1. May 22, 2006

### Al68

Why do most explanations of the Twins Paradox claim that the twin on the space ship 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 space ship 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 travelled 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

2. May 22, 2006

### JesseM

It is not necessary to assume the travelling twin turns around when he reaches a marker that's at rest relative to the earth, you could just as easily have the travelling 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 travelling 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 travelling 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 travelling 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).
No. If the travelling twin is the one who accelerates to turn around, the travelling twin will always have aged less when they reunite. If you disagree, perhaps we should go through the math for a simple numerical example?

3. May 22, 2006

### 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.

4. May 22, 2006

### Al68

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

We will specify the distance travelled 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

Last edited by a moderator: May 22, 2006
5. May 22, 2006

### clj4

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:

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.

Last edited by a moderator: Apr 22, 2017
6. May 22, 2006

### 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 space ship 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 travelled 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

Last edited by a moderator: May 22, 2006
7. May 22, 2006

### clj4

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

8. May 22, 2006

### Al68

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 refering 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

9. May 22, 2006

### clj4

Yes, I did. It is incorrect.

You are welcome, I think we are done.

10. May 22, 2006

### 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 irrelevent 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

Last edited by a moderator: May 22, 2006
11. May 22, 2006

### pervect

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

12. May 22, 2006

### Al68

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

13. May 22, 2006

### robphy

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
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.)

Last edited by a moderator: Apr 22, 2017
14. May 22, 2006

### Al68

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.

Thanks,
Alan

15. May 22, 2006

### pervect

Staff Emeritus
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

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
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.

#### Attached Files:

• ###### bornrigid.jpg
File size:
9.4 KB
Views:
77
16. May 23, 2006

### 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

Last edited by a moderator: May 23, 2006
17. May 23, 2006

### pervect

Staff Emeritus
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.

18. May 23, 2006

### 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 travelled as "length contracted" while the observer on the ship does not. The distance travelled 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

Last edited by a moderator: May 23, 2006
19. May 23, 2006

### pervect

Staff Emeritus
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 recieve 6 signals, because I will recieve 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 recieved at 2 second intervals, signals emitted at 1 month intervals are recieved 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 recieve 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 recieve 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, therfore the Earth clock must have aged more (to be able to send out more signals).

Here's a space-time diagram

$$\] \unitlength .5mm \begin{picture}(90,300)(0,0) { \linethickness{0.3mm} \multiput(0,0)(0.12,0.2){750}{\line(0,1){0.2}} \linethickness{0.3mm} \multiput(0,10)(0.12,0.12){125}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,20)(0.12,0.12){250}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,30)(0.12,0.12){375}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,40)(0.12,0.12){500}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,50)(0.12,0.12){625}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,60)(0.12,0.12){750}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,300)(0.12,-0.2){750}{\line(0,-1){0.2}} \linethickness{0.3mm} \multiput(0,70)(0.12,0.12){750}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,80)(0.12,0.12){708}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,90)(0.12,0.12){667}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,100)(0.12,0.12){625}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,110)(0.12,0.12){625}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,120)(0.12,0.12){583}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,130)(0.12,0.12){542}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,140)(0.12,0.12){500}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,150)(0.12,0.12){500}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,160)(0.12,0.12){458}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,170)(0.12,0.12){417}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,180)(0.12,0.12){375}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,190)(0.12,0.12){375}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,200)(0.12,0.12){333}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,210)(0.12,0.12){292}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,220)(0.12,0.12){250}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,230)(0.12,0.12){250}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,240)(0.12,0.12){208}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,250)(0.12,0.12){167}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,260)(0.12,0.12){125}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,270)(0.12,0.12){125}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,280)(0.12,0.12){83}{\line(1,0){0.12}} \linethickness{0.3mm} \multiput(0,290)(0.12,0.12){42}{\line(1,0){0.12}} } \end{picture} $$$ Last edited: May 23, 2006 20. May 23, 2006 ### robphy Assuming that I have understood the situation you posed... $$$ \def\qbz#1#2#3#4#5#6{\qbezier(#1,#2)(#3,#4)(#5,#6)} \unitlength 2mm \begin{picture}(87.5,85)(0,0) \qbz{ 46.88}{ 14.38}{ 59.38}{ 36.88}{ 71.88}{ 59.38} \qbz{ 60.00}{ 15.00}{ 72.50}{ 37.50}{ 85.00}{ 60.00} \qbz{ 37.50}{ 20.00}{ 62.50}{ 32.50}{ 87.50}{ 45.00} \qbz{ 60.00}{ 60.00}{ 72.50}{ 37.50}{ 85.00}{ 15.00} {\linethickness{1pt} \qbz{ 60.00}{ 6.00}{ 60.00}{ 45.50}{ 60.00}{ 85.00} \qbz{ 72.50}{ 37.50}{ 66.31}{ 37.50}{ 60.12}{ 37.50}} \put(62,22){\makebox(0,0)[cc]{\theta}} \put(67,36){\makebox(0,0)[cc]{\theta}} \put(66,40){\makebox(0,0)[cc]{{\color{red}5}}} \put(64.38,31.88){\makebox(0,0)[cc]{}} \put(57,13){\makebox(0,0)[cc]{A}} \put(76,38){\makebox(0,0)[cc]{B}} \put(44,13){\makebox(0,0)[cc]{C}} \put(57,38){\makebox(0,0)[cc]{D}} \put(62,30){\makebox(0,0)[cc]{E}} \put(56,27){\makebox(0,0)[cc]{L}} \put(57,60){\makebox(0,0)[cc]{R}} \end{picture} \]$$

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.

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.

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 .

Last edited: May 23, 2006