Geek or Dork?

There are a bazillion alleged “paradoxes” in special relativity, all based on exactly the same fallacy, but I might have just invented a brand-new one—-where “invented” is shorthand for “confused the hell out of myself for a while”. When I finally got up and drew a picture (as opposed to lying in bed with my eyes closed doing something that felt like thinking), it became clear that, sure enough, it was the same old fallacy again (how could it not have been?), but in a new enough guise that someone reading this might find it amusing.

So: A circular train (front of the locomotive attached to the rear of the caboose) sits on a circular track. At some point, the train accelerates and starts traveling around the track. Because the train is moving, I (an observer stationary relative to the track) should see it shrink. But the track doesn’t shrink. So the train can’t stay on the track, and gets pulled inward, ending up inside the track. On the other hand, the passengers say the track has shrunk, so they should expect to get pushed outside the track. How can everyone be right?

Yes, the train is moving in a circle and is therefore not in an inertial frame. No, that has nothing to do with resolving this. I won’t spoil the fun by posting the answer just yet. I had a very “d’oh!” moment when I got it.

I’m not sure whether I’m a geek for figuring this out or a dork for not seeing it instantly.

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17 Responses to “Geek or Dork?”


  1. 1 1 Shane

    Are you certain this can be resolved without breaking the system up into small slices with their own non-inertial reference frame?

    I’m kinda bouncing ideas around in my head at the moment but from my basic understanding wouldn’t the problem be that all the observers would measure the same track radius but have different values for the circumference?

  2. 2 2 Douglas Colkitt

    Don’t keep reading if you want to figure it out, but I think it goes like this.

    Let’s say the train has 360 cars (1-360), and I’m in car 1. I observer the track shrinking relative to me, but the further away a car is from my angle the faster it’s moving relative to to my instantaneous inertial reference frame. So Car 2 is moving slowly relative to me, while car 181 is moving very fast relative to me. In fact because it’s moving in the opposite direction at the same speed (relative to the observer in the center) it’s more dilated than the track itself. Thus the train cars on the opposite end of the circle will shrink even more than the track, so the total length of the train will equal out to the length of the tracks.

  3. 3 3 Ben

    In GR:

    I once invented a perpetual motion heat engine, based on raising and lowering a compressible gas through a gravitational field.

    I was excited until I noticed that the effect was entirely due to a simplifying assumption I made in the calculation. Once I made an exact calculation the effect disappeared. Shame…

    Still, it was an exciting couple of hours!

    Cheers,
    Ben

  4. 4 4 Dave

    Why would it shrink?

    [I honestly feel like I'm the dumbest of all the readers of this blog - which is a werid feeling for me because objectively I'm pretty smart amongst my peers who aren't too shabby themselves]

  5. 5 5 Josh

    So I’m guessing that the answer is not that the track is moving for you as a stationary observer but it moving for the riders…that would be too easy right.

  6. 6 6 Josh

    I meant to say the track is _not_ moving for the stationary observer… But is moving for the passengers…the iPhone messes me up! But too simple a solution to be correct I think for this.

  7. 7 7 GregS

    Ha! When I was a grad student, a fellow TA came to me with a similar problem. A clever engineering student had asked him what happens on a treadmill (or some other kind of conveyor belt) if the belt is moving at relativistic speed. Does it contract and squeeze the rollers that hold it in place? Obviously not, but I never did manage to think of a solution. (I promptly forgot about it.) Thanks for posting this; I think I have it figured out.

  8. 8 8 Josh

    I somehow missed the part about the front of the train being connected to the caboose. And now the problem seems more complex. Could the answer somehow lie in the fact that people can view different parts of the track At different times by rotating their heads?

  9. 9 9 Joe Z

    If you’re observing from the center of the circular track, the train is moving slower for you than for the passengers on the train. Like riding the inside of a carousel vs. the outside. The passengers would also have a horizon reference whereas the observer wouldn’t.

    I’m still sorting this. I’m I on the right track?

  10. 10 10 Michael

    Does the train get thinner rather than shorter?

  11. 11 11 Neverfox

    Are you referring to the fallacy described here.

  12. 12 12 GregS

    It seems that the radius of the track would not contract, but the circumference would. (Incidentally, I’ve found out that this problem is called “Ehrenfest’s Paradox”.) I convinced myself I had the answer when realizing that the radius doesn’t change, but that I didn’t have the real answer because the circumference DOES change.
    The thing that usually resolves this type of paradox is the realization that “things which are simultaneous in one frame are not simultaneous in another frame.” Maybe I’m missing something, but I don’t see how that saves us. I’d be pleased to see Professor Landsburg’s answer after this thread has played out.

  13. 13 13 Al V.

    @Douglas Colkitt, if you are in a car of the train, none of the cars are moving relative to you, are they? If I’m in car 1, car 181 is directly opposite me across the circle, and stays there. Thus, I would think the other cars are stationary relative to me.

    I think, although it doesn’t make logical sense, that GregS is right. If I’m on the train, the tracks will appear shortened in the direction they are traveling relative to me, but since the radius is perpendicular to my direction of travel, the radius would not change. It’s hard for me to visualize, though.

  14. 14 14 Douglas Colkitt

    @AIV

    Yes car 181 is moving relative to my inertial reference frame. If you’re talking about the rotating reference frame then it is stationary. But remember a rotational reference frame is not inertial, ie it is accelerating. Hence the reason why someone on a Tilt-A-Whirl feels centripetal force.

    The easiest way to think about this problem is to think about the train car’s instantenous inertial frame. That is to do the calculations assuming each car on the train continues to move with the same velocity and no acceleration. The inertial vector for any given car is the tangent of the circle, therefore every car has a different inertial frame, and the further cars are away from each other the greater the difference between their inertial frames become.

  15. 15 15 Douglas Colkitt

    Also as for the observer in the center, here’s the most elegant way I can think of to explain.

    Instead of thinking of a circle, let’s think on a regulat N-Gon with a seperate car travelling along each side. (I think a square’s conceptually easiest). The observer on each side on the N-Gon is a inertial reference frame relative to the track, because he is travelling in a straight line correponding to the side he’s on.

    The both the observer on the car and the observer on the track will observe the other one shrinking in length relative to the other one. This is nothing new, it’s the classic relatively scenario of two spaceships passing each other in space and both saying that the other one’s meter sticks are too short.

    Now let’s figure out for the observer in the center what the distance is between each of the space ships on each side. If the car and polygon side are the same length when both are at rest a track-stationary observer will see the cars shorten more and more as it they approach the speed of light.

    For simplicity sake let’s say each car has dillated to half the length of the side of the track it’s on. This can be represented by circumscribing a smaller N-gon inside the N-gon where the middle of each side forms the new points. Now if you start with a square you have just shurnk by a lot, but as you go to an Octogon the dilatted train is almost as big as the original tracks. As N -> Infinity, the size of the dilatted train (or circumscribed N-Gon) approaches the size of the original N-gon, which is what you get when you have a circle.

  16. 16 16 dave

    just because you see something shrink doesnt mean it has actually shrunk.
    i dont understand the shrinking myself. is it like the doppler effect for light?
    who would ride a train that just went in a circle, anyway?
    dork. ;]

    @dave dont worry about your relative intelligence to these geeks. i just imagine that im stationary and they are all on a train that is moving around a circular track. =]

  17. 17 17 Neil

    I’m a dork still waiting for that d’oh palm slap moment. I cannot get past the assertion that we can ignore the non-inertial frame. The passengers of the train “know” that it is they who are moving in a circle and not the tracks by the fact that they expeience the orthogonal acceleration.

    I have long puzzled over a similar non-inertial frame paradox–a version of the twin paradox. Isolate two caesium clocks from the outside environment and send one on a one-G roundtrip to the nearest star. According to the Lorentz calculation, the travelling clock shows far less elapsed time upon reuniting. My question–how did the travelling clock “know” it was the travelling clock so it can show less elapsed time? Both clocks were in a sealed one-G environment.

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