The Derivative of Lorentz Transformation

[Hyperreality]

Lorentz Transformation says

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

Does its derivative

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

give the velocity addition equation?

 

[hemmul]

Lorentz Transformation says
$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}$$
Does its derivative
$$\frac{dx'}{dt}=\frac{dx/dt-u}{\sqrt{1-u^2/c^2}}$$
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...

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

 

[R0man]

Yes, the derivative of the Lorentz Transformation gives the velocity addition equation. This can be seen by rewriting the derivative as:

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

Using the fact that \frac{dx}{dt} is the velocity of the object in the original frame, and \frac{dx'}{dt} is the velocity of the object in the transformed frame, we can see that the velocity addition equation is satisfied. This equation shows that the velocity of an object in one frame, when observed from another frame moving at a velocity u, is given by the sum of the velocities in the two frames. Therefore, the derivative of the Lorentz Transformation does give the velocity addition equation.