How to Teach Special Relativity is a famous article by John Bell where he advocates that the way we teach relativity does not give good results. He describes an experiment now known as Bell’s spaceship paradox (even if Bell did not invent it):

In Bell’s version of the thought experiment, two spaceships, which are initially at rest in some common inertial reference frame are connected by a taut string. At time zero in the common inertial frame, both spaceships start to accelerate, with a constant proper acceleration g as measured by an on-board accelerometer. Question: does the string break – i.e. does the distance between the two spaceships increase?

Considered a difficult problem

The correct answer is that the string does break, even if the spaceships appear to always be at the same distance from one another as seen from an observer who did not accelerate with the spaceships. Yet, according to Wikipedia:

Bell reported that he encountered much skepticism from “a distinguished experimentalist” when he presented the paradox. To attempt to resolve the dispute, an informal and non-systematic canvas was made of the CERN theory division. According to Bell, a “clear consensus” of the CERN theory division arrived at the answer that the string would not break.

In other words, this problem was considered hard by a majority of serious physicists at the time Bell raised the question in 1976. I would venture to say that this remains the case today, except that this particular paradox is probably well known now. But the teaching of special relativity has not changed. This is how we explain the paradox today. I have a lot of admiration for John Baez in general, his “blog” is even in the sidebar of this one. But with all due respect, the explanation of the paradox posted on his web site is utterly complicated (I know he gives credit to someone else for it, but by hosting it on his web site, I would say that he condones it).

It should be easy

This particular formulation of the paradox was not known to me until someone recently asked me if the string would break. Using my little technique, it took me less than one minute to have the correct answer, without looking it up, obviously, but also without any computation or complicated diagram. Here is the mental diagram I used (click to see it in high resolution):

On this diagram, time is represented horizontally, and the two space ships are represented by the green and red curves, which are identical but separated by a distance along the vertical “spatial” axis. The distance at rest is represented by the blue arrow. The distance as measured between the two ships after they started moving is measured by the green and red arrows. The distance as measured “from the ground” is along the vertical axis, and remains constant.

Remember the only trick is that a “cosine” contraction on this diagram corresponds to a dilatation in relativity and conversely. On the diagram, the red and green arrows are obviously shorter than the blue arrow. The contraction factor is the cosine of the angle between these arrows and the vertical (space) axis, which is the same as the angle between the red or green curve and the horizontal axis. Therefore, relativity predicts that the distance between the two ships, as seen from the ship, will increase. Specifically, it increases by a factor usually denoted “gamma” (but which I prefer writing as the cosine of an angle myself), which can also be seen as a hyperbolic cosine, and which plays in Minkowski geometry the exact same role as the cosine in the Euclidean diagram above. You can find the precise mathematical relationship here.

Consequently, the string will break.

Accelerated solids in relativity

Another interesting observation one can make from the diagram is that you cannot draw a straight line that is perpendicular to both curves. What is “space” for one ship is not just “space” for the other. You need to draw a curved line between the two rockets if you want to always be perpendicular to the local “time” direction. In other words, the “time” direction for the string is not constant along the way, so all parts are not moving at the same speed. Someone sitting anywhere on the string will see other parts of the string move relative to him. That’s another way to explain why the string will break.

You can easily verify that this problem exists for any kind of accelerated solid. All parts of an accelerated solid in special relativity do not move at the same speed.

My own puzzle

Here is the interesting other thing that I realized within this short moment of reflection: there is a way for the two ships to accelerate “identically” (for a suitable definition of identically which remains to be given) so that the string will not break. Can you find it?

C’est tout de même dur à avaler.
Comment voulez vous qu’on enseigne cela à des élèves de Terminale?

Jean-Claude Carrière, in Entretiens sur la multitude du monde with Thibault Damour


24 thoughts on “How To Teach Special Relativity

  1. It’s amazing how you didn’t even look to see why a growing number of scientists around the world are saying special relativity will be dead in 10 years. Look before you leap! You didn’t even look into the idea that relativity is wrong. You are relegating yourself to the few “intellectuals” clinging to a dying, useless, and wrong theory.Just go on repeating what people tell you without investigating. Wonderful!

  2. Dear Moose,As my reply to your earlier comment should have made clear, I did read the web site, found what theory it was referring to (autodynamics), and checked that the theory clearly contradicted too many known experimental facts to be valid.Just look at the autodynamics web site. You will easily find by clicking on the “AD in one page” link that it prides itself on the following, among other things: No Length Contraction – No Time Dilation – No Twin Paradox – No Failure of Simultaneity – No Heisenberg Uncertainty PrincipleFor each of these, the experimental track record is pretty heavy by now. So this autodynamics theory is as easily refuted as a theory claiming that heavier-than-air flight is impossible. It is fantasy, not science. The twin paradox, for instance, is verified several times a day on the lifetime of particles in accelerators: the particles do last longer when they travel fast.I realize that special relativity, if not taught correctly, feels unintuitive. I really tried hard to show that it is not complicated, and can actually become pretty intuitive. I myself never need a computation to predict a qualitative result. That technique is the result of my investigation.This teaching method also worked pretty well with a number of people, including kids. But apparently, with you, I failed. I only hope it’s because you did not read my web page. Otherwise, sorry, I hope you can find a better teacher.

  3. The string will NOT break!!The distance between ships will increase, but the string will not break? Why? On the diagram the area between the two curves stands for ‘string’. If you cut a loaf of bread in different directions you always end up with bread, but of different lengths!!

  4. Dear Erre,The area between the two curves is where the string could be, it is not the string itself. In the relativistic case, the length of any curve connecting the two curves is greater than the length of the string. So the string will break.In the classical case, put a string between the two curves: it will slack.See the Wikipedia article for an interesting discussion.Christophe

  5. Dear Christophe, Will break? No! Comments: 1/ (The area between the two curves is where the string could be, it is not the string itself). The area contains all the space-time events of the rope. Wherever you cut through the area you have a total of rope-space-time events, physical facts, which in every case equals a real ‘rope’. 2/(In the relativistic case, the length of any curve connecting the two curves is greater than the length of the string.) Correct.3/ (So the string will break)NO! The longer string is made of completely different space-time events than the shorter rope. 4/(In the classical case, put a string between the two curves: it will slack.)NO. There is no ‘pulling’ of a ‘classical’ rope to a greater length! The longer rope is made of different space-time events out of the 4-dimensional existence of al the space-time events of the rope.Erre

  6. Dear Erre,I believe that you are in error.What you are trying to do is to say that I can keep the blue arrow vertical, and distance between the two curves remains constant. Therefore, there is a “slice” connecting the two curves that has a constant length and contains rope atoms.Place yourself along the red curve in my graph, and look how your blue-parallel curve looks seen from that end… Now, you are measuring the length of your rope looking into the future. Does that sound right to you?Christophe

  7. Dear Christophe, (I believe that you are in error).Oh no, sorry, I am 100 % sure I am correct… But I do agree that things a that little bit complicated because the moving string stays in your reference frame the same lenth, and therefore in de sting’s reference frame the string has to get longer (but NOT break!) … Normally it’s the other way around: the moving string stays the same length in it’s own reference frame, but not in yours where the rope’s length gets shorter. I think you ‘read’ the space-time diagram not correctly… Let me explain it a different way.As time goes by your reference frame is a vertical line that stays vertical and moves with time from left to right. The vertical line indicates the space-time events that are simultaneous at one specific moment in your world. Hence what you say§ is correct: (What you are trying to do is to say that I can keep the blue arrow vertical, and distance between the two curves remains constant. Therefore, there is a “slice” connecting the two curves that has a constant length and contains rope atoms.)What is the rope in your world of simultaneous events? What is ‘rope’ on that vertical line? Correct, the vertical cut through the area between the two curves. Because in the exercise as per definition the moving rope stays the same length, the vertical length between de ropes stays the same. But what happens in the reference frame of the rope (and that of the pilots…)? Their reference frame is a line following your oblique red line. In their reference frame the space-time events are not the same as those in your reference frame. In fact we both have a ‘different’ rope in our respective reference world of simultaneous events!! Rope-events that are simultaneous for you, are not simultaneous for the pilots, and vice-versa. Therefore what you say doesn’t make sense: (Place yourself along the red curve in my graph, and look how your blue-parallel curve looks seen from that end… Now, you are measuring the length of your rope looking into the future. Does that sound right to you?) Erre

  8. Dear Erwin,You wrote: “the moving string stays in your reference frame the same lenth, and therefore in de sting’s reference frame the string has to get longer (but NOT break!) …”Well, you can redefine the exercise if you want. In John Bell’s formulation, the string is fixed size and inelastic. Therefore, saying that it becomes longer in its reference frame is equivalent to saying it breaks.Of course, you can replace the string with some rubber band, and then the rubber band may simply become longer, at least to some extent.You also wrote: [The rope’s] reference frame is a line following your oblique red line. In their reference frame the space-time events are not the same as those in your reference frame.Precisely. The notion of simultaneity for the red end is along the red arrow. The notion of simultaneity for the green end is along the green arrow. Neither is parallel to the blue arrow. So neither notion allows you to build a set of rope atoms that is as short as the rope at rest.Finally, regarding: Therefore what you say doesn’t make sense: It makes sense. I am only observing that from the point of view of the red pilot, you cannot look at rope atoms along a line parallel to the blue arrow, because that would be akin to looking into the future. I was simply pointing out your error.As to being 100% sure you are correct, or believing that I read the diagram incorrectly… I happen to reach what is generally considered the correct solution. Between trusting you or John Bell, my choice goes to John Bell, in particular if I agree with him ;-)RegardsChristophe

  9. If the diagram is a reflection of what happens, could you tell me, and/or show on the diagram, WHERE exactly the rope breaks -‘breaking’ means that from then on there is an area between the two spaceship curves with no longer rope space-time event points…- ?

  10. Dear Erre,The part in the diagram where the rope breaks is the part where the rope would have to become too long. Say we decide that the rope breaks if elongated by more than 1%. Then as soon as you cannot join the red and green curves along a curve that is locally perpendicular to the local time line without exceeding that 1% budget, the rope will break.RegardsChristophe

  11. Dear Chrisophe, Two comments:(PS capital letters are no shouting, only to stress some views. Italics or underlining don’t appear as such in the posted text…). What you say about breaking after f.e. 1% can be correct if you stay in your reference frame with a immobile rope and then stretch the rope. However, what happens for the pilots is a complete shift of space-time events which makes a longer rope possible without breaking (different simultaneous events which total up to a longer total length). If you take an area of rope space-time events, however you cut it through, you end up with rope whatever the length.But, you know, I am all of a sudden a little bit confused. I am still of the opinion that you claim that the rope will break in the pilot’s frame, BUT NOT IN YOUR GROUND FRAME, and therefore you still believe there is paradox. For me there is no paradox. There is no rope breaking in both frames, or there is rope breaking in both frames. I only wanted tot stress that if in your frame the rope does NOT break (as I believe you claim but am I wrong?), it will not break in the moving frame either! If you say that the rope breaks in the pilot’s reference frame of simultaneous events you have to accept that it breaks in your ground reference frame. You have to. Suppose the rope breaks (and f.e. disappears or evaporates) in your diagram at the red line. This would mean that there would be no longer rope space-time events in the area at the right side of the red line (still between the two curves). However, this would mean that if you cut with a vertical line (your reference frame at a later stage) through that area -right part of diagram the- you have NO rope in your world of simultaneous events either! So if the rope DOES break and disappear, then there would be no longer any rope between the curves, and therefore no paradox because no rope in BOTH reference frames… I said, “if the rope DOES break”… “if” …. Well, I claimed that the rope would NOT break neither in their frame nor in your ground frame, and therefore no paradox. Believe it or not, I will now agree it indeed DOES break, BUT… in BOTH frames (still no paradox!) because I take into account the following reasoning:It might not be clear in some websites you refer to (and therefore I misread the exercise…), but we know that in your ground restframe not only the moving (or with same accelleration) spaceships ‘shrink’, also the rope shrinks (moving lengths shrink). If by definition you want to keep the total length in your restframe invariable you have to inform the pilots that they have to adjust their respective accelleration and/or speed to lengthen in their restframe the distance between the airplanes. This will indeed break the rope. But it also means that the rope breaks in your ground restframe. Let me phrase it this way: Suppose a train approaches at a constant speed. It is shorter in my ground frame because it moves. In order to reinstore the original length of the train the trainpassengers have to plull/push/stretch their train until finally the train will have the requested total length in my ground frame. But in ends up in broken pieces in their AND my frame. So the rope or train WIIL break, in BOTH reference frames. No paradox. Erre

  12. Dear Erwin,Comments accept some HTML tags: <i>italics</i> or <b>bold</i> in particular.Of course, the rope will break in all referentials. I never claimed otherwise. I only called the problem a “paradox” because that’s the standard terminology, not because there is a real paradox.RegardsChristophe

  13. I addressed this previously on an “alt.relativity” post regarding acceleration. If the FRONT of a 1 light year long ship accelerated at > 1 light year per year per year, it would never receive a signal from > the rear of the ship. Since rods and spaceships maintain their> structural> integrity through molecular forces, with a limiting speed of c, and> since> the molecules at the rear of the ship whose forward end is> accelerating at> 1 light year per year per year cannot interact with the molecules at> the> front of the ship, there’s no net molecular force holding the front> and> rear of the ship together.> > Since there’s no net molecular force holding the front and rear of the> ship together, the slightest bit of turbulence will separate the ends.> The ship MUST come apart, per the pole- barn paradox. > > You’re right that the front and rear could accelerate at different> rates,> and still be one light year apart after they stopped accelerating, but> if> they exceed the rates I gave, they won’t remain a single coherent> object.> > Of course, this limit on acceleration is only a very high theoretical> upper bound, and there are other bounds on the size of the ship, such> as the force> of gravity, and tensile strength of the material, which would place> much lower> bounds on the force of acceleraton a material body could withstand. > > With a ship under constant acceleration, from the point of view of the> initial rest frame, the distance traveled from the initial rest> frame’s> point of view will be> X=(c^2/a) cosh(at/c) – (c^2/a), the initial rest frame’s time will be> T=(c/g) sinh(at/c), and the velocity of the ship will be> > at/(sqrt(1 +(at/c)^2)).> > > Let c=1, g= 1 light year per year per year.> Get your WINDOWS scientific calculator from accessories, > punch 1, punch that hyp button, punch cos, and you will> see that cosh 1 = 1.543. Since the actual distance traveled is> cosh(at/c) – 1, the front of the rocket will have traveled> 1.543 – 1= 0.543 light years if its rest frame acceleration is > 1 light year per year per year.> > The initial rest frame’s time will be sinh (1) = 1.175 years . > > The velocity will be 1*1)/sqrt(1 + (1)^2) > or 0.7017 c. From the Fitzgerald-Lorentz contraction, the length of> the ship will have contracted to 0.7071 of its “at rest” length. The> rear of the vessel will have to have advanced .543 light years to> keep up,> plus have gained an additional 0.2929 light years to keep the> spaceship> from stretching apart, for a total of 0.8359 light years in a year. That should be a gain of 0.8359 light years in the INITIAL REST FRAMEin1.175 of the INITIAL REST FRAME’s years.The acceleration of the rear of the 1 light year long ship would haveto be2 light years per year per year to keep up with the 1 light year peryear acceleration of the front of the ship.A. McIntire

  14. this is a great more thing just to verify.after the acceleration stops, thedistance between the ships is theLo in the comoving frame and thestandard lorentz contracted distancein the lab that correct?perhaps you can point me to how thateffects ehrenfests paradox.thanksjpd

  15. Hi Anonymous,You wrote: after the acceleration stops, thedistance between the ships is theLo in the comoving frame and thestandard lorentz contracted distancein the lab frame. is that correct?Actually, it is not correct. Try to apply the technique I described carefully to see if you can come up with the correct answer. Please do not read on, the answer is below…If you look at the distance between the two curves on the euclidean space, you see that it is shorter than the original distance between the two curves. This is a contraction in euclidean space, and therefore will be a dilation in Minkowski space. This is the reason the non-elastic string breaks: it is not long enough.To answer the question in full, however, one must define precisely when the acceleration stops. Does it stop after the same time as measured from within the ship, or at the same moment measured in the original reference frame, or at a time specified through some other protocol (e.g. one ship signals the other)? The final distance between the two ships will depend on that choice.

  16. hi, thanks for the reply,i was thinking if both curves became parallel to the x-axis at the same time (same x value) ,then the distance they measurebecomes the same as the distance they measure before any acceleration.the ‘same x-value’ would have to bemeasured by identical, syncronized clocks brought on both spaceships.if thats true, then the ‘paradox’is only a ‘noninertial paradox’ andwikipedia needs to fix a paragraph. this distance between them should be lorentz contracted with respectto the lab ehrenfest paradox question is related in that if you do it noninertially(assume it survives accelerationand measure circumference only afteracceleration) it seems that the circumference of a rotatingdisc should be contracted.thanksjpd

  17. Hi anonymous jpd,Sorry for the delay, I’m a bit slow on everything these days (lots of travelling).First, I know what the Ehrenfest paradox is, but I completely forgot to answer that. What happens in that case corresponds to the following analogy in Euclidean space.Imagine that you have a cylinder made of some wire mesh, with some wires running horizontally along the x axis (the analogous of the time axis). Start with the wires parallel, all horizontal along the x axis. The distance between the wires is always the same. Now imagine that you twist one end of the cylinder.Now, if you want the wires to remain straight, the distance between the wires cannot remain the same from one end of the cylinder to the other. They will become closer to one another in the middle, and it will look a bit like this. Alternatively, you can keep the wires “parallel” to one another, but only if you accept that they are no longer straight.Here, “straight” corresponds to the non-accelerated case, the original definition of rigidity where everything is both straight and kept at identical distance over the horizontal (time) axis.I’m not sure if that makes sense without a drawing, but I don’t have much time for a drawing right now… Hope the Wikipedia picture link above helps.Back to the original question about the Bell paradox. I am not sure what you mean with “noninertial paradox”, but it’s true that Bell’s paradox involves accelerated motion. However, the effect remains even after acceleration stopped.However, I have to disagree with the statement that the distance between the ships has to Lorentz-contract. The Lorentz-contraction only occurs in the direction of motion. The big idea of Bell was to construct a case where relativistic effects were observed perpendicular to motion, thus confusing everybody. The neat result is that instead of observing a contraction, you observe a dilation in that case, which is why a nonelastic rope breaks.To understand the difference between the “normal” Lorentz contraction and the Bell scenario, compare the picture illustrating the “train paradox” in and the picture illustrating this post.You see that in the train paradox, we see distances expand in Euclidean space (specifically, d is greater than d’ in the picture), which turns into a Lorentz contraction. By contrast, in the Bell paradox, the two curves get closer to one another (the red arrows in this picture are shorter than the blue one), therefore we observe an increase of the distance between the ships in the relativistic case.I hope that this makes sense…

  18. hi again,thanks, that is all good info.i am still going over it.please feel free to ignoreif i am not making sense, but heres one more try at what i meanby it being a noninertial rest the spaceships measurea distance d between theman observer at rest also measures d.they accelerate for a while then stop accelerating, as determinedby identical syncronized clockson the ships.the spaceships must measure the original distance d between them.otherwise do another thoughtexperiment where it is one bigspaceship with engines at both ends.after accelerating then stopping acceleration, the length of the ship measured in its own reference frame should be the original noninertial the same argument, the rest framedistance must be the lorentz contracted d.i think this all related to Bornrigidity and how no strictly rigid object can accelerate.just sort of thinking out loud.i do need to carefully go overthe links you sent. thanksjpd

  19. Hi JPD,You make sense, but I think that you are missing an essential point: what makes the Bell paradox “different” is that it’s a case of transverse distance measurement, instead of a longitudinal one.RegardsChristophe

  20. Hi jpd,OK, my wording was not exactly crystal clear here. The choice of “longitudinal” vs. “transverse” was really poor. Let me try again with a more detailed explanation.Let’s compare the main picture in the post with the picture describing the “train paradox” in my relativity page.In the train case, we measure the length along the red line, which is parallel to the local space axis. What is taken as a reference is the length of the train in its own referential (in blue). That’s what I called “longitudinal” measurement.By contrast, in Bell’s case, if you look at the picture, you will see that what you consider as a reference now is the distance in the local referential, which means that the relativistic effect is observed by considering a “space” axis that is not aligned with yours, that’s what I called “transverse”.There are no perfect choices of word, this one was particularly awful 😉

  21. I know this is an old post and I'm way over my head here.But the rope is inelastic in space, does that have an effect in time? In the ground frame the rope is never stretched.Is the rope elastic in the time dimension?

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s