TITLE: Why are points considered to be zero dimensional?
QUESTION [0 upvotes]: Aside
This is a silly observation born out of my own curiosity. I'd love some perspective from more mathematically minded folks.
Please try to evaluate my reasoning, and expand on which assumptions are incorrect or misguided.
General question
Why are points considered to be zero dimensional? Is there a distinction between zero dimensional and dimensionless that I am missing?
Assumptions

A single point cannot exist in the absence of at least one dimension.
The point has no meaning beyond its distance from zero along that dimension.

Conclusion

A single point is not dimensionless.
Every point is coupled to some dimension(s), and its meaning is generated via its distance to zero along those dimension(s).
In this sense, every point has content, as it assumes a value on a dimension. A dimensionless point would have no capacity to assume a value.

REPLY [1 votes]: Here's just one viewpoint.
You may be interested in the difference between affine dimension and linear dimension. For flat things:

The affine dimension is the least number of points needed to specify the thing.
The linear dimension is the number of orthogonal directions available to someone living on the thing.

For example, since there's a unique line through any two points, ergo a line has affine dimension $2$. But if you're living on a line, there's only one direction along which you can move back and forth, so it has linear dimension $1$.
In general, the linear dimension is always $1$ less than the affine dimension.
Since a point takes $1$ point to specify, its affine dimension is $1$. Thus its linear dimension is $0$. Since "dimension" usually means "linear dimension", hence a point can be considered $0$-dimensional.


TITLE: When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C.
QUESTION [0 upvotes]: When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.)
(a) What is the temperature of the drink after 40 minutes?
°C
(b) When will its temperature be 15°C?
min
My work is here: I not sure I am doing it right.

REPLY [2 votes]: Notice, we have $$T=Ce^{kt}+T_s$$
Setting $T=T_0$ (initial temperature of cold drink) at $t=0$ then we have $$T_0=Ce^{0}+T_s\iff C=T_0-T_s$$
Now, we have $$T=(T_0-T_s)e^{kt}+T_s$$ $$e^{kt}=\frac{T-T_s}{T_0-T_s}$$
$$t=\frac{1}{k}\ln\left(\frac{T-T_s}{T_0-T_s}\right)\tag 1$$
Given condition: Temperature rise of cold drink from $\color{red}{T_0=5^\circ \ C\to T=10^\circ \ C}$ in time $t=25\ minutes$ & $\color{red}{T_s=20^\circ \ C}$ substituting values in (1) we get $$25=\frac{1}{k}\ln\left(\frac{10-20}{5-20}\right)$$
$$k=\frac{1}{25}\ln\left(\frac{2}{3}\right)$$
a) Temperature $\color{red}{T}$ after $t=40 \ minutes$, setting corresponding values in (1), we get
$$40=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{T-20}{5-20}\right)$$ $$\ln\left(\frac{20-T}{15}\right)=\frac{40}{25}\ln\left(\frac{2}{3}\right)$$ $$\frac{20-T}{15}=\left(\frac{2}{3}\right)^{8/5}$$
$$\color{red}{T}=20-15\left(\frac{2}{3}\right)^{8/5}\approx \color{blue}{12.15^\circ\ C}$$
b) Time $\color{red}{t}$ when the temperature is $\color{red}{T=15^\circ C}$, setting corresponding values in (1), we get
$$t=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{15-20}{5-20}\right)$$
$$=\frac{25 \ln\left(\frac{1}{3}\right)}{\ln\left(\frac{2}{3}\right)}$$
$$\color{red}{t}\approx \color{blue}{67.74\ minutes}$$


TITLE: How to solve linearization to approximate $\sqrt[3]{61}$. to the nearest ten thousandth
QUESTION [0 upvotes]: I'm not sure what the nearest ten-thousandth place means in the question above.
How can I solve this problem step by step?

REPLY [4 votes]: To use linearization you need to identify an appropriate function and a point to linearize that function.
The function $f(x) = x^{1/3}$ seems appropriate, and you can evaluate $f(64)$ directly. The linearization of $f$ at $a = 64$ is given by
$$y = f(64) + f'(64)(x-64).$$
Since $f'(x) = \dfrac 13 x^{-2/3}$ you have $f(64) = 4$ and $f'(64) = \dfrac 1{3 \cdot 16}$ so that the approximation with $x=61$ is
$$y = 4 + \dfrac 1{3 \cdot 16} (61-64) = 4 - \frac 1{16} = 3.9375.$$
The actual value is pointed out in the comments, so the linearization doesn't give you enough decimal places. Since you are in Calc 2 you have probably seen Taylor's theorem, so instead you should try the second order approximation
$$y = f(64) + f'(64)(x-64) + \frac{f''(64)}{2}(x-64)^2$$ and see what that yields.


TITLE: Why can you not use equation of motion $v^2 - u^2 = 2as$ when attempting escape velocity questions?
QUESTION [1 upvotes]: There is the this question:
Imagine you are projecting a ball with $3/4$th the escape velocity from the surface of the earth. What is the farthest distance will it reach it from the centre of earth? ($R=$ Radius of the earth)
The answer given in my workbook is $16R/7$
I could solve this question using the conservation of energy theorem. Why can I not use the third equation of motion
$$v^2 - u^2 = 2as~?$$
$s=$ distance from the surface of the earth
I tried solving through the equation of motion method but my answer is widely different.

REPLY [6 votes]: That equation is for constant acceleration motions. The value of $g$ decreases with altitude according to $g=\frac{GM}{r^2}$. In this problem, the decrease in $g$ is significant, as you could expect the projectile to reach a great height.


TITLE: Newtons Cooling Law estimate constant
QUESTION [2 upvotes]: Newton's law of cooling states that $\frac{dT}{dt}= -k(T-T_s)$ if $T(0)=100$, $T(3)=75$ and $T_s=25$ then how would you use linear approximation to estimate the value of k?
Can you use the Mean Value Theorem which tells you that at a point c $\in  (0,3)$ the slope of the tangent is $\frac{-25}{3}$ ?

REPLY [0 votes]: Here's one way to answer the question - it depends on a little bit of not quite common knowledge and may not count as linear approximation.
You know the "$1/3$ life" of the cooling process is $3$ seconds. That's how long it takes to get $1/3$ of the way from $100$ to $25$.
That means
$$
e^{3k} = 1/3
$$
so
$$
3k = -\ln 3 \approx -1.1
$$
and
$$
k \approx -1.1/3 \approx -0.37.
$$
The not quite common knowledge is the value of
$$ \ln 3  = 1.0986\ldots \approx 1.1 .
$$
(More people know $\ln 2 \approx 0.7$, which leads to the "rule of $70$" for doubling time when compounding interest. This approximation tells you the "rule of $110$" for tripling.)


text:
\section{Ideal polyhedra and approximation of laminations.}

In this section, we show how to approximate a pair of measured laminations that strongly fill in $\HH^2$ by a sequence of pairs of ``polyhedral'' laminations, that is, measured laminations on ideal polygons with support contained in a union of disjoint diagonals. Those polyhedral laminations will be constructed so that they satisfy a result on the dihedral of ideal polyhedra in $AdS^3$, so as to provide a sequence of polyhedra with dihedral angles converging, in a proper sense, to the pair of laminations. We will then show that this sequence of ideal polyhedra converges to the convex hull of an acausal curve, as needed for the proof of Theorem \ref{tm:bending}

\subsection{Ideal polyhedra with prescribed dihedral angles}

We first define what we mean by a ``polyhedral'' measured lamination on $\HH^2$.

\begin{defi}
  A measured lamination on $\HH^2$ is {\em polyhedral} if its support is the disjoint union of a finite set of complete geodesics.
\end{defi}

This notion is related to that of {\em ideal polyhedron} in $AdS^3$. It is quite simple to define ideal polyhedra in the projective model of $AdS^3$, see Section \ref{ssc:AdS}: it is a polyhedron with all vertices on the boundary $\partial AdS^3$, but such that the complement of the vertices is contained in $AdS^3$. It follows from this definition that all edges and all faces must be space-like, see \cite{idealpolyhedra}. (There is also a similar notion of hyperideal polyhedron, see \cite{hyperideal}).

\begin{defi}
  Let $P$ be an ideal polyhedron in $AdS^3$, and let $P_-$ and $P_+$ be its past and future boundary components.
  Each of those two pleated ideal polygons defines a polyhedral measured lamination. We denote these two laminations by $\lambda_-(P)$ and $\lambda_+(P)$ respectively.
  We denote by $V(P)$ the $n$-tuple of vertices of $P$, cyclically oriented in the order in which they occur on any acausal meridian containing them.
\end{defi}

By construction, $\lambda_\pm(P)$ have a support which is a disjoint union of diagonals, that is, complete geodesics connecting two vertices.

\begin{theorem}\label{thm:approx}
Let $\lambda_-$, $\lambda_+\in \cML$ be a pair of laminations that strongly fill.
There exists a sequence $(P_n)_{n\in \N}$ of ideal polyhedra and a sequence $u_n$ of functions from $V(P_n)$ to $\RP^1$ such that $((u_{n})_*(\lambda_-(P_n)))_{n\in \N}$ and $((u_{n})_*(\lambda_+(P_n)))_{n\in \N}$ converge to $\lambda_-$ and $\lambda_+$, respectively.
\end{theorem}

The convergence considered here is in the sense of weak-$*$ convergence on measures in $\RP^1\times \RP^1\setminus \Delta$, see section \ref{ssc:laminations}.

In the proof of Theorem \ref{thm:approx}, the main tool to exhibit ideal polyhedra is \cite[Theorem 1.4]{idealpolyhedra} which we recall below.

Let $\Gamma$ be a weighted graph, embedded in the 2-dimensional sphere. For an edge $e$ of this graph, we denote by $\theta(e)$ its weight. To $\Gamma$, we associate the dual graph $\Gamma^*$. Given an edge $e$ of $\Gamma$, we denote by $e^*$ the dual edge.
We say that a polyhedron $P(\Gamma)$ in $AdS^3$ is the realization of $\Gamma$ if there is a bijection between the edges of $\Gamma$ and the edges of $P(\Gamma)$ taking the weight of $e\in E(\Gamma)$ to the exterior dihedral angle of the corresponding edge in $P(\Gamma)$. \cite[Theorem 1.4]{idealpolyhedra} states that $\Gamma$ can be realized if and only if the following criteria are fulfilled.
\begin{enumerate}
\item The graph $\Gamma$ has a Hamiltonian path.
\item For any edge $e$ of the Hamiltonian path, we have $\theta(e)<0$, and $\theta(e)>0$ otherwise.
\item If $e^*_1,\cdots,e^*_k$ bound a face of $\Gamma^*$, then
\[\sum_{i=1}^k \theta(e^*_i)=0.\]
\item If $e^*_1,\cdots,e^*_k$ is a simple continuous path in $\Gamma^*$, which does not bound a face of $\Gamma^*$, then
\[\sum_{i=1}^k \theta(e^*_i)>0.\]
\end{enumerate}
Moreover the polyhedron $P(\Gamma)$ realizing $\Gamma$ enjoys the following property: the vertices can all be connected by a polygonal curve composed of the edges separating two faces for which the exterior normals have opposite time orientation. Those edges are the ones coming from the Hamiltonian path.
We refer to this polygonal curve as the {\em equator} of $P(\Gamma)$.

In light of this result, our goal becomes to construct a weighted graph satisfying properties (1) to (4) above.

\subsection{Approximation of a pair of laminations}

Let us start by an elementary construction.

We fix $n\in\mathbb{N}$, and we consider the disk $D(o,n)$ of radius $n$ centered at a fixed point $o\in\mathbb{H}^2$ . We remove from $\lambda_-$ and $\lambda_+$ the leaves which does not intersect $D(o,n)$. We denote those truncated laminations by $\lambda_-^n$ and $\lambda_+^n$; our aim is to approximate them by polyhedral laminations. Making an arbitrarily small perturbation of $\lambda_-$ and $\lambda_+$, we assume that those truncated laminations are \textit{rational}: they consist in disjoint union of isolated leaves and the measure is a finite sum of Dirac masses. Moreover, we can assume (and will use below) that the weight of each leaf is also a rational number. The total weight of those truncated lamination is finite, it equals the sum of the intersections of the circle $C(o,n)$ with the laminations $\lambda_+$ and $\lambda_-$. We denote this total weight by $\Lambda_n$.

We now arrange $k$ points on the circle $\mathbb{S}^1=\partial_\infty\mathbb{H}^2$ according to the following procedure. The number $k$ will be chosen later as a function of $n$. We use the Poincaré disk model of $\mathbb{H}^2$. We first set $a_1=i$.

Assuming that $a_j$ has been constructed, we construct $a_{j+1}$ in such a way that the leaves of $\lambda_-^n$ and $\lambda_+^n$ limiting in the interval $\left(a_ja_{j+1}\right)$ have a total weight at most $\frac{\Lambda_n}{k}$ and that the leaves of $\lambda_-^n$ and $\lambda_+^n$ limiting in the interval $\left(a_ja_{j+1}\right]$ have a total weight at least $\frac{\Lambda_n}{k}$.
The vertex $a_{j+1}$ is then split in two vertices, one still named $a_{j+1}$ and another one say $a'_{j+1}$ which is very close to $a_{j+1}$ in the counterclockwise direction, and any weight of leaves ending at $a_{j+1}$ in excess of $\frac{\Lambda_n}{k}$ is moved to $a'_{j+1}$. We refer to this process as leaf-splitting.

The endproduct of this construction is $k$-tuple of points $a_1, \cdots, a_k$, arising in this cyclic order, such that the total weight of the leaves ending in $(a_j,a_{j+1}]$ is always exactly $\Lambda_n/k$.

We now consider the following graph $\Gamma_0$:
\begin{itemize}
\item $\Gamma_0$ has a vertex $v_j$ for each of the $k$ points $a_j$.
\item The edges between $v_j$ and $v_{j'}$ correspond to the leaves of $\lambda_-^n$ and $\lambda_+^n$ from the interval $\left(a_ja_{j+1}\right]$ to $\left(a_{j'}a_{j'+1}\right]$, if there is any such leaf.
\item The adjacent vertices $v_j$ and $v_{j+1}$ are related by an edge of weight $-\frac{\Lambda_n}{2k}$.
\end{itemize}
Roughly speaking the graph $\Gamma_0$ is the union of $\lambda^n_-$ and $\lambda^n_+$ to which we pulled back the endpoints of the leaves in the interval $\left[a_{j}a_{j+1}\right)$ to the single vertex $v_j$.
Remark that, since the laminations $\lambda^n_-$ and $\lambda^n_+$ are assumed to be rational, we can arrange that each of the points $a_j$ and hence $v_j$ is the endpoint of some leaf in $\lambda^n_-$ or $\lambda^n_+$.

\begin{lemma}\label{lem:laminations_graph}
There is an infinite number of integers $k$ so that the graph $\Gamma_0$ is exactly the union of $\lambda^n_-$ and $\lambda^n_+$, maybe with leaves splitted, to which we have added the equatorial edges.
\end{lemma}

\begin{proof}
Recall that we have assumed that the lamination $\lambda^n_-$ and $\lambda_n^+$ are rational. We then denote the weights of the leaves by $\frac{p_1}{q_1},\cdots,\frac{p_m}{q_m}$. We now choose $k$ to be an integer multiple of $\frac{q_1 q_2\cdots q_m }{\Lambda_n}$, say $k=\alpha\frac{q_1\cdots q_m}{\Lambda_n}$. It follows that the weight of any leaf is a integer multiple of $\frac{\Lambda_n}{k}$ (indeed $\frac{p_i}{q_i}=\alpha p_i\cdot q_1\cdots \hat{q_i}\cdots q_m\frac{\lambda_n}{k}$). The leaf-splitting procedure gives rise to $2\alpha p_i\cdot q_1\cdots \hat{q_i}\cdots q_m$ vertices in $\Gamma_0$, close to the endpoints of the leaf of weight $\frac{p_i}{q_i}$ and edges, all of which having the weight $\frac{\Lambda_n}{k}$ between the vertices, and close to the original leaf.

\end{proof}

The graph $\Gamma_0$ does not yet encode the 1-skeleton and dihedral angle of the polyhedron we are looking for. We need to construct a slight modification of $\Gamma_0$ to which we can apply \cite[Theorem 1.4]{idealpolyhedra}. Nevertheless, $\Gamma_0$ is the starting point of our construction: it obviously satisfies properties (1), (2) and (3). Throughought the modifications of $\Gamma_0$, it will be easy to see that those three properties persist.

The ``slight modification'' will be understood in the following sense.

\begin{defi}
  We consider the set of weighted graphs $\Gamma=(V;E;\theta)$ in $\mathbb{H}^2$ such that:
  \begin{itemize}
  \item the vertices are on $\partial_\infty\mathbb{H}^2$,
  \item the set of edges is bipartite $E=E_+\cup E_-$,
  \item  both $(E_+,\theta_{|E_+})$ and $(E_-,\theta_{|E_-})$ are measured geodesic laminations.
  \end{itemize}
  We endow the set of such graphs with the topology coming from the topology on the space of measured laminations.
\end{defi}

We denote by $\lambda^{k,n}_-$ and $\lambda^{k,n}_+$ the laminations associated to $\Gamma_0$.

We can now modify $\Gamma_0$ to satisfy the requirements of \cite[Theorem 1.4]{idealpolyhedra} through the following three steps.

\vspace{0.5cm}

\paragraph{{\bf 1st step: coloring the vertices.}}

The first part of the modification for the weighted graph $\Gamma$ is to notice that its vertices
can be separated in two groups, depending on whether they are endpoints of an edge in $E_+$ of in $E_-$.

\begin{lemma}\label{lem:endpoints}
For any $n\in\mathbb{N}$, there is a number $\delta(n)>0$ with the following property. Let $a_+\in\mathbb{S}^1$ be the endpoint of a geodesic $F_+\in\lambda^n_+$ and $a_-\in\mathbb{S}^1$ be the endpoint of a geodesic $F_-\in\lambda^n_-$. Then
\[d(a_-,a_+)>\delta(n)\]
(where $d$ denotes the visual distance on $\partial_\infty \HH^3$ associated to the center $o$).
\end{lemma}

The proof will use the following simple lemma.

\begin{lemma} \label{lem:parallel}
  Let $\lambda_-,\lambda_+$ be two bounded measured geodesic lamination on the hyperbolic plane that strongly fill. There exists $\varepsilon>0$ and $L>0$ such that if $\gamma_+$ and $\gamma_-$ are geodesic segments in the support of $\lambda_+$ and $\lambda_-$, respectively, of length at least $L$, then one of the endpoints of $\gamma_+$ is at distance at least $\varepsilon$ from $\gamma_-$.
\end{lemma}

\begin{proof}
  For clarity we consider $\gamma_-$ and $\gamma_+$ as oriented, and call $\gamma_{\pm,0}$ and $\gamma_{\pm,1}$ their endpoints. We assume by contradiction that $d(\gamma_{-,0},\gamma_{+,0})\leq \varepsilon$ and that $d(\gamma_{-,1},\gamma_{+,1})\leq \varepsilon$.

  Any leaf of $\lambda_+$ intersecting $\gamma_-$ must intersect either $[\gamma_{-,0},\gamma_{+,0}]$ or $[\gamma_{-,1},\gamma_{+,1}]$. However the strict filling hypothesis imply that the intersection of $\gamma_-$ with $\lambda_+$ is bounded from below by a constant which goes to infinity with $L$. So the sum of the intersections with $\lambda_+$ of $[\gamma_{-,0},\gamma_{+,0}]$ and of $[\gamma_{-,1},\gamma_{+,1}]$ is bounded from below by a constant going to infinity as $L\to \infty$. This however contradicts the fact that $\lambda_+$ is bounded, if $L$ is large and $\varepsilon$ is small.
\end{proof}

\begin{cor} \label{cr:parallel}
  For each $n>0$, there exists $\varepsilon_n>0$ as follows. Let $F_-$ and $F_+$ be leaves of $\lambda_-$ and $\lambda_+$, respectively, with endpoints $a_-$ and $a_+$ on $\partial \HH^2$, and let $b_-, b_+$ be the first intersection of $F_-, F_+$ with $C(o,n)$, when starting from $a_-, a_+$. Then either the visual metric from $o$ between $a_-$ and $a_+$ is at least $\varepsilon_n$, or the visual metric from $o$ between $b_-$ and $b_+$ is at least $\varepsilon_n$.
\end{cor}

\begin{proof}
  This clearly follows from Lemma \ref{lem:parallel}, since otherwise, for $\varepsilon_n$ small enough, the segments of $F_-$ and $F_+$ of length $L$ starting from their first intersections with $C(o,n)$ towards the outside of $D(o,n)$ would contradict that lemma.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:endpoints}]
  We consider $a_-=F_-(+\infty)$ and $a_+=F_+(+\infty)$ as in the statement of Lemma \ref{lem:endpoints}, at distance less than $\varepsilon_n$. It follows by Corollary \ref{cr:parallel} that the points $b_-$ and $b_+$ which are the first intersections of $F_-$ and $F_+$ with $C(o,n)$ are at distance at least $\varepsilon_n$. In fact, with the same type of argument, we can consider a rescaled visual metric on $C(0,n)$ and we may as well assume that the points $b_-$ and $b_+$ are at distance at least $C\varepsilon$ for a constant $C>1$ which will be adjusted later.

  We consider two cases.

  First case: the segments of $F_-$ and $F_+$ between $a_-$ (resp. $a_+$) and $C(o,n)$ intersect at a point $p$. Since $d_o(b_-,b_+)\geq Cd_o(a_-, a_+)$, the hyperbolic distance between $D(o,n)$ and $p$ is bounded from below by a function $F(n)$ of $n$ which is bounded away from 0 by a constant depending on $C$. Let $\gamma$ be a geodesic segment going from $p$ to a point of $C(o,n)$, staying within the triangle boundedy by $C(o,n)$ and by the segments of $F_-$ and $F_+$ between $p$ and $C(o,n)$. $\gamma$ has length at least $F(n)$ and therefore, as $n\to \infty$ and by choosing $C$ big enough, must have large intersection with either $\lambda_-$ or $\lambda_+$ (because $\lambda_-$ and $\lambda_+$ strongly fill). But any leave of $\lambda_-$ or $\lambda_+$ intersecting $\gamma$ must intersect $C(o,n)$, a contradiction. See figure \ref{fig:notation1}.

\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}
\definecolor{ffqqqq}{rgb}{1,0,0}
\begin{figure}[!h]
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm]
\clip(-4.3,-3.78) rectangle (8.84,6.3);
\draw [shift={(1.5973181254381705,8.1258075823761)},line width=0.5pt]  plot[domain=4.029956078586286:5.384183684640062,variable=\t]({1*9.001858067944205*cos(\t r)+0*9.001858067944205*sin(\t r)},{0*9.001858067944205*cos(\t r)+1*9.001858067944205*sin(\t r)});
\draw [shift={(1.6503294117647058,7.604176470588235)},line width=0.8pt]  plot[domain=3.982907677116938:5.554881166812981,variable=\t]({1*7.007424392046069*cos(\t r)+0*7.007424392046069*sin(\t r)},{0*7.007424392046069*cos(\t r)+1*7.007424392046069*sin(\t r)});
\draw [line width=3.6pt,color=ffqqqq] (1.08,0.62)-- (1.06,-0.86);
\draw (-2.92,-0.04) node[anchor=north west] {$ \partial_\infty\mathbb{H}^2 $};
\draw (-3.9,2.4) node[anchor=north west] {$ B(o,n) $};
\draw (-3.9,3.4) node[anchor=north west] {$ F_- $};
\draw (3.5,2.6) node[anchor=north west] {$ F_+ $};
\draw (0.70,0.78) node[anchor=north west] {$ \gamma $};
\draw (0.36,-0.86) node[anchor=north west] {$ a_+ $};
\draw (1.5,-0.94) node[anchor=north west] {$ a_- $};
\draw [shift={(-5.593748101465419,-4.05670102767756)},line width=2pt]  plot[domain=0.4225681039466382:1.3427547327428608,variable=\t]({1*7.757535873271658*cos(\t r)+0*7.757535873271658*sin(\t r)},{0*7.757535873271658*cos(\t r)+1*7.757535873271658*sin(\t r)});
\draw [shift={(7.267323507617761,-3.444333807712094)},line width=2pt]  plot[domain=1.7059915715599123:2.761271550453879,variable=\t]({1*7.028468340494026*cos(\t r)+0*7.028468340494026*sin(\t r)},{0*7.028468340494026*cos(\t r)+1*7.028468340494026*sin(\t r)});
\draw (0.12,0.84) node[anchor=north west] {$b_-$};
\draw (1.54,0.78) node[anchor=north west] {$b_+$};
\begin{scriptsize}
\draw [fill=uuuuuu] (1.082344930721114,-0.10586002218428667) circle (2pt);
\draw[color=uuuuuu] (1.3,0.00) node {$p$};
\end{scriptsize}
\end{tikzpicture}
\caption{Lemma \ref{lem:endpoints} Case 1}
\label{fig:notation1}
\end{figure}

  Second case: those segments do not intersect. We then apply exactly the same argument, but with $p$ replaced by the intersection between the segments $[a_-,b_+]$ and $[a_+,b_-]$, see figure \ref{fig:notation2}.
\end{proof}

\definecolor{xdxdff}{rgb}{0.49,0.49,1}
\definecolor{ffqqqq}{rgb}{1,0,0}
\begin{figure}[!h]
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\clip(-4.3,-2.7) rectangle (7.4,6.3);
\draw [shift={(1.59,7.38)}] plot[domain=3.97:5.44,variable=\t]({1*8.43*cos(\t r)+0*8.43*sin(\t r)},{0*8.43*cos(\t r)+1*8.43*sin(\t r)});
\draw [shift={(1.76,7.43)}] plot[domain=3.92:5.62,variable=\t]({1*6.62*cos(\t r)+0*6.62*sin(\t r)},{0*6.62*cos(\t r)+1*6.62*sin(\t r)});
\draw [shift={(-11.09,-1.68)}] plot[domain=0.06:0.71,variable=\t]({1*11.72*cos(\t r)+0*11.72*sin(\t r)},{0*11.72*cos(\t r)+1*11.72*sin(\t r)});
\draw [shift={(15.77,-2.67)}] plot[domain=2.49:3.03,variable=\t]({1*14.16*cos(\t r)+0*14.16*sin(\t r)},{0*14.16*cos(\t r)+1*14.16*sin(\t r)});
\draw [line width=3.6pt,color=ffqqqq] (1.22,0.84)-- (1.12,-1.04);
\draw [shift={(5.91,-3.71)}] plot[domain=2.28:2.67,variable=\t]({1*5.96*cos(\t r)+0*5.96*sin(\t r)},{0*5.96*cos(\t r)+1*5.96*sin(\t r)});
\draw [shift={(-5.23,-4.28)}] plot[domain=0.44:0.76,variable=\t]({1*7.65*cos(\t r)+0*7.65*sin(\t r)},{0*7.65*cos(\t r)+1*7.65*sin(\t r)});
\draw (-2.92,-0.1) node[anchor=north west] {$ \partial_\infty\mathbb{H}^2 $};
\draw (-2.6,2.94) node[anchor=north west] {$ D(o,n) $};
\draw (-1.18,5.2) node[anchor=north west] {$ F_+ $};
\draw (2.92,4.98) node[anchor=north west] {$ F_- $};
\draw (0.82,0.8) node[anchor=north west] {$ \gamma $};
\draw (0.7,-1.3) node {$a_+$};
\draw (1.8,-1.3) node {$a_-$};

\draw (1.4,-0.1) node {$a$};
\end{tikzpicture}
\caption{Lemma \ref{lem:endpoints} Case 2}
\label{fig:notation2}
\end{figure}

By construction, all vertices of $\lambda^{k,n}_+$ (resp. $\lambda^{k,n}_-$) are arbitrarily close, for $k$ large enough, from the endpoint of a leave of $\lambda^n_+$ (resp. $\lambda^n_-$). It follows that for $k$ large enough, the statement of Lemma \ref{lem:endpoints} also applies (with a slightly different function $\delta$) to the endpoints of $\lambda^{k,n}_+$ and $\lambda^{k,n}_-$.

Since any vertex $v_j$ of $\Gamma_0$ is the endpoint of a leaf in either $\lambda^{k,n}_-$ or $\lambda^{k,n}_+$, but not both according to the previous argument, it follows that each vertex has a well defined type ($+$ or $-$) depending on which type of leaf it terminates.

\vspace{0.5cm}

\paragraph{{\bf 2nd step: A graph $\Gamma_1$ which satisfies the weaker condition (4')}}

Property (4) reduces
to the following statement: let $\gamma$ be a geodesic path in $\mathbb{H}^2$ connecting the interval $\left(v_1v_2\right)$ to $\left(v_{p}v_{p+1}\right)$. Then the total weight of the edges of $\lambda_-^n$ and $\lambda_+^n$ it crosses is bigger than $\frac{\Lambda_n}{k}$. (The reduction comes from the fact that any closed path in $\Gamma^*$ intersecting the equator in two points can be ``straightened'' into the double cover of a geodesic. The weight of the intersection of this closed path with the equator is $-\frac{\Lambda_n}{k}$, so condition (4) is satisfied if and only if the sum of the intersections with all the geodesics in $\lambda_-^n$ and $\lambda_+^n$ is larger than $\frac{\Lambda_n}{k}$.)

The first modification of $\Gamma$ is done according to the following lemma

\begin{lemma}\label{lem:4'}
There exists a weighted graph $\Gamma_1$ arbitrarily close to $\Gamma_0$ which satisfies properties (1), (2), (3) and the following weakening of property (4):
\newline
(4') Let $\gamma$ be a geodesic path in $\mathbb{H}^2$. Then
\[i(\gamma,\lambda^{k,n}_-)+i(\gamma,\lambda^{k,n}_+)>0.\]
\end{lemma}

\begin{proof}
Note that two cases are easily understood: if $\gamma$ does not enter the disk $D(o,n)$ and joins the two intervals $(v_p,v_{p+1})$ to $(v_{p',}v_{p'+1})$, then $\gamma$ must cross every edge emanating between $v_{p+1}$ and $v_{p'}$, with total weight $|p'-p|\Lambda_n/k$. On the opposite, if the length of the intersection of $\gamma$ with the disk $D(o,n)$ is large enough, then the conclusion follows from the strong filling hypothesis. Nevertheless, it may happen that, for some $p$ and $p'$, a geodesic joining $(v_p,v_{p+1})$ to $(v_{p',}v_{p'+1})$ does not cross either $\gamma^{k,n}_-$ nor $\gamma^{k,n}_+$. In this case, we wish to add an edge between $v_p$ and $v_{p'}$ of small weight. We can indeed choose a lamination to which we add an edge since $\gamma$ cannot be close to both $\lambda^{k,n}_-$ and $\lambda_+^{k,n}$ (Lemma \ref{lem:endpoints}).
\end{proof}

We still denote by $\lambda^{k,n}_-$ and $\lambda^{k,n}_+$ the geodesic laminations associated to the modified graph $\Gamma_1$.
\vspace{0.5cm}

\paragraph{{\bf 3rd step: From (4') to (4)}} The last step is to prove that property (4') in fact implies property (4) for an arbitrarily close weighted graph.

\begin{lemma}\label{lem:splitting}
Let $\Gamma_1$ be the graph constructed in step 2. We can find a graph $\Gamma$ arbitrarily close to $\Gamma_1$ satisfying properties (1) to (4).
\end{lemma}

\begin{proof}
For any pair of vertices $p$ and $p'$, we know that the geodesic joining $(v_p,v_{p+1})$ to $(v_{p',}v_{p'+1})$ has positive mass. Denote my $m=m(k)$ the smallest intersection:
\[m=\inf_{p,p'}\left\lbrace i(\gamma,\lambda^{k,n}_-)+i(\gamma,\lambda^{k,n}_+)\right\rbrace\]
where, for each $p,p'$, $\gamma$ is a geodesic joining $(v_p,v_{p+1})$ to $(v_{p',}v_{p'+1})$. (Note that the intersection number does not depend on which geodesic is chosen with endpoint in those intervals.)

We now split the set of vertices according to the following procedure. Any vertex $v_p$ is replaced by the set $v_p^{i_1},\cdots,v_p^{i_p}$ on $\mathbb{S}^1$ so that $v_p^{i_1},\cdots,v_p^{i_p}$ are cyclically ordered and arbitrarily close. We replace the equatorial edges by edges, for any $j$, from $v_p^{i_j}$ to $v_p^{i_{j+1}}$ of equal weight. We choose $i_p$ so that the weight of any edge $(v_p^{i_j},v_p^{i_{j+1}})$ is less that $m/2$.

The resulting graph satisfies condition (4) and is arbitrarily close to $\Gamma_1$ since the $v_{i_j}$ are arbitrarily close to the former $v_p$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:approx}]
For each integers $n$, we consider the graph $\Gamma_0$, with $k$ large enough so that $\Gamma_0$ satisfies the hypothesis of Lemma \ref{lem:laminations_graph}. We modify it once with the process described in Lemma \ref{lem:4'}, adding leaves of weights $\frac{1}{n}$ so that we get a graph satisfying properties (1) -- (4'). Then we add vertices as described in lemma \ref{lem:splitting}. The resulting graph satisfies properties (1) -- (4) and has a number of vertices $K=K(n)$. We can now apply \cite[Theorem 1.4]{idealpolyhedra} for each integer $n\in\mathbb{N}$ we get an ideal polyhedron $P_n$ in $AdS^3$ with $K$ vertices.

The fact that $\lambda_+(P_n)$ converge to $\lambda_+$ and $\lambda_-(P_n)$ converge to $\lambda_-$ as $n\to \infty$ follows directly from the definition of the topology on $\cML$ and Lemma \ref{lem:laminations_graph}.
\end{proof}

We now associate to those truncated laminations, a discrete approximation of the quasi-circle we are looking for. Indeed, let $P_n$ be the ideal polyhedra we constructed in the previous theorem. Its vertices form a subset $V_n\subset \partial AdS^3$, which are naturally ordered since each can be connected to the next by a space-like geodesic. We will see in the next section that those subsets $V_n$ can be considered as a ``discrete homeomorphism from $\RP^1$ to $\RP^1$'', that is, an increasing function from a finite subset of $\RP^1$ to $\RP^1$, and that it is ``uniformly quasi-symmetric'' in a natural sense.

config: arxiv


TITLE: What causes pressure in a liquid?
QUESTION [2 upvotes]: Say there is a mass inside a liquid. What causes the force on its upside downwards and the force on its downside upwards? It seems logical that the downward force is due to the weight of the fluid above it, is it similar with the upward force? Just collision of liquid particles?

REPLY [1 votes]: Consider a vessel of any shape.
At any 2 or more points is the vessel that are coplanar,
no matter what the shape of the vessel, the pressure will be equal.
Just collision of liquid particles?

If an object is immersed in the liquid, pressure will act on all directions on it from the liquid's particles.
The net pressure (the pressure acting on both sides of the object cancels each other) acting on the  on an object at a depth $d$ in the liquid is given by $$P=d\rho g+P_0$$ where $P_0$ is the atmospheric pressure acting on the free surface and $\rho$ is the density of the liquid.
is it similar with the upward force?

There will be an upwards pressure from the liquid's particles due to the buoyant force too.


TITLE: Differential of Gauss map
QUESTION [2 upvotes]: Let $S$ we a regular surface, differential of Gauss map is $\mathrm{d} \mathrm{N}_{\mathrm{p}}: \mathrm{T}_{\mathrm{p}}(\mathrm{S}) \rightarrow \mathrm{T}_{\mathrm{p}}(\mathrm{S})$.
To evaluate the differential at some point, we can choose a curve $\alpha(t) = x(u(t),v(t))$ on $S$ as a image of plane curve $(u(t),v(t))$ under the chart map $\textbf{x}$.
So we have :
$$\begin{aligned}
d N_{p}\left(\alpha^{\prime}(0)\right) &=d N_{p}\left(\mathbf{x}_{u} u^{\prime}(0)+\mathbf{x}_{v} v^{\prime}(0)\right) \\
&=\left.\frac{d}{d t} N(u(t), v(t))\right|_{t=0} \\
&=N_{u} u^{\prime}(0)+N_{v} v^{\prime}(0)
\end{aligned}$$

The first equality and the third one is easy, what I don't understand is the second one, we know in general $df_p(v) = \frac{d}{dt}_{t=0}(f\alpha(t))$ so for here shouldn't it be $\left.\frac{d}{d t} N(x(u(t), v(t)))\right|_{t=0}$.

This is the result from Do Carmo's differential geometry textbook page 142-143.

The second question is  the book says in particular $d
   N_{p}\left(\mathbf{x}_{u}\right)=N_{u} \text { and } d
   N_{p}\left(\mathbf{x}_{v}\right)=N_{v}$. I don't know how to get it. My idea is to choose plane curve $(u(t),v(t))$ vary only along the u-axis so $N_u =dN_p(x_u)$ in this case.Is my interpretation right?

REPLY [1 votes]: You are correct in your first bullet point. For instance, for the second equality he could have wrote $=\frac{d}{d t} N\left(t)\right|_{t=0}$  which would be true in the sense that $\textbf{N}$ depends on $t$ (in the case of our curve), but this way of writing it wouldn't have been useful to us. Equally he could have wrote $\frac{d}{d t} \textbf{N}\left(\textbf{x}(u(t), v(t)))\right|_{t=0}$ as you say, (which is the most explicit form as it shows all dependencies of variables).
But  Do Carmo chooses to write it the way he does because we care about how $\textbf{N}$ changes as we change the parameters $u$ and $v$ (i.e. $N_u$ and $N_v$).
Then finally in your second bullet point, the answer follows from the fact that $d N_{p}$ is a linear operator in conjunction with the fact that $d N_{p}\left(\mathbf{x}_{u} u^{\prime}(0)+\mathbf{x}_{v} v^{\prime}(0)\right) = N_{u} u^{\prime}(0)+N_v v^{\prime}(0)$.