A moving rod; two Lorentz boosts compared with one

[Hiero]

@Ibix Wow! I thought you were giving an analogy in the first post, I didn't realize you were describing minkowski diagrams! That is quite a powerful visualization, very well made and captures the heart of the issue. I'm really impressed by this perspective, thank you.

 

[PeterDonis]

All I have is the 'hand wavey' reasoning that the perpendicular space should not get mixed in with the plane of the boost.

Yes, the rotation must be in the plane of the two boost directions, i.e., the x-y plane. (Why? Think about the composition of the two boosts; what transformation matrix components can be affected?) That means it must be about the z axis. The article @Dale linked to basically says the same thing, just with a lot more math.

 

[sweet springs]

The point is, from the proper frame, we are boosting off at some angle with the rod's length, (not zero and not perpendicular) and as far as I can tell this necessarily causes the rod to be rotated when changing frames (whereas the analysis with an intermediate frame A told me that the rod remains along x).

By boost of arbitrary direction, Lorentz contract works on boost direction of rod, so it has to rotate. Similarly if we put three rods assembled as XYZ axis or a cube box such an arbitrary direction boost, the cube becomes a moving parallelepiped.

Say rod proper frame is R, Lorentz transformations A→R and A→B keep rods vertical, but R→(A)→B does not.

 

[Dale]

The thing which bothers me is how should two vectors which are parallel to an intermediate vector not be parallel to each other?

The rod is not a vector. It is a worldsheet. Different frames select completely different lines on that worldsheet, and as a result the direction of the rod does not transform as a vector.

For flat spacetime and for actual vectors, two vectors which are parallel to an intermediate vector are indeed parallel to each other in all frames.

 

[Hiero]

The rod is not a vector.

I didn't mean that. The three vectors I was speaking of were the unit vectors in the x, x', and x'' directions; x vector is parallel to x' vector and x'' vector is parallel to x' (all this in the x' frame though).

I've come to grips with it though because all this parallel-ness is measured in the x' frame in which x'' and x are actually parallel. They only become non-parallel in the other two frames which doesn't violate that transitive rule of logic because they're no longer parallel to an intermediate vector in those frames.

 

[Dale]

The three vectors I was speaking of

Oh, I also thought you were speaking of four vectors.