# A spool in Einstein's theory of special relativity

I would like to know how can we model a simple spool that unrolls at speed close to light (in the frame of the wire remaining on the ground AND in the frame of the translatory motion of the rolling spool)? (See attached image)

The spool unrolls in the wire's frame, but what about the spool where time should be dilated?

If we count the number of turns, isn't there a problem by comparing the two frames, once the spool unwound? (See attached image)

According to different interpretations of relativity, either the spool does not make the same number of turns in each frames (which is curious), or it rotates faster in the frame of the dilated time (which is contradictory).

I try to understand my mistake:

A: person or frame of a thread left on the ground /B: person or frame in rectilinear motion following the spool

Lets put 4 breaths (equivalent to the time) of an individual per turns in his referential. Imagine that a huge spool makes a single full turn at speed 0.999c.

Isn't B's breathing more slowly compared with A (and in A)? The number of turns should be the same. Is the number of breaths less than 4 for B in B? If the breathing of B in B is normal (proper time), has the spool not stopped before? Or did it turn faster (perception of a faster speed for B in B than A in A) to finish its turn? (Both last questions are linked and any answer would be paradoxical)

I hope my thought experiment is interesting and will lead you to question relativity.

• Seems to me that this is at least related to the Ehrenfest paradox. Not sure what the downvote is for, in that case. – Kyle Kanos Dec 5 '16 at 0:54
• It seems very close to Ehrenfest paradox, but the question is a little different. The frame B is translational, so Ehrenfest should not come in question. I would say it is more close to Langevin paradox. – Copernic Dec 5 '16 at 1:00
• Look at the "Relativistic Trolley Paradox" in the Am. J. Phys, June 2016. There are two resolutions with either Lorentz contracted radius of the spool or with constant one. – Albert Jan 5 '17 at 13:27
• "I hope my thought experiment is interesting and will lead you to question relativity" Doesn't that suggest an invitation for non-mainstream answers? The answers are based on accepted theories, but the question isn't phrased as a doubt about the present concepts, in my opinion. – user191954 Jun 25 '18 at 9:33

Unspooled wire's frame: It takes a long time for the spool to make one turn. It takes a long time for all the wire to unwind. The wire on the spool is contracted in an interesting way, for example at the point where the wire leaves the spool the contraction is zero, while at the opposite side of the spool the length contraction is the largest.

Spool hub's frame: It takes a short time for the spool to make one turn. It takes a short time for all the wire to unwind. The wire on the spool is contracted, and also the unspooled wire is contracted. Nothing too weird here.

And here's a mathematical version:

Unspooled wire's frame: It takes gamma times more time for the spool to make one turn, compared to the hub's frame. It takes gamma times more time for all the wire to unwind, compared to hub's frame. The wire on the spool is contracted in an interesting way, for example at the point where the wire leaves the spool the contraction is zero, while at the opposite side of the spool the length contraction is proportional to gamma(v), where v is the result of relativistic addition of the hub's velocity in the unspooled wire's frame and the wires velocity in the hub's frame.

• You wrote "Unspooled wire's frame: It takes a long time for the spool to make one turn." AND "Spool hub's frame: It takes a short time for the spool to make one turn." THEN the spool has turned quicker in the dilated frame (which is contradictory). A perception of quicker speed in a dilated frame, is that possible? – Copernic Dec 5 '16 at 10:47
• I mean "TIME dilated". – Copernic Dec 5 '16 at 10:56
• The spool is a clock, it time dilates like a clock, so a very fast moving spool is a very slowly turning spool, just like a very fast driving car has very slowly turning wheels, and motor, and the thing that shows the traveled distance. It's very simple! :) Details of wheel motion may be a bit complicated though. – stuffu Dec 5 '16 at 11:36
• "so a very fast moving spool is a very slowly turning spool". I disagree, it has no sense. Do not mix real perception of movement and illusion of a movement perception. When sometimes people see a fast wheel turning in the other way, it still turns in the right way. – Copernic Dec 6 '16 at 18:32

I'm not sure what your question is. When you account for the travel time of the light coming from the wire, the spool will see the appropriate amount of physical progression as it should within the relatively smaller time frame as designated by SR.

I hesitate to answer this because it's completely standard and must have been asked and answered a hundred times before. But since the question has been auto-bumped to the homepage, I hope this will finish it off:

Clearly the spool makes the same number of turns in each frame. If the spool starts rolling at noon (according to both its internal clock and the clock of the observer on the ground), then it stops rolling at (say) 2:00 according to the ground observer, after having turned (say) 120 times and at time (say) 1:00 according to the observer who rides the spool, also having turned the same 120 times. The observer on the spool says that the spool takes 30 seconds to complete a turn; the observer on the ground says the spool takes 60 seconds to complete a turn. In other words, the observer on the ground says that the spool unrolls at half-speed.

• 4 rotations/seconds for a riding observer in a time dilated frame and 2 rotations/seconds for the standing observer (no time dilation). A quicker speed in a time dilated frame??? – Copernic Feb 18 '17 at 18:01
• @Copernic: I do not understand your question. Of course both frames are equally "time dilated" relative to each other. But the spool is a clock, so it must run faster in its own frame than in any other --- in this case 4 rotations per second, as opposed to 2. – WillO Feb 18 '17 at 18:35

Yes, it seems like a paradox, because in the frame of the "wire on the ground" due to dilation of moving clock rotation of a spool must slow down, as the hub of the spool moves faster and faster. When the hub of the spool reaches velocity very close to that of light, rotation almost complete stops and the wire should tear.

It seems like a paradox, because the wire cannot tear in one frame and remain intact in another.

Resolution of the paradox is in relativistic kinematic. Rim of the spool Lorentz – contracts.

The rest length of the rim of the spool must remain constant. This means that the rim Lorentz contracts, and that the radial extension of the spool contracts accordingly. The result is that the spool become infinitely small in the limit that the spool moves with the velocity of light.

If $v$ is velocity on the rim in the rest frame $K$ of the spool, we have $\Omega=v/R$, where $R=R_0/\gamma$ is the contracted radus of the rotating spool, and $R_0$ is its radius when it is at rest. The angular velocity of the rotating spool is then

$\Omega = \gamma v /R_0$

Hence, in this case the angular velocity $\Omega$ must approach an infinitely great value in $K$ when the speed of the wire approaches that of light. As observed in the “unspooled” wire frame $K'$, the distance between the marks on the wire each time a point on a rim of the spool leaves it is

$l'=\gamma 2 \pi R = 2\pi R_0$

and this distance is independent of the speed of the spool, even if the radius of the spool decreases with increasing velocity, because the distance between the marks depends upon the rest length of the rim of the spool and not its Lorentz contracted length. Also in this frame the angular velocity of the spool remains finite even if the spool has a vanishing radius when the velocity of the spool approaches that of light,

$\Omega'=\gamma^{-1} \Omega = v/R_0$

and hence $\lim\limits_{v \to c} \Omega' =c/R_0$, which is finite.