The Derivative of Lorentz Transformation

  • #1
202
0

Main Question or Discussion Point

Lorentz Transformation says

[tex]x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}[/tex]

Does its derivative

[tex]\frac{dx'}{dt}=\frac{dx/dt-u}{\sqrt{1-u^2/c^2}}[/tex]

give the velocity addition equation?
 
Last edited:

Answers and Replies

  • #2
hemmul
Hyperreality said:
Lorentz Transformation says
[tex]x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}[/tex]
Does its derivative
[tex]\frac{dx'}{dt}=\frac{dx/dt-u}{\sqrt{1-u^2/c^2}}[/tex]
give the velocity addition equation?
Note, that Lorentz transformations link the coordinates in two RFs: S and S'. Velocity in S is dx/dt, while that in S' will be dx'/dt'
The derivatives like dx'/dt or dx/dt' - do not make any sense - because they consist of values from different RFs...
So to derive the velocity formula, you have to find first the expressions for dx and dt, and then divide them one on the other. Regrouping the terms it is possible to express it via velocity in the neighbour RF and their relative velocities...

[tex]v=\frac{dx}{dt}}[/tex]
[tex]dx'=\frac{dx-u*dt}{\sqrt{1-u^2/c^2}}[/tex]
[tex]dt'=\frac{dt-u*dx/c^2}{\sqrt{1-u^2/c^2}}[/tex]
[tex]v'=\frac{dx'}{dt'}=\frac{v-u}{1-u*v/c^2}[/tex]
 

Related Threads on The Derivative of Lorentz Transformation

Replies
1
Views
562
Replies
16
Views
5K
Replies
28
Views
2K
  • Last Post
2
Replies
27
Views
3K
Replies
4
Views
569
Replies
6
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
15K
Top