Solving the ladder paradox: Minkowski distance leads to a contradiction

Question

Consider the ladder paradox;

Let the length of the barn in its fixed frame be $80m$, and consider the ladder's reference frame which is moving toward the barn with a speed $v$, and in ladder's frame the length of the barn is $20m$.

Now, since both doors of the barn are closed simultaneously in the barn's fixed frame, if we were to write the Minkowski distance between these events, i.e the events that the doors are shut, we get

$$c^2 \Delta t_B^2 - \Delta x_B ^2 = c^2 \Delta t_L ^2 - \Delta x_L^2$$ where $\Delta t_B^2 = 0, \quad \Delta x_B^2 = (80m)^2, \quad \Delta x_L^2 = (20m)^2$, but when we plug these values in the the equation above, we get

$$\Delta t_L^2 < 0,$$ which is not possible.

So where is my mistake in my reasoning? I mean since the doors of the barn are moving with speed $-v$ relative the ladder, the distance between the doors should be smaller, due to length contraction, but this leads to a weird thing as I have explained above.

Answer 1

The inconsistency is when you assume the $\Delta x_L$ equal to the contracted length of the barn. The contracted length is what you measure with simultaneity in the ladder's reference frame, however here you assume simultaneity in the barn's reference frame. As for both temporal and spatial difference in ladder's frame you have to apply the Lorentz transformation.

Answer 2

Michele Grosso has given you the right answer, but let me spell this out: In the ladder's frame, the barn is moving (say leftward). And in that frame, the time sequence is this:

First Door 1 (the door on the right) closes. At that moment, Door 2 is 20 meters to the left of Door 1.

Then some time passes, during which time the entire barn (including Door 2) moves leftward.

Then Door 2 closes, at a location that is considerably more than 20m to the left of the location where Door 1 closed. So your calculation of $\Delta x_L$ is wrong.

Answer 3

As both previous answers have correctly identified, your mistake is in assuming that $\Delta x_L=20$. In your scenario $\gamma = 4$ which implies that $v=0.968 c$. With that information it is easy to apply the Lorentz transform. For $(t_B c,x_B)=(0,80)$ transforming to the ladder frame gives $(t_L c,x_L)=(310,320)$. Plugging those values back into the spacetime interval formula indeed gives the correct value for the spacetime interval. So $\Delta x_L=320\ne 20$

Hopefully with 3 answers all identifying the same mistake you will be more willing to accept that as correct.