- #1

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**a**).

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

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- Thread starter snoopies622
- Start date

- #1

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

- #2

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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

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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

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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

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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.

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