# Acceleration doesn't cause the Twin Paradox?

Acceleration doesn't "cause" the Twin Paradox?

In a recent review of a physics textbook, the reviewer is critical of the author of the book because the the author doesn't correct the persistent notion of many students that it is the acceleration of one of the twins that "causes"[reviewer's quotes] the differential aging in the twin paradox.

Suppose we have a set of twins, Eartha and Stella. Stella accelerates in a ship to nearly the speed of light and lands on a planet 30 light years away. Immediately upon landing,Stella sends a picture of herself to Eartha, the stay-at-home twin. What would Eartha say upon receiving the image of her twin? Eartha would say that her sister looks exactly like the day she left!

All Stella did was accelerate to near the speed of light.There was no meet-up back on earth. There wasn't even a turn around or any change in direction. Just acceleration and deceleration

My Question: Why is this persistent notion in error?
If acceleration didn't cause the differential aging, what did?

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Matterwave
Gold Member

There's a few sources which just say "well one of the twins accelerated, so use GR for this". That's maybe what the reviewer was criticizing? In effect, it is the acceleration that breaks the symmetry of the problem and allows a non-ambiguous result to be reached. But one certainly does not need GR to analyze the twin paradox. Whichever twin switched reference frames will be the one who turns out to be older.

Whichever twin switched reference frames will be the one who turns out to be older.
You mean younger?

Matterwave
Gold Member

Right, younger. >.>

I wrote that without thinking. <.<

Indeed acceleration is sufficient but not necessary - you need a way to break the symmetry, and acceleration is one way to do so. Consider the case where the the universe is compact and thus finite, say a torus, that allows its observers to travel in one direction and then arrives back where he started. See here for more details. This gives rise to the so-called "cosmological twin paradox", see this thread and this thread for detailed discussion. I just quote the scenario for clarity here:

Take two twins, one twin stays put and the other is accelerated to
9.999…%c she passes her sister and they synchronise clocks. She continues at constant velocity along a geodesic path and circumnavigates the universe. She eventually meets up with her sister again.

She has aged only 50 years in this near light-speed voyage. However her sister has aged 10 billion years!

Or is it the other way round? How do you tell?

PAllen
2019 Award

In a recent review of a physics textbook, the reviewer is critical of the author of the book because the the author doesn't correct the persistent notion of many students that it is the acceleration of one of the twins that "causes"[reviewer's quotes] the differential aging in the twin paradox.

Suppose we have a set of twins, Eartha and Stella. Stella accelerates in a ship to nearly the speed of light and lands on a planet 30 light years away. Immediately upon landing,Stella sends a picture of herself to Eartha, the stay-at-home twin. What would Eartha say upon receiving the image of her twin? Eartha would say that her sister looks exactly like the day she left!

All Stella did was accelerate to near the speed of light.There was no meet-up back on earth. There wasn't even a turn around or any change in direction. Just acceleration and deceleration

My Question: Why is this persistent notion in error?
If acceleration didn't cause the differential aging, what did?
Note that you have two accelerations in your scenario: to reach near c, and to land on a planet. They are not essential to your interesting question (in fact they refute it if the goal is to show inessentiality of acceleration).

Let's suppose stella passes right by a space station near earth, and stella and eartha look at each other, seeing they are about the same age (and they syncrhonize wristwatches). Then suppose stella sends eartha a self image as stella passes the distant planet. Further, let's suppose eartha sends stella a self image from a time such that it happens to reach stella at a time after stella passes the planet that is very slightly less than the time it took (per stella) to reach the planet. This means, stella would conclude eartha sent her image 'at the same time' per stella, that stella passed the planet. Eartha, meanwhile, by direct (delayed) communication with the planet verifies stella's image was sent when stella passed the planet.

What do they see? Eartha sees a picture of stella that has aged only a month (for example), though sent after 30 years (per Eartha), and arriving after 60 years per Eartha. Stella gets a picture of Eartha after (for example) two months (planet passed after one month, per stella). But Eartha's picture (say, based on wristwatch in image) has aged less than an hour since stella left eartha! (It would be a good exercise for you to work out why this is so).

Thus there is symmetrical time dilation, and no paradox, when all motion is inertial. Acceleration of at least one twin is necessary for a twin paradox in SR. Further, they have to get back together to have a mutually agreed on age difference. As to what part of difference in path through spacetime is responsible for the age difference, I hold that is a completely meaningless question. In that sense (only) I would say you can't say acceleration caused the age difference, but you can say it enabled it.

PAllen
2019 Award

Indeed acceleration is sufficient but not necessary - you need a way to break the symmetry, and acceleration is one way to do so. Consider the case where the the universe is compact and thus finite, say a torus, that allows its observers to travel in one direction and then arrives back where he started. See here for more details. This gives rise to the so-called "cosmological twin paradox", see this thread and this thread for detailed discussion. I just quote the scenario for clarity here:
Actually, what this cute example shows is that (if you allow flat but topologically nontrivial spacetime), acceleration is neither necessary nor sufficient. It is not sufficient, because if you symmetric acceleration, there is no differential aging (each accelerates away and back, with identical thrust profile). It is not necessary due to (and only due to) nontrivial topology.

robphy
Homework Helper
Gold Member

from an earlier post of mine
"First, is it reasonable to say that
it is during his acceleration at C
that twin 2 suddenly loses time or age, and that
this loss at C causes the final age difference at B?
It is just as unreasonable to blame the acceleration at C
for the total age difference of the twins
as it would be,
in the case of a triangle ACB in the ordinary Euclidean plane,
to say that the larger length of the path ACB,
as compared to the straight path AB,
is caused by a sudden gain of length at the corner C."

The Clock Paradox in Relativity Theory
Alfred Schild,
The American Mathematical Monthly, Vol. 66, No. 1 (Jan., 1959) (pp. 1-18)
http://www.jstor.org/pss/2309916

Actually, what this cute example shows is that (if you allow flat but topologically nontrivial spacetime), acceleration is neither necessary nor sufficient. It is not sufficient, because if you symmetric acceleration, there is no differential aging (each accelerates away and back, with identical thrust profile). It is not necessary due to (and only due to) nontrivial topology.
Yes. You are right. I should have been more careful :tongue:

What about the claim by Dr. Mendel Sachs that the twin paradox is itself not a valid interpretation?

I should note that a post asking for clarification on how, given two paths relative to a given inertial frame, the integral of time along the path integral does not correspond to aging was deleted. I asserted that the path with the most acceleration relative to that frame should age more.

I guess Mendel Sachs did not like my definition of aging as a biological measure of the rate of time, and thus subject to SR.

PAllen
2019 Award

What about the claim by Dr. Mendel Sachs that the twin paradox is itself not a valid interpretation?

I should note that a post asking for clarification on how, given two paths relative to a given inertial frame, the integral of time along the path integral does not correspond to aging was deleted. I asserted that the path with the most acceleration relative to that frame should age more.

I guess Mendel Sachs did not like my definition of aging as a biological measure of the rate of time, and thus subject to SR.
I was actually in college when Sachs first proposed his thesis on the twin paradox (circa 1971). It caused a lot discussion and snickers.

Sachs is/was a serious scientist, but on this, his position is crank, and in decades of writing has not swayed any to his side.

The integral of proper time along world line (with any amount or lack of proper acceleration) is, by definition, the time experience by any physical process following that world line. In both SR and GR, this a definition leading to predictions. Any observation counter to this would be disproof of relativity. There are no such observations, so far as I know.

I was actually in college when Sachs first proposed his thesis on the twin paradox (circa 1971). It caused a lot discussion and snickers.

Sachs is/was a serious scientist, but on this, his position is crank, and in decades of writing has not swayed any to his side.

The integral of proper time along world line (with any amount or lack of proper acceleration) is, by definition, the time experience by any physical process following that world line. In both SR and GR, this a definition leading to predictions. Any observation counter to this would be disproof of relativity. There are no such observations, so far as I know.
Thank you for confirming my thoughts. This appears to make his sites a haven for SR doubters. It seems the flaw in logic is outlined by TataKai in the following forum post.

http://mendelsachs.com/cgi-bin/forum/Blah.pl?b-MSFORUM4/m-1155004653/

Here it is argued that Einsteins postulates are inconsistent, thus SR is in error. It would seem to me that these postulates are not well defined, leaving the status of SR inconclusive based on the invalid arguments deriving from these postulates.

Unfortunately, it appears to me that those who refute SR based on this line of argument fail to observe that SR can also be derived based on the following, clearly consistent, postulates:

1) An inertial reference frame is Euclidean.
2) Any trajectory moving with a constant velocity in any reference frame defines the origin of another reference frame.
3) The transformations between frames form a mathematical group.

This then limits the form of the transformation to a single undetermined parameter, 1/c^2, that has been determined to be a precise constant using many methods beyond the optical methods implied by Einsteins postulates. In fact, this can be used to clarify what is intended by Einstein's postulates as theorems.

Electromagnetic/optical phenomenon were the first system where we reached an accuracy able to distinguish this value from 0, and refute the Galilean transformations.

Returning to the twin problem in SR, this derivation would require us to first fix a reference frame. Further, the twins trajectories must have identical initial and final coordinates for a comparison to be meaningful. Each twin will have their age reduced based on how much time dilation is observed in the given frame along the trajectory they followed.

This delta in age must be Lorentz invariant. Otherwise the theory WOULD be inconsistent.

Hello starfish99:
The Twin Paradox outcome is not dependent on having an acceleration phase. An age difference can also be demonstrated by other means of symmetry breakage.

Let’s say that observers A and B are separated and are approaching each other on a collision course with constant relative velocity. Observer A takes the initiative of synchronizing their clocks as follows. When A’s clock reads T1 he sends a signal (at light speed) to B. B’s receipt of that signal resets his clock to zero and immediately sends a return signal to A, who receives it at his time T2. With the assumption that both light signals traveled at the same speed, A concludes that from his point of view, B reset his clock to zero when A’s clock was at (T1+T2)/2, the midpoint of the T1-to-T2 interval. That being so, A resets his clock at T2 to the time (T2-T1)/2. As far as A is concerned the two clocks have been synchronized.

Nevertheless, when A and B finally collide (or, hopefully, pass one another) they compare clocks and find that A’s clock has advanced more than B’s clock (of course by an amount predicted by SR for their relative velocity.)

As you might expect, if B initiates the synchronization process, the aging outcome would be inverted. And all this without any acceleration.

It does no good to ask why time behaves this way. Such questioning falls in the same category as “Why are there only three spatial dimensions?” (if indeed that’s all there are.)

The current answer to such questions can only be: “That’s the structure of spacetime.”

ghwellsjr
Gold Member

Let’s say that observers A and B are separated and are approaching each other on a collision course with constant relative velocity. Observer A takes the initiative of synchronizing their clocks as follows. When A’s clock reads T1 he sends a signal (at light speed) to B. B’s receipt of that signal resets his clock to zero and immediately sends a return signal to A, who receives it at his time T2. With the assumption that both light signals traveled at the same speed, A concludes that from his point of view, B reset his clock to zero when A’s clock was at (T1+T2)/2, the midpoint of the T1-to-T2 interval. That being so, A resets his clock at T2 to the time (T2-T1)/2. As far as A is concerned the two clocks have been synchronized.
Clocks that are moving with respect to one another cannot be synchronized since they tick at different rates. They can only be reset to the same time when they are colocated. By the time A responds to the signal from B, B's clock will no longer be at the same time. I don't know what you think this accomplishes.

George:
Yes, clocks in motion relative to each other will not remain synchronized. That’s the gist of SR. That’s what accounts for the age difference, both in my example and in the standard twin paradox problem. Both have the same result for this same reason.

In one case (the standard twin paradox problem) they momentarily synchronize their clocks and mutually agree that they had done so because they were together at the synch moment. In my example only A can claim the synch moment , and only by inference.

For A to conclude that B’s clock read zero when A’s own clock read (T1+T2)/2 is most reasonable. And for A to reset his clock to (T2-T1)/2, so that it would have read zero at what he considered to be a simultaneous event with B’s clock-zero, gives him an easy way to compare elapsed times when they meet. In either case A’s synch moment is fleeting, and serves only to make it easier to compare the clocks’ elapsed times between events.

The “mechanism” that brings about the aging difference is the same in both the standard twin problem and the one I showed here.
My purpose was only to put to rest the idea that acceleration is a necessary ingredient.

I was actually in college when Sachs first proposed his thesis on the twin paradox (circa 1971). It caused a lot discussion and snickers.

Sachs is/was a serious scientist, but on this, his position is crank, and in decades of writing has not swayed any to his side.

The integral of proper time along world line (with any amount or lack of proper acceleration) is, by definition, the time experience by any physical process following that world line. In both SR and GR, this a definition leading to predictions. Any observation counter to this would be disproof of relativity. There are no such observations, so far as I know.
Actually, we could have wrapped up the discussion with PAllen's post, because it tells the story in a nutshell. I'll just go ahead and add the space-time diagram for what PAllen has just said (but leaving out acceleration details). I've included a couple of hyperbolic calibration curves to help keep track of the proper times for the twins. The traveling twin takes 10 years going out and 10 years returning, while the stay-at-home twin sits there and ages 40 years.

I think the reviewer of the text book referred to in post #1 was frustrated because he felt that it is the path taken through space-time that should be the point of focus when talking about the twin paradox.

JDoolin
Gold Member

I have a question.

Why are there no textbooks that actually DO the Lorentz Transformation?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like from the traveling twin's point of view on the OUTBOUND trip?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like on the RETURN trip?

Is there some kind of conspiracy, or is it just considered "wrong" to do it, for some reason?

I have a question.

Why are there no textbooks that actually DO the Lorentz Transformation?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like from the traveling twin's point of view on the OUTBOUND trip?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like on the RETURN trip?

Is there some kind of conspiracy, or is it just considered "wrong" to do it, for some reason?
Is this what you are looking for? Or did you want to see space-time diagrams for the traveling twin's rest system?

ghwellsjr
Gold Member

I have a question.

Why are there no textbooks that actually DO the Lorentz Transformation?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like from the traveling twin's point of view on the OUTBOUND trip?

Why do we never actually take the event-mapping one-to-one, and see what the space-time diagram looks like on the RETURN trip?

Is there some kind of conspiracy, or is it just considered "wrong" to do it, for some reason?
Since you can analyze any scenario from any Frame of Reference, why bother with doing a Lorentz Transformation?

The Twin Paradox is so easy to analyze from a frame in which the one twin remains at rest. You just apply Einstein's super simple time dilation formula for a moving clock, τ=t√(1-β2), you plug in the speed, β, as a fraction of the speed of light that the traveler goes at and the time, t, in the rest frame that he is gone and you get his age, τ, when he returns. So if he's traveling at 0.8c and he's gone for 10 years (5 years out and 5 years back), he'll be only 6 years older when he gets back compared to the 10 years of his brother:

τ=t√(1-β2)=10√(1-0.82)=10√(1-0.64)=10√(0.36)=10(0.6)=6

Now if you want to use the Lorentz transformation to analyze this from a frame in which the traveling twin is at rest during the outbound portion of the trip, you will have to first assign events to the first frame. I prefer to only include t and x in the form [t,x] and use units of t in years and x in light-years.

So we start with both twins at the origin of our frame, [0,0].

Next we calculate where the traveling twin will be after 5 years at 0.8c which is 4 light-years away, [5,4].

Meanwhile the other twin is at [5,0].

Then 5 years later, both twins are reunited at [10,0].

To do any Lorentz Transforms, we start by calculating gamma for beta of 0.8:

γ=1/√(1-β2)=1/√(1-0.82)=1/√(1-0.64)=1/√(0.36)=1/0.6=1.667

Now it's fairly easy to transform the first two events into the rest frame of the traveler using the simplified Lorentz Transform. In fact, the first one is the origin which is also the origin of any other frame. But the second event for the traveler is:

t'=γ(t-xβ)=1.667(5-4*0.8)=1.667(5-3.2)=1.667(1.8)=3 years

x'=γ(x-tβ)=1.667(4-5*0.8)=1.667(4-4)=1.667(0)=0 light-years

But the event for the stay-at-home twin is:

t'=γ(t-xβ)=1.667(5-0*0.8)=1.667(5-0)=1.667(5)=8.333 years

x'=γ(x-tβ)=1.667(0-5*0.8)=1.667(0-4)=1.667(-4)=-6.667 light-years

Now the event for the traveler looks good because since he is at rest in this frame, his position remains at 0 and his time is 3 years, half of the accumulated age that we calculated earlier (since he is half-way through his 6 year trip, according to him).

But what about the event for the stay-at-home twin? Those numbers don't make any sense at all, do they? But they do if we remember that these are coordinates in a different frame. If we want to know how much the stay-at-home twin aged up to this point, we have to use the time dilation formula on the coordinate time to get his proper time which is 8.333 times 0.6 which is 5 years.

Now we want to bring the twins back together using the last of the events:

t'=γ(t-xβ)=1.667(10-0*0.8)=1.667(10-0)=1.667(10)=16.667 years

x'=γ(x-tβ)=1.667(0-10*0.8)=1.667(0-8)=1.667(-8)=-13.333 light-years

Again, the stay-at-home twin has a coordinate time of 16.667 years but if we multiply this by 0.6 we get 10 years.

The traveling twin is a little more complicated because we don't know off hand what his speed is but it's not too hard to calculate if we just take the difference in the last two events for him, [16.667,-13.333] and [3,0] which is [13.667,-13.333]. This means he has traveled 13.333 light-years in 13.667 years for a speed of 0.9756c. Plugging this into the time dilation formula, we get an accumulated age of:

τ=13.667√(1-0.97562)=13.667√(1-0.9518)=13.667√(0.0482)=13.667(0.2195)=3 years

Wasn't that fun?

JDoolin
Gold Member

Wasn't that fun?
Hi ghwellsjr.

Yes, that was fun. I like your problem set-up. We have three major events
e1: departure (x,t)=(0,0)
e2: turnaround (x,t)=(4,5)
e3: return (x,t)=(0,10)

You also have defined another event (x,t)=(0,5), for which (x',t')= (-6.667,8.333). Then you said: "Those numbers don't make any sense at all, do they? But they do if we remember that these are coordinates in a different frame." I totally agree; but there's a lot more to say about that. Why doesn't it make sense. Why does it make sense? How does it make sense? How long is the outbound twin's reference frame relevant to the outbound twin? For only three years. But this event happens at t=8.333 years! How far away is it going to happen? 6.667 light-years from the origin.

We should also figure out when and where this event happens according to the inbound twin's reference frame.

Here is my calculation of the coordinates of e1, e2, and e3 in the outbound and inbound frames:

e1': departure (x,t)=(0,0)
e2': turnaround (x,t)=(0,3)
e3': return (x,t)=(-13.333, 16.667)

We can calculate the necessary change in velocity by figuring Δx/Δt between event 2 and event 3.

And in the return-frame, we have
e1'': departure (x,t)=(-13.333,-10.667)
e2'': turnaround (x,t)=(0,3)
e3'': return (x,t)=(0,6)

My point is, those numbers DO make sense if you show the space-time diagram in the other frames. However, none of the textbooks on relativity actually SHOW the space-time diagrams in the other frames, so the reader is always left with these loose ends, wondering if it really makes sense, or just accepting the authority of the author, who claims it makes sense.

I know it does make sense, but that's because I've gone through the effort of actually looking at it from the different frames. But I've never seen ANY relativity texts actually go through the effort of transforming the coordinates of the events to other reference frames, and showing how those coordinates DO make sense.

When an author says "it doesn't make sense" does he mean
• "it doesn't make sense to me," or
• "it doesn't make sense to most people" or
• "it really makes no sense, i.e. it is really wrong"
• It makes no sense, i.e. it is meaningless;
• "it does not make sense to our primitive human brains, but it is mathematically correct."
• other?

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JDoolin
Gold Member

So which event coordinates "do make sense," and which ones "don't make sense?"

In the earth frame:
e1: departure (x,t)=(0,0)
e2: turnaround (x,t)=(4,5)
e3: return (x,t)=(0,10)

In the outbound frame
e1': departure (x,t)=(0,0)
e2': turnaround (x,t)=(0,3)
e3': return (x,t)=(-13.333, 16.667)

In the return frame
e1'': departure (x,t)=(-13.333,-10.667)
e2'': turnaround (x,t)=(0,3)
e3'': return (x,t)=(0,6)

It appears that everyone agrees that events e1, e2, e3 all make sense, e1' and e2' both make sense, and e2'' and e3'' both make sense.

However, the coordinates of the two remaining events seem to be in question:
e3': return (x,t)=(-13.333, 16.667)
e1'': departure (x,t)=(-13.333,-10.667)

What is the opinion of the General Relativity Experts? Do the coordinates of those events "make sense" or "not make sense?"

ghwellsjr
Gold Member

Hi ghwellsjr.

Yes, that was fun. I like your problem set-up. We have three major events
e1: departure (x,t)=(0,0)
e2: turnaround (x,t)=(4,5)
e3: return (x,t)=(0,10)

You also have defined another event (x,t)=(0,5), for which (x',t')= (-6.667,8.333). Then you said: "Those numbers don't make any sense at all, do they? But they do if we remember that these are coordinates in a different frame." I totally agree; but there's a lot more to say about that. Why doesn't it make sense. Why does it make sense? How does it make sense?
Why are there no textbooks that actually DO the Lorentz Transformation?
I'm trying to provide an answer. Part of that answer is that when you transform some significant events in one frame to another frame, they are no longer significant. The event in question is the one for the stay-at-home twin that is simultaneous with the turn-around event for the traveler. When we transform these two events into the frame in which the traveler is at rest during the outbound portion of the trip, they are no longer simultaneous and therefore don't represent anything significant to the traveler. But they do represent something significant to us once we remember, as I previously said, that they are coordinates in a different frame. I showed how we can use the time dilation formula to convert the coordinate time to the proper time and get back the age difference of the home twin. What we are really doing is dividing the coordinate time, 8.333, by gamma, 1.667, to get the proper time of 5 years and we could also divide the coordinate distance, -6.667 by 1.667 to get the proper distance, -4 light-years.
How long is the outbound twin's reference frame relevant to the outbound twin? For only three years. But this event happens at t=8.333 years! How far away is it going to happen? 6.667 light-years from the origin.
Yes, since this is still not a significant event to the traveling twin, what we really need to do is consider how far away the home-twin is after traveling for 3 years at -0.8c. That would be -2.4 light-years. That is the symultaneous event (-2.4,3) of the home-twin at the turn around event of the traveling twin in the rest frame of the traveling twin. (I hope no one is getting confused by the interchange of x and t in our two conventions.)

All of this is illustrating the relativity of simultaneity. But my point is that simply doing Lorentz transforms of the events in one frame to another frame is not enough. You have to do extra work to figure out the significant events. I'm not sure students learning Special Relativity are going to grasp all this in a beginning course, especially, as I said earlier, it's so easy to understand almost any scenario in one frame.
We should also figure out when and where this event happens according to the inbound twin's reference frame.

Here is my calculation of the coordinates of e1, e2, and e3 in the outbound and inbound frames:

e1': departure (x,t)=(0,0)
e2': turnaround (x,t)=(0,3)
e3': return (x,t)=(-13.333, 16.667)

We can calculate the necessary change in velocity by figuring Δx/Δt between event 2 and event 3.

And in the return-frame, we have
e1'': departure (x,t)=(-13.333,-10.667)
e2'': turnaround (x,t)=(0,3)
e3'': return (x,t)=(0,6)
I need some help understanding what you have done here. First off, what is the value of β? Where is the origin? What is the starting frame that you are converting into the double-prime frame?
My point is, those numbers DO make sense if you show the space-time diagram in the other frames. However, none of the textbooks on relativity actually SHOW the space-time diagrams in the other frames, so the reader is always left with these loose ends, wondering if it really makes sense, or just accepting the authority of the author, who claims it makes sense.
Can you draw the spacetime diagrams that will help it make sense? I don't see how it can help because as I pointed out significant events in one frame may not be significant in another frame.
I know it does make sense, but that's because I've gone through the effort of actually looking at it from the different frames. But I've never seen ANY relativity texts actually go through the effort of transforming the coordinates of the events to other reference frames, and showing how those coordinates DO make sense.

When an author says "it doesn't make sense" does he mean
• "it doesn't make sense to me," or
• "it doesn't make sense to most people" or
• "it really makes no sense, i.e. it is really wrong"
• It makes no sense, i.e. it is meaningless;
• "it does not make sense to our primitive human brains, but it is mathematically correct."
• other?
I stated what I meant in my original post: it doesn't make sense until "we remember that these are coordinates in a different frame".

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ghwellsjr
Gold Member

What is the opinion of the General Relativity Experts? Do the coordinates of those events "make sense" or "not make sense?"
General Relativity Experts? What has this got to do with General Relativity?

After you explain how you got the double-prime coordinates, if it's still relevant, I'll comment on the rest of your post.

JDoolin
Gold Member

Is this what you are looking for? Or did you want to see space-time diagrams for the traveling twin's rest system?
Sorry I missed that question this morning bobc2.

Yes, space-time diagrams for the traveling twin's rest systems. (plural)

JDoolin
Gold Member

I need some help understanding what you have done here. First off, what is the value of β? Where is the origin? What is the starting frame that you are converting into the double-prime frame?
So there are two β values, 0.8 for the outbound trip, and -0.97561 for the inbound trip. For the first boost, I used (0,0) for the origin. For the second boost, I converted directly from the outbound reference frame to the inbound reference frame. I used (x=0,t=3) for the origin, and used a "translate"-"boost"-"translate back" strategy.

I was trying to follow along closely with what you did, with a little more matrices.

Here is a line-by-line explanation.

Code:
e1 = {x1, t1} = {0, 0};
e2 = {x2, t2} = {4, 5};
e3 = {x3, t3} = {0, 10};
This defines the original events in the original reference frame.

Code:
eventList = Transpose[{e1, e2, e3}];
(*This puts all of the relevant events into a matrix form:
$$eventList=\begin{pmatrix} 0 & 4 & 0\\ 0 & 5 & 10 \end{pmatrix}$$*)

Code:
MatrixForm[eventList]  (*Shown as Out[184]*)
(*This simply sent the output. Mathematica hides the output if you use a semicolon, and shows the output if you leave off the semi-colon*)

Code:
\[Beta] = .8;
\[Gamma] = 1/Sqrt[1 - \[Beta]^2];
LT = \[Gamma]*{{1, -\[Beta]}, {-\[Beta], 1}};
This is the Lorentz Transformation Matrix $$LT = \left( \begin{array}{cc} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{array} \right)$$

Code:
outBoundEventList = Chop[LT.eventList];
This line performs the Lorentz Transformation. The dot (.) operator does matrix multiplication. The Chop function takes any value that is closer to zero than 10^-15 and zeros it out. It basically gets rid of a rounding error.

Code:
MatrixForm[outBoundEventList](*shown as Out[189]*)
Shows the result of the operation

Code:
reCenteredOutBoundList = Chop[
outBoundEventList - {{0, 0, 0}, {3, 3, 3}}];
This translates all of the relevant events back in time 3 years (0,-3). Note the minus sign, putting the turn-around event at the origin. The goal here is to go through a process of "translate"-"boost"-"translate back" so that the transformation is done around the correct origin.
Code:
MatrixForm[reCenteredOutBoundList] (*Shown as Out[172]*)
{{X1, X2, X3}, {T1, T2, T3}} = reCenteredOutBoundList;
I'm just preparing, here, to calculate the velocity for the return trip. We have six numbers in the reCenteredOutBoundList, and we want four of them. This is the quick way to get them.

Code:
\[Beta]2 = (X3 - X2)/(T3 - T2) (*Shown as Out[174]*)
This calculates the velocity change needed for the return trip. This is mathematically, exactly the same thing you did. It came out to be the same value that you got.

Code:
\[Gamma]2 = 1/Sqrt[1 - \[Beta]2^2];
LT2 = \[Gamma]2*{{1, -\[Beta]2}, {-\[Beta]2, 1}};
This finds the Lorentz Transformation matrix needed for the return trip.

Code:
reCenteredInBoundList = Chop[LT2.reCenteredOutBoundList];
inBoundList = reCenteredInBoundList + {{0, 0, 0}, {3, 3, 3}};
This adds three to the hour on all of the events. This is the third step in "Translate"-"Boost"-"Translate back"

Code:
MatrixForm[inBoundList] (*Shown as Out[179]*)
That gives the relevant event coordinates in the final frame.