Another Twins Paradox question

[Al68]

Why do most explanations of the Twins Paradox claim that the twin on the spaceship ages less because he is the one who undergoes acceleration and/or changes direction, causing asymmetry between the points of view of each twin? It seems clear to me that the twin on the spaceship would age less even if we ignore acceleration, or if we use a variation of the Twin Paradox where there is no acceleration of either twin during the experiment.

The Twins Paradox is asymmetrical in a very important way that has nothing to do with acceleration, and that is rarely even mentioned. The turnaround point is stipulated to be a certain distance from earth, as measured from earth, typically a distant star assumed to be at rest relative to earth. Each twin uses the distance between Earth and this distant star, and the elapsed time of the trip between them in their calculations, then they compare them to each other.

Importantly, these two objects (earth and the distant star) are both at rest relative to the twin on earth, and they are both in motion relative to the twin on the space ship. Without showing all the math, this fact alone accounts for the asymmetry between the respective twins.

I would also note that it is possible to come up with a version of the Twin Paradox in which the twin in the space ship, who undergoes acceleration and turns around to return to earth, ages more than the twin on earth, according to SR. This could be done by stipulating the distance traveled as the distance between two objects at rest relative to the ship, but in motion relative to earth. Perhaps using a rigid rod a few light years long being pulled by the ship.

I didn't include the text of the Twins Paradox here, and I left out the math for the sake of brevity, but I am really looking for comments from someone already familiar with both.

Thanks,
Alan

 

[JesseM]

The Twins Paradox is asymmetrical in a very important way that has nothing to do with acceleration, and that is rarely even mentioned. The turnaround point is stipulated to be a certain distance from earth, as measured from earth, typically a distant star assumed to be at rest relative to earth.

It is not necessary to assume the traveling twin turns around when he reaches a marker that's at rest relative to the earth, you could just as easily have the traveling twin moving away from the Earth at 0.9c, and a marker moving away from the Earth at 0.8c with a head start, and the traveling twin could turn around when he catches up to the marker. The difference in speed between the marker and the Earth is irrelevant to the problem, all that matters is that you can figure out the distance of the traveling twin from the Earth at the moment he reaches the marker and turns around, in any inertial frame (there's no need to specify the distance in terms of the Earth's rest frame, you could equally well specify it in terms of the the distance in the inertial frame where the traveling twin was at rest during the non-accelerating portion of the outbound leg of the trip...of course you can always translate the distance in one frame to the distance in another, and you'll always get the same answer about their ages upon reuniting no matter whose frame you specify the distance in).

I would also note that it is possible to come up with a version of the Twin Paradox in which the twin in the space ship, who undergoes acceleration and turns around to return to earth, ages more than the twin on earth, according to SR. This could be done by stipulating the distance traveled as the distance between two objects at rest relative to the ship, but in motion relative to earth. Perhaps using a rigid rod a few light years long being pulled by the ship.

No. If the traveling twin is the one who accelerates to turn around, the traveling twin will always have aged less when they reunite. If you disagree, perhaps we should go through the math for a simple numerical example?

 

[Ich]

No, the twin paradox is not merely a mathematical problem. No matter how you measure distances, the accelerating twin will be younger.
Real, measurable acceleration is not necessary. What matters is that one twin changes velocity (thas is, direction in spacetime) which makes him take a detour (through spacetime). Just as a car taking a detour will read more kilometers, the twinn will read less seconds.
This works also when the twin swings by a distant star or transfers his clock reading to some third person as he passes him.

 

[Al68]

Thanks for the replies. OK, here is an example which should suffice:

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth. From the twin on the ship's point of view, he will turn around after 11.5 years, total round trip time is 23 years. From the twin on Earth's point of view, he will see the buoy arrive after 5.75 years, total time passed for him when ship returns is 11.5 years.

I notice here, just like in the more common version, the twins will disagree about when the ship turned around, and how far apart they were at the time. And the turnaround point was specified as a certain distance as measured in the frame of the twin on the ship. Also notice that, the twin on Earth is actually present at the turnaround point in this example, instead of the other way around. It looks to me like this example has the opposite result, even though I haven't changed which twin accelerated. But I switched around the real source of the asymmetry.

I also notice that if you put this on a diagram, and ignore acceleration, it could look like the twin on Earth changed frames instead of the twin on the ship.

Also, if he were to ignore G-forces, a passenger on the ship could claim that the Earth accelerated to turn around instead of the ship. I only point this out because a lot of sources claim that these G-forces determine who ages less.

JesseM, you are right that it doesn't matter that the turnaround marker is at rest relative to earth, I misspoke. It is important that the distance to it is specified as measured from the rest frame of earth.

JesseM, since you offered, I would appreciate it if you could check my math, since I did this in a hurry. And of course point out any other mistakes.

Thanks,
Alan

 

[clj4]

Thanks for the replies. OK, here is an example which should suffice:

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth. From the twin on the ship's point of view, he will turn around after 11.5 years, total round trip time is 23 years. From the twin on Earth's point of view, he will see the buoy arrive after 5.75 years, total time passed for him when ship returns is 11.5 years.

I notice here, just like in the more common version, the twins will disagree about when the ship turned around, and how far apart they were at the time. And the turnaround point was specified as a certain distance as measured in the frame of the twin on the ship. Also notice that, the twin on Earth is actually present at the turnaround point in this example, instead of the other way around. It looks to me like this example has the opposite result, even though I haven't changed which twin accelerated. But I switched around the real source of the asymmetry.

I also notice that if you put this on a diagram, and ignore acceleration, it could look like the twin on Earth changed frames instead of the twin on the ship.

Also, if he were to ignore G-forces, a passenger on the ship could claim that the Earth accelerated to turn around instead of the ship. I only point this out because a lot of sources claim that these G-forces determine who ages less.

JesseM, you are right that it doesn't matter that the turnaround marker is at rest relative to earth, I misspoke. It is important that the distance to it is specified as measured from the rest frame of earth.

JesseM, since you offered, I would appreciate it if you could check my math, since I did this in a hurry. And of course point out any other mistakes.

Thanks,
Alan

Here is a very good explanation:

http://sheol.org/throopw/sr-ticks-n-bricks.html

Note that there is no reference to acceleration. NEVERTHELESS, it is the twin that changed course (turned his rocket around) thet JUMPED frames. The problem is not symmetric in that the Earth DID NOT jump frames. So, the twin in the rocket "skipped" some "bricks" (units of time) when he jumped frames. Hope that this answers your question about the lack of symmetry.

I think that the problem lies with your statement:

I also notice that if you put this on a diagram, and ignore acceleration, it could look like the twin on Earth changed frames instead of the twin on the ship.

You cannot do that. It is clear that the twin in the rocket switched courses, not the Earth. The two are not interchangeable because of...acceleration! The twin in the rocket will feel the acceleration whether he turns around or he goes in a big loop. The two frames (Earth and rocket) are not equivalent because the rocket is not an inertial frame. So you can't declare them interchangeable as you did above.

 

[Al68]

clj4, I didn't say the Earth actually turned around, or that it was correct to see it that way, just that someone could see it that way. My point is that in my example, the twin on the spaceship that acually accelerates, ages more than the twin on Earth according to SR, because I changed it around to make it asymmetric the other way around without changing which twin accelerated.

Another way to put things is that, in both my example above, and the common Twins Paradox, it is the twin who views the distance traveled as "length contracted" that ends up aging less. This is the only asymmetry that is apparent in the math alone, and that is predicted by SR. At least Einstein's 1905 paper never said that time contraction was related to acceleration in any way. It is only some later sources that try to use acceleration to explain the asymmetry. But not all, as you pointed out, clj4.

Thanks,
Alan

 

[clj4]

clj4, I didn't say the Earth actually turned around, or that it was correct to see it that way, just that someone could see it that way. My point is that in my example, the twin on the spaceship that acually accelerates, ages more than the twin on Earth according to SR, because I changed it around to make it asymmetric the other way around without changing which twin accelerated.

Thanks, Alan

Look at the diagrams on the link. They cannot be switched around. The point is that "no one could see it that way"

 

[Al68]

Look at the diagrams on the link. They cannot be switched around. The point is that "no one could see it that way"

OK, clj4, I would have to agree that it would be incorrect to see it that way.

BTW, did you read my example of where the twin who accelerates ages more?

It is my example I was referring to, that if one put it on a diagram, it could mislead some people into thinking that the Earth changed directions. I did not mean to say that they would be correct.

Thanks,
Alan

 

[clj4]

OK, clj4, I would have to agree that it would be incorrect to see it that way.

BTW, did you read my example of where the twin who accelerates ages more?

Yes, I did. It is incorrect.

It is my example I was referring to, that if one put it on a diagram, it could mislead some people into thinking that the Earth changed directions. I did not mean to say that they would be correct.

Thanks,
Alan

You are welcome, I think we are done.

 

[Al68]

OK, can anyone clear up what is wrong with my example. I am open to the possibility that I made a huge error. I would just like to know where.

Also, I cannot claim to have come up with this example myself. Although the details are different, I remember a similar example from an article explaining how acceleration was irrelevant to the Twins Paradox. And there are many reputable sources that state that the fact that the twin who accelerates in the Twins Paradox is also the one who ages less is entirely coincidental. This seems obvious to me.

Here is an article at Scientific American, which has a great explananation of the Twins Paradox, and points out partly what I mean:

http://sciam.com/print_version.cfm?articleID=000BA7D8-2FB2-1E6D-A98A809EC5880105

Thanks,
Alan

 

[pervect]

Thanks for the replies. OK, here is an example which should suffice:

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest.

Here is the problem with your example. Perfectly rigid bodies don't exist in SR.

 

[Al68]

Here is the problem with your example. Perfectly rigid bodies don't exist in SR.

pervect,

It's obviously an assumption that the rod would be rigid. The same assumption Einstein made repeatedly in his 1905 SR paper when he speaks of "rigid rods". Einsteins use of this term (rigid rod) in describing length contraction is the reason I chose to use it.

Thanks,
Alan

 

[robphy]

I also notice that if you put this on a diagram, and ignore acceleration, it could look like the twin on Earth changed frames instead of the twin on the ship.

Also, if he were to ignore G-forces, a passenger on the ship could claim that the Earth accelerated to turn around instead of the ship. I only point this out because a lot of sources claim that these G-forces determine who ages less.

Technically, you can't "ignore acceleration" altogether since your traveller has changed from one inertial frame to another. You, however, can execute this acceleration instantaneously so that no proper time ticks off (i.e. so that you don't have to "integrate over a non-inertial segment of arc-length").

If one were to construct all of the lines of simultaneity of the traveller, one would find that the lines cross in some places (which amounts to assigning two sets of coordinates for the same event). This amounts to a faulty coordinate system for Minkowski spacetime.
On the other hand, the inertial observer on Earth has no such problem.

Another way to characterize what is happening in the last paragraph is that there is no Lorentz transformation that will straighten out the kink in the noninertial traveller's worldline. So, that traveller over the course of his non-inertial trip is not equivalent to an inertial one.

For diagrams of these situations, look at
the video on
http://physics.syr.edu/courses/modules/LIGHTCONE/twins.html
the applet on
http://physics.syr.edu/courses/modules/LIGHTCONE/java/TwinParadox.html
and the diagrams and videos on
http://physics.syr.edu/courses/modules/LIGHTCONE/LightClock/ (See the last diagram which shows the kink and the "break" in continuity of the lines of simultaneity. The second to the last video explicitly constructs the clock ticks.)

 

[Al68]

Originally Posted by Al68:
[I also notice that if you put this on a diagram, and ignore acceleration, it could look like the twin on Earth changed frames instead of the twin on the ship.

Also, if he were to ignore G-forces, a passenger on the ship could claim that the Earth accelerated to turn around instead of the ship. I only point this out because a lot of sources claim that these G-forces determine who ages less.]

Technically, you can't "ignore acceleration" altogether since your traveller has changed from one inertial frame to another. You, however, can execute this acceleration instantaneously so that no proper time ticks off (i.e. so that you don't have to "integrate over a non-inertial segment of arc-length").

If one were to construct all of the lines of simultaneity of the traveller, one would find that the lines cross in some places (which amounts to assigning two sets of coordinates for the same event). This amounts to a faulty coordinate system for Minkowski spacetime.
On the other hand, the inertial observer on Earth has no such problem.

Another way to characterize what is happening in the last paragraph is that there is no Lorentz transformation that will straighten out the kink in the noninertial traveller's worldline. So, that traveller over the course of his non-inertial trip is not equivalent to an inertial one.

Obviously I should not have included these two statements in my post. I only meant to say that these ideas could look right to someone. It is obviously not true that the twin on Earth turned around. And since these two statements only detract from the points presented, as evidenced by the responses, I apologize for including them.

I would only ask that someone review my alternate version of the Twins Paradox, shown again below, and point out any mistake, or correct it and provide reasons for the corrections.

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth. From the twin on the ship's point of view, he will turn around after 11.5 years, total round trip time is 23 years. From the twin on Earth's point of view, he will see the buoy arrive after 5.75 years, total time passed for him when ship returns is 11.5 years.

I notice here, just like in the more common version, the twins will disagree about when the ship turned around, and how far apart they were at the time. And the turnaround point was specified as a certain distance as measured in the frame of the twin on the ship. Also notice that, the twin on Earth is actually present at the turnaround point in this example, instead of the other way around. It looks to me like this example has the opposite result, even though I haven't changed which twin accelerated. But I switched around the real source of the asymmetry.

Thanks,
Alan

 

[pervect]

pervect,

It's obviously an assumption that the rod would be rigid. The same assumption Einstein made repeatedly in his 1905 SR paper when he speaks of "rigid rods". Einsteins use of this term (rigid rod) in describing length contraction is the reason I chose to use it.

Thanks,
Alan

Interesting (and IMO unfortunate) that Einstein used this notion in his 1905 paper, because a rigorous defintion of a rigid body (actually, rigid motion - "Born-rigidity") wasn't developed until 1909 (by Born).

See for instance

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

Special Relativity

Born rigidity: in 1909, Born proposed a Lorentz-invariant definition of "rigid body". Pauli's monograph on relativity [1] gives a nice summary of Born's notion, and the responses it drew from Ehrenfest, Herglotz, Noether, and von Laue. (Pais's Einstein bio suggests that Born's 1909 paper may have helped set Einstein on the road to Riemannian geometry [2].)

We know already that rigidity and SR don't mix--just think of the Barn and the Pole! How could a physicist like Born, mathematically sophisticated, have made such an elementary error? A simple remark by Pauli clarifies things considerably:

If thus the concept of a RIGID BODY has no place in relativistic mechanics, it is nevertheless useful and natural to introduce the concept of a RIGID MOTION of a body. We shall denote those motions as rigid for which Born's condition (*) is satisfied.

Born thought he was defining a rigid body, but Pauli's rephrasing saves the mathematics while improving the physics. We have no rigid rods in SR, but if you accelerate every atom of an ordinary rod in just the right way, you can move the rod rigidly. And Born's definition is Lorentz-invariant.

Note the remark in the FAQ that "The rigid body has no place in relativistic mechanics". Unfortunately, this concept has confused a lot of people, including yourself. It would be far, far better to avoid "rigid bodies" in relativity, even if you can (unfortunately) point to Einstein as having originated the notion at an early time.

There is a reasonably extensive discussion of Born-rigidity at

http://www.mathpages.com/home/kmath422/kmath422.htm

if you paramaterize the motion in coordinate terms (t,x) the hyperbolic functions get eliminated, and the coordinate equations of motion are

x^2 - t^2 = constant

Let's go on to your example:

you write

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth.

you can see from the attached diagram - or similar diagrams at
http://www.mathpages.com/home/kmath422/kmath422.htm

that the far end of the ship does NOT start deaccelerating "at the same time" as the near end of the ship in the Earth frame.

The far end of the ship and the near end of the ship start deacceelrating "at the same time" in the SHIP frame, but NOT the Earth frame.

The attached diagram shows a ship slowing down from .9c to rest at an accaeleration of 1 ly/y^2 (roughly 1g).

The diagram is drawn in the Earth frame of reference.

The three curves on the diagram represent three points on the Born-rigid body that represents the ship. Your example was 10 LY, the diagram I drew was for 3 points, each 1 light year from each other.

The straight blue line on the diagram denotes the instant at which each particle of the body starts deacelerating.

The important point to notice is that while the blue line (the start of deacceleration) appears to start at "the same time" in the frame of the moving body, it definitely does not appear to start at the same time in the Earth frame.

 

[Al68]

Pervect,

My example assumes the rigid rod to exist (hypothetically) the same way Einstein did. I am not arguing that it does or could exist, and it doesn't matter, just like Einstein's SR doesn't require rigid rods to actually exist, even though he assumes they do for the purpose of illustrating SR.

If you prefer, there are other ways to present my example without using a rigid rod. It is not important to the story. What is important is that the buoy is at rest relative to the ship, with or without a rigid rod. Maybe the buoy could have its own power and just turn around when it reaches earth. The details aren't important, there are plenty of ways to keep the buoy and the ship at rest relative to each other.Thanks,
Alan

 

[pervect]

Eliminating the rigid rod would certainly simplify the discussion.

Basically, if you have a ship A that goes out, turns around, and comes back, the fact that a second ship B is following the first at some "constant distance" is going to be totally irrelevant to the proper time elapsed on A's clock. I'm not sure why this isn't totally obvious to you.

B can follow A, or B can not follow A - it won't affect A's clock one bit.

The important point is that in flat space-time (where SR is applicable) if A starts out at the same point as some inertial observer C, follows a path which is not inertial, then re-unites with C at some later date, A's clock will have less time on it than C's clock.

 

[Al68]

It seems like a lot of people are telling me I'm wrong, but that wasn't my question. My question is: What part of my example is wrong and why?

A lot of answers have said that I'm wrong when I say acceleration is irrelevent. But they say the reason I'm wrong is that acceleration is relevent. That's like saying the sky is blue because it's not a different color than blue. While it may be true, it doesn't help much.

I only use the buoy so that the observer on Earth sees the distance traveled as "length contracted" while the observer on the ship does not. The distance traveled is the distance between the ship and the buoy. This is the opposite of the common version. Instead of the distance being stipulated in the frame of earth, and length contracted from the ship's point of view, it is the other way around.

This is the difference I wanted comments on, since this is what I think causes the asymmetry in both my example and the Twins Paradox.

Thanks,
Alan

 

[pervect]

It seems like a lot of people are telling me I'm wrong, but that wasn't my question. My question is: What part of my example is wrong and why?

Your notion of a rigid body is wrong. I went to a fair amount of effort to explain, provide references for, and create a diagram for how a Born-rigid body actually appears and why your intiutive notion of a rigid body (which is one of the main sources of your problem) was incorrect.

I realize that this was a rather complicated, so when you wanted to "skip it", I said OK, because hyperbolic trig functions aren't everyone's bag.

Now it looks like your backsliding again, and don't want to provide the simpler example that you promised.

Well, if you want to know where you went wrong, then re-read my previous rather long post where I explained where you went wrong. The short version of where you made a mistake is that you ignored the relativity of simultaneity - simultaneity in the spaceship frame is NOT the same as simultaneity in the Earth frame.

In addition, consistency arguments show that you must be wrong. These arguments are popular and being made because they are easier to follow than the detailed diagrams. Unfortunately you don't seem to be listening.

Trust us, you ARE wrong.

I think it's better in this case to be pro-active, rather than re-active.

Rather than explain "why you are wrong", why not look at a post that tells you the right way to work the problem? Then, if you are serious, you can hopefully look over your own work and see where it went wrong. Especially since you've had so many clues.

---->[CLUE: LOOK AT RELATIVITY OF SIMULTANEITY!]<-----

Being pro-active is easier to write, easier to debug, and probably better all-around than getting tied up in the details of explaning exactly where you made an error, esp. as you seem to be rather attached to your own viewpoint and wanting to defend it (as opposed to sitting down and studying how other people have worked the problem to get the right and well-known answer).

The #1 simplest way of solving the problem is this. (It's known as the k-calculus approach, though it doesn't use calculus at all).

Suppose the Earth sends signals at a rate of 1 message a month

There will be a relativistic doppler shift factor k given by the well-known formula

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

freq-recieved/freq-emitted = \sqrt{\frac{1+\beta}{1-\beta}}

Let's let \beta = 3/5, i.e. our velocity be 3/5 of the speed of light, or v=.6c.

Then on the outgoing trip, the signal will be redshifted, and by the formula, \beta = -3/5, giving a doppler shift factor k of 1/2

Thus if I take a 12 month outbound journey, I will receive 6 signals, because I will receive signals at half the frequency with which they were sent.

(You may be used to seeing doppler shift used only for high frequency signals, but it works just as well for low frequency ones. If signals are redshifted 2:1, signals emitted at 1 second intervals are received at 2 second intervals, signals emitted at 1 month intervals are received every 2 months, etc. It doesn't matter what the base frequency is, only the ratio).

Now, I turn around. We recompute k from the new value of \beta=3/5, and get k=2. Our blueshift factor on the inbound trip will be 2.

It will be generally true that the doppler shift factor on the inbound trip will just be the reciprocal of the doppler shift factor k on the outbound trip. (You can see this by inspecting the formula).

If it took me 12 months to go out at .6 c, it will take me 12 months to get back.

During those 12 months, I will receive 24 signals, due to the blueshift factor of 2.

Now, look at the tally. I've been on a 24 month journey, but the Earth has sent 6 + 24 = 30 signals, each of which was spaced 1 month apart by the Earth clock. My clock is reading less elapsed time than the Earth clock - the Earth clock is reading 30 months, and my clock is reading only 24 months.

We can write the formula as

earth signals received = (k*(T/2) + 1/k*(T/2) ) / T = (k + 1/k) /2

where T/2 is half my trip time, so T is my total trip time

Here k and 1/k are the doppler shift values from the formulas I posted.

If you do the math to substitute the known values of k in terms of \beta, you get

time dilation = \frac{1}{\sqrt{1-\beta^2}}

Note that it *does not matter* to this argument if I'm towing a buoy, being followed by a stray cat, or whatever. I am not turning around at any specific point in space, I'm turning around after a specific amount of time (by my watch) has gone by, and I know that it will take me just as long to get back as it took for me to go out.

The Earth is emitting signals, I'm receiving them, and during the round-trip I receive every signal that the Earth emits at some point in my journey. By using the simple fact that the doppler shift is constant (this is known as Bondi k-calculus), I can easily deduce that I can (and must) receive more signals from the Earth than ticks from the clock that I carry, therefore the Earth clock must have aged more (to be able to send out more signals).

[add]
Here's a space-time diagram

<br /> \]<br /> \unitlength .5mm<br /> \begin{picture}(90,300)(0,0) {<br /> \linethickness{0.3mm}<br /> \multiput(0,0)(0.12,0.2){750}{\line(0,1){0.2}}<br /> \linethickness{0.3mm}<br /> \multiput(0,10)(0.12,0.12){125}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,20)(0.12,0.12){250}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,30)(0.12,0.12){375}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,40)(0.12,0.12){500}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,50)(0.12,0.12){625}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,60)(0.12,0.12){750}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,300)(0.12,-0.2){750}{\line(0,-1){0.2}}<br /> \linethickness{0.3mm}<br /> \multiput(0,70)(0.12,0.12){750}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,80)(0.12,0.12){708}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,90)(0.12,0.12){667}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,100)(0.12,0.12){625}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,110)(0.12,0.12){625}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,120)(0.12,0.12){583}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,130)(0.12,0.12){542}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,140)(0.12,0.12){500}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,150)(0.12,0.12){500}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,160)(0.12,0.12){458}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,170)(0.12,0.12){417}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,180)(0.12,0.12){375}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,190)(0.12,0.12){375}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,200)(0.12,0.12){333}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,210)(0.12,0.12){292}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,220)(0.12,0.12){250}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,230)(0.12,0.12){250}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,240)(0.12,0.12){208}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,250)(0.12,0.12){167}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,260)(0.12,0.12){125}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,270)(0.12,0.12){125}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,280)(0.12,0.12){83}{\line(1,0){0.12}}<br /> \linethickness{0.3mm}<br /> \multiput(0,290)(0.12,0.12){42}{\line(1,0){0.12}}<br /> }<br /> \end{picture}<br /> \[<br />

 

[robphy]

Assuming that I have understood the situation you posed...

\]<br /> \def\qbz#1#2#3#4#5#6{\qbezier(#1,#2)(#3,#4)(#5,#6)}<br /> \unitlength 2mm<br /> \begin{picture}(87.5,85)(0,0)<br /> \qbz{ 46.88}{ 14.38}{ 59.38}{ 36.88}{ 71.88}{ 59.38}<br /> \qbz{ 60.00}{ 15.00}{ 72.50}{ 37.50}{ 85.00}{ 60.00}<br /> \qbz{ 37.50}{ 20.00}{ 62.50}{ 32.50}{ 87.50}{ 45.00}<br /> \qbz{ 60.00}{ 60.00}{ 72.50}{ 37.50}{ 85.00}{ 15.00}<br /> <br /> {\linethickness{1pt}<br /> \qbz{ 60.00}{ 6.00}{ 60.00}{ 45.50}{ 60.00}{ 85.00} <br /> \qbz{ 72.50}{ 37.50}{ 66.31}{ 37.50}{ 60.12}{ 37.50}}<br /> <br /> \put(62,22){\makebox(0,0)[cc]{$\theta$}}<br /> \put(67,36){\makebox(0,0)[cc]{$\theta$}}<br /> \put(66,40){\makebox(0,0)[cc]{{\color{red}5}}}<br /> \put(64.38,31.88){\makebox(0,0)[cc]{}}<br /> \put(57,13){\makebox(0,0)[cc]{A}}<br /> \put(76,38){\makebox(0,0)[cc]{B}}<br /> <br /> \put(44,13){\makebox(0,0)[cc]{C}}<br /> \put(57,38){\makebox(0,0)[cc]{D}}<br /> <br /> \put(62,30){\makebox(0,0)[cc]{E}}<br /> \put(56,27){\makebox(0,0)[cc]{L}}<br /> <br /> \put(57,60){\makebox(0,0)[cc]{R}}<br /> <br /> \end{picture}<br /> \]



Working in the inertial frame of the earth...

AR is the Earth worldline.
AB is the ship worldline.
CD is the buoy worldline. For now, ignore any "rigid rod" pulling on the buoy. Instead just take CD parallel to AB. That is, the buoy is traveling inertially along CD with the same velocity as the ship along AB.

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth.

Okay...
AB and CD have velocity v=\tanh\theta=\frac{\sqrt{3}}{2}=0.866.
Note that \gamma=\cosh\theta=\cosh(\mbox{arctanh}\ v)=2 and \beta\gamma=\sinh\theta=\tanh\theta\cosh\theta=(\frac{\sqrt{3}}{2})(2)=\sqrt{3}.
BL is the "proper distance between the buoy and the ship" = 10 ly.
BD is the "distance [measured by the earth] between the Earth and the turn around event B".

Note the Minkowski right triangle with legs LD and LB and hypotenuse DB... so BD\cosh\theta=BL.. thus BD=BL/\cosh\theta=(10)/(2)=5. Agreed.

From the twin on the ship's point of view, he will turn around after 11.5 years, total round trip time is 23 years. From the twin on Earth's point of view, he will see the buoy arrive after 5.75 years, total time passed for him when ship returns is 11.5 years.

To determine the ship's proper time along AB...
consider the Minkowski right triangle ADB with legs AD and DB and hypotenuse AB. (In passing... in this frame, ADB happens to also be a Euclidean right triangle [with, of course, a different measure for \theta and the hypotenuse AB]).

Since DB=AB\sinh\theta, we have AB=DB/\sinh\theta=(5)/(\sqrt{3})=2.886751345 y... but not the 11.5y you claim. By symmetry [i.e. a similar analysis], we double the ship's proper time to obtain the total round-trip time.

To determine the earth-time between events A and D...
use the same triangle
Since AD=AB\cosh\theta, we have AD=(5/\sqrt{3}) (2)=10/\sqrt{3}=5.773502690 y. So, as above, the round-trip time of the ship is 2(10/\sqrt{3})=11.54700538 y according to the earth. This agrees with your "5.75 y" and your "11.5y".

postscript...
Since the buoy is traveling inertially, it doesn't make the turn when the ship turns at B. For the buoy to somehow turn with the ship, one has to correctly model the "interaction" between the buoy and the ship. As has been mentioned above, the notion of a "rigid body" in relativity is not trivially carried over from Newtonian physics. One aspect of this is the fact that "the buoy turning at D" is not causally related to "the ship turning at B". In addition, according to the ship, events B and D are not simultaneous .

 

[Al68]

Robphy,

Thanks for your response, it was exactly what I was looking for. I notice something very interesting, though.

According to your results, from Earth's point of view, the distance traveled (one way) is 5 ly, and the elapsed time is 5.77 years, so the relative velocity is (5)/(5.77)= 0.866c. Sounds good so far.

From the ship's point of view, the distance traveled (one way) is 10 ly, and his elapsed time is 2.89 years, so his velocity is (10)/(2.89)=3.46c. Not so good.

Am I doing something wrong here, or is it neccessarily true that the twin with the most mileage will have the most elapsed time, since v=d/t and t=d/v in each frame?

Thanks,
Alan

 

[robphy]

From the ship's point of view, the distance traveled (one way) is 10 ly, and his elapsed time is 2.89 years, so his velocity is (10)/(2.89)=3.46c. Not so good.

Am I doing something wrong here, or is it neccessarily true that the twin with the most mileage will have the most elapsed time, since v=d/t and t=d/v in each frame?

Thanks,
Alan

From the ship's point of view, the distance traveled BY THE SHIP from A to B is zero.

From the ship's point of view, the Earth is moving. How fast is the Earth moving? Well, at the event B. The Earth is a distance BE away according to the ship. Note that ABE is a Minkowski right-triangle with legs BE and BA and hypotenuse AE... and that the velocity of the Earth is BE/BA. (By similar triangles with ADB we can see that v=\tanh\theta... but let's not do that.) We know BA=5/\sqrt{3}. To find BE, use the M-right-triangle BDE with hypotenuse BE. Since BD=BE\cosh\theta, we have BE=BD/\cosh\theta=(5)/(2). So, finally, BE/BA=( 5/2 )/ (5 /\sqrt{3} )= \sqrt{3}/2=0.866. Thus, as expected, their relative speeds are equal.

 

[Al68]

Robphy,

So, according to the ship's point of view, v=0.866c, and t= 2.89 years (one way), the distance to Earth at the turnaround point would be d = vt = (0.866)(2.89) = 2.5 light years. So from his point of view, the buoy is 4 times as far from him as the earth. Why would he choose to turn around at this point? This seems to contradict what was given in my example, that the ship would turn around after traveling 10 light years as measured in his own frame. ie, when he judged the buoy to reach earth. This would not be the same as when the twin on Earth judged the buoy to reach earth, since the event is not simultaneous in both frames.

I realize that I should have worded my example better.

Thanks,
Alan

P.S. Your math (for calculating elapsed time for each twin) seems to be consistant with the ship turning around at a turnaround point which is 5 light years from earth, as measured in Earth's frame, and 2.5 light years away, as measured in the ship's frame. ie, same as the Twins Paradox. So your math is entirely self consistant, but it assumes that the ship will choose to turn around after traveling 2.89 years. In my example, he simple would not choose to turn around at this point. He would turn around after traveling at 0.866c a distance of 10 light years as measured in his own frame. I know I worded my example poorly, but my goal was to have the distance traveled to be measured in the ship's frame, and length contracted only in Earth's frame. Instead of the other way around like in the Twins Paradox.

 

[JesseM]

Thanks for the replies. OK, here is an example which should suffice:

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth. From the twin on the ship's point of view, he will turn around after 11.5 years, total round trip time is 23 years. From the twin on Earth's point of view, he will see the buoy arrive after 5.75 years, total time passed for him when ship returns is 11.5 years.

As others pointed out, the rod cannot stay rigid if it accelerates. But perhaps we could consider that instead of towing the rod, the ship is simply moving alongside a rod which is moving inertially and is at rest relative to the ship during the outbound leg of the trip, and continues to move inertially in the same direction even after the ship turns around. So in the rod's frame, the Earth is moving away at 0.866025404c throughout the entire process, while the ship is at rest until the Earth is 10 light years away from it (which will take a time of 10/0.866025404 = 11.54700538 years), then the ship accelerates to (2*0.866025404)/(1.75) = 0.989743319c in the same direction as the Earth (using the formula for addition of velocities in relativity, under the assumption that the ship is moving towards the Earth at 0.866025404c in the Earth's rest frame). The Earth is still moving away at 0.866025404c in the rod's frame, so according to the rod the time for the ship to catch up with the Earth can be found by solving 0.989743319t = 0.866025404t + 10 for t, giving t = 80.82903757 years.

So the outbound leg takes 11.5470 years in the rod's frame, the inbound leg takes 80.8290 years in the rod's frame. During the outbound leg, the rod observes the Earth's clock to be ticking at half the normal rate while the ship is at rest so it ticks at the normal rate, so 5.7735 years pass on the Earth's clock while 11.5470 years pass on the ship's clock between the moment they depart and the moment the ship turns around, as seen in the rod's frame. Then during the inbound leg, the Earth's clock continues to tick at half the normal rate, elapsing an additional 40.4145 years, while the ship's clock ticks at \sqrt{1 - 0.989743319^2} = 0.14285714 the normal rate, elapsing 80.8290*0.14285714 = 11.5470 years. So in the rod's frame, you predict that when the ship and the Earth reunite, the Earth's clock has elapsed 5.7735 + 40.4145 = 46.188 years, while the ship's clock has elapsed 11.5470 + 11.5470 = 23.094 years. So in this frame, you still predict that the ship's clock has elapsed exactly half the amount of time as the Earth's clock between the time they depart and the time they reunite, and the specific amounts of time elapsed on each clock are the same as you'd predict in the Earth's frame too (I can show that if you like--you have to take into account both the fact that the rod is only 5 light-years long in the Earth's frame due to Lorentz contraction, and the fact that the event of the ship turning around is not simultaneous with the event of the buoy on the back end of the rod passing the Earth in this frame, as it was in the rod's own rest frame).

 

[robphy]

Robphy,

So, according to the ship's point of view, v=0.866c, and t= 2.89 years (one way), the distance to Earth at the turnaround point would be d = vt = (0.866)(2.89) = 2.5 light years. So from his point of view, the buoy is 4 times as far from him as the earth. Why would he choose to turn around at this point? This seems to contradict what was given in my example, that the ship would turn around after traveling 10 light years as measured in his own frame. ie, when he judged the buoy to reach earth. This would not be the same as when the twin on Earth judged the buoy to reach earth, since the event is not simultaneous in both frames.

I realize that I should have worded my example better.

Thanks,
Alan

P.S. Your math (for calculating elapsed time for each twin) seems to be consistant with the ship turning around at a turnaround point which is 5 light years from earth, as measured in Earth's frame, and 2.5 light years away, as measured in the ship's frame. ie, same as the Twins Paradox. So your math is entirely self consistant, but it assumes that the ship will choose to turn around after traveling 2.89 years. In my example, he simple would not choose to turn around at this point. He would turn around after traveling at 0.866c a distance of 10 light years as measured in his own frame. I know I worded my example poorly, but my goal was to have the distance traveled to be measured in the ship's frame, and length contracted only in Earth's frame. Instead of the other way around like in the Twins Paradox.

Your problem was stated unambiguously (boldfaced for emphasis)

We will specify the distance traveled in the frame of the twin on the space ship. The ship is towing a buoy behind on a rigid steel rod which is 10 light years long measured at rest. The ship leaves Earth at 0.866c and will turn around when the twin on Earth sees the buoy reach earth. He will, of course, have to be familiar with SR to figure out when to turn around. The rod will be 5 light years long as seen by the twin on earth.

although that is apparently not what you wanted.

So, I now offer a revised diagram (not drawn to scale):


\[<br /> \def\qbz#1#2#3#4#5#6{\qbezier(#1,#2)(#3,#4)(#5,#6) }<br /> \unitlength 2mm<br /> \begin{picture}(87.5,85)(0,0)<br /> \qbz{ 46.98}{ 14.38}{ 59.48}{ 36.88}{ 71.98}{ 59.38}<br /> \qbz{ 60.00}{ 15.00}{ 72.50}{ 37.50}{ 85.00}{ 60.00}<br /> \qbz{ 37.50}{ 26.50}{ 62.50}{ 39.00}{ 87.50}{ 51.50}<br /> \qbz{ 60.00}{ 77.25}{ 72.50}{ 54.75}{ 85.00}{ 32.25}<br /> <br /> {\linethickness{1pt}<br /> \qbz{ 60.00}{ 6.00}{ 60.00}{ 45.50}{ 60.00}{ 85.00} <br /> \qbz{ 72.50}{ 37.50}{ 66.31}{ 37.50}{ 60.12}{ 37.50}<br /> \qbz{ 80.50}{ 46.20}{ 71.31}{ 46.20}{ 60.12}{ 46.20}}<br /> <br /> <br /> \put(61,20){\makebox(0,0)[cc]{$\theta$}}<br /> \put(67,36){\makebox(0,0)[cc]{{\color{red}5}}}<br /> \put(67,41){\makebox(0,0)[cc]{{\color{red}10}}}<br /> \put(64,38.5){\makebox(0,0)[cc]{$\theta$}}<br /> \put(72,45){\makebox(0,0)[cc]{$\theta$}}<br /> \put(61,41){\makebox(0,0)[cc]{$\theta$}}<br /> <br /> \put(64.38,31.88){\makebox(0,0)[cc]{}}<br /> \put(57,13){\makebox(0,0)[cc]{A}}<br /> \put(77.5,48){\makebox(0,0)[cc]{B}}<br /> <br /> \put(44,13){\makebox(0,0)[cc]{C}}<br /> <br /> \put(75,39){\makebox(0,0)[cc]{M}}<br /> \put(59,39){\makebox(0,0)[cc]{L}}<br /> <br /> \put(59,48){\makebox(0,0)[cc]{P}}<br /> \put(57,80){\makebox(0,0)[cc]{R}}<br /> <br /> \end{picture}<br /> \]

As before,
v=\tanh\theta=\frac{\sqrt{3}}{2}=0.866
\gamma=\cosh\theta=\cosh(\mbox{arctanh}\ v)=2
\beta\gamma=\sinh\theta=\tanh\theta\cosh\theta=(\frac{\sqrt{3}}{2})(2)=\sqrt{3}

BL is the "proper distance between the buoy and the ship" = 10 ly.
Since ML is the hypotenuse of M-right-triangle MBL with legs BL and BM, we have ML=BL/\cosh\theta=(10)/(2)=5.

AB is the proper time elapsed from A-at-earth to the turn around-event at B. AB is the leg of M-right-triangle ABL with legs BL and AB and hypotenuse AL. Since BL/AB=\tanh\theta, we have AB=BL/\tanh\theta=(10)/(\sqrt{3}/2)=11.54700538{\rm\ y}. By symmetry, the total round trip according to the ship is twice this... 23.09401076 y.

Since BL=AL\sinh\theta, we have AL=BL/\sinh\theta=(10)/(\sqrt{3})=5.773502690{\rm\ y}. Note, however, that AL is not congruent to LR. So, the total round trip according to the Earth is NOT twice this... i.e. not 11.54700538 y.. The quantity to double is AP or RP. With M-right-triangle APB with legs AP and PB and hypotenuse AB, we have AP=AB\cosh\theta=[(10)/(\sqrt{3}/2)] [2]=(40)/\sqrt{3}=23.09401076{\rm\ y}, yielding a total round trip according to the Earth to be 46.18802152 y.

As a check, since AP=AL+LP, we have LP=AP-AL=(40)/\sqrt{3}-(10)/\sqrt{3}=(30)/\sqrt{3}=17.32050807{\rm\ y}. You can also get LP using the M-right-triangle LPB.

 

[Al68]

Robphy,

Thanks for your response. And I apologize for the error in my wording in my example. I don't have the world's best writing skills.

So, in your last post, you have:

According to the ship, it traveled a total of 23 years at 0.866c, total distance of 20 ly.
According to earth, the ship traveled a total of 46 years at 0.866c, total distance of 40 ly?

I know I ask a lot. So maybe I should give up on this for now.

The main thing I wanted to do was come up with a variation of the Twins Paradox in which the distance traveled is specified in the frame of the ship, and length contracted only from the point of view of the earth. Instead of the other way around like in the original. I obviously didn't do a very good job.

Can you think of any way to do this? And I appreciate all your efforts to show all the math, but if we know the relative velocity and the distance traveled as measured by each twin, the simple equation v=d/t will tell us the elapsed time for each twin.

Thank You,
Alan

 

[JesseM]

The main thing I wanted to do was come up with a variation of the Twins Paradox in which the distance traveled is specified in the frame of the ship, and length contracted only from the point of view of the earth. Instead of the other way around like in the original. I obviously didn't do a very good job.

If the ship accelerates, it can't have a single "frame" in which the usual laws of SR still work, because these laws only work in inertial frames. You can use SR to analyze the twin paradox from the point of view of an inertial frame in which the ship is at rest during the outbound leg, or at rest during the inbound leg, but not both.

 

[pervect]

but if we know the relative velocity and the distance traveled as measured by each twin, the simple equation v=d/t will tell us the elapsed time for each twin.

There is a reason Rob is drawing all those diagrams. The above formula will not always work. The reason it won't always work is that clocks may not be synchronized. In fact, clocks that are synchronized in one frame of reference will not be synchronized in another, see for instance Einstein's example online at:

http://www.bartleby.com/173/9.html

In short, "simultaneity" is relative. Note that in Rob's diagram in post #25, for instance, the line of simultaneity for the ship, BL, forms an angle theta with respect to the line of simultaneity for the Earth, BP.

 

[Al68]

If the ship accelerates, it can't have a single "frame" in which the usual laws of SR still work, because these laws only work in inertial frames. You can use SR to analyze the twin paradox from the point of view of an inertial frame in which the ship is at rest during the outbound leg, or at rest during the inbound leg, but not both.

Jesse,

OK, I agree, but a ruler at rest with the ship will stay the same length to the twin on the ship. Even when he changes frames. And the length of that ruler will be length contracted the exact same amount as measured by Earth during both legs of the trip. The same would be true of any distance between any two points (at rest with respect to the ship) measured by the ship. The ship's bathtub, for example will always be the same size to the twin on the ship. And it would appear the same size to Earth during each leg of the journey, even though the ship switched frames.

So it should be possible to specify a distance in the (outgoing) inertial frame of the ship which would be the same distance in the (incoming) inertial frame of the ship, and that would be seen by Earth as length contracted by the same amount during both legs of the trip, since the relative velocity is the same for both legs. I thought I could use the length of one of Einstein's "rigid rods" for this purpose, since he used them for a similar purpose. Then I thought I could use a powered buoy that could stay in the ship's frame by it's own power. Sounds easy, since the buoy could easily figure out when to start and turnaround, since it would be in the same frame as the ship at all times. But if I elaborate too much, only the very patient would read it. If I try to be brief and simple, people will assume I don't know basic SR and explain it to me (obviously). Or assume that I believe "rigid rods" exist other than hypothetically. Or assume I don't know you can't use v=d/t indescriminately between frames.

Pervect,

I only meant the equation v=d/t would be used seperately by each respective twin. For example, v', d', and t' could be for the twin on the ship, so v'=d'/t' and v=d/t. We just have to make sure that the three terms are not mixed and matched between frames.

Thanks,
Alan

 

[JesseM]

Jesse,

OK, I agree, but a ruler at rest with the ship will stay the same length to the twin on the ship. Even when he changes frames. And the length of that ruler will be length contracted the exact same amount as measured by Earth during both legs of the trip.

No, because when the ruler accelerates it cannot stay rigid, the material it is made out of will be compressed or stretched, which would be cause changes in length in addition the Lorentz contraction.

The same would be true of any distance between any two points (at rest with respect to the ship) measured by the ship. The ship's bathtub, for example will always be the same size to the twin on the ship.

The bathtub can also not stay rigid when the ship accelerates. And what method is the observer on the ship using to measure the bathtub during the acceleration? Remember that measuring an object's length involves not just having a measuring-rod, but also having a definition of simultaneity, so you can say something like "at the same moment the back end of the object was at the 1-meter mark, the front end was at the 3-meter mark, so the object is 2 meters long."

And it would appear the same size to Earth during each leg of the journey, even though the ship switched frames.

During the inertial legs of the journey, sure.

So it should be possible to specify a distance in the (outgoing) inertial frame of the ship which would be the same distance in the (incoming) inertial frame of the ship, and that would be seen by Earth as length contracted by the same amount during both legs of the trip, since the relative velocity is the same for both legs. I thought I could use the length of one of Einstein's "rigid rods" for this purpose, since he did.

He only used rigid rods in the context of inertial motion. If you like you could have two inertial rods moving alongside each other, one which is at rest with respect to the ship during the outbound leg, one which is at rest with respect to the ship during the inbound leg, and you could have the ship's acceleration be instantaneous, with him switching from measuring distances with one ruler to the other immediately after the acceleration. But you also have to worry about simultaneity--does the ship observer also switch from using the definition of simultaneity in the first rod's rest frame to the definition of simultaneity in the second rod's rest frame at the moment he makes the switch? If so, there will be a discontinuous jump in what he says the Earth clock reads "right now" at the moment he makes the switch. And if not, whose definition of simultaneity should he use?

But if I elaborate too much, only the very patient would read it.

Go ahead, elaborate--I'll read it. Be sure to make it clear which frame's definition of simultaneity the ship-observer would be using at each moment on the trip, though.