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import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
from IPython.display import HTML
HTML('../style/course.css') #apply general CSS
We have seen in $\S$ 2.13 ➞, three main spherical trigonometry relationships between angles and arcs in a spherical triangle. Applying each of these equations, we will derive the equation A used in $\S$ 4.1 ➞.
Let's start from Figure 4.1.3, dislayed in $\S$ 4.1 ➞. We will consider the following spherical triangle STZ (as in Figure A1 right). From $\S$ 2.13 ➞, we have the following relationships:
The quantity $s$, $t$, and $z$ are the arcs opposite to the vertices S, T and Z. Given the definition of these quantities and classical trigonometric identities, we can already state that:
$\cos s=\sin L_a$
$\cos t=\sin \delta$
$\cos z=\sin E$
Similarly, the $\sin$ of $s$, $t$ and $z$ also transform into the $\cos$ of $L_a$, $\delta$ and $E$. We will now derive each line of the transform matrix of Eq. x using the spherical trigonometry rules applies to the triangle STZ.
Left-hand side: $\cos b=\cos t= \sin \delta$
Right-hand side: $\cos a \cos c + \sin a \sin c \cos \hat{B}= \cos z \cos s + \sin z \sin s \cos \mathcal{A} = \sin \mathcal{E} \sin L_a + \cos \mathcal{E} \cos L_a \cos \mathcal{A}$
We have our first relationship:
Left-hand side: $\sin b \sin \hat{A}= \sin t \sin (-H) = - \cos \delta \sin H$
Right-hand side: $\sin \hat{B} \sin a = \sin \mathcal{A} \sin z = \sin \mathcal{A} \cos \mathcal{E}$
We have our second relationship:
Left-hand side: $\sin b \cos \hat{A} = \sin t \cos(-H)= \cos \delta \cos{H}$
Right-hand side: $\cos a \sin c - \sin a\cos c\cos \hat{B}= \cos z \sin s - \sin z \cos s \cos \mathcal{A} = \sin \mathcal{E} \cos L_a - \cos \mathcal{E} \sin L_a \cos \mathcal{A}$
We have our last relationship:
Rearranging in a matrix form, and correct line order, we have:
Similarly to A.1, we will derive three equations by applying spherical trigonometry identities in the triangle $\text{S}_\text{c} \text{S} \text{Z}$
and
$$l^2+m^2+n^2=1$$Left-hand side: $\cos b=\cos \theta = n$
Right-hand side: $\cos a \cos c + \sin a \sin c \cos \hat{B}= \cos (\frac{\pi}{2}-\delta) \cos (\frac{\pi}{2}-\delta_0) + \sin (\frac{\pi}{2}-\delta) \sin (\frac{\pi}{2}-\delta_0) \cos \Delta \alpha = \sin \delta \sin \delta_0 + \cos \delta \cos \delta_0 \cos \Delta \alpha$
We have our first relationship:
Left-hand side: $\sin b \sin \hat{A}= \sin \theta \sin \psi = l$
Right-hand side: $\sin \hat{B} \sin a = \sin \frac{\pi}{2}-\delta \sin \Delta \alpha = \cos \delta \sin \Delta \alpha$
We have our second relationship:
Left-hand side: $\sin b \cos \hat{A} = \sin \theta \cos \psi= m$
Right-hand side: $\cos a \sin c - \sin a\cos c\cos \hat{B}= \cos (\frac{\pi}{2} - \delta) \sin (\frac{\pi}{2} - \delta_0) - \sin (\frac{\pi}{2} - \delta) \cos (\frac{\pi}{2} - \delta_0) \cos \Delta \alpha =\sin \delta \cos \delta_0 - \cos \delta \sin \delta_0 \cos \Delta \alpha$
We have our third relationship:
Let's summarize the results
From the ($l$, $m$, $n$) coordinates of one source, one can compute back the ($\alpha$,$\delta$) coordinates by combinations of the $(i)$, $(ii)$, $(iii)$ relatioships.
$\require{cancel}$
\begin{eqnarray} \frac{(i)}{(ii)} &\Leftrightarrow& \frac{\cancel{\cos{\delta}} \sin\Delta \alpha}{\cancel{\cos{\delta}} \sin \delta_0 \cos\Delta \alpha} &=& \frac{l}{\sin \delta \cos \delta_0 - m}&\Leftrightarrow& \frac{\tan \Delta \alpha}{\sin\delta_0} = \frac{l}{\sin\delta\cos\delta_0-m}= (i')\\ \frac{(i)}{(iii)} &\Leftrightarrow& \frac{\cancel{\cos{\delta}} \sin\Delta \alpha}{\cancel{\cos{\delta}} \cos \delta_0 \cos\Delta \alpha} &=& \frac{l}{n-\sin \delta \sin \delta_0} &\Leftrightarrow& \frac{\tan \Delta \alpha}{\cos \delta_0}=\frac{l}{n-\sin\delta\sin\delta_0} = (ii') \\ \end{eqnarray}$\require{cancel}$ \begin{eqnarray} (i') \sin \delta_0 = (ii') \cos \delta_0 &\Leftrightarrow & \frac{\cancel{l}\sin\delta_0}{\sin\delta\cos\delta_0 -m} = \frac{\cancel{l}\cos\delta_0}{n-\sin\delta\sin\delta_0} \\ &\Leftrightarrow& \sin\delta \cos^2\delta_0-m\cos\delta_0 = m\sin\delta_0 - \sin\delta\sin^2\delta_0 \\ &\Leftrightarrow& \sin\delta \left[ \underbrace{\cos^2 \delta_0 + \sin^2 \delta_0}_{=1} \right] = n\sin\delta_0+m\cos\delta_0 \\ &\Leftrightarrow& \boxed{\delta = \arcsin (n\sin\delta_0 + m\cos\delta_0)} \end{eqnarray}
Now, we reinject the expression of $\delta$ into $(i')$ or $(ii')$ which display a $\sin \delta$ term.
$\require{cancel}$ \begin{eqnarray} \frac{\tan \Delta \alpha}{\sin\delta_0} &=& \frac{l}{(n\sin\delta_0+m\cos\delta_0)\cos\delta_0-m} \Leftrightarrow \\ \tan \Delta \alpha &=& \frac{l \bcancel{\sin\delta_0}}{n\bcancel{\sin\delta_0}\cos\delta_0 +m \underbrace{(\cos^2\delta_0-1)}_{\sin^{\bcancel{2}}\delta_0}} \Leftrightarrow \\ \tan (\alpha - \alpha_0) &=& \frac{l}{n\cos\delta_0 + m\sin\delta_0} \Leftrightarrow \\ &\Leftrightarrow& \boxed{\alpha = \alpha_0 + \arctan \left[ \frac{l}{n\cos\delta_0 + m\sin\delta_0} \right]} \end{eqnarray}
Starting from the expression of $u$ and $v$.
$\lambda u = X \sin H + Y \cos H$
$\lambda v= -X \sin \delta \cos H + Y \sin\delta\sin H + Z \cos\delta$
\begin{eqnarray} v &=&\frac{1}{\lambda} -X \sin \delta \cos H + Y \sin\delta\sin H + Z \cos\delta \\ \Leftrightarrow v -\frac{Z}{\lambda} \cos \delta &=& \frac{Y}{\lambda}\sin\delta\sin H - \frac{X}{\lambda} \sin\delta \cos H \\ \Leftrightarrow \frac{v -\frac{Z}{\lambda} \cos \delta}{\sin \delta} &=& \frac{Y}{\lambda}\sin H - \frac{X}{\lambda} \cos H \\ \Leftrightarrow \left[ \frac{v -\frac{Z}{\lambda} \cos \delta}{\sin \delta} \right]^2 &=& \left[ \frac{Y}{\lambda}\right]^2 \sin^2 H + \left[ \frac{X}{\lambda}\right]^2 \cos^2 H - \frac{2XY\cos H\sin H}{\lambda^2}\\ \Leftrightarrow \left[ \frac{v -\frac{Z}{\lambda} \cos \delta}{\sin \delta} \right]^2 &=& \left[ \frac{Y}{\lambda}\right]^2 + \left[ \frac{X}{\lambda}\right]^2 \underbrace{-\left[ \frac{Y}{\lambda}\right]^2 \cos^2 H - \left[ \frac{X}{\lambda}\right]^2 \sin^2 H - \frac{2XY\cos H\sin H}{\lambda^2}}_{\left[ \frac{X\sin H}{\lambda} + \frac{Y\cos H}{\lambda}\right]^2=u^2} \\ \Leftrightarrow u^2 + \left[ \frac{v -\frac{Z}{\lambda} \cos \delta}{\sin \delta} \right]^2 &=& \left[ \frac{Y}{\lambda} \right]^2 + \left[ \frac{X}{\lambda} \right]^2 \end{eqnarray}
In Sec. 2.12 ➞ the infinitesimal solid angle $d\Omega$ was defined in terms of polar coordinates. The infinitesimal solid angle $d\Omega$ can also be expressed in direction cosine coordinates by using the Jacobian transformation. First the relationship between $(\theta,\phi)$ and $(l,m)$ must be established. It should be clear from Fig. A.3(b) ⤵ and Fig. A.3(c) ⤵ that: \begin{eqnarray} \phi &=& \tan^{-1}\frac{\rho m}{\rho l} = tan^{-1}\frac{m}{l}=f(l,m)\\ \theta &=& \sin^{-1}\frac{\rho \sqrt{l^2+m^2}}{\rho} = \sin^{-1} \sqrt{l^2+m^2}=g(l,m).\nonumber \end{eqnarray}
Now by the Jacobian transformation $\boldsymbol{J}$: \begin{equation} d\Omega = \sin \theta d\theta d\phi = \sin(g(l,m)) \left| \boldsymbol{J} \right| dl dm, \end{equation} where \begin{equation} \left|\boldsymbol{J}\right|= \left| \begin{array}{cc} \frac{\delta\theta}{\delta l} & \frac{\delta\theta}{\delta m} \\ \frac{\delta\phi}{\delta l} & \frac{\delta\phi}{\delta m} \end{array} \right|, \end{equation} and \begin{eqnarray} \frac{\delta \theta}{\delta l} &=&\frac{l}{\sqrt{1-l^2-m^2}\sqrt{l^2+m^2}},\\ \frac{\delta \theta}{\delta m} &=& \frac{m}{\sqrt{1-l^2-m^2}\sqrt{l^2+m^2}},\nonumber\\ \frac{\delta \phi}{\delta l} &=& \frac{-\frac{m}{l^2}}{1+\frac{m^2}{l^2}},\nonumber\\ \frac{\delta \phi}{\delta m} &=& \frac{\frac{1}{l}}{1+\frac{m^2}{l^2}},\nonumber \end{eqnarray} so that \begin{eqnarray} \left|\boldsymbol{J}\right| &=& \frac{1}{\sqrt{1-l^2-m^2}\sqrt{l^2+m^2}(1+\frac{m^2}{l^2})}+\frac{\frac{m^2}{l^2}}{\sqrt{1-l^2-m^2}\sqrt{l^2+m^2}(1+\frac{m^2}{l^2})},\\ &=& \frac{1}{\sqrt{1-l^2-m^2}\sqrt{l^2+m^2}},\nonumber \end{eqnarray} and \begin{equation} d\Omega = \sin(g(l,m)) \left| \boldsymbol{J} \right| dl dm = \frac{dldm}{\sqrt{1-l^2-m^2}}=\frac{dldm}{n}. \end{equation}
The delayed product identities of cos and sin are given by: \begin{eqnarray} \cos(x-y)\cos(x) &=& \frac{1}{2}\cos y + \frac{1}{2}\cos(2x-y)\\ \sin(x-y)\sin(x) &=& \frac{1}{2}\cos y - \frac{1}{2}\cos(2x-y)\\ \sin(x-y)\cos(x) &=& -\frac{1}{2}\sin y + \frac{1}{2}\sin(2x-y)\\ \cos(x-y)\sin(x) &=& \frac{1}{2}\sin y + \frac{1}{2}\sin(2x-y). \end{eqnarray}
The delayed product identities are derived from the standard trigonometric identities. We can derive the first identiy as follow:
\begin{eqnarray} \cos(x-y)\cos(x) &=& [\cos x\cos y + \sin x \sin y]\cdot \cos x\\ &=& \cos^2 x\cos y + \cos x\sin x \sin y\nonumber\\ &=& \bigg(\frac{1+\cos 2x}{2}\bigg)\cdot \cos y + \frac{\sin 2x \sin y}{2}\nonumber\\ &=& \frac{1}{2}\cos y + \frac{\cos2x \cos y}{2} + \frac{\sin2x \sin y}{2}\nonumber\\ &=& \frac{1}{2}\cos (y) + \frac{1}{2}\cos(2x-y).\nonumber \end{eqnarray}The remaining delayed product identities are derived similarly.