Length contraction and the pole-in-a-barn paradox

Required math: algebra, geometry

Required physics: basics of relativity

Two famous results of relativity are time dilation and length contraction. We looked at time dilation in an earlier post, so we’ll examine length contraction here. Although it is possible to analyze length contraction using space-time diagrams, it is actually easier to derive the effect from the Lorentz transformations, which are

\displaystyle   t_{2} \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-v^{2}}}(t_{1}-vx_{1})\ \ \ \ \ (1)
\displaystyle  x_{2} \displaystyle  = \displaystyle  \frac{1}{\sqrt{1-v^{2}}}(x_{1}-vt_{1})\ \ \ \ \ (2)
\displaystyle  y_{2} \displaystyle  = \displaystyle  y_{1}\ \ \ \ \ (3)
\displaystyle  z_{2} \displaystyle  = \displaystyle  z_{1} \ \ \ \ \ (4)

These transformations apply to the special case of inertial frames that are aligned so that observer {O_{2}} travels with speed {v} along {O_{1}}‘s {x_{1}} axis, with the corresponding {y} and {z} axes parallel in the two systems.

Since the quantity {1/\sqrt{1-v^{2}}} occurs so often in relativity, it is usual for it to be given the shorthand notation

\displaystyle  \gamma\equiv\frac{1}{\sqrt{1-v^{2}}} \ \ \ \ \ (5)

so the Lorentz transformations become

\displaystyle   t_{2} \displaystyle  = \displaystyle  \gamma(t_{1}-vx_{1})\ \ \ \ \ (6)
\displaystyle  x_{2} \displaystyle  = \displaystyle  \gamma(x_{1}-vt_{1})\ \ \ \ \ (7)
\displaystyle  y_{2} \displaystyle  = \displaystyle  y_{1}\ \ \ \ \ (8)
\displaystyle  z_{2} \displaystyle  = \displaystyle  z_{1} \ \ \ \ \ (9)

Now suppose a rod of length {l} is at rest on the {x_{1}} axis. How long does this rod appear to observer {O_{2}}?

The key point in answering this seemingly simple question is that {O_{2}} must measure the length of the rod in such a way that the two events defining the location of the two endpoints of the rod occur at the same time in {O_{2}}‘s frame. That is, if we require both events to occur at a fixed time {t_{2}=0}, from the Lorentz transformation, we must satisfy the condition

\displaystyle   t_{2} \displaystyle  = \displaystyle  0=\gamma(t_{1}-vx_{1})\ \ \ \ \ (10)
\displaystyle  t_{1} \displaystyle  = \displaystyle  vx_{1} \ \ \ \ \ (11)

Since the two frames coincide at {t_{1}=t_{2}=x_{1}=x_{2}=0}, this condition is satisfied for the measurement of one of the end points. At the other end point, {x_{1}=l}, so {t_{1}} must be {t_{1}=vl}, and using the second Lorentz transformation

\displaystyle   x_{2} \displaystyle  = \displaystyle  \gamma(l-v^{2}l)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \gamma l(1-v^{2})\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{l}{\gamma} \ \ \ \ \ (14)

using the definition of {\gamma}. Since {\gamma\ge1}, we see that the length of the rod is measured to be shorter by {O_{2}} which gives rise to the phenomenon of length contraction.

A popular ‘paradox’ that involves length contraction is the pole-in-a-barn problem. Suppose there is a pole of length 20 m and a barn of length 15 m (both measured in their rest frames). If an athlete picks up the pole and runs at a speed {0.8} (giving {\gamma=5/3}) from left to right towards the barn, then (assuming the barn has a front and back door in line with the athlete’s path) to an observer in the barn’s frame of reference, since the pole is contracted to a length of 12 m, it will fit entirely inside the barn. However, in the athlete’s frame of reference, it is the barn which appears contracted to a length of 9 m so the pole will not fit inside the barn. Who is right?

The catch is that the problem is phrased in a way which ignores the concept of simultaneity. In each case, the observer must measure the lengths of the pole and the barn at the same time in their respective frames. This can be seen by using a space-time diagram and the Lorentz transformations. We’ll examine the Lorentz transformations first, and work out the problem in general, assuming that the coordinate systems coincide at the event where the right end of the pole enters the left end of the barn. We’ll look specifically at three events:

  • Event A: the left (back) end of the pole enters the left end of the barn;
  • Event B: the location of the right end of the pole at the same time as event A, as seen by the athlete;
  • Event C: as event B, except in the frame of the barn.

To work out the coordinates of A, it is easiest to start in the frame of the athlete. The pole’s right end passed the left end of the barn at {t_{2}=0}, the pole is length {l} and the athlete is moving at speed {v}, so to him it will take a time {l/v} for the entire length of the pole to pass into the barn. Since the pole is at rest in this frame and its right end has coordinate 0, the left end has coordiante {-l}. Thus the coordinates of A in the athlete’s frame are

\displaystyle  A_{a}=\left(\frac{l}{v},-l\right)=(25,-20) \ \ \ \ \ (15)

We can now use the Lorentz transformations to convert this to the barn’s frame. Remember that the barn is moving at speed {-v} relative to the athlete so we need to use {-v} in the Lorentz formulas. Plugging in the coordinates, we get

\displaystyle  A_{b}=\left(\gamma\left(\frac{l}{v}-lv\right),0\right)=(15,0) \ \ \ \ \ (16)

Note as a check that the {x_{b}} coordinate is zero, since the left end of the barn is fixed at that point in the barn’s frame. Note also that in the barn’s frame, event A occurs earlier than in the athlete’s frame.

Now we want to find event B. In the athlete’s frame, we want the location of the front of the pole at the same time as event A, that is at {t_{a}=l/v}. Since the pole is at rest in that frame and its front end defines the location {x_{a}=0}, this is just

\displaystyle  B_{a}=\left(\frac{l}{v},0\right)=(25,0) \ \ \ \ \ (17)

Using the Lorentz transformations, we find the coordinates of B in the barn’s frame:

\displaystyle  B_{b}=\left(\gamma\frac{l}{v},\gamma l\right)=\left(41.67,33.3\right) \ \ \ \ \ (18)

Note that both observers agree that event B occurs well outside the right end of the barn, since the barn is 15 m long in its rest frame, and event B occurs at {x_{b}=33.3} m. In the athlete’s frame, the barn is only 9 m long and the pole is 20 m long, so the right end of the pole will be 11 m outside the right end of the barn.

However, although events A and B occur at the same time to the athlete, they clearly do not occur at the same time to the barn ({t_{b}=15} for event A and {t_{b}=41.67} for event B). So this leads us to event C: the location and time of the right end of the pole at the same time as event A, as measured in the barn’s frame. Since the pole appears to be contracted to length {l/\gamma} in this frame, we get

\displaystyle  C_{b}=\left(\gamma\left(\frac{l}{v}-lv\right),\frac{l}{\gamma}\right)=(15,12) \ \ \ \ \ (19)

Applying Lorentz, we get the coordinates in the athlete’s frame (this time we’re converting to an observer with speed {+v}, so we use {+v} in the Lorentz formulas):

\displaystyle  C_{a}=\left(\frac{l}{v}-vl,0\right)=(9,0) \ \ \ \ \ (20)

Again, both observers agree that the right end of the pole is inside the barn at event C, however the athlete believes that the left end of the pole has not yet entered the barn. Remember that the left end of the pole enters the barn at event A, and for that event {t_{a}=25}, so event C’s time of 9 is well before event A.

The events are summarized in the space-time diagram (remember that the scales along the athlete’s axes are not the same as those along the barn’s axes; you need to use the invariant hyperbolas (which we haven’t drawn here) to calibrate the athlete’s axes):

The diagram shows the situation in the barn’s frame, so the ends of the barn are represented by the world line of the vertical axis for the left end and the blue line for the right end. The world lines of the ends of the pole are the two solid red lines. The dashed line AB is the position of the pole as seen by the athlete (that is, it corresponds to events A and B above). The solid green line AC shows the pole as seen by the barn’s frame (corresponding to events A and C above). Thus to the athlete, if both ends of the pole are always viewed at the same time in his frame, the pole will never fit inside the barn, but in the barn’s frame, viewing the two ends of the pole at the same time does allow it to fit inside the barn. So in a sense, both conclusions are correct; it just depends on which frame you are in.

Actually, the phrasing of the question as to whether the pole fits inside the barn is misleading, since the pole really only can be said to fit inside the barn if they are both at rest, and in this case it clearly won’t fit. The conclusion that the pole somehow fits inside if it is moving very fast is misleading, since the the line AC in the diagram cuts across a set of lines parallel to AB, each of which represents a constant time in the althlete’s frame, so in a sense, the barn observer is sampling a set of infinitesimal slices of the pole, each selected from a different time in the athlete’s frame (the rest frame of the pole). Such a sampling process isn’t really a true reflection of the physical pole itself, so in that sense, the pole doesn’t actually fit inside the barn.

2 thoughts on “Length contraction and the pole-in-a-barn paradox

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