# The Derivative of Lorentz Transformation

#### Hyperreality

Lorentz Transformation says

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

Does its derivative

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

Last edited:
Related Special and General Relativity News on Phys.org
H

#### hemmul

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

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

### 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