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!

 

[Kick-Stand]

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?

Duh, since $\mathbf{v},\mathbf{w}$ are orthogonal I can just choose the y axis to correspond to either $\mathbf{w}$ or $-\mathbf{w}$, whichever way ensures a right handed coordinate system.

 

[Kick-Stand]

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?

So only after Taylor expanding $\gamma_v, v\gamma_v, \gamma_w, w\gamma_w$ for $v,w\ll 1$ and only keeping terms up to second order (e.g. $1, v, w, v^2, w^2, vw$) do I get a rotation matrix, e.g. $R^T R = I$ and det(R) = 1 to second order, with my rotation matrix given by

$$
R = \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & \pm vw & 0 \\
0 & \mp vw & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$

where the sign in the $\pm$ and $\mp$ is based on whether $\mathbf{w}$ points in the same direction as $\mathbf{e}_2$ or in the opposite direction. Kind of anti-climactic to get such a simple rotation matrix to second order lol.

 

[Ibix]

You might want to have a quick look at Wigner rotation, which says that the composition of two boosts can be rewritten as a single boost and a rotation. So I think you can see what's happening here by compositing the two "plus" boosts and, separately, the two "minus" boosts. Then when you composite those two transforms they almost cancel for low velocities (they would cancel exactly in the Galilean case), and the residual transform is dominated by their rotations.

 

[Kick-Stand]

You might want to have a quick look at Wigner rotation, which says that the composition of two boosts can be rewritten as a single boost and a rotation. So I think you can see what's happening here by compositing the two "plus" boosts and, separately, the two "minus" boosts. Then when you composite those two transforms they almost cancel for low velocities (they would cancel exactly in the Galilean case), and the residual transform is dominated by their rotations.

Any book to recommend with a good derivation of the result? Wikipedia is awful IMO.

 

[PeroK]

Any book to recommend with a good derivation of the result? Wikipedia is awful IMO.

You could always figure it out for yourself. The Wigner rotation is essentially caused by a convention that underlies the Lorentz Transformation:

I Post in thread 'Time Dilation along Multiple Axes'

Feb 22, 2022

Alrighty then, and for your example in the previous post I would do the same but for the $x$ axis. And the vector $n$ would be like a position vector stemming from the origin point of the First Reference Frame, along which the "boost" is occurring.

I think I get it now.

Many, many thanks @ergospherical !

Note that there is a subtlety here. If the boost is along a coordinate axis (e.g. the x-axis), then the two frames can agree upon the x-y-z axes. This is because at time $t = 0, t' = 0$, the points $(0, 1, 0)$ and $(0, 0, 1)$ transform to the points ##(0, 1...

• PeroK

• 
•