to know the time between two of its readings in another frame, the time dilation formula has the following form:

(coordinate time in frame where clock is moving) = (time as measured by clock itself)*gamma

If you're dealing with two events that don't both happen at a single position coordinate in either of the two frames, then you can't use the time dilation formula at all! Instead you should use the more general time conversion formula from the Lorentz transformation, which works for time and distance intervals as well as time and distance coordinates:

t' = gamma*(t - vx/c^2)

This clearly displays the twin paradox using Simple Relativity by itself without going into "world lines" or \tau time coordinates or whatever. You do have to ignore the acceleratio/deceleration aspect but so did all the other posts on this blog.

I think the reasons above show why it's a good idea to talk about the proper time \tau along a given twin's worldline, rather than just coordinate times, since the twin that turns around doesn't have any single inertial rest frame.

 

[simonh]

OK, so I'm a complete physics wannabe, an armchair one even - considering getting involved in this wonderful world and the concepts do appear to be making some sort of sense. Now while I can quite happily wrap my head around the concepts, I don't understand the why and as such a visual aid would help me.

Now slap me if this is a theoretical impossibility, but it would make my understanding so much clearer!

If A has a telescope and is watching B throughout the entire 40 years (with perhaps the exception of restroom breaks), does B appear to be moving uniformly slower throughout the entire process, or would there be a difference between the departing speed and the return speed from A's perspective.

Or in essence, is the time dilation as a result of B's actions or B's actions comparative to A.

I have a sense of dread as I hit the reply button lol! (Be kind!)

 

[simonh]

In fact scratch that, i read the diagram incorrectly. B does appear to move slower to A than B does to B is what I get from that. But it does however raise another question, is the transition on A of 25.3 years visible to B as one massive time 'burst'?

 

[Dale]

Hi simonh, welcome to PF!

As far as what is actually visible to each twin, that is usually called the "Doppler shift" analysis. There is a good write up about it here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html
and please also see figure 2 here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler

 

[sylas]

In fact scratch that, i read the diagram incorrectly. B does appear to move slower to A than B does to B is what I get from that. But it does however raise another question, is the transition on A of 25.3 years visible to B as one massive time 'burst'?

Not really. There is rather a massive distance burst that is actually "visible".

Consider this. Twin A stays at rest. From the perspective of twin A, moves out 6 light years at 60% lightspeed, then reverses direction and returns at 60% light speed. This takes 20 years total. I'm picking numbers here to give nice round calculations. The gamma factor is 1.25

What twin A actually SEES, however, is a bit different because of the time delay in light signals.

Twin A sees twin B reverse direction at a point 6 light years distance, and this is seen after 16 years. That's the 10 years B takes for the trip, plus the 6 years it takes for the light to get back. B gets back in another 4 years after that.

So for 16 years, A can see B receding into the distances, and for 4 years A can see B coming back. A knows the turn around was 6 light years away, and so knows that B took 10 years out and 10 years coming back.

What does B see? B sees A receding into the background at 60% light speed. Furthermore, B experiences only 8 years until the turn around. Just before turn around, B is receiving signals from A that are (from B's perspective) coming from a distance of 3 light years away.

How do we get this? Well, the light from A left 3 years previously, which is 5 years after they separated. In five years, at 60% light speed, A has moved 3 lightyears to the rear.

Now B turns around. What changes in what B SEES? B is now in a new different rest frame. In this frame, the signals currently being received from A are not from 3 light years distance. They are from 12 light years distance! And what B sees is that A suddenly becomes much smaller in the sky (being further away). If A sent a signal to B which was received at the moment B turns around, the event of sending that signal is something that occurred 3 years ago in one frame, and 12 years ago in another frame. But it is the same event for A. A is seen to be 4.8 years older than at departure (calculate it) and this event is now 12 years ago rather than 3 years ago. You do not suddenly see A aging. You rather see A at the same age, but that was longer ago, in your new rest frame.

Cheers -- sylas

 

[simonh]

Thanks for the replies guys! Will give me some good material to mull on while I try to get to sleep tonight, even if that last paragraph does give me nightmares :) The concepts seem simple enough after reading 7 pages of this, but as is evident, I'm missing key pieces!

 

[stevmg]

Did I start all that again?

Sorry, Sports Fans.

Just answer just the one question... Twin B ages 2 years, Twin A ages 20 years... yes?

 

[sylas]

Did I start all that again?

Sorry, Sports Fans.

Just answer just the one question... Twin B ages 2 years, Twin A ages 20 years... yes?

In your example, with twin B traveling at +/- 0.9949874371 c relative to an inertial twin A: yes.

 

[stevmg]

A) Exactly, what is "proper time?" Keep the definition simple and NOT RECURSIVE, i.e. - non tautological. Does it mean that in the x, y, z and -ct coordinates, all clocks agree (-ct is constant.) If t were a specific value, then all 3-tuples of x, y and z that calculate to this specific t are merely from the infinite number of F.O.R.'s that relate to this particular t?

B) In the example above (we ignore the acceleration/deceleration to 0.9949874371*c) there is no acceleration or deceleration or curvilinear motion, so aren't both F.O.R.'s "inertial?" F.O.R. A (the earth) and F.O.R. B (the spaceship.)

 

[yuiop]

A) Exactly, what is "proper time?" Keep the definition simple and NOT RECURSIVE, i.e. - non tautological. Does it mean that in the x, y, z and -ct coordinates, all clocks agree (-ct is constant.) If t were a specific value, then all 3-tuples of x, y and z that calculate to this specific t are merely from the infinite number of F.O.R.'s that relate to this particular t?

Let us say a test particle moves from event 1 with coordinates (t1,x1,y1,z1) to event 2 with coordinates (t2,x2,y2,z2) then the proper time is the time interval recorded by a clock attached to the test particle.

Note: From here onwards I am using t, x, y and z as shorthand for (t2-t1), (x2-x1), (y2-y1) and (z2-z1) or (\Delta t, \Delta x, \Delta y, \Delta z)

In flat Minkowski space and considering inertial motion only, the proper time (tau or \tau) is defined mathematically (using units of c=1) as:

\tau = \sqrt{(t^2-x^2-y^2-z^2)}

Different observers will disagree on the values of t,x,y and z for a given particle but they all agree on the value of tau.

For example let us say one observer is watching a particle that moves a distance x=0.8 in time t=1.0 (so it has a velocity of 0.8c relative to his rest frame) then the proper time is:

\tau = \sqrt{(t^2-x^2-y^2-z^2)} = \sqrt{(1-0.8^2-0-0)} = 0.6 s

Another observer is moving with the particle (i.e. at rest with the particle so the velocity of the particle in his reference frame is zero) then he calculates the proper time as:

\tau = \sqrt{(t^2-x^2-y^2-z^2)} = \sqrt{(0.6^2-0-0-0)} = 0.6 s

 

[yuiop]

B) In the example above (we ignore the acceleration/deceleration to 0.9949874371*c) there is no acceleration or deceleration or curvilinear motion, so aren't both F.O.R.'s "inertial?" F.O.R. A (the earth) and F.O.R. B (the spaceship.)

Yes, BUT the spaceship does not remain in one inertial reference frame. For part of the journey the spaceship is in an inertial reference frame that a velocity +v reltive to the Earth and on the way back the spaceship is a different inertial reference frame that has a velocity -v relative to the Earth. That is a crucial difference.

 

[stevmg]

Interesting how my calculations still work out - but that's because the paths have the same distance and velocities back and forth (except for the negative (-v) and negative x (-x).

They wouldn't work out if twin B took a different path back [say like a half semicirlce where going out was along the diameter and the return was along the arc, even if you ignore the circular path (General Relativity) and just go by the longer distance.] There's where your \tau comes into play.

I am going to have to read selected texts about this to gain a better feel for what's what as I am out of my league at this point.

Thanks to all, Jesse, Dale, kev and sylas

 

[stevmg]

To DaleSpam or JesseM or anybody–

Help me with this. I will keep it simple. I just finished an excellent short text by Dr. Richard Wolfson of Middlebury College, Simply Einstein – Relativity Demystified (2002) and it was excellent. He discusses the twin paradox.

He posits a scenario of twins being born and one being rushed away to a star some 20 light years away. He treats the problem as if there were two spaceships. One blasting past the Earth at 0.8c when the twins were born (and immediately carrying, say, twin B) and a second spaceship returning to Earth from the star which is 20 light years away from Earth in the Earth-star time frame at 0.8c immediately after twin B arrives. The trip, looking from the Earth resting frame takes 25 light years. The trip from the outgoing spaceship time frame takes 15 light-years. Gamma is, of course SQRT(1 – 0.8)^2 = 0.6. He adds the two times up, the 25 years out and 25 years back and gets 50 years. He adds the two 15 years in spaceship time and gets 30 years – a 20 year difference in age.

He also states that the distance via the spaceship frame is 12 light years and notes that 12 light years/15 years = 4/5 = 0.8c, which agrees with the velocity of the spaceship.

He states because of the supposed symmetry (as one of you once mentioned in a prior post) that there is the paradox of the twin B being younger from the Earth frame and twin A being younger from the spaceship frame. He then goes on to say that, in fact, there is NO symmetry as twin B has to accelerate, fly at 0.8c, slow down, stop, turn around, accelerate again to -0.8c and come back and slow down again.

He later in his book uses the space-time diagram to go into this again and it looks somewhat like this (we use only one dimension – x for distance and t for time):
..t
..|
C|\ ... you’ll have to imagine the 45-degree light line as it is hard to draw with typographical characters
..|. \
..|. . \D
..|. . /
..|. /
A|/___________x

A is the event of the birth of the twins. AC represents the world-line of twin A who just sits there in her own frame of reference. AD represents twin B’s outward journey to the star (20 light years) and DC represents twin B’s return journey to Earth.

This is as simple a diagram that can be presented.

Can you go through the math of the “proper time” [is that “ds” in the equation ds/dt = SQRT[c^2 – (dx/dt)^2]?]. In this case I assumed the “tau” is t (thus dt/dtau = 1 so the first term in this “reverse” Pythagorean expression is c^2) and I ignored the y and z axes as we were only discussing the x-axis (and the t axis.) I cannot understand how, geometrically, ADC (the path of the rocket in space-time) is shorter than AC (sitting still.)

I’ve looked in different books and cannot find it anywhere, just allusions to it, with no specific examples or calculations.

 

[JesseM]

Can you go through the math of the “proper time” [is that “ds” in the equation ds/dt = SQRT[c^2 – (dx/dt)^2]?]. In this case I assumed the “tau” is t (thus dt/dtau = 1 so the first term in this “reverse” Pythagorean expression is c^2) and I ignored the y and z axes as we were only discussing the x-axis (and the t axis.) I cannot understand how, geometrically, ADC (the path of the rocket in space-time) is shorter than AC (sitting still.)

It's because proper time in spacetime is not calculated the same way as length in space. In a 2D space, if the endpoints of a straight segment of a path are at coordinates (x₁, y₁) and (x₂, y₂), then according to the pythagorean theorem, the length of that segment would be \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} (if you don't want to worry about specific coordinates and you know the distance along the x-axis \Delta x and the distance along the y-axis \Delta y between the two endpoints, the formula becomes \sqrt{\Delta y^2 + \Delta x^2 }). On the other hand, in a 2D spacetime with a spatial dimension x and a time dimension t, if you have a constant-velocity segment (like a straight-line segment in your diagram) with endpoints (x₁, t₁) and (x₂, t₂), then the proper time along this segment is \sqrt{(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2}, or if we use units where c=1 (like light-years and years), it'd be \sqrt{(t_2 - t_1)^2 - (x^2 - x_1)^2} = \sqrt{\Delta t^2 - \Delta x^2 }. You can see that this is different from the pythagorean formula because it has a minus in the middle rather than a plus. That change means that unlike in 2D geometry where a straight-line path is always the shortest distance between two points, in spacetime a constant-velocity path between two events (like the event of the twins departing from one another and the event of the twins reuniting) is always the one with the largest proper time.

Doing the calculations for your specific example:

He posits a scenario of twins being born and one being rushed away to a star some 20 light years away. He treats the problem as if there were two spaceships. One blasting past the Earth at 0.8c when the twins were born (and immediately carrying, say, twin B) and a second spaceship returning to Earth from the star which is 20 light years away from Earth in the Earth-star time frame at 0.8c immediately after twin B arrives. The trip, looking from the Earth resting frame takes 25 light years. The trip from the outgoing spaceship time frame takes 15 light-years. Gamma is, of course SQRT(1 – 0.8)^2 = 0.6. He adds the two times up, the 25 years out and 25 years back and gets 50 years. He adds the two 15 years in spaceship time and gets 30 years – a 20 year difference in age.

Here we have three constant-velocity segments as you described:

A is the event of the birth of the twins. AC represents the world-line of twin A who just sits there in her own frame of reference. AD represents twin B’s outward journey to the star (20 light years) and DC represents twin B’s return journey to Earth.

The segments are AC, AD, and DC. If we're calculating things in the rest frame of twin A, then in this frame \Delta x for AC is zero (since twin A doesn't change positions between the two endpoints) and \Delta t for AC is 50 years. So, the proper time on this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{50^2 - 0^2 } = 50 years. Meanwhile, in this same frame if we look at segment AD, \Delta x is 20 light-years while \Delta t is 25 years, so the proper time of this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{25^2 - 20^2 } = \sqrt{225} = 15 years. And for DC, in this frame \Delta x is again 20 light-years while \Delta t is again 25 years, so the proper time on this segment is 15 years as well. Of course you could repeat the calculations of the proper time in a different frame using x',t' coordinates where the \Delta x' and \Delta t' for each segment would be different, but the proper time on each segment would remain the same.

 

[stevmg]

1. I keep getting gamma screwed up. It is [SQRT(1 - v²/c²)]^-1. I apologize for that. I also don't know how to get all that what I wrote under the square root sign in this blog. I tried what you did but it doesn't come out correctly.

2. I just remember three things:

a - Items are longest" in their "rest" frame (when they are not moving.) That applies to distances as well such as the 20 lt-year distance between the Earth and the star as Dr. Wolfson proposed.

b - Time is most "rapid" in its "rest frame" (where there is no motion of the rest frame.) Hence, in Dr. Wolson's example, the trip is 25 years in the Earth-star frame but only 15 years in either spaceship frame (to and fro) which are moving at 0.8c.

c - Mass (not a player here) is "least" or "smallest" in the rest frame and increases as the frame moves. Dr. Wolfson never goes into this in his book.

Thus \gamma is always a ratio:

\gamma = length. at. rest/length. in. motion
\gamma = duration. at. rest/duration. in. motion
1/(\gamma) = mass. at. rest/mass. in. motion

Sorry for all those "."'s but the words run onto each other unless I insert them - Ooops! Just saw the "white spaces and dots selection in the ?Latex? Reference (what's this, we dealing with rubber gloves?) Oh, well...

The reason why Dr. Wolfson chose 0.8c is that it is a neat square root as 1 - 0.8² = 1 - 0.64 which = 0.36 and \sqrt{0.36} = 0.6. Hey! Look at that! I got the "\sqrt{}" to work - for a simple expression.

Now, the coup de grace

In your example below which I quote from you, you actually show how a world-line is calculated in explainable terms. I see clearly why the AC length (the twin that sits still) is less than the ADC length (the twin that moved) - it is because of the negative sign summation under the square root. This sort of makes the world line diagram counter-intuitive as the "longer" path is the "shorter" path in proper time. That is freakin' weird. Is there a way of looking at that diagram to make it look not so weird?

It's because proper time in spacetime is not calculated the same way as length in space. In a 2D space, if the endpoints of a straight segment of a path are at coordinates (x₁, y₁) and (x₂, y₂), then according to the pythagorean theorem, the length of that segment would be \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} (if you don't want to worry about specific coordinates and you know the distance along the x-axis \Delta x and the distance along the y-axis \Delta y between the two endpoints, the formula becomes \sqrt{\Delta y^2 + \Delta x^2 }). On the other hand, in a 2D spacetime with a spatial dimension x and a time dimension t, if you have a constant-velocity segment (like a straight-line segment in your diagram) with endpoints (x₁, t₁) and (x₂, t₂), then the proper time along this segment is \sqrt{(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2}, or if we use units where c=1 (like light-years and years), it'd be \sqrt{(t_2 - t_1)^2 - (x^2 - x_1)^2} = \sqrt{\Delta t^2 - \Delta x^2 }. You can see that this is different from the pythagorean formula because it has a minus in the middle rather than a plus. That change means that unlike in 2D geometry where a straight-line path is always the shortest distance between two points, in spacetime a constant-velocity path between two events (like the event of the twins departing from one another and the event of the twins reuniting) is always the one with the largest proper time.

Doing the calculations for your specific example:

Here we have three constant-velocity segments as you described:

The segments are AC, AD, and DC. If we're calculating things in the rest frame of twin A, then in this frame \Delta x for AC is zero (since twin A doesn't change positions between the two endpoints) and \Delta t for AC is 50 years. So, the proper time on this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{50^2 - 0^2 } = 50 years. Meanwhile, in this same frame if we look at segment AD, \Delta x is 20 light-years while \Delta t is 25 years, so the proper time of this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{25^2 - 20^2 } = \sqrt{225} = 15 years. And for DC, in this frame \Delta x is again 20 light-years while \Delta t is again 25 years, so the proper time on this segment is 15 years as well. Of course you could repeat the calculations of the proper time in a different frame using x',t' coordinates where the \Delta x' and \Delta t' for each segment would be different, but the proper time on each segment would remain the same.

"The proper time along this segment is SQRT[(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2]" Is that the same as s, or, to say it another way:

s² = (t_2 - t_1)² - (1/c²)*(x_2 - x_1)²

I've seen that equation s² = ya-de-da in the past and is "s" proper time?

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

I am a retired physician (still have a license) with a math major and Masters that teaches college here and there and have time in my life now to learn about things that I have seen before but never had the time to study - after all - when I was doctoring, wouldn't you have preferred that I read medical articles (which I did) rather than study theoretical physics, which I didn't.

 

[yuiop]

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

I have that book too and the cover and title is misleading. The cartoon ilustration on the front and the promise of "laymen's explanations" on the back lead one to believe it going to be a clearly explained text but not far into it you discover that that author assumes everyone was born knowing what Kronecker delta function is. It certainly isn't a simple introductory text to relativity if you do not have pretty advanced math skills already and pretty much omits the Michelson Morley experiment and discussions of things like the twin's paradox, presumably because the author thinks those subjects are self explanatory.

 

[atyy]

Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

Yes. In physics, you can choose any reference frame you want, no matter how weird. But you pay the price for the weirdness, in that the laws of physics will look weird, even though they haven't changed at all. It's the same as if you define green as red, and red as green. Then grass is red, but the grass hasn't changed, just that you have chosen a weird nomenclature for colours. Of course, once you know the laws of physics in anyone reference frame inertial or not, you know the laws of physics in any other reference frame, just as once you know that grass is green in a particular colour nomenclature, then you know the colour of grass in any colour nomenclature.

In Newtonian physics and special relativity, there are a special class of reference frames called inertial reference frames, in which the laws of physics look nice, or at least have their standard textbook form. The fact that there is a class of frames that are inertial, rather than only one such frame is due to symmetries in the laws of physics. The difference between Newtonian physics and special relativity is that the symmetry is Galilean in the former, but Lorentzian in the latter. In the context of the twin paradox, twin A is special because she is always at rest in a particular inertial frame. In the broader context of special relativity, twin A is not special compared to another person who is moving at constant velocity relative to her, since that person is at rest in another inertial reference frame.

 

[matheinste]

1.
Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

You can look at it from the point of view of twin A, who is at rest in an inertial reference frame throughout. For the simple case of a constant and equal speed of outward and inward journeys, and assuming instantaneous accelerations at the end points and turning point, the time passed for B is twice the time dilated value of time passed for A during the outward journey of B. If finite accelerations are used, leading to non-linear speed profiles, then integration is required of B's instantaneous co-moving inertial frames.

Matheinste.

 

[JesseM]

In your example below which I quote from you, you actually show how a world-line is calculated in explainable terms. I see clearly why the AC length (the twin that sits still) is less than the ADC length (the twin that moved) - it is because of the negative sign summation under the square root. This sort of makes the world line diagram counter-intuitive as the "longer" path is the "shorter" path in proper time. That is freakin' weird. Is there a way of looking at that diagram to make it look not so weird?

I don't know of any way to represent it so paths with longer proper time actually have a longer spatial length on the diagram...I think you just kind of have to remember that spacetime geometry works differently than spatial geometry.

I've seen that equation s² = ya-de-da in the past and is "s" proper time?

s² is the square of the spacetime interval which is defined a little differently than proper time...whereas proper time squared would be equal to \Delta t^2 - (1/c^2)*\Delta x^2, the square of the spacetime interval s is equal to \Delta x^2 - c^2 \Delta t^2. So s² is basically just -c² times the proper time squared, with s having units of distance rather than time. For any two points in spacetime, the separation between them is either "spacelike" which means s² is positive, "timelike" which means it's negative, or "lightlike" which means it's zero. Any two events on the worldline of a slower-than-light object have a timelike separation, any two events on the worldline of a light ray moving in a single direction in a vacuum have a lightlike separation, and if two events have a spacelike separation then no particle moving at or slower than light can have both events on its worldline (a spacelike separation also means there is one inertial frame where the two events occurred simultaneously at different positions in space, and s would be the spatial distance between them in that frame). A light cone consists of all the points with a timelike or lightlike separation from the event on the "tip".

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

https://www.amazon.com/dp/0716723271/?tag=pfamazon01-20 by A.P. French was my college relativity book, I remember it being pretty clear.

Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

You don't need to use twin A's frame to analyze this problem, you can use any inertial frame and you'll still get the same answer. Check out my post #36 on this thread where I talked about how you could analyze a twin paradox scenario from the perspective of two different frames, one where the inertial twin "Stella" was at rest, and another where the non-inertial twin "Terence" was at rest during the outbound leg of his trip (but not the inbound leg).

I am a retired physician (still have a license) with a math major and Masters that teaches college here and there and have time in my life now to learn about things that I have seen before but never had the time to study - after all - when I was doctoring, wouldn't you have preferred that I read medical articles (which I did) rather than study theoretical physics, which I didn't.

Well, if you'd been my doctor I'd have preferred you read medical articles! ;) But yeah, studying things on your own that you didn't get to learn in school can be a great experience...

 

[DrGreg]

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

You might like to try General Relativity from A to B by Robert Geroch, University of Chicago Press, 1978, ISBN 0-226-28864-1. The preface says it is based on lectures given to non-science undergraduates. Despite the name, most of the book is about Special Relativity (i.e. without gravity), but carefully written so that when gravity is introduced at the end, nothing previously said turns out to be incorrect. It's a moderately slim paperback, with only light-weight maths in it.

A substantial number of sample pages are available at Google Books.

 

[matheinste]

You might like to try General Relativity from A to B by Robert Geroch, University of Chicago Press, 1978, ISBN 0-226-28864-1. The preface says it is based on lectures given to non-science undergraduates. Despite the name, most of the book is about Special Relativity (i.e. without gravity), but carefully written so that when gravity is introduced at the end, nothing previously said turns out to be incorrect. It's a moderately slim paperback, with only light-weight maths in it.

A substantial number of sample pages are available at Google Books.

Yes, this is an excellent book which, up to chapter 7 deals simply but thoroughly with the very basics of spatial geometry and the geometry of the spacetime interval. Don't be tempted to skip the bits you think you are already familiar with.

Matheinste.

 

[JesseM]

This wikibook is also a pretty good free intro, especially the chapter on spacetime which talks about the spacetime interval, light cones, and how to interpret spacetime diagrams (more on spacetime diagrams in the next chapter too, along with a discussion of how they can be used to understand the twin paradox).

 

[stevmg]

Thank you all -

JesseM
Kev
Atyy
Matheinste
DrGreg
Sylas

For your inputs on thisd matter. I know a hell of a lot more now than before,including the math (which I am relatively good at.) I will have toperuse it more slowly to digest it.

JesseM's example of "Stella & ..." is really weird but, as was pointed out, if one takes different frames of reference, a price is to be paid in the interpretation.

By my example of twin A and B. Using the ADC path of twin B with its inherrent accelerations and decelerations makes it the "less clean" choice and by JesseM's "Stella" example, doesn't change anything in the end (when they meet up) after all as they would agree no matter how weird the timing appears en route.

I opted for the Geroch textbook at Amazon as it appeared to be the most reasonable. Of course I downloaded the Wiki .pdf book on relativity which I will print up.

 

[JesseM]

JesseM's example of "Stella & ..." is really weird but, as was pointed out, if one takes different frames of reference, a price is to be paid in the interpretation.

By my example of twin A and B. Using the ADC path of twin B with its inherrent accelerations and decelerations makes it the "less clean" choice and by JesseM's "Stella" example, doesn't change anything in the end (when they meet up) after all as they would agree no matter how weird the timing appears en route.

If you try to use the ADC path you would have to use a non-inertial frame, where normal rules like the time dilation equation would no longer apply. Note that in my analysis, that's not what I did--instead I analyzed things using the inertial frame where Stella was at rest during the outbound leg, but not during the inbound leg. You could imagine a third observer traveling alongside Stella during the outbound leg, who didn't accelerate when Stella did but just kept on moving inertially, so that the analysis could be from this observer's rest frame.

 

[stevmg]

Thanks for the input. I walked through the explanations you gave (even in the old post #36) and it is slowly coming together. I've changed my mind and think I'll go for your college textbook because it sure seemed to explain things to you pretty damn well.

I was a math major and have a Masters in it (primarily statistics) and am used to thinking of math in "real world" scenarios. I did take a course in topology which blew my mind but I got a "A."

I do remember this one item from topology - Take any point on the globe and note its temperature and barometric pressure (t₁, p₁). On any great circle though that point there is at least one other point with an equal temperature and barometric pressure, i.e., t₁ = t₂ and p₁ = p₂. It's true but is hard to prove. Has something to do with continuous functions.

 

[stevmg]

To JesseM

Regarding Post 36 of “What happens biologically during time dilation?”

It seems like you use the reduction factor of 0.8 twice…It works out but I don’t get it.

- Stella is presumed at rest so Terence is moving to the left at 0.6c. Correct?

- Stella starts her leftward movement when Terence is ?light-years away from Stella.? I understand from the first example (Terence at rest, Stella moving at 0.6c to the right) that the time elapsed in Terence’s frame is 10 years (outbound) and 10 years inbound, thus Stella has moved out to 10*0.6 = 6 light-years. This calculation is obvious.

The rest, below, becomes a bit more oblivious

- In the second example in which Stella is at rest and Terence moves to the left at 0.6c you are given nothing. All you know is that Terence moves left at 0.6c but you do not know for what distance or for how long. It looks like you are using the 6 light-years from the first example (Terence stationary) to calculate this distance. Can you do that? Can you use the distance from the stationary F.O.R. of Terence assumed in the first example in this approach to this example assuming Stella is stationary? From there you “contract” the distance from Stella to Terence (which you calculated in the first example where Terence remained stationary) to be 4.8 light years by using the (1/gamma) = 0.8 as a length contraction factor [6 light-years*0.8 = 4.8 light-years.] Then you back calculate the elapsed time as 8 years by dividing the 4.8 light-years by 0.6c = 8 years. In whose frame is the 8 years – Stella or Terence? What justification do you have for assuming the 6 light-years from the first example (Terence stationary) is correct for this alternate look at the same problem (Stella initially stationary?) Also, the 6.0*0.8 = 4.8 light-years is true for whose frame? Is it Stella’s (sitting still) or is it Terence’s (who is in motion?)

- Moving right along: Now, you then “re-contract” [or "time-dilate"] the 8 years you just calculated above (6.0*0.8)/0.6]*(0.8) = 6.4 years. Where did that come from? Whose frame is that happening in – again, Stella (she’s stationary) or Terence (moving right along?) Haven’t they both been “time-dilated” by now?

- Now, you claim that at this point, after the 8 years or 6.4 years or whatever Stella blasts off to the left at 0.88235c which you calculated by the relativistic velocity addition formula. Now, who is that relative to - Stella’s original F.O.R. or Terence’s?

- The arithmetic comes out as advertised – Stella = 8 + the second 8 years which you demonstrated = 16 and Terence = 6.4 + 13.6 = 20 which is what was the Terence stationary approach.

I have tried different speeds (such as 08c or 0.5c) and your method works out but what I am after is that you match the various distances, elapsed times and speeds which you discussed with the appropriate F.O.R.’s and also to “justify” your use of the 6 light-years initially (in this Stella-stationary approach) as the true “distance” between Terence and Stella such that when this magical distance is achieved, Stella begins her leftward 0.88235c gallop towards the slower but still moving Terence who continues moseying left at 0.6c.

Steve G

 

[starthaus]

- I speak 23 languages superfluously.)

You mean "fluently", right? :-)
You don't have to buy a book in order to get a good, simple explanation of the "Twins paradox".
Just read this wiki entry.

 

[JesseM]

To JesseM

Regarding Post 36 of “What happens biologically during time dilation?”

It seems like you use the reduction factor of 0.8 twice…It works out but I don’t get it.

- Stella is presumed at rest so Terence is moving to the left at 0.6c. Correct?

- Stella starts her leftward movement when Terence is ?light-years away from Stella.? I understand from the first example (Terence at rest, Stella moving at 0.6c to the right) that the time elapsed in Terence’s frame is 10 years (outbound) and 10 years inbound, thus Stella has moved out to 10*0.6 = 6 light-years. This calculation is obvious.

The rest, below, becomes a bit more oblivious

- In the second example in which Stella is at rest and Terence moves to the left at 0.6c you are given nothing. All you know is that Terence moves left at 0.6c but you do not know for what distance or for how long. It looks like you are using the 6 light-years from the first example (Terence stationary) to calculate this distance. Can you do that?

Sure, remember that both frames have to agree about local physical events, like what object Stella was next to when she accelerated. And I specified that we could imagine that Terence had a measuring-rod moving along with him, with one end next to Terence's position and the other end 6 light years away in Terence's frame; if it's true in Terence's frame that Stella accelerates when she reaches the end of this measuring rod, it must be true in the other frame as well. Just read this part again:

In Terence's frame, remember that Stella accelerated when she was 6 light-years away from Earth, so we can imagine she turns around when she reaches the far end of a measuring-rod at rest in Terence's frame and 6 light-years long in that frame, with Terence sitting on the near end; in the frame we're dealing with now, the measuring-rod will therefore be moving along with Terence at 0.6c, so it'll be shrunk via length contraction to a length of only 0.8*6 = 4.8 light-years.

From there you “contract” the distance from Stella to Terence (which you calculated in the first example where Terence remained stationary) to be 4.8 light years by using the (1/gamma) = 0.8 as a length contraction factor [6 light-years*0.8 = 4.8 light-years.]

I don't directly contract the distance from Stella to Terence--I contract the actual length of the physical measuring-rod, and then point out that since Terence is at one end and Stella is at the other when she accelerates, then this contracted length must also be the distance between Terence and Stella in this frame at the moment Stella accelerates.

Then you back calculate the elapsed time as 8 years by dividing the 4.8 light-years by 0.6c = 8 years. In whose frame is the 8 years – Stella or Terence?

This is the frame where Stella was at rest during the outbound leg. In this frame Terence was moving away at 0.6c, and we know by the above argument involving the measuring-rod that he had reached a distance of 4.8 light years from Stella when Stella accelerated, so it must have taken him 8 years to get out this distance in this frame.

What justification do you have for assuming the 6 light-years from the first example (Terence stationary) is correct for this alternate look at the same problem (Stella initially stationary?)

It isn't correct! 4.8 light-years is the correct distance between Stella and Terence when Stella accelerates in this frame, not 6 light years as in Terence's frame. Again, this is just because Stella accelerates when her position coincides with the end of the measuring-rod (a local event that all frames must agree on), and if the measuring rod has a length of 6 light years in Terence's frame where the rod is at rest, then according to the length contraction equation it must have a shorter length of 4.8 light years in this frame.

Also, the 6.0*0.8 = 4.8 light-years is true for whose frame? Is it Stella’s (sitting still) or is it Terence’s (who is in motion?)

Stella's frame (or more specifically the inertial frame where Stella was at rest during the outbound phase). The measuring-rod is at rest and 6 light years long in Terence's frame, it is moving at 0.6c in Stella's frame and therefore is length-contracted to 4.8 light years.

Moving right along: Now, you then “re-contract” [or "time-dilate"] the 8 years you just calculated above (6.0*0.8)/0.6]*(0.8) = 6.4 years. Where did that come from? Whose frame is that happening in – again, Stella (she’s stationary) or Terence (moving right along?) Haven’t they both been “time-dilated” by now?

Everything in the second paragraph of my explanation, the one that begins "Now let's analyze the same situation in a different inertial frame--namely, the frame where Stella was at rest during the outbound leg of her trip", was meant to be from the perspective of Stella's frame. In this frame's coordinates, Terence was moving away from Stella at 0.6c until he reached a distance of 4.8 light years (when the other end of the rod moving along with him was next to Stella), so in this frame this must have taken a coordinate time of 4.8/0.6 = 8 years. Stella is at rest in this frame, so her clock keeps pace with coordinate time, meaning 8 years have passed on her clock between the moment Terence departed and the moment she is lined up with the end of the measuring-rod moving along with Terence (the same moment she accelerates)--she experiences no time dilation up until then in this frame. On the other hand, since Terence is moving at 0.6c, his clock is dilated by a factor of 0.8 relative to coordinate time, therefore if a coordinate time of 8 years passes between Terence leaving Stella and Stella accelerating, Terence's clock must only elapse 6.4 years between the times of these two events in this frame, just based on the time dilation equation.

- Now, you claim that at this point, after the 8 years or 6.4 years or whatever Stella blasts off to the left at 0.88235c which you calculated by the relativistic velocity addition formula. Now, who is that relative to - Stella’s original F.O.R. or Terence’s?

Again it's all in Stella's original frame. We already knew Stella's speed in Terence's frame, it was 0.6c in both directions. As explained here, the velocity addition formula tells you that if some object A is moving at speed v in some direction the frame of B, and B is moving at speed u in the same direction in the frame of C, then the speed of object A in the frame of C will be (u + v)/(1 + uv/c^2). In this case we know Stella (object A) was moving at speed 0.6c to the left in the frame of Terence during her inbound trip, and Terence (object B) was moving at 0.6c to the left in the frame of an inertial observer (object C) who saw Stella at rest during her outbound trip, which means in the frame of this observer, during her inbound trip Stella must have had a speed of (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c.

I have tried different speeds (such as 08c or 0.5c) and your method works out but what I am after is that you match the various distances, elapsed times and speeds which you discussed with the appropriate F.O.R.’s and also to “justify” your use of the 6 light-years initially (in this Stella-stationary approach) as the true “distance” between Terence and Stella such that when this magical distance is achieved, Stella begins her leftward 0.88235c gallop towards the slower but still moving Terence who continues moseying left at 0.6c.

6 light years was just the starting assumption of the distance in Terence's rest frame when Stella accelerated. It doesn't need to be justified since it's just how I originally defined the problem from the perspective of Terence's frame, you could easily have picked any other distance/time until Stella accelerated in this frame, if you preferred. What did need to be justified was the idea that if Stella accelerated at a distance of 6 light years from Terence in Terence's frame, then that implies that Stella accelerated at a distance of 4.8 light years in the frame of the inertial observer who saw Stella at rest during the outbound leg. I justified this by imagining a measuring-rod at rest relative to Terence, with one end lining up with Terence and the other end being 6 light-years away in Terence's rest frame, so that the other end lined up with Stella at the moment she accelerated.

 

[Dale]

Hi stevemg,

Since you have a decent math background you may want to look at this from the four-vector formulation where each event is given coordinates (ct,x,y,z) and different reference frames are related to each other simply with a http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form" (the Lorentz transform is a linear transform).

So, for example we have three events which are described in Terrance's rest frame as follows using units of years and light-years:
A (Stella departs from Terrance): (0,0,0,0)
B (Stella turns around): (10,6,0,0)
C (Stella returns to Terrance): (20,0,0,0)
and the elapsed times are given by the Minkowski norms:
Stella: |B-A|+|C-B|=16
Terrance: |C-A|=20

The coordinates in the frame where Stella is initially at rest are obtained by the Lorentz transform:
A': (0,0,0,0)
B': (8,0,0,0)
C': (25,-15,0,0)
and the elapsed times are given by the Minkowski norms:
Stella: |B'-A'|+|C'-B'|=16
Terrance: |C'-A'|=20

 

[stevmg]

Hello again, DaleSpam:

Thanks for the matrix notation but, unfortunately, for me, although this works it does not give me a “feel” as to why it works. It is quite abstract, even though I have a math background.

After reviewing JesseM’s explanations above, it does work out as advertised in his previous explanations.

I understand now how he is able to initially use the 6 lt-yr distance between Stella and Terence in the Stella stationary example. That is because with Stella stationary, the first half of the trip, represented by Terence moving to the left at 0.6c is the mirror image of Stella moving to the right at 0.6c from the first example (Terence stationary.) Thus, the elapsed distance on that leg is the same (6 lt-yr). Then, JesseM goes through his magical calculations to produce the 4.8 lt-yr distance and 8 yr quantities that result when looked at by Stella because of the 1/gamma factor. The second application of the 0.8 factor is a little tricky. Here, JesseM states that because Terence is still moving at 0.6c to the left, in his time frame of reference, his elapsed time is only 8years*0.8 = 6.4 years (and we are going to need that 6.4 eventually.)

He then states that at this point, when the distance between Stella and Terence is 4.8 lt-yr Stella jumps to 0.88235c by “catching up” to Terence [by use of the velocity addition formula (0.6 + 0.6)/(1 + 0.6^2)]. Then, using simple algebra and a 4.8 lt-yr distance, the “time” of 17 years is calculated [4.8/(0.88235 – 0.60]. Again, by time dilation, that translates to 17*0.22235 = 13.6 years. Adding 13.6 + the 6.4 from above we get 20. Presto! Terence still advances 20 years.

As far as Stella goes, we have the 8 years from the third paragraph above + 17*SQRT(1 – 0.88235^2) = 8 years for a total of 16 years, and, again, Stella ages as predicted before.

Miracles will never cease!

Wow! Was that a workout!

Steve G

 

[stevmg]

When I say 23 languages superfluously I mean just that - one word out of each! And DaleSpam can testify to all that my English (originally from New York) isn't so good either. And, being Italian, without using my hands, I am almost a total mute!

 

[Dale]

Thanks for the matrix notation but, unfortunately, for me, although this works it does not give me a “feel” as to why it works. It is quite abstract, even though I have a math background.

Fair enough. I have dealt with linear algebra quite a bit in the context of computer graphics and geometry, so I guess I had a good understanding of rotation and shear and other similar linear matrix operations. Anyway, the 4-vectors and the matrix notation is what finally made things "click" for me, but I have learned that it doesn't work for everyone like it did for me.

 

[stevmg]

JesseM -

Great elucidation of my questions. When I used the word "justify" I really meant "on what basis" (which you have answered.)

Keep up the great works, eeryone.

To DaleSpam - I will now try to review my matrix multiplication. I am a little rusty (actually, a whole loty rusty) on coordiante subraction for absolute didtances or "proper time" or whatever. I know in the time-space equation one cannot do calculations as one would do in ordinary cartesion 2 0r 3 dimension space. As you expalined that this sort of a reverse Pythagorean theorem.

 

[stevmg]

DaleSpam:

I remembered how to obtain distances from two coordinates in n-space (of course, the t-vector is a + and you subtract the sum of squares of the cartesian distances from the time-squared (everything in terms of light speed) to obtain the "proper time." You had given that formula some time ago. If you use x_1, x_2, x_3 and x_4 as the space time coordinates you must use x_1 = -ct)

Example 1

A (Stella departs from Terrance): (0,0,0,0)
B (Stella turns around): (10,6,0,0)
C (Stella returns to Terrance): (20,0,0,0)

Stella: |B-A|+|C-B|=16
Terrance: |C-A|=20

|B - A| = SQRT[(10 - 0)^2 - [(6 - 0)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(64) = 8
|C - B| = SQRT[(20 - 10)^2 - [(0 - 6)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(64) = 8
8 + 8 = 16 Q.E.D.
|C - A| = SQRT[(20 - 0)^2 - [(0 - 0)^2 + (0 - 0)^2 +(0 - 0)^2)]] = SQRT(400) = 20
Q.E.D.

Example 2

A': (0,0,0,0)
B': (8,0,0,0)
C': (25,-15,0,0)

Stella: |B'-A'|+|C'-B'|=16
Terrance: |C'-A'|=20

|B' - A'| = SQRT[(0 - 8)^2 - [(0 - 0)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(64) = 8
|C' - B'| = SQRT[(25 - 8)^2 -[(-15 - 0)^2 + (0 - 0)^2 + (0 - 0)^2]]= SQRT(64) = 8
8 + 8 = 16
|C' - A'| = SQRT[(25 - 0)^2 - [(-15 - 0)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(400) = 20
Q.E.D.

Actually, I DO see this better this way (the way you were trying to explain it in your post) than the long winded way that I tried to explain it before. It was the matrix multiplication which tied me up and was making me see the "trees" instead of the "forest" and get bogged down. I should have used the Lorentz transforms directly (even without matrix multiplication) to obtain the various coordinates as you demonstrated.

 

[Dale]

Excellent! That is exactly right. I'm glad it helped.

 

[stevmg]

To those of us who DIDN'T get what DaleSpam was saying, Dale uses the Lorentz transforms (which can be done without matrix multiplication to transform the following base coordinates given by the problem:)

The assumption was that Stella left Terence initially, moved to the right at 0.6c for 10 years, then turned around and came back at 0.6c for 10 years. The space time coordinates describing these events are thus (TERENCE STATIONARY - at this point there is NO transformation - these are the initial coordinates from stationary Terence):

A (Stella departs from Terrance): (0,0,0,0)
B (Stella turns around): (10,6,0,0)
C (Stella returns to Terrance): (20,0,0,0)

By using the Minkowski norms, Dale calculated the elapsed or proper time for Stella and Terence, thus:
Stella: |B-A|+|C-B|=16
Terrance: |C-A|=20

|B - A| = SQRT[(10 - 0)^2 - [(6 - 0)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(64) = 8
|C - B| = SQRT[(20 - 10)^2 - [(0 - 6)^2 + (0 - 0)^2 + (0 - 0)^2]] = SQRT(64) = 8
8 + 8 = 16 years elapsed time for Stella
|C - A| = SQRT[(20 - 0)^2 - [(0 - 0)^2 + (0 - 0)^2 +(0 - 0)^2)]] = SQRT(400) = 20 years elapsed time for Terence

Now Dale transforms the above Terence coordinates to Stella coordinates by using the Lorentz transformation which I will briefly go through below. Remember we use for t, 0, 10 and 20; for x we use 0, 6 and 0. So now we have to find t' and x' which are the Stalla coordinates.

A' (Terence departs from Stella) t' = 0 x' = 0 therefore this is represented by (0, 0, 0, 0). Our A' coordinate is thus (0, 0, 0, 0)
B' (Stella turns around) t' = [10 - 0.6*6/SQRT[1 - 0.6^2] = 6.4/0.8 = 8
x' = (6 - 0.6*10)/0.8 = 0. Thus our B' coordinate is (8, 0, 0, 0)
C' x' = (0 - 0.6*20)/0.8 = -15. t' = (t - vx/c^2)/0.8. But x = 0, thus x' = 20/0.8 = 25. Hence our C' coordinate is (25, -15, 0, 0)

The Minkowski norms (or "proper times") are calculated as shown in post 90 just above. We see by both sets of coordinates, the Minkowski norm for Stella is 16 and for Terence is 20. We don't even "count" the general relativity effect of acceleration and deceleration.

So, we see there is no paradox as there is no way to look at it that will make Terence younger than Stella as long as we remain in the "time-like" portion of the light cone. Of course, by current technology, it is impossible to get to the space-like portion where all kinds of weird stuff can happen.

That was a workout for me to understand and it took me a month. It won't come easy.

Steve G