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.

  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos May 16 at 16:58
  • $\begingroup$ Also related. $\endgroup$ – J.G. May 16 at 18:45
  • $\begingroup$ Related : General matrix Lorentz transformation $\endgroup$ – Frobenius May 16 at 21:15
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    $\begingroup$ 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. $\endgroup$ – Nat May 16 at 22:46

It is not possible. Two boosts are, however, equivalent to one boost combined with one rotation! This surprising rotation is known as a Thomas-Wigner rotation.

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

$$[J_i,J_j]=\epsilon_{ijk}J_k$$ $$[K_i,K_j]=-\epsilon_{ijk}J_k \,\text{(!)}$$ $$[J_i,K_j]=\epsilon_{ijk}K_k$$

  • $\begingroup$ Thank you for the explanation and the link. $\endgroup$ – Kekks May 16 at 16:42
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    $\begingroup$ 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"? $\endgroup$ – Feynmans Out for Grumpy Cat May 16 at 17:12
  • $\begingroup$ @DvijMankad I’m not sure what you are asking, but my understanding of Lie algebras probably isn’t strong enough to answer even if you clarify it. $\endgroup$ – G. Smith May 16 at 17:21
  • $\begingroup$ 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? $\endgroup$ – Feynmans Out for Grumpy Cat May 16 at 18:02
  • $\begingroup$ @DvijMankad That’s an interesting question. If someone doesn’t answer it here, please consider posting it as a question. $\endgroup$ – G. Smith May 16 at 19:21

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