text:
# Change font size and style in custom environment

I am using a textbook template from tstextbook. This template has several custom environments (e.g. theorem, exercise etc.) which are outlined in a box and have title styling.

    \newenvironment{boxtheorem}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=black!5,colframe=tsorange]}
{\end{tcolorbox}}

\newtheoremstyle{styletheorem}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{tsorange}}{\;}{0.25em}
{\small\sffamily\color{tsorange}\thmname{#1}
\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3) \newline}}

\theoremstyle{styletheorem}
\newtheorem{envtheorem}{Derivation}[chapter]

\newenvironment{theorem}
{\begin{boxtheorem}\begin{envtheorem}}
{\end{envtheorem}\end{boxtheorem}}

This environment looks something like this:

I want to essentially copy this environment which will serve as a "historical figure" section where a small picture of a person and bio will be displayed. I have found that wrapfigure doesn't work well inside environments, but I think I have found a workaround using minipage.

Now I have the following code for my custom environment:

\newenvironment{boxhistory}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsgrey1!50,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newtheoremstyle{stylehistory}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{black}\thmname{#1}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} #3 \newline}}

\theoremstyle{stylehistory}
\newtheorem{envhistory}{Historical Figure:}[chapter]

\newenvironment{history}
{\begin{boxhistory}\begin{envhistory}}
{\end{envhistory}\end{boxhistory}}

Wrapping the environment in minipage gets the figure wrapping to work:

\begin{history}[Pavel Cherenkov]
\begin{minipage}{\linewidth}
\begin{wrapfigure}{l}{0.28\textwidth}
\centering
\end{wrapfigure}
\footnotesize Cherenkov was born in 1904 to Alexey Cherenkov and Mariya Cherenkova in the small village of Novaya Chigla. This town is in present-day Voronezh Oblast, Russia.

In 1928, he graduated from the Department of Physics and Mathematics of Voronezh State University. In 1930, he took a post as a senior researcher in the Lebedev Physical Institute. That same year he married Maria Putintseva, daughter of A.M. Putintsev, a Professor of Russian Literature. They had a son, Alexey, and a daughter, Yelena.

Cherenkov was promoted to section leader, and in 1940 was awarded the degree of Doctor of Physico-Mathematical Sciences. In 1953, he was confirmed as Professor of Experimental Physics. Starting in 1959, he headed the institute's photo-meson processes laboratory. He remained a professor for fourteen years. In 1970, he became an Academician of the USSR Academy of Sciences.

Cherenkov died in Moscow on January 6, 1990, and was buried in Novodevichy Cemetery.
\end{minipage}
\end{history}

And it looks something like this:

My questions are the following:

1. How can I get the text inside the box to be small, something like \footnotesize but defined in the environment?
2. How can I get natural indentation? I want each paragraph to indent.
3. How can I increase the spacing between the title and first line?

I would like these changes to be defined in the environment if possible. A bonus would be a better way to get word wrap around the figure to work in the environment. Thanks for any suggestions!

Here is the complete code:

%% tstextbook.cls
%% Copyright 2016 Typesetters.se
%
% This work may be distributed and/or modified under the
% conditions of the LaTeX Project Public License, either version 1.3
% of this license or (at your option) any later version.
%   http://www.latex-project.org/lppl.txt
% and version 1.3 or later is part of all distributions of LaTeX
% version 2005/12/01 or later.
%
% This is a free LaTeX text book template from Typesetters.se.
% See tstextbook.tex for an example of how to use this class.

\NeedsTeXFormat{LaTeX2e}
\ProvidesClass{tstextbook}[2017/02/19 Free LaTeX text book template from Typesetters.se]

% Language
%--------------------------------------------------------------------------
%\RequirePackage[swedish]{babel}
\RequirePackage[utf8]{inputenc}

% Geometry
%--------------------------------------------------------------------------
\RequirePackage[top=2.5cm,bottom=2.5cm,left=3cm,right=3cm,
marginparwidth=1.1cm,marginparsep=2mm]{geometry}
%\geometry{asymmetric}

% Fonts
%--------------------------------------------------------------------------
\RequirePackage{kpfonts}

% Colors
%--------------------------------------------------------------------------
\RequirePackage{xcolor}
\definecolor{tssteelblue}{RGB}{70,130,180}
\definecolor{tsorange}{RGB}{255,138,88}
\definecolor{tsblue}{RGB}{23,74,117}
\definecolor{tsforestgreen}{RGB}{21,122,81}
\definecolor{tsyellow}{RGB}{255,185,88}
\definecolor{tsgrey}{RGB}{226,236,217}
\definecolor{tsgrey1}{RGB}{208,206,206}

% Graphics
%--------------------------------------------------------------------------
\RequirePackage{graphicx}
\graphicspath{{images/}}
\RequirePackage{tikz}
\RequirePackage[pages=some]{background}
\backgroundsetup{scale=1,color=black,opacity=1.0,angle=0,
contents={\includegraphics[width=\paperwidth,height=\paperheight]{cover_cloud_chamber}}}
\usepackage[font=small,labelfont=bf]{caption}

% Header and footer styling
%--------------------------------------------------------------------------
\RequirePackage{fancyhdr}
\fancyhf{}
\renewcommand{\footrulewidth}{0pt}

% Section styling
%--------------------------------------------------------------------------
\RequirePackage{sectsty}
\sectionfont{\sf\color{tssteelblue}\sectionrule{0ex}{0pt}{-1ex}{1pt}}
\reversemarginpar
\@mparswitchfalse%

% Chapter styling
%--------------------------------------------------------------------------
\makeatletter

% Numbered chapters (with minitoc)
\begin{tikzpicture}[remember picture,overlay]
\node at (current page.north west)
{
\begin{tikzpicture}[remember picture,overlay]
% Background image
\node[anchor=north west,inner sep=0pt] at (0,0)
\draw[anchor=west, inner sep=0pt] (-0.1cm,-1.5cm) node
[black,
ultra thick,
fill=tsorange!10,
fill opacity=.6,
inner sep=10pt]
(0,0)
{\parbox[t][1.6cm][t]{\paperwidth}
{\huge\bfseries\sffamily\flushleft\thechapter. #1}};
% Chapter contents
\draw[anchor=west,inner sep=0pt] (8.8cm,-7cm) node
[tssteelblue,
ultra thick,
fill=white,
fill opacity=.8,
draw=tssteelblue,
draw opacity=1,
line width=1.5pt,
rounded corners,
inner sep=0pt]
(0,0)
{\hspace{1cm}\parbox[t][][t]{11cm}
{\vspace{-0.5cm}\huge\bfseries\sffamily\flushleft
\textcolor{black}{\sffamily\minitoc}}};
\end{tikzpicture}
};
\end{tikzpicture}
\vspace{7.5cm}
}

% Unnumbered chapters (without minitoc)
\begin{tikzpicture}[remember picture,overlay]
\node at (current page.north west)
{
\begin{tikzpicture}[remember picture,overlay]
% Background image
\node[anchor=north west,inner sep=0pt] at (0,0)
\draw[anchor=west,inner sep=0pt] (-0.1cm,-1.5cm) node
[black,
ultra thick,
fill=tsorange!10,
fill opacity=.6,
inner sep=10pt]
(0,0)
{\parbox[t][1.6cm][t]{\paperwidth}
{\huge\bfseries\sffamily\flushleft #1}};
\end{tikzpicture}
};
\end{tikzpicture}
\vspace{8cm}
}

% Minitoc styling
%--------------------------------------------------------------------------
\RequirePackage{minitoc}
\setcounter{minitocdepth}{1}
\dominitoc[n]
\nomtcrule
\nomtcpagenumbers
\renewcommand{\mtifont}{\large\sffamily}
\renewcommand{\mtcfont}{\small\sffamily}
\renewcommand{\mtcSfont}{\small\sffamily}
\renewcommand{\mtcSSfont}{\small\sffamily}
\renewcommand{\mtcSSSfont}{\small\sffamily}

%--------------------------------------------------------------------------
\RequirePackage{titletoc}
\setcounter{tocdepth}{1}

\titlecontents{chapter}[0cm]
{\color{tssteelblue}\contentslabel[\Large\thecontentslabel]{1.25cm}}{}
{\color{tssteelblue}\normalsize\sffamily\bfseries
\;\titlerule*[.5pc]{.}\;\thecontentspage}

\titlecontents{section}[0cm]
{\contentslabel[\thecontentslabel]{1.25cm}}{}
{\sffamily\hfill\color{black}\thecontentspage}[]

% Front matter
%--------------------------------------------------------------------------
\newcommand{\tsbook}[9]{

\pagestyle{empty}

\def\@tstitle{#1}
\def\@tsauthor{#2}
\def\@tscoverdesign{#3}
\def\@tsyear{#4}
\def\@tsartno{#5}
\def\@tsisbn{#6}
\def\@tsedition{#7}
\def\@tspublished{#8}
\def\@tsprinted{#9}

% Title page
\BgThispage
\begin{center}
\null
\smallskip

\color{white}
\sffamily

%\Huge
%\textsc{\@tstitle}

%\bigskip
%\smallskip

%\LARGE
%\textsc{\@tsauthor} %\\[1em]
\end{center}
%\Large
%\textsc{\@tspublished}
%\end{center}

\pagebreak

\null
\vfill
\begin{flushleft}
Copyright \copyright \ \@tsyear \ \@tsauthor \\
\vspace{5mm}
\textbf{Copying prohibited} \\
\vspace{5mm}
All rights reserved. No part of this publication may be reproduced
or transmitted in any form or by any means, electronic or
mechanical, including photocopying and recording, or by any
information storage or retrieval system, without the prior written
permission of the author.\\
%\vspace{5mm}
%Art. No \@tsartno \\
%ISBN \@tsisbn \\
%Edition \@tsedition \\
\vspace{5mm}
Cover image by \@tscoverdesign \\
%Printed in \@tsprinted
\end{flushleft}

\tableofcontents

\cleardoublepage
\pagestyle{fancy}
\thispagestyle{empty}
}

% Boxes for theorems
%--------------------------------------------------------------------------
\newenvironment{boxtheorem}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=black!5,colframe=tsorange]}
{\end{tcolorbox}}

\newenvironment{boxdefinition}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=yellow!10!white,boxrule=0pt,frame hidden,
borderline west={1.5mm}{-2mm}{tsforestgreen}]}
{\end{tcolorbox}}

{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsyellow!20,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newenvironment{boxquestion}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsorange!25,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newenvironment{boxhomework}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsblue!10,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newenvironment{boxexample}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsforestgreen!10,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newenvironment{boxsolution}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsforestgreen!10,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

\newenvironment{boxhistory}
{\begin{tcolorbox}
[enhanced jigsaw,breakable,pad at break*=1mm,
colback=tsgrey1!50,boxrule=0pt,frame hidden]}
{\end{tcolorbox}}

% Styling of theorems
%--------------------------------------------------------------------------
\RequirePackage{amsthm}
\usepackage{enumitem}

\newtheoremstyle{styletheorem}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{tsorange}}{\;}{0.25em}
{\small\sffamily\color{tsorange}\thmname{#1}
\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3) \newline}}

\newtheoremstyle{styledefinition}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{tsforestgreen}\thmname{#1}
\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3) \newline}}

\newtheoremstyle{styleexercise}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{black}\thmname{#1}
\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3) \newline}}

\newtheoremstyle{styleexercise1}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{black}\thmname{#1}
%\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3) \newline}}

\newtheoremstyle{styleexercise2}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{black}\thmname{#1}
%\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} (#3)}}

\newtheoremstyle{stylehistory}
{0pt}{0pt}{\normalfont}{0pt}{\small\bf\sffamily\color{black}}{\;}{0.25em}
{\small\sffamily\color{black}\thmname{#1}
%\nobreakspace\thmnumber{\@ifnotempty{#1}{}\@upn{#2}}
\thmnote{\nobreakspace\the\thm@notefont\sffamily\bfseries\color{black} #3 \newline}}

\theoremstyle{styledefinition}
\newtheorem{envdefinition}{Definition}[chapter]
\theoremstyle{styletheorem}
\newtheorem{envtheorem}{Derivation}[chapter]
\newtheorem{envlemma}{Lemma}[chapter]
\theoremstyle{styleexercise1}
\newtheorem{envquestion}{Question:}[chapter]
\theoremstyle{styleexercise}
\newtheorem{envhomework}{Homework Exercise}[chapter]
\newtheorem{envexample}{Example}[chapter]
\theoremstyle{stylehistory}
\newtheorem{envhistory}{Historical Figure:}[chapter]

% Environments for theorems
%--------------------------------------------------------------------------
\newenvironment{definition}
{\begin{boxdefinition}\begin{envdefinition}}
{\end{envdefinition}\end{boxdefinition}}

\newenvironment{theorem}
{\begin{boxtheorem}\begin{envtheorem}}
{\end{envtheorem}\end{boxtheorem}}

\newenvironment{corollary}
{\begin{boxtheorem}\begin{envtheorem}}
{\end{envtheorem}\end{boxtheorem}}

\newenvironment{lemma}
{\begin{boxtheorem}\begin{envlemma}}
{\end{envlemma}\end{boxtheorem}}

\newenvironment{question}
{\begin{boxquestion}\begin{envquestion}}
{\end{envquestion}\end{boxquestion}}

\newenvironment{homework}
{\begin{boxhomework}\begin{envhomework}}
{\end{envhomework}\end{boxhomework}}

\newenvironment{example}
{\begin{boxexample}\begin{envexample}}
{\end{envexample}\end{boxexample}}

\newenvironment{solution}
{\begin{boxsolution}\begin{envsolution}}
{\end{envsolution}\end{boxsolution}}

\newenvironment{history}
{\begin{boxhistory}\begin{envhistory}}
{\end{envhistory}\end{boxhistory}}

% Other environments
%--------------------------------------------------------------------------
\newenvironment{summary}
{\hspace{10pt}\par\small\it
\begin{list}{}{\leftmargin=35pt\rightmargin=25pt}
{\end{list}\vspace{-0.5cm}}

\newenvironment{remark}
{\vspace{0.5cm}\noindent\small\it
\marginpar{\vspace{-6mm}\includegraphics[width=1.0cm]{idea}}}
{\vspace{0.5cm}}

% Bibliography
%--------------------------------------------------------------------------
\RequirePackage[backend=biber]{biblatex}

% Index
%--------------------------------------------------------------------------
\RequirePackage{makeidx}
\makeindex

% Packages for mathematics and algorithms
%--------------------------------------------------------------------------
\RequirePackage{algorithm}
\RequirePackage{algpseudocode}
\RequirePackage{xfrac}
\RequirePackage{mathtools}
\RequirePackage{wasysym}
\RequirePackage{siunitx}
\newcommand\bmmax{2}
\RequirePackage{bm}
\usepackage{cases}
\usepackage{wrapfig}
\usepackage{cutwin}

% Other useful packages
%--------------------------------------------------------------------------
\RequirePackage{blindtext}
\RequirePackage[most]{tcolorbox}
\RequirePackage{booktabs}
\RequirePackage{url}

\documentclass{tstextbook}
\begin{document}

A charged particle moving with uniform velocity can NOT generate radiation in a vacuum.  However, when a charged particle moves in a dielectric medium with refractive index $\textit{n}$, it becomes possible to irradiate!  This was indeed discovered experimentally by the Russian physicist, Pavel Cherenkov in 1934, and later explained theoretically by Ilya Tamm and Igor Frank in 1937. They received the Nobel Prize for this discovery in 1958.
\begin{history}[Pavel Cherenkov]
\begin{minipage}{\linewidth}
\begin{wrapfigure}{l}{0.28\textwidth}
\centering
\end{wrapfigure}
Cherenkov was born in 1904 to Alexey Cherenkov and Mariya Cherenkova in the small village of Novaya Chigla. This town is in present-day Voronezh Oblast, Russia.

In 1928, he graduated from the Department of Physics and Mathematics of Voronezh State University. In 1930, he took a post as a senior researcher in the Lebedev Physical Institute. That same year he married Maria Putintseva, daughter of A.M. Putintsev, a Professor of Russian Literature. They had a son, Alexey, and a daughter, Yelena.

Cherenkov was promoted to section leader, and in 1940 was awarded the degree of Doctor of Physico-Mathematical Sciences. In 1953, he was confirmed as Professor of Experimental Physics. Starting in 1959, he headed the institute's photo-meson processes laboratory. He remained a professor for fourteen years. In 1970, he became an Academician of the USSR Academy of Sciences.

Cherenkov died in Moscow on January 6, 1990, and was buried in Novodevichy Cemetery.
\end{minipage}
\end{history}
\end{document}

• Welcome to TeX.SE! Can you please make your code snippet be compilable? Then we do not have to guess what you are doing ... Oct 13 at 22:59
• @Mensch sorry, I added the complete code! Oct 13 at 23:09


text:
# Completing the Square when a ≠ 1

A quadratic equation is an equation that contains a squared variable as its highest power on any variable. The general form of a quadratic equation is:

ax2 + bx + c = 0

Where a, b, and c are constants and a ≠ 0. In other words there must be a x2 term.

Some examples are:

x2 + 3x - 3 = 0
4x2 + 9 = 0 (Where b = 0)
x2 + 5x = 0 (where c = 0)

One way to solve a quadratic equation is by completing the square.

ax2 + bx + c = 0 → (x- r)2 = S

Where r and s are constants.

PART I of this topic focused on completing the square when a, the x2-coefficient, is 1. This part, PART II, will focus on completing the square when a, the x2-coefficient, is not 1.

Let's solve the following equation by completing the square:

2x2 + 8x - 5 = 0
 Step 1: Write the equation in the general form ax2 + bx + c = 0. This equation is already in the proper form where a = 2 and c = -5. 2x2 + 8x - 5 = 0 Step 2: Move c, the constant term, to the right-hand side of the equation. c = -5 2x2 + 8x = 5 Step 3: Factor out a from the left-hand side. This changes the value of the x-coefficient. a = 2 2(x2 + 4x) = 5 Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation. The expression is x2 + 4x. Divide the x-coefficient by two and square the result. x2 + 4x x-coefficient = 4 $\frac{4}{2}={2}\to {r}$ (2)2 = 4 Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the left-hand side the total value added is a x result. So this value must also be added to the right-hand side. 2(x2 + 4x + 4) = 5 + 2(4) Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the x-coefficient of the parenthetical expression divided by 2 as found in Step 4. 2(x + 2)2 = 13 Now that the square has been completed, solve for x. Step 7: Divide both sides by a. ${\left(x+2\right)}^{2}=\frac{13}{{2}}$ Step 8: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative. $x+2=±\sqrt{\frac{13}{2}}$ Step 9: Solve for x. $x=-2±\sqrt{\frac{13}{2}}$
Example 1:      3x2 = 6x + 7
 Step 1: Write the equation in the general form ax2 + bx + c = 0. Where a = 3 and c = -7. 3x2 - 6x - 7 = 0 Step 2: Move c, the constant term, to the right-hand side of the equation. c = -7 3x2 - 6x = 7 Step 3: Factor out a from the left-hand side. This changes the value of the x -coefficient. a = 3 3(x2 - 2x) = 7 Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation. The expression is x2 - 2x. Divide the x-coefficient by two and square the result. x2 - 2x x -coefficient = -2 $\frac{-2}{2}={-}{1}\to {r}$ (-1)2 = 1 Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the left-hand side the total value added is a x result. So this value must also be added to the right-hand side. 3(x2 - 2x + 1) = 7 + 3(1) Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the x-coefficient of the parenthetical expression divided by 2, as found in Step 4. 3(x - 1)2 = 10 Now that the square has been completed, solve for x. Step 7: Divide both sides by a. ${\left(x-1\right)}^{2}=\frac{10}{{3}}$ Step 8: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative. $x-1=±\sqrt{\frac{10}{3}}$ Step 9: Solve for x. $x=1±\sqrt{\frac{10}{3}}$
Example 2:      5x2 - 0.6 = 4x
 Step 1: Write the equation in the general form ax2 + bx + c = 0. Where a = 5 and c = 0.6. 5x2 - 4x - 0.6 = 0 Step 2: Move c, the constant term, to the right-hand side of the equation. c = -0.6 5x2 - 4x = 0.6 Step 3: Factor out a from the left-hand side. This changes the value of the x-coefficient. a = 5 5(x2 - 0.8x) = 0.6 Step 4: Complete the square of the expression in parentheses on the left-hand side of the equation. The expression is x2 - 0.8x. Divide the x-coefficient by two and square the result. x2 - 0.8x x-coefficient = -0.8 $\frac{-0.8}{2}={-}{0.4}\to {r}$ (-0.4)2 = 0.16 Step 5: Add the result from Step 4 to the parenthetical expression on the left-hand side. Then add a x result to the right-hand side. To keep the equation true what is done to one side must also be done to the other. When adding the result to the parenthetical expression on the left-hand side the total value added is a x result. So this value must also be added to the right-hand side. 5(x2 - 0.8x + 0.16) = 0.6 + 5(0.16) Step 6: Rewrite the left-hand side as a perfect square and simplify the right-hand side. When rewriting in perfect square format the value in the parentheses is the x-coefficient of the parenthetical expression divided by 2 as found in Step 4. 5(x - 0.4)2 = 1.4 Now that the square has been completed, solve for x. Step 7: Divide both sides by a. ${\left(x-0.4\right)}^{2}=\frac{1.4}{{5}}=0.28$ Step 8: Take the square root of both sides of the equation. Remember that when taking the square root on the right-hand side the answer can be positive or negative. $x-0.4=±\sqrt{0.28}$ Step 9: Solve for x. $x=0.4±\sqrt{0.28}$

 Related Links: Math algebra Factoring Quadratic Equations when a equals 1 Factoring Quadratic Equations when a ≠ 1 Algebra Topics

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### 26 2 Electromagnetic Wave Velocity Conceptual Academy

Peruse the Table of Videos to explore our video library as aligned to the Conceptual Physics textbook To the Student: You'll need a Course ID from your instructor to register After signing in you'll be brought to your profile page

### Electromagnetic Therapy Miami Floatation

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### (pdf) Parabolic Equation Modeling Of Electromagnetic Wave

The parabolic equation is an efficient numerical solution for electromagnetic wave propagation In order to address the difficulties in predicting electromagnetic wave propagation in the maritime

### Electromagnetic Energy Red Hook Central Schools

Electromagnetic Energy Energy Energy: The ability to do work Earth has two sources of heat energy: External source: The energy provided by the sun Internal source: The decay of radioactive substances and compression inside the Earth Electromagnetic Energy Electromagnetic Energy: a type of energy that is given off in the form of transverse waves from all matter not at absolute zero How

### Electromagnetic Floater : 6 Steps (with Pictures

Electromagnetic Floater: This Instructable will show you how to make a device that can float almost any object with a magnet in it It is much like the floating globes you can buy except it works by balancing the forces of permanent magnets with electromagnets rather th

### Lightweight Electromagnetic Interference Shielding

Motivated by the increase of stress over electromagnetic pollution issues arising from the fast growing development and need for electronic and electrical devices the demand for materials with high electromagnetic interference (EMI) shielding performance has become urgently Considering the energy consumption in real applications lightweight EMI shielding materials has been attentive in

### 7 2 Electromagnetic Radiation Geog 160: Mapping Our

Most remote sensing instruments measure the same thing: electromagnetic radiation Electromagnetic radiation is a form of energy emitted by all matter above absolute zero temperature (0 Kelvin or 273 Celsius) X rays ultraviolet rays visible light infrared light heat microwaves and radio and television waves are all examples of electromagnetic energy

### Full Wave Vector Electromagnetic Scattering From Two

A solution to scattering from two dimensional (2D) surfaces with arbitrary roughness based on the extended boundary condition method (EBCM) is presented The bistatic scattering cross section (SCS) is obtained from the superposition of the upgoing Floquet modes This method places no limitation on the roughness scale of the surface and produces both co pol and cross pol components of fields

### Citeseerx — Monte Carlo Simulation Of Electromagnetic

CiteSeerX Document Details (Isaac Councill Lee Giles Pradeep Teregowda): The fast multipole method fast Fourier transform (FMM FFT) method is developed to compute the scattering of an electromagnetic wave from a two dimensional rough surface The resulting algorithm computes a matrixvector multiply in O(NlogN) operations This algorithm is shown to be efficient than

### The Electromagnetic Spectrum Reflection Absorption And

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### 13 6 Flotation Conceptual Academy

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### Monte Carlo Simulation Of Electromagnetic Scattering From

Abstract: The fast multipole method fast Fourier transform (FMM FFT) method is developed to compute the scattering of an electromagnetic wave from a two dimensional (2 D) rough surface The resulting algorithm computes a matrix vector multiply in O(N log N) operations This algorithm is shown to be efficient than another O(N log N) algorithm the multilevel fast multipole algorithm (MLFMA

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### Monte Carlo Simulation Of Electromagnetic Scattering From

For the rougher surface backscattering enhancement is clearly observable as a pronounced peak in the backscattering direction of the computed bistatic scattering coefficient For the smoother surface the Monte Carlo results compare well with the results of the approximate Kirchhoff theory KW Electromagnetic scattering KW Numerical analysis

### Published June 15 2012 Flotation Drying Of Coated Grades

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text:
# What additional conditions are there to make POVM measurement same as projective measurement?

We know that POVMs are applied in the more general cases where the system is not necessarily closed. So mathematically, how does going from open to closed system change the scenario in case of POVM so that it becomes a projective measurement?

• I don't understand the question. What do you mean exactly with "how does going from open to closed change the scenario"?
– glS
Feb 16 at 9:25
• @glS.: We can't apply projective measurement when the system is not closed. In an open system the evolution of a state can't be described by unitary operation. We often hear that a general measurement will be projective measurement if it follows the additional condition that $$M_{m}^\daggerM_{m'}= \delta_{m,m'}M_{m}$$. I want to know if there is some relation holds between these two. Feb 16 at 11:51
• Are you aware of the concept of a Naimark dilation? Feb 16 at 11:59
• @Rammus.: Heard first time in my life a min ago from you. Feb 16 at 12:01
• sure you can do projective measurements when the system is not closed. The system being "open" or "closed" refers to its dynamics, i.e. how it evolves. That doesn't affect the types of measurements you can perform on the state itself. About the relation between POVMs and projective measurements, have you seen quantumcomputing.stackexchange.com/q/7004/55? See also quantumcomputing.stackexchange.com/q/12275/55
– glS
Feb 16 at 12:26

An $$m$$-outcome positive-operator value measure or POVM consists of $$m$$ positive operators $$A_1,\ldots,A_m$$ on $$\mathbb{C}^n$$ such that $$\sum_{i=1}^m A_i=I.$$ Each $$A_i$$ is positive iff $$A_i=B_i^*B_i$$ for some operator $$B_i$$ on $$\mathbb{C}^n$$, thus each POVM is a projection-valued measure or PVM iff each $$B_i$$ is an orthogonal projection, if so then $$B_i=B_i^*=B_i^2$$ and $$B_1,\ldots,B_m$$ is a PVM since $$\sum_{i=1}^n B_i=\sum_{i=1}^n B_i^2=\sum_{i=1}^n B_i^*B_i=\sum_{i=1}^n A_i=I.$$ As mentioned in the comments, Naimarks dilation theorem shows that for any POVM on $$\mathbb{C}^n$$ there is an isometry $$V:\mathbb{C}^n \rightarrow \mathbb{C}^{N}$$ where $$N\geq n$$ and each positive operator $$A_i$$ gets mapped to an orthogonal projection $$\widetilde{A_i}$$ in the larger space where the "completeness relation" of summing to the identity still holds, thus turning any POVM into a PVM on the larger space.


text:
# Simple Partial Differentiation

1. Jan 6, 2010

### Chewy0087

1. The problem statement, all variables and given/known data
Hey >.< another simple problem;

find $$\frac{\partial u}{\partial y} , \frac{\partial v}{\partial y}, \frac{\partial w}{\partial y}$$

given that

u - v + 2w = x + 2z
2u + v - 2w = 2x - 2z
u - v + w = z - y

3. The attempt at a solution

I don't know, this question is really bugging me, i understand that the whole idea of partial differentiation is only to have to deal with the change of one variable, so why are you given the first two equations? if you assume that everything apart from y & u are constant then of course

$$\frac{\partial u}{\partial y} = -1$$

however this isn't the answer, i thought maybe you can't assume that they're constant, because u is dependant on them, however this becomes an insanely hard/impossible simultaneous equation if you try to solve for all of their changes!

could someone give me a hint? i think i'm missing the point of this question (done all of the other questions in the exercise so far)

and of course, once again thank you for help you give me! =D

p.s answers are 0, 2 & 1

Last edited: Jan 6, 2010
2. Jan 6, 2010

### Dick

Take d/dy of all three equations. You'll get a pretty simple set of three equations in the three unknowns that you are looking for. Since it looks like x,y and z are your independent variables, dx/dy=dz/dy=0.

3. Jan 6, 2010

### Chewy0087

Hm, but how do you know that x, y and z are the independant variables, just because they ask for u, v and w's change with respect to y? Also i suppose that that means I CAN assume like you said that dx/dy = dz/dy = 0, but can't assume that dw/dy = 0?

Why can i simply choose to do that? =P (sorry if that's a long explainy question, if it is just forget it and i'll try to find a link)

and thanks alot sir, the help is invaluable (trying to self-teach this, you can tell how it's going xP)

4. Jan 6, 2010

### mugaliens

http://en.wikipedia.org/wiki/Partial_differential" [Broken]may help.

If not, consider this: Partial derivatives force all variables except one to be constants at some determined point in order to see the behavioral (derivative aka slope) result of the variable in question. However, instead of assigning constants to the other variables, partials continue to solve for the variable in question mathematically, so as to provide for a derivative solution at all values of the other variables.

I'm not a mathematician, though, so I hope one comes along, soon...

Last edited by a moderator: May 4, 2017
5. Jan 6, 2010

### Dick

It is sort of implied. It should have been explicitly stated. If you are taking partial derivatives with respect to y, that implies there are other independent variables around. Since x,y,z are consecutive letters in the alphabet I would assume they are the independent variables and that u,v and w are to be regarded as functions of x,y and z. I.e. u(x,y,z), v(x,y,z) and w(x,y,z). As I said, I think they really should have stated that if they didn't. I'm just reading between the lines.

6. Jan 6, 2010

### Chewy0087

I see, it's just common sense really, i guess if i had more experience answering these questions it would have been more obvious to me...much obliged Dick (i managed to get the answer out)! =D ps i voted for you in the polls a couple of weeks ago :D

and i have read that link...it's not VERY helpful but i agree I need to read more into the theory, i'm just learning to compute right now (not very good)


text:
PIRSA:13100074

# Bootstrapping the O(N) Vector Models

### APA

Simmons-Duffin, D. (2013). Bootstrapping the O(N) Vector Models. Perimeter Institute. https://pirsa.org/13100074

### MLA

Simmons-Duffin, David. Bootstrapping the O(N) Vector Models. Perimeter Institute, Oct. 15, 2013, https://pirsa.org/13100074

### BibTex

          @misc{ pirsa_13100074,
doi = {10.48660/13100074},
url = {https://pirsa.org/13100074},
author = {Simmons-Duffin, David},
keywords = {Quantum Fields and Strings},
language = {en},
title = {Bootstrapping the O(N) Vector Models},
publisher = {Perimeter Institute},
year = {2013},
month = {oct},
note = {PIRSA:13100074 see, \url{https://pirsa.org}}
}

David Simmons-Duffin Institute for Advanced Study (IAS)

Collection
Talk Type
Subject

## Abstract

We study the conformal bootstrap for 3D CFTs with O(N) global symmetry. We obtain rigorous upper bounds on the scaling dimensions of the first O(N) singlet and symmetric tensor operators appearing in the \phi_i x \phi_j OPE, where \phi_i is a fundamental of O(N). Comparing these bounds to previous determinations of critical exponents in the O(N) vector models, we find strong numerical evidence that the O(N) vector models saturate the bootstrap constraints at all values of N. We also compute general lower bounds on the central charge, giving numerical predictions for the values realized in the O(N) vector models. We compare our predictions to previous computations in the 1/N expansion, finding precise agreement at large values of N.


text:
### Sidebar

wiki_manual:create_a_wiki_page

# Create a Wiki page

Hereinafter you can find a collection of useful information on how to create content in VieVS-Wiki.

## Permissions and users' rights

Basically, the content of VieVS-Wiki is publicly available, without registration or logging on - everyone has read access.

To created or modify content in VieVS-Wiki, you need extended users' rights. Only registered VieVS-users get the right to create or modify pages. For more information on how to get VieVS and to become a VieVS-user follow the link: Get VieVS.

If you have already become a VieVS user, feel free to register to VieVS-Wiki. You find the Register button in the top right corner of this page. You will get a username and a password to logging in and the site admin will give you the rights to create and modify pages.

If there are any questions and/or suggestions, please use the contact form to contact the VieVS-Wiki admin.

## Wiki Manual

### General DokuWiki manual

The official DokuWiki Manual is provided here: DokuWiki Manual.

If you intend to use VieVS-Wiki (at least if you intend to create/modify pages), you should have a look at it!

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### Formatting Syntax

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### Formatting mathematical expressions

VieVS-Wiki supports TeX(LaTeX)-syntax for mathematical expressions. For more information on the mathematics-syntax in TeX we refer to respective (online-)resources (e.g. Wiki books).

The TeX mathematics support is provided by a DokuWiki plugin called MathJax.

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This is text is indented by two spaces.
<code>
First line:  This is preformatted code all spaces are preserved: like    <-this
Second line: This is preformatted code all spaces are preserved: like    <-this
</code>

⇒ Results in:

First line:  This is preformatted code all spaces are preserved: like    <-this
Second line: This is preformatted code all spaces are preserved: like    <-this
<file>
This is pretty much the same, but you could use it to show that you quoted a file.
</file>

⇒ Results in:

This is pretty much the same, but you could use it to show that you quoted a file.

### Pagenames, namespaces and filenames

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##### Publically available content

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screenshot_gui_scheduling.png

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Example: Link to the headline History and technological developments on the page public:vlbi_fundamantals:introduction

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Appearance on the wiki page:

### Create a new Wiki page

To create a new Wiki page, just create an empty link to the non-existing page. The empty link appears in red color. If you click an the empty link, you will reach a new empty wiki page. Klick the Create page button on the right sidebar to finally create this page. Now the new page is ready to be edited and filled with contend.

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== Ray, 1999 ==
Ray, R.D. (1999). A global ocean tide model From TOPEX/Poseidon altimetry:
GOT99.2. NASA Technical Memorandum TM-209478, 58.

By means of the headline == Ray, 1999 == you are able to cite from the bibliography entry by using the syntax for internal links to particular headlines:

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## Installed Plugins

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• MathJax: To be able to use LaTeX Synthax for math. expressions.
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text:
# Why would thermal imaging be used to locate the Chandrayaan-2 lander?

Tweets and news suggest that Chandrayaan-2 lander may have been located, using thermal imaging by the Chandrayaan-2 orbiter.

• K Sivan, chief of the Indian Space Research Organisation, said today (Sept. 8) that the Vikram lander was located by Chandrayaan-2 and efforts to restore contact the probe will continue for at least 14 days, according to a Times of India report.

"We have found the location of Lander Vikram on [the] lunar surface and Orbiter has clicked a thermal image of Lander," Sivan told the ANI news service in an interview, adding that attempts to communicate with the lander are ongoing.

• n an exclusive interview, ISRO Chairman K. Sivan told India Today TV that the Chandrayaan 2 orbiter had sent back a thermal image of the location of the Vikram lander, and they're now attempting to contact it.

• Asian News International: tweet

Indian Space Research Organisation (ISRO) Chief, K Sivan to ANI:We've found the location of #VikramLander on lunar surface&orbiter has clicked a thermal image of Lander. But there is no communication yet. We are trying to have contact. It will be communicated soon. #Chandrayaan2

• K. Sivan, the director of the Indian Space Research Organization, told national news outlets that a thermal image had been taken by the Chandrayaan-2 mission’s orbiter. He said it was still unclear whether the lander was damaged, though he expected it had experienced a “hard landing.”

“We are trying to establish a contact,” he was quoted as saying by Asian News International.

The thermal image from the orbiter has not been released publicly. A spokesman for the space agency did not respond to requests for comment on Sunday.

Question: Why a thermal image? Is there expected to be better contrast than in visible light? Does Chandrayaan-2 not have a visible light camera? Wouldn't the diffraction-limited resolution at thermal infrared (5-20 microns) be much worse than in the visible (0.4-0.7 microns)?

• the ANI report (non tweet): aninews.in/news/national/general-news/… – amI Sep 9 '19 at 15:55
• While it appears to be a false report it does make sense. The lander would have a bunch of fuel on board, rockets have a considerable tendency to fireball if they break up. If it went down in a shady spot I would expect that spot to be warmer for a bit. – Loren Pechtel Sep 10 '19 at 2:53

It was not a thermal image at all.

It is an optical image that has been captured by Orbiter of the lander spot and not thermal image as reported by others media houses. OHRC is same like our human eye. Since it consists of only one spectral band in visible region, so Image will be of gray scale, Not color image.

ISRO Sources confirms me that "Lander found 500m away from the actual landing spot. Seen from OHRC (Optical High Resolution Camera) image and it is intact. Trying to send commands from Bangalore control center." @ABPNews

There are many fake images going viral claiming to be of "Lander Spot on Lunar surface" clicked by Orbiter. FYI, No such image has been released by @isro so far.

It is noticed that accounts in the name of Kailasavadivoo Sivan is operational on many Social media. This is to clarify that Dr. K Sivan, Chairman, ISRO does not have any personal accounts.

For official accounts of ISRO, please see https://bit.ly/2lHe9GT

There are some articles trying to "sell" what looks like a thermal image of the lander (I would prefer not to link it), but in one case the image shows almost half of the Moon, so has no resolution to represent what it says it does, and a second one is some kind of closeup (if real at all), possibly made just after separation of the lander.

• Well that certainly explains a lot! – uhoh Sep 9 '19 at 13:36
• I've added a supplementary answer. Thanks for all the great links! – uhoh Sep 9 '19 at 13:50

As is pointed out there, there seems to be an abundance of false social media accounts claiming to be K. Silvan.

From today's Sep 09, 2019 post: Regarding Social Media Accounts in the name of Kailasavadivoo Sivan, Chairman ISRO:

It has been noted that social media accounts in the name of Kailasavadivoo Sivan (along with photographs of K. Sivan) is operational and active on social media platforms. This is to clarify that K. Sivan, Chairman, ISRO does not have any personal account on any social media platforms. Hence all the information on all such accounts are not authentic.

The official account of ISRO on all social media platforms are as follows:

According to the latest Sep 07, 2019 update: https://www.isro.gov.in/update/07-sep-2019/chandrayaan-2-latest-update Chandrayaan-2 has an extremely nice visible light telescope with a 30 centimeter aperture! At 0.5 microns wavelength, $$1.22 \lambda / d$$ would provide a resolution of about 0.4 arcseconds. At a minimum distance of 100 km that would be a resolution of about 20 cm.

Chandrayaan-2 mission was a highly complex mission, which represented a significant technological leap compared to the previous missions of ISRO, which brought together an Orbiter, Lander and Rover to explore the unexplored south pole of the Moon. Since the launch of Chandrayaan-2 on July 22, 2019, not only India but the whole world watched its progress from one phase to the next with great expectations and excitement. This was a unique mission which aimed at studying not just one area of the Moon but all the areas combining the exosphere, the surface as well as the sub-surface of the moon in a single mission. The Orbiter has already been placed in its intended orbit around the Moon and shall enrich our understanding of the moon’s evolution and mapping of the minerals and water molecules in the Polar Regions, using its eight state-of-the-art scientific instruments. The Orbiter camera is the highest resolution camera (0.3m) in any lunar mission so far and shall provide high resolution images which will be immensely useful to the global scientific community. The precise launch and mission management has ensured a long life of almost 7 years instead of the planned one year. The Vikram Lander followed the planned descent trajectory from its orbit of 35 km to just below 2 km above the surface. All the systems and sensors of the Lander functioned excellently until this point and proved many new technologies such as variable thrust propulsion technology used in the Lander. The success criteria was defined for each and every phase of the mission and till date 90 to 95% of the mission objectives have been accomplished and will continue contribute to Lunar science , notwithstanding the loss of communication with the Lander.

It's unusual for the New York Times to get things wrong, but reading carefully they are quoting other news outlets, not K. Silvan directly.


text:
# Ethnicity

Share of ethnicity XY is 26%, which is 1/9 more than in the prewar period.

What was the share of that ethnicity in the pre-war period?

Result

p =  23.4 %

#### Solution:

$26/100 \cdot x = (1+1/9)p/100 \cdot x \ \\ 26= (1+1/9)p \ \\ p = \dfrac{ 26 }{ 1+1/9} \ \\ p = 23.4 \%$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Showing 0 comments:
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text:
# Principled bsdf Subsurface is causing weird white dots in cycles render mode

As soon as I enable the Subsurface in the Principled BSDF, it's just painting a lot of mostly white, small dots on the mesh, no matter what color I use as subsurface color. I tried changing all the parameters but it didn't help. It works just fine with Eevee for some reason but not with Cycles. The mesh is made out of a nurbs path and modelled to look like hair. Any help is appreciated. Edit: Im using a subsurface radius of 0.5

• Hello ! Could you add a screenshot of the problem ? Remember that subsurface values are supposed to be in meters (or blender units) so you should use very low values to get a realistic effect. Make sure your object is to-scale, too. Mar 1 at 22:24
• @Gorgious Hi, thanks for your help. I have added two pictures; one with subsurface and the other without. Additionally, I have included a picture of the Principled BSDF. Im using a subsurface radius of 0.5.
– Mash
Mar 2 at 16:34
• What do you mean Im using a subsurface radius of 0.5? You put 0.5 on all three channels? Default is 1:0.2:0.1. Also the SS value is more often around 0.03, definitely never 0.1, that's ten times too much. Why do you use SSS for hair ? Mar 2 at 17:18
• @Gorgious Well, I guess I don't really have to use it, but I'm just wondering why it's not working. Changing the values isn't doing anything noticeable.
– Mash
Mar 2 at 22:29
• Well my first comment still stands. A real hair strand is fractions of a millimiter in diameter. here you are setting a subsurface value of 100 millimeters ! You should read a bit on how subsurface scattering works Mar 3 at 7:25


text:
An introduction to equal temperament tuning.

You have been given a (theoretical) keyboard. Inconveniently, this keyboard has not been tuned. You’d like to play it, and therefore decide to tune it yourself. Luckily, the mechanics of tuning this keyboard are easy… you just need to work out which frequency to assign to each note…

## The octave doubling rule

You try to recollect any relevant info you might have picked up over the years.. and remember the following:

A note an octave up has double the frequency.

In other words, if note $C5$ has frequency $f$, then note $C6$ will have frequency $2f$. As far as you remember, this should apply to all the notes on the keyboard.

It follows from this that the note an octave below must have half the frequency. So $C4$ would have a frequency of $f/2$.

Great! But where to go from here?..

## The first frequency

You allow your eyes to wander up and down the $88$ key (theoretical) keyboard…

You realise that you could pick a relatively middling key, and assign to this a relatively middling frequency.

You choose $A4$ to be your middling key. You then play around with a tone generator, and decide that you perceive $\sim 440Hz$ to be a fairly middling frequency.

You turn on your keyboard, and program in $freq(A4) = 440Hz$.

Great! You have now tuned one note!

### All the As

Now that you have decided on a frequency for $A4$, you realise that you can work out the frequencies of all the other $A$s, using the octave doubling rule:

$A4 \leftarrow 440 Hz$

$A5 \leftarrow 2*440 Hz$

$A6 \leftarrow 4*440 Hz$

And:

$A3 \leftarrow 440/2 Hz$

$A2 \leftarrow 440/4 Hz$

And so on.

Great! Now you have tuned all of your $A$s!

### Join the dots

You have all the $A$s tuned, but you’re not sure what to do about the other notes.

You wonder if a graph might help, so you painstakingly draw out the following:

Looking at this graph, you realise that drawing a curve through the points would assign a frequency to each key on the keyboard. However, drawing a curve looks somewhat difficult..

On a whim, you decide to plot the points again, but this time using logarithmic graph paper:

Aha! Now you can use a ruler to draw a straight line through the points:

If you want to find the frequency of a certain key, you can read it off the graph!

You wonder if the octave doubling rule still applies to any note, or just to the $A$s. You read a few values from the graph, and it seems like it does apply to all notes! “Hurrah!” you shout.

Great! By reading from the graph, you can now tune all the notes on your piano!

## Finding the expression

Unfortunately, you don’t enjoy reading values from the graph… it’s slow and imprecise.

“If only I had a mathematical expression for this curve”, you think to yourself, “then I could accurately compute the frequencies for this keyboard!”…

You write out the frequencies of the $A$ keys on your keyboard:

$A$ key number ($n_a$) frequency ($freq$) in $Hz$
1 55
2 110
3 220
4 440
5 880
6 1760
7 3520

After some time spent looking at the sequence, you notice that $880$ is $440 * 2^1$, that $1760$ is $440 * 2^2$, that $3520$ is $440 * 2^3$, and so on. You realise that you can write an expression for frequency in terms of $n_a$:

$freq = 440 * 2^{n_a-4}$

You put this into a table:

$A$ key number ($n_a$) frequency ($freq$) in $Hz$ expression
1 55 $440 * 2^{-3}$
2 110 $440 * 2^{-2}$
3 220 $440 * 2^{-1}$
4 440 $440 * 2^0$
5 880 $440 * 2^1$
6 1760 $440 * 2^2$
7 3520 $440 * 2^3$

You now have an expression for the $A$s, but what about the keys in between the $A$s? You wonder if you can modify the numbering $n_a$, to allow for other keys… recalling that $A4$ is normally labelled as key $49$ on the keyboard, $A5$ is $61$, etc.. You write a table, showing how $n_a$ maps to the standard key number $n$:

 $A$ key number ($n_a$) standard key number ($n$) 1 13 2 25 3 37 4 49 5 61 6 73 7 85

You see that you can express $n$ in terms of $n_a$:

$n = n_a*12 + 1$

You then re-arrange for $n_a$:

$n_a = (n - 1) / 12$

You realise you can sub this into your expression for freq:

$freq = 440 * 2^{n_a-4}$

$freq = 440 * 2^{(n - 1)/12 -4}$

$freq = 440 * 2^{(n - 1 - 4*12)/12 }$

$freq = 440 * 2^{(n - 1 - 48)/12 }$

$freq = 440 * 2^{(n - 49)/12}$

Using this expression, you compute the frequency for all $88$ keys on your keyboard accurately and conveniently.

You celebrate because you have successfully tuned your keyboard.

## Summary

You started with the octave doubling rule.

You then selected a key on the keyboard, and assigned it a frequency (a middling frequency for middling key).

This allowed you to fit a curve which gave a sensible tuning for the whole keyboard.

You then found a mathematical expression for this curve.

The tuning system you created is called twelve-tone equal temperament [1]. It is “the most common tuning system since the 18th century”, but not the only tuning system available [2].

The tuning of real acoustic pianos is generally based on twelve-tone equal temperament, but it deviates from this theoritical tuning. To find out more, read up on the Railsback curve [3].

## Preserved frequency ratios

A property of twelve-tone equal temperament is that a major third sounds like a major third wherever it is played on the piano. A fifth sounds like a fifth wherever it is played on the piano. A fourth sounds like a fourth wherever it is played on the piano.. As you move a given interval up and down, the pitch changes, but the interval always sounds like that interval. Tuning systems that are not equal temperament do not have this property; a third might sound sharp at some places on the keyboard, and flat in other places. Let’s look at why equal temperament has this property.

Pick key $n$ on the keyboard.

Now pick a key that is $k$ semitones above $n$. This is key $n+k$.

We can write the frequency ratio for the two notes as:

$r = \frac{freq(n+k)}{freq(n)}$

Let’s plug in our expression for $freq(n)$ from above:

$r = \frac{440 * 2^{(n+k-49)/12}}{440 * 2^{(n - 49)/12}}$

And simplify:

$r = \frac{2^{(n+k-49)/12}}{2^{(n - 49)/12}}$

$r = 2^{(n+k-49)/12 - (n-49)/12}$

$r = 2^{(n+k-49-n+49)/12}$

$r = 2^{k/12}$

The $n$s have cancelled out! The ratio of frequencies, $r$, only depends on the number of semitones between the two notes, $k$. This explains why in equal temperament, any interval (e.g. a third), has the same character even as you move it up and down the keyboard: the $n$ may change, but the $k$ stays the same, and hence the ratio of frequencies stays the same.

Other tuning systems, such as just intonation [5], lack this property, meaning that a given interval can sound more like the interval, or less like the interval, as you move it up and down the keyboard, because the frequency ratio doesn’t stay constant. For certain notes and intervals these tuning systems may sound ‘better’ than equal temperament, but then for other notes and intervals the sound may be ‘worse’.

## References

[1] https://en.wikipedia.org/wiki/Equal_temperament

[2] https://en.wikipedia.org/wiki/Musical_tuning#Systems_for_the_twelve-note_chromatic_scale

[3] https://en.wikipedia.org/wiki/Piano_acoustics#The_Railsback_curve

[4] https://en.wikipedia.org/wiki/Piano_key_frequencies

[5] https://en.wikipedia.org/wiki/Just_intonation


text:
Centripetal Force

Written by Jerry Ratzlaff on . Posted in Relativity

Centripetal force is when an object keeps traveling along a axis of rotation and this force is acting always towards the center.

formula

$$F_c = \frac { m v^2 } { r }$$

$$F_c = m a_c$$

Where:

$$F_c$$ = centripetal force

$$m$$ = mass

$$v$$ = velocity

$$r$$ = radius of circular path

$$a_c$$ = centripetal acceleration

Solve for:

$$v = \sqrt { \frac { F_c r } { m } }$$

$$r = \frac { m v^2 } { r }$$

Tags: Equations for Force


text:
# Ten Small But Important Things To Observe In How To Take Care Of Short Hair African American

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text:
## 数値解析セミナー

開催情報 火曜日　16:30～18:00　数理科学研究科棟(駒場) 056号室 齊藤宣一 http://www.infsup.jp/utnas/

### 2015年03月20日(金)

13:30-15:00   数理科学研究科棟(駒場) 122号室
Gadi Fibich 氏 (Tel Aviv University)
Asymmetric Auctions (English)
[ 講演概要 ]
Auctions are central to the modern economy, both on-line and off-line. A fundamental result in auction theory is that when bidders are symmetric (identical), then under quite general conditions, all auctions are revenue equivalent. While it is known that this result does not hold when bidders are asymmetric, the effect of bidders' asymmetry is poorly understood, since asymmetric auctions are much harder to analyze.

In this talk I will discuss the mathematical theory of asymmetric auctions. I will focus on asymmetric first-price auctions, where the mathematical model is given by a nonstandard system of $n$ nonlinear ordinary differential equations, with $2n$ boundary conditions and a free boundary. I will present various analytic and numerical approaches for this system. Then I will present some recent results on asymptotic revenue equivalence of asymmetric auctions.

Joint work with A. Gavious and N. Gavish.


text:
How to check if two sequences of setoid members are mutual rotations? [closed]

Real world background: There are 2 independent sources of polygon definitions, with arbitrary parameters. Meaning each source has it's own floating point precision, starting point and orientation of vertices.

The problem is to find an algorithm, that would receive those 2 polygon definitions and by a certain margin declare them equal or not. If a cutting machine would receive those 2 inputs, would it cut out the same shapes? (Which is btw. the real motivation behind the problem.)

E.g.: If

polygonA = ( (1.0001, 1.0002), (2.0000, 1.0001), (1.5001, 1.9998) )

polygonB = ( (1.9998, 0.9999), (0.9999, 1.0002), (1.5002, 2.0001) )

and there is precision margin of 0.001 (i.e. 1 mm) then polygonA would be equal to polygonB.

This problem can formally formulated as follows:

Assume we have a set $X$ and two sequences $S_1 = (a_1, a_2, \ldots,a_n)$ and $S_2 = (b_1, b_2, \ldots,b_n)$, where $a_i \in X, b_i \in X, \forall i \in [1..n]$.

We define, that two sequences are congruent and denote them $S_1 \cong S_2$ if $\exists k,l \in [1..n]$ such that $a_{k+i \pmod{n}}=b_{l+i \pmod{n}}, \forall i \in [1..n]$. (Plainly said, they would define the same cycle.) If $S_1 \cong S_2$ then we say that $S_1$ is a rotation of $S_2$ and vice-versa.

The task is to find an algorithm that determines if $S_1 \cong S_2$.

Problem

In the case that $X$ is a total order set the task can be solved quite easily with a linear algorithm. (Finding a lexicographical extremum of all rotations for both sequences and then comparing those extrema.)

The question is, is there linear algorithm if $X$ is a setoid? I.e. members of $X$ can be checked only for equivalence. If not, does at least exist an algorithm with a better complexity than brute force?

closed as unclear what you're asking by Raphael♦Apr 11 '14 at 6:43

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Do we have a way to hash elements of $X$, via some hash function? This is less than a comparison operator, but more than just being able to test two elements of $X$ for equality. – D.W. Apr 11 '14 at 5:18
• This is a dump of a problem, not a question. If you have a specific question regarding the wording of the problem or about concrete steps in your own attempts at solving the problem, feel free to edit accordingly and we can reopen the question. See here for a relevant discussion. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? – Raphael Apr 11 '14 at 6:42
• @Raphael: I thought that the question is very specific. "Does a linear algorithm exist that for a defined input provides a defined output?" Definitions are all stated in the description, again quite specific (unless I have to define what a set, sequence, natural numbers or modular arithmetic are). Could you please state where you see an ambiguity, i.e. what is to "specify"? – Zipper Apr 11 '14 at 10:02
• @D.W. No hash function is available for $X$. Unfortunately the environment of the problem doesn't even provide data of a setoid. With some tweaks, the $X$ can be viewed as a setoid, although with the price that algorithms provide invalid results for certain inputs. I'm aware of that and I'm accepting such trade-off. – Zipper Apr 11 '14 at 10:12
• Have you read my comment? ("unclear what you're asking" is just a close reason for lack of a better fitting one) – Raphael Apr 11 '14 at 10:15

It might depend on $n$. If $n$ is prime then the following algorithm works in linear time. Compare $a_1$ to all elements in $a$ and to all elements in $b$. The resulting $a_1$ patterns must be congruent, and if they are, then they are congruent uniquely. Once you know the shift, it is easy to check whether the original sequences are congruent.
When $n$ is composite then you might run into trouble since there are several shifts possible. Still, it seems possible but messy to get a linear or quasilinear time algorithm.
• $n$ is an arbitrary natural number and sadly, in the actual data the $n$ is mostly an even number on top of it. :-/ – Zipper Apr 11 '14 at 10:10


text:
• Create Account

### #ActualKhatharr

Posted 12 February 2013 - 09:53 PM

rhuala, on 12 Feb 2013 - 18:49, said:

int i, a[10];
for(i=0;i<15;i++)
a[i]=i;

That's not a memory leak. That's a buffer overrun. You can avoid it by not using magic numbers:
const size_t ARRAY_LENGTH = 10; //or whatever
int ary[ARRAY_LENGTH];
for(int i=0; i < ARRAY_LENGTH; ++i) {
ary[i] = i;
}

Quote
One technique I use to track them down or errors in general is I write a for loop to a high number say 100,000 then I keep calling that function. If it has a memory leak it will usually crash or I examine Task Manager-> Performance and watch the Free Memory shrink.

No bueno. This will only catch leaks that always happen, and it's too hard to distinguish from normal allocation behavior if leaks are small. Use a utility like VLD instead, and use smart pointers and RAII to make leakage a non-issue.

Quote
Another one, that I think causes a crash:

BitBlt(hdc, -5, -5 , 100 , 100, hdcPic, 0, 0, SRCCOPY);
When using GDI, does blitting outside the screen resolution cause a memory leak?

No. It clips the blit according to the target context.

You're confused about what a memory leak is, or at least about what causes them.

This is a memory leak:
int main() {
int* derp = new int;
return 0;
}

This is a harmless one, but it's a leak because memory is allocated and not freed. You avoid memory leaks by strictly following proper memory management practices like RAII.

The examples you gave were concerning buffer overrun behavior. You avoid overruns by limiting buffer actions by always implementing bounds checking. Write loops in such a way that they are made aware of - and are unable to pass beyond - the end of the buffer, etc. Use the secure CRT functions rather than the original ones.

ProvenDantheman, on 12 Feb 2013 - 18:59, said:
Also, make sure you only call delete once, or if there is a possible instance of you calling delete twice, set the variable to NULL when you delete it. Deleting a NULL pointer has no effect.

Obscuring a bug is not the same as fixing a bug. Your program should always be aware of its allocations and react intelligently. NULL deletion has its uses but abusing it to avoid implementing allocation control is like taking aspirin instead of going to the hospital after cutting your arm off.

### #4Khatharr

Posted 12 February 2013 - 09:46 PM

int i, a[10];
for(i=0;i<15;i++)
a[i]=i;

That's not a memory leak. That's a buffer overrun. You can avoid it by not using magic numbers:

const size_t ARRAY_LENGTH = 10; //or whatever
int ary[ARRAY_LENGTH];
for(int i=0; i < ARRAY_LENGTH; ++i) {
ary[i] = i;
}

One technique I use to track them down or errors in general is I write a for loop to a high number say 100,000 then I keep calling that function. If it has a memory leak it will usually crash or I examine Task Manager-> Performance and watch the Free Memory shrink.

No bueno. This will only catch leaks that always happen, and it's too hard to distinguish from normal allocation behavior if leaks are small. Use a utility like VLD instead, and use smart pointers and RAII to make leakage a non-issue.

Another one, that I think causes a crash:

BitBlt(hdc, -5, -5 , 100 , 100, hdcPic, 0, 0, SRCCOPY);

When using GDI, does blitting outside the screen resolution cause a memory leak?

No. It clips the blit according to the target context.

You're confused about what a memory leak is, or at least about what causes them.

This is a memory leak:

int main() {
int* derp = new int;
return 0;
}

This is a harmless one, but it's a leak because memory is allocated and not freed. You avoid memory leaks by strictly following proper memory management practices like RAII.

The examples you gave were concerning buffer overrun behavior. You avoid overruns by limiting buffer actions by always implementing bounds checking. Write loops in such a way that they are made aware of - and are unable to pass beyond - the end of the buffer, etc. Use the secure CRT functions rather than the original ones.

### #3Khatharr

Posted 12 February 2013 - 09:43 PM

rhuala, on 12 Feb 2013 - 18:49, said:

int i, a[10];
for(i=0;i<15;i++)
a[i]=i;

That's not a memory leak. That's a buffer overrun. You can avoid it by not using magic numbers:
const size_t ARRAY_LENGTH = 10; //or whatever
int ary[ARRAY_LENGTH];

for(int i=0; i < ARRAY_LENGTH; ++i) {
ary[i] = i;
}

rhuala, on 12 Feb 2013 - 18:49, said:
One technique I use to track them down or errors in general is I write a for loop to a high number say 100,000 then I keep calling that function. If it has a memory leak it will usually crash or I examine Task Manager-> Performance and watch the Free Memory shrink.

No bueno. This will only catch leaks that always happen, and it's too hard to distinguish from normal allocation behavior if leaks are small. Use a utility like VLD instead, and use smart pointers and RAII to make leakage a non-issue.

Quote

Another one, that I think causes a crash:

BitBlt(hdc, -5, -5 , 100 , 100, hdcPic, 0, 0, SRCCOPY);

When using GDI, does blitting outside the screen resolution cause a memory leak?

No. It clips the blit according to the target context.

You're confused about what a memory leak is, or at least about what causes them.

This is a memory leak:
int main() {
int* derp = new int;
return 0;
}
It's a harmless one, since the page will be released after the program finishes, but it's a leak because memory is allocated and not freed. You avoid memory leaks by strictly following proper memory management practices like RAII.

The examples you gave were concerning buffer overrun behavior. You avoid overruns by limiting buffer actions by always implementing bounds checking. Write loops in such a way that they are made aware of - and are unable to pass beyond - the end of the buffer, etc. Use the secure CRT functions rather than the original ones.

### #2Khatharr

Posted 12 February 2013 - 09:36 PM

int i, a[10];

for(i=0;i<15;i++)

a[i]=i;

That's not a memory leak. That's a buffer overrun. You can avoid it by not using magic numbers.
const size_t ARRAY_LENGTH = 10; //or whatever
int ary[ARRAY_LENGTH];

for(int i=0; i < ARRAY_LENGTH; ++i) {
ary[i] = i;
}

One technique I use to track them down or errors in general is I write a for loop to a high number say 100,000 then I keep calling that function. If it has a memory leak it will usually crash or I examine Task Manager-> Performance and watch the Free Memory shrink.

No bueno. This will only catch leaks that always happen, and it's too hard to distinguish from normal allocation behavior if leaks are small. Use a utility like VLD instead, and use smart pointers and RAII to make leakage a non-issue.

Another one, that I think causes a crash:

[source]
BitBlt(hdc, -5, -5 , 100 , 100, hdcPic, 0, 0, SRCCOPY);

When using GDI, does blitting outside the screen resolution cause a memory leak?

No. It clips the blit according to the target DC.

You're confused about what a memory leak is, or at least about what causes them.

This is a memory leak:
int main() {
int* derp = new int;
return 0;
}
It's a harmless one, since the page will be released after the program finishes, but it's a leak because memory is allocated and not freed. You avoid memory leaks by strictly following proper memory management practices like RAII.

The examples you gave were concerning buffer overrun behavior. You avoid overruns by limiting buffer actions by always implementing bounds checking. Write loops in such a way that they are made aware of - and are unable to pass beyond - the end of the buffer, etc. Use the secure CRT functions rather than the original ones.

### #1Khatharr

Posted 12 February 2013 - 09:32 PM

rhuala, on 12 Feb 2013 - 18:49, said:
[Source]
int i, a[10];

for(i=0;i<15;i++)
a[i]=i;

That's not a memory leak. That's a buffer overrun. You can avoid it by not using magic numbers.
[Source]
const size_t ARRAY_LENGTH = 10; //or whatever
int ary[LENGTH];

for(int i=0; i < ARRAY_LENGTH; ++i) {
ary[i] = i;
}

rhuala, on 12 Feb 2013 - 18:49, said:
One technique I use to track them down or errors in general is I write a for loop to a high number say 100,000 then I keep calling that function. If it has a memory leak it will usually crash or I examine Task Manager-> Performance and watch the Free Memory shrink.

No bueno. This will only catch leaks that always happen, and it's too hard to distinguish from normal allocation behavior if leaks are small. Use a utility like VLD instead, and use smart pointers and RAII to make leakage a non-issue.

rhuala, on 12 Feb 2013 - 18:49, said:
Another one, that I think causes a crash:

[Source]
BitBlt(hdc, -5, -5 , 100 , 100, hdcPic, 0, 0, SRCCOPY);

When using GDI, does blitting outside the screen resolution cause a memory leak?

No. It clips the blit according to the target DC.

You're confused about what a memory leak is, or at least about what causes them.

This is a memory leak:
int main() {
int* derp = new int;
return 0;
}
It's a harmless one, since the page will be released after the program finishes, but it's a leak because memory is allocated and not freed. You avoid memory leaks by strictly following proper memory management practices like RAII.

The examples you gave were concerning buffer overrun behavior. You avoid overruns by limiting buffer actions by always implementing bounds checking. Write loops in such a way that they are made aware of - and are unable to pass beyond - the end of the buffer, etc. Use the secure CRT functions rather than the original ones.

PARTNERS


text:
# Why has Earth's core not become solid?

The Earth is billions of years old, yet its core has not yet cooled down and become solid. Will this happen in the foreseeable future?

• Isn't the inner core solid? – ProfRob Jul 24 '16 at 8:21
• yes, the inner core is solid. – Roghan Arun May 6 '20 at 15:53

I generally take "foreseeable future" to mean "in my lifetime" in which case the answer is No. However, if you are really asking, "What does science say about the solidification1 of earth's core?" then we can answer this.

In the core of the planet, we actually have this pretty picture:

The "solid inner core" is pretty much pure iron at a nice pretty temperature of about 5700 K (strangely close to the sun's surface temperature), which seems to be greater than iron's melting point of about 1800 K, but the Clausius-Clapeyron relation shows that melting points change with an increase in pressure, so all is well there.

The reason the heat has been retained in the core is that the only processes to move the heat are convective transport and thermal conduction, the latter being a fairly slow process (aside: it's the convective currents that cause the magnetic field). Degeun et al (2013) state,

...the dynamical time-scale of the thermal convection in the inner core [is] $\sim$1 My or more.

Which, though shorter than the age of the planet, is still pretty long. But the inner core is also growing at a rate of about 0.5 mm/year (possible paywall) and has been doing so for about a billion years. Though unreferenced, this Earth.StackExchange answer suggests that the core has cooled only 250 K since it was formed (a rate of 55 K/Gyr). At that rate, "...it would take something like 91 billion years to cool to 0 K." So no, not in our lifetime for sure!

Note also that the mantle is being partially heated by radioactive decays of Uranium-238, Uranium-235, Thorium-232, and Potassium-40, all of which have half-lives of greater than 700 million years (up to about 14 billion years for Thorium). This will protect us for some time from the complete cooling of the core.

So, fortunately for our tiny planet, the core will stay nice and toasty warm for a few more billion years (my estimation). Unfortunately for the planet, the sun will go red giant phase and eat it up before the core cools.

I'm being liberal with this word. As I state elsewhere, the core is solid. OP intends it to mean a cold lump of iron rather than the toasty lump of iron that we have

• " (a rate of 55 K/Gyr). At that rate, "...it would take something like 91 billion years to cool to 0 K". That assumes a linear relation. That is not reasonable for such systems. It is more like exponential (if the "heat flow" is proportional to the temperature difference), and it would take much much longer (infinity in fact for 0 K - a more reasonable temperature of 300 K could be assumed). – Peter Mortensen May 2 '19 at 1:10
• @PeterMortensen that is probably a more appropriate assumption; I just quoted a value I found there. If I get a chance later today I can edit it in. Though if you want to edit it in yourself, feel free (that is what the edit button is for ;)) – Kyle Kanos May 2 '19 at 11:01

Why has Earth's core not become solid? Will this happen in the foreseeable future?

• Why has Earth's core not become solid?
A lot of residual heat from the Earth's formation remains inside the Earth. Heat is released as iron and nickel freeze onto the solid inner core, which slows this freezing process down. The Earth is big. Moon and Mars are much smaller, and they still have liquid outer cores.

• Will this happen in the foreseeable future?
No.

The answer to the second question is no. It might happen in a few billion years, then again it might not happen at all. The Sun will turn into a red giant in about five billion years or so and might expand to engulf the Earth. If that happens, the Earth will be gone.

The core answer to the first question is that a lot of residual heat from the Earth's formation still remains inside the Earth, and a good fraction of this residual heat is concentrated in the Earth's core.

An argument against the Earth's core freezing any time soon (e.g., a billion years is "soon") is that the Moon (Williams 2006) and Mars (Yoder 2003) still have liquid outer cores. The Moon and Mars are much smaller than is the Earth. The square-cube law suggests that a large body such as the Earth will have an even slower core cooling rate than those much smaller bodies. Going against this, the Moon and Mars have stagnant lid convection systems while Earth has an active plate tectonics system. However, that our Moon still has a liquid core has to argue against the the Earth's core freezing any time soon.

One key to answering this question is the rate at which heat energy crosses the core-mantle boundary. Estimates of this heat flux vary widely, from less than 2 terawatts to over 15 terawatts. The possible sources of this heat flux are:

• Latent heat of freezing as molten iron and nickel in the outer core freeze onto the inner core.
• Secular cooling of the Earth's inner and outer core.
• Gravitational heating as lighter elements in the outer core are rejected from the freezing process.
• Radiogenic heating from decay of radioactive elements in the Earth's core.

The final item is highly controversial. Most geochemists and geophysicists are of the opinion that there are but negligible amounts of radioactive elements (235U, 238U, 232Th, 40K) in the Earth's core. Uranium, thorium, and potassium are strongly lithophilic elements. Moreover, they are incompatible elements. Compared to chondrites, these elements are concentrated in the Earth's crust, depleted in the upper mantle, and are likely slightly depleted in the lower mantle. These elements just shouldn't be in the core.

Recent geoneutrino detection experiments (KamLAND 2011) bolster these views, at least with regard to 238U and 232Th. (The neutrinos produced by 235U and 40K are of too low an energy to be detected by existing detectors.) That uranium and and thorium are not present in the Earth's core is consistent with results from the these geoneutrino observations.

Nonetheless, several recent papers (e.g., Lay 2008) argue for a very high core-mantle boundary (CMB) heat flux. This presents several problems. One is that it means the Earth's inner core is very young, a billion years old or less (Labrosse 2001). Others argue that the inner core is at least two billion years old (Stacey 1999), and possibly even older. A related problem: If the Earth's inner core is very young, what powered the geodynamo prior to the formation of the inner core? Another problem is that this high heat flux means implies a non-negligible amount of radiogenic heating in the Earth's core.

One way around this dilemma is that the Earth's core does contain a non-negligible amount of radioactive elements, particularly 40K (Nimmo 2004). Others argue that even if it is possible for potassium to enter the core, there just isn't enough potassium in the Earth to explain a large heat flux from the core (Lassiter 2006).

An alternative is that a considerable amount of heat is generated at the CMB, in the D'' layer. The recent discovery of post-perovskite might provide a mechanism. There could well be a deep magma ocean or a partial melt at the base of the mantle (Labrosse 2007, Stixrude 2009). As uranium, thorium, and potassium are incompatible elements, this deep magma ocean would naturally contain concentrated amounts of those elements.

References

• Isn't the inner core solid? – ProfRob Jul 24 '16 at 8:24
• It occurs to me that radiological heating of the mantle affects the rate of heat transfer across the core-mantle boundary even if there is no appreciable decay heating in the core proper. Not sure how that feeds into "The final item is highly controversial.". – dmckee --- ex-moderator kitten Sep 6 '17 at 14:48
• Does the radiation from elements in the crust also penetrate towards the Earth's center and provide some heating effect? – StayOnTarget Sep 2 '18 at 11:46
• Is CMB = core-mantle boundary? – PM 2Ring Nov 2 '18 at 16:24
• Can you introduce the abbrev. CMB in your answer? – Peter Mortensen May 2 '19 at 1:42

Hot stuff on the surface of the Earth cools fast because air can carry away the heat pretty easily by convection. This heat is eventually radiated back to space.

The outer core, on the other hand, touches the hot mantle. Due to the relatively small temperature difference, there's not much heat transfer to the mantle. The mantle itself does slowly cool, but this is at the subduction zones in the ocean (which are relatively few compared to the entire surface area of the mantle), as the rocky crust insulates the mantle.

Besides that, radioactive decay is a significant contributor to keeping the outer core molten.

So, no, the core won't cool significantly in the forseeable future.

1. Earth is big, so takes a long time to cool down. Note that Mars, being smaller, has already cooled down to this point.
2. The heat from radioisotope decay over something with the mass/area ratio of earth is significant.
• @dj_mu: The initial heat is lost by black body radiation from the surface. The energy of this heat is proportional to the planet mass, but the loss rate is proportional to the surface area. Small planets like Mars have less mass per surface, so cool faster. Radioisotope heating power is also proportional to mass or volume, but is also lost proportional to surface area. Mars looses again compared to Earth, as does our moon. – Olin Lathrop Oct 11 '13 at 16:03
• Mars has not cooled down to this point. It still has a liquid core. See Yoder, et al. (2003), "Fluid core size of Mars from detection of the solar tide," Science 300.5617 : 299-303. – David Hammen Dec 22 '14 at 21:43

Earth's inner core is already solid. However not because it is cold, but because of high pressures. Now for whether Earth outer core will freeze, well radioactive decay heats up Earth's outer core. Pressure also prevents Earth outer core from losing heat. Of course a part of Earth's outer core is solidifying. Lighter elements are solidifying on the surface of the outer core as the outer core loses heat to the mantle. This causes dynamical patterns and is one reason for why the Earth has a magnetic field. As this happens molten iron moves through these solidified lighter elements. Amplifying the magnetic field coming from deeper in the outer core.

Don't forget the gravitational title forces of the moon and the sun combined, that stretch and cool the earth, also generate heat within the planet.

Example, take a plastic spoon or something like that and start to slowly bend it – bend it back and forth slowly so it doesn't break and it will slowly start to get soft and as it gets soft, bend it back faster and faster until it actually gets hot enough to cause a first real first-degree burn in the spot that's been bending the most.

If we didn't have the moon, we would've ended up like Mars. Mars does not have any real moons to speak of – it has two large asteroids. If Mars had a moon similar to ours, it would've kept the center of the planet liquid and hot, which would have its magnetic field and prevented the atmosphere from being blown away by solar winds....

Mars once had a huge ocean and an atmosphere but then lost it… And the only way to lose that is to lose your magnetic field… And the only way to lose that is for the core of the planet to cool.

The amount of radioactivity based on the materials in our planet earth are not enough to keep our planet as warm and hot as it has been underneath the crust... Certainly radioactive decay has played a role in keeping the center of our planet warm… But the two main factors are high pressure and density combined with the title forces of the sun and moon. When the sun and moon are in opposition, the planet is slightly stretched out and is more oval-shaped...when the sun and moon are at 90° angles of the planet that pulls on the planet in a different manner… And when the sun and the moon are on the same side of the planet, that pulls that side of that side of the planet even further towards the sun and the moon.

So our planet is constantly being kneaded like bread dough....and if you ever kneaded bread, you know that the act of doing it keeps the dough warm.

Plus don't forget all of this tugging and pulling creates a huge amount of friction… And as we all know when you rub your hands together and generate friction – they get warm. The same thing with our planet.

Just do an Internet search for "what if the earth had no moon" and you will find many scientific articles that say we would be just like Mars or close to it nowadays. I'm not saying a moon is required for a planet to stay warm, but in our case, it is the most contributing factor to prevent our core from being cold by now.

• That isn't true. Tidal heating of the earth by the Moon is negligable. See this book for the details. – John Rennie Sep 28 '15 at 16:13
• I read that book and it doesn't really have scientific merit in my opinion… I have read plenty of scientific articles that basically contradict that entire book – David Dutton Sep 28 '15 at 16:17


text:
#### Koç University Mathematics Department Seminars

Subdigraph Counts in Random k Nearest Neighbor Digraphs
Özet : We study the number of copies of a given digraph in k nearest neighbor (kNN) digraph of the data from a random point process in $\mathbb{R}^d$. Based on the asymptotic theory for functionals of point sets under homogeneous Poisson process (HPP) and uniform binomial process, we provide a general result for the asymptotic behavior of subdigraph counts and as corollaries, we obtain asymptotic results for the number of vertices with fixed indegree, the number of shared kNN pairs and the number of reflexive kNN's in a kNN digraph. Moreover, we use the method to analyze kNN contingency tables for several classes under HPP.