# Combining two Lorentz boosts

Is it possible to express two Lorentz boosts $$A_x(\beta)$$ and $$A_y(\beta)$$ along the x/y-axis as one boost described by $$A(\overrightarrow \delta)$$?

To answer this, I start by defining $$\theta \equiv \arctan\left(\beta\right)$$ and $$\beta \equiv v/c ,$$ then: \begin{align} A_x(\beta) &= \begin{bmatrix} \cosh(\theta) & \sinh(\theta) & 0 \\ \sinh(\theta) & \cosh(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}\\[10px] A_y(\beta) &= \begin{bmatrix} \cosh(\theta) & 0 & \sinh(\theta) \\ 0 & 1 & 0 \\ \sinh(\theta) & 0 & \cosh(\theta) \end{bmatrix} \\[10px] A(\overrightarrow \delta) &= \begin{bmatrix}\gamma & \gamma \delta^1 & \gamma \delta^2 \\ \gamma \delta^1 & 1+B_{11} & B_{12} \\ \gamma \delta^2 & B_{21} & 1+B_{22} \end{bmatrix} \\[10px] B_{ij} &= \frac{\gamma -1}{\delta^2}\beta_i\delta_j \end{align}

I first calculated: $$A_y(\beta)*A_x(\beta) = \begin{bmatrix} \cosh^2(\beta) & \cosh(\beta)*\sinh(\beta) & \sinh(\beta) \\ \sinh(\beta) & \cosh(\beta) & 0 \\ \cosh(\beta)* \sinh(\beta) & \sinh^2(\beta) & \cosh(\beta) \end{bmatrix}$$ then since $$B_{ij}=B_{ji} ,$$ I get $$\sinh^2\left(\theta\right) = 0 \quad\Rightarrow\quad \theta=0 \quad\Rightarrow\quad v=0 \,.$$

That would mean, that it is not possible to express two subsequent boosts in $$x\text{-}$$ and $$y\text{-}$$ direction as a combined boost. This does seem a bit odd to me and I wonder if I made a mistake in my computation.

• – Cosmas Zachos May 16 '19 at 16:58
• – J.G. May 16 '19 at 18:45
• – Frobenius May 16 '19 at 21:15
• Welcome to SE.Physics! I edited in a bit of formatting. Please feel free to edit it to make any fixes. In particular, I read the definitions after the quote block as your own work rather than as something provided by the problem statement, which I hope was a correct interpretation. – Nat May 16 '19 at 22:46

This rotation arises because two infinitesimal boosts do not commute; their commutator is an infinitesimal rotation. In terms of rotation generators $$J_i$$ and boost generators $$K_i$$, the Lorentz algebra is
$$[J_i,J_j]=\epsilon_{ijk}J_k$$ $$[K_i,K_j]=-\epsilon_{ijk}J_k \,\text{(!)}$$ $$[J_i,K_j]=\epsilon_{ijk}K_k$$
• This is really interesting! I never gave enough time to appreciate the asymmetry between the commutation relations of $J_i$s being closed among $J_i$s and the commutation relations of $K_i$s not being so. Many thanks to @Kekks as well :) Is there an interesting imprint of this asymmetry when we re-write $\mathbb{so}(1,3)$ as $\mathbb{su}(2)\times\mathbb{su}(2)$ where $J_i$ and $K_i$ insert the formulae rather "democratically"? – Dvij D.C. May 16 '19 at 17:12
• I just meant that when we go to $\mathbb{su}(2)\times\mathbb{su}(2)$, we construct its generators out of the $K_i$s and $J_i$s as $A_i=\frac{J_i+iK_i}{2}$ and $B_i=\frac{J_i-iK_i}{2}$ and then, these $A_i$s form their own group and so do $Bi$s and they seem on an equal footing unlike $J_i$s and $K_i$s. I was (probably naively) wondering if some imprint of the fact that $J_i$s and $K_i$s are not on equal footing might show up somewhere in this latter formalism. Maybe in relation to chirality? – Dvij D.C. May 16 '19 at 18:02