The next problem is, as I understand, a spin on the ladder paradox which is extensively documented everywhere. There are two rockets ($$A$$ and $$B$$), of length $$L_p=1$$m which are traveling in opposite parallel trajectories (the distance of which are negligible). From the frame of $$A$$, $$B$$ is getting closer at a speed of $$u$$, and vice versa.

According to $$A$$, when the tail of $$B$$ reaches the front of $$A$$, a missile is launched from tail $$A$$ to rocket $$B$$. It is clear, according to the text, that the missile will fail, due to the length contraction of $$B$$. However, if we look at this from $$B$$, we have that the missile would hit $$B$$ since $$A$$ is now shorter.

The exercise asks me to point which alternative (a or b) is correct. In the ladder paradox, to study the simultaneity of two events on the $$A$$ system you first have to write them down and then Lorentz-transform then to see how $$B$$ perceives it. I consider 5 events, with front $$B$$ as the position of B:

1. Both fronts coincide, at (0,0)
2. Tail $$B$$ coincides with front $$A$$.
3. A missile is shot from tail $$A$$.
4. Front $$B$$ coincides with tail $$A$$.
5. Both tails coincide.

from which, I think, only 2. and 3. are relevant (right?). Both events have $$(ct,x)$$ coordinates $$(ct,x)|_2 =(\frac{L_p}{\gamma\beta},\frac{L_p}{\gamma})$$ and $$(ct,x)|_3 =(\frac{L_p}{\gamma\beta},L_p)$$ (right?). Is it valid to choose events which happen at different points/bodies simultaneously? From here do I have to Lorentz-transform the coordinates and find that they happen separately?

How do I solve the question of if the missile hits $$B$$ or not?

• I think this is more of a variation on the pole in the barn "paradox" rather than the ladder.
– user226006
Commented Sep 12, 2019 at 20:03
• In Wikipedia they are treated as the same. en.wikipedia.org/wiki/Ladder_paradox Commented Sep 12, 2019 at 20:19

I suggest you understand this problem by finding some graph paper and constructing spacetime diagrams from the reference frames of both A and B. You'll have four worldlines to contend with, one each for the head and tail of A and B. In A's reference frame, the worldlines for A's head and tail will be vertical, and the horizontal (that is, simultaneous for A) distance between the slanted worldlines for B's head and tail will be shorter than the rest length for B due to length contraction; simultaneous events in A's frame will appear on the same horizontal spacelike axis. In B's reference frame the situation is reversed, including the critical point that events which A believes are simultaneous are aligned on a slanted spacelike axis (labeled $$x'$$ in most explanations for spacetime diagrams).
I'll save you a little guessing and suggest that you have your two ships approach each other, relative to you, at $$\beta = v/c = 1/2$$. That makes their relative speeds $$\beta_\text{rel} = (\beta_1+\beta_2)/(1+\beta_1\beta_2)$$ and relativistic factors $$\gamma_\text{rel} = (1-\beta_\text{rel}^2)^{-1/2}$$ into nice rational numbers, and you can construct the spacetime diagram by counting squares on graph paper rather than getting lost in a bunch of algebra and arithmetic.