Question 39.4: The Pole-in-the-Barn Paradox The twin paradox, discussed ear......

Conceptualize From your everyday experience, you would be surprised to see a 15-m pole fit inside a 10 -m barn, but we are becoming used to surprising results in relativistic situations.

Categorize The pole is in motion with respect to the ground observer so that the observer measures its length to be contracted, whereas the stationary barn has a proper length of 10 \mathrm{~m}. We categorize this example as a length contraction problem.

Analyze Use Equation 39.9

L=\frac{L_p}{\gamma}=L_p \sqrt{1-\frac{v^{2}}{c^{2}}}  (39.9)

to find the contracted length of the pole according to the ground observer:

L_{\text {pole }}=L_{p} \sqrt{1-\frac{v^{2}}{c^{2}}}=(15 \mathrm{~m}) \sqrt{1-(0.75)^{2}}=9.9 \mathrm{~m}

Therefore, the ground observer measures the pole to be slightly shorter than the barn and there is no problem with momentarily capturing the pole inside it. The “paradox” arises when we consider the runner’s point of view.

Use Equation 39.9 to find the contracted length of the barn according to the running observer:

L_{\text {barn }}=L_{p} \sqrt{1-\frac{v^{2}}{c^{2}}}=(10 \mathrm{~m}) \sqrt{1-(0.75)^{2}}=6.6 \mathrm{~m}

Because the pole is in the rest frame of the runner, the runner measures it to have its proper length of 15 \mathrm{~m}. Now the situation looks even worse: how can a 15-m pole fit inside a 6.6-m barn? Although this question is the classic one that is often asked, it is not the question we have asked because it is not the important one. We asked, “Does the runner make it safely through the barn?”

The resolution of the “paradox” lies in the relativity of simultaneity. The closing of the two doors is measured to be simultaneous by the ground observer. Because the doors are at different positions, however, they do not close simultaneously as measured by the runner. The rear door closes and then opens first, allowing the leading end of the pole to exit. The front door of the barn does not close until the trailing end of the pole passes by.

We can analyze this “paradox” using a space-time graph. Figure 39.12 \mathrm{a} is a space-time graph from the ground observer’s point of view. We choose x=0 as the position of the front doorway of the barn and t=0 as the instant at which the leading end of the pole is located at the front doorway of the barn. The world-lines for the two doorways of the barn are separated by 10 \mathrm{~m} and are vertical because the barn is not moving relative to this observer. For the pole, we follow two tilted worldlines, one for each end of the moving pole. These world-lines are 9.9 \mathrm{~m} apart horizontally, which is the contracted length seen by the ground observer. As seen in Figure 39.12a, the pole is entirely within the barn at some time. Figure 39.12 \mathrm{~b} shows the space-time graph according to the runner. Here, the world-lines for the pole are separated by 15 \mathrm{~m} and are vertical because the pole is at rest in the runner’s frame of reference. The barn is hurtling toward the runner, so the world-lines for the front and rear doorways of the barn are tilted to the left. The world-lines for the barn are separated by 6.6 \mathrm{~m}, the contracted length as seen by the runner. The leading end of the pole leaves the rear doorway of the barn long before the trailing end of the pole enters the barn. Therefore, the opening of the rear door occurs before the closing of the front door.

From the ground observer’s point of view, use the particle under constant velocity model to find the time after t=0 at which the trailing end of the pole enters the barn:

(1) t=\frac{\Delta x}{v}=\frac{9.9 \mathrm{~m}}{0.75 c}=\frac{13.2 \mathrm{~m}}{c}

From the runner’s point of view, use the particle under constant velocity model to find the time at which the leading end of the pole leaves the barn:

(2) t=\frac{\Delta x}{v}=\frac{6.6 \mathrm{~m}}{0.75 c}=\frac{8.8 \mathrm{~m}}{c}

Find the time at which the trailing end of the pole enters the front door of the barn:

(3) t=\frac{\Delta x}{v}=\frac{15 \mathrm{~m}}{0.75 c}=\frac{20 \mathrm{~m}}{c}

Finalize From Equation (1), the pole should be completely inside the barn at a time corresponding to c t=13.2 m. This situation is consistent with the point on the c t axis in Figure 39.12 \mathrm{a} where the pole is inside the barn. From Equation (2), the leading end of the pole leaves the barn at c t=8.8 \mathrm{~m}. This situation is consistent with the point on the c t axis in Figure 39.12 \mathrm{~b} where the rear doorway of the barn arrives at the leading end of the pole. Equation (3) gives c t=20 m, which agrees with the instant shown in Figure 39.12 \mathrm{~b} at which the front doorway of the barn arrives at the trailing end of the pole.