Show that the Lorentz boosts, $\begin{pmatrix}\gamma & -\beta\gamma\\ -\beta\gamma & \gamma\end{pmatrix}$ form a one-parameter Lie group.

Question

In the following post, I am questioning certain parts of the author's solution for the beginning of a proof that the Lorentz boosts form a one-parameter Lie group:

Show that the Lorentz boosts, \begin{pmatrix}\gamma & -\beta\gamma\\ -\beta\gamma & \gamma\end{pmatrix} form a one-parameter Lie group. Is this group Abelian?

Here is the author's solution:

The Lorentz boosts, $\Lambda$, are defined by $$\Lambda(\beta)=\begin{pmatrix}\gamma & -\beta\gamma\\ -\beta\gamma & \gamma\end{pmatrix}=\gamma(\beta)\begin{pmatrix}1 & -\beta\\ -\beta & 1\end{pmatrix}$$ where $\gamma(\beta)=\frac{1}{\sqrt{1-{\beta}^2}},\quad \beta=\frac{v}{c}, \quad$ and $c$ is the speed of light, so $\lvert v \rvert\lt c$. To show that these matrices form a Lie group, we must demonstrate that these matrices under matrix multiplication satisfy the group axioms.

Closure. The product of two boosts $\Lambda(\beta )$ and $\Lambda({\beta}^{\prime})$ is $$\begin{align}\Lambda(\beta )\Lambda({\beta}^{\prime})&=\gamma({\beta})\gamma({\beta}^{\prime})\begin{pmatrix}1 & -\beta\\ -\beta & 1\end{pmatrix}\begin{pmatrix}1 & -{\beta}^\prime\\ -{\beta}^\prime & 1\end{pmatrix}\\&=\gamma({\beta})\gamma({\beta}^{\prime})\begin{pmatrix}1+\beta{\beta}^\prime & -\beta-{\beta}^\prime\\ -\beta-{\beta}^\prime & 1+\beta{\beta}^\prime\end{pmatrix}\end{align}$$ For this product to have the form $$\Lambda(\beta^{\prime\prime})=\gamma(\beta^{\prime\prime})\begin{pmatrix}1 & -{\beta}^{\prime\prime}\\ -{\beta}^{\prime\prime} & 1\end{pmatrix}\tag{1}$$ we must require that $$\color{red}{\gamma(\beta^{\prime\prime})=\gamma(\beta)\gamma({\beta}^\prime)\left(1+\beta{\beta}^\prime\right)}\tag{2}$$ $${\begin{align}\color{red}{{\beta}^{\prime\prime}\gamma({\beta}^{\prime\prime})=\gamma(\beta)\gamma({\beta}^\prime)\left(\beta+{\beta}^\prime\right)}\tag{3}&\\=[\cdots]\end{align}}$$

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I would like to understand the reasoning behind the need for the red equations, $(2)$ and $(3)$.

Taking equation $(2)$ for now, and plugging it back into $(1)$, $$\begin{align}\Lambda(\beta^{\prime\prime})&=\gamma(\beta^{\prime\prime})\begin{pmatrix}1 & -{\beta}^{\prime\prime}\\ -{\beta}^{\prime\prime} & 1\end{pmatrix}\\&=\gamma(\beta)\gamma({\beta}^\prime)\left(1+\beta{\beta}^\prime\right)\begin{pmatrix}1 & -{\beta}^{\prime\prime}\\ -{\beta}^{\prime\prime} & 1\end{pmatrix}\\&=\gamma(\beta)\gamma({\beta}^\prime)\begin{pmatrix}1 +\beta{\beta}^\prime & -\left(1 +\beta{\beta}^\prime\right){\beta}^{\prime\prime}\\ -\left(1 +\beta{\beta}^\prime\right){\beta}^{\prime\prime} & 1 +\beta{\beta}^\prime\end{pmatrix}\\&=\gamma(\beta)\gamma({\beta}^\prime)\left(1+\beta{\beta}^\prime\right)\begin{pmatrix}1 & -{\beta}^{\prime\prime}\\ -{\beta}^{\prime\prime} & 1\end{pmatrix}\end{align}$$
So I see why the factor of $\color{red}{\left(1+\beta{\beta}^\prime\right)}$ was needed in equation $(2)$, but there was no way I could intuit the need for this $\color{red}{\left(1+\beta{\beta}^\prime\right)}$ factor without direct substitution as I did above. Is there some deeper technique going on here that eludes me, like matrix diagonalization?

The other equation (in red) I am questioning is $(3)$ and I really don't understand what the author is trying to do with this. So, if anyone has any insights or thought's on why this is being done please let me know.

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N.B. There is more to equation $(3)$ than I have written above, but I didn't want to ask too many questions at once, so for now I am just questioning the need for $\color{red}{{\beta}^{\prime\prime}\gamma({\beta}^{\prime\prime})=\gamma(\beta)\gamma({\beta}^\prime)\left(\beta+{\beta}^\prime\right)}$.

Answer 1

Set the components of the matrix equal to one another.

On one hand, the product of the two boosts is equal to:

$$\begin{bmatrix}\gamma_1 \gamma_2 (1+\beta_1\beta_2) & - \gamma_1\gamma_2 (\beta_1 + \beta_2) \\ - \gamma_1\gamma_2 (\beta_1 + \beta_2) & \gamma_1 \gamma_2 (1+\beta_1\beta_2)\end{bmatrix}$$

on the other hand, for closure, you want this product to have the form of a boost for some unknown value of $\beta_3$ and $\gamma_3(\beta_3)$:

$$\begin{bmatrix}\gamma_3 & -\gamma_3 \beta_3 \\ -\gamma_3\beta_3 & \gamma_3\end{bmatrix}$$

Setting these two matrices equal to one another, you find constraints on $\gamma_3$ and $\beta_3$:

$$\gamma_3 = \gamma_1\gamma_2(1+\beta_1\beta_2)\\ -\gamma_3\beta_3 = -\gamma_1\gamma_2(\beta_1+\beta_2) \\ \Rightarrow \beta_3 = \frac{\beta_1+\beta_2}{1+\beta_1\beta_2}$$

So you can solve for $\gamma_3$ and $\beta_3$ in terms of the original parameters $\beta_1,\beta_2,\gamma_1,\gamma_2$ and show that $\gamma_3$ is indeed the Lorenz factor corresponding to $\beta_3$, i.e. you can prove that, as required, $$\gamma_3 = \frac{1}{\sqrt{1-\beta_3^2}}$$

so the product of two boosts is itself a boost corresponding to $\beta_3,\gamma_3$.