Boosting four-acceleration

In an instantaneous co-moving inertial frame, the four-acceleration vector reduces to (0,a).

Why then does applying a Lorentz transformation to the above vector not produce the general form?

 

Yes, that makes sense to me. The problem I'm having is one of imagination. I have always pictured the Lorentz transformation as a kind of rotation. So I imagine a (0,a) vector being rotated from one position to another, and a one-to-one correspondence between a vector in the v=0 frame and the set of the same vector in all the other frames.

I know it works for four-momentum. That is, if one starts with the specific v=0 case [itex] m_0 (c,0) [/itex] and Lorentz transforms it, it turns into the general case
[tex] \gamma m_0 (c,\bf {v}) [/tex]

I will give it more thought.

 

Well, if you graph xi ([itex] i^2 = -1 [/itex]) on the horizontal axis versus ct on the vertical, then it's a rotation, and the angle is the rapidity. I suspect that some find that idea objectionable - it's misleading in some ways - but I still like it.

Correction: the angle is rapidity times i. Pretty weird, yes.