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\large

\section*{Recap of \textit{Synechococcus} post-translational oscillator}

\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{pto.png}
\caption{Model for the post-translational oscillator, from Rust 2007.}
\end{center}
\end{figure}

\begin{itemize}
\item Recall that the post-translational oscillator at the heart of the cyanobacterial clock cycles through four states characterized by peaks in four phosphorylation states of the protein KaiC:
\begin{enumerate}
\item U-KaiC: no phosphorylation (``dawn")
\item T-KaiC: threonine phosphorylated (``noon")
\item D-KaiC: both phosphorylated (``dusk")
\item S-KaiC: serine phosphorylated (``midnight")
\end{enumerate}
\item KaiA promotes KaiC's kinase activity. The accumulation of S-KaiC (in conjunction with KaiB) inhibits KaiA.
\item A general model for this clock is described by:

\begin{eqnarray*}
\frac{dT}{dt} & = & k_{UT} U + k_{DT} D -\left(  k_{TD} + k_{TU} \right) T\\
\frac{dD}{dt} & = & k_{TD} T + k_{SD} S - \left(k_{DT} + k_{DS} \right) D\\
\frac{dS}{dt} & = &k_{DS} D +  k_{US} U - \left( k_{SU} + k_{SD} \right) S\\
k_{xy} & = & k_{xy}^0 + \frac{k_{xy}^A A(S)}{K + A(S)}\\
A(S) & = & \max \left(0, \left[ \textrm{KaiA} \right] - 2 S \right)
\end{eqnarray*}

This includes every forward and backward reaction along the loop (though some rate constants are measured to be very low). First-order kinetics are assumed for all phosphorylation and dephosphorylation reactions.

\item A few features seen here that are very common in oscillators:
\begin{itemize}
\item A process driving forward progression.
\item A negative feedback loop that kicks in after a delay (either explicity temporal or resulting from nonlinearity)
\end{itemize}
\item The model presented does not include a positive feedback loop. Notice that most oscillations are relatively sinusoidal, like a harmonic oscillator. Later in this lecture, we will argue that oscillators which contain both positive and feedback loops behave like relaxation oscillators, with strikingly fast transitions.
\end{itemize}

\section*{Entrainment}

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{adp_falling.png}
\caption{Change in ATP:ADP ratio upon a dark pulse, from Rust 2011.}
\end{center}
\end{figure}

\begin{itemize}
\item Unexpected periods of darkness during the subjective day lead to decreases in the ATP:ADP ratio. Typically, ATP makes up 90\% of the combined pool of these two molecules; after an eight-hour pulse of darkness, the ratio falls to about 40\%.

\item Since ATP serves as the phosphate group donor for KaiC's phosphorylation reactions, this increases the tendency for KaiC to be in an unphosphorylated state.

\item The dependence on the ATP:ADP ratio is easily studied in vitro, where more ADP can be spiked in (or ATP can be regenerated by adding pyruvate kinase). The phase of the clock is reset to a lower overall phosphorylation (primarily U-KaiC and S-KaiC) after a period when the ATP:ADP ratio is low. This is sensible because that phosphorylation pattern corresponds to the period between midnight and dawn.

\end{itemize}
\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{adp_feedback.png}
\caption{Model for the effect of ATP:ADP ratio on the PTO, from Rust 2011.}
\end{center}
\end{figure}
\begin{itemize}

\item The model described above was modified to account for the entrainment system as follows:

\[k_{phos} = k_{phos}^A \left( \frac{A(S)}{K + A(S)} \right) \left( \frac{[\textrm{ATP}]}{[\textrm{ATP] + $K_{rel}$[ADP}]} \right) \]

The dephosphorylation reactions do not appear to depend on the ATP:ADP ratio when observed experimentally, so they remain dependent only on $A(S)$.

\item This model follows a limit cycle that maintains the appropriate period for various steady-state daylight ATP:ADP ratios, but shifts phase in response to sudden changes in the ratio.

\item If the dark pulse is delivered around subjective dusk, the phase is modestly advanced; if delivered around subjective dawn, the phase lags. The magnitude of the phase shift is largest around mid-day, when kinase activity would otherwise have peaked: this results in a large phase advance.
\end{itemize}

\section*{Phase response curve (PRC)}

\begin{itemize}
\item One way to quantify the phase shift due to perturbation is with a \textit{phase response curve}. The change in phase (positive if advanced, negative if retarded) is plotted against the timing of the perturbation. These plots are rather popular because they can be made even if nothing mechanistic is understood about the system.

\item Example for circadian rhythm and ``light therapy" illumination common for seasonal affective disorder. Effect on the cycle shifts sharply overnight (exposure in the evening retards the cycle, and exposure in the early morning advances it; little effect during the day).

\item This example emphasizes that the ideal phase response curve is not necessarily flat at zero; entrainment requires the ability to shift phase and a PRC can be more or less well-suited to adjusting the phase in response to stimuli.

\end{itemize}

\section*{\textit{Synechococcus elongatus} transcriptional-translational feedback loop}

\begin{itemize}
\item The mechanism for entrainment described above provides a means for response to natural fluctuations in the ATP:ADP ratio presumably caused by increased availability of ATP during photosynthetic periods.
\item Others have also proposed that the clock could entrain on temperature (another environmental cue that reflects recent lighting conditions). The cyanobacterial clock period is fairly robust to changes in temperature ($\pm 5$ degrees Celsius, a wider range than most marine varieties encounter during the circadian cycle; \textit{Synechococcus elongatus} is a freshwater cyanobacterium, however); the phase, however, can be shifted by abrupt temperature changes of the same range.
\item When temperature and light-dark cycling are offset in phase by 12 hours, the cyanobacterial clock syncs to the light-dark cycling, suggesting that illumination (via ATP:ADP ratios?) is the physiologically-predominant form of entrainment.
\item Unfortunately, the cyanobacterial clock can also be phase-shifted by means which are presumed to be unintentional. For example, overexpression of KaiC for a few hours can reset the phase to subjective dawn. Indeed the in vitro post-translational oscillator can be broken by varying the stoichiometry of the components.
\item KaiA and KaiBC are transcribed from two different operons. KaiA's rate of translation is effectively constant; however, KaiBC transcription and translation vary with the circadian cycle. The problem of keeping ratios constant is therefore exacerbated by the variation in cyanobacterial growth rate with media conditions (e.g. division times ranging from 14-34 hours in common media).
\item A response regulator, RpaA, is activated through phosphorylation by a histidine kinase (SasA) downstream of the PTO. It then binds at hundreds of genes throughout the genome to drive expression of its targets, including \textit{kaiBC}.
\item In $\Delta$rpaA strains, \textit{kaiBC} expression is four-fold lower than normal, breaking the PTO.
\end{itemize}

\begin{figure}
\begin{center}
\includegraphics[width=\textwidth]{rpa.png}
\caption{Role of RpaA in regulation of kaiBC and other targets, from Markson 2013.}
\end{center}
\end{figure}

\begin{itemize}
\item The RpaA loop helps to maintain stoichiometry because RpaA is activated through phosphorylation just before the majority of KaiC is double-phosphorylated, a condition that occurs around dusk normally, but could also reflect an aberrantly high KaiA:KaiC ratio. Under this condition RpaA would help restore the normal stoichiometry by causing more KaiBC to be expressed.
\item In an upcoming discussion paper (after the break), you'll find that in addition to keeping the clock operational, the transcriptional feedback loop helps prevent phase drift of the PTO.

\end{itemize}

\section*{Simple harmonic oscillator}

\begin{itemize}
\item We now return to the notion that there are two general categories of oscillators driving biological clocks. The first characterizes oscillators that lack a positive feedback loop and show oscillatory patterns similar to harmonic oscillators.

\item Some oscillators have greater stability in amplitude, phase, and/or frequency against likely perturbations than others.

\item Consider the harmonic oscillator system:

\[  \frac{d^2 x}{dt^2} + a \frac{dx}{dt} + bx = 0 \]

\item In order to apply stability analysis techniques we already know, we could e.g. convert this into a linear system by substitutition:

\begin{eqnarray*}
\frac{dx}{dt}  & = & y\\
\frac{dy}{dt} + a y + b x & = & 0\\
\frac{dy}{dt} & = & -b x - a y\\
\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} & = & \begin{pmatrix} 0 & 1  \\ -b & -a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}
\end{eqnarray*}

\item If $a=0$, we have stable oscillations. Arbitrarily choosing $b=1$ and initial conditions of $x(0)=0$ and $y(0)=1$ gives a pair of sinusoidal functions with a phase offset of 90 degrees, e.g.
\[ x(t) = \sin t \hspace{3 cm} y(t) = \cos(t) \]

\item Consider what happens when we perturb $x$, say, by decreasing its magnitude soon after $x$ reaches its nadir. This will cause the slope of $y$ to decrease prematurely, and though the system will continue to oscillate at the same frequency, its magnitude will be permanently decreased and its phase will be shifted.

\item Even worse, a non-zero value of $a$ would cause these oscillations to either grow or decay.
\end{itemize}

\section*{van der Pol oscillator}
\begin{itemize}
\item Not all oscillators suffer from these problems. Consider the van der Pol oscillator with $\mu \gg 1$:

\[ \frac{d^2 x}{dt^2} + \mu \left( x^2 -1 \right)\frac{dx}{dt} + x = 0 \]

\item Notice that

\[ \frac{d^2 x}{dt^2} + \mu \left( x^2 -1 \right)\frac{dx}{dt} = \frac{d}{dt} \left[ \frac{dx}{dt} + \mu \left( \frac{x^3}{3} - x \right) \right] \]

\item To simplify this system, we substitute\footnote{Following Strogatz example 7.5.1}:

\begin{eqnarray*}
&F(x) = \frac{x^3}{3} - x \hspace{3 cm} w = \frac{dx}{dt} + \mu F(x)&\\
&\frac{dx}{dt} = \mu F(x) - w&\\
& \frac{dw}{dt} =  \frac{d^2 x}{dt^2} + \mu \left( x^2 -1 \right)\frac{dx}{dt} = -x&
\end{eqnarray*}

\item Finally, substituting $y=w/\mu$:
\begin{eqnarray*}
\frac{dx}{dt} & = & \mu \left(F(x) - y \right)\\
 \frac{dy}{dt} & = & -\frac{x}{\mu}
\end{eqnarray*}

\item By plotting we can see that the system will spend the majority of its time along the cubic nullcline $y=F(x)$, where it moves slowly, then zap quickly to the other half of the plane. This builds intuition that the oscillator functions on two distinct time scales, which indeed we can see from a simulation of the time evolution of the system.

\item It is possible to show that perturbations do not change the amplitude (or frequency) of the oscillations, unlike the simple harmonic oscillator studied above.

\item Perturbations of $x$ will still cause a phase shift. Intuitively, however, we predict that the impact on phase will be small unless the sign of $x$ changes during the perturbation. (Otherwise the oscillator will quickly snap back to a similar position along the nullcline.)

\item Conceptually-similar biological models include the Fitzhugh-Nagumo simplification of Hodgkin and Huxley's action potential model.

\item In general, oscillators based on a fast positive feedback and slow negative feedback loop will behave like relaxation oscillators, while oscillators based on negative feedback loops with a long delay are more smoother (``sinusoidal").

\end{itemize}

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{gonze.png}
\caption{Image taken directly from the \href{http://homepages.ulb.ac.be/~dgonze/TEACHING/osc_design.pdf}{lecture notes} of Didier Gonze.}
\end{center}
\end{figure}

\begin{itemize}

\item Another interesting feature of the vdPo and oscillators that combine positive and negative feedback loops is that their frequencies can be tuned while maintaining constant amplitude (Tsai et al., 2008).
\end{itemize}

\section*{Isochrons}
\begin{itemize}
\item To explore the notion that perturbations will have less impact on phase for van der Pol oscillators, we introduce the concept of \textit{isochrons} (some authors call them isochrones).

\item Consider a point $x_0$ which does not lie on the limit cycle, but whose trajectory will approach the limit cycle as $t \to \infty$. Since $x_0$ is not on the limit cycle, its trajectory is not periodic, formally speaking.

\item However. there is some corresponding point $y_0$ on the limit cycle whose trajectory will be indistinguishable from that of $x_0$ as $t \to \infty$. We could define $x_0$ to have the ``same phase" as $y_0$.

\item Using this definition, we can define the \textit{isochrons} of $y_0$ as the set of all points that have the same phase as $y_0$. Finding isochrons analytically is a daunting task, but MATLAB is pleased to oblige us.

\item What we see for the van der Pol oscillator is that in most parts of the cycle, the isochrons are almost horizontal. (The more so as $\mu$ increases.) This suggests that perturbations of $x$ will not greatly change the phase of the oscillator unless the $y$-axis is crossed, or the perturbations come at one of the few points in the cycle where the isochrons are nearly vertical.

\end{itemize}

\section*{Coupled oscillators and phase stabilization}

\begin{itemize}
\item Christiaan Huygens is credited with performing the first analysis of coupled oscillators in 1665. He discovered that two pendulums mounted on the same support beam would synchronzie, evidently through vibrations of the beam.
\item Synchronization is an effective means to overcome error in frequency and phase of otherwise-independent oscillators. If you are a former MA 19a student, you have likely seen examples, including synchronization of firefly flashing.
\item One rather humorous example are male Japanese tree frogs, which ribbit periodically.When two frogs are close to one another, they coordinate their calls to be approximately 180 degrees out of phase from one another. Adding a third frog results in one of two stable patterns: two frogs calling at the same time, or all three frogs calling at phases offset by 120 degrees.
\item We'll model the change in phases over time with $\omega$ as the innate frequency and phase differences $\phi = \theta_1 - \theta_2$ and $\psi = \theta_2 - \theta_3$.
\begin{eqnarray*}
\frac{d\theta_1}{dt} & = & \omega + H(- \phi) + H(- \phi - \psi) = \omega - H(\phi) - H(\phi + \psi)\\
\frac{d\theta_2}{dt} & = & \omega + H(\phi) + H(-\psi) =  \omega + H(\phi) - H(\psi)\\
\frac{d\theta_3}{dt} & = & \omega + H(\phi + \psi) + H(\psi)\\
\frac{d\phi}{dt} & = & -2 H(\phi) - H(\phi + \psi) + H(\phi)\\
\frac{d\psi}{dt} & = & H(\phi) - H(\phi + \psi)  - 2 H(\phi)\\
\end{eqnarray*}
where $H(\theta)$ describes the change of phase due to a phase difference, which is assumed to be $2\pi$ periodic and odd (so that the phase changes go in opposite direction if the phase difference changes sign).

\item  $H(\theta)= a \sin x + b \sin 2x$ fulfills our requirements for $H$: if we plot the trajectories for this system, we'll see that the frogs can be made to ribbit at equal intervals (either thirds or halfs, depending on the initial conditions.
\end{itemize}

\begin{lstlisting}
function [] = frogs()
    initial_phases = [0.1, 0.2];
    time_interval = [0, 20];
    global a b
    a = -1;
    b = -0.1;
    [timepoints, concentrations] = ode45(@chain1ddt, ...
        time_interval, initial_phases);
    plot(timepoints, concentrations(:,1), '-b', 'LineWidth', 3); hold on;
    plot(timepoints, concentrations(:,2), '-r', 'LineWidth', 3);
    title(sprintf('phi_0=%0.2f, psi=%0.2f',initial_phases(1),initial_phases(2)));
    xlabel('Time');
    ylabel('Phase differences');
    h = legend('\phi','\psi','Location','SouthEast');
    set(gca, 'FontSize', 24);
    set(h, 'FontSize', 24);
end

function changes_in_values = chain1ddt(time, current_values)
    phi = current_values(1);
    psi = current_values(2);
    change_in_phi = - 2*H(phi) + H(psi) - H(phi+psi);
    change_in_psi = - 2*H(psi) + H(phi) - H(phi+psi);
    changes_in_values = [change_in_phi; change_in_psi];
end

function returned_value = H(phase_difference)
    global a b
    returned_value = a * sin(phase_difference) + b * sin(2*phase_difference);
end
\end{lstlisting}

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{frogs.pdf}
\caption{Timecourse of $\phi=\theta_2 - \theta_1$ and $\psi = \theta_3 - \theta_2$ for $a=-1$, $b=-1$ and (left) $\phi(0) = 0$ and $\psi(0) = 1.7$ or (right) $\phi(0)=0.1$ and $\psi(0) = 0.2$.}
\end{center}
\end{figure}
\begin{itemize}
\item If we perturb the phase of one frog, the system will tend to restore its relative phase (unless the perturbation is so large that the system approaches the other stable solution).
\item Similar models can also explain synchronization between coupled oscillators with slightly different frequencies. Some fireflies, for example, can sync to artificial light pulses with $\pm$15\% differences in period. (Fireflies of one species can even tune their baseline frequency for improved long-term matching.)
\item In this case the coupling was negative (driving the frog's timing apart), but for positive coupling we'd expect synchronization and stabilization of phase.
\end{itemize}

\section*{\textit{Dictyostelium discoideum} aggregation}

\begin{itemize}

\item Another example of a system of coupled oscillators is a cellular slime mold community coordinating aggregation. While the phase of this clock does not appear to be linked to an external environmental cue, individual cells can entrain on the extracellular products of already-oscillating neighbors, establishing synchrony between cells in the population.
\item \textit{Dictyostelium} normally live as individual amoebae, but when they begin to run out of food, these amoeba aggregate together to form a motile slug that can travel much faster than any cell alone. The goal of this movement is for the cells to reach a position from which their spores could be disbursed, hopefully to germinate in more favorable conditions. The slug is capable of traveling a few millimeters upward through the leaf litter, where it forms an upright stalk, at the tip of which will form a ball of spores.
\item One issue with this life strategy is that it requires cells to signal to one another that they are starving, identify the center of their community, and move toward that point to form an aggregate.
\item The major chemical player in communication between cells in this system is cAMP. The external concentration of cAMP is found to oscillate in aggregating populations, and with each peak, cells exhibit contractions toward the center of the group.
\item After each peak, both intracellular and extracellular cAMP are hydrolyzed. Inside the cell, this process is carried out by the phosphodiesterase REG A.
\item When cAMP is added exogenously at the incorrect phase of the cycle, the phase is shifted; when it is continuously kept high, the cycle is eliminated entirely.
\item Cells detect cAMP through a high affinity receptor called CAR1. There are two main effects of signalling through CAR1: activation of ACA, an adenylyl cyclase, and activation of ERK2, which inhibits REG A.
\end{itemize}
\begin{figure}
\begin{center}
\includegraphics[width=0.5\textwidth]{maeda.png}
\caption{cAMP-based oscillator in \textit{Dictyostelium}, from Maeda 2004.}
\end{center}
\end{figure}
\begin{itemize}
\item What I have described so far is a positive feedback loop that would tend to increase intra- and extracellular cAMP indefinitely. The negative feedback loop in this case is that cAMP activates a protein called protein kinase A (PKA), which represses both ERK2 and ACA.
\item The following set of differential equations, modified from Laub and Loomis (1998) by Maeda et al. (2004), produce an oscillating system for an appropriate choice of parameter values:

\begin{eqnarray*}
\frac{d [\textrm{ACA}]}{dt} & = & k_1  [\textrm{CAR1}] - k_2 [\textrm{PKA}]  [\textrm{ACA}]\\
\frac{d [\textrm{PKA}]}{dt} & = & k_3  [\textrm{cAMPint}] - k_4 [\textrm{PKA}] \\
\frac{d[ \textrm{ERK2}]}{dt} & = & k_5  [\textrm{CAR1}] - k_6 [\textrm{PKA}]  [\textrm{ERK2}]\\
\frac{d [\textrm{REG A}]}{dt} & = & k_7  - k_8 [\textrm{REG A}]  [\textrm{ERK2}]\\
\frac{d [\textrm{cAMPint}]}{dt} & = & k_9  [\textrm{ACA}] - k_{10} [\textrm{REG A}] [\textrm{cAMPint}]\\
\frac{d[ \textrm{cAMPext}]}{dt} & = & k_{11}  [\textrm{ACA}] - k_{12}  [\textrm{cAMPext}]\\
\frac{d [\textrm{CAR1}]}{dt} & = & k_{13}  [\textrm{cAMPext}]  - k_{14} [\textrm{CAR1}] \\
\end{eqnarray*}

\end{itemize}

\begin{center}
\includegraphics[width=0.7\textwidth]{slime.pdf}
\end{center}

\begin{itemize}

\item Although neighboring cells can phase-lock, on the larger scale of the population, moving fronts of external cAMP appear, ultimately forming spirals. Cells within the field of one spiral will ultimately form a slug together.
\item External cAMP also binds to a different G protein receptor to ultimately activate phospholipase C, which will produce IP$_3$, causing calcium oscillations that lead to directional pseudopod extension.
\end{itemize}

\section*{Summary}

\begin{itemize}
\item When we return after break, we'll continue our discussion of clocks and oscillations, launching into natural clocks whose oscillators depend on transcriptional feedback loops.
\item A majority of synthetic clocks rely on such feedback loops, which have been easier to construct than protein-protein interactions.
\item However, you will see that coupling has also been used to synchronize synthetic clocks such as the one developed in Jeff Hasty's group at UCSD.
\end{itemize}

\end{document}

lectures/Lecture 19 - Smooth, Relaxation, and Coupled Oscillators/lecture notes/lecture 19 notes.tex


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\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}

%----------------------------------------------------------------------------------------
%	ENVIRONMENTS
%----------------------------------------------------------------------------------------
% The incantations below define how theorem environments
% (\begin{theorem} ... \end{theorem}) and their likes will look like.
\newtheoremstyle{plainsl}% <name>
  {8pt plus 2pt minus 4pt}% <Space above>
  {8pt plus 2pt minus 4pt}% <Space below>
  {\slshape}% <Body font>
  {0pt}% <Indent amount>
  {\bfseries}% <Theorem head font>
  {.}% <Punctuation after theorem head>
  {5pt plus 1pt minus 1pt}% <Space after theorem headi>
  {}% <Theorem head spec (can be left empty, meaning `normal')>

% Environments which make the text inside them slanted:
\theoremstyle{plainsl}
  \newtheorem{theorem}{Theorem}[section]
  \newtheorem{proposition}[theorem]{Proposition}
  \newtheorem{lemma}[theorem]{Lemma}
  \newtheorem{corollary}[theorem]{Corollary}
  \newtheorem{conjecture}[theorem]{Conjecture}
% Environments that don't:
\theoremstyle{definition}
  \newtheorem{definition}[theorem]{Definition}
  \newtheorem{example}[theorem]{Example}
  \newtheorem{exercise}[theorem]{Exercise}
  \newtheorem{examples}[theorem]{Examples}
  \newtheorem{algorithm}[theorem]{Algorithm}
  \newtheorem{question}[theorem]{Question}
 \theoremstyle{remark}
  \newtheorem{remark}[theorem]{Remark}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\newenvironment{fineprint}{\begin{small}}{\end{small}}

%----------------------------------------------------------------------------------------
%	METADATA
%----------------------------------------------------------------------------------------
\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\myid}{00000000} % ENTER YOUR UMN ID HERE
\newcommand{\mymail}{dgrinber@umn.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{1} % ENTER THE NUMBER OF THIS PSET HERE

%----------------------------------------------------------------------------------------
%	HEADER AND FOOTER
%----------------------------------------------------------------------------------------
\ihead{Solutions to homework set \#\psetnumber} % Page header left
\ohead{page \thepage\ of \pageref{LastPage}} % Page header right
\ifoot{\myname, \myid} % left footer
\ofoot{\mymail} % right footer

%----------------------------------------------------------------------------------------
%	TITLE SECTION
%----------------------------------------------------------------------------------------
\title{
\normalfont \normalsize
\textsc{University of Minnesota, School of Mathematics} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Math 4281: Introduction to Modern Algebra, \\
Spring 2019:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{\myname}

\begin{document}

\maketitle % Print the title

\begin{center} % Delete this if you want to save space!
{\large due date: \textbf{Friday, 8 February 2019} at the beginning of class, \\
or before that by email or canvas.

Please solve \textbf{at most 4 of the 6 exercises}!}
\end{center}

%----------------------------------------------------------------------------------------
%	EXERCISE 1
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 1: Mutual divisibility is rare}

\subsection{Problem}

Let $a$ and $b$ be two integers such that $a \mid b$ and
$b \mid a$.
Prove that $\abs{a} = \abs{b}$.

\subsection{Solution}

[...]

%----------------------------------------------------------------------------------------
%	EXERCISE 2
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 2: Congruence means equal remainders}

\subsection{Problem}

Let $n$ be a positive integer.
Let $u$ and $v$ be two integers.
Prove that $u \equiv v \mod n$
        % While it's called "congruent", the LaTeX command for
        % the symbol is "\equiv".
if and only if $u \% n = v \% n$.
        % To get the "%" sign, you need to type "\%".
        % Just typing "%" starts a comment.

\subsection{Solution}

[...]

%----------------------------------------------------------------------------------------
%	EXERCISE 2
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 3: Even and odd}

\subsection{Problem}

Let $u$ be an integer.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $u$ is even if and only if $u \% 2 = 0$.

\item[\textbf{(b)}]
Prove that $u$ is odd if and only if $u \% 2 = 1$.

\item[\textbf{(c)}]
Prove that $u$ is even if and only if $u \equiv 0 \mod 2$.

\item[\textbf{(d)}]
Prove that $u$ is odd if and only if $u \equiv 1 \mod 2$.

\item[\textbf{(e)}]
Prove that $u$ is odd if and only if $u + 1$ is even.

\item[\textbf{(f)}]
Prove that exactly one of the two numbers $u$ and $u + 1$ is even.

\item[\textbf{(g)}]
Prove that $u \tup{u+1} \equiv 0 \mod 2$.

\item[\textbf{(h)}]
Prove that $u^2 \equiv -u \equiv u \mod 2$.

\end{enumerate}

\subsection{Solution}

[...]

%----------------------------------------------------------------------------------------
%	EXERCISE 4
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 4: Factorials 102}

\subsection{Problem}

\begin{enumerate}

\item[\textbf{(a)}]
Prove that
\[
\dfrac{1! \cdot 2! \cdot \cdots \cdot \tup{2n}!}{n!}
= 2^n \cdot \prod_{i=1}^n \tup{\tup{2i-1}!}^2
\qquad \text{for each } n \in \NN .
\]

\item[\textbf{(b)}]
Prove that
\[
\sum_{k=0}^n \dfrac{1}{k! \cdot \tup{k+2}}
= 1 - \dfrac{1}{\tup{n+2}!}
\qquad \text{for each } n \in \NN .
\]

\end{enumerate}

\subsection{Solution}

[...]

%----------------------------------------------------------------------------------------
%	EXERCISE 5
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 5: Binomial coefficients 102}

\subsection{Problem}

Prove that
\[
\dfrac{\tup{ab}!}{a! \tup{b!}^a}
= \prod_{k=1}^a \dbinom{kb-1}{b-1}
\]
for all $a \in \NN$ and all positive integers $b$.

\subsection{Solution}

[...]

%----------------------------------------------------------------------------------------
%	EXERCISE 6
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 6: Binomial coefficients and coprimality}

\subsection{Problem}

It is well-known (see, e.g., \cite[Proposition 3.20]{detnotes})
that $\dbinom{n}{k} \in \ZZ$ for all $n \in \ZZ$ and $k \in \NN$.
(This is not at all clear from the definition of $\dbinom{n}{k}$;
it is saying that the product of any $k$ consecutive integers
is divisible by $k!$.
The case of $k = 2$ is the statement of Exercise 3 \textbf{(g)}.)
Thus, we can study the divisibility of binomial coefficients
by various integers.
There are hundreds of theorems about this; this exercise is
about one of them.

Let $a$ and $b$ be two coprime positive integers.

\begin{enumerate}

\item[\textbf{(a)}]
Prove that $\dfrac{a}{a+b} \dbinom{a+b}{a} = \dbinom{a+b-1}{a-1}$
and $\dfrac{b}{a+b} \dbinom{a+b}{a} = \dbinom{a+b-1}{b-1}$.

\item[\textbf{(b)}]
Prove that if $h \in \QQ$ satisfies $ah \in \ZZ$ and $bh \in \ZZ$,
then $h \in \ZZ$.
(This is where the coprimality of $a$ and $b$ comes into play.)

\item[\textbf{(c)}]
Prove that $a+b \mid \dbinom{a+b}{a}$.

\item[\textbf{(d)}]
Find a counterexample to the claim of part \textbf{(c)} if
$a$ and $b$ are allowed to not be coprime.

\end{enumerate}

\subsection{Solution}

[...]

\begin{thebibliography}{99999999}                                                                                %

% Feel free to add your sources -- or copy some from the source code
% of the class notes ( http://www.cip.ifi.lmu.de/~grinberg/t/19s/notes.tex ).

% This is the bibliography: The list of papers/books/articles/blogs/...
% cited. The syntax is: "\bibitem[name]{tag}Reference",
% where "name" is the name that will appear in the compiled
% bibliography, and "tag" is the tag by which you will refer to
% the source in the TeX file. For example, the following source
% has name "GrKnPa94" (so you will see it referenced as
% "[GrKnPa94]" in the compiled PDF) and tag "GKP" (so you
% can cite it by writing "\cite{GKP}").

\bibitem[GrKnPa94]{GKP}Ronald L. Graham, Donald E. Knuth, Oren Patashnik,
\textit{Concrete Mathematics, Second Edition}, Addison-Wesley 1994.\\
See \url{https://www-cs-faculty.stanford.edu/~knuth/gkp.html} for errata.

\bibitem[Grinbe19]{detnotes}Darij Grinberg,
\textit{Notes on the combinatorial fundamentals of algebra},
10 January 2019. \\
\url{http://www.cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf}
\\
The numbering of theorems and formulas in this link might shift
when the project gets updated; for a ``frozen'' version whose
numbering is guaranteed to match that in the citations above, see
\url{https://github.com/darijgr/detnotes/releases/tag/2019-01-10} .

\end{thebibliography}

\end{document}


%\documentclass{emulateapj}
%\documentclass[letterpaper,12pt,preprint]{aastex}
\documentclass[usenatbib]{mn2e}
\bibliographystyle{mn2e}

% packages
\usepackage{amssymb,amsmath,amsbsy}
\usepackage{booktabs}
\usepackage[caption=false]{subfig}
\usepackage{color}

% commands
\newcommand{\given}{\,|\,}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\transpose}[1]{{#1}^{\mathsf{T}}}
\newcommand{\inverse}[1]{{#1}^{-1}}
\newcommand{\msun}{\mathrm{M}_\odot}

\newcommand{\project}[1]{\textsl{#1}}
\newcommand{\superfreq}{\project{SuperFreq}}

% TO DO
\usepackage{graphicx}
\newcommand{\apwtodo}[1]{{\color{red} APW: (\MakeUppercase{#1})}}
\newcommand{\djdtodo}[1]{{\color{green} DJD: (\MakeUppercase{#1})}}

%Dan's Math Definitions
\newcommand{\scripty}[1]{\ensuremath{\mathcalligra{#1}}}
\def\sr{\scripty{r}}
\def\Mach{\mathcal{M}}
\def\bin{\rm{bin}}
\def\Mdot{\dot{M}}
\def\Msun{ M_{ \rm{\odot} } }

\def\xb{\bar{x}}
\def\yb{\bar{y}}

\begin{document}

\title{Cavity search: a parameter study of the importance of chaos for feeding black hole binaries}
\author[D. J. D'Orazio, Adrian M. Price-Whelan]{Daniel J. D'Orazio$^1$, Adrian M. Price-Whelan$^1$
    \thanks{dorazio@astro.columbia.edu; adrn@astro.columbia.edu}\\
     $^1$Department of Astronomy, Columbia University, 550 West 120th Street, New York, NY 10027
}

%\author{
%Daniel J. D'Orazio\altaffilmark{\colum},
%Adrian M. Price-Whelan\altaffilmark{\colum}
%}

% Affiliations
%\newcommand{\colum}{1}
% \newcommand{\adrn}{2}

%\altaffiltext{\colum}{Department of Astronomy,
   %                   Columbia University,
      %                550 W 120th St.,
         %             New York, NY 10027, USA}
% \altaffiltext{\adrn}{To whom correspondence should be addressed: adrn@astro.columbia.edu}

\maketitle

\begin{abstract}
%Danny boy
% Context
The interaction of a binary and a thin gasesous disk lies at the heart
of a number of important astrophysical phenomena. These include the
formation of planetary systems and the fate of massive black hole
binaries at the centers of galactic nuclei.
%The masses and orbital distributions of planetary systems are
%dictated by the interaction of a planet-star binary with a
%proto-planetery disk. The
%merger rates and electromagnetic signatures of massive black hole
%binares are influenced by gas which is torqued to the centers of
%galactic nuclei upon merger. Their fate is tied to the level
%of orbital evolution and mass feeding influenced by the coupled nature
%of binary+disk evolution.
The salient features of binary+disk interactions are captured largley
by the gravitaitonal dynamics of a disk of particles in the plane of
the binary.
%Aims
Here we gain insight into binary+disk dynamics by studying orbits of
test particles in the binary plane via the restricted three body
problem.
%Methods
We investigate the resonant structure of orbits over a range of binary
mass ratios and orbital eccentricities relvant to astrophysical
systems.
% Results
We find some cool chaos crap related to Dan's other paper on CBD
transitions and also some dynamics stuff related to Adrian's work.
% Conclusions

\end{abstract}

%\keywords{
%stuff
%}

\section{Introduction}\label{sec:introduction}

\section{Methods}\label{sec:methods}

\subsection{Equations of Motions}
%\section{Equations of Motions}
The equations of motion for a test particle in the plane of an
eccentric binary are described by the elliptical, restricted three
body problem (ER3Bp). We write the ER3Bp as two coupled second order
ODE's in the non-uniformly rotating, isotropically pulsating, frame of
the eccentric binary ($\bar{x}$, $\bar{y}$),
\begin{equation}
\begin{array}{c}
\ddot{\xb} - 2\dot{\yb} = \frac{\partial{U}}{\partial{\xb}}  \left( 1 + e \cos{f} \right)^{-1} \nonumber \\ \nonumber \\
\ddot{\yb} + 2\dot{\xb} = \frac{\partial{U}}{\partial{\yb}}  \left( 1 + e \cos{f} \right)^{-1}
\label{Eqmotion}
\end{array}
\end{equation}
for binary with true anomaly $f$, eccentricity $e$, mean motion $\equiv 1$,
and $ \ \dot{} \ \equiv d/df$. Because the independent variable is
$f$, there is no need to solve for $f$ in terms of the time $t$ unless
converting the solution back to the non-rotating frame.

The ($\bar{x}$, $\bar{y}$) are dimensionless coordinates of the
massless third particle which are constructed by dividing the
dimensional coordinates by the time changing binary separation
\begin{equation}
\begin{array}{c}
\xb = \xb^* \frac{1 + e \cos{f}}{a ( 1-e^2) } \nonumber \\ \nonumber \\
\yb = \yb^* \frac{1 + e \cos{f}}{a ( 1-e^2) }
\end{array}
\end{equation}
where $*$ denotes the dimensional variable.

The position of the primaries is found from dividing the dimensional
position over time by the binary position and is always fixed at the
same values as in the circular R3B, \textit{i.e.}
\begin{equation} \nonumber
\xb_2 = \xb^*_2 \frac{1+e \cos{f}}{a (1-e^2)} =   \frac{a}{1+q} \frac{(1-e^2)}{1+e \cos{f}}  \frac{1+e \cos{f}}{a (1-e^2)}  = \frac{1}{1+q}
\end{equation}
where $q = M_2/M_1$, $M_2 < M_1$.Similarly $\xb_p = -q/(1+q)$.

There is no longer a conserved quantity in the ER3B, but there is an
analogue to the Jacobi constant which is time dependent,
\begin{equation}
C_{eJ} = \frac{2 U}{1 + e \cos{f}} - \left( \dot{\xb}^2 + \dot{\yb}^2 \right) - 2 e \int^f_{f_0}{\frac{ U \sin{f} }{(1 + e \cos{f})^2 } \ df}
\end{equation}
and can be used to check the accuracy of the integrator (not yet
implemented).

\subsection{Set-up}
To set up initial velocity profiles recall that the independent
variable is no longer the time, but the true anomaly $f$. This can be
taken into account most simply by using the Virial theorem with the
pseudo potential of Eq. \ref{Eqmotion}. Assuming only azimuthal
initial velocities,
\begin{equation}
\begin{array}{c}
\bar{v_x} = -v_{\phi} \frac{\yb_0}{R} \quad \bar{v_x} = v_{\phi} \frac{\xb_0}{R} \\ \nonumber \\ \nonumber
v_{\phi} =   \left[ \sqrt{ \frac{\mu_1}{\bar{r}_1}  +  \frac{\mu_2}{\bar{r}_2} }  \right] \left( 1 + e \frac{\xb}{R} \right)^{-1}  - R\\ \nonumber \\ \nonumber
\bar{r}^2_1 =  (\xb + \mu_2)^2 + \yb^2  \qquad \bar{r}^2_2 =  (\xb - \mu_1)^2 + \yb^2  \\ \nonumber \\ \nonumber
R^2 = \xb^2 + \yb^2 \\ \nonumber \\ \nonumber
\mu_1 = \frac{1}{1+q} = 1-\mu_2  \qquad \mu_2 = \frac{q}{1+q}
\end{array}
\end{equation}
where we subtract the speed of the rotating frame from the azimuthal
velocity. Since the angular frequency of the rotating frame is unity
in our coordinates we need only subtract the particle specific
velocity $R$.

%If we were going to convert back into the time coordinate, we would need to use
%\begin{equation}
%df  = n \sqrt{1-e^2} \left( \frac{a}{r}\right)^2 dt = \frac{n (1 + e \cos{f})^2}{(1-e^2)^{3/2}}  dt
%\end{equation}
%where the last line substitutes the time dependent binary separation $r$.

To convert back to time t, and unscaled distances we multiply the
coordinates by the time dependent binary separation and solve for
$t(f)$ in the usual manner.

Describe potential and things varied (mass ratio, viscosity,
eccentricity, what else?)\\ Lets start with mass ratio - look for
Linblad resonances, and see what happens when linear stability is
lost.

Orbits are integrated with ...

\subsection{Numerical determination of the fundamental frequencies}\label{sec:freqs}

Regular orbits in Hamiltonian systems may be represented in a special set of coordinates known as angle-action variables \citep[e.g.,][]{goldstein80}: in these coordinates, the position variables---the angles---increase linearly with time with rates set by a set of fundamental frequencies, and the momentum variables---the actions---are integrals of motion. More specifically, for a regular orbit represented in coordinates $(x_n, v_n)$, there exists a transformation $(x_n, v_n)\rightarrow(\theta_n, J_n)$ such that
\begin{align}
	\theta_n(t) &= \theta_n(0) + \Omega_n\,t\\
	J_n &= {\rm const.}
\end{align}
The $\Omega_n$ are known as the fundamental frequencies and there are $N$ such frequencies for non-resonant orbits in systems with $N$ degrees of freedom.

\apwtodo{Need to use a different symbol here for integer vector}
A resonant orbit is an orbit for which there exists at least one relation such that\footnote{The repeated indices imply summation.} $n_n\,\Omega_n = 0$ where $n_n$ is a vector of integers. \apwtodo{More theory-speakery here methinks...}

[How to numerically determine the frequencies using:]
\begin{equation}
	x(t) = \sum a_k\,e^{i \, \omega_k \, t}
\end{equation}

[How to detect chaos with this shite and measure rate of frequency diffusion]

\section{Conclusions}\label{sec:conclusions}

%\acknowledgements
\section{Acknowledgments}
APW is supported by a National Science Foundation Graduate Research Fellowship under Grant No.\ 11-44155. DJD is supported by a National Science Foundation Graduate Research Fellowship under Grant No. DGE1144155
This work was supported in part by the National Science Foundation under Grant No. PHYS-1066293.
This research made use of Astropy, a community-developed core Python package for Astronomy \citep{astropy13}.
This work additionally relied on Columbia University's \emph{Hotfoot} and \emph{Yeti} compute clusters, and we acknowledge the Columbia HPC support staff for assistance, especially Mr. Alex Bergier.

%\bibliographystyle{apj}
%\bibliography{refs}

\end{document}


\documentclass[main.tex]{subfiles}
\begin{document}

\section{Sheet 7}

\subsection{Photons travelling in the Schwarzschild metric}

\subsubsection{A different proof for the conservation of the component of the velocity of a geodesic along a Killing vector field (complement)}

Here I present a different proof to what was done in the lectures for the fact that the component of the 4-velocity along the Killing vector field is conserved.
This is not necessary to know for the exam, do skip this section if it does not interest you.

If the metric does not depend on the coordinate \(\widetilde{\alpha }\), then \(\partial_{\widetilde{\alpha }} g_{\mu \nu } = 0\). So, let us differentiate covariantly the vector \(\xi_{\mu } = g_{\mu \nu } \delta^{\nu}_{\widetilde{\alpha }}\).
It will be apparent later that differentiating the lower-index vector field gives us the interesting property.
We get
%
\begin{align}
  \nabla_{\mu } \xi_{\nu } =
  g_{\nu \sigma } \nabla_{\mu } \xi^{\sigma }
  =  g_{\nu \sigma }
  \qty(\cancelto{}{\partial_{\mu } \xi^{\sigma }} + \Gamma^{\sigma }_{\mu \rho } \xi^{\rho })
  = \Gamma_{\nu \mu \widetilde{\alpha }}
\,,
\end{align}
%
since the only component which survives the contraction with \(\xi \) is the one along \(\widetilde{\alpha} \); also, we lowered an index of the Christoffel symbols with the metric.

The explicit expression for the lower indices Christoffel symbols is
%
\begin{align}
  \Gamma_{\nu \mu \widetilde{\alpha }}
  = \frac{1}{2} \qty(\cancelto{}{g_{\mu \nu , \widetilde{\alpha }}} +
  g_{\mu \widetilde{\alpha }, \nu }
  - g_{\mu \widetilde{\alpha }, \nu })
\,,
\end{align}
%
since by hypothesis any derivative of the metric along \(\widetilde{\alpha}\) is zero. So, we can directly see that the object \(\nabla_{\mu } \xi_{\nu }\) is antisymmetric in its indices: this can be written as
%
\begin{align}
  \nabla_{(\mu } \xi_{\nu )} = 0
\,,
\end{align}
%
and is called \emph{Killing's equation}. We have shown that is equivalent to the metric not depending on the coordinate \(x^{\widetilde{\alpha}}\).

Now, we can quickly prove the conservation a component of the 4-velocity of a geodesic along the Killing vector field: we just need to differentiate \(u^{\mu } \xi_{\mu }\) with respect to the arc parameter.
Recall that geodesics are defined by the equation
%
\begin{align}
 a^{\mu } =  \dv[]{}{s} u^{\mu } = u^{\nu } \nabla_{\nu } u^{\mu } = 0
\,.
\end{align}
%

We find
%
\begin{align}
  u^{\nu }\nabla_{\nu }\qty(u^{\mu } \xi_{\mu })
  = u^{\nu } u^{\mu } \nabla_{\nu } \xi_{\mu } + \xi_{\mu } u^{\nu } \nabla_{\nu } u^{\mu } = 0
\,,
\end{align}
%
where both terms are zero: the first because it is the contraction of an antisymmetric object with a symmetric one, and the second one because of the geodesic equation.

\subsubsection{Conserved quantities in Schwarzschild motion}

The Schwarzschild metric is given by
%
\begin{subequations}
\begin{align}
  g_{\mu \nu } = \left[\begin{array}{cccc}
  -(1-\frac{2GM}{r}) & 0 & 0 & 0 \\
  0 & (1-\frac{2GM}{r})^{-1} & 0 & 0 \\
  0 & 0 & r^2 & 0 \\
  0 & 0 & 0 & r^2 \sin^2\theta
  \end{array}\right]
\,
\end{align}
\end{subequations}
%
in the coordinates \((t, r, \theta , \varphi )\). The vector fields \(\xi_{(t)}^{\mu } = (1, \vec{0})\) and \(\xi_{(\varphi )}^{\mu } = (0,0,0,1)\) in these coordinates are Killing vector fields, since the metric does not depend on \(t\) or \(\varphi \).

So, the following quantities are conserved in geodesic motion parametrized as \(x^{\mu }(\lambda )\): \footnote{There is a typo in the exercise sheet: a \(G\) is missing in the definition of \(e\).}
%
\begin{align}
  e = -u^{\mu } g_{\mu \nu } \xi^{\nu }_{(t)}
  = -u^{t} g_{tt } \times 1
  = \dv{t}{\lambda } \qty(1 - \frac{2GM}{r})
\,
\end{align}
%
and
%
\begin{align}
  l = u^{\mu } g_{\mu \nu } \xi^{\nu }_{(\varphi )}
  = u^{\varphi } g_{\varphi \varphi } \times 1
  = \dv{\varphi }{\lambda } r^2 \sin^2\theta
\,,
\end{align}
%
which for motion on the \(xy\) plane, for which \(\theta = \pi /2\), reduces to \(l = r^2 \dv*{\varphi }{\lambda }\).

\subsubsection{Photons escaping a black hole}

The equation of motion can be derived from the normalization of the photon's four velocity: The equation \(u^{ \mu } u_{\mu }= 0 \) can be  written as
%
\begin{align}
    \qty(\dv{t}{\lambda })^2 g_{tt} +
    \qty(\dv{r}{\lambda })^2 g_{rr} +
    \qty(\dv{\theta }{\lambda })^2 g_{\theta \theta } +
    \qty(\dv{\varphi }{\lambda })^2 g_{\varphi \varphi } = 0
\,,
\end{align}
%
but the term \(\dv*{\theta }{\lambda }\) is zero if we assume the motion to be in the \(xy \) plane, while the velocity components along \(t\) and \(\varphi \) can be written in terms of the integrals of motion: \(\dv*{t}{\lambda } = e \qty(1 - 2GM/r)^{-1}\) and \(\dv*{\varphi }{\lambda } = l r^{-2}\). So, we find
%
\begin{align}
  - \qty(1 - \frac{2GM}{r})^{-2+1} e^2 + \qty(1 - \frac{2GM}{r})^{-1} \qty(\dv{r}{\lambda })^2 + l^2 r^{-4} r^2 = 0
\,,
\end{align}
%
we divide through by \(-g_{tt}\) and find:
%
\begin{align}
    -e^2 + \qty(\dv{r}{\lambda })^2+ \frac{l^2}{r^2} \qty(1 - \frac{2GM}{r}) = 0
    \,,
\end{align}
%
or, dividing through by \(l\):
%
\begin{align}
    -\frac{e^2}{l^2} + \frac{1}{l^2}\qty(\dv{r}{\lambda })^2+ \frac{1}{r^2} - \frac{2GM}{r^3} = 0
\,.
\end{align}

We can give names to the terms in this equation: we call
%
\begin{align}
  V _{\text{eff}} (r) \equiv \frac{1}{r^2} - \frac{2GM}{r^3}
\,
\end{align}
%
the \emph{effective potential}, and
%
\begin{align}
  b^2 = \frac{l^2}{e^2}
\,
\end{align}
%
the \emph{impact parameter} (unjustified for now). Then, the equation is in the form
%
\begin{align}
  \frac{\dot{r}^2}{l^2}  + V _{\text{eff}} (r) = \frac{1}{b^2}
\,,
\end{align}
%
where we denoted derivation with respect to \(\lambda \) with a dot.
So, we can study the motion of the photon as if it were 1-dimensional.

To study the problem, it is convenient to use the rescaled adimensional radial coordinate \(R = r / 2GM\). In this variable, the effective potential (which I will denote as just \(V\) hereafter) looks like:
%
\begin{align}
  V (R) = (2GM)^{-2} \qty(R^{-2} - R^{-3})
\,,
\end{align}
%
so we can readily differentiate it to find its stationary points: there is only one, the equation is  \(V^{\prime }(R) \propto -2R^{-1} + 3R^{-2} =0 \), which is satisfied by \(R = 3 / 2\).

\begin{figure}[ht]
    \centering
    \includegraphics[width=0.7\textwidth]{figures/photon_effective_potential.pdf}
    \caption{Plot of the function \(R^{-2} - R^{-3}\).}
    \label{fig:effective-potential}
\end{figure}

The term \(b^{-2}\) is the maximum value which can be attained by the LHS: we can call it the total energy, the kinetic and potential contributions must add up to it.
Also, the kinetic term is always positive.
So, if the total energy is less than the maximum of the potential, the photon is constrained to stay on either side of the potential barrier around \(R = 3/2\). The potential there equals
%
\begin{align}
  V(3/2) = (2GM)^{-2} \qty((3/2)^{-2} - (3/2)^{-3})
  = \frac{4/27}{(2GM)^2} = \frac{1}{27 (GM)^2}
\,,
\end{align}
%
so the condition of the photon being above the potential barrier is
%
\begin{subequations}
\begin{align}
  E _{\text{tot}} &> V _{\text{max}}  \\
  \frac{e^2}{l^2} &> \frac{1}{27 G^2M^2}  \\
  \frac{l^2}{e ^2} &< 27G^2M^2
\,,
\end{align}
\end{subequations}
%
and under this condition, if the photon initially has positive \(\dv*{r}{\lambda }\) it will remain as such, since everything is continuous and (if the strict inequality is satisfied) we can never have \(\dv*{r}{\lambda }=0\).

If instead we had \(l^2/e^2 < 27G^2M^2\) and the photon was initially travelling away from the black hole, there would come a point for which \(\dv*{r}{\lambda }\) would equal zero, and then it would become negative, since the photon could not go away from the center anymore.

This can be understood graphically by drawing horizontal lines of constant total energy in the potential diagram.

\subsubsection{A basis for a stationary observer}

If our observer's coordinates are \((t, r_{*}, \pi /2 , \varphi_{*} )\), that is, it is at rest with respect to our spatial coordinates but it is not following geodesic motion, then it will have nonzero 4-acceleration. So, since we know that velocity and acceleration are orthogonal, we can form our coordinate system as \((u^{\mu }, a^{\mu } / \sqrt{ a^{\rho } a_{\rho }}, e_{\theta }, e_{\varphi })\), where the angular basis vectors are simply normalized vectors in the \(\theta \) and \(\varphi \) directions.

This method has the advantage of being easily generalizable to find a comoving basis for any non-geodesic motion of our observer.

Their 4-velocity must be normalized so that \(u^{\mu } u_{\mu }= -1\): so
%
\begin{subequations}
\begin{align}
  u^{\mu } = \left[\begin{array}{c}
  \dv{t}{\tau } \\
  0 \\
  0 \\
  0
  \end{array}\right]
  =
  \left[\begin{array}{c}
  1/\sqrt{1 - \frac{2GM}{r}} \\
  0 \\
  0 \\
  0
  \end{array}\right]
\,,
\end{align}
\end{subequations}
%
then we can compute the 4-acceleration: it will be
%
\begin{subequations}
\begin{align}
  a^{\mu } &= u^{\nu } \nabla_{\nu } u^{\mu }
  = u^{\nu } \qty(\partial_{\nu } u^{\mu } + \Gamma^{\mu }_{\nu \rho } u^{\rho })  \\
  &= \frac{1}{\sqrt{1 - \frac{2GM}{r}}} \qty(\cancelto{}{\partial_{t} u^{\mu }} + \Gamma^{\mu }_{tt} u^{t})  \\
  &= \frac{1}{\sqrt{1-\frac{2GM}{r}}}
  \frac{1}{\sqrt{1-\frac{2GM}{r}}} \frac{A'}{2B} \delta^{\mu }_{r}
\,,
\end{align}
\end{subequations}
%
where \(A = 1/B = (1 - 2GM/r)\) are the coefficients of the Schwarzschild metric, with respect to which the Christoffel symbols are expressed in \eqref{eq:schwarzschild-christoffel}.
The \(\delta^{\mu }_{r}\) signifies that the only component which survives is the radial one.
The two square root terms simplify the \(B\), so we get that the only nonzero component of the 4-acceleration is:
%
\begin{align}
  a^{r} = \frac{A^{\prime }}{2} =
  \frac{1}{2} \dv{}{r} \qty(1 - \frac{2GM}{r})
  = \frac{GM}{r^2}
\,.
\end{align}
%
The actual value of this component is not actually useful to us, since we need a normalized basis.
It can, however, be used to illustrate the equivalence principle: in the weak-field limit where the metric is approximately the Minkowski one the value of this component approaches the modulus of the 3-acceleration.

Therefore, we will use a vector parallel to \(a^{\mu }\) but of length defined by the normalization \(e_{r} \cdot e_{r} = 1\). So, we get for our basis:
%
\begin{subequations}
\begin{align}
  \begin{cases}
    (e_{t})^{\mu } = u^{\mu } = \left[\begin{array}{cccc}
    (1-2GM/r)^{-1/2}, & 0, & 0, & 0
  \end{array}\right]^{\top} \\
    (e_{r})^{\mu } = a^{\mu } / \sqrt{a_{\rho } a^{\rho }}
  = \left[\begin{array}{cccc}
    0, & (1-2GM/r)^{1/2}, & 0, & 0
  \end{array}\right]^{\top}  \\
  (e_{\theta })^{\mu } = \left[\begin{array}{cccc}
  0, & 0, & 1/r, & 0
  \end{array}\right]^{\top} \\
  (e_{\varphi })^{\mu } = \left[\begin{array}{cccc}
  0, & 0, & 0, & 1/(r \sin \theta )
  \end{array}\right]^{\top}
\end{cases}
\,,
\end{align}
\end{subequations}
%
where the last vector is written for a generic angle \(\theta \), but the sine is equal to one if \(\theta = \pi /2\).
This basis satisfies the property \(e_{\alpha } \cdot e_{\beta } = \eta_{\alpha \beta }\).

We will denote vectors written with respect to this basis with a subscript \(c\).

\subsubsection{Critical angle of launch}

In the flat frame, the 4-velocity of the photon launched at \(\theta = \pi /2\) looks like:
%
\begin{subequations}
\begin{align}
  u^{\mu } _{\text{ph}} = \left[\begin{array}{c}
  1 \\
  \cos \psi  \\
  0 \\
  \sin \psi
  \end{array}\right]_c
\,,
\end{align}
\end{subequations}
%
which means that in the Schwarzschild frame it looks like
%
\begin{subequations}
\begin{align}
  u^{\mu }_{\text{ph}} = \left[\begin{array}{c}
  \frac{1}{\sqrt{1-\frac{2GM}{r}}} \\
  \cos \psi \sqrt{1 - \frac{2GM}{r}} \\
  0 \\
  \frac{\sin \psi}{r}
  \end{array}\right]
\,.
\end{align}
\end{subequations}

This means that the parameter \(l = r \sin \psi \), while \(e = \sqrt{1-2GM/r}\).

The inequality to satisfy for the photon to be able to escape is:
%
\begin{align}
  \qty[\frac{e^2}{l^2}]_{r_{*}} \geq \frac{1}{27G^2M^2}
\,,
\end{align}
%
which translates to
%
\begin{align}
  \frac{1 - \frac{2GM}{r_{*}}}{r_{*}^2 \sin^2 \psi } \geq \frac{1}{27G^2M^2}
\,,
\end{align}
%
or
%
\begin{align}
  r_{*}^2 \sin^2 \psi \leq 27G^2M^2 \qty(1 - 2GM/r_{*})
  \,,
\end{align}
%
\begin{figure}[ht]
  \centering
  \includegraphics[width=0.7\textwidth]{figures/critical_psi.pdf}
  \caption{Critical angle \(\psi \) in terms of the adimensional radial coordinate \(R = r/ 2GM\).}
  \label{fig:critical-psi}
\end{figure}
%
which once again can be expressed in terms of the adimensional radial coordinate \(R = r/2GM\): we find
%
\begin{align}
  \sin^2 \psi \leq \frac{27}{4 R_{*}^2} \qty(1 - \frac{1}{R_{*}})
  \,,
\end{align}
%
so the critical angle can be computed by making the relation explicitly in terms of \(\psi \), which gives the plot shown in figure \ref{fig:critical-psi}, for the equation
%
\begin{align}
  \psi = \arcsin \sqrt{\frac{27}{4 R_{*}^2} \qty(1 - \frac{1}{R_{*}})}
\,.
\end{align}
%

So, as we move closer to the horizon, the range of angles at which we can throw our photon and still have it escape decreases, until finally we can only throw it straight forward otherwise it will fall back in. After \(r = 2GM\), not even that is enough.

All communication with the outside world is lost.

\subsection{Light motion in a Schwarzschild geometry}

\subsubsection{Equation of motion}

This was done during the lectures, but I will recall it here.

We start from the equation of motion of a photon, a reframing of \(u^2=0\):
%
\begin{align}
  \frac{1}{l^2} \qty(\dv{r}{\lambda })^2 + V _{\text{eff}} (r) = \frac{1}{b^2} = \frac{e^2}{l^2}
\,.
\end{align}

We want to write this as a differential equation for the radius in terms of the angle \(\varphi \): so, we use the expression of the first integral
%
\begin{align}
  l = \dv{\varphi }{\lambda } r^2 \implies \dv{\varphi }{\lambda } = \frac{l}{r^2} \implies \dv{}{\lambda } = \frac{l}{r^2} \dv{}{\varphi }
\,.
\end{align}
%

So, we can rewrite
%
\begin{align}
  \qty(\dv{r}{\lambda })^2= \frac{l^2}{r^{4}} \qty(\dv{r}{\varphi })^2
\,,
\end{align}
%
which allows us to write the equation as
%
\begin{align}
  \frac{1}{r^{4}} \qty(\dv{r}{\varphi })^2 + V _{\text{eff}}(r) = \frac{1}{b^2}
\,,
\end{align}
%
since the \(l^2\) simplifies.

Now, notice that if we define \(u = 1/r\) we find
%
\begin{align}
  \dv{u}{\varphi } = - \frac{1}{r^2} \dv{r}{\varphi }
\,,
\end{align}
%
which conveniently simplifies the \(r^{-4}\): expressing everything with respect to \(u\) we get
%
\begin{align}
  \qty(\dv{u}{\varphi })^2 +  \qty(u^2-2GMu^3) = \frac{1}{b^2}
\,.
\end{align}

Now, we just need to differentiate everything to eliminate the constant term and find
%
\begin{align}
  2 u' u'' + 2 u u' - 6 GM u^2 u' = 0
\,,
\end{align}
%
where we denoted derivatives with respect to \(\varphi \) with primes.
One solution is \(u'=0\): a circular orbit, which we are not interested in now (and which is unstable\dots).

Otherwise, we can simplify a factor \(2 u'\): we find
%
\begin{align}
  u'' + u = 3 GMu^2
\,
\end{align}
%
as we wanted to show.

\subsubsection{Zero black hole mass solution}

If \(M=0\), our diffential equation is simply \(u''+u=0\), a harmonic oscillator.
Our boundary condition is \(r (\varphi = 0, \pi ) \rightarrow \infty \), which means \(u \rightarrow 0\) in those cases.

Also, one can geometrically see that at any radius \(\sin \varphi = b/r\) where \(b\) is the impact parameter.
Therefore, \(bu = \sin \varphi \) or \(u = \sin(\varphi ) / b \). This actually is already a solution to our differential equation!

We are giving two boundary conditions, therefore the solution \(u = \sin(\varphi ) / b \) is unique.

\subsubsection{Small mass deflection}

We now insert a mass \(M\), and consider a solution which is a perturbation to the sinusoidal one: our proposed solution is \(u = b^{-1} (\sin \varphi + w)\), and we will only look at the first order in \(w\). Substituting into \(u'' + u = 3GM u^2\) we get
%
\begin{align}
  - \sin \varphi + w'' + \sin \varphi + w = \frac{3GM}{b} \qty(\sin^2 \varphi + \cancelto{}{2w  \sin \varphi} + \cancelto{}{w^2})
\,,
\end{align}
%
where we kept only terms which are of first order in either \(GM\) or \(w\), since both are small relative to the scale of the problem. Also, we simplified a common factor of \(b\).
So the differential equation for \(w\) is
%
\begin{align}
  w'' + w = \frac{3GM}{b} \sin^2 \varphi
\,.
\end{align}

This can be solved with an \emph{ansatz} in the form \(w = A + B \sin^2\varphi\).
Its derivatives are \(w' = 2B \sin \varphi \cos \varphi = B \sin(2\varphi )\), and \(w'' = 2B \cos(2 \varphi )= 2 B \qty(1 - 2\sin^2\varphi )\).

Plugging this in we find
%
\begin{align}
  2 B (1 - 2 \sin^2 \varphi) + A + B \sin^2 \varphi = \frac{3GM}{b} \sin^2 \varphi
\,,
\end{align}
%
so equating the terms in \(\sin^2\varphi  \) and the constant ones we get
%
\begin{subequations}
\begin{align}
  \begin{cases}
    2B+A = 0 \\
    -3B = 3GMb^{-1}
  \end{cases}
\,,
\end{align}
\end{subequations}
%
so our perturbation is \(w = 2 GM b^{-1} (1- \sin^2 \varphi / 2)\), therefore the full solution is
%
\begin{align}
  u(\varphi ) = \frac{1}{b} \qty(\sin \varphi + \frac{2GM}{b} \qty(1 - \frac{\sin^2 \varphi }{2}))
\,.
\end{align}
%

What follows is an alternative derivation, more complicated than plainly discarding the second-order \(\sin^2  \varphi  \) term and expanding the sine to first order, which gives \(\varphi \sim -2GM/b\) right away. I did it in this way mostly to check that it still works.

We want to solve the equation of the particle coming in from radial infinity at an angle \(\varphi _{\text{in}}\), and leaving at an angle \(\varphi _{\text{out}}\) towards radial infinity. This means that we are seeking two solutions to \(r = \infty \implies u = 0\), respectively near \(\varphi = 0\) and \(\varphi = \pi \). This can be written as a second degree equation in terms of the variable \(x = \sin \varphi \), and the adimensionalized impact parameter \(b/ 2GM = d\), by which we multiply everything to get:
%
\begin{align}
  x^2 - 2dx - 2 = 0
\,,
\end{align}
%
which can be solved as
%
\begin{align}
  x =d \pm \sqrt{d^2 + 2}
  = d \qty(1 \pm \sqrt{1 + \frac{2}{d^2}})
\,,
\end{align}
%
so we have two solutions: one near \(x=0\), one near \(x=2d\). The second is meaningless, since \(d \gg 1 \) but \(x \leq 1\). Expanding around \(d = \infty\) we find:
%
\begin{align}
  x = d \qty(1 - \qty(1 + \frac{1}{2} \frac{2}{d^2})) = -\frac{d}{2}  \frac{2}{d^2} = -\frac{1}{d} = - \frac{2GM}{b} = \sin \varphi
\,.
\end{align}
%

So, we want solutions to this near \(\varphi = 0\) and \(\varphi = \pi \), and we are working up to first order in \(\varphi \). Then, we have
%
\begin{align}
  \varphi _{\text{in}} = - \arcsin \qty(\frac{2GM}{b})
  \sim - \frac{2GM}{b}
\,
\end{align}
%
and
%
\begin{align}
  \varphi _{\text{out}}  = \pi + \arcsin \qty( \frac{2GM}{b}) \sim \pi + \frac{2GM}{b}
\,.
\end{align}
%
The deflection can be calculated by assuming \( \varphi _{\text{out}} - \varphi _{\text{in}} = \pi + \delta \varphi \), which gives
%
\begin{align}
  \delta \varphi \approx \frac{4GM}{b}
\,.
\end{align}
%

\subsubsection{Newtonian prediciton and Eddington observations (complement)}

The Newtonian prediction, which is computed using the Keplerian formula for eccentricity (and, notably, simplifying the ``mass'' of a light particle) is instead:\footnote{\url{https://arxiv.org/pdf/physics/0508030.pdf}}
%
\begin{align}
  \delta \varphi \approx \frac{2GM}{b}
\,.
\end{align}

During the solar eclipse of 1919, Sir Eddington\footnote{\url{https://royalsocietypublishing.org/doi/abs/10.1098/rsta.1920.0009}} made two observations: one gave
%
\begin{align}
  \delta \varphi = \qty(\SI{4.5(3)}{}) \frac{GM}{b}
\,,
\end{align}
%
and another gave
%
\begin{align}
  \delta \varphi = \qty(\SI{3.7(7)}{}) \frac{GM}{b}
\,.
\end{align}
%

\subsection{Orbit at 7GM}

\subsubsection{Orbital radius}

The radius of a \(7G M_{\odot}\) orbit is equal to
%
\begin{subequations}
\begin{align}
  7GM_{\odot}/c^2 \approx\\
  7 \times \SI{6.67e-11}{kg m^{3} s^{-2}} \times \SI{2e30}{kg} \times \qty(\SI{3e8}{ms^{-1}})^{-2} \approx \\
  \approx \SI{1.0e4}{m}
\,,
\end{align}
\end{subequations}
%
or around \SI{10}{km}.

\subsubsection{Proper period}

The equation governing a circular orbit around a BH can be derived from the normalization of the 4-velocity \(u \cdot u = -1\): in terms of \(u = 1/r\) it is
%
\begin{align}
  u_c = \frac{GM}{l^2} + 3 GM u_c^2
\,,
\end{align}
%
and in our case we know that \(u_c = 1/7GM\), substituting in we find
%
\begin{align}
  \frac{1}{7GM} = \frac{GM}{l^2} + \frac{3GM}{49 (GM)^2}
\,,
\end{align}
%
or
%
\begin{align}
  l^2 = (GM)^2 \qty(\frac{1}{7} - \frac{3}{49})^{-1}
  = 7 (GM)^2 \qty(1 - \frac{3}{7})^{-1} = \frac{49}{4} (GM)^2
\,,
\end{align}
%
so \(l = 7GM/2\). Now, recall the definition of this first integral:
%
\begin{align}
  l = \dv{\varphi }{\tau } r^2
\,,
\end{align}
%
into which we can substitute the expression we found:
%
\begin{align}
  \frac{7GM}{2} (7GM)^{-2} = \frac{1}{14GM} =  \dv{\varphi }{\tau } = \omega_{\text{proper}}
\,,
\end{align}
%
which gives us the angular velocity as measured by the orbiting observer. We actually have it in units of $1/\SI{}{m^3 s^{-2}}$, so we will need to multiply by \(c^3\) to get inverse seconds.

This gives us a pulsation of
%
\begin{align}
  \omega _{\text{proper}} = \frac{c^3}{14GM_{\odot}} \approx
  \SI{1.44e4}{rad/s}
\,,
\end{align}
%
which corresponds to a period of
%
\begin{align}
  T _{\text{proper}} = \frac{2\pi}{\omega _{\text{proper}}} \approx \SI{4.36e-4}{s} = \SI{436}{\micro s}
\,.
\end{align}

\subsubsection{Period at infinity}

Now, to compute the period for an observer at infinity: from the regular equation of motion of a massive observer we have
%
\begin{align}
  \frac{e^2-1}{2} = \frac{1}{2} \qty(\dv{r}{\tau })^2+ \underbrace{\qty(-\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{r^3})}_{V _{\text{eff}}}
\,,
\end{align}
%
but since the orbit is circular \(r\) is constant so the derivative vanishes, and we can substitute in our expressions for \(r\) and \(l\) to find out what \(e\) is: we get
%
\begin{subequations}
\begin{align}
  \frac{e^2-1}{2} &= - \frac{GM}{7GM} + \frac{(7GM/2)^2}{2\times (7GM)^2} - \frac{GM (7GM/2)^2}{(7GM)^3}  \\
&= - \frac{1}{7}+ \qty(\frac{7}{2})^2 \frac{1}{2\times 7^2} - \qty(\frac{7}{2})^2 \frac{1}{7^3}  \\
&= -\frac{3}{56}
\,,
\end{align}
\end{subequations}
%
therefore \(e = 5 \sqrt{7} / 14\). Now recall the definition of \(e\):
%
\begin{align}
  e = \dv{t}{\tau } \qty(1 - \frac{2GM}{r})
\,,
\end{align}
%
so
%
\begin{align}
  \frac{T_{ \infty }}{T _{\text{proper}}} =
  \dv{t}{\tau } = \frac{5 \sqrt{7}}{14} \qty(1 - \frac{2GM}{7GM})^{-1} = \frac{\sqrt{7}}{2} \approx 1.323
\,,
\end{align}
%
therefore the period as measured by the outside observer is
%
\begin{align}
  T_{ \infty } \approx \SI{577}{\micro s}
\,.
\end{align}
%

Do note that this is \emph{not} the same as plainly applying the gravitational redshift formula as \(T_1 / T_2 = \sqrt{ g_{tt}^{1}/ g_{tt}^{2}}\): that gives a smaller ratio. The Doppler effects of the orbital speed do not average out over a period: this is treated in more depth in section \ref{sec:general-time-dilation}.

An alternative way to derive \(\dv*{t}{\tau }\) is to use \(\Omega \): we defined it as \(\dv*{\varphi }{t}\), and proved that it is given by
%
\begin{align}
  \Omega^2 = \qty(\dv{\varphi }{\tau } \dv{\tau }{t})^2
  = \frac{GM}{r^3}
\,.
\end{align}

It allows us to write the 4-velocity as
%
\begin{align}
  u^{\alpha } = \qty(\dv{t}{\tau }, 0, 0, \dv{\varphi}{\tau })^{\top} = \dv{t}{\tau } \qty(1, 0, 0, \Omega )^{\top}
\,.
\end{align}

Now, we can write the proper time interval starting from the line element:
%
\begin{align}
  - \dd{s^2} = \dd{\tau^2}
  = g_{00} \dd{t^2} - g_{33} \dd{\varphi^2}
  = \dd{t^2} \qty(g_{00} - g_{33} \Omega^2)
\,,
\end{align}
%
which we can calculate, since we know all the parameters: we find
%
\begin{align}
  \frac{\dd{\tau^2 }}{ \dd{t^2}} = \qty(1 - \frac{2GM}{r} - r^2 \frac{GM}{r^3})
  = \qty(1 - \frac{3GM}{r})
\,.
\end{align}

This gives
%
\begin{align}
  \dd{t}  = \frac{\dd{\tau }}{\sqrt{1 - \frac{3GM}{r}}}
  \sim \dd{\tau } \qty(1 + \frac{3GM}{2r})
\,.
\end{align}
%

\subsubsection{Proper acceleration}

The orbit is a geodesic, so a pointlike observer feels no acceleration.

\subsubsection{Adimensionalized parameters of motion and a general formula for time dilation (complement)} \label{sec:general-time-dilation}

We can make the expressions above more clear in terms of adimensionalized coordinates: we will use \(R = r/(2GM)\), \(L = l / (2GM)\), while \(e\) is already adimensional.
Under the assumption of circularity for the orbit, the expression for \(L\) can be calculated to be:
%
\begin{align}
  L  =\frac{R}{\sqrt{2 R - 3}}
\,,
\end{align}
%
while the one for \(e\) is:
%
\begin{align}
  e^2 = \qty(1 + \frac{L^2}{R^2}) \qty(1 - \frac{1}{R})
\,.
\end{align}

Also, the definition for \(e\) is
%
\begin{align}
  e = \dv{t}{\tau } \qty(1 - \frac{1}{R})
\,,
\end{align}
%
therefore we can write
%
\begin{subequations}
\begin{align}
  \dv{t}{ \tau } = e \qty(1-\frac{1}{R})^{-1} &= \qty(1 - \frac{1}{R})^{-1+1/2} \qty(1+ \frac{1}{2 R - 3})^{1/2} \\
  &= \frac{\sqrt{1 + \displaystyle \frac{1}{2R-3}}}{\sqrt{ 1- \displaystyle \frac{1}{R}}}
  = \sqrt{ \frac{2R-3+1}{2R-3} \frac{R}{R-1}}
  = \sqrt{\frac{R}{R - 3/2}}
\,.
\end{align}
\end{subequations}

With this, we can see that in the limit of \(R \rightarrow \infty\) we have the following effect:
%
\begin{align}
  \dv{t}{\tau } \sim \sqrt{1 + \frac{1}{R} + \frac{1}{2R-3} }
  \sim 1 + \frac{1}{2} \bigg( \underbrace{\frac{1}{2R}}_{\text{gravitational}} + \underbrace{\frac{1}{R}}_{\text{Doppler}} \bigg)  = 1 + \frac{3}{4R}
\,,
\end{align}
%
where we have a contribution both from the angular momentum, and from the gravitational redshift; the gravitational contribution is larger than the Doppler one in the limit (specifically for any \(R>2\)), and the ratio between the contributions approaches \(2\).

Also, we can see that we have a divergence of this time dilation at \(R \rightarrow 3/2\):
I suspect that this corresponds to the velocity needed to stay in orbit at constant \(R\) diverges towards 1 as \(R \rightarrow 3/2\).

In our specific case, the formula comes out to be
%
\begin{align}
  \dv{t}{\tau } = \underbrace{\sqrt{\frac{7}{5}}}_{\text{grav}} \underbrace{\sqrt{\frac{5}{4}}}_{\text{Dopp}} = \sqrt{\frac{7}{4}}
\,.
\end{align}

\subsubsection{Tidal acceleration (complement)}

A pointlike particle would feel no acceleration, however it would be quite unconfortable to be in that orbit if one happens not to be pointlike.
As for the first part of the exercise, this is not in the exercise sheet, so do skip it if you are in a hurry. I think it is quite interesing though.

Let us compute the tidal effects on an extended observer.

They are described by the equation
%
\begin{align}
  \dv[2]{\xi^{\mu }}{\tau } = R^{\mu }_{\nu \rho \sigma } u^{\nu } u^{\rho } \xi^{\sigma }
\,,
\end{align}
%
where \(u^{\mu }\) is the 4-velocity of a geodesic, \(\xi^{ \mu } \) is a small deviation in starting position for the geodesic such that the geodesic starting from there almost has the same tangent vector. For now we assume it to be true, in the next section we give justification for it.

We can restrict ourselves to radial geodesic deviation: then we only need to compute the \(R^{r}_{\nu \rho r}\) components of the Riemann tensor.
I might do the full computation, but I found a source\footnote{\url{https://physics.stackexchange.com/questions/295814/non-zero-components-of-the-riemann-tensor-of-the-schwarzschild-metric}} which gives the nonzero components:
%
\begin{subequations}
\begin{align}
  R^{r}_{ttr} &= \frac{2GM}{r^3} \qty(1 - \frac{2GM}{r}) \\
  R^{r}_{\theta \theta r} &= \frac{GM}{r} \\
  R^{r}_{\varphi \varphi r} &= \frac{G M \sin^2\theta}{r}
\,,
\end{align}
\end{subequations}
%
and since the orbital motion is assumed to happen on the \(\theta = \pi /2\) plane, the formula simplifies to
%
\begin{subequations}
\begin{align}
  \dv[2]{\xi^{r}}{\tau } &= R^{r}_{ttr} u^{t} u^{t} \xi^{r}
  + R^{r }_{\varphi \varphi r} u^{\varphi } u^{\varphi } \xi^{r}  \\
  &= \qty(\frac{e^2}{(1-2GM/r)^2} \frac{2GM}{r^{3}}\qty(1 - \frac{2GM}{r}) +  \frac{l^2}{r^{4}} \frac{GM}{r}) \xi^{r}  \\
  &= \qty(\frac{e^2}{1-2GM/r} + \frac{l^2}{r^2} ) \frac{2GM}{r^3} \xi^{r}
\,,
\end{align}
\end{subequations}
%
which holds in general: in our specific case, at \(r = 7GM \) we find
%
\begin{subequations}
\begin{align}
  \dv[2]{\xi^{r}}{\tau } &= \qty(\frac{7}{5} \qty(\frac{5 \sqrt{7}}{14})^2 + \qty(\frac{7GM/2}{7GM})^2 ) \frac{2GM}{(7GM)^3}  \xi^{r}  \\
  &= \frac{3}{343} \frac{1}{(GM)^2} \xi^{r}
\,,
\end{align}
\end{subequations}
%
so the object multiplying \(\xi^{r}\), in units of inverse square seconds, is
%
\begin{align}
  \frac{3}{343} \frac{c^{6}}{(GM)^2} \approx \SI{356e6}{s^{-2}}
  \approx \qty(\SI{18.9}{kHz})^2
\,.
\end{align}

The acceleration which would be experienced from one side of the other of a meter-long observer would then be of around \SI{36}{M}\(g\).

\subsubsection{A proof for the geodesic deviation formula (complement)}

The starting separation between geodesics is not, properly speaking, a \emph{vector} since it does not belong to the tangent space, however we can be sloppy and approximate it as a tangent vector.
Then it is a fact from differential geometry that under an assumption of vanishing torsion (which we always make anyways in GR) and if the vector fields \(u^{\mu }\) and \(\xi^{\mu }\) commute (which they do) the following holds:
%
\begin{align} \label{eq:lie-bracket-commutation}
  \xi^{\mu } \nabla_{\mu } u^{\nu } = u^{\mu } \nabla_{\mu } \xi^{\nu }
\,.
\end{align}

So, we want to see by how much the two initially-close geodesics diverge: in order to do this, we compute the second derivative of \(\xi^{\mu }\) with respect to proper time: recall that the derivative with respect to proper time is also the Lie derivative along the 4-velocity: \(\dv*{}{\tau } = u^{\mu } \nabla_{\mu }\).

So, this second derivative looks like
%
\begin{align}
  \dv[2]{\xi^{ \mu } }{\tau } = u^{\alpha } \nabla_{\alpha } \qty(u^{\beta } \nabla_{\beta } \xi^{\mu })
  = u^{\alpha } \nabla_{\alpha } \qty( \xi^{\beta }\nabla_{\beta } u^{\mu } )
\,
\end{align}
%
by the formula shown above. Now, we can apply the Leibniz rule: we find
%
\begin{align}
  \dv[2]{\xi^{\mu }}{\tau } = u^{\alpha } \qty( \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  + \xi^{\beta } \nabla_{\alpha } \nabla_{\beta } u^{\mu } )
\,,
\end{align}
%
and using the fact that\footnote{A trick for remembering this: it would be intuitive to just write the lower indices as \(\alpha \beta \gamma \), but the last two are antisymmetric and must correspond to the antisymmetrization of the covariant derivatives in the LHS.} \([\nabla_{\alpha }, \nabla_{\beta }] V^{\mu } = R^{\mu }_{ \gamma \alpha \beta  } V^{\gamma }\) we can commute the covariant derivatives by inserting a Riemann tensor: we get
%
\begin{align}
  \dv[2]{\xi^{\mu }}{\tau } = u^{\alpha } \qty( \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  + \xi^{\beta } \nabla_{\beta }\nabla_{\alpha } u^{\mu }
  + \xi^{\beta } R^{\mu }_{ \gamma \alpha \beta  } u^{\gamma } )
\,,
\end{align}
%
and now we have gotten the term which will remain in the end: let us bring it to the front, then we will show that the rest of the expression is null. We get
%
\begin{align}
  \dv[2]{\xi^{\mu }}{\tau } =
  R^{\mu }_{ \gamma \alpha \beta  } u^{\gamma } u^{\alpha } \xi^{\beta }
  +
  u^{\alpha } \qty( \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  + \xi^{\beta } \nabla_{\beta }\nabla_{\alpha } u^{\mu } )
\,.
\end{align}
%

To show that the rest of the expression is zero, the thing to remember is that we want to use the fact that our curve is a geodesic, which is expressed as
%
\begin{align}
  u^{\alpha } \nabla_{\alpha } u^{\mu } = \dv{}{\tau } u^{\mu } = 0
\,,
\end{align}
%
so we apply the Leibniz rule backward, to find:
%
\begin{subequations}
\begin{align}
  & u^{\alpha } \qty( \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  + \xi^{\beta } \nabla_{\beta }\nabla_{\alpha } u^{\mu } )= \\
  =& u^{\alpha } \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  + \xi^{\beta } \nabla_{ \beta } \qty( \cancelto{}{u^{\alpha } \nabla_{\alpha } u^{\mu }})
  - \xi^{\beta } \qty(\nabla_{\alpha } u^{\mu } ) (\nabla_{\beta } u^{\alpha })  \\
  =& u^{\alpha } \qty(\nabla_{\alpha } \xi^{\beta }) \qty(\nabla_{\beta } u^{\mu })
  - u^{\beta } \nabla_{\beta }\xi^{\alpha } ( \nabla_{\alpha } u^{\mu }) = 0
\,,
\end{align}
\end{subequations}
%
where in the last step we applied again the commutation relation \eqref{eq:lie-bracket-commutation}, and finally recognised that the terms were equal up to a relabeling of indices.

\end{document}