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Documentation for the resolution of Arnold's equation #2375

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Jun 3, 2020

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@pleroy pleroy commented Nov 20, 2019

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pleroy added a commit that referenced this pull request Nov 23, 2019
Separate out the code changes from #2375
@@ -108,7 +110,27 @@ \section*{Solution of Euler's equation}
value of $T$ .\label{figGT}}
\end{figure}

In the rest of this section, we derive (corrected) formul{\ae} for the three cases described above.
In the rest of this section, we describe our notation and derive (corrected) formul{\ae} for the three cases described above.
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We are using XeTeX, and it is not the nineties anymore; just write ӕ.

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Done.

journaltitle = {Monthly Notices of the Royal Astronomical Society},
number = {4},
pages = {3620--3632},
title = {Orientation and rotational parameters of asteroid 4179 Toutatis: new insights from Chang′e-2's close flyby},
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Chang PRIME e?
Replace that with an APOSTROPHE ', which LaTeX will turn into the more appropriate RIGHT SINGLE QUOTATION MARK (or we could directly use everywhere but this is best done in a separate pull request).

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Done, but note that the PRIME comes from the article title.

\cite{Celledoni2007} uses a dimensionless formulation where $\norm\vm = 1$, and absolute values for $I_{jh}$ and $\gD_j$.
We prefer to use a dimensionful formulation where $\norm\vm = G$, and to avoid absolute values. Thus we define:
\begin{align*}
I_{jh} &\DefineAs I_j - I_h &\gD_j &\DefineAs G^2 - 2 T I_j &B_{jh} &\DefineAs \sqrt\frac{I_j \gD_h}{I_{jh}}} \\
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Add a rad2 in the definition of 𝛥𝑗.

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Done.

B_{31}^2 + B_{13}^2 = \frac{\gD_1 I_3 - \gD_3 I_1}{I_{31}} = G^2
\]

Physically, $I_{jh}$ has the dimension of an inertial momentum $\squareBrackets{L^2 M}$. $G$ has the dimension of an angular
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s/inertial momentum/moment of inertia/.

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Done.

momentum $\squareBrackets{L^2 M T^{-1} A}$. $T$ has the dimension of an energy $\squareBrackets{L^2 M T^{-2}}$.
$\gD_j$ has the same dimension as $G^2$. $B_{jh}$ has the same dimension as
$\sqrt{\gD_h}$, i.e., the same dimension as $G$. $\gl_1$ and $\gl_3$ have the
same dimension as the quotient $\frac{G}{I_j}$, i.e., $\squareBrackets{T^{-1} A}$ which is appropriate for their usage.
\subsection*{Case (i)}
Case (i) of the solution of Euler's equation in section 2.2 of \cite{Celledoni2007} is:
\[
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Done.

to map $\vm$ onto $\VectorSymbol{e_1}$ or $\VectorSymbol{e_3}$. In the rest of this section we detail the calculations used to compute $\mathscr S$,
$\mathscr P_t$ and $\gy\of{t}$.

With the introduction of $\mathscr S$, we are effectively introducing a new base $\mathscr B^p$ for the ``preferred'' principal axes of the body, and
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Explicitly mention that, with 𝒮, 𝜎 = 1 (and likewise 𝜎′, 𝜎″) from now on; the only allusion to that above is specific to m3 and case (ii).

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Done.

\]
and, for the angle $\gy\of{t}$ of the rotation $\mathscr Y_t$ around $\VectorSymbol{e_1}$:
\[
\TimeDerivative{\gy}\of{t} = \frac{2 T + G m_1 / I_1}{G + m_1} + 4 G \frac{p_1 \TimeDerivative{p_0} - p_0 \TimeDerivative{p_1}}{G + m_1}
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rad2

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After T?

\]
and the angle to:
\[
\TimeDerivative{\gy}\of{t} = \frac{2 T + G m_1 / I_1}{G + m_1} = \frac{G^2 - \gD_1 + G m_1}{I_1\pa{G + m_1}} = \frac{G}{I_1} - \frac{\gD_1/I_1}{G + m_1} =
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rad2

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After T?

\[
\gy\of{t} = \frac{G}{I_1}t + \frac{G I_{13}}{\gl I_1 I_3}
\EllipticPi\of{\JacobiAmplitude\of{\gl t - \gn}, \frac{I_1 I_{32}}{I_3 I_{12}}, k} -
\InverseTrigonometricTangent\of{\sqrt\frac{I_2 I_{31}}{I_3 I_{21}} \JacobiSC\of{\gl t - \gn, k}}
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s/𝑘/𝑘-1/

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Same.

Using equation (\ref{integraldn}) with $a = B_{13}/G$ and simplifying the various coefficients we obtain:
\[
\gy\of{t} = \frac{G}{I_1}t + \frac{G I_{13}}{\gl I_1 I_3}
\EllipticPi\of{\JacobiAmplitude\of{\gl t - \gn}, \frac{I_1 I_{32}}{I_3 I_{12}}, k} -
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s/𝑘/𝑘-1/

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I don't think so, I think you meant around line 560.

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Also, I noticed that the second parameter of am(u, k) was missing in a few places.

@eggrobin eggrobin added the LGTM label Jun 3, 2020
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pleroy commented Jun 3, 2020

retest this please

@pleroy pleroy merged commit d925509 into mockingbirdnest:master Jun 3, 2020
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