windpipe33u

2022-04-23

Show that the Lorentz boosts,

$\left(\begin{array}{cc}\gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{array}\right)$

form a one-parameter Lie group.

form a one-parameter Lie group.

Friegordigh7r7

Beginner2022-04-24Added 16 answers

Step 1

Set the components of the matrix equal to one another.

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

$\left[\begin{array}{cc}{\gamma}_{1}{\gamma}_{2}(1+{\beta}_{1}{\beta}_{2})& -{\gamma}_{1}{\gamma}_{2}({\beta}_{1}+{\beta}_{2})\\ -{\gamma}_{1}{\gamma}_{2}({\beta}_{1}+{\beta}_{2})& {\gamma}_{1}{\gamma}_{2}(1+{\beta}_{1}{\beta}_{2})\end{array}\right]$

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 ${\gamma}_{3}\left({\beta}_{3}\right)$:

$\left[\begin{array}{cc}{\gamma}_{3}& -{\gamma}_{3}{\beta}_{3}\\ -{\gamma}_{3}{\beta}_{3}& {\gamma}_{3}\end{array}\right]$

Setting these two matrices equal to one another, you find constraints on $\gamma}_{3$ and $\beta}_{3$:

${\gamma}_{3}={\gamma}_{1}{\gamma}_{2}(1+{\beta}_{1}{\beta}_{2})$

${\gamma}_{3}={\gamma}_{1}{\gamma}_{2}(1+{\beta}_{1}{\beta}_{2})$

$-{\gamma}_{3}{\beta}_{3}=-{\gamma}_{1}{\gamma}_{2}({\beta}_{1}+{\beta}_{2})$

$\Rightarrow {\beta}_{3}=\frac{{\beta}_{1}+{\beta}_{2}}{1+{\beta}_{1}{\beta}_{2}}$

So you can solve for $\gamma}_{3$ and $\beta}_{3$ in terms of the original parameters $\beta}_{1},\text{}{\beta}_{2},\text{}{\gamma}_{1},\text{}{\gamma}_{2$ 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 $bat{a}_{3},\text{}{\gamma}_{3}$

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