# Generalizing the Galilean law of addition of velocities using the Lorentz transformation [closed]

I am reading about how to generalize the Galilean law of addition of velocities using the Lorentz transformation, but I am confused about one step.

Here, I have the following equations for Lorentz transformation: $$\overline{t} = \frac{t}{\sqrt{1-v^2}} - \frac{vx}{\sqrt{1-v^2}}$$ $$\overline{x} = \frac{-vt}{\sqrt{1-v^2}} + \frac{x}{\sqrt{1-v^2}}$$ $$\overline{y} = y$$ $$\overline{z} = z$$

Suppose a particle has speed W in the $$\overline{x}$$ direction of $$\overline{O}$$, so $$\frac{\Delta\overline{x}}{\Delta\overline{t}} = W$$. In the other frame O, its velocity is $$\frac{\Delta\overline{x}}{\Delta\overline{t}} = W'$$. Then, we can deduce $$\Delta x$$ and $$\Delta t$$ from the Lorentz transformation. Suppose $$\overline{O}$$ moves with velocity v with respect to O, then the above equations imply that $$\Delta x = \frac{\Delta\overline{x} + v\Delta\overline{t}}{\sqrt{1-v^2}}$$ $$\Delta t = \frac{\Delta\overline{t} + v\Delta\overline{x}}{\sqrt{1-v^2}}$$ I don't see how these two equations are derived. From the equations, I can only derive $$\Delta \overline x = \frac {\Delta x - v\Delta t}{\sqrt{1-v^2}}$$ $$\Delta \overline t = \frac {\Delta t - v\Delta x}{\sqrt{1-v^2}}$$

• You have two linear equations for two unknowns. Solve them. Commented Mar 2 at 6:46
• Oh I see. Thanks for the comment!@Ghoster
– Gene
Commented Mar 2 at 7:00