# Relative Lorentz factor for two shells

Shell 1 is travelling at a speed $v_1$ with Lorentz factor $\Gamma_1$, and shell 2 at speed $v_2$ with Lorentz factor $\Gamma_2$. Just before the two shells meet, the relative lorentz factor $\Gamma_r$ is given by: $$\gamma_r = \Gamma_1 \Gamma_2 (1-v_1v_2/c^2)$$

as detailed in equation 141 of page 102 here.

Can anyone explain why this should be so?

## 2 Answers

Use the Lorentz Boost parameter ($\lambda$) for each shell, where

$$v/c=\tanh(\lambda)$$ $$\gamma = \cosh(\lambda)$$ $$\beta \gamma = \sinh(\lambda)$$

For boosts in the same direction the boost parameters just add or subtract. So $\lambda_1 -\lambda_2$ is the boost with which one shell is approaching the other. Consider the trigonometric identity

$$\cosh(\lambda_1 -\lambda_2)= \cosh(\lambda_1) \cosh(\lambda_2)- \sinh(\lambda_1) \sinh(\lambda_2)$$ $$= \cosh(\lambda_1) \cosh(\lambda_2)(1- \tanh(\lambda_1) \tanh(\lambda_2))$$

Then putting $\gamma$ and $v/c$ in for cosh and tanh gives

$$\gamma_{12}=\gamma_1 \gamma_2 (1-(v_1/c)(v_2/c))$$

Using Lorentz Boost parameters makes most Special Relativity problems much easier!

Lorentz Transformation $\:O x y z \longrightarrow O_1 x_1 y_1 z_1\:$ and its inverse $\:O_1 x_1 y_1 z_1 \longrightarrow O x y z \:$, $\:\left(y_1 = y, z_1 = z\right)\:$

$$\begin{bmatrix} x_1\\ \\ \\ c t_1 \end{bmatrix} = \gamma_1 \begin{bmatrix} 1 & -\dfrac{v_1}{c}\\ & \\ -\dfrac{v_1}{c} & 1 \end{bmatrix} \begin{bmatrix} x\\ \\ \\ c t \end{bmatrix} \Longleftrightarrow \begin{bmatrix} x\\ \\ \\ c t \end{bmatrix} =\gamma_1 \begin{bmatrix} 1 & \dfrac{v_1}{c}\\ & \\ \dfrac{v_1}{c} & 1 \end{bmatrix} \begin{bmatrix} x_1\\ \\ \\ c t_1 \end{bmatrix} \tag{01}$$ Lorentz Transformation $\:O x y z \longrightarrow O_2 x_2 y_2 z_2\:$, $\:\left(y_2 = y, z_2 = z\right)\:$ and its composition with $\:O_1 x_1 y_1 z_1 \longrightarrow O x y z \:$

$$\begin{bmatrix} x_2\\ \\ \\ c t_2 \end{bmatrix} = \gamma_2 \begin{bmatrix} 1 & -\dfrac{v_2}{c}\\ & \\ -\dfrac{v_2}{c} & 1 \end{bmatrix} \begin{bmatrix} x\\ \\ \\ c t \end{bmatrix} =\gamma_2 \begin{bmatrix} 1 & -\dfrac{v_2}{c}\\ & \\ -\dfrac{v_2}{c^2} & 1 \end{bmatrix} \gamma_1 \begin{bmatrix} 1 & \dfrac{v_1}{c}\\ & \\ \dfrac{v_1}{c} & 1 \end{bmatrix} \begin{bmatrix} x_1\\ \\ \\ c t_1 \end{bmatrix} \tag{02}$$ So,

Lorentz Transformation $\:O_1 x_1 y_1 z_1 \longrightarrow O_2 x_2 y_2 z_2\:$, $\:\left(y_2 = y_1, z_2 = z_1\right)\:$ $$\begin{bmatrix} x_2\\ \\ \\ \\ \\ c t_2 \end{bmatrix} = \gamma_1 \gamma_2\left(1-\dfrac{v_1 v_2}{c^2} \right) \begin{bmatrix} 1 & - \dfrac {\left(v_2 - v_1\right)}{c\left(1-\dfrac{v_1 v_2}{c^2}\right)}\\ & \\ - \dfrac {\left(v_2 - v_1\right)}{c\left(1-\dfrac{v_1 v_2}{c^2}\right)} & 1 \end{bmatrix} \begin{bmatrix} x_1\\ \\ \\ \\ \\ c t_1 \end{bmatrix} \tag{03}$$ that is the velocity of $\:O_2 x_2 y_2 z_2\:$ relative to $\:O_1 x_1 y_1 z_1\:$ is

$$v_{2\: relative\; to\: 1}= v_{21}=\dfrac {v_2 - v_1}{1-\dfrac{v_1 v_2}{c^2} } \tag{04}$$ with $\:\gamma-$factor $$\gamma_{21}=\left[1-\left(\dfrac { v_{21}}{c}\right)^{2}\right]^{-\frac{1}{2}} =\left[1-\left(\dfrac{\dfrac {v_2 - v_1}{1-\dfrac{v_1 v_2}{c^2}}}{c}\right)^{2}\right]^{-\frac{1}{2}}= \gamma_1 \gamma_2\left(1-\dfrac{v_1 v_2}{c^2} \right) \tag{05}$$

There exists a quantity called rapidity defined by $\:\cosh \zeta = \gamma \:$, that is

$$\zeta =\cosh^{-1}\gamma =\cosh^{-1}\left[\left(1-\dfrac{v^2}{c^2}\right)^{-\frac{1}{2}}\right] \tag{06}$$

Of course $$\tanh \zeta = \dfrac{v}{c} \tag{07}$$

Having in mind that $$\cosh \zeta = \cos \left(i\zeta\right)\;, \quad \sinh \zeta = -i\sin \left(i\zeta\right) \tag{08}$$

equation (01) is expressed as

$$\begin{bmatrix} x_1\\ \\ i c t_1 \end{bmatrix} = \begin{bmatrix} \cos \left(i\zeta_1 \right) & - \sin \left(i\zeta_1 \right)\\ & \\ \sin \left(i\zeta_1 \right) & \cos \left(i\zeta\right) \end{bmatrix} \begin{bmatrix} x\\ \\ i c t \end{bmatrix} \tag{09}$$

or defining like an angle $$\theta_1 = i\zeta_1 \tag{10}$$

$$\begin{bmatrix} x_1\\ \\ i c t_1 \end{bmatrix} = \begin{bmatrix} \cos \theta_1 & - \sin \theta_1 \\ & \\ \sin \theta_1 & \cos \theta_1 \end{bmatrix} \begin{bmatrix} x\\ \\ i c t \end{bmatrix} \tag{11}$$

which reminds us a rotation through the (imaginary) angle $\:\theta_1 \:$.

Then equation (05) for the $\:\gamma-$factor is the well-known identity for the $\:\cos \left(\theta_2 - \theta_1\right)\:$

$$\cos \left(\theta_2 - \theta_1\right)=\cos \theta_2 \cos \theta_1 + \sin \theta_2 \sin \theta_1 \tag{12}$$