TITLE: Probability - a clerk controlling customers
QUESTION [0 upvotes]: It is known that a certain clerk checks on 15 out of 100 customers.
  What is the probability that from a group of 12 customers, exactly 3 will be
  checked on?

I haven't got a faintest idea how to solve this task - could you give me any suggestions? I tried drawing a "tree", but it wasn't too effective.

REPLY [3 votes]: You want to select exactly 3 persons from the lot of 15 people who are being controlled by the clerk and the rest of 9 from the other 85 customers. Thus the probability is
$$\frac{{15 \choose 3}{85 \choose 9}}{{100 \choose 12}}$$
The denominator shows all the possible ways of selecting 12 persons from a group of 100 people.


TITLE: Convert specific humidity to relative humidity but without having all variables available?
QUESTION [0 upvotes]: I was wondering if there is a way to obtain relative humidity from specific. My constraint is that I am not trying to solve a problem by hand. I am downloading data for a specific region, in order to construct a multi model for climate modelling. Unfortunately, I do not have access to any dataset that I want. The following variables have available datasets which I can use:
-specific humidity (Dimensionless, Amount of moisture in the air divided by amount of air plus moisture at that location.)
-maximum and minimum air temperature
-mean daily wind speed at 10m
-vapour pressure (hPa)
-sum of precipitation (mm/day)
-potential evapotranspiration from a crop canopy (mm/day)
-total global radiation (KJ/m2/day)
Is there any equation that I can use to obtain relative humidity?

REPLY [0 votes]: "Vapour pressure" usually means the equilibrium vapour pressure of water, which is the partial pressure of water vapour in air when the air "holds" its maximum amount of water.  In other words, if the partial pressure of water vapour is equal to the equilibrium vapour pressure, then you are at 100% relative humidity.   Note that equilibrium vapour pressure varies rather strongly with temperature.
In a more general case, the relative humidity of an air sample is defined as the ratio of the partial pressure of water vapour in the sample to the equilibrium vapour pressure of water at that temperature:
$$
\text{relative humidity} = \frac{\text{partial pressure of H}_2\text{O}}{\text{equilibrium vapour pressure}}
$$
So if you could figure out the partial pressure of the water vapour, you could then infer the relative humidity.
If you knew the overall atmospheric pressure at a given time, and you knew the fraction of the air (either by mass or by moles) that was water vapour, you could use Dalton's Law to figure out the partial pressure of the water vapour, and thereby infer the relative humidity.  But if you just know the equilibrium vapour pressure of water and the specific humidity, without knowing the total atmospheric pressure, there's no way to get the relative humidity;  a given specific humidity can correspond to different partial pressures of water vapour, depending on the overall atmospheric pressure.

That said, it is conceivable that the data set is providing the partial pressure of water vapor as "vapor pressure".  If this is the case, then you know the numerator of the equation above and you would need to figure out the denominator.  Thankfully, the equilibrium vapour pressure of water is a well-studied quantity.  It does vary strongly with temperature; but so long as you have the temperature at which the measurement was made, then it is straightforward to figure out the equilibrium vapour pressure (either via a lookup table & interpolation, or via one of the approximate formulas linked in the above article) and then to find the relative humidity.


TITLE: Component sequence vs subsequence
QUESTION [2 upvotes]: I'm having a little difficulty wrapping my head around the difference between these two terms, so could someone verify if this is correct? I've struggle to find an answers or reference in a book or online.
Component sequence:
If you have a set $X \subset \mathbb{R}^m$ and a sequence $x \in X^\infty$ such that
$$x = (x_1, x_2, \dots ) \in X^\infty = (\mathbb{R}^m)^\infty$$
Each $x_i$ is itself a vector of $m$-terms,
e.g., $x_1 = (x_{1,1}, x_{1,2}, ... x_{1,m})$, or for another notation: $x_1 = (x_1(1), x_2(1), ... x_m(1))$
i.e., each term of the sequence is itself an m-tuple. Then, a component sequence is defined as the $k-th$ component of each term $x(n)$, and hence, there are $m$ of these component sequences, a sequence for each component.
But what does that actually mean, practically? If I have a sequence:
$$ S = (x_1, x_2, x_3 , ...)$$
where $x_1 = (1,3,5,7,9)$, is one such component sequence of $S$ just $(1,3,5,7,9)$?
As opposed to a subsequence which could be $(x_1, x_3, x_5, ...) $ ?

REPLY [3 votes]: Component sequences are projections of your sequences on the one dimensional axes of your space. Each component sequence is an infinite sequence of picking the sequence of terms in one dimension  of the original sequence .
For example if $$\{(1,4,5),(3,2,4),(5,3,6),...\}$$ is a sequence in $R^3$, then you have three component sequences namely $$\{1,3,5,....\}, \{4,2,3,....\},\{5,4,6,...\}$$


text:
\begin{document}
\title[Intrinsic distances on complete Reinhardt domains]{Comparison of the Carath\'eodory-distance, the K\"ahler-Einstein distance and the Kobayashi-distance on complete Reinhardt domains}

\author{Gunhee Cho}
\address{Department of Mathematics\\
University of Connecticut\\
196 Auditorium Road,
Storrs, CT 06269-3009, USA}

\email{gunhee.cho@uconn.edu}

\begin{abstract}
In this paper, we consider the comparison among natural intrinsic distances on bounded, complete Reinhardt domains. First, we show that on bounded, complete Reinhardt domains $G^n_{\epsilon}$ in $\mathbb{C}^n$, the Carath\'eodory-distance and the distance of the complete K\"ahler-Einstein metric are different. Second, we prove that if the Carath\'eodory-distance and the Kobayashi-distance are different on $G^n_{\epsilon}$, then there exist infinitely many strongly pseudoconvex sub-domains of $G^n_{\epsilon}$ such that those domains admit a sequence of points which makes the Carath\'eodory-distance and the Kobayashi-distance different. In particular, this sequence goes towards the boundary. Third, we prove that the Kobayashi-distance $k_{G^n_{\epsilon}}$ and the distance function $l_{G^n_{\epsilon}}$ are different for sufficiently small $\epsilon>0$. We also investigate the interesting behavior of the Kobayashi-distance $k_{G^n_{\epsilon}}$ when $\epsilon \rightarrow 0$. We use our results to conclude that there are no holomorphic retractions to simply-connected open Riemann surfaces containing two points in our class of examples.
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction and results}

In this paper, we compare intrinsic distances on bounded, complete Reinhardt domains.

For a complex manifold $M$, we say $d$ on $M$ is a pseudo-distance if it serves as a topological distance but there could be points $x\neq y$ with $d(x,y)=0$. Also, we say a pseudo-distance on a complex manifold is intrinsic if the pseudo-distance possesses the distance decreasing property by holomorphic maps. The classical Schwarz-Pick lemma implies the Poincare-distance is intrinsic on the unit-disk $\mathbb{D}$. The followings are examples of intrinsic pseudo-distances on general complex manifolds :

The Carath\'eodory pseudo-distance $c_{M}$ on $M$ is defined by
\[c_{M}(x,y):=\sup_{f\in \Hol(M,\mathbb{D})} \rho_{\mathbb{D}}(f(x),f(y)).\]
Here, $\rho_{\mathbb{D}}(a,b)$ denotes the integrated Poincare distance on the unit disk $\mathbb{D}$.

The Kobayashi pseudo-distance $k_{M}$ on $M$ is defined by
\[k_{M}(x,y):=\inf_{f_i \in \Hol(\mathbb{D},M)}\left\{\sum_{i=1}^{n} \rho_{\mathbb{D}}(a_i,b_i) \right\}\]
where $x=p_0 < ... < p_n =y, f_i(a_i)=p_{i-1}, f_i(b_i)=p_i $.
Note that the Carath\'eodory pseudo-distance $c_{M}$ and the Kobayashi pseudo-distance $k_{M}$ are minimal and maximal among intrinsic pseudo-distances respectively.

Also, we define the distance function $l_M$ by
\[l_{M}(x,y):=\inf\left\{ \rho_{\mathbb{D}}(a,b) | f \in \Hol(\mathbb{D},M),  f(a) =x, f(b)=y\right\}\]
People sometimes call $l_{M}$ the Lempert's function. For general complex manifolds, $l_M$ does not satisfy the triangle inequality and is not even continuous as a function on $M\times M$.

For general complex manifolds, the following relation is true :
\[l_{{M}} \geq k_{M} \geq c_{M}\geq 0. \]

By the classical Lempert's theorem (See \cite{Lem81}), it is well known that for a bounded convex domain $M$ in $\mathbb{C}^n$, the Carath\'eodory-distance $c_{M}$ and the distance function $l_M$ are the same. For general pseudoconvex domains, however, intrinsic pseudo-distances might not be equal. There are some interesting examples of this (See \cite{MJPP90}, \cite{VM18}, \cite{NNPPPJTWZ08}, \cite{J-P}). It is also known that the Carath\'eodory pseudo-distance and the Kobayashi pseudo-distance can be the same for non-convex pseudoconvex domains (See \cite{JANJY04}, \cite{CC04}). In particular, one interesting observation is the following relations among three intrinsic distances; If a bounded weakly pseudoconvex domain $M$ in $\mathbb{C}^n$ admits the canonical complete K\"ahler-Einstein metric which has the holomorphic sectional curvature bounded from above by $-1$, the following relation is true :
\[k_{M} \geq d^{{KE}}_{M} \geq c_{M}>0. \]
Here, we denote $d^{{KE}}_{M}$ be the distance induced by the complete K\"ahler-Einstein metric on $M$. To obtain $d^{{KE}}_{M} \geq c_{M}$, one can apply the Schwarz-Yau lemma (Theorem~\ref{th:SC}), and we provide the proof in the part of the Lemma~\ref{lem:JP}, and to get $k_{M} \geq d^{{KE}}_{M}$, see p34 in \cite{Kob98}.

Our original motivation for this paper was to extend the result in \cite{DY17}. From \cite{DY17}, D.Wu and S.T.Yau showed that on the complete noncompact simply-connected K\"ahler manifold $M$ with negative sectional curvature ranges, the Kobayashi-distance, the distance of K\"ahler-Einstein metric, the distance of Bergman metric, and the distance of given K\"ahler metric are all uniformly equivalent (See Corollary 7 in \cite{DY17}). It remains to study the fourth invariant metric, the Carath\'eodory-metric on $M$. The study of the Carath\'eodory-metric on $M$ is closely related to the long-standing problem about the existence of a bounded non-constant holomorphic function on $M$. To understand the Carath\'eodory-distance, we wanted to get concrete examples which distinguish the Carath\'eodory-distance from other intrinsic distances. We also provide some relevant comments in Section 2 and Section 3. During our investigation of examples, we made contributions to answer for the following open problem (also see \cite{JP13} and \cite{Kob98}) :

For $\epsilon>0$, define $D^2_{\epsilon}:=\left\{ z=(z_1,z_2)\in {\mathbb{D}}^2:\left| z_1 z_2 \right|<\epsilon \right\}$. We found that the distance related information for the diagonal entries of $D^2_{\epsilon}$ is an open problem posed by M. Jarniki and P. Pflug (See \cite{JP13}, the picture in p86 and the problem 2.10 in p105). Only the explicit formula of the Carath\'eodory- distance of sub-regions of $D^2_{\epsilon}$ which are closed to each axis in $D^2_{\epsilon}$ are known, but the information of the diagonal entries of $D^2_{\epsilon}$ is still unknown. In this paper, we distinguish the intrinsic distances for the diagonal entries of $D^2_{\epsilon}$, especially the Carath\'eodory-distance and the distance of the canonical complete K\"ahler-Einstein metric. Also, the approach in this paper can be extended to the higher dimensional pseudoconvex domains.

We got inspiration from J-P Vigue's paper (See \cite{J-P}). Due to his work, it was known that with a clever choice of one point with origin on the diagonal entries on $\left\{ z=(z_1,z_2)\in\mathbb{C}^{2}:\left| z_1 \right|+\left| z_2 \right|<1, \left| z_1 z_2 \right|<\frac{1}{16} \right\}$, he could find the difference between the Carath\'eodory-distance and the distance of the Carath\'eodory-Reiffen metric. By extending this idea, we could distinguish the Carath\'eodory-distance from the distance of the complete K\"ahler-Einstein metric (Theorem~\ref{th:CC} and Theorem~\ref{th:CC2}). In particular, this gives the relevant information for the problem 2.10 of \cite{JP13}.

 Here are the results :
\begin{thm-Intro}\label{th:CC}
	Let $0<r<1$ and $G$ be a bounded pseudoconvex domain in $\mathbb{C}^1$ such that $r\mathbb{D}\subset G \subset \mathbb{D}$. For $n\geq2$, let $0<\epsilon<\min\{\frac{1}{n}r^2,(\frac{r}{n})^{\frac{n}{n-1}}\}$. Define $G^n_{\epsilon}:=\left\{ z=(z_1,...,z_n)\in{G}^{n}:\left| \Pi^{n}_{i=1}z_i \right|<\epsilon \right\}$. Denote the canonical complete K\"ahler-Einstein metric on $G^n_{\epsilon}$ by $\omega_{KE}$, and let $d^{{KE}}_{G^n_{\epsilon}}$ be the distance induced by $\omega_{KE}$. Then for any $x\in G$ which satisfies $(x,...,x)\in G^n_{\epsilon}$ and $\frac{n\epsilon}{r}<|x|$, we have
	\[c_{G^n_{\epsilon}}((0,...,0),(x,...,x)) <  d^{{KE}}_{G^n_{\epsilon}}((0,...,0),(x,...,x)).\]
\end{thm-Intro}
From the Theorem C in \cite{ZF94}, it was known that the product manifold $G^n$ cannot be negatively curved. Thus we consider $G^n_{\epsilon}$ which is not the product manifold $G^n$. We can see that $G^n_{\epsilon}$ is the analytic polyhedron so that it is a bounded weakly pseudoconvex domain. Furthermore, $G^n_{\epsilon}$ is the complete Reinhardt domains, i.e., for any $z=(z_1,...,z_n)\in G^n_{\epsilon}$ and $\lambda=(\lambda_1,...,\lambda_n)\in \overline{\mathbb{D}}^n$, the point $(\lambda_1 z_1,...,\lambda_n z_n)$ belongs to $G^n_{\epsilon}$.

The proof of Theorem~\ref{th:CC} will be presented in Section 4.

Although Theorem~\ref{th:CC2} seems almost same with the Theorem~\ref{th:CC}, the situation is different, because the assumption in the theorem below requires that the n-product $(r\mathbb{D})^n$ is completely in $G^n_{\epsilon}$. This embedding gives the different proof from Theorem~\ref{th:CC}, but this approach does not work for $n=2$.
\begin{thm-Intro}\label{th:CC2}
	Let $0<r<1$ and $G$ be a bounded pseudoconvex domain in $\mathbb{C}^1$ such that $r\mathbb{D}\subset G \subset \mathbb{D}$. For $n\geq3$, let $\epsilon:=r^n$. Define $G^n_{\epsilon}:=\left\{ z=(z_1,...,z_n)\in{G}^{n}:\left| \Pi^{n}_{i=1}z_i \right|<\epsilon \right\}$. Denote the canonical complete K\"ahler-Einstein metric on $G^n_{\epsilon}$ by $\omega_{KE}$, and let $d^{{KE}}_{G^n_{\epsilon}}$ be the distance induced by $\omega_{KE}$. Then for any $x\in G$ which satisfies $(x,...,x)\in G^n_{\epsilon}$ and $|x|>\sqrt{2}n (2n)^{\frac{1}{2n}}r^{\frac{n^2-n-1}{n}}=\frac{\sqrt{2}n (2n)^{\frac{1}{2n}}}{{r}^{\frac{n+1}{n}}}\epsilon$, we have
	\[c_{G^n_{\epsilon}}((0,...,0),(x,...,x)) <  d^{{KE}}_{G^n_{\epsilon}}((0,...,0),(x,...,x)).\]
\end{thm-Intro}

From the definition of $G^n_{\epsilon}$, the necessary condition to have $x\in G$ as described in Theorem~\ref{th:CC2} is $r^{n^2-2n-1}<\frac{1}{ \sqrt{n 2^{n+1}}n^n}$. The proof of Theorem~\ref{th:CC2} will be presented in Section 5.

In both Theorem~\ref{th:CC} and Theorem~\ref{th:CC2}, we used the existence of the complete K\"ahler-Einstein metric on a bounded pseudoconvex domain as given in the main theorem in \cite{NMY83}. The reader might expect that one can eliminate the boundness assumption of $G$ to prove Theorem~\ref{th:CC} and Theorem~\ref{th:CC2} if there is the complete K\"ahler-Einstein metric on that complex manifold $G^n_{\epsilon}$. In that case, however, our proof is not working, because the step 4 in proofs of Theorem~\ref{th:CC} and Theorem~\ref{th:CC2} do not work.

In particular, if $\epsilon>0$ satisfies $\epsilon<\min\{\frac{1}{n}r^2,(\frac{r}{n})^{\frac{n}{n-1}}\}$, then
\[D^n_{\epsilon}:=\left\{ z=(z_1,...,z_n)\in{\mathbb{D}}^n:\left| \Pi^{n}_{i=1}z_i \right|<\epsilon \right\}\]
serves as a concrete example for Theorem~\ref{th:CC}. Also, we can get the same conclusion by taking different pair of $(r,\epsilon)$ to apply Theorem~\ref{th:CC2} for $D^n_{\epsilon}$ when $n\geq3$ if $r$ satisfies $r<\min\{(\frac{1}{ \sqrt{n 2^{n+1}}n^n})^{\frac{1}{n^2-2n-1}},(\frac{1}{ \sqrt{2}n(2n)^{\frac{1}{2n}}})^{\frac{n}{n^2-n-1}}\}$.

Next, the following theorem tells us that we can distinguish the Carath\'eodory-distance from the Kobayashi-distance on strongly pseudoconvex subdomains of $G^n_{\epsilon}$ if the Carath\'eodory-distance and the Kobayashi-distance are different on $G^n_{\epsilon}$.

\begin{thm-Intro}\label{th:GH}
	Given $G^n_{\epsilon}$ as in Theorem~\ref{th:CC} or Theorem~\ref{th:CC2}, write $G^n_{\epsilon}$ as the countable union of an ascending sequence $\{S_{l} | l \in \mathbb{N}\}$ of bounded, strongly pseudoconvex domains with smooth boundaries. Suppose there are points $p=(p,...,p),q=(q,...,q) \in G^n_{\epsilon}$ such that
	\[ c_{G^n_{\epsilon}}(p,q) < k_{G^n_{\epsilon}}(p,q).  \]
	 Then there exists $l_0$ such that whenever $l\geq l_0$, there exists a sequence $\{r_m=(r_m,...,r_m)\in S_{l}\}$ (depending on $l$) which goes towards the boundary of $G^n_{\epsilon}$ as $m \rightarrow \infty$ such that \[ c_{S_{l}}(r_m,.) \neq k_{S_{l}}(r_m,.).  \]

\end{thm-Intro}

We can produce the comparison of the Carath\'eodory-distance with the Kobayashi-distance on strongly pseudoconvex subdomains by applying Theorem~\ref{lem:SV}, since those subdomains possess the $C^k$ boundary, $k\geq 2$. The proof of Theorem~\ref{th:GH} will be presented in Section 6.

Finally, let us compare the Kobayashi-distance $k_{G^n_{\epsilon}}$ and the distance function $l_{G^n_{\epsilon}}$.

On a complex manifold $M$, for each $m\in \mathbb{N}$, define
\[k^{(m)}_{M}(x,y):=\inf_{f_i \in \Hol(\mathbb{D},M)}\left\{\sum_{i=1}^{m} \rho_{\mathbb{D}}(a_i,b_i) \right\}\]
where $x=p_0 < ... < p_m =y, f_i(a_i)=p_{i-1}, f_i(b_i)=p_i $. Then we can observe that
\[l_{M}=k^{(1)}_{M}\geq k^{(m)}_{M}\geq {\lim}_{l\rightarrow \infty}k^{(l)}_{M}=k_{M}. \]

We say that $M$ is taut if $\Hol(\mathbb{D}, M)$ is normal. i.e., every sequence of $\Hol(\mathbb{D}, M)$ has a subsequence which converges uniformly on compact subsets of $\mathbb{D}$. It is well known that the distance function $l_{M}$ on a taut domain is continuous (see p119 in \cite{JP13}). It follows that $l_{G^n_{\epsilon}}$ is continuous, since $G^n_{\epsilon}$ in Theorem~\ref{th:CC} and Theorem~\ref{th:CC2} are taut domains (see p252 in \cite{Kob98}). The following theorem tells us the interesting behavior of $k^{(m)}_{G^n_{\epsilon}}$ with $m\in \mathbb{N}$.

\begin{thm-Intro}\label{co:DD}
Given $G^n_{\epsilon}$ as in Theorem~\ref{th:CC} or Theorem~\ref{th:CC2}, let $x,y\in r\mathbb{D}-0$. Then for $x=(x,...,x,0)$ and $y=(0,y,...,y)$ in $G^n_{\epsilon}$, we have
\begin{align*}
& {\lim}_{\epsilon\rightarrow 0}  k^{(m)}_{G^n_{\epsilon}} (x,y)={\lim}_{\epsilon\rightarrow 0}k^{(2)}_{G^n_{\epsilon}} (x,y) \\
& \leq \rho_{\mathbb{D}}(\frac{1}{r}x,0)+\rho_{\mathbb{D}}(0,\frac{1}{r}y)< {\lim}_{\epsilon\rightarrow 0}  l_{G^n_{\epsilon}} (x,y)=\infty
\end{align*}
for any $m\geq 2$.
\end{thm-Intro}

The proof of Theorem~\ref{co:DD} will be presented in Section 7.
Here, we can choose $x=(x,...,x,0)$ and $y=(0,y,...,y)$ in $G^n_{\epsilon}$ without considering the choice of $\epsilon>0$, due to the zero component of $x$ and $y$. Thus we can make $\epsilon>0$ as small as possible.

One interesting application of our results is the non-existence of holomorphic retractions to simply-connected open Riemann surfaces on those domains. The useful equivalent descriptions that the Carath\'eodory-distance $c_{G^n_{\epsilon}}$ and the distance function $l_{G^n_{\epsilon}}$ are matched on taut domains are known (Theorem~\ref{th:HR}).

\begin{cor-Intro}\label{th:CCCH}
	Given $G^n_{\epsilon}$ as in Theorem~\ref{th:CC} or Theorem~\ref{th:CC2}, consider any pair $x,y$ in $G^n_{\epsilon}$ such that $c_{G^n_{\epsilon}}(x,y) < l_{G^n_{\epsilon}}(x,y)$ and any simply-connected open Riemann surface $V$ in $G^n_{\epsilon}$ which contains $x,y$. Then there is no holomorphic retraction $p : G^n_{\epsilon} \rightarrow V$.
\end{cor-Intro}

To get Corollary~\ref{th:CCCH}, one can take the pair $x=(x,...,x,0),y=(0,y,...,y)\in G^n_{\epsilon}$ where $x,y\in r\mathbb{D}-0$ from Theorem~\ref{co:DD}.

\section{Schwarz lemmas, a boundary estimate and holomorphic retractions}
In this section, we collect the borrowed theorems that we use to prove our results. The following two generalized Schwarz lemma due to Yau will be used to compare the Carath\'eodory-distance and the distance of the K\"ahler-Einstein metric.

\begin{thm}[the Schwarz-Yau lemma, \cite{Y78}]\label{th:SC}
	Let $(M,g)$ be a complete K\"ahler manifold with Ricci curvature bounded from below by a negative constant $K_1$. Let $(N,h)$ be another Hermitian manifold with holomorphic bisectional curvature bounded from above by a negative constant $K_2$. If there is a non-constant holomorphic map $f$ from $M$ to $N$, we have
	\[f^{*}h\leq \frac{K_1}{K_2}g. \]
\end{thm}

To apply above theorem, one can use the upper bound of holomorphic sectional curvature of $(N,h)$ instead of the upper bound of bisectional curvature if $N$ is a complete Riemann surface. The following different version of Schwarz lemma is useful if one wants to use the complete K\"ahler-Einstein metric on the codomain.

\begin{thm}[The volume version of the Schwarz-Yau lemma, \cite{Y78}]\label{th:VSC}
	Let $(M,g)$ be a complete K\"ahler manifold with scalar curvature bounded from below by a negative constant $K_1$. Let $(N,h)$ be another Hermitian manifold with Ricci curvature bounded from above by a negative constant $K_2$. Then the existence of a non-degenerate holomorphic map $f$ from $M$ to $N$ implies that $K_1\leq 0$ and
	\[f^{*}dV_N\leq \frac{K_1}{K_2}dV_M. \]
\end{thm}

The proof of Theorem~\ref{th:GH} uses the following boundary estimate.
\begin{thm}[Theorem 1 in \cite{SV89}]\label{lem:SV}
	Let $M \Subset \mathbb{C}^n$ be a bounded, strongly pseudoconvex domain with $C^2$ boundary. For each $\delta>0$, there exists a compact set $K_{\delta}\Subset M$ depending only on $\delta>0$ such that for any $z\in M - K_{\delta}$ and $w \in M$,
	\[ k_{M}(z,w)\leq (1+\delta)c_{M}(z,w). \]
\end{thm}
The theorem above is interesting in its own way because this theorem tells us that the Carathe\'odory-distance is not much different from the Kobayashi-distance. We believe that a similar conclusion might hold when $M$ is a complete simply-connected K\"ahler manifold with negative sectional curvature ranges.

The following theorem is used to conclude Corollary~\ref{th:CCCH}.
\begin{thm}[Proposition 11.1.7 in \cite{JP13}]\label{th:HR}
	Let $M \subset \mathbb{C}^n$ be a taut domain. Then the following are equivalent:
	\begin{enumerate}
		\item $c_{M}=l_{M}$.
		\item For any $x,y \in M$ with $x\neq y$, there exists a holomorphic function $f : \mathbb{D} \rightarrow M$ and a holomorphic function $p : M \rightarrow \mathbb{D}$ such that $x,y \in f(\mathbb{D})$ and $p \circ f=id_{\mathbb{D}}$.
		\item For any $x,y \in M$ with $x\neq y$, there exists a holomorphic embedding $f : \mathbb{D} \rightarrow M$ and a holomorphic retraction $p : M \rightarrow f(\mathbb{D})$ such that $x,y \in f(\mathbb{D})$.
	\end{enumerate}
\end{thm}

We used the third statement to formulate Corollary~\ref{th:CCCH}, but the second statement is also interesting. We found the statement from \cite{JW03} that for any complex manifold $M$, for any $x,y \in M$ with $x\neq y$, there exists a holomorphic function $f : \mathbb{D} \rightarrow M$ such that $x,y \in f(\mathbb{D})$. Hence one can formulate the statement of non-existence of a holomorphic function $p : M \rightarrow \mathbb{D}$ such that $x,y \in f(\mathbb{D})$ and $p \circ f=id_{\mathbb{D}}$.

\section{Comparison the Carath\'eodory distance with the K-E distance}

The proof of Theorem~\ref{th:CC} can be reduced to showing the following lemma. This lemma gives the comparison of the Carath\'eodory pseudo-distance and the distance induced by the K\"ahler-Einstein metric. Note that by Montel's theorem, for any two points in a complex manifold $M$, we can always achieve the extremal map $f\in \Hol(M, \mathbb{D})$ with respect to the Carath\'eodory pseudo-distance.
In below lemma, $\gamma_{\mathbb{D}}$ is the Poincare metric on the unit disk.

\begin{lemma}\label{lem:JP}
	Let $M$ be a bounded pseudoconvex domain in $\mathbb{C}^n$, and $a,b \in M$, $a\neq b$. Denote the canonical complete K\"ahler-Einstein metric on $M$ by $\omega_{KE}$, and let $d^{{KE}}_{M}$ be the distance induced by $\omega_{KE}$. Suppose there exists $f\in \Hol(M, \mathbb{D})$ that is extremal for $c_{M}(a,b)$ such that
	\[ \gamma_{\mathbb{D}}(f(a),f'(a)X) <  \sqrt{(\omega_{KE}(a)(X,X))}, X \in \mathbb{C}^n-\left\{ 0\right\}.  \]
	Then
	\[c_{M}(a,b) <  d^{{KE}}_{M}(a,b).\]
	\begin{proof}
		By assumption, continuity of metrics gives $\epsilon, \delta$ such that
		\[ \gamma_{\mathbb{D}}(f(z),f'(z)X) + \epsilon \| X\| \leq \sqrt{(\omega_{KE}(z)(X,X))}, z\in \mathbb{B}(a,\delta) \Subset M, X\in \mathbb{C}^n.   \]
		On the other hand, by the Schwarz-Yau lemma, we have
		\[ \gamma_{\mathbb{D}} (f(z), f'(z)X) \leq \sqrt{(\omega_{KE}(z)(X,X))}, z\in M, X\in \mathbb{C}^n.  \]
		Let $\alpha : [0,1] \rightarrow M$ be any piecewise $C^1$-curve joining $a$ and $b$. Denote by $t_0$ the maximal $t\in[0,1]$ such that $\alpha([0,t])\subset \overline{\mathbb{B}(a,\delta)}$. Denote the arc-length of $\alpha$ with respect to the K\"ahler-Einstein metric by $L_{KE}(\alpha)$. Then
		\[ L_{KE}(\alpha)= \int_{0}^{t_0}\sqrt{\omega_{KE}(\alpha(t))({\alpha'(t),\alpha'(t)})}dt + \int_{t_0}^{1}\sqrt{\omega_{KE}(\alpha(t))({\alpha'(t),\alpha'(t)})dt}  \]
		\[ \geq \int_{0}^{1}\gamma_{\mathbb{D}}(f(\alpha(t)), f'(\alpha(t))dt + \epsilon \int_{0}^{t_0} \| \alpha'(t)\| dt \]
		\[ \geq L_{\gamma_{\mathbb{D}}}(f\circ \alpha) +\epsilon \delta \geq \rho_{\mathbb{D}}(f(a),f(b)) + \epsilon \delta = c_{M}(a,b)+\epsilon \delta.  \]
		Hence $d^{{KE}}_{M}(a,b) \geq c_{M}(a,b)+\epsilon \delta$.

	\end{proof}
\end{lemma}
Notice that in the proof of the lemma above, we used the Schwarz-Yau lemma which only gives the one-side inequality. This yields the difficulty to prove that the Carath\'eodory-distance is equivalent to other intrinsic distances since one needs to have different techniques to get the reverse inequality unless the pseudoconvex domain is bounded.

\section{Proof of Theorem~\ref{th:CC} : $c_{G^n_{\epsilon}} \neq   d^{{KE}}_{G^n_{\epsilon}}$, $n\geq2$}
\begin{proof}

Step 1, we will establish the basic settings for the proof. Fix $\epsilon>0$, with the global coordinate $(z_1,...,z_n)\in G^n_{\epsilon}$ in $\mathbb{C}^{n}$, let $\{\frac{\partial}{\partial{z_i}} | i=1,...,n\}$ be the basis on $T^{1,0}_{0}G^n_{\epsilon}$. Without loss of generality, we may assume that $\{\frac{\partial}{\partial{z_i}}| i=1,...,n\}$ are orthonormal with repect to the Euclidean metric and orthogonal with respect to $\omega_{KE}(0)$. From now on, we will assume that $n=2$ and the same proof works for general $n$. With the usual global coordinates $(z_1,z_2)\in G^2_{\epsilon}$, denote $\frac{\partial}{\partial{z_1}}=X_1$, $\frac{\partial}{\partial{z_2}}=X_2$. Also, we may assume that $0<\omega_{KE}(0)(X_1,\overline X_1)\leq\omega_{KE}(0)(X_2,\overline X_2)$.
For $z=(z_1,z_2)$, define $\chi(z):=\frac{1}{\epsilon}{z_1}{z_2}\in \Hol(G^2_{\epsilon},\mathbb{D})$. Then by the distance-decreasing property, $c_{\mathbb{D}}(\chi(0,0),\chi(x,x))\leq c_{G^2_{\epsilon}}((0,0),(x,x))$, so we get
\begin{equation}\label{eq:first}
c_{\mathbb{D}}(0,\frac{1}{\epsilon}{x}^{2})\leq c_{G^2_{\epsilon}}((0,0),(x,x)).
\end{equation}

Step 2, we want to notice that $\epsilon < \frac{1}{4}r^2$ implies $\frac{2\epsilon}{r}<r$ and
for any $x \in G$ satisfying $\frac{2\epsilon}{r}<|x|<r$, $(x,x)\in G^2_{\epsilon}$. We may assume $x>0$ and we will use that there exists $x \in r\mathbb{D}$ which satisfies
\begin{equation}\label{eq:second}
x > \frac{2\epsilon}{r}.
\end{equation}

Step 3, we will control the extremal map with respect to the Carath\'eodory-distance $f : G^2_{\epsilon} \rightarrow \mathbb{D}$ such that
\begin{equation}\label{eq:third}
c_{G^2_{\epsilon}}((0,0),(x,x))=c_{\mathbb{D}}(f(0,0),f(x,x)).
\end{equation}
After acting on the unit disk by an automorphism, we may assume that $f(0,0)=0$. Also we can replace $f$ by the symmetrization map $\frac{1}{2}({f(x,y)+f(y,x)})$.
Around the origin, one can write $f$ as a power series $\sum_{i=1}^{2} a_i z_i+\sum_{p=m}^{\infty}P_m(z_1,z_2)=\sum_{i=1}^{2} a_i z_i+f_2(x,y)$, where each $P_m(z_1,z_2)$ is a homogeneous polynomial of degree $m$. Since we use $\frac{1}{2}({f(x,y)+f(y,x)})$, we can replace $ a_1 z_1 + a_2 z_2$ by  $a(z_1 + z_2)$ for some real number $a$ (after acting on the unit disk by an automorphism if necessary). Then by the Schwarz-Yau Lemma, $f^* \gamma_{\mathbb{D}} \leq \sqrt{\omega_{KE}}$. In particular, we get
\[a \leq \sqrt{\omega_{KE}(0)(X_1,\overline X_1)}.\]

Step 4, we will show there exists a hypersurface of $G^2_{\epsilon}$ given by  \[\sqrt{\omega_{KE}(X_1,\overline X_1)(0)}(z_1+z_2)=1.\] To show this, take $0<R\leq1$ such that $G$ is contained in $R\mathbb{D}$ and the boundary of $G$ touches the boundary of  $R\mathbb{D}$. Then by using a projection map, we can define the holomorphic map from $(G^2_{\epsilon}, \omega_{KE})$ to $(R\mathbb{D},h)$, where $h$ is the Poincare metric on $R\mathbb{D}$. Then by the Schwarz-Yau lemma, $h(0)\leq R^2\omega_{KE}(0)$. This implies
\begin{equation}\label{eq:fifth}
1\leq \frac{1}{R} \leq R\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}.
\end{equation}
In particular, $1 \leq R\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}$ in \eqref{eq:fifth} implies we can take $(z_1,z_2)\in G^2_{\epsilon}$ which satiesfies $\sqrt{\omega_{KE}(X_1,\overline X_1)(0)}(z_1+z_2)=1$ by taking one component by almost R and the other component by almost zero.

Step 5, we will show $a < \sqrt{\omega_{KE}(0)(X_1,\overline X_1)}$. Assume   $a=\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}$. Since there exists a hypersurface of $G^2_{\epsilon}$ given by  $\sqrt{\omega_{KE}(X_1,\overline X_1)(0)}(z_1+z_2)=1$, thus we can choose $(u_1,u_2)\in G^2_{\epsilon}$ such that $u_1 + u_2=\frac{1}{\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}}$. Define $g \in \mathbb{D} \rightarrow G_{\epsilon}$ by $\lambda \mapsto (\lambda u_1,\lambda u_2)$. Then $f \circ g (0)=0$ and $(f \circ g)' (0)=1$. Thus $f \circ g = id_{\mathbb{D}}$ by the classical Schwarz's lemma. This gives $f_2(\lambda u_1, \lambda u_2)=0$. Since $\lambda\in \mathbb{D}$ is arbitrary and with other choices of $(u_1,u_2)$, we can take a small open set in $G_{\epsilon}^2$ such that $f_2=0$ on this open set. Thus by the identity theorem, $f_2\equiv0$ on $G^2_{\epsilon}$. Hence $f(z_1,z_2)=a(z_1+z_2)$. Then by \eqref{eq:first} and \eqref{eq:third},
\[ c_{\mathbb{D}}(0,2\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}x)=c_{G^2_{\epsilon}}((0,0),(x,x))\geq c_{\mathbb{D}}(0,\frac{1}{\epsilon}{x}^{2}). \]
Thus we obtain $x \leq 2\epsilon \sqrt{(\omega_{KE}(0)(X_1,\overline{X_1})}$. On the other hand, since the restriction of $f : G^2_{\epsilon} \rightarrow \mathbb{D}$ from $G^2_{\epsilon}$ to $r\mathbb{D}\subset G$ gives the holomoprhic map from $r\mathbb{D}$ to $\mathbb{D}$, the classical Schwarz lemma implies $ar\leq1$. In particular, we have
\[2\epsilon\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}=2\epsilon a\leq \frac{2\epsilon}{r}<r.   \]

However, by \eqref{eq:second}, we also have $x > \frac{2\epsilon}{r}\geq2\epsilon \sqrt{(\omega_{KE}(0)(X_1,\overline{X_1})}$, which is impossible. Hence $a<\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}$.

Step 6, we will finish the proof. $a < \sqrt{\omega_{KE}(0)(X_1,\overline X_1)}$ implies
\begin{equation}\label{eq:fourth}
\left| f'(0)X_i \right| <  \sqrt{{\omega_{KE}(0)(X_i,\overline{X_i})}}, i=1,2.
\end{equation}
Since we showed the above inequality for $X_1$ and $X_2$, which are orthogonal to the Euclidean metric and $\omega_{KE}(0)$ at the same time, the same inequality holds for arbitrary tangent vectors $X \in \mathbb{C}^n-\left\{ 0\right\}$. Consequently, \eqref{eq:fourth} implies the assumption in Lemma~\ref{lem:JP}, and we are done.
\end{proof}

\section{Proof of Theorem~\ref{th:CC2} : $c_{G^n_{\epsilon}} \neq   d^{{KE}}_{G^n_{\epsilon}}$, $n\geq3$}
\begin{proof}

 For fixed $\epsilon>0$, with the global coordinate $(z_1,...,z_n)\in G^n_{\epsilon}$ in $\mathbb{C}^{n}$, let $\{\frac{\partial}{\partial{z_i}} | i=1,...,n\}$ be the basis on $T^{1,0}_{0}G^n_{\epsilon}$. Without loss of generality, we may assume that $\{\frac{\partial}{\partial{z_i}}| i=1,...,n\}$ are orthonormal with repect to the Euclidean metric and orthogonal with respect to $\omega_{KE}(0)$. Denote $\frac{\partial}{\partial{z_1}}=X_1$, ..., $\frac{\partial}{\partial{z_n}}=X_n$. Also, we may assume that $0<\omega_{KE}(0)(X_1,\overline X_1)$ is the minimum among all $\omega_{KE}(0)(X_i,\overline X_i)$'s.

 The rest of the proof of Theorem~\ref{th:CC2} is exactly same with the proof of Theorem~\ref{th:CC} except for the role of \eqref{eq:second} in Step 2. Thus It suffices to show that
for any $x \in G$ satisfying $|x|>\sqrt{2}n (2n)^{\frac{1}{2n}}r^{\frac{n^2-n-1}{n}}=\frac{\sqrt{2}n (2n)^{\frac{1}{2n}}}{{r}^{\frac{n+1}{n}}}\epsilon$,
\[|x|>n\epsilon{\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}}.\]
To show the claim, consider the n-product $(r\mathbb{D})^n$ in $G^n_{\epsilon}$. Since the scalar curvature with respect to the product metric on $(r\mathbb{D})^n$ is $-\frac{2n}{r^2}$ and the ricci curvature on $G^n_{\epsilon}$ is $-1$, it follows from the volume version of the Schwarz-Yau lemma,

\[2^n{\Pi^{n}_{j=1}\omega_{KE}(0)(X_j,\overline X_j)}\leq \frac{2^{2n+1}n}{r^{2n+2}}.  \]
To obtain the right hand side in above, we used the fact that the formula of the volume element of polydisk of radius $r$ in $\mathbb{C}^n$ is given by $(2r)^{2n} \Pi^{n}_{j=1}\frac{\sqrt{-1}dz^{j}\wedge d\overline{z}^{j}}{(r^2-|z_j|^2)^2}$.
Since $\omega_{KE}(0)(X_1,\overline X_1)$ is the minimum among all $\omega_{KE}(0)(X_i,\overline X_i)$', we get
\[n\epsilon\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}\leq \frac{\sqrt{2}n (2n)^{\frac{1}{2n}}}{{r}^{\frac{n+1}{n}}}\epsilon.\]
Thus we get $|x| > n\epsilon {\sqrt{\omega_{KE}(0)(X_1,\overline X_1)}}$ if $|x|>\frac{\sqrt{2}n (2n)^{\frac{1}{2n}}}{{r}^{\frac{n+1}{n}}}\epsilon$. Hence we established the claim. Thus for any $x\in G$ which satisfies $(x,...,x)\in G^n_{\epsilon}$ and $x>\sqrt{2}n (2n)^{\frac{1}{2n}}r^{\frac{n^2-n-1}{n}}=\frac{\sqrt{2}n (2n)^{\frac{1}{2n}}}{{r}^{\frac{n+1}{n}}}\epsilon$, we have $x > n\epsilon \sqrt{(\omega_{KE}(0)(X_1,\overline{X_1})}$ (Here, we assumed that $x$ is the real number.) Then the rest of the proof follows as in the proof of Theorem~\ref{th:CC}.
\end{proof}
\section{Proof of Theorem~\ref{th:GH} : $c_{S_{l}} \neq k_{S_{l}}$}
\begin{proof}
We will assume $n=2$. The same argument works for general $n$. Suppose there are two points $p=(p,p),q=(q,q) \in G^2_{\epsilon}$ such that
\[ c_{G^2_{\epsilon}}(p,q) < k_{G^2_{\epsilon}}(p,q).  \]
Write $G^2_{\epsilon}=\bigcup^{\infty}_{l=1}S_{l}$, where $\{S_{l} | l \in \mathbb{N}\}$ is an ascending sequence  of bounded, strongly pseudoconvex domains with smooth boundaries. By the continuity of the distances, there exists $l_0$ such that whenever $l\geq l_0$, $c_{S_l}(p,q) < k_{S_l}(p,q)$.
Then we can find $\delta>0$ such that $(1+\delta)c_{S_{l}}(p,q) < k_{S_{l}}(p,q).$ Then by Theorem~\ref{lem:SV}, there exists a compact set $K_{\delta}$ in $S_{l}$ such that
\[ k_{S_{l}}(a,b) < (1+\delta) c_{S_{l}}(a,b), a\in S_{l}, b \in S_{l}-K_{\delta}.  \]
Define the map $f$ on the diagonal entries of $S_{l}$, by assigning $x=(x,x)\in S_{l} \mapsto (1+\delta) c_{S_{l}}(p,x)-k_{S_{l}}(p,x)$. Then $f(q)<0$ and $f(s)>0$ for some $s=(s,s)\in S_{l}$ such that $s=(s,s)\notin K_{\delta}$. Then by the intermediate value theorem, there exists $r=(r,r)\in S_{l}$ such that $f(r)=0$. In particular,
\[ c_{S_{l}}(p,r) < k_{S_{l}}(p,r).  \]
By repeating this process, we can construct a monotone sequence $\{r_m:=(r_m,r_m)\}$ in $S_{l}$ such that $c_{S_{l}}(r_m,) \neq k_{S_{l}}(r_m,.)$ for each $m$ and this sequence goes towards the boundary. If this sequence converges before reaching the boundary, then there exists a compact subset $K$ in $S_{l}$ such that $K$ contains the sequence $\{ r_m \in S_{l} \}$. Then we can apply the previous argument again to produce another point $r'=(r',r') \notin K$ such that $c_{S_{l}}(r',.) \neq k_{S_{l}}(r',.)$, which is a contradiction. This finishes the proof of Theorem~\ref{th:GH}.
\end{proof}

\section{Proof of Theorem~\ref{co:DD} : $k^{(m)}_{G^n_{\epsilon}} =k^{(2)}_{G^n_{\epsilon}}  \neq l_{G^n_{\epsilon}} $ for any $m\geq2$ as $\epsilon \rightarrow 0$}
\begin{proof}
We will assume $n=2$. The same argument works for general $n$.
For each $\epsilon>0$, let $x=(x,0),y=(0,y),0=(0,0)$.  Since we know $r\mathbb{D} \subset G$, the holomorphic maps $z\in \mathbb{D} \rightarrow (rz,0)\in G^2_{\epsilon}$ and $z\in \mathbb{D} \rightarrow (0,rz)\in G^2_{\epsilon}$ implies
\[k^{(2)}_{G^2_{\epsilon}}(x,y)\leq \rho_{\mathbb{D}}(\frac{1}{r}x,0)+\rho_{\mathbb{D}}(0,\frac{1}{r}y). \]

Notice that the above inequality does not depend on the choice of $\epsilon>0$. Hence we also get
\[{\lim}_{\epsilon\rightarrow 0}k^{(2)}_{G^2_{\epsilon}}(x,y)\leq \rho_{\mathbb{D}}(\frac{1}{r}x,0)+\rho_{\mathbb{D}}(0,\frac{1}{r}y). \]
We now claim that $l_{G^2_{\epsilon}}(x,y) \rightarrow \infty$ as $\epsilon \rightarrow 0$. Suppose not, then there is a uniform bound $M$ and a sequence of positive real numbers $(r_n)$ which converges to $0$ such that $l_{G^2_{r_n}}(x,y)\leq M$ for all $n$. Hence there exists a sequence $(f_n:=(g_n,h_n)\in \Hol(\mathbb{D},G^2_{r_n}))$ such that $f_n (0)=x$, $f_n(\sigma_n)=y$ for $\sigma_n \in (0,1)$ and $\rho_{\mathbb{D}}(0,\sigma_n)\leq 2M$ for each $n$. By the uniform bound $2M$, we can apply Montel's theorem, so $(f_n)$ contains a subsequence $(f_n)$ (using the same notation) which converges uniformly on each compact subset of $\mathbb{D}$ to $f=(g,h)\in \Hol(\mathbb{D},G^2)$. Since $(\sigma_n)$ is a bounded sequence with respect to the Poincare distance, we can assume further that $(\sigma_n)$ converges to $\sigma$ in $\mathbb{D}$. Since $\rho_{\mathbb{D}}(0,\sigma_n)\leq 2M$, we have $0\leq \sigma \leq tanh(2M) <1$. Since $\left|g_n h_n\right|\leq r_n$ for all $n\in \mathbb{N}$, $gh \equiv 0$. By the Identity theorem, $g \equiv 0$ or $h \equiv 0$. Since $g_n(0)=x$, $g \not \equiv 0$. But $h(\sigma)=\lim_{n\rightarrow \infty} h_{n}(\sigma_n)=y$. This contradiction establishes our claim.

Now, we will show ${\lim}_{\epsilon\rightarrow 0}k^{(m)}_{G^2_{\epsilon}}(x,y)={\lim}_{\epsilon\rightarrow 0}k^{(2)}_{G^2_{\epsilon}}(x,y)$ for any $m\geq3$. Suppose there exists $m\geq3$ such that ${\lim}_{\epsilon\rightarrow 0}k^{(m)}_{G^2_{\epsilon}}(x,y) < {lim}_{\epsilon\rightarrow 0}k^{(m-1)}_{G^2_{\epsilon}}(x,y)$. Then there exists a sequence of positive real numbers $(r_n)$ which converges to $0$ and m number of sequences $(f^{l}_n:=(g^{l}_n,h^{l}_n)\in \Hol(\mathbb{D},G^2_{r_n}))$, $l=1,...,m$ such that
\[{\lim}_{n\rightarrow \infty}\sum_{l=1}^{m} \rho_{\mathbb{D}}(0,b^{l}_n) = {\lim}_{\epsilon\rightarrow 0} k^{(m)}_{G^2_{\epsilon}}(x,y) \]
where $x=p^{0}_n < ... < p^{m}_n =y, f^{l}_n(0)=p^{l-1}_n, f^{l}_n(b^{l}_n)=p^{l}_n$. By using ${\lim}_{\epsilon\rightarrow 0}k^{(m-1)}_{G^2_{\epsilon}}(x,y)$ as a uniform bound, we may assume that all sequences $(b^{l}_n)$, $(p^{l}_n)$ and $f^{l}_n=(g^{l}_n,h^{l}_n)$ are convergent. Then we get
\[\sum_{l=1}^{m} \rho_{\mathbb{D}}(0,b^{l}) = {\lim}_{\epsilon\rightarrow 0} k^{(m)}_{G^2_{\epsilon}}(x,y)\]
where $(b^{l})$ is the limit of $(b^{l}_n)$. We can notice that each limit $p^{l}$ of $(p^{l}_n)$ should lie on some axis ${r \mathbb{D}}\subset G$ in $G^2_{0}$. Let $f^{l}=(g^{l},h^{l})\in \Hol(\mathbb{D},G^2)$ be the limit functions of $(f^{l}_n=(g^{l}_n,h^{l}_n))$. Then $g^{l}h^{l} \equiv 0$ implies $g^{l}$ or $h^{l}$ must be vanished for each $l$. Then there exists $l\geq2$ such that $f^{l}(b^{l})=p^{l}=(0,0)$. This forces that $p^{l},p^{l-1},p^{l-2}$ must lie on the same axis ${r \mathbb{D}}\subset G$ in $G^2_{0}$. Regarding each $g^{l}$ and $h^{l}$ as mappings from $\mathbb{D}$ to $\mathbb{D}$, the intrinsic property by those maps and the triangle inequality of the Poincare-distance imply
\[\rho_{\mathbb{D}}(0,b^{l-1})+\rho_{\mathbb{D}}(0,b^l)\geq \rho_{\mathbb{D}}(p^{l-2},p^{l-1})+\rho_{\mathbb{D}}(p^{l-1},p^{l})\geq \rho_{\mathbb{D}}(p^{l-2},p^{l})=:\rho_{\mathbb{D}}(0,{b'}^{l}). \]
Hence we can reduce m chains to m-1 chains. Consequently,
\begin{align*}
& {\lim}_{\epsilon\rightarrow 0} k^{(m-1)}_{G^2_{\epsilon}}(x,y) > {\lim}_{\epsilon\rightarrow 0} k^{(m)}_{G^2_{\epsilon}}(x,y)=\sum_{l=1}^{m} \rho_{\mathbb{D}}(0,b^{l}) \\
& \geq \sum_{l=1}^{m-1} \rho_{\mathbb{D}}(0,{b'}^{l})\geq {\lim}_{\epsilon\rightarrow 0} k^{(m-1)}_{G^2_{\epsilon}}(x,y). \label{eq:last}
\end{align*}

This is impossible, and it completes the proof.
\end{proof}

\subsection*{Acknowledgements}
This work was partially supported by NSF grant DMS-1611745. I would like to thank my academic advisor Damin Wu for many insights and helpful comments.

config: arxiv


TITLE: Does a photon attract or repel matter?
QUESTION [4 upvotes]: So, I know experimentally that matter and light interact. And I can calculate cross sections in Quantum Electrodynamics. Yet a fundamental intuition about how light and matter interact is completely missing in my understanding.
If a photon were (for a brief instant) next to a proton, would they attract? Repel? How about an electron? Is there some kind of potential $V(x_2-x_1)$ which can approximate the interaction between light in a particle-like state and matter? If we can do it for protons and electrons, which are both particles and waves, why can't we also do it for photons?
Maybe this is going to be categorized as too broad. If that's so, let's just narrow it down to: Does a photon (a gaussian wavepacket of light) attract a proton, or repel it? If the answer depends on the distance, is there must be an optimal distance between them like in the Van der Waals potential?:

REPLY [1 votes]: I think this question needs to be asked very carefully to get a precise answer. But, the zeroth order answer should be clear: the photon has no electric charge or magnetic moment, so it will neither attract nor repel another electric charge.
One interpretation of this question, given in the comments to another answer, is to ask about the trajectory of a photon passing near an atom.
Well, in fact, even here we need to be careful. When I hear "trajectory" I think of a one dimensional line moving through space; and it makes sense to think of the impact parameter (distance of closest approach) of such a line with an atom. But an EM wave, or a photon wavepacket, is not (generally) of this form.
If the EM wavelength (1 over the photon momentum) is larger than the size of the atom, then the atom will look approximately like a point particle, compared to a traveling plane wave. First, note there is no useful sense in which we can think of the EM wave / photon as having a 1 dimensional trajectory and ask for its distance with respect to an atom in this limit. We should think of a wave extending over space, scattering off of a point source. This is the limit where we expect more or less elastic scattering to apply. There is an interaction, but we describe this interaction in terms of the scattered wave produced at infinity, and not a trajectory.
If the EM wavelength (1 over the photon momentum) is smaller than the size of the atom, then we can think of a trajectory, if we imagine building a wavepacket of high energy radiation. But in this limit, where by definition the photon has a larger energy than the binding energy of the atom, we certainly do not expect an elastic scattering process to occur. Instead, the EM radiation will be absorbed by the atom and re-emitted with a different energy with some probability, or possibly even ionize the atom. If the photon is super-well localized and not on a collision course with the atom, it will have a small probability to interact; the trajectory won't be "dragged toward" or "pushed away from" the atom.
Another possible interpretation of this question is whether one can derive an "effective potential" analogous to the Coulomb potential for photon-charged particle interactions. To address this, one should remember that how Coulomb potential can be derived from quantum field theory: one starts with the full relativistic scattering amplitude, takes the non-relativistic limit, and matches with the scattering amplitude with a potential in non-relativistic quantum mechanics. The problem is that there is no way to take the non-relativistic limit for a photon.
One could ask what happens with a massive, neutral particle -- say a neutron, or imagine the photon really did have a tiny mass. In this case, you can go through the above exercise. You will find that neutral particles will have a zero effective potential in the non-relativistic limit.

REPLY [1 votes]: In photonic crystals, there are bandgaps because there is a difference in energy for standing electromagnetic waves, depending on where the nodes are. The energy is lower when the antinodes (regions with stronger electric fields) are in matter with larger $\varepsilon_r$.
So there is an interaction between light and matter. I am a simple experimentalists and would not dare to say even where the photons are in such a case. Quantum electrodynamics can only treat very simple cases. It cannot really deal with solids or even with larger atoms.