In STR the composition of non parallel movements is in general non-associative. The formula is

$\displaystyle\bar{u}\oplus\bar{v}= \frac{\bar{u}+\bar{v}_{\|}+\bar{v}_\bot/\gamma}{1+\bar u\cdot\bar v/c^2}$, where $\;\displaystyle\bar{v}_{\|}=\frac{\bar u\cdot\bar v}{|\bar u|^2}\cdot \bar u\;$ and $\;\bar{v}_\bot=\bar v - \bar{v}_{\|}\;$ and $\displaystyle\;\gamma=\frac{1}{\sqrt{1-|\bar u|^2/c^2}}$

If $\bar{v}_{ij}$ is the velocity between the systems $S_i$ and $S_j$ for $i,j=1,2,3,4\;$ where $\bar{v}_{13}=\bar{v}_{12}\oplus\bar{v}_{23}$ and $\bar{v}_{24}=\bar{v}_{23}\oplus\bar{v}_{34}$, then in general, $\bar{v}_{12}\oplus\bar{v}_{24}\ne\bar{v}_{13}\oplus\bar{v}_{34}$.

$\require{AMScd}$ \begin{CD} S_1 @>\bar{v}_{12}>> S_2\\ @V \bar{v}_{13} V \qquad\swarrow\bar{v}_{23}V @VV \bar{v}_{24} V\\ S_3 @>>\bar{v}_{34}> S_4 \end{CD}

  • Does that mean that there is no unique velocity between $S_1$ and $S_4$?
  • Are Lorentz transformations more adequate representations of motion, than the more intuitive velocities?


  1. (Wikipedia on velocity addition 15 rows down):

Also it is not associative and

$\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w}) = (\mathbf{u} \oplus \mathbf{v})\oplus \mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$ where "gyr" is the mathematical abstraction of Thomas precession

  1. Paper of Abraham Ungar p. 4 formula (10).

  • 2
    $\begingroup$ Do you have a reference stating that it's non-associative, or is it a conclusion you drew on your own from the formula? If the latter I wonder if it's some error in the mathematical manipulations, perhaps you could show them. $\endgroup$
    – Hypnosifl
    Nov 23, 2014 at 17:12
  • 1
    $\begingroup$ The wiki article doesn't give a reference, if it's well known could you provide one? And if it's easy to check, could you give a numerical example? Often looking at specific examples helps clarify the physical meaning, which is what I think you're asking about. For example, how did we ensure that $\bar{v}_{34}$ and $\bar{v}_{24}$ represent the velocity of the same physical system $S_4$ relative to two other systems $S_3$ and $S_2$? Did we first define the velocities of both $S_4$ and $S_3$ relative to $S_2$, and then use the velocity addition to find $S_4$ relative to $S_3$? $\endgroup$
    – Hypnosifl
    Nov 23, 2014 at 18:01
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    $\begingroup$ @Hypnosifl: No, composition of Lorentz transformations is associative, but Einstein addition of velocities is not. $\endgroup$
    – Lehs
    Nov 23, 2014 at 18:31
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    $\begingroup$ Someone else can probably give a fuller answer than I can, but it seems to be to do with the fact that in more than 1 spacial dimension the Lorenz boost are not a closed subgroup of the proper lorentz group. In general the product of 2 boosts in different direction is a combination of a boost and a rotation. This means if you want to find the relative velocity of 2 frame from their relative velocities with some intermediate frames, you have to take account of the fact that you may have rotated your coordinates along the way, and clearly when that rotation occurs is going to matter. $\endgroup$ Nov 23, 2014 at 18:43
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    $\begingroup$ @Hypnosifl: Amazingly, no. The composition of two boosts is a boost followed by a rotation. $\endgroup$
    – ACuriousMind
    Nov 23, 2014 at 22:35

2 Answers 2


Are Lorentz transformations more adequate representations of motion, than the more intuitive velocities?

Yes. The non-associativity that bothers you simply arises because there is no group of three dimensional boosts. Confined to one dimension, boosts form a rather lovely one parameter subgroup of the Lorentz group $SO^+(1,3)$. So everything "works", associativity in particular. This one parameter group is even commutative, like all one parameter Lie groups.

When you try to reproduce this elegance in three spatial dimensions, it doesn't work. Two boosts compose in general to a boost and a rotation. The "Thomas precession" is key here: this is the rotation that must in general be present in the polar decomposition of a Lorentz group member.

When you expand your system to include rotations, you get the whole group $SO^+(1,3)$ and thus recover associativity.

Does that mean that there is no unique velocity between $S_1$ and $S_4$?

Kind of. You need a boost and a rotation to represent the transformation between $S_1$ and $S_4$ and, moreover, the boost in the polar decomposition is different depending on whether you write the latter as:

$$B(v_{2\,4})\,B(v_{1\,2}) = R\left(\mathrm{gyr}(v_{2\,4},\,v_{1\,2})\right)\,B(v_{1\,2}\oplus v_{2\,4})$$

or whether you write it as:

$$B(v_{2\,4})\,B(v_{1\,2}) = B(v_{2\,4}\oplus v_{1\,2})\,R\left(\mathrm{gyr}(v_{2\,4},\,v_{1\,2})\right)$$

where $B(v)$ is the boost with velocity vector $v$ and $R\left(\mathrm{gyr}(v_{2\,4},\,v_{1\,2})\right)$ the Thomas precession rotation. Note that the rotation in each polar decomposition is the same, but the two boosts are different because $\oplus$ is not commutative: you can prove that, for any rotation $R$ and boost $B$ that $R^T\,B\,R = R^{-1}\,B\,R = B^\prime$ where $B^\prime$ is another boost (this follows from the braiding relationship $\exp(\mathrm{ad}(X)) = \mathrm{Ad}(e^X)$ and the commutation relationships for the Lorentz Lie algebra). But if you stick to a consistent ordering (e.g the rotation always on the left) in the polar decomposition, the boost matrix is uniquely defined.

Your basic problem with the diagram is that you can't draw a commutative diagram in the way you have with only "velocity" arrows because that would be predicated on the end points' full, unique specification by relative velocities. You would need one more node in your commutative diagram:

Lorentz Transformations are boosts composed with rotations

where system 5 is at rest relative to system 4, but rotated by the appropriate rotation $R_{54}$

The Lorentz Transformation Wiki page in the "Boost in any direction" section shows you how to relate the $\mathrm{gyr}$ and Lorentz transformation notation.

  • $\begingroup$ Suppose S1-S4 are inertial coordinate systems, and we know the coordinate transformation from S1 to S2, the transformation from S2 to S3, and the transformation from S3 to S4. Given a coordinate transform from Si to Sj, doesn't this give us a single correct answer for the velocity of the origin of Si in Sj? If so, then although you say "kind of" to the question about no unique velocity between S1 and S4, isn't it the case that there would be a unique coordinate transformation from S1 to S4 (even if it wasn't a pure boost), and thus a unique truth about the velocity of S1's origin in S4? $\endgroup$
    – Hypnosifl
    Nov 24, 2014 at 5:36
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    $\begingroup$ @Hypnosifl That's what I meant by "kind of": the velocity is different depending on which way around you write your polar decomposition. But yes, in principle, if you keep your polar decomposition a consistent order, what you say is true. The two different velocities are quite simply related, one being the conjugation of the other by the rotation. $\endgroup$ Nov 24, 2014 at 5:40
  • $\begingroup$ @Hypnosifl Also, I should have said this sooner: of course, if $S_4$ is to be a well defined frame, then of course there is a unique single Lorentz transformation between the two. That's why the OP's commutative diagram can't work: there are two different $S_4$'s, one of which becomes $S_5$ in my updated diagram. Also, the velocity of the origin in both frames $S_4$ and $S_5$ relative to $S_1$ is the same. $\endgroup$ Nov 24, 2014 at 7:09
  • $\begingroup$ I suspect that Ungar's formula is inadequate even though it's algebraically consistent. I postulated $\bar v_{13}=\bar v_{12}\oplus\bar v_{23}$ but if $S_1$ measure the velocity to $S_2$ and $S_3$, and $S_2$ measure the velocity to $S_3$ I think that $S_1$ would disagree either with the information from $S_2$ or with Ungar's formula. As far as I know all experiments confirming Ungar/STR refers to parallel movements. I agree that the composition of motion should be associative. $\endgroup$
    – Lehs
    Nov 24, 2014 at 8:47
  • $\begingroup$ @Lehs There should be a way to make Ungar-like formulas work, indeed the Wiki page gives an explicit way to calculate the Lorentz transformations. I'm guessing that you can derive them as a closed form of the Campbell Baker Hausdorff theorem in much the same way that one can do for $SO(3)$ and $SU(2)$ in papers such as Kenth Engø-Monsen, "On the BCH Formula in $SO(3)$", BIT 2001, 41, #3, pp. 629–632 which work beginning with the appropriate Rodrigues formula for the matrix exponential $\endgroup$ Nov 24, 2014 at 9:02

How to understand non-associative composition of velocities in STR?

Special relativity introduces a weirdness about how your axes can be related to other observers' axes: if your axes are aligned with observer A's axes and theirs are aligned with observer B then special relativity (i.e. the Lorentz transformations) say that B's axes will be rotated with respect to yours.


Adding two velocities means taking into account both magnitudes and directions, but direction depends on choice of axes, so velocity-addition only makes sense if the two systems of axes are consistently specified: the velocity-addition formula is a formula for adding velocities from two systems whose axes are aligned.

The sentiment has been expressed in the comments on the question that composition should be associative, i.e. that three velocities should be able to be added in either order. However the addition formula only applies to two sets of axes that are aligned and STR says that three sets of axes cannot be aligned.

When adding more than two velocities you have to be careful, instead of just using relative velocites, you may need (depending on exactly what the expression involves, see below) to rotate each additional velocity to counteract the misalignment of axes caused by STR, so that each application of the addition formula is applied to two velocities given with respect to the same axes alignment.


Applying a rotation to the third velocity you get $\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w}) = (\mathbf{u} \oplus \mathbf{v})\oplus R\mathbf{w}$ for some rotation R.

Examining what the elements of that equation represent you'll see why only the right hand side has a rotation in it: This equation applies to a situation with 4 observers O,A,B,C with relative velocities OA=u AB=v, BC=w. A axes aligned with O, B with A, and C with B. STR says C will be rotated to A, B rotated to O, C rotated to O. So relative velocity of OB is $u\oplus v$, and AC is $v\oplus w$. Adding $u$ to $(v\oplus w)$ is ok because O and A are aligned, but adding $(u\oplus v)$ to $w$ needs a rotation because O and B are not aligned.

Rotation notation:

The amount of rotation R needed of the relative velocity $\mathbf{w}$ depends on the velocity of the observer that the $\mathbf{w}$ is relative to and is denoted by Ungar as $\mathrm{gyr}[\mathbf{u},\mathbf{v}]$

$\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w}) = (\mathbf{u} \oplus \mathbf{v})\oplus \mathrm{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}$


The above discussion of non-associativity involved 4 observers with 3 given velocities, but you don't need 3 velocities to see non-uniqueness: it comes into play with just 2 velocities:

STR says composition of two boosts is not equivalent to a single boost, but to a rotation and a boost. A Lorentz transformation between frames is applied to coordinates. If you apply the rotation first then the direction of the required boost from the rotated coordinates will be different than if you applied a boost first: 3 observers O,A,B with velocites OA=u, AB=v, then the velocty of B from O's unrotated coordinates will be $u\oplus v$ but B's axes will be rotated. Alternatively rotate O's axes first and then boost with velocity $v\oplus u$ to get to B. $u\oplus v$ and $v\oplus u$ are not equal, the velocity-addition is non-commutative.

Velocity addition in 1D, 2D, 3D:

If all the velocities are along one line then there is no rotation and the addition formula reduces to $\mathbf{u}\oplus \mathbf{v}={{u+v} \over {1+(uv/c^2)}}$ which is commutative and associative.

The formula you give in your question uses parallel and perpendicular components which is 2-dimensionsal, but if an expression involves three velocites then the third velocity could be in a different plane so the 3D formula must be used:

$\mathbf{u}\oplus \mathbf{v}=\frac{1}{1+\frac{\mathbf{u}\cdot\mathbf{v}}{c^2}}\left\{\mathbf{u}+\frac{1}{\gamma_\mathbf{u}}\mathbf{v}+\frac{1}{c^2}\frac{\gamma_\mathbf{u}}{1+\gamma_\mathbf{u}}(\mathbf{u}\cdot\mathbf{v})\mathbf{u}\right\}$

  • $\begingroup$ I think this is the answer and that $|u\oplus(v\oplus w)|=|(u\oplus v)\oplus w|$. It seems that transformations and composition of transformations is more adequate than velocities and composition of velocities. $\endgroup$
    – Lehs
    Nov 25, 2014 at 8:14
  • $\begingroup$ Well |u+v|=|v+u| and |u+(v+w)|=|(u+v)+Rw| but |u+(v+w)| is not equal to |(u+v)+w|. $\endgroup$
    – Snor
    Nov 25, 2014 at 8:30
  • $\begingroup$ Are you sure? Some concrete calculations indicate equality. $\endgroup$
    – Lehs
    Nov 25, 2014 at 8:49
  • $\begingroup$ No, you are right about that too. $\endgroup$
    – Lehs
    Nov 25, 2014 at 9:13

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