Pole and Barn Paradox w/ Spacetime Interval

I'm having trouble with a pole and barn paradox problem. The problem is as follows:

A pole vaulter is running with a pole at $v=\frac{\sqrt3}{2}c$. Her pole has a proper length of $L$. She runs into a barn with proper length $\frac{L}{2}$ with doors on the front and back. When the pole vaulter runs into the barn, a farmer tries to close both front and back doors at the same time, but only for an instant, and then reopens them.

What is the expression for the time interval of the door closings in the pole vaulter's frame?

So I know that $\gamma=2$, so in the pole vaulter's frame, the barn has length $\frac{L}{4}$, and in the farmer's frame, the pole has length $\frac{L}{2}$.

In order to find the time interval, I tried using the spacetime interval. In the farmer's frame, the time interval between the two doors closing is 0. This means that the spacetime interval between the two events is $$(\Delta s) ^2=-\left(\frac{L}{2}\right)^2$$

Then, I equate this to the spacetime interval from the pole vaulter's frame. $$(\Delta s) ^2=(c\Delta t)^2-(\Delta x)^2 = (c\Delta t)^2 - \left(\frac{L}{4}\right)^2 = -\left(\frac{L}{2}\right)^2$$

But solving for $\Delta t$ gives a complex number, when I should be getting a real solution. Why am I getting this result? Does it have to do with how I am selecting my $\Delta x$ for the pole vaulter?

Often, I think a nice way to untangle the mess we find ourselves in is by appealing to the all-knowing God of special relativity, blessed be her name, Lorentztransformalia.

Let event $A$ denote the front door closing, and let event $B$ denote the rear door closing. Without loss of generality, we assume that we've chosen our coordinates so that in the farmer's frame, event $A$ has spacetime coordinates $(ct_A, x_A) = (0,0)$, and event $B$ has spacetime coordinates $(ct_B, x_B) = (0, L/2)$.

Using the standard Lorentz boost, we find that the coordinates of the first event in the vaulter's frame are again $(ct_A', x_A') = (0,0)$ because linear transformations always map the zero vector to itself, while

$$\begin{pmatrix} ct_B' \\ x_B'\end{pmatrix} = \begin{pmatrix} \gamma & -\gamma\beta \\ -\gamma\beta & \gamma\end{pmatrix} \begin{pmatrix} 0 \\ L/2 \end{pmatrix} = \begin{pmatrix} 2 & -\sqrt{3} \\ -\sqrt{3} & 2\end{pmatrix} \begin{pmatrix} 0 \\ L/2 \end{pmatrix} = \begin{pmatrix} -\sqrt{3}L/2 \\ L\end{pmatrix}$$

In particular, the spatial separation of the events in the farmer's frame is

$$x_B' - x_A' = L \neq L/4$$

hmmm...so it seems the original assumption that the spatial separation of these events equals $L/4$ in the vaulter's frame is not correct.

• As usual thinking about 'length contraction' gets confounded because length measurements assume that thing happen simultaneously which they do in only one of a pair of frames in motion relative one another. But given the nature of the title, the point may be to find the result without appealing to the Lorentz transform, which means constructing the space-time diagram (which is, of course, equivalent in information content). – dmckee May 5 '16 at 4:17
• @dmckee Yes indeed. Given that both the time and space separations change in unintuitive ways, it seems the original method of the OP isn't so straightforward to implement without additional information that amounts to just transforming. Maybe there's something more clever that I'm missing... – joshphysics May 5 '16 at 4:41

You caught that the barn doors do not close and open at the same time in the vaulter's reference frame, but missed the issue that they are moving in her frame.

So, the difference in position between where they were when they flashed shut will not be L/4, but L/4 + vΔt.

Include that, and you should be able to solve it.