Product of two boosts and then two inverse boosts is rotation matrix

[Kick-Stand]

Homework Statement

Let $\Lambda(\mathbf{v})$ be a Lorentz boost associated with the three velocity $\mathbf{v}$ and set $R = \Lambda(\mathbf{v})\Lambda(\mathbf{w})\Lambda(-\mathbf{v})\Lambda(-\mathbf{w})$ where $\mathbf{v}\cdot\mathbf{w} = 0$ and $v,w\ll 1$. Then $R$ is a rotation matrix.

Relevant Equations

For $\mathbf{v} = v_1 \mathbf{e}_1 + v_2\mathbf{e}_2 + v_3\mathbf{e}_3$ and $\gamma = (1-v^2)^{-1/2}$, where we use natural units with $c=1$, the boost associated to $\mathbf{v}$ is given by


$$
\Lambda(\mathbf{v}) & = \begin{pmatrix}
\gamma & -\gamma v_1 & -\gamma v_2 & -\gamma v_3 \\
-\gamma v_1 & 1 + (\gamma-1)\frac{v_1^2}{v^2} & (\gamma-1)\frac{v_1 v_2}{v^2} & (\gamma-1)\frac{v_1 v_3}{v^2} \\
-\gamma v_2 & (\gamma-1)\frac{v_1 v_2}{v^2} &1 + (\gamma-1)\frac{v_2^2}{v^2} & (\gamma-1)\frac{v_2 v_3}{v^2} \\
-\gamma v_3 & (\gamma-1)\frac{v_1 v_3}{v^2} & (\gamma-1)\frac{v_2 v_3}{v^2} &1 + (\gamma-1)\frac{v_3^2}{v^2} \\
\end{pmatrix}
$$


while for $v\ll 1$ we have $\gamma = 1 + \frac{1}{2} v^2 + O(v^4)$ and $v\gamma = v + O(v^3)$.

So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption $v,w\ll 1$, then do three ugly matrix multiplications and get some horrifying kludge for $R$ and show that the product of $R$ and its transpose is the identity matrix with det(R)=1?

Without loss of generality I made $\mathbf{v}$ point along the x-axis and since $\mathbf{v}\cdot\mathbf{w} = 0$ I set $w_1 = 0$ to make the boosts a little easier to deal with. But it still comes out a pretty ugly kludge even with $\Lambda(\mathbf{v})$ a very simple boost matrix. Is this the way I need to do the problem or am I missing something that makes this easy instead of brute forcing like a dummy?

 

[Kick-Stand]

Ugh what am I doing wrong in the homework statement and relevant equations? Double pound for inline latex and double dollar signs for displayed latex right?

 

[Ibix]

Ugh what am I doing wrong in the homework statement and relevant equations?

Nothing. LaTeX at PF has been dodgy since the last XenForo upgrade and it just doesn't work in some places. Probably the best thing to do is quote your entire OP and remove the quote tags, like this.

OP's question was:
Homework Statement: Let $\Lambda(\mathbf{v})$ be a Lorentz boost associated with the three velocity $\mathbf{v}$ and set $R = \Lambda(\mathbf{v})\Lambda(\mathbf{w})\Lambda(-\mathbf{v})\Lambda(-\mathbf{w})$ where $\mathbf{v}\cdot\mathbf{w} = 0$ and $v,w\ll 1$. Then $R$ is a rotation matrix.
Relevant Equations: For $\mathbf{v} = v_1 \mathbf{e}_1 + v_2\mathbf{e}_2 + v_3\mathbf{e}_3$ and $\gamma = (1-v^2)^{-1/2}$, where we use natural units with $c=1$, the boost associated to $\mathbf{v}$ is given by


$$
\Lambda(\mathbf{v}) = \begin{pmatrix}
\gamma & -\gamma v_1 & -\gamma v_2 & -\gamma v_3 \\
-\gamma v_1 & 1 + (\gamma-1)\frac{v_1^2}{v^2} & (\gamma-1)\frac{v_1 v_2}{v^2} & (\gamma-1)\frac{v_1 v_3}{v^2} \\
-\gamma v_2 & (\gamma-1)\frac{v_1 v_2}{v^2} &1 + (\gamma-1)\frac{v_2^2}{v^2} & (\gamma-1)\frac{v_2 v_3}{v^2} \\
-\gamma v_3 & (\gamma-1)\frac{v_1 v_3}{v^2} & (\gamma-1)\frac{v_2 v_3}{v^2} &1 + (\gamma-1)\frac{v_3^2}{v^2} \\
\end{pmatrix}
$$


while for $v\ll 1$ we have $\gamma = 1 + \frac{1}{2} v^2 + O(v^4)$ and $v\gamma = v + O(v^3)$.

So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption $v,w\ll 1$, then do three ugly matrix multiplications and get some horrifying kludge for $R$ and show that the product of $R$ and its transpose is the identity matrix with det(R)=1?

Without loss of generality I made $\mathbf{v}$ point along the x-axis and since $\mathbf{v}\cdot\mathbf{w} = 0$ I set $w_1 = 0$ to make the boosts a little easier to deal with. But it still comes out a pretty ugly kludge even with $\Lambda(\mathbf{v})$ a very simple boost matrix. Is this the way I need to do the problem or am I missing something that makes this easy instead of brute forcing like a dummy?

 

[Ibix]

Immediate observation: you correctly identified that you can set $\mathbf{v}$ to point along x, but you only set $w_1$ to zero?

 

[Kick-Stand]

Immediate observation: you correctly identified that you can set $\mathbf{v}$ to point along x, but you only set $w_1$ to zero?

Is that not the most general condition on $\mathbf{w}$ for
$$
vw_1 = \mathbf{v}\cdot\mathbf{w}=0
$$
to hold? Of course assuming $v\neq 0$.

 

[Ibix]

Sure. But you already set the coordinate system in a clever way to simplify the representation of v - why not make the same move for w?

 

[Ibix]

Then have a think about what type of transformation the composition of two boosts must be.

 

[Kick-Stand]

Then have a think about what type of transformation the composition of two boosts must be.

Ugh I'll have to do it tomorrow, I have to be up for work in 4 1/2 hours lol. Thanks.

 

[Ibix]

Sleep well!