# How do u get Eq3 from Eq2&Eq1 [closed]

I keep getting the wrong results by subbing eq1 into eq2 It’s from Kleppner’s an introduction to mechanics 2nd edition

## closed as off-topic by Bob D, ZeroTheHero, tpg2114♦May 20 at 10:49

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The quoted text effectively combined two Lorentz transformations into one single transformation to identify the factor $$\beta$$, which is related to the velocity by $$\beta=\frac{v}{c}$$, thus the relativistic velocity addition formula. There are a few mistakes, so I'll give you the full proof (this is equivalent to the text).

The first Lorentz transformation can be expressed by a matrix

$$\Lambda_1 = \begin{bmatrix} \gamma_1 & -\gamma_1 \beta_1 \\ -\gamma_1 \beta_1 & \gamma_1 \\ \end{bmatrix}, \begin{bmatrix}u_1'\\u_4'\end{bmatrix} = \Lambda_1 \begin{bmatrix}u_1\\u_4\end{bmatrix}.$$

Similarly for the second transformation we have

$$\Lambda_2 = \begin{bmatrix} \gamma_2 & -\gamma_2 \beta_2 \\ -\gamma_2 \beta_2 & \gamma_2 \\ \end{bmatrix}, \begin{bmatrix}u_1''\\u_4''\end{bmatrix} = \Lambda_2 \begin{bmatrix}u_1'\\u_4'\end{bmatrix}.$$

The matrix representation for the combined boost is therefore $$\Lambda_3 = \Lambda_2 \Lambda_1 = \begin{bmatrix} \gamma_1\gamma_2 (1+\beta_1\beta_2) & -\gamma_1\gamma_2(\beta_1+\beta_2) \\ -\gamma_1\gamma_2(\beta_1+\beta_2) & \gamma_1\gamma_2(1+\beta_1\beta_2) \\ \end{bmatrix}.$$

From this it is clear that

$$\gamma_3 = \gamma_1\gamma_2(1+\beta_1\beta_2)$$

and

$$\beta_3 = \frac{\beta_1+\beta_2}{1+\beta_1\beta_2}.$$