# Prove that infinitesmal angular displacements commute & non-infinitesimal ones don't?

Recently, I were taught about uniform circular motion in which I learn that non-infinitesimal angular displacement are non-commutative and infinitesimal angular displacement are commutative.

I easily understood intuition behind non-infinitesimal angular displacement are non-commutative but was unable to prove it mathematically and same goes for infinitesimal angular displacement.

Secondly, it makes me uneasy that infinitesimal angular displacement are commutative but their sum loses that property.Any reason's why?

Can you prove it so mathematically how sum of infinitesimal angular displacement loses it's property of commutivity?

FYI:I'm a highschool student(someone edited this out).SO can you you please give me a little simpler explanation because I curretnly don't have any exposure to Lie ALgebra

• This is a mathSE problem, and I think you should post it there, as you want a proof, with a better chance of an answer. This page phys.virginia.edu/LectureDemo/Manual/M13a.html shows you why normal rotations don't commute. Also this is related physics.stackexchange.com/q/48345
– user108787
Aug 28, 2016 at 15:56
• Your question title doesn't seem to be a clear sentence. I suggest to reformulate it more clearly - it significantly highten your chance to get an useful answer. Aug 28, 2016 at 18:00
• There are probably terms in some kind of expansion that goes arbitrarily close to zero the smaller the displacement is.
– Emil
Aug 28, 2016 at 18:04
• For physical significance of non-commutative rotations, see e.g. this Phys.SE answer. Aug 29, 2016 at 12:16
• @count_to_10 : Thankyou sir for your suggestion but I think only a mod can move the questions from one forum to another? Aug 29, 2016 at 15:29

There is no such thing as infinitesimal transformation. This is just a mental shortcut which is frequently used, but (as often is the case with mental shortcuts) causes confusion of these who don't know what is really involved. When we talk of infinitesimal transformations we always mean that there is some set of transformations which depends smoothly on some set of parameters $\alpha_i$. For simplicity of notation I assume that all parameters $\alpha_i=0$ coresponds to identity transformation. Rotations are nice example, and the parameters $\alpha$ can be taken as some angles. Any such transformation can be expanded in power series $$R(\vec{\alpha}) = 1 - i \vec \alpha \cdot \vec J + \mathrm{higher \ order \ terms},$$ where $J_i= \left. i \frac{d}{d \alpha_i} R(\vec{\alpha}) \right|_{\vec{\alpha}=0}$. These $J$ are called generators of the group of transformations. Set of all generators is called Lie algebra. What we call infinitesimal transformation is first order expansion $1-i \vec{\alpha} \cdot \vec J$ or even the generator $\vec J$. It is not really a rotation in any meaningful way, to have a rotation you need to have all (infnite number of) terms in the power series. Now you can calculate commutator of two rotations $R(\vec{\alpha})$ and $R(\vec{\beta})$ and you will find that it is zero in the first order in parameters $\vec{\alpha}$ and $\vec{\beta}$. You can say that the commutator of infinitesimal transformations is small in higher order in parameters than the infinitesimal transformation. Therefore it will be very close to zero if $\vec{\alpha}$ and $\vec{\beta}$ are small.

EDIT: This is an attempt to put this answer in simple terms. I will leave the old version of the answer also here, because it might be useful for someone in the future.

The essence what I was trying to explain with formulas and some more fancy mathemathical machinery is as follows. There isn't anything like "infinitesimal transformation". This is only a mental shortcut which is convenient for some reasons. Every real rotation has an axis of rotation and finite angle. If you take two rotations such that this angle is very small and compose them in two different orders, then the difference will be even smaller. That's why some people say that if angles of rotation becomes infinitely small then the rotations are comutative. This is misleading and, in my opinion, useless. This fits in general theme of thinking that there is such thing as an "infinitely small number" or an "infinitesimal". If you have studied some calculus, you might have encountered this already. I emphasise: there is not such thing as an infinitesimal number. Any number which is positive and smaller or equal than all other numbers is zero. The same is true for angles or rotations, differences in the values of a function as argument is shifted slightly, or whatever else. Whenever people speak of "infinitesimal" something they are really talking about "very very small something" and then using this smallness to make some approximations. This is the same as in high school formula $\sin x \approx x$ for $x$ small. This is a very good and simple approximation if $x$ is very small, therefore it is quite useful. However it is never strictly true that $\sin x =x$ for $x \neq 0$.

• Very simple and good answer. Aug 29, 2016 at 12:08
• Respected Sir,thank you for taking out your precious time to help me out .Can you please me a little simpler explanation as I haven't studied Linear ALgebra(that's why i specified my background u someone edited that out).Can I prove that same using only Calc + Algebra(whatever taught at highchool level) Aug 30, 2016 at 10:07
• @Xasel I added new, simplified part of the answer. I hope it's clearer now. To stress it once again: there is really no such thing as an infinitesimal rotation, and this whole notion is just simplification (which is very useful if you know what it really means). Therefore statements such as "infinitesimal rotations are commutative" are meaningless as they are stated - so you are completely correct here. What is really meant is that for two rotations $R_1,$ $R_2$, $R_1R_2-R_2 R_1$ is VERY small if angles of $R_1$ and $R_2$ are small. Aug 30, 2016 at 12:42
• Aha! I didn't notice that::They are the infinitesimals of same order .Thank you very much sir .One more thing that bugs me is that when we approximate (uniform)circular motions as 2D. Ain't angles are commutative on 2D plane? Aug 30, 2016 at 15:11
• If you want to count the "order" of infinitesimals then $R_1$ and $R_2$ are first order and $R_1 R_2 - R_2 R_1$ is of second order. This is very informal statement though. It is true that in two dimensions rotations are commutative. It is easy to understand that it must be the case, since there is only one possible "axis" of rotation. Aug 30, 2016 at 23:08

The 3x3 matrices which do small rotations of a vector by the angles $\theta_x, \theta_y, \theta_z <<1 \quad radian$ about the x, y, and z axis respectively (put your right thumb along the positive axis and push the vector counter-clockwise with your fingers) are: $$M_x(\theta_x)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -\theta_x \\ 0 & \theta_x & 1 \end{bmatrix}\quad M_y(\theta_y)= \begin{bmatrix} 1 & 0 & \theta_y \\ 0 & 1 & 0 \\ -\theta_y & 0 & 1 \end{bmatrix} \quad M_z(\theta_z)= \begin{bmatrix} 1 & -\theta_z & 0 \\ \theta_z & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ Now you can calculate the commutatator $$[M_x(\theta_x), M_y(\theta_y)]=\left[\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -\theta_x \\ 0 & \theta_x & 1 \end{bmatrix} , \begin{bmatrix} 1 & 0 & \theta_y \\ 0 & 1 & 0 \\ -\theta_y & 0 & 1 \end{bmatrix}\right]=\begin{bmatrix} 0 & -\theta_x \theta_y & 0 \\ \theta_x \theta_y & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Notice that the commutator gives a non-zero matrix. The magnitudes of it's components are much much smaller than either $\theta_x$ or $\theta_y$ when $\theta_x<<1$ and $\theta_y<<1$. Thus it might seem to us that rotations commuted for such small angles. These rotations about x, y, and z form the Rotation Group O(3).

Your intuition “it makes me uneasy that infinitesimal angular displacement are commutative but their sum loses that property” was absolutely correct. If infinitesimal rotations were commutative, the group would not be O(3), and my representation of the rotations by the above $M_x, M_y, M_z$ matrices would be incorrect.

• Small correction - the commutator is relevant to the generators, not the group elements (the rotation matrices). If I add two rotation matrices, I don't get a rotation matrix back. The quantity you want to look at for the test of whether a group is abelian is: $$B^{-1}A^{-1}BA,$$ ie try to walk in a closed square. In other words, if you rewrite the matrices $$M_i = I + \theta_i J_i$$ then $$M_x^{-1}M_y^{-1}M_xM_y \approx I + \theta_x\theta_y [J_x, J_y]$$ to first order in both of the $\theta$s with $I$ the identity matrix. Aug 29, 2016 at 16:31
• @Sean - Yes, you are absolutely correct. I interpreted the question as asking specifically for the commutator of angular displacements, so I kept it simple and didn't introduce generators. As you suggest, the group of rotations and the Lie algebra of its generators would be very useful/enlightening for the questioner to learn. Aug 29, 2016 at 18:22
• Respected Sir,thank you for taking out your precious time to help me out .Can you please me a little simpler explanation as I haven't studied Linear ALgebra(that's why i specified my background u someone edited that out).Can I prove that same using only Calc + Algebra(whatever taught at highchool level). Aug 30, 2016 at 10:08