# Time in the Lorentz transformation

#### asdf1

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

Related Special and General Relativity News on Phys.org

Staff Emeritus
Gold Member
Dearly Missed
asdf1 said:
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

asdf1 said:
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 [Broken]
http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf [Broken]

Last edited by a moderator:

#### robphy

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

#### learningphysics

Homework Helper
robphy said:
It should be $$+\sinh(\theta)$$.
(The determinant has to be 1.)
Or rather they should both be $$-\sinh(\theta)$$ I believe.

#### robphy

Homework Helper
Gold Member
learningphysics said:
Or rather they should both be $$-\sinh(\theta)$$ I believe.
Yes, of course :tongue2: , considering the original post. Thanks.

thanks! :)

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving