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 the both doors of the barn is closed simultaneously in 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.

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.

• But, Lorentz transformations are consequeneces of Minkowski metric, so I should not neet to do that to derive what I'm trying to derive. There has to be another way. – onurcanbektas Mar 9 '18 at 2:32
• No. The two principles of SR (special relativity), that is the equivalence of inertial reference frames and the constancy of the speed of light plus assumptions of homogeneity and isotropy of vacuum allow for the Lorentz transformation. Then based on the Lorentz transformation you can construct the metric of Minkowski spacetime with the requirement that the four-distance is an invariant. – Michele Grosso Mar 9 '18 at 15:30

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.

• Well, don't get me wrong, but as long as you don't give me why that has to be the case, I cannot even comment on this answer, i.e "this is what happens ..." is not a good argument among physicists. – onurcanbektas Mar 14 at 4:33
• @onurcanbektas : You're not being asked to take anything on faith. All of the above is easy to check by writing down the Lorentz transformation, as Michele Grosso has already said. – WillO Mar 15 at 4:36