Poem: Orbits

Orbits
By Sergio Tafur - 18 April, 2021
Composed while listening to Aníron, by Enya
https://youtu.be/iMyo8I8AKmY

----

Like celestial Thenard spheres 
Grasped by a cosmic force
We orbited round the other 
On an ineluctable course

Our light from afar we saw
Where preempting our arrival 
Our nucleons did excite

And you arrived
My skin ablaze
And you spoke

Your words angelic

And my heart leapt
And we collide

Leaving a part 
A swaying nova

A million burgundy suns 
Whose spots shone through

Birth and death in an eternal blink

Diamonds remaining in white-stars 
Surrounded by Dyson spheres 
Bathing in energy of the greater good

Traversing the uknown
Awake in the eve of a new realization
Celestial orbs along their Waltz

Where do orbits lead?

----

Dyson Sphere Reference:
https://en.m.wikipedia.org/wiki/Dyson_sphere

Thenard's Blue Reference:
https://www.syntaxofcolor.com/post/the-discovery-of-thenard-blue-cobalt-blue

Why is Napier's number, e=2.71828... so important?

I was recently reminded of the reason why Napier's number, \[e\approx 2.71828...\] is so important to calculus, and therefore most (if not all) of science and engineering, not to mention all the experiences/opportunities one may seek to understand as part of the human condition.

Simply put, the importance of e in these areas stems from the attempt at taking the derivative of an exponential function. That is, the derivative \[f'(x)=\frac{da^x}{dx}\]. Taking this derivative explicitly yields, \[\frac{da^x}{dx}=\lim_{\Delta x \rightarrow 0}\frac{a^{x+\frac{\Delta x}{2}}-a^{x-\frac{\Delta x}{2}}}{\Delta x}\]. The expression can be simplified by factoring out the common factor of \[a^x\], to get \[\frac{da^x}{dx}=a^{x}\lim_{\Delta x \rightarrow 0}\frac{a^{\frac{\Delta x}{2}}-a^{-\frac{\Delta x}{2}}}{\Delta x}\]. In addition, making use of the properties of limits, \[\frac{da^x}{dx}=a^{x}\left ( \lim_{\Delta x \rightarrow 0}a^{-\frac{\Delta x}{2}}\right )\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )\] Where since, \[\left ( \lim_{\Delta x \rightarrow 0}a^{-\frac{\Delta x}{2}}\right )=1\], \[\frac{da^x}{dx}=a^{x}\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )\], and\[\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )=\frac{da^{x}}{d(x)}|_{x=0}=f'(0)\] one can see that taking this derivative might lead to a result that is counter intuitive. In fact, this seems kind of counter-intuitive and circular, since, according to this result, it would appear that one would need to know  the derivative of the exponential function to evaluate the derivative of the exponential function. \[\frac{d a^x}{dx}=f'(x)=f'(0)f(x)\]. One can get around this issue by looking for a solution where \[f'(0)=1\]. In fact, this condition is exactly what Napier's number yields. But don't just take anyone's word for it. This is provably so. That is, if for some number t you let, \[\frac{1}{t}=(a^{\Delta x}-1)\], then \[a^{\Delta x}=1+\frac{1}{t}\] and \[\Delta x = \log_a \left ( \frac{1}{t} + 1 \right ) \], and the question the becomes one where one should seek the base of the logarithm which yields unity. Since the previous limit can be assessed by approaching it from the positive and negative sides of the number 0. \[\lim_{\Delta x \rightarrow 0_\pm}\equiv \lim_{t\rightarrow\pm\infty}\]. In this way, one can relate the expression for the derivative of the exponential function to the limit of the number t via, \[\frac{da^x}{dx}=a^{x}\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )=a^{x}\lim_{t\rightarrow \infty}\frac{1+\frac{1}{t}-1}{\log_a \left ( \frac{1}{t}+1\right )}\].

It would seem fortunate that since \[\lim_{t\rightarrow \infty}\frac{1+\frac{1}{t}-1}{\log_a \left ( \frac{1}{t}+1\right )}=\left [ \log_a \lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^\frac{1}{t}\right ] ^ {-1}\] one can take advantage of the definition of Napier's number as the base of the logarithm to resolve this question, \[e=\lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^\frac{1}{t}\] . 

Using these arguments, it is then abundantly clear that the derivative of any exponential function for any base is then explicitly dependent on Napier's number through the expression, \[\frac{da^x}{dx}=a^{x}\left [ \log_a \lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^{\frac{1}{t}}\right ] ^{-1}\] which when written explicitly in terms of e is, \[\frac{da^x}{dx}=a^{x}\frac{1}{\log_a e}\]. 

To complete the story, for the case where a is defined to be 2.71828...=e, the derivative and the function are then provably equal: \[\frac{de^x}{dx}=e^{x}\frac{1}{\log_e e}= e^{x}\]

Back to the Rotation Matrices

Here I am interested in playing with the idea of two vectors, \(\vec{u}\) and \(\vec{v}\), which define a three dimensional space, enumerated by reals \(\mathbb{R}^{3}\). I am also interested in playing with their rotational transformations and further progress the discussion in my previous post.

To this end, I will look at how to define general rotation matrices for a third unit vector, \(\hat{B}_{\parallel}\), that bisects these two original vectors. I will also look at how a three dimensional coordinate system may be constructed by defining two additional unit vectors, \(\hat{B}_{\perp}\) and \(\hat{B}_{\hat{n}}\), that result from operating on the bisecting vector with the generally defined rotation matrices.

I will start by making the following assumptions:

1. The space defined by the two vectors, \(\vec{u}\) and \(\vec{v}\), is locally euclidean in the three dimensional space they define.
2. The unit vector \(\hat{B}_{\parallel}\) can be expressed by the column matrix \(\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}\).

The second point implies the existence of an additional two or more linearly independent vectors which can be described by similar column vectors. 

The reason I am restricting the discussion to three dimensions is due to the aparent observation (which may be true or not) that we live in a space which is three dimensional and, as a result, I am interested in working in three dimensions. For now, I will leave the discussion of dimensional spaces greater than three to the following statement: One can use similar ideas as the ones presented herein to construct higher dimensional spaces. 

In the future, and for more playful reasons, I will be interested in the spaces between seven and eight dimensional space. This is due to the fact that, for a locally euclidean n-dimensional space, the volume of a sphere (n-sphere) in that space is maximized somewhere between seven and eight dimensional space:

$$V_n \left(R \right)=\frac{\pi^{ \frac{n}{2}}}{\Gamma \left(\frac{n}{2}+1 \right)}R^{n}$$

Above, \(\Gamma \left(\frac{n}{2}+1 \right)\) is the gamma function.

Ok, now to get started!

Say that there are two intersecting line segments, of length \(\left\|u\right\|\) and \(\left\|v\right\|\), somewhere in n-dimensional space at an angle \(\theta\) from one another.

Bisecting the angle \(\theta\) and defining a new line segment along the angle of bisection \(\alpha\) (I call these derived virtual line segments when working on physics problems), we can consider that the original two line segments \(\left\|u\right\|\) and \(\left\|v\right\|\) have some sort of direction relative to  each other at the point of intersection.

... more to come later ...

Back to the Dot (Inner) Product

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.

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.

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)