- #1

- 753

- 15

**a**).

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter snoopies622
- Start date

- #1

- 753

- 15

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

- #2

- 1,227

- 2

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.

- #3

- 753

- 15

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 general result cannot be obtained from a special result.

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.

- #4

- 1,352

- 90

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.

- #5

- 753

- 15

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.

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

Last edited:

Share: