Finite rotations and infinitesimal rotations

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  • #1
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Hi
I am using Kleppner and it states that finite rotations do not commute but infinitesimal rotations do commute. I follow the logic in the book but i don't understand the concept. Surely a finite rotation consists of many , many infinitesimal rotations and if they commute why doesn't the finite rotation ? Can anyone explain this using words not equations ?
Thanks
 

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  • #2
jbriggs444
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Hi
I am using Kleppner and it states that finite rotations do not commute but infinitesimal rotations do commute. I follow the logic in the book but i don't understand the concept. Surely a finite rotation consists of many , many infinitesimal rotations and if they commute why doesn't the finite rotation ? Can anyone explain this using words not equations ?
Thanks
Two infinitesimal rotations commute. But if you stack infinitely many of them up, the result is finite and they don't.

Possibly this won't appeal to you, but...

If you have an infinite series that is "conditionally convergent" such as 1, -1/2, 1/3, -1/4, 1/5, -1/6, then you can rearrange the terms and get the resulting series to converge to any number you like. So even though ordinary numbers are finitely commutative, they are not infinitely commutative.
 
  • #3
wrobel
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good question indeed.

Surely a finite rotation consists of many , many infinitesimal rotations
infinitesimal rotations do commute
Just try to translate these slippery phrases to the precise mathematical language.
Unfortunately, physics textbooks frequently admit mathematical dishonest.
 
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  • #4
vanhees71
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This is very misleading. If it were true the rotation group should be Abelian, which it definitely is not, and that's why the angular-momentum components do not "commute". In the here discussed classical case the most efficient way to formulate symmetries is the Hamilton version of Hamilton's principle, where you have Poisson brackets defining a Lie algebra on phase-space functions. The three components of angular momentum obey the Lie algebra of the rotation group, and they don't have vanishing Poisson brackets and that's why the Lie group, which you get from the Lie algebra through the exponential of the corresponding Lie derivative, defined by the Poisson brackets, and since the angular momentum components do not commute so don't rotations around different rotation axes.
 
  • #5
dyn
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Thanks for your replies. The last one was over my head. I am hoping for a "simple" explanation of the following - if infinitesimal rotations commute and act as vectors I can add them together. So I can add 2 together , 3 together and so on and the resultant rotations will still commute. But if I add "too many" of them together they stop commuting and don't act as vectors
Thanks
 
  • #6
vanhees71
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Rotations around axes in different directions do not commute but those around one single axis do. There is no contradiction here. You can always build a finite rotation from many rotations around the same axis with smaller and smaller angles, and all these rotations commute. As I said all this is the relation between the Lie algebra of the rotation group (built by infinitesimal rotations) and the corresponding Lie group (finite rotations). The link between them is the (matrix) exponential function.
 
  • #7
Stephen Tashi
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I am hoping for a "simple" explanation of the following - if infinitesimal rotations commute and act as vectors I can add them together.

What is Kleppner's definition of an "infinitesimal rotation"? You say it is a vector; is it also a matrix? Are you talking about commuting with respect to addition of vectors or commuting with respect to multiplication of matrices - or something else?
 
  • #8
dyn
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Kleppner talks about commuting with respect to addition of vectors. Also the infinitesimal rotations are around different axes but they commute but as soon as the rotations are finite they do not commute
 
  • #9
Stephen Tashi
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Kleppner talks about commuting with respect to addition of vectors. Also the infinitesimal rotations are around different axes but they commute but as soon as the rotations are finite they do not commute

But in what sense do "rotations" not commute? Considered as vectors, they do commute (under addition). How does Kleppner define a rotation?
 
  • #10
Stephen Tashi
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Whatever Kleppner says, if you look up "infinitesmal rotation" in online sources, I think you'll find that infinitesimal rotations are defined in the context of linear transformations of vectors. So more than the concept of "vector" is involved. The concept of commuting vs non-commuting rotations does not refer to the question of whether vectors commute under the operation of addition in a vector space. By definition of vectors in a vector space, they must commute under the operation of vector addition in that space.

Physical phenomena such forces, masses, rotations etc. can be modeled by mathematical structures in various ways. Some times we automatically associate a particular physical phenomena (e.g. force) with a particular mathematical structure (e.g. vector). However, different aspects of a physical concept such as "rotation" can be modeled by different mathematical structures (e.g. linear transformation, element of a group, nxn matix, etc.) I suppose there can be a physical notion of what it means for two physical rotations to commute or not commute. But to answer a mathematical question about whether things commute or don't commute, we need a specific definition of what mathematical structures and operations are involved.
 
  • #11
dyn
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Thanks for your replies. I was looking for a "simple" explanation of the original question I posed but it seems that there is no "simple" explanation
 
  • #12
Stephen Tashi
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I was looking for a "simple" explanation of the original question I posed but it seems that there is no "simple" explanation

The first step to a simple explanation is to state the question precisely. That requires understanding what is meant by a "rotation" and what it means for rotations to commute or not to commute. Vectors commute under vector addition. So interpreting what you book claims about rotations not commuting requires grasping a concept that is different than the addition of vectors.
 
  • #13
vanhees71
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But in what sense do "rotations" not commute? Considered as vectors, they do commute (under addition). How does Kleppner define a rotation?
Rotations do not commute. Just take a rotation ##\hat{R}_1(\varphi)## and ##\hat{R}_3(\vartheta)## around the 1- and 3-axis and calculate the matrix products in different order to see that.

It's clear that the Lie algebra of "infinitesimal rotations" build a vector space (as any Lie algebra), but there's also an additional structure, the Lie product. If you use a matrix representation for the Lie algebra, which is natural here since you can define the Lie algebra in terms of matrices, it's the commutator of the infinitesimal rotation matrices, and those do not commute, and that's why finite rotations around different axis don't commute either.

A basis of the Lie algebra are the matrices ##\hat{t}_i## with ##(t_i)_{jk}=-\epsilon_{ijk}##. An infinitesimal rotation with the angle ##\delta \varphi## around an axis given by the unit vector ##\vec{n}## is given by ##\delta \phi \vec{n} \cdot \hat{\vec{t}}##. And you get the finite rotation around this axis by the differential equation
$$\mathrm{d}_{\varphi} \hat{R}(\vec{n},\varphi)=(\vec{n} \cdot \hat{\vec{t}}) \hat{R}(\vec{n},\varphi).$$
The solution is the matrix-exponential function
$$\hat{R}(\vec{n},\varphi)=\exp(\varphi \vec{n} \cdot \hat{\vec{t}}).$$
It's easy to see that rotation matrices around different axes don't commute, because also the generators ##\vec{n}_1 \cdot \hat{\vec{t}}## and ##\vec{n}_2 \cdot \hat{\vec{t}}## commute if and only if ##\vec{n}_2=\pm \vec{n}_1##.
 
  • #14
hutchphd
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Rotations do not commute. Just take a rotation ##\hat{R}_1(\varphi)## and ##\hat{R}_3(\vartheta)## around the 1- and 3-axis and calculate the matrix products in different order to see that.

It's clear that the Lie algebra of "infinitesimal rotations" build a vector space (as any Lie algebra), but there's also an additional structure, the Lie product. If you use a matrix representation for the Lie algebra, which is natural here since you can define the Lie algebra in terms of matrices, it's the commutator of the infinitesimal rotation matrices, and those do not commute, and that's why finite rotations around different axis don't commute either.

A basis of the Lie algebra are the matrices ##\hat{t}_i## with ##(t_i)_{jk}=-\epsilon_{ijk}##. An infinitesimal rotation with the angle ##\delta \varphi## around an axis given by the unit vector ##\vec{n}## is given by ##\delta \phi \vec{n} \cdot \hat{\vec{t}}##. And you get the finite rotation around this axis by the differential equation
$$\mathrm{d}_{\varphi} \hat{R}(\vec{n},\varphi)=(\vec{n} \cdot \hat{\vec{t}}) \hat{R}(\vec{n},\varphi).$$
The solution is the matrix-exponential function
$$\hat{R}(\vec{n},\varphi)=\exp(\varphi \vec{n} \cdot \hat{\vec{t}}).$$
It's easy to see that rotation matrices around different axes don't commute, because also the generators ##\vec{n}_1 \cdot \hat{\vec{t}}## and ##\vec{n}_2 \cdot \hat{\vec{t}}## commute if and only if ##\vec{n}_2=\pm \vec{n}_1##.
Here is where I get confused. If I take your formal solutions for two rotations
$$\hat{R}(\vec{n}_1,\varphi)=\exp(\varphi \vec{n}_1 \cdot \hat{\vec{t}}).$$ $$\hat{R}(\vec{n}_2,\vartheta)=\exp(\vartheta \vec{n}_2 \cdot \hat{\vec{t}}).$$
If I take the limit that each rotation is infinitesimal and formally expand the exponentials keeping linear order (in the cross product) do the results not commute? Clearly I am not facile with this...
 
  • #15
vanhees71
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You have to note that the exponential is not in linear order in the ##\hat{t}_j##. Even if you linearize the two exponentials the resulting matrices do not commute though the commutator is at 2nd order in the "small parameters". It's simply not true that "infinitesimal rotations commute" if these are not rotations around the same axis. The rotation group is NOT ABELIAN!

In other words: you cannot conclude from approximations of an expression to a fixed order in some expansion in powers of a small parameter about the exact expression. Though two rotations commute at linear order in the rotation angles they don't at higher orders!

No matter what the book says, obviously it's misleading at bet or even wrong at worst!
 
  • #16
hutchphd
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Pardon my lack of rigor but does one not usually blithely ignore the terms that are second order in the "small parameters". I am really asking here ...not just to be argumentative...I am a lousy mathematician!
 
  • #17
wrobel
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Physical phenomena such forces, masses, rotations etc. can be modeled by mathematical structures in various ways. Some times we automatically associate a particular physical phenomena (e.g. force) with a particular mathematical structure (e.g. vector). However, different aspects of a physical concept such as "rotation" can be modeled by different mathematical structures (e.g. linear transformation, element of a group, nxn matix, etc.) I suppose there can be a physical notion of what it means for two physical rotations to commute or not commute. But to answer a mathematical question about whether things commute or don't commute, we need a specific definition of what mathematical structures and operations are involved.
sounds as if properties of a physical phenomenon depend on what mathematical tool we use to describe it
 
  • #18
Stephen Tashi
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No matter what the book says, obviously it's misleading at bet or even wrong at worst!

With the goal of answering the OP in mind, the second edition of An Introduction to Mechanics by Kleppner and Kolenkow can be found at various sites online. Glancing over it, the text does not introduce Lie Groups or Lie Algebras. It does everything by what I'd call vector calculus.

On pages 329-330 of the book (page 351-352 of the PDF version), it says:

Note 8.1 Finite and Infinitesimal Rotations
In this Note we shall demonstrate that the order in which rotations are made is important if the rotations are large—finite rotations do not commute—but not if they are small—infinitesimal rotations do commute. By an infinitesimal rotation we mean one for which all powers of the rotation angle beyond the first can be neglected. Commutatitivity is important because it allows us to treat small angular displacements as components of a vector.

On page 331:
The equality of Eqs. (3) and (4) indicates that the result of two infinitesimal rotations can be found by evaluating the effects of the rotations independently. To first order, the effect of rotating r = r î through Δα about z is to generate a y component rΔα ĵ. The effect of rotating r through Δβ about y is to generate a z component, −rΔβ k̂. The total change in r to first order is the sum of the two effects, Δr = rΔα ĵ − rΔβ k̂, in agreement with Eq. (5).

The mathematical question requires interpreting what the book means by "evaluating the effects".
 
  • #19
Stephen Tashi
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And you get the finite rotation around this axis by the differential equation
$$\mathrm{d}_{\varphi} \hat{R}(\vec{n},\varphi)=(\vec{n} \cdot \hat{\vec{t}}) \hat{R}(\vec{n},\varphi).$$

This illustrates an aspect of the physical notion of rotation not present in some mathematical models of rotation. Mathematically, one can speak of rotating a vector as multiplying it by a (constant) matrix. You multiply and - Zap! - the vector is rotated. By contrast, an aspect of the physical notion of rotating an object is that it passes through a series of intermediate positions on the way to its final position.
 
  • #20
vela
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Thanks for your replies. The last one was over my head. I am hoping for a "simple" explanation of the following - if infinitesimal rotations commute and act as vectors I can add them together. So I can add 2 together , 3 together and so on and the resultant rotations will still commute. But if I add "too many" of them together they stop commuting and don't act as vectors.
If you combine a finite number of infinitesimals, you still get an infinitesimal. To get a finite rotation, you have to combine an infinite number of infinitesimal rotations. It's the transition from finite to infinite where you lose the commutativity.

You may just need to rely on the math here. I don't think you're going to find a simple or intuitive explanation. Mathematicians will tell you that just because a statement is true for finite cases, you can't assume it's true for the infinite case. There are other examples that I can't recall at the moment, but I don't think it's unusual.
 
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  • #21
vanhees71
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sounds as if properties of a physical phenomenon depend on what mathematical tool we use to describe it
No, but it depends on the correct use of the mathematical tools!
 
  • #22
vanhees71
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If you combine a finite number of infinitesimals, you still get an infinitesimal. To get a finite rotation, you have to combine an infinite number of infinitesimal rotations. It's the transition from finite to infinite where you lose the commutativity.

You may just need to rely on the math here. I don't think you're going to find a simple or intuitive explanation. Mathematicians will tell you that just because a statement is true for finite cases, you can't assume it's true for the infinite case. There are other examples that I can't recall at the moment, but I don't think it's unusual.
No, it's very simple!

Again: rotations around a fixed axis commute, no matter whether they are finite or you use an infinitesimal approximation.

As an alternative to my previous derivation in #13 of the matrix-exponential which maps the Lie algebra to the Lie group you may consider a rotation by a finite angle ##\varphi## around an axis given by the unit vector ##\vec{n}## (the usual right-hand rule applies).

Make ##N## a large natural number. Then ##\varphi/N## can be made arbitrarily small and you can use the linear approximation
$$\hat{R}(\vec{n},\varphi/N)=(\hat{1} + \vec{n} \cdot \hat{\vec{t}} \varphi/N).$$
Now
$$\hat{R}(\vec{n},\varphi_1+\varphi_2)= \hat{R}(\vec{n},\varphi_1) \hat{R}(\vec{n},\varphi_2),$$
which also shows formally that the rotations around one fixed axis commute. This equation can be used to write
$$\hat{R}(\vec{n},\varphi)=\hat{R}(\vec{n},N \varphi/N) = [\hat{R}(\vec{n},\varphi/N)]^N.$$
In the limit ##N \rightarrow \infty## the error made when using the linear approximation for ##\hat{R}(\vec{n},\varphi/N)## becomes arbitrarily small and thus you can write
$$\hat{R}(\vec{n},\varphi) = \lim_{N \rightarrow \infty} \left (\hat{1} + \vec{n} \cdot \hat{\vec{t}} \frac{\varphi}{N} \right) ^{N}=\exp(\vec{n} \cdot \hat{\vec{t}} \varphi).$$
 
  • #23
vanhees71
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This illustrates an aspect of the physical notion of rotation not present in some mathematical models of rotation. Mathematically, one can speak of rotating a vector as multiplying it by a (constant) matrix. You multiply and - Zap! - the vector is rotated. By contrast, an aspect of the physical notion of rotating an object is that it passes through a series of intermediate positions on the way to its final position.
Nowhere is anything about physical motion involved here. It's a purely mathematical derivation of the relation between Lie algebra and Lie group.
 
  • #24
vanhees71
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With the goal of answering the OP in mind, the second edition of An Introduction to Mechanics by Kleppner and Kolenkow can be found at various sites online. Glancing over it, the text does not introduce Lie Groups or Lie Algebras. It does everything by what I'd call vector calculus.
The result is an utterly confused student! One should do things as simple as possible but not simpler (Einstein).
 
  • #25
dyn
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The result is an utterly confused student! One should do things as simple as possible but not simpler (Einstein).

Thank you for your efforts in trying to explain this. The problem is that when you study mechanics at the level of Kleppner , you have not yet studied Lie algebra so are looking to try and understand things in terms of vectors and vector addition
 
  • #26
vanhees71
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It's a great opportunity to introduce the idea of the Lie algebra for this quite intuitive example of the rotation group. Note that you can start with rotation around a fixed axis here, which is even an Abelian group! If you don't want to do this, as obviously this textbook, you should stay away from incomplete and confusing statements about infinitesimal rotations, though you cannot avoid it without leaving out large parts of interesting mechanics, i.e., rotational motion, including angular velocity and finally rigid-body dynamics. For that you don't necessarily need Lie-algebra and Lie-group theory but simply time derivatives of rotational matrices, and this can be nicely treated without confusing statements about infinitesimal rotations and their commutators by just taking derivatives!
 
  • #27
wrobel
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Actually kinematics and dynamics of rigid body do not require neither Lie group theory in general nor even any of its elements . This theory is used in some very special and advanced topics ( integrability for example, Poincare equations etc). Moreover this is true for the basic concepts such as angular velocity, equations of motion of the rigid body. All these things can be completely rigorously deduced without Lie groups.
 
  • #28
vanhees71
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As I said, you can treat the rigid body without any reference to Lie groups, though you hiddenly use nothing else than that ;-)).
 
  • #29
vanhees71
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$$\newcommand{\uvec}[1]{\underline{#1}}$$
With Lie-group theory that's very easy to understand. You have the rotation ##\hat{D}(t)## between the body-fixed basis and the space-fixed basis. Then you have
$$\hat{D}(t)=\exp(\vec{\varphi} \cdot \hat{\vec{t}})$$
and
$$\hat{D}(t+\mathrm{d} t)=\hat{D}(t) [\hat{1}+\mathrm{d} t \dot{\vec{\varphi}} \cdot \hat{\vec{t}}],$$
from which you get
$$\dot{\hat{D}}(t)=\dot{\vec{\varphi}} \cdot \hat{\vec{t}} = \hat{D} \vec{\omega} \cdot \hat{\vec{t}},$$
Then for any vector (unprimed: space-fixed components, primed: body-fixed components).
$$\mathrm{d}_t \uvec{V}=\mathrm{d}_t (\hat{D} \uvec{V}') = \hat{D} \dot{\uvec{V}} + \dot{\hat{D}} \uvec{V}' = \hat{D} (\dot{\uvec{V}}'+\vec{\omega} \cdot \hat{\vec{t}} \uvec{V}').$$
From this you get the body-fixed components of ##\dot{\vec{V}}##
$$\mathrm{D}_t \uvec{V}'=\hat{D}^{-1} \dot{\uvec{V}} = \dot{\uvec{V}}' + \vec{\omega} \cdot \hat{\vec{t}} \uvec{V}'.$$
To bring this in the usual notation, just remember that the here used Cartesian basis of the Lie algebra is given by
$$(\hat{t}_i)_{jk}=-\epsilon_{ijk} \; \Rightarrow \; (\vec{\omega} \cdot \vec{\hat{t}} \uvec{V}')_j=-\epsilon_{ijk} \omega_i' V_{k}'=+\epsilon_{jik} \omega_i' V_k'=(\uvec{\omega}' \times \uvec{V}')_{j},$$
i.e.,
$$\mathrm{D}_t \uvec{V}'=\dot{\uvec{V}}'+\uvec{\omega}' \times \uvec{V}'.$$
 
  • #30
Stephen Tashi
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when you study mechanics at the level of Kleppner , you have not yet studied Lie algebra so are looking to try and understand things in terms of vectors and vector addition
As I've said before, in discussing commuting or non-commuting, the first step is to define the mathematical process that is under consideration. Kleppner (and several of the respondents) have not defined it (nor have "rotation" and "infinitesimal rotation" been defined). Based on my scan of the Kleppner text, it considers "rotation" to be a rotation (of some object) about an axis by a given angle and mathematically represented by a 3-D vector pointing along the axis (in the "right hand" direction) with a magnitude equal to the angle in question. An "infinitesimal rotation" is only defined in an intuitive way - subject to all the controversies and interpretations that surround the notion of infinitesimals in calculus.

Kleppner's remarks about commuting or non-commuting refer to two different mathematical processes of "evaluating effects".

Situation 1) We want to evaluate the position of a point on a body initially at (x0,y0,z0) after we perform several rotations. Can we do this by adding the vector representations of the rotations together and calculating the final position as if this sum represents a single rotation? No, that doesn't work. The order in which the rotations are performed does matter.

Situation 2) We want to evaluate the velocity of a point on a body that is mounted on an apparatus that is simultaneously rotating it about two different axes that pass through a common point in space. (For example, a rotisserie turning a roast mounted on a turntable whose axis intersects the rod of the rotisserie.) A given angular velocity of a known point about a known axis let's us calculate a 3-D velocity vector for the point. One way to do the calculation is to find the 3-D velocity vector for the point about each axis and add the two 3-D velocity vectors together. Another way is to add the two angular velocity vectors together and use the result to represent a single angular velocity about a single axis. Then calculate a single 3-D velocity vector from that result.

As to how situation 2) relates to "infinitesmal rotations", people can have a variety of opinions, depending on how they think about infinitesimals. Generally speaking velocities of a quantity q are often represented using notation that employs a "dq". How you want to think about such things is largely a matter of taste. There are fields of mathematics where such notation is rigorously defined, but that's not the approach used in most physics texts.

A more sophisticated answer to your question is given at: https://physics.stackexchange.com/q...-velocities-are-vectors-while-rotations-arent
 
  • #31
wrobel
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With Lie-group theory that's very easy to understand.
Oh! Now I see what you call Lie group theory. This is not yet Lie group theory this is just some trivial argument. Lie group theory is a serious mathematical branch. By the way, your formulas are too cumbersome. Particularly there is no need to refer on coordinate representations every time. Everything is much simpler.

Let ##u## be any vector frozen into a rigid body. The rigid body moves so that ##u=u(t)##. Define a linear operator $$A(t)u(0)=u(t).\qquad (!)$$ It is a unitary operator it's clear:
$$AA^*=I,\qquad (*)$$
Actually, formulas (*) and (!) are the definition of a rigid body.

Differentiate formula (!):
$$\dot u(t)=\dot A(t)u(0)=\dot A(t)A^{-1}(t)u(t)=\dot A(t)A^*(t)u(t).\qquad (**)$$
The operator ##\Omega=\dot AA^*## is skew-symmetric: ##\Omega^*=-\Omega##. Indeed, just take ##\frac{d}{dt}## from both sides of (*).

And finally, any skew-symmetric operator in ##\mathbb{R}^3## can unequally be presented with the help of (pseudo) vector such that ##\Omega x=\omega\times x,## and correspondingly formula (**) takes the form ##\dot u(t)=\omega\times u(t)##
 
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  • #32
vela
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No, it's very simple!
Simple for you, but probably not to the OP! Personally, I think talk of Lie groups, Lie algebra, and the like is too advanced for an I-level thread.
 

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