The basic variant of this paradox makes sense to me but I have problems in proving mathematically variation of this non-paradox. So lets imagine that we have barn and pole which is stationary to the barn and they both have rest length $L_0$. Moreover, we are given team of people who are rushing towards the barn with speed $v$. When they reach the barn, they grab the pole and continue. We are asked to find what will be the length of the pole according to Team Pole if Team Barn says that the members of Team Pole grab the pole simultaneously. So as Team Barn says that the pole is grabbed simultaneously, there is no stretching or compression from their perspective-only Lorentz contraction. Solving the equations from their perspective I get that they will claim that the pole's length is $L_0/\gamma$ and from here according to Team Pole the length must be $\gamma(L_0/\gamma)=L_0$. So far so good but lets now solve first for Team pole. Team Barn says that they have grabbed the pole simultaneously so from team pole's perspective-they have grabbed the pole with $\Delta t=(v L_0/c^2)\gamma$. this leads us to conclusion that there is stretching of the pole from Team Pole's perspective. Solving all the equations I get: $$L_{new length}=L_0/\gamma+v(v L_0/c^2)\gamma=L_0 \gamma$$ which is in contradiction with the first answer. My question is where is my mistake in calculating the new length first from Team Pole's perspective. I suppose that there must be something wrong with the calculations in the second way because the first way makes sense in all the perspectives.


I think you are considering two different situations:

1) Team Pole passes through the barn at constant velocity, simultaneously (in Barn's frame) grab the pole, and continue on with the same constant velocity.

In this case, your second calculation is correct. The pole's length does not change from Team Barn's perspective. The pole remains length $L_0$ because Team Pole stretches the pole enough (in their frame) to cancel out the length contraction (in Barn frame). In Team Pole's frame the pole is length $L_0\gamma$.

2) The pole is rigid - it remains length $L_0$ in its rest frame at all times.

Then Team Pole has to adjust their positions as they pick up the pole - so they pick it up but then must move closer together as they accelerate the pole, in order to keep the pole its proper length in its own rest frame. In this case the pole is eventually length $L_0/\gamma$ in the Barn frame, and length $L_0$ in the Team Pole frame.

In both cases Team Pole can't instantly accelerate the pole up to their speed - this is not unique to special relativity and you would have the same issue grabbing a heavy object while running past it at 10 km/h. However the details of this process aren't necessary to resolve the paradox.

Another thing to keep in mind is that rigid objects don't really exist in special relativity - rigidity is a concept that requires simultaneity (of the motion), but there's always a frame in which the object is distorted as one end of it starts moving before the other.


So as Team Barn says that the pole is grabbed simultaneously, there is no stretching or compression from their perspective-only Lorentz contraction.

Actually, this can't be the case since instantaneous contraction would contradict special relativity.

In the barn inertial frame of reference, the pole is initially at rest and has a length of $L_0$.

If an identical pole were moving with speed $v$ relative the barn, the moving pole would have length $L = \frac{L_0}{\gamma_v}$.

This is, I believe, uncontroversial.

Now, if I understand your thought experiment correctly, at time $t=0$, the pole at rest is instantly accelerated to $v$ (somehow) by Team Pole in such a way that the length of the pole instantaneously changes from $L_0$ to $L$ as observed by Team Barn.

I conclude this is the case since it is specified that

When they reach the barn, they grab the pole and continue. We are asked to find what will be the length of the pole according to Team Pole if Team Barn says that the members of Team Pole grab the pole simultaneously.

But this is impossible according to SR.

Assume for simplicity that the position of the left end of the rod is given by:

$$x_L = 0 + vt\cdot u(t)$$

where $u(t)$ is the unit step function. In other words, the left end of the pole is at rest at $x=0$ for $t\lt 0$ and has speed $v$ for $t \gt 0$. The left end of the pole is instantly accelerated to $v$ at $t = 0$.

It follows that, assuming the result bolded above, the right end of the pole instantaneously changes position from $x_R = L_0$ to $x_R = L$ at $t=0$ in the reference frame of the barn.

But this requires that, according to Team Barn, the matter at the right end of the pole travel with infinite speed 'backwards' from $x = L_0$ to $x = L$ which is inconsistent with SR.

Thus, we must conclude that, assuming SR is correct, the instantaneous contraction of the pole is impossible.

  • $\begingroup$ It's not impossible for the pole to fail to Lorentz-contract in relativity--it can do that just fine, if its material stretches like a rubber band in its own rest frame. In fact most forms of acceleration will cause some form of material stretching/compression in the object's rest frame, the only exception being Born rigid acceleration, which requires that different points along the object experience different proper accelerations (different G-forces would be measured by an accelerometer at each point). $\endgroup$ – Hypnosifl Oct 5 '14 at 22:35
  • $\begingroup$ @Hypnosifl, I don't think you understand my answer. I did not state that it is impossible for the pole to fail to Lorentz-contract. As I clearly wrote, it is impossible, in the context of SR, for the rod to instantly contract as observed by Team Barn. Do you disagree that this is impossible? $\endgroup$ – Alfred Centauri Oct 5 '14 at 22:39
  • $\begingroup$ Apologies, I did misunderstand. I read the bolded part you were quoting too quickly and just noticed the part about "no stretching or compression", but misunderstood what was meant by that--I missed that the quote was suggesting there'd be no stretching or compression in the pole rest frame, but that there would be instantaneous compression in the barn frame, which as you say is impossible. $\endgroup$ – Hypnosifl Oct 5 '14 at 22:46

There's going to be a distortion of the pole during the acceleration period that goes away. The pole bearers should observe that they are holding a pole with length $L_{0}$, assuming that it's rigid. Note that it will start out length contracted from their potential, though, so during your $\Delta t$, they will stretch the pole out to its rest length.

  • $\begingroup$ this distortion, you are talking about, is presented in the formula, the question is : Where is the mistake in it? $\endgroup$ – Blake Oct 5 '14 at 17:19
  • $\begingroup$ @user1163511: I don't understand why the stretching of the pole necessarily has to be $v\Delta t$ $\endgroup$ – Jerry Schirmer Oct 5 '14 at 17:30
  • $\begingroup$ Also, I don't think that your algebra is right. $\endgroup$ – Jerry Schirmer Oct 5 '14 at 17:31
  • $\begingroup$ in fact it is $v\Delta t$ (i am taking part in a course and there a similar problem is solved there) and indeed i am pretty sure that my algebra is right, the problem is either in $\Delta t$ or in more fundamental level $\endgroup$ – Blake Oct 5 '14 at 18:40

If you were to try to grab a real pole at both ends at a significant fraction of the speed of light, the result would be similar to a nuclear explosion. The final state would be a bunch of ionized gas expanding a tens of thousands of km/s from the center of the system. As Jerry Schirmer said, you can't simply neglect the dynamics of the impact in this case. Even at lower velocities, you would basically just rip off the ends of the pole, while the middle piece would try to stay in place because of inertia. As a consequence one can't model this with just one coordinate system, but one needs an infinity of accelerated ones, which are all moving relative to each other at different velocities as the pole accelerates unevenly. This problem is already central for the design of rockets, which are being compressed along their axis by the acceleration and the resulting "weight" of the propellant column. If the acceleration becomes too large, the bottom of the fuel tank will burst.

  • $\begingroup$ ok, but lets not calculate what will be the contraction of the pole before it explodes, can't we forget about explosions for a moment $\endgroup$ – Blake Oct 5 '14 at 20:11
  • $\begingroup$ @user1163511: We can even forget that the universe is relativistic and then your "paradox" will go away all by itself. In theory you can make any random assumptions that you like and you will get any result that you could ever wish for... in the real world, however, you can't work around the fact that you have introduced a very strongly accelerated system into your "experiment", which means that CORRECT theory has to account for it. $\endgroup$ – CuriousOne Oct 5 '14 at 20:22
  • $\begingroup$ @CuriousOne - I think you are being too dismissive--there is no fundamental inconsistency between the laws of relativity and the notion of a pole whose material simply stretches rather than exploding when all points along it are accelerated simultaneously in the frame where it's initially at rest. If you like, we could even imagine that the "pole" is actually made of a collection of small rockets which are not even in contact with one another, and that they all independently accelerate simultaneously in the frame where they are initially at rest--none of problem's math would change. $\endgroup$ – Hypnosifl Oct 5 '14 at 20:50
  • $\begingroup$ @Hypnosifl: There is a huge difference between an accelerate system and an inertial system in SR, though, and we can't just "forget" about it, as the paradox tries to do. As you can see, you are already forced to change the setup of the paradox from a finite number of "grabbers" to an infinite collective. That, of course, does not just have to exist around the pole, but we have to extend it to infinity, which is probably equivalent to the treatment of accelerating coordinate systems in SR. That none of the problem's math changes is, of course, an assumption that is the source of the paradox. $\endgroup$ – CuriousOne Oct 5 '14 at 20:56
  • $\begingroup$ @CuriousOne - Assuming the material of the pole was such that it could stretch without exploding or breaking, the result of a large but finite set of grabbers grabbing at different points on the pole simultaneously (in the pole's initial rest frame) would only differ from the result predicted for an infinite set by a small amount, which could be made arbitrarily small by increasing the number--surely you've seen plenty of physics problems that are idealized, but make sense as limiting cases? (like perfectly elastic collisions, or frictionless sliding, for macroscopic objects) $\endgroup$ – Hypnosifl Oct 5 '14 at 21:42

The scenario quoted in the question is obviously nonsense; however, accepting it in that spirit, let's assume for the sake of clarity that the members of team pole are spread along the pole when they grab it, and that the pole stretches without resistance when grabbed.

From team pole's perspective, the team is more widely spread than the length of the barn/pole. Their spatial spread is 𝛾L0 in their frame of reference. The first member of the team will have left the barn carrying the front of the pole before the last member enters the barn to grab the tail end of it.

So, from the perspective of team pole, the pole is stretched so that its length is now the same was the original spatial separation of the team, namely 𝛾L0.

Therefore from the perspective of the people in the barn, the pole will appear to be its original length after it has been grabbed, as it will appear to them to be 𝛾L0/𝛾 = L0.

The 'paradox' arises because of an oversight in the sixth sentence of the question. To team barn there seems to be no stretching, but that is because the pole is stretched and at the same time subject to Lorenz contraction. Therefore the length of the pole continues to be 𝐿0 to team barn, while to team pole it is 𝛾L0.


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