Deriving Lorentz transformation

  • #1

Main Question or Discussion Point

Hello,

I have few question for deriving the Lorentz transformation (LT):

While deriving the LT, we draw a graph as x,y,z in one frame of reference and x',y',z' in the other frame of reference as S and S' as two frames of reference.

Now the factor ct comes in, which is the flash of light. Now we do:

x^2+y^2+z^2=c^2t^2.

(1) The ct factor is speed of light multiplied by time as ct, right?
(2) the square factor of x^2+y^2........How do we calculate? Are we using Pythagoras' theorem in this case?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,848
1,645
You have to draw the trajectory of the light-ray in the primed and unprimed frames. i.e. you need two diagrams. One for the unprimed observer and one for the primed observer.

Usually the frames are set up so the trajectory is just a line in the unprimed frame and triangular in the primed frame. Which is how Pythagoras comes in.

Work out the equation for the time period between emmission and reception of the light pulse in each frame.
Classically you would assume the two observers agree about the time period but disagree about the speed of light (one of them would say, "yeah, but you're the one moving.") In relativity, the observers agree about the speed of light which means they must disagree about the time period. So put the speed of light the same in both cases and compare the lightspeeds.
 
  • Like
Likes 1 person

Related Threads on Deriving Lorentz transformation

  • Last Post
4
Replies
78
Views
4K
  • Last Post
Replies
4
Views
2K
Replies
6
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
2
Replies
27
Views
3K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
22
Views
1K
  • Last Post
Replies
4
Views
948
Replies
4
Views
556
Top