Time in the Lorentz transformation

[asdf1]

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

 

[selfAdjoint]

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

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.

 

[Perspicacious]

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

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
http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf

 

[robphy]

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

 

[learningphysics]

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

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

 

[robphy]

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

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

 

[asdf1]

thanks! :)