# Talk:Euler's rotation theorem

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## quaternions

Why is this article more about quaternions than it is about Euler rotation? One reference to quaternion algebra is adequate. The topic has been obfuscated by a writer who wants to steer readers. 65.175.151.8 15:08, 24 October 2006 (UTC)Tim Golden BandTechnology.com

they are intimately connected, same thing in different disguises. IMHO, the section on quaternions could use some elaboration and clarification. Mct mht 01:42, 8 February 2007 (UTC)

## unclear about number of parameters and quaternion significance

• Technically one needs only 3 numbers to represent an arbitrary rotation, since the rotation axis can be normalized, with its magnitude representing the angle (I suspect that this closer to what Euler's rotation theorem states)
yes, you're right. maybe that'll get corrected. Mct mht 01:42, 8 February 2007 (UTC)
• The article makes it sound as if any set of 4 numbers representing a rotation is called a quaternion. This is untrue, since a quaternion has a specific mathematical meaning, and is used to represent 3D rotation in a specific way.
right there also. current version can certainly be improved. rotations corresponds to unit quaternions, in a "1 to 2" way, meaning unit quaternions is a double cover of SO(3). Mct mht 01:42, 8 February 2007 (UTC)

## Show calculation of axis and angle of rotation

Should we show how to calculate the angle and axis of rotation? I have the formula here:

As a corollary of the theorem follows that the trace of a rotation matrix (R) is equal to:

$Trace[R] = 2 * cos(\theta) +1$

so s can be written as:

$s = \arccos{\left[ \frac{Trace(R) - 1}{2} \right]} = \arccos{\left[ \frac{r_{00} + r_{11} + r_{22} - 1}{2} \right]}$

Second, the unique axis of rotation is defined as:

$\vec{k} = \frac{1}{2 \sin{(\theta)}} * \left( \begin{matrix} r_{21}-r_{12} \\ r_{02}-r_{20} \\ r_{10}-r_{01} \end{matrix} \right)$

and the magnitude:

$|\vec{k}| = \frac{1}{2 \sin{(\theta)}} * \sqrt{(r_{21}-r_{12})^2 + (r_{02}-r_{20})^2 + (r_{10}-r_{01})^2}$

BUT I need a citation for these values...

## Excursion into matrix theory

The statement

An m×m matrix A has m orthogonal eigenvectors if and only if A is normal.

is false. The statement should read

An m×m matrix A can be diagonalised if and only if A is normal. An orthogonal set of m eigenvectors can be obtained if A is hermitian. The eigenvectors of a rotation matrix are not in general mutually orthogonal.

Really? "a matrix is normal if and only if its eigenspaces span Cn and are pairwise orthogonal with respect to the standard inner product of Cn" (Normal matrix#Consequences). Boris Tsirelson (talk) 10:33, 19 April 2010 (UTC)

## The matrix proof is wrong

The proof shows that a rotation matrix has an invariant vector. Period. It does not show that "in 3D space, any two Cartesian coordinate systems with a common origin are related by a rotation about some fixed axis". When saying "An algebraic proof starts from the fact that a rotation is a linear map in one-to-one correspondence withi a 3×3 rotation matrix R", you are already assuming what you want to show. That is, that the cartesian coordinates are related by a rotation about an axis. --190.188.3.11 (talk) 16:22, 23 June 2010 (UTC)

Two cartesian coordinates are related by a rotation matrix whose columns are the vectors of the second frame. The matrix has a real eigenvector v and is linear, therefore any vector k.v is also a eigenvector. This eigenspace is the rotation axis.--94.126.240.2 (talk) 07:36, 25 June 2010 (UTC)
You say that "Two cartesian coordinates are related by a rotation matrix whose columns are the vectors of the second frame." It seems this should be mentioned in the article, and proved too. Although it may seem obvious that an isometry S such that S (0) = 0 is linear, this has to be shown. So it seems that several things are being assumed because they are "obvious", making the proof not rigorous.--190.188.3.11 (talk) 04:12, 2 July 2010 (UTC)
I can't see any problems with it. The only thing it asserts is that the rotation can be represented by a matrix with the property
$\mathbf{R}^\mathrm{T}\mathbf{R} = \mathbf{R}\mathbf{R}^\mathrm{T} = \mathbf{E}$,
i.e. its inverse is its transpose. This is non-obvious but is easy enough to show by e.g. building the matrix up from rotations about three orthogonal axes. All rotations can be generated this way, otherwise gimbals would not work. Other than that the only thing established is that the determinant is +1 and not -1, but that follows from considering the volume-preserving nature of rotations. From these properties the unit root is found algebraically, and the existence of a fixed vector deduced from the theory of eigenvectors. --JohnBlackburnewordsdeeds 11:41, 2 July 2010 (UTC)
I do not have all these things very clear, but the article does not define what a rotation is. What is a rotation? The algebraic proof shows that a rotation matrix has an invariant vector; it does not show that two cartesian coordinates are related by a rotation. Reflextions don´t also preserve volume? As far as I know, that the determinant is 1 is shown using continuity arguments. The article as written does not seem to be correct.

--190.188.3.11 (talk) 20:24, 7 August 2010 (UTC)

The definition is in the first sentence: a rotation is "any displacement of a rigid body such that a point on the rigid body remains fixed". It also links to rotation group and rotation matrix for more algebraic treatments. Then in the first section it gives Euler's definition.
I think it gives it this general definition to make it clear that the definition does not assume there's a fixed axis: Euler's theorem is a statement that such an axis always exists, an a proof of it is a proof of the existence of such an axis, so it cannot be assumed. So a very general definition of rotation is needed. From this and from properties that you can assume (such as a rotation is always volume preserving, so the matrix has determinant 1) the theorem is proved.--JohnBlackburnewordsdeeds 20:55, 7 August 2010 (UTC)

## Incorrect rendering of alpha character in Safari browser

In the derivation that appears in the article, the Greek character alpha appears as one of the variables. Despite the fact that alpha is a common Greek character, it displays as an italicized "a" in the Safari browser, as I can confirm by copying the character and pasting it into Microsoft Word. The same process in Firefox gives the correct alpha character. Since both the letter a and alpha appear in the derivation, the derivation can be quite confusing when all alphas are replaced with the letter "a." This confusion can be minimized by replacing the letter alpha with the LaTex markup $\alpha$. It doesn't appear that this character is rendered as a PNG image, because the font size appears exactly the same as the other characters. In the Manual of Style (mathematics), it specifies that "if you enter a very simple formula using the mathematical notations like $L^p$ this will (in the default used by most users) not be displayed using a PNG image but using HTML, like this: Lp." Jdlawlis (talk) 04:49, 10 May 2011 (UTC)

It appears as a letter alpha to me. Here is a compact paragraph it appears in, copied in full:
Now Euler needs to construct point O in the surface of the sphere that is in the same position in reference to the arcs aA and αA. If such a point exists then:
The 'α' looks quite different from the 'a' in both the edit window and when rendered. It's worth remembering that many things control how things are rendered. The choice of browser (i.e. browser, OS/Platform and browser version), browser and OS settings, what fonts you have, the skin specified at Special:Preferences#preftab-1 and the Math formatting options on the same page. The latter is the one that varies the appearance of maths the most: it can mean that almost all [itex] tags are rendered as text or none are. It can even disable their interpretation altogether.
What this means is you can't make assumptions about how it will appear on other users machines from your own. In particular if you see a problem in an article, but the markup seems perfectly OK (and α, a Greek letter, is far from exotic or difficult for browsers to render) and is like many other articles, then it is more likely your setup needs adjusting. It is also why it says in the section of the MOS you highlight:
"Either form is acceptable, but do not change one form to the other in other people's writing."
--JohnBlackburnewordsdeeds 13:34, 10 May 2011 (UTC)

Fair enough. I'm only trying to highlight something that I find confusing. I don't think that I'm the only person who will run into this issue, but you have made your position clear. Jdlawlis (talk) 15:26, 10 May 2011 (UTC)

## Proposed edits to this article

I feel this is an important theorem that can be presented in a somewhat simpler way. I will developed this in the near future and hope that you find it to be of value.Prof McCarthy (talk) 01:17, 23 June 2011 (UTC)

## Demonstration with geometric algebra.

I agree with the observations made by 190.188.3.11 in " The matrix proof is wrong ". It is at least incomplete, that is one should indeed prove that any two cartesian coordinate systems (with same orientation) are related by a rotation.

I will try to prove that with " geometric algebra " (I know, some people here will make a grimace ... ).

Let us consider a unit sphere which can slide on itself. After some displacement the extremities of an $(e_1,e_2,e_3)$ reference frame will be situated at $(f_1,f_2,f_3)$. We must be more precise : that means we define arbitrarily two new points $(f_1,f_2)$ with the only condition $\mathbf(f_1). \mathbf(f_2) =0$, which assures that we respect conservation of distances and angles. Of course $f_3$ is determined by ${f_1} \times {f_2} = {f_3}$.

Each point of the sphere can of course be defined by its (invariant) coordinates relative to both frames, by definition of the displacement of a rigid body.

We know that a rotation in $(R^3)$ can be characterized by a rotor like :

$(1) \qquad R=\cos(\theta /2)+ I a \sin(\theta /2) \qquad \qquad R^{\dagger}=\cos(\theta /2)- I a \sin(\theta /2)$

where $a$ is the unit vector of the rotation axis, and $\theta$ the rotation angle.

What we have to do now is to verify that the knowledge of the couples $(e_1,f_1)$ and $(e_2,f_2)$ is necessary and sufficient to determine uniquely the rotor and thus the axis and rotation angle.

We must have :

$(2) \qquad f_1=R^{\dagger} e_1 R \qquad \qquad f_2=R^{\dagger} e_2 R \qquad \qquad f_1.f_2=0$

We can write successively :

$(3) \qquad R f_1=e_1 R$

$(4) \qquad (\cos(\theta /2)+ I a \sin(\theta /2))f_1=e_1(\cos(\theta /2)+ I a \sin(\theta /2))$

$(5) \qquad \cos(\theta /2)f_1+\sin(\theta /2) I (a.f_1+a \wedge f_1)=\cos(\theta /2)e_1+\sin(\theta /2) I (a.e_1-a \wedge e_1)$

$(6) \qquad \cos(\theta /2)(f_1-e_1)+ \sin(\theta /2)I a \wedge (f_1+e_1)+\sin(\theta /2)I a . (f_1-e_1)=0$

and :

$(7) \qquad \cos(\theta /2)(f_2-e_2)+ \sin(\theta /2)I a \wedge (f_2+e_2)+\sin(\theta /2)I a . (f_2-e_2)=0$

By splitting in pseudoscalar and vectorial parts we get:

$(6 bis) \qquad a.(f_1-e_1)=0 \qquad \qquad \cos(\theta /2)(f_1-e_1)+\sin(\theta /2)I a \wedge (f_1+e_1)=0$

$(7 bis) \qquad a.(f_2-e_2)=0 \qquad \qquad \cos(\theta /2)(f_2-e_2)+\sin(\theta /2)I a \wedge (f_2+e_2)=0$

As a consequence of the scalar relations we can write :

$(8)\qquad a=\lambda I (f_1-e_1)\wedge(f_2-e_2)$

where $\lambda$ is the inverse of the norm of $(f_1-e_1)\wedge(f_2-e_2)$.

We may calculate : $\lambda=[(f_1-e_1)^2(f_2-e_2)^2-(f_1-e_1).(f_2-e_2)]^{(-1/2)}$

As for the vectoriel relations, they should be coherent. After replacing $a$ by its value and simplifying in (6bis) and (7bis) we get :

$(6ter) \qquad \cos(\theta /2)-\lambda \sin(\theta /2)(f_2-e_2).(f_1+e_1)=0$

$(7ter) \qquad \cos(\theta /2)+\lambda \sin(\theta /2)(f_1-e_1).(f_2+e_2)=0$

These equations coincide if :

$(9)\qquad -(f_2-e_2).(f_1+e_1)=(f_1-e_1).(f_2+e_2)$

That is true if and only if the condition $\mathbf (f_1) . \mathbf (f_2) =0$ is verified.

Thus we get:

$(10) \qquad \cot(\theta /2)=\lambda (f_2-e_2).(f_1+e_1)=\lambda (f_2 .e_1-f_1 .e_2)$

and the rotation is unambiguously determined.

Obviously the same demonstration can be made starting with non-orthogonal points and the condition $f_1.f_2=e_1.e_2$ .

Chessfan (talk) 20:02, 20 January 2012 (UTC)

## Euler pole

As far as I know, the single rotation axis is typically called "Euler axis". I guess that the "Euler pole" is not the Euler axis, but just the fixed point around which the body rotates. See this image, for instance: http://www.earth.northwestern.edu/people/seth/202/new_2004/euler_pole.html Paolo.dL (talk) 16:30, 2 May 2013 (UTC)