Twin Paradox aging question

DrGreg

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

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

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

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

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:
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.
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.
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.
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.
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.
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.
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.

DaleSpam

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 matrix multiplication (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!

DaleSpam

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.

DaleSpam

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