Why is addition of velocities under a Lorentz-transformation associative and commutative for when boost are in same direction? I'm working with Lorentz transformations, and one of the problems I encountered needed a derivation of the velocity of a particle $u$ in a stationary reference frame S, in the variables $u'$, which is the velocity measured of the particle of an observer moving with another reference frame S', that's moving with a velocity $v$ to the right.
Either way, the problem is one dimensional with respect to spatial dimensions, meaning, we're only dealing with $x$-coordinates.
I managed to find it to be $$u = \frac{u' + v}{1 + \frac{u' v}{c^2}}.$$ Then I encounted a question where I needed to argue why velocity addition under the Lorentz transformation is commutative, and associative, in the case when we have a boost in only one direction. I did some research and found out that this isn't always the case for when we have boosts in multiple directions (since a boost can be described my a matrix, two boosts after one another (that're not in the same direction) is a product of two matricies, for which the order matters).
How can one intuitively explain why it's associative and commutative under my given conditions?
 A: *

*Composition $\circ$ is a manifestly associative operation.


*Concerning commutativity, one may show that the restricted Lorentz group $$SO^+(1,1)\cong (\mathbb{R},+)$$ of boosts in 1+1D is isomorphic to the abelian group $(\mathbb{R},+)$ of rapidities $w=\tanh^{-1} \frac{v}{c}$.


*By the way, the full Lorentz group $O(1,1)$ in 1+1D is non-abelian.


*Similarly, the restricted Lorentz group in higher dimensions is non-abelian. See also e.g. this Math.SE post.
A: You can view two boosts along the same axis as simply being "scaled" versions of each other: they have the same sense, just in a different amount. That is, if $\Lambda$ is a boost matrix, then intuitively any other boost in the same direction is always $\Lambda^a$ for some $a\in\mathbb R$ (for once, not index notation—this $a$ is an exponent). Matrices in general don't commute, true (but their multiplication is always associative...), but powers of the same matrix always commute with each other, and indeed $\Lambda^a\Lambda^b=\Lambda^{a+b}$ as you'd expect (assuming the powers are defined, which they always are for boosts.) (The exponents are proportional to the hyperbolic angles/rapidities mentioned in other answers—I am leaving the math implicit.)
One way to see that composing boosts must give you a boost in the same direction, with no rotation or change in the direction of motion, is that either of these things would be asymmetrical. Two boosts along the same direction define only an axis, not a plane, so you can't get a rotation without having a preference for clockwise vs counterclockwise or something similarly asymmetrical. Similarly, the direction of the composed boost cannot change, because no direction (other than the shared axis of the boosts) is more special than any other.
A: For parallel boosts we can treat space as $1$-dimensional, so Lorentz boosts are in $2$ dimensions. They are hyperbolic rotation matrices, of the form $$\left(\begin{array}{rl}
\cosh w & \sinh w\\
\sinh w & \cosh w
\end{array}\right),$$which multiply viz.$$\left(\begin{array}{rl}
\cosh w_{1} & \sinh w_{1}\\
\sinh w_{1} & \cosh w_{1}
\end{array}\right)\left(\begin{array}{rl}
\cosh w_{2} & \sinh w_{2}\\
\sinh w_{2} & \cosh w_{2}
\end{array}\right)=\left(\begin{array}{rl}
\cosh\left(w_{1}+w_{2}\right) & \sinh\left(w_{1}+w_{2}\right)\\
\sinh\left(w_{1}+w_{2}\right) & \cosh\left(w_{1}+w_{2}\right)
\end{array}\right).$$As @Qmechanic's second point notes, you're just adding $w$s.
