Here I am interested in taking the inner product of two vectors with respect to the eigen-basis of the rotation matrices.
To calculate the dot product between two vectors one first needs two vectors, \(\vec{u}\) and \(\vec{v}\) at an angle \(\angle uOv\), denoted by \(\theta\), from each other. It is possible to bisect the angle \(\theta\) and define a unit bisection vector \(\hat{B}_{\parallel}\) at an angle \(\alpha=\frac{\theta}{2}\) from both \(\vec{u}\) and \(\vec{v}\). Since the two vectors, \(\vec{u}\) and \(\vec{v}\) define a plane, we are then free to choose an eigen-vector of the rotation matrix:
$$
\mathbf{R}^{3}_{\parallel}\left(\phi_{\parallel}\right)=
\begin{pmatrix}
1 & 0 & 0 \\
0 & \cos\phi_{\parallel} & -\sin\phi_{\parallel}\\
0 & \sin\phi_{\parallel} & \cos\phi_{\parallel}
\end{pmatrix}
$$
to be the unit bisecting vector \(\hat{B}_{\parallel}\) parallel to the plane. From the previous post the reader should be aware that there are three such eigen-vectors. Note that I have included an unknown angle \(\phi\) to take into consideration that the plane defined by the two vectors \(\vec{u}\) and \(\vec{v}\) can be oriented by any angle \(\phi\) around the bisecting unit vector, \(\hat{B}_{\parallel}\).
The vector perpendicular to \(\hat{B}_{\parallel}\), denoted by \(\hat{B}_{\perp}\), is then defined as an eigen-vector of the rotation matrix:
$$
\mathbf{R}^{3}_{\perp}\left(\phi_{\perp}\right)=
\begin{pmatrix}
\cos\phi_{\perp} & 0 & \sin\phi_{\perp} \\
0 & 1 & 0\\
-\sin\phi_{\perp} & 0 & \cos\phi_{\perp}
\end{pmatrix}
$$
The inner product of the two vectors, \(\vec{u}\) and \(\vec{v}\), with regards to the eigen-basis of \(\mathbf{R}^{3}_{\perp}\left(\phi_{\perp}\right)\) and \(\mathbf{R}^{3}_{\parallel}\left(\phi_{\parallel}\right)\) is then given by
$$
\vec{u}\cdot\vec{v}=
\left\|\vec{u}\right\| \left\|\vec{v}\right\| \cos^{2}\left(\alpha\right) \mathbf{X}^{\left(\phi_{\parallel}\right)}_{\parallel}\cdot\mathbf{X}^{\left(\phi_{\parallel}\right)}_{\parallel}
-
\left\|\vec{u}\right\| \left\|\vec{v}\right\| \sin^{2}\left(\alpha\right)
\mathbf{X}^{\left(\phi_{\perp}\right)}_{\perp}\cdot\mathbf{X}^{\left(\phi_{\perp}\right)}_{\perp} \\
=
\left\|\vec{u}\right\| \left\|\vec{v}\right\|
\left[\cos^{2}\left(\frac{\theta}{2}\right)
-
\sin^{2}\left(\frac{\theta}{2}\right)
\right] \\
=
\left\|\vec{u}\right\| \left\|\vec{v}\right\| \cos\left(\theta\right)
$$
This dot product by itself is already interesting when written in terms of the eigen-vectors \(\mathbf{X}^{\left(\phi_{\parallel}\right)}_{\parallel}\) and \(\mathbf{X}^{\left(\phi_{\perp}\right)}_{\perp}\) due to the introduction of the negative sign in front of the term corresponding to the perpendicular components.
Of course, by inspection this makes perfect sense. That is, we would usually say the unit vector for the perpendicular components point toward each other.
However, to be more rigorous, one needs to find the relationship between the eigen-vectors of \(\mathbf{R}^{3}_{\perp}\left(\phi_{\perp}\right)\) and how they relate to the rotation of the bisecting unit vector, \(\hat{B}_{\parallel}\).
As a prelude to further discussion, in physics, one might also be interested in how quickly such a rotation of the bisecting unit vector occurs.
Wednesday, March 7, 2012
Tuesday, March 6, 2012
Rotation Matrices (Cont.)
All of these rotation matrices have the same eigen-values:
$$
\lambda_{i}\in\{1,\pm i\}
$$
though they have different eigen-vectors:
$$
\mathbf{X}^{\left(\pm\right)}_{\hat{n}}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
e^{\frac{-i\pi}{4}} & e^{\frac{-i\pi}{4}} & 0 \\
\mp e^{\frac{i\pi}{4}} & \pm e^{\frac{i\pi}{4}} & 0 \\
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}}
\end{pmatrix} \\
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\parallel}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}} \\
\pm e^{\frac{-i\pi}{4}} & \pm e^{\frac{-i\pi}{4}} & 0 \\
- e^{\frac{i\pi}{4}} & e^{\frac{i\pi}{4}} & 0
\end{pmatrix} \\
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\perp}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
e^{\frac{-i\pi}{4}} & e^{\frac{-i\pi}{4}} & 0 \\
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}} \\
\pm e^{\frac{i\pi}{4}} & \mp e^{\frac{i\pi}{4}} & 0
\end{pmatrix}
$$
What's interesting is assigning some sort of coordinate system to these eigen-vectors. For example, one could say that the third eigen-vector of the set of eigen-vectors for \(\mathbf{X}^{\left(\pm\right)}_{\hat{n}}\) corresponds to \(\hat{z}\), that is to say:
$$
\mathbf{X}^{\left(\pm\right)}_{\hat{n}}= \begin{pmatrix}
0\\
0\\
1
\end{pmatrix}=\hat{z}
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\perp}= \begin{pmatrix}
0\\
1\\
0
\end{pmatrix}=\hat{y}
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\parallel}= \begin{pmatrix}
1\\
0\\
0
\end{pmatrix}=\hat{x}
$$
But, you may ask yourself, we have other eigen-vectors, what about those? What do they correspond to? Or, more importantly, what can I use them for?
Well, that's the part that's interesting. I've run out of time and will continue this later tonight.
P.S. Finding the eigen-vectors and eigen-values is easily verified with qtoctave or other linear algebra software.
$$
\lambda_{i}\in\{1,\pm i\}
$$
though they have different eigen-vectors:
$$
\mathbf{X}^{\left(\pm\right)}_{\hat{n}}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
e^{\frac{-i\pi}{4}} & e^{\frac{-i\pi}{4}} & 0 \\
\mp e^{\frac{i\pi}{4}} & \pm e^{\frac{i\pi}{4}} & 0 \\
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}}
\end{pmatrix} \\
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\parallel}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}} \\
\pm e^{\frac{-i\pi}{4}} & \pm e^{\frac{-i\pi}{4}} & 0 \\
- e^{\frac{i\pi}{4}} & e^{\frac{i\pi}{4}} & 0
\end{pmatrix} \\
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\perp}=\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}}
\begin{pmatrix}
e^{\frac{-i\pi}{4}} & e^{\frac{-i\pi}{4}} & 0 \\
0 & 0 & \sqrt{2}e^{\frac{-i\pi}{4}} \\
\pm e^{\frac{i\pi}{4}} & \mp e^{\frac{i\pi}{4}} & 0
\end{pmatrix}
$$
What's interesting is assigning some sort of coordinate system to these eigen-vectors. For example, one could say that the third eigen-vector of the set of eigen-vectors for \(\mathbf{X}^{\left(\pm\right)}_{\hat{n}}\) corresponds to \(\hat{z}\), that is to say:
$$
\mathbf{X}^{\left(\pm\right)}_{\hat{n}}= \begin{pmatrix}
0\\
0\\
1
\end{pmatrix}=\hat{z}
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\perp}= \begin{pmatrix}
0\\
1\\
0
\end{pmatrix}=\hat{y}
$$
$$
\mathbf{X}^{\left(\pm\right)}_{\parallel}= \begin{pmatrix}
1\\
0\\
0
\end{pmatrix}=\hat{x}
$$
But, you may ask yourself, we have other eigen-vectors, what about those? What do they correspond to? Or, more importantly, what can I use them for?
Well, that's the part that's interesting. I've run out of time and will continue this later tonight.
P.S. Finding the eigen-vectors and eigen-values is easily verified with qtoctave or other linear algebra software.
A=[0,-1,0;1,0,0;0,0,1]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
A=[0,1,0;-1,0,0;0,0,1]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
A=[1,0,0;0,0,-1;0,1,0]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
A=[1,0,0;0,0,1;0,-1,0]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
A=[0,0,1;0,1,0;-1,0,0]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
A=[0,0,-1;0,1,0;1,0,0]
det(A)
eig(A)
[EVECT,EVAL]=eig(A)
Rotation Matrices
So I recently took interest in having fun with rotation matrices by analyzing two vectors, their inner products, their cross products, and their bisecting vectors after my previous post. Basically, the rotation matrices I am interested in are:
$$
\mathbf{R}^{3}_{\hat{n}}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
0 & \mp 1 & 0 \\
\pm 1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}\\
\mathbf{R}^{3}_{\parallel}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & \mp 1\\
0 & \pm 1 & 0
\end{pmatrix}\\
\mathbf{R}^{3}_{\perp}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
0 & 0 & \pm 1 \\
0 & 1 & 0\\
\mp 1 & 0 & 0
\end{pmatrix}
$$
I'ts kinda late, so I'll finish explaining why this is interesting tomorrow.
$$
\mathbf{R}^{3}_{\hat{n}}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
0 & \mp 1 & 0 \\
\pm 1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}\\
\mathbf{R}^{3}_{\parallel}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & \mp 1\\
0 & \pm 1 & 0
\end{pmatrix}\\
\mathbf{R}^{3}_{\perp}\left(\pm\frac{\pi}{2}\right)=
\begin{pmatrix}
0 & 0 & \pm 1 \\
0 & 1 & 0\\
\mp 1 & 0 & 0
\end{pmatrix}
$$
I'ts kinda late, so I'll finish explaining why this is interesting tomorrow.
Monday, March 5, 2012
Rotations: transformations that preserve length of and angle between, at least, two vectors
Wikipedia has a nice article on the SO(3) group where they start by stating that:
Rotation Group SO(3)
$$
\vec{u}\cdot\vec{v} =
\frac{1}{2}\left(
\left\|\vec{u} + \vec{v}\right\|^{2} - \left\|\vec{u}\right\|^{2} - \left\|\vec{v}\right\|^{2}
\right)
$$
However, the reader may not be aware of the fact that:
$$
\vec{u}\cdot\vec{v} = \left\|\vec{u}\right\|\left\|\vec{v}\right\|\cos\left(\theta\right)
$$
or
$$
\cos\left(\theta\right)=\frac{\vec{u}\cdot\vec{v}}{\left\|\vec{u}\right\|\left\|\vec{v}\right\|}
$$
where one can explicitly see that, by definition of the dot product, if the coordinate system of \(\vec{u}\) and \(\vec{v}\) undergoes any rotation, where the dot product is conserved, and we know that the lengths of \(\vec{u}\) and \(\vec{v}\) are conserved, then the angle \(\theta\), that is the angle \(\angle uOv\), can not change and is, therefore, also conserved.
(Really, when I saw that Wikipedia ignored readers that didn't know the definition of the dot (inner) product, I felt this was just a good excuse to play with latex in blogger. Though it left me wanting to be able to post graphs & vector diagrams in my blog, a la LaTeX.)
Rotation Group SO(3)
"length-preserving transformation in R3 preserves the dot product, and thus the angle between vectors"They then continue by stating that:
$$
\vec{u}\cdot\vec{v} =
\frac{1}{2}\left(
\left\|\vec{u} + \vec{v}\right\|^{2} - \left\|\vec{u}\right\|^{2} - \left\|\vec{v}\right\|^{2}
\right)
$$
However, the reader may not be aware of the fact that:
$$
\vec{u}\cdot\vec{v} = \left\|\vec{u}\right\|\left\|\vec{v}\right\|\cos\left(\theta\right)
$$
or
$$
\cos\left(\theta\right)=\frac{\vec{u}\cdot\vec{v}}{\left\|\vec{u}\right\|\left\|\vec{v}\right\|}
$$
where one can explicitly see that, by definition of the dot product, if the coordinate system of \(\vec{u}\) and \(\vec{v}\) undergoes any rotation, where the dot product is conserved, and we know that the lengths of \(\vec{u}\) and \(\vec{v}\) are conserved, then the angle \(\theta\), that is the angle \(\angle uOv\), can not change and is, therefore, also conserved.
(Really, when I saw that Wikipedia ignored readers that didn't know the definition of the dot (inner) product, I felt this was just a good excuse to play with latex in blogger. Though it left me wanting to be able to post graphs & vector diagrams in my blog, a la LaTeX.)
A more physical example.
Another example:
$$\hat{H}\left|\Psi\right>=i\hbar\partial_{t}\left|\Psi\right>$$
I am really looking forward to having fun with this blog :-D.
$$\hat{H}\left|\Psi\right>=i\hbar\partial_{t}\left|\Psi\right>$$
$$\hat{H}\left|\Psi\right>=i\hbar\partial_{t}\left|\Psi\right>$$
I am really looking forward to having fun with this blog :-D.
Sunday, March 4, 2012
Adding Latex Math to blogger.
I have been randomly checking how to add math to my blogger page in hopes of better communicating with readers. I recently came upon a link that uses www.watchmath.com to accomplish this task. The basic steps, as of this date to get this working are:
1. Add an HTML/JavaScript footer Gadget to the blog for interpretation of latex code in the document body.
2. If new to latex and ease of learning is important, you can include a WYSIWYG equation editor by adding another HTML/JavaScript footer.
3. Thank the great folks over at MathJax, CodeCogs, watch math, and Aam Sudrajat's Blog for the tips. For formating code to a bloggable format see: http://formatmysourcecode.blogspot.com/
A brief example can be seen here:
$$\alpha \beta \gamma$$
Update: There was an error with Aam's page and the solution can be found here: http://mytechmemo.blogspot.com/2012/02/how-to-write-math-formulas-in-blogger.html
1. Add an HTML/JavaScript footer Gadget to the blog for interpretation of latex code in the document body.
<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
2. If new to latex and ease of learning is important, you can include a WYSIWYG equation editor by adding another HTML/JavaScript footer.
<script type="text/javascript" src="http://latex.codecogs.com/editor.js"></script><p align="center" style="margin-top: 0; margin-bottom: 0"><font face="Arial">
Sample:
<a href="javascript:OpenLatexEditor('testbox','latex','')">
<img src="http://www.codecogs.com/gif.latex?\sum_{i=1}^{n}{(X_i - \overline{X})^2}"
align="middle" /></a></font></p>
<p align="center" style="margin-top: 0; margin-bottom: 0"><b><font face="Arial">
<a href="javascript:OpenLatexEditor('testbox','latex','')">
Launch
Editor</a></font></b></p>
<p align="center" style="margin-top: 0; margin-bottom: 0">
<textarea id="testbox" rows="10" cols="20"></textarea></p>
<p align="center" style="margin-top: 0; margin-bottom: 0">
<a href="http://www.codecogs.com" target="_blank">
<img src="http://www.codecogs.com/images/poweredbycc.gif" border="0"
title="CodeCogs - An Open Source Scientific Library"
alt="CodeCogs - An Open Source Scientific Library" /></a>
<p style="margin-top: 0; margin-bottom: 0" align="center"><font size="1"><font face="Arial">
<a href="http://a2mstats.blogspot.com/">Aam Sudrajat</a></font> </font>
</p>
3. Thank the great folks over at MathJax, CodeCogs, watch math, and Aam Sudrajat's Blog for the tips. For formating code to a bloggable format see: http://formatmysourcecode.blogspot.com/
A brief example can be seen here:
$$\alpha \beta \gamma$$$$\alpha \beta \gamma$$
Update: There was an error with Aam's page and the solution can be found here: http://mytechmemo.blogspot.com/2012/02/how-to-write-math-formulas-in-blogger.html
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