Why Does Lorentz Transformation Not Yield the General Form of Four-Acceleration?

[snoopies622]

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?

 

[lalbatros]

That's because the co-moving frame has a changing speed,
and because a general result cannot be obtained from a special result.
The simple form (0,a) is valid only for a small lapse of time.

Using the full expression for the acceleration at any time in the once-co-moving frame,
would allow you to derive the expression in any other frame.
The general expression is Lorentz invariant.

 

[snoopies622]

...a general result cannot be obtained from a special result.

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 m_0 (c,0) and Lorentz transforms it, it turns into the general case
\gamma m_0 (c,\bf {v})

I will give it more thought.

 

[nitsuj]

I have always pictured the Lorentz transformation as a kind of rotation.

Are you able to explain that? Most people seem to picture & refer to it as a rotation, but I can't envision it.

 

[snoopies622]

Well, if you graph xi (i^2 = -1) 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.