Time in the Lorentz transformation

  • Thread starter asdf1
  • Start date
734
0
where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?
 

selfAdjoint

Staff Emeritus
Gold Member
Dearly Missed
6,764
5
asdf1 said:
where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?
From
[tex] \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) [/tex]

where [tex]v = tanh(\theta)[/tex] and c is taken to be 1.
 
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

Science Advisor
Homework Helper
Insights Author
Gold Member
5,378
657
It should be [tex]+\sinh(\theta)[/tex].
(The determinant has to be 1.)
 

learningphysics

Homework Helper
4,099
5
robphy said:
It should be [tex]+\sinh(\theta)[/tex].
(The determinant has to be 1.)
Or rather they should both be [tex]-\sinh(\theta)[/tex] I believe.
 

robphy

Science Advisor
Homework Helper
Insights Author
Gold Member
5,378
657
learningphysics said:
Or rather they should both be [tex]-\sinh(\theta)[/tex] I believe.
Yes, of course :tongue2: , considering the original post. Thanks.
 
734
0
thanks! :)
 

Related Threads for: Time in the Lorentz transformation

  • Posted
Replies
12
Views
3K
Replies
4
Views
1K
Replies
46
Views
3K
Replies
7
Views
1K
Replies
41
Views
5K

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

Hot Threads

Top