Answer
See the detailed answer below.
Work Step by Step
Let's assume that the farmer's frame is $\rm S$ and the pole's frame is $\rm S'$, where $\rm S'$ moves at a speed of $v=0.888c$ relative to $\rm S$.
$\bullet$ Now let's start with the framer's point of view, according to his reference frame $\rm S$.
The farmer sees that the pole has shrunk to a length of
$$L =16\sqrt{1-\dfrac{v^2}{c^2}}=16\sqrt{1-\dfrac{0.866^2c^2}{c^2}}=\bf 8.001\;\rm m$$
So, from his point of view, he can close the two doors simultaneously without breaking the pole.
Now we have two events, closing the front door when the whole pole is inside the barn at which the pole's end is just at the front door.
The coordinate of this event is $\rm (0\;m,0\;s)$.
The second event which occurs simultaneously with this first event is closing the back door, so the coordinate of this event is $\rm (10\;m,0\;s)$.
According to the farmer, the front end of the pole is 2 m away from the back door and the back end of the pole is at the front door.
We chose the front door to be our origin for the farmer frame, so the front of the pole is at $\rm (8\;m,0\;s)$.
Now we need to find the coordinate of the front pole when it hits the back door after moving these 2 meters.
We know its position, so we need to find the time,
$$t =\dfrac{x}{t}=\dfrac{2 }{0.866\times 3\times 10^8}=\bf 7.7\;\rm ns$$
As an abstract, we have now 3 events,
1- Closing the front door at $(x_1,t_1)=\rm (0\;m,0\;s)$.
2- Closing the back door at $(x_2,t_2)=\rm (10\;m,0\;s)$.
3- The front of the pole hits the back door at $(x_3,t_3)=\rm (10\;m,7.7\; ns)$.
$\bullet$$\bullet$ Now let's start with the pole vaulter's point of view, according to his reference frame $\rm S'$.
The first event has the same coordinates,
1'- Closing the front door at $(x'_1,t'_1)=\rm (0\;m,0\;s)$.
The second event occurs at
$$x_2'=\gamma(x-vt)=\left[1-\dfrac{v^2}{c^2}\right]^{-1/2}(x_2-vt_2)$$
$$x_2'= \left[1-\dfrac{0.866^2c^2}{c^2}\right]^{-1/2}(10-0.866c (0))$$
$$x_2'=\bf 20\;\rm m$$
$$t_2'=\gamma\left(t_2-\dfrac{vx_2}{c^2}\right)$$
$$t_2'= \left[1-\dfrac{0.866^2c^2}{c^2}\right]^{-1/2}\left(0-\dfrac{(0.866c)(10)}{c^2}\right)=\bf -57.7\;\rm ns$$
This means that according to the pole vaulter's frame $\rm S'$, the back door closed before the front door by a time of 57.7 ns. And it closed at $x'=20$ mw here the barn length, according to $\rm S'$, is 5 m long which means that the back door is at $x'=20$ m, and the front door is then at $x'=15$ m.
The length of the pole, according to this frame, is 16 m which means that the front of the pole is 1 m inside the front door and away 4 m from the back door.
This 4 meters are shrunk to 2 m in $\rm S$ frame, $L=4\sqrt{1-\dfrac{0.866^2c^2}{c^2}}=\bf 2\;\rm m$
2'- Closing the back door at $(x'_2,t'_2)=\rm (20\;m,-57.7\;ns)$.
After the moment the right door closes, the pole moves toward it for 57.7 ns and this means that the front of the pole is now at
$$\Delta x'=vt=0.866c(57.7)\rm n=\bf 15\;\rm m$$
which is the same distance to be at the back end of the pole at $t'_1=0$ ns.
According to $\rm S'$, the back door moves toward the front end of the pole and will hit it at $x'=16$ m.
This means that it will travel 4 meters through a time of
$$\Delta t_3=\dfrac{4}{0.866\times 3\times 10^8}=\bf 15.4\;\rm ns$$
where the door closes at -57.7 ns, hence the door hits the front end of the pole at
$$t_3'=15.4-57.7=\bf 42.3\;\rm ns$$
3'- The front of the pole hits the back door at $(x'_3,t'_3)=\rm (16\;m,42.3\; ns)$.
$\bullet\bullet\bullet$ Now let's back to frame $\rm S$; to test the third event $(x'_3,t'_3)$,
$$x_3'=\gamma(x_3-vt_3)$$
$$x_3'= \left[1-\dfrac{0.866^2c^2}{c^2}\right]^{-1/2}(10-[0.866\times 3\times 10^8\times 7.7\times 10^{-9}])=\bf 16\;\rm m$$
$$t_3'=\gamma\left(t_3-\dfrac{vx_3}{c^2}\right)$$
$$t_3'= \left[1-\dfrac{0.866^2c^2}{c^2}\right]^{-1/2}\left(7.7\times 10^{-9}-\dfrac{(0.866c)(10)}{c^2}\right)=\bf -42.3\;\rm ns$$
Now we have a perfect match.
Therefore, from all the above,
- The farmer can close the two doors without breaking the pole since to him the pole is shrunk to less than the length of the barn.
- The pole vaulter finds that the farmer can close both doors without breaking the pole because event 2, which is closing the back door occurs, before event 1, which is closing the front door.
- These two events [closing the two doors] are not causally related, so these two can occur in a different order in the two reference frames.
