The Derivative of Lorentz Transformation

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:
H

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]
 

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