Time in the Lorentz transformation

1. Aug 22, 2005

asdf1

where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?

2. Aug 22, 2005

Staff Emeritus
From
$$\left( \begin{array}{c} x' \\ t' \end{array} \right) = \left( \begin{array}{cc} cosh(\theta) & -sinh(\theta) \\ sinh(\theta) & cosh(\theta) \end{array} \right) \left( \begin{array}{c} x \\ t \end{array} \right)$$

where $$v = tanh(\theta)$$ and c is taken to be 1.

3. Aug 22, 2005

Perspicacious

There are many elegant ways to derive the Lorentz transformation:

http://www.everythingimportant.org/relativity/
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000043000005000434000001 [Broken]
http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf [Broken]

Last edited by a moderator: May 2, 2017
4. Aug 22, 2005

robphy

It should be $$+\sinh(\theta)$$.
(The determinant has to be 1.)

5. Aug 22, 2005

learningphysics

Or rather they should both be $$-\sinh(\theta)$$ I believe.

6. Aug 22, 2005

robphy

Yes, of course :tongue2: , considering the original post. Thanks.

7. Aug 26, 2005

thanks! :)