homework and exercises - Speed of a falling pencil

If you balance a pencil of length $d$ on its tip, and let it fall, how do you compute the final velocity of its other end just before it touches the ground?

(Assume the pencil is a uniform one dimensional rod)

Answer

The amount of kinetic energy in the pencil just before it touches the ground is equal to the gravitational potential energy it lost while falling. That quantity is easy to compute, being simply m*g*h, where h=d/2 (the center of mass of the pencil started at height d/2 and ends at height 0). If we assume the tip of the pencil hasn't moved (this becomes a much more complex problem if there is friction between the tip and the table, etc) then all of that energy is now rotational kinetic energy, which is equal to 1/2*I*ω. I is the moment of inertia (1/12*m*d^2) and ω is the angular velocity. You know d, the mass terms cancel out, and once you solve for angular velocity you can determine linear velocity of the end.

particle physics - Derivation of Gell-Mann Okubo relation for mesons

In SU(3) quark model of hadronic structure one assumes that mass splitings between hadrons is due to difference between masses of $s$ quark and $u,d$. This is modeled by perturbation Hamiltonian $$ \delta H=\frac{m_s-m}{3}(1-3 Y),$$ where $m$ is mass of $u,d$ and $Y$ is hypercharge. In particular in fundamental representation in basis $u,d,s$ this matrix has the form $$\delta H =\mathrm{diag(0,0,m_s-m)}.$$ In eigenbasis of hypercharge one immedietely gets the expected values of this operator and from that corrections to energy to first order in perturbation theory. This yields correct formulas for baryon multiplets: mass differences are approximately proportional to differences in hypercharge, $$ M=a+b Y,$$ with $a,b$ some constants. However in lecture notes from the class in particle physics I attended different approach is used for mesons. My teacher uses the fact that $Y$ is an eighth element of $(1,1)$ irreducible representation of SU(3) and then claims to have used Clebsch-Gordan coefficients for SU(3) to obtain the following formula: $$ M= a'+ b' Y + c' \left( I(I+1) -\frac{1}{4} Y^2 \right), $$ with $a',b',c'$ some constants. From that using assumption that $b'=0$ because mass is the same for particle-antiparticle pairs it is quite easy to get the celebrated Gell-Mann Okubo relation (actually one gets this for masses rather than their squares, but it is closer to truth if we put squares by hand) $$ 4 M_K ^2 = M_{\pi}^2 +3 M_{\eta}^2. $$ I don't understand why in this case we can't just explicitly evaluate the $Y$ operator to get the usual relation which holds for baryons. In Perkins it is written that this GMO relation is empirical rather than derived from SU(3) model. How should I understand this?

Answer

This is a good question that puzzled theorists for a while, until the modern understanding of chiral symmetry breaking in QCD clarified itself. The crucial thing to note is that the quadratic formula you are quoting is valid, and necessary, for pseudoscalar mesons only---the abnormally light pseudogoldstone bosons of spontaneously broken chiral symmetry. By contrast, if you tried to evaluate the formula for the vector meson octet, instead, i.e. the ρ(775), ω(783), φ(1020), with the ω-φ, suitably unmixed to take out the singlet, and the K*(896) s, the linear formula would be pretty good, as the ρ would not punish you as badly as the π!

The complete theoretical explanation is in Dashen's formula for the masses of pseudogoldstone bosons, and is neatly summarized in section 5.5 of T. P. Cheng's & L. F. Li's tasteful book. If you were a glutton for detail, you might opt for S. Weinberg's (1996) The Quantum Theory of Fields (v2. Cambridge University Press. ISBN 978-0-521-55002-4. pp. 225–231).

The basic idea of Dashen's formula (often also referred to as Gell-Mann-Oakes-Renner (1968) doi:10.1103/PhysRev.175.2195 in the sloppy shorthand of chiral perturbation theory. It is a blending of a current algebra Ward identity with PCAC, $m_\pi^2 f_\pi^2=-\langle 0|[Q_5,[Q_5,H]]|0\rangle$) is that the square of the mass of the pseudoglodstone boson is proportional to the explicit breaking part of the effective lagrangian, here linear in the quark masses, as you indicated.

That is, for example, naively, the pion mass, which should have been zero for massless quarks, now picks up a small value $m_\pi^2 \sim m_q \Lambda^3/f_\pi^2$, where $m_q$ is the relevant light quark mass in the real world QCD Lagrangian, which explicitly breaks chiral symmetry; $f_\pi$ is the spontaneously broken chiral symmetry constant, about 100MeV; and Λ the fermion condensate value ~ 250MeV. That is to say, the square of the mass of the pseudogoldston is the coefficient of the second derivative of the effective lagrangian (it pulls two powers of the goldston out of the chiral vacuum with strength $f_\pi^2$) and so the commutator of the QCD lagrangian w.r.t. two chiral charges. Normally, that would be zero, but if there is a small quark mass term, it snags, so you get the quark mass term provide a quark bilinear times a quark mass, the v.e.v. of the bilinear amounting to Λ cubed.

The GM-O formula served to explain flavor SU(3) breaking half a century ago in terms of "octet dominance" (code for the strong hypercharge Y), effectively your operator δH with the trivial identity term taken out, before quarks were invented, and, more importantly, taken seriously. (There was a strange hiatus of almost a decade in which everybody was thinking in terms of quarks, but it was thought to be flakey to admit it! But George Zweig had no fear.). With the advent of quarks, lattice gauge theory appreciation of chiral symmetry breaking, and finally chiral perturbation theory, such abstract formulas are needlessly obscure, cumbersome, and "magical", and mostly old-timers and science historians spend time on them. Calculators just calculate now.

particle physics - What is the symmetry of the pion triplet ($pi^{-}, pi^{0}, pi^{+}$)?

Under the entry "Isospin" in Wikipedia, it is written:

The pions are assigned to the triplet (the spin-1, $\mathbf{3}$, or adjoint representation) of $SU(2)$

Why is the symmetry not $SU(3)$ since there are three particles? And in what circumstance do we have an $SU(3)$ symmetry?

Answer

$\newcommand{\BK}[3]{\left|{#1},{#2}\right\rangle_{#3}} \newcommand{\BKB}[3]{\mathbf{\left|{#1},{#2}\right\rangle_{\boldsymbol{#3}}}} \newcommand{\FR}[2]{{\textstyle \frac{#1}{#2}}} \newcommand{\BoldExp}[2]{{#1}^{\boldsymbol{#2}}} \newcommand{\CMRR}[2] { \begin{bmatrix} #1 \\ #2 \end{bmatrix} } \newcommand{\MM}[4] { \begin{bmatrix} #1 & #2\\ #3 & #4 \end{bmatrix} } \newcommand{\MMM}[9] { \begin{bmatrix} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \\ \end{bmatrix} } \newcommand{\CMRRRR}[4] { \begin{bmatrix} #1 \\ #2 \\ #3 \\ #4 \end{bmatrix} } \newcommand{\CMRRR}[3] { \begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix} } \newcommand{\RMCC}[2] { \begin{bmatrix} #1 & #2 \end{bmatrix} } \newcommand{\RMCCC}[3] { \begin{bmatrix} #1 & #2 & #3 \end{bmatrix} } \newcommand{\RMCCCC}[4] { \begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix} } \newcommand{\OSS}[1] {\overset{\boldsymbol{\sim}}{#1}} \newcommand{\BoldSub}[2]{{#1}_{\boldsymbol{#2}}} \newcommand{\OSB}[1] {\overset{\boldsymbol{-\!\!\!\!\!-}}{#1}}$

These pions are mesons, composite particles of a quark $\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace}$ and an antiquark $\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace}$ :
\begin{equation} \begin{array}{cccccccc} &\boldsymbol{\lbrace}\boldsymbol{u},\boldsymbol{d}\boldsymbol{\rbrace} \!\!\!\!\!&\boldsymbol{\otimes}& \!\!\!\!\boldsymbol{\lbrace}\OSB{\boldsymbol{u}},\overline{\boldsymbol{d}}\boldsymbol{\rbrace} & \!\!\boldsymbol{=}\!\! & \boldsymbol{\lbrace}\boldsymbol{\omega}\boldsymbol{\rbrace}& \!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{\lbrace}\BoldExp{\boldsymbol{\pi}}{-},\BoldExp{\boldsymbol{\pi}}{0},\BoldExp{\boldsymbol{\pi}}{+}\boldsymbol{\rbrace} & \\ & \boldsymbol{2}\!\!\!\!\! & \boldsymbol{\otimes} & \!\!\!\!\OSB{\boldsymbol{2}} & \!\!\boldsymbol{=}\!\!&\boldsymbol{1}&\!\!\!\!\boldsymbol{\oplus}\!\!&\boldsymbol{3}& \end{array} \tag{01}\label{eq01} \end{equation} \begin{align} &\left\{ \boldsymbol{\omega} = \sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}+\boldsymbol{d}\overline{\boldsymbol{d}} \right)\hphantom{=\,}\right\} \quad \,\text{the singlet }\boldsymbol{1} \tag{02.1}\label{eq02.1}\\ &\left. \begin{cases} \BoldExp{\boldsymbol{\pi}}{-} =\boldsymbol{d}\OSB{\boldsymbol{u}} \\ \BoldExp{\boldsymbol{\pi}}{0} =\sqrt{\tfrac{1}{2}}\left(\boldsymbol{u}\OSB{\boldsymbol{u}}-\boldsymbol{d}\overline{\boldsymbol{d}} \right)\\ \BoldExp{\boldsymbol{\pi}}{+} =\boldsymbol{u}\overline{\boldsymbol{d}} \end{cases}\right\}\quad \text{the triplet }\boldsymbol{3} \tag{02.2}\label{eq02.2} \end{align} The subspaces $\;\boldsymbol{1},\boldsymbol{3}\;$ are invariant under the isospin group $\;SU(2)$.

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EDIT

responds to a comment by the OP owner :

This explanation is fine. But I still have a puzzlement. While the three pions ($\pi^{-}, \pi^{0}, \pi^{+}$) have an $SU(2)$ symmetry, why do the three quarks ($u,d,s$) have an $SU(3)$ [not $SU(2)$] symmetry? More generally, given three similar particles, how do we know whether they have an $SU(2)$ symmetry or an $SU(3)$ symmetry?

We must not confuse the number $\;n\;$ of the symmetry group $\;SU(n)\;$ with the number $\;m\;$ of the resulting $\;m-$plets (singlets,doublets,triplets,...nonets, etc).

In the following three examples the number $\;n\;$ of the symmetry group $\;SU(n)\;$ is the number of the $\;n\;$ independent $\;n-$dimensional systems we put together to build a composite system.

$\color{blue}{\textbf{Example A :}}$ If we put together a particle $\;\alpha\;$ of spin angular momentum $\;j_{\alpha}=\frac12\;$ with a particle $\;\beta\;$ of spin angular momentum $\;j_{\beta}=\frac12\;$ then the resulting multiplets is a singlet of angular momentum $\;j_{1}=0\;$ and a triplet of angular momentum $\;j_{2}=1\;$ \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{ed-01}\label{eqed-01} \end{equation} Now let apply the following $\;SU(2)\;$ transformations to the systems $\;\alpha,\beta\;$ (particles) respectively \begin{align} ^{\bf 2}U_{\bf \alpha} & = \MM{\hphantom{\boldsymbol{-}}g_{\bf \alpha}}{h_{\bf \alpha}}{\vphantom{h^{\boldsymbol{*}}_{\bf \beta}}\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}}{g^{\boldsymbol{*}}_{\bf \alpha}}_{\bf a} \,,\quad g_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \alpha}\boldsymbol{+}h_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \alpha}=1 \tag{ed-02a}\label{eqed-02a}\\ ^{\bf 2}U_{\bf \beta} & = \MM{\hphantom{\boldsymbol{-}}g_{\bf \beta}}{h_{\bf \beta}}{\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \beta}}{g^{\boldsymbol{*}}_{\bf \beta}}_{\bf b} \,,\quad g_{\bf \beta}g^{\boldsymbol{*}}_{\bf \beta}\boldsymbol{+}h_{\bf \beta}h^{\boldsymbol{*}}_{\bf \beta}=1 \tag{ed-02b}\label{eqed-02b} \end{align} In the composite system this is a $\;SU(4)\;$ transformation, the product of the two ones above

\begin{equation} ^{\bf 4}U_{ f} = \left(^{\bf 2}U_{\bf \alpha}\right)\boldsymbol{\otimes}\left(^{\bf 2}U_{\bf \beta}\right) = \MM{\hphantom{\boldsymbol{-}}g_{\bf \alpha}}{h_{\bf \alpha}}{\vphantom{h^{\boldsymbol{*}}_{\bf \beta}}\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}}{g^{\boldsymbol{*}}_{\bf \alpha}}_{\bf a}\!\!\! \boldsymbol{\otimes} \MM{\hphantom{\boldsymbol{-}}g_{\bf \beta}}{h_{\bf \beta}}{\boldsymbol{-}h^{\boldsymbol{*}}_{\bf \beta}}{g^{\boldsymbol{*}}_{\bf \beta}}_{\bf b}\!\!\! = \begin{bmatrix} \hphantom{\boldsymbol{-}}g_{\bf \alpha}g_{\bf \beta} & \hphantom{\boldsymbol{-}}g_{\bf \alpha}h_{\bf \beta} & \hphantom{\boldsymbol{-}}h_{\bf \alpha}g_{\bf \beta} & h_{\bf \alpha}h_{\bf \beta} \\ \boldsymbol{-}g_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & \hphantom{\boldsymbol{-}}g_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}h_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & h_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} \\ \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}g_{\bf \beta} & \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}h_{\bf \beta} & \hphantom{\boldsymbol{-}}g^{\boldsymbol{*}}_{\bf \alpha}g_{\bf \beta} & g^{\boldsymbol{*}}_{\bf \alpha}h_{\bf \beta} \\ \hphantom{\boldsymbol{-}}h^{\boldsymbol{*}}_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}h^{\boldsymbol{*}}_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} & \boldsymbol{-}g^{\boldsymbol{*}}_{\bf \alpha}h^{\boldsymbol{*}}_{\bf \beta} & g^{\boldsymbol{*}}_{\bf \alpha}g^{\boldsymbol{*}}_{\bf \beta} \end{bmatrix}_{\bf e} \tag{ed-03}\label{eqed-03} \end{equation}

But the $\;SU(2)\;$ transformations in \eqref{eqed-02a},\eqref{eqed-02b} represent rotations in the real space $\;\mathbb{R}^{3}\;$ wherein both particles live, so they must be identical (we would not rotate one system differently from the other) \begin{equation} ^{\bf 2}U_{\bf \alpha} =\,^{\bf 2}U_{\bf \beta}=\, ^{\bf 2}U = \MM{\:\:g}{h}{\boldsymbol{-}h^{\boldsymbol{*}}}{\:\:g^{\boldsymbol{*}}} \,,\quad gg^{\boldsymbol{*}}\boldsymbol{+}hh^{\boldsymbol{*}}=1 \tag{ed-04}\label{eqed-04} \end{equation} so that \eqref{eqed-03} yields \begin{equation} ^{\bf 4}U_{ f} = \left(^{\bf 2}U_{\bf \alpha}\right)\boldsymbol{\otimes}\left(^{\bf 2}U_{\bf \beta}\right) =\left(^{\bf 2}U\right)^{\boldsymbol{\otimes}2} = \begin{bmatrix} \:g^{2} & \:\:gh & \:hg & \!\!\!h^{2} \\ \boldsymbol{-}gh^{\boldsymbol{*}} & \hphantom{\boldsymbol{-}}gg^{\boldsymbol{*}} & \boldsymbol{-}hh^{\boldsymbol{*}} & hg^{\boldsymbol{*}}\\ \boldsymbol{-}h^{\boldsymbol{*}}g & \,\boldsymbol{-}h^{\boldsymbol{*}}h & \hphantom{\boldsymbol{-}}g^{\boldsymbol{*}}g & g^{\boldsymbol{*}}h \\ \hphantom{\boldsymbol{-}}h^{\boldsymbol{*}2} & \:\:\boldsymbol{-}h^{\boldsymbol{*}}g^{\boldsymbol{*}} & \:\:\boldsymbol{-}g^{\boldsymbol{*}}h^{\boldsymbol{*}} & g^{\boldsymbol{*}2} \end{bmatrix}_{\bf e} \tag{ed-05}\label{eqed-05} \end{equation}

This matrix expressed in the basis of the irreducible direct sum \eqref{eqed-01} is \begin{equation} ^{\bf 4}\OSS{U}_{ f}= \begin{bmatrix} \begin{array}{c|ccc} \:\: 1 \:\: &\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&g^{2}& \sqrt{2} g h & h^{2} \\ \rule [-3ex]{0pt}{6ex}& -\sqrt{2} g h^{\boldsymbol{*}} & \left(g g^{\boldsymbol{*}}-h h^{\boldsymbol{*}}\right) & \sqrt{2} g^{\boldsymbol{*}} h \\ \rule [-3ex]{0pt}{6ex}& \left(h^{\boldsymbol{*}}\right)^{2} & - \sqrt{2}g^{\boldsymbol{*}} h^{\boldsymbol{*}} & \left(g^{\boldsymbol{*}}\right)^{2} \end{array} \end{bmatrix}_{\:\mathbf{f}} = \begin{bmatrix} \begin{array}{c|ccc} ^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}&\rule [0ex]{20pt}{0.0ex}&\rule [-2.5ex]{0pt}{6.0ex} \rule [0ex]{16pt}{0ex}& \rule [0ex]{16pt}{0ex}\\ \hline \rule [-3ex]{0pt}{6ex}&\rule [0.0ex]{50pt}{0.0ex}& \rule [0.0ex]{50pt}{0.0ex} &\rule [0.0ex]{50pt}{0.0ex}\\ \rule [-3ex]{0pt}{6ex}& & ^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}} & \\ \rule [-3ex]{0pt}{6ex}& & & \end{array} \end{bmatrix}_{\:\mathbf{f}} \tag{ed-06}\label{eqed-06} \end{equation} where $\:^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}\:$ and $\:^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}\:$ are special unitary matrices in the spaces of the singlet and of the triplet respectively given by \begin{equation} ^{\mathbf{1}}U_{\boldsymbol{\left[1\right]}}= \begin{bmatrix} 1 \end{bmatrix} \quad \in SU(1)\equiv \{1\} \tag{ed-07}\label{eqed-07} \end{equation}

\begin{equation} ^{\mathbf{3}}U_{\boldsymbol{\left[2\right]}}= \begin{bmatrix} g^{2}& \sqrt{2} g h & h^{2} \rule [-3ex]{0pt}{6ex}\\ -\sqrt{2} g h^{\boldsymbol{*}} & \left(g g^{\boldsymbol{*}}-h h^{\boldsymbol{*}}\right) & \sqrt{2} g^{\boldsymbol{*}} h \rule [-3ex]{0pt}{6ex}\\ \left(h^{\boldsymbol{*}}\right)^{2} & - \sqrt{2}g^{\boldsymbol{*}} h^{\boldsymbol{*}} & \left(g^{\boldsymbol{*}}\right)^{2} \rule [-3ex]{0pt}{6ex} \end{bmatrix} \quad \in SU(3) \tag{ed-08}\label{eqed-08} \end{equation} So if we apply the $\;SU(2)\;$ transformation $\:^{\bf 2}U\:$ of \eqref{eqed-04} on both spaces in the product of the lhs of \eqref{eqed-01} then the spaces of the terms of the direct sum of the rhs side of the same equation remain invariant, the singlet \eqref{eq02.1} invariant under $\;SU(1)\;$ (more exactly unchanged) and the triplet \eqref{eq02.2} transformed under $\;SU(3)\;$ remaining in its invariant space.

We say that the symmetry group is $\;SU(2)$, NOT $\;SU(1)\;$ or $\;SU(3)\;$ of the resulting multiplets.

Reference link : Total spin of two spin-1/2 particles.

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$\color{blue}{\textbf{Example B :}}$ The quark model of baryons consisting of three quarks. So, suppose we know the existence of three quarks only : $\boldsymbol{u}$, $\boldsymbol{d}$ and $\boldsymbol{s}$. Under full symmetry (the same mass) these are the basic states, let
\begin{equation} \boldsymbol{u}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \qquad \boldsymbol{d}= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \qquad \boldsymbol{s}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \tag{ed-09}\label{eqed-09} \end{equation} of a 3-dimensional complex Hilbert space of quarks, say $\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$. A quark $\boldsymbol{\xi} \in \mathbf{Q}$ is expressed in terms of these basic states as \begin{equation} \boldsymbol{\xi}=\xi_1\boldsymbol{u}+\xi_2\boldsymbol{d}+\xi_3\boldsymbol{s}= \begin{bmatrix} \xi_1\\ \xi_2\\ \xi_3 \end{bmatrix} \qquad \xi_1,\xi_2,\xi_3 \in \mathbb{C} \tag{ed-10}\label{eqed-10} \end{equation} Let take 2 more quarks in order to construct baryons from 3 quarks \begin{equation} \boldsymbol{\eta}=\eta_1\boldsymbol{u}+\eta_2\boldsymbol{d}+\eta_3\boldsymbol{s}= \begin{bmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{bmatrix} \:, \qquad \boldsymbol{\zeta}=\zeta_1\boldsymbol{u}+\zeta_2\boldsymbol{d}+\zeta_3\boldsymbol{s}= \begin{bmatrix} \zeta_1\\ \zeta_2\\ \zeta_3 \end{bmatrix} \tag{ed-11}\label{eqed-11} \end{equation} A baryon state $\:T\:$ in the product space \begin{equation} \mathbf{B}=\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}=\mathbf{Q}\boldsymbol{\otimes}\mathbf{Q}\boldsymbol{\otimes}\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{C}^{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{C}^{\boldsymbol{3}}=\mathbb{C}^{\boldsymbol{27}} \tag{ed-12}\label{eqed-12} \end{equation} is the product of the states of above 3 quarks \begin{equation} T=\boldsymbol{\xi}\boldsymbol{\otimes}\boldsymbol{\eta}\boldsymbol{\otimes}\boldsymbol{\zeta} \tag{ed-13}\label{eqed-13} \end{equation} The final result of a full analysis is \begin{equation} \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} \tag{ed-14}\label{eqed-14} \end{equation} that is the space of states of a baryon is the direct sum of a singlet $\;\boldsymbol{1}$, a decuplet $\;\boldsymbol{10}$, a mixed symmetric octet $\;\boldsymbol{8'}$ and a mixed anti-symmetric octet $\;\boldsymbol{8}$.

Now applying a $\;SU(3)\;$ transformation $\;^{\bf 3}U\;$ on the 3-dimensional space $\mathbf{Q}\equiv \mathbb{C}^{\boldsymbol{3}}$ results in a $\;SU(27)\;$ transformation $\;^{\bf 27}U\;$ on the 27-dimensional space $\;\mathbf{B}\;$ of equation \eqref{eqed-12} \begin{equation} ^{\bf 27}U = \left(^{\bf 3}U\right)\boldsymbol{\otimes}\left(^{\bf 3}U\right)\boldsymbol{\otimes}\left(^{\bf 3}U\right) =\left(^{\bf 3}U\right)^{\boldsymbol{\otimes}3} \tag{ed-15}\label{eqed-15} \end{equation} The space of each $\;m-$plet remains invariant and a state in this $\;m-$plet is transformed under a $\;SU(m)\;$ transformation, where $\;m=1,10,8,8$. But

We say that the symmetry group is $\;SU(3)$, NOT $\;SU(1)\;$ or $\;SU(10)\;$ or $\;SU(8)\;$ of the resulting multiplets.

Reference link : Symmetry in terms of matrices.

quantum mechanics - Why are Only Real Things Measurable?

Why can't we measure imaginary numbers? I mean, we can take the projection of a complex wave to be the "viewable" part, so why are imaginary numbers given this immeasurable descriptor? Namely with operators in quantum mechanics, why must measurable quantities be Hermitian, and consequently real?

Answer

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators $B$ and $C$ can be packed into a normal operator (2). We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition (1).

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility to use normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see e.g. this Phys.SE post and links therein.

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$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\frac{A+A^{\dagger}}{2}$ and $C=\frac{A-A^{\dagger}}{2i}$.

gravity - Can the mass of an orbiting object and the object being orbited be determined by the distance and orbit velocity alone?

This question is the inverse of: "Could an object orbit while moving at twice the speed, but at the same distance, if it had half the mass?"

I'm curious about the nature of orbits, but am not well enough versed in mathematics to understand Kepler's laws well. I have been wondering if the mass of a planet and a star it orbits could be determined based solely on the distance and speed of the orbit, or if the ability to orbit at a given speed/distance was based relatively on the mass of both objects (i.e., we could determine the ratio of the mass of the two objects, but not the actual mass).

Answer

No, if you only know the distance between the objects and the relative orbital velocity of the planet, you cannot determine its mass. In fact, if you only know the distance and velocity at one particular moment, you don't have enough information to determine the orbit.

Suppose we know the distance $r$ and the relative orbital velocity $\vec{v}=(v_r,v_T)$ of a planet at a given moment. Here, $v_r=\dot{r}$ is the radial velocity component, and $v_T$ the tangential component. The orbit of the planet has two constants of motion: the specific orbital energy $E$ and the specific relative angular momentum $h$: $$ \begin{align} E &= \frac{1}{2}v_{r}^2 + \frac{1}{2}v_{T}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\ h^2 &= r^2\,v^2_{T} = \mu a(1-e^2), \end{align} $$ where $a$ is the semi-major axis of the orbit, $e$ is the orbital eccentricity, and $\mu = G(M_\text{p} + M_\text{s})$, with $M_\text{p}$ the mass of the planet and $M_\text{s}$ the mass of the star. So we have two equations and three unknowns $(\mu,a,e)$, in other words we need additional information to solve them.

For instance, if we assume that the orbit is circular, then $e=0$, and we can solve for $\mu$ and $a$. Another possibility is that we know the distance and velocity at two instances $t_1$ and $t_2$, then $$ \frac{1}{2}v_{r}^2(t_1) + \frac{1}{2}v_{T}^2(t_1) - \frac{\mu}{r(t_1)}=\frac{1}{2}v_{r}^2(t_2) + \frac{1}{2}v_{T}^2(t_2) - \frac{\mu}{r(t_2)}, $$ and we can solve for $\mu$, and, from the first two equations, we know $a$ and $e$. A third possibility is that we know the orbital period $T$ of the planet. In that case, we can use Kepler's Third Law: $$ T^2 = (2\pi)^2\frac{a^3}{\mu}, $$ which, in combination with the first two equations, yields $(\mu,a,e)$. In any case though, we can only derive $\mu = G(M_\text{p} + M_\text{s})$, i.e. we only know the sum of the masses (which will be dominated by the mass of the star).

If you want to derive the mass of the planet, you need to know the motion of the planet and the star with respect to their common centre of mass. If $(r_\text{p},v_{r,\text{p}},v_{T,\text{p}})$ and $(r_\text{s},v_{r,\text{s}},v_{T,\text{s}})$ are the position and velocity of the planet and the star with respect to their common centre of mass, then $$ \begin{align} E_\text{p} &= \frac{1}{2}v_{r,\text{p}}^2 + \frac{1}{2}v_{T,\text{p}}^2 - \frac{\mu_\text{p}}{r_\text{p}}= -\frac{\mu_\text{p}}{2a_\text{p}},\\ h^2_\text{p} &= r^2_\text{p}\,v^2_{T,\text{p}} = \mu_\text{p} a_\text{p}(1-e^2),\\ E_\text{s} &= \frac{1}{2}v_{r,\text{s}}^2 + \frac{1}{2}v_{T,\text{s}}^2 - \frac{\mu_\text{s}}{r_\text{s}}= -\frac{\mu_\text{s}}{2a_\text{s}},\\ h^2_\text{s} &= r^2_\text{s}\,v^2_{T,\text{s}} = \mu_\text{s} a_\text{s}(1-e^2), \end{align} $$ and in addition, if we know $T$, $$ T^2 = (2\pi)^2\frac{a^3}{\mu} = (2\pi)^2\frac{a^3_\text{p}}{\mu_\text{p}} = (2\pi)^2\frac{a^3_\text{s}}{\mu_\text{s}}. $$ In principle, these are five equations with five unknowns $(\mu_\text{p},\mu_\text{s},a_\text{p},a_\text{s},e)$ (actually, these are six equations, but they're not independent; also note that the eccentricities of the orbits are the same). Also $$ \begin{align} r &= r_\text{p} + r_\text{s},\\ a &= a_\text{p} + a_\text{s},\\ M_\text{p}r_\text{p} &= M_\text{s}r_\text{s},\\ M_\text{p}a_\text{p} &= M_\text{s}a_\text{s},\\ \mu_\text{p} &= \frac{GM_\text{s}^3}{(M_\text{p} + M_\text{s})^2},\\ \mu_\text{s} &= \frac{GM_\text{p}^3}{(M_\text{p} + M_\text{s})^2}. \end{align} $$ Once the equations are solved, you can derive $a$ and $\mu$ (using Kepler's Third Law again). So you have $M_\text{p}/M_\text{s}$ and $M_\text{p} + M_\text{s}$, so that you can derive $M_\text{p}$ and $M_\text{s}$ separately.

In practice, it is usually too difficult to measure $r_\text{p}$ and $r_\text{s}$, because the centre of mass will be very close to the centre of the star. But by measuring the velocities at two instances $t_1$ and $t_2$, we can treat $r_\text{p}$ and $r_\text{s}$ as additional unknowns and calculate them as well.

general relativity - Curved space-time VS change of coordinates in Minkowski space

I'm looking for a rather intuitive explanation (or some references) of the difference between the metric of a curved space-time and the metric of non-inertial frames.

Consider an inertial reference frame (RF) with coordinates $\bar x^\mu$, in flat spacetime $\eta_{\mu \nu}$ (Minkowski metric).

1. 
   

   If I have well understood, on one hand, I can go to an accelerated RF by change of coordinates $x^\mu(\bar x)$. The metric is given by: $$\tag{1}g_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}$$

   
   


2. 
   

   On the other hand, I know that a curved space-time with metric $q_{\mu \nu}$ cannot be transformed to Minkowski $\eta_{\mu \nu}$ by coordinate transformation. In other words there does NOT exist any coordinate $x^\mu(\bar x)$ such that (in the whole coordinate patch): $$\tag{2}q_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}\qquad \leftarrow \text{(does not exists in curved space)}$$

   
   

So far, everything is more or less ok... But my question is:

1. 
   

   What is the difference between $q_{\mu \nu}$ and $g_{\mu \nu}$? I mean, in both cases a particle would "feel" some fictitious forces (in which I include the weight force due to the equivalence principle).

   
   


2. 
   

   What physical situation can $q_{\mu \nu}$ describe and $g_{\mu \nu}$ cannot?

   
   

I additionally know that by change of coordinates $q_{\mu \nu}$ is locally Minkowski. But still, I can't see clearly the difference.

electromagnetism - If vacuum is an insulator, then why do charges flow inside a vacuum tube?

Since vacuum or free space is used as insulators in capacitors, how is it possible for charges to flow inside a vacuum tube?

homework and exercises - Earthing the plates (one or both) of a parallel plate capacitor — How can one make use of the fact that the potential of the plate is zero?

Please take a look at this question :

In the figure, plate $A$ has $100 \times 10^{-6}$ C charge, while plate $B$ has $60 \times 10^{-6}$ C charge. Find the values of $q_{1}, q_{2}, q_{3}, q_{4}$ when

1. 
   

   Both switches are open (no doubts here)

   
   
   


2. 
   

   Only switch $S_{1}$ is closed

   
   


3. 
   

   Switch $S_{2}$ is also closed / both switches are open

   
   

The answers provided are (all values in $10^{-6}$ C)

1. 
   

   (This one is easy) $q_1 = 80, q_2 = 20, q_3 = -20, q_4 = 80$

   
   


2. 
   

   (This one is a doubt) $q_1 = 0, q_2 = -60, q_3 = 60, q_4 = 0$

   
   


3. 
   

   (This one is also a doubt) $q_1 = q_2 = q_3 = q_4 = 0$

   
   

My approach for such problems

Basically, there are three things at work here.

1. 
   

   Guass's law helps us to show that the charges on the inner/facing surfaces (2 & 3) will be equal in magnitude and opposite in sign.

   
   


2. 
   

   We know that the electric field inside the plate (assuming that it has some thickness), a conductor, is zero under electrostatic conditions. This helps us to show that the charge on the outer surface of a plate will be equal to the algebraic sum of the rest of the charges, irrespective of whether or not the plate is earthed.

   
   



3. 
   

   Thirdly, and finally, when a plate is earthed, its potential becomes zero.

   
   

I know that the third point is an important clue to solve such problems, but I'm having trouble making use of this clue.

Also, notice the pattern in the answers key — whenever a plate is earthed, the charge on its outer surface becomes zero — I've seen this in Manny problems. Why is this so? I know that in colloquial language, it is easy to say that the earth just neutralises the excess charge on the outer surface of the plate. But, I can't understand it terms of something concrete, like Guass's law. Also, if this is indeed the case, why wouldn't the earth neutralise all of the charge on the plate?

Answer

The really important thing about ground isn't that it has potential 0: we can define the zero of the potential scale to be wherever we want. The really important thing about it is that it can give or accept large amounts of charge while remaining at more or less the same potential. (You can think of it as a really gigantic capacitor.) For an ordinary capacitor, $\frac{dV}{dq} = \frac{1}{C}$. For ground, $\frac{dV}{dq} \approx 0$.

The pattern where a plate connected to ground has no charge on its outside surface is because of this property of ground.

When there is an electric field in a region of space, there is energy stored in that space with an energy density proportional to the square of that electric field. That is the reason why like charges want to spread out: they want to reduce the strength of the field they produce.

When a plate is not connected to ground, charge collects on its outside surface. This charge produces an electric field that fills space. When it is connected to ground, however, a new lower energy configuration becomes available: The charges that formerly lived on the outside surface can flow to ground at a very low energy cost. And the energy gain is quite large: all the energy that used to be stored in the electric field around the plate has been released.

You should be able to solve many problems of this type by keeping the following three principles in mind:

1) The energy density of the electric field at a point in space is proportional to the square of the electric field at that point in space.

2) The actual charge distribution will generally be the one for which the total amount of energy stored in the electric field is minimized.

3) A plate connected to ground can take on whatever charge is most convenient for minimizing the energy in the electric field.

logical deduction - The way to Acarien, with Knights and Knaves

Once again, you're on your way to Acarien. You find yourself in a land of Knights and Knaves, who understand your language but don't speak it. You understand nor speak their language, but you do know their words for yes and no, "yok" and "pom". Unfortunately, you don't know which is which. And of course, as always, Knights always tell the truth, while knaves always lie.

You find yourself at a fork in the road, unsure which road to take, when you see someone coming down one of the roads you are considering. You don't know if it's a knight or a knave.

Is there a question you can ask, one single question, answerable with "yok" or "pom", that tells you which way to go?

---

I do not know if such a question exists, so answers can be either a question that works, or proof that no such question exists.

Answer

Let's test the "If I were to ask you "is left is the correct path?" would you say "yok"?" question.

$$ \begin{array}{ccc|c} \text{opponent} & \text{yok} & \text{Acarien} & \text{answer} \\ \hline \text{knave} & \text{yes} & \text{left} & \text{yok} \\ \text{knight} & \text{yes} & \text{left} & \text{yok} \\ \text{knave} & \text{no} & \text{left} & \text{yok} \\ \text{knight} & \text{no} & \text{left} & \text{yok} \\ \hline \text{knave} & \text{yes} & \text{right} & \text{pom} \\ \text{knight} & \text{yes} & \text{right} & \text{pom} \\ \text{knave} & \text{no} & \text{right} & \text{pom} \\ \text{knight} & \text{no} & \text{right} & \text{pom} \end{array}$$

As shown above I've tried all possible combinations of answerer type, language, and correct path. They all seem to work well.

mathematics - Thirty genuine and seventy fake coins

In the country Curgonia, there are many types of fake coins and only a single type of genuine coins. The weights of these coins satisfy the following conditions:

• All genuine coins have the same weight


• Every fake coin is heavier than any genuine coin


• No two fake coins have the same weight

Cosmo puts 100 coins on the table and tells Fredo: "Thirty of these coins are genuine and seventy of them are fake." Then Cosmo leaves the room. On the table, there is a balance with two pans (but there are no weights).

Question: What is the smallest possible number of weighings that guarantee Fredo to identify at least one genuine coin?

Answer

As many have already stated, the best you can do is

seventy weighings.

Suppose the genuine coins have weight 0, and the other coins have weights of distinct powers of two. Then any try on the old balance will always just tip the scale to the side of the heaviest fake coin, since it weighs more than all the other coins on the scale put together.

Now suppose the scale somehow magically marked the heaviest coin on the scale with a big red X everytime you weighed some coins. Now, there is no use using the coin with an X again. Also, you've learned absolutely nothing about the other coins. Hence, even with this extra magic, the only information you can gain each time is to eliminate one coin as being fake. Thus, it takes in the worst case seventy weighings to eliminate all the fake coins and find at least one genuine one.

homework and exercises - Radioactive Decay

Problem:Nuclei of a radioactive element $\Bbb X$ having decay constant $\lambda$ , ( decays into another stable nuclei $\Bbb Y$ ) is being produced by some external process at a constant rate $\Lambda$.Calculate the number of nuclei of $\Bbb X$ and $\Bbb Y$ at $t_{1/2}$

I tried to create an equation for rate of change of the number of nuclei a:

$$\dfrac{dN_{X}}{dt}=\Lambda-N_X\lambda $$

I did that because in simple decay $\dfrac{dN}{dt}=-\lambda N$ holds and here it's also being produced by rate. But after integration should we write $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{\lambda N_0-\Lambda}\Bigg)=-\lambda t$$ or $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{N_0}\Bigg)=-\lambda t$$ First one because limit was on $N: (N_0\to N)$ And next what to substitute for $t$ (ie. what is $t_{1/2}$? $ln2/\lambda$ or something else?)

Also how to do it for $\Bbb Y$? Just write $$\dfrac{dN_Y}{dt}=\lambda N_x $$?

Answer

The first of your equations is correct. You can see this in two ways. First, just look at the dimensions. In general, the argument of a logarithm should be dimensionless; only your first option is. Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$. You should be able to reproduce the standard decay equation: \begin{equation} N_X(t) = N_0\, e^{-\lambda\, t}~. \end{equation} In your first equation, the factors of $\lambda$ on the left-hand side cancel, and you get this result. With your second equation, you would get $N_X(t) = \frac{N_0}{\lambda}\, e^{-\lambda\, t}$. So that must be wrong.

As for what $t_{1/2}$ is, surely it must just be the half-life of $\mathbb{X}$ (with no creation). In particular, if $\Lambda$ is large enough, $N_X$ will actually grow, so there is no time at which half of the material is left. Since $\mathbb{Y}$ is stable, you can assume there's no relevant half-life there.

Also, your expression for $N_Y$ is correct. It's a slightly harder integration, but not too bad.

mathematics - Use 2 0 1 and 8 to make 67

Use $2$ $0$ $1 $ and $8$ to make the number $67$

RULES

1. 
   

   You must use all 4 digits. Only the digits 2, 0, 1, and 8 can be used. You can make multi-digit numbers out of the numbers. Examples: 20, 82, 2.8

   
   


2. 
   

   The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power. You may use the ^ operation if you use a digit, for example, [(8 + 1)^2 - 0!] is acceptable (if you're trying to get 80), because 2, 0, 1, and 8 is used. However, [20 ^ 2 / 8 + 1] can't be used to get 51 because it uses an extra 2.

   
   



3. 
   

   Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, or truncate functions.

   
   


4. 
   

   +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.

   
   

Please no brute-force methods. Good luck.

I see that there are many answers. I like #3 because it doesn't use more than 2 factorials, but the most upvoted one gets the credit.

Wow! You guys are good!

Answer

How about:

$(8!!!!+1)\times2+0!\\=(8\cdot4+1)\times2+1\\=(32+1)\times2+1\\=33\times2+1\\=66+1\\=67$

and similarly, with the digits in order:

$20!!!!!!!!!!!!!!!!! - 1 + 8$, where $20!^{(17)}=20\cdot3=60$.

riddle - We're found in vans, in bags, on bikes - we're found in books, in maps, online

We're found in vans, in bags, on bikes,
Most everywhere that people dwell.
Dogs may hate us, but we bring
Important tidings, good or bad.

We're found in books, in maps, online,

Most everywhere that info dwells.
China may hate us, but we're here:
Even in this riddle, you know.

What are we?

Answer

I'm guessing the answer is:

Letters?

We're found in vans, in bags, on bikes,

Postal vans, postmen carry bags and ride bikes, and all of these have letters

Most everywhere that people dwell.

And postal services are present in most populated areas

Dogs may hate us, but we bring

Dogs chasing postmen is a long-running trope

Important tidings, good or bad.

Letters carry news (e.g. in the past, news of a child being born, or a death)

We're found in books, in maps, online,

Letters of the alphabet

Most everywhere that info dwells.

A lot of information is written down with letters (most, because some information is pictorial)

China may hate us, but we're here:

Chinese writing uses logograms, which are pictorial representations of words (similar to Egyptian hieroglyphics); these are not letters

Edit - @Apep pointed out in the comments that this probably refers to China banning the letter 'N', which does fit the riddle much better.

Even in this riddle, you know.

Letters make up this riddle

riddle - what Am I Find Out?

I am a 5 letters word.

If you remove the first letter I am a form of energy.

Remove two and I'm needed to live.

Scramble the last 3 and you can drink me down.

What am I?

Answer

What am I?

wheat or cheat

I am a 5 letters word.

Both wheat and cheat contain 5 letters.

If you remove the first letter I am a form of energy.

Heat

Remove two and I'm needed to live.

You need to eat to survive.

Scramble the last 3 and you can drink me down.

You can drink tea which is eat scrambled.

mystery - Murder of the President - Part 3

This is Part 3 of the Murder of the President brainteaser/riddle. If you have not already, you might want to see the answer for Murder of the President - Part 1 posted by Nit and the answer for Murder of the President - Part 2 posted by Joe Z. Each part will give you a clue and you must solve it. Use all knowledge you have of cryptography, ciphers, past puzzles, etc. You should also use Google. This case is meant to take place in the present day, so all politicians, celebrities, places, etc. are who they are now. Please post your answers in spoiler tags.

Here's the riddle:

You open up the Bible to Romans 7:6 which reads

⁶ But now, by dying to what once bound us, we have been released from the law so that we serve in the new way of the Spirit, and not in the old way of the written code.

On the page after this verse, there are 8 sticky notes, some typed, some handwritten, all coded (for this part, all typed notes will be bold, and all handwritten ones will be italics). At the top of the page, there is some writing in marker.

Every ⁴/₁₆

WDFPSSDOQWXLVMKYSGFACRVLKJJPDDDHMNNAZXZBQWSEXNXTKJDIXXXC

Then the notes read:

#1 Romans
UDQF BNEQRS TBNT VT QHJGRDH

---

#2 Snake
VCEP SNACPW YWWP NLBU NTKFFNY

---

#3 Cow
FWBFZ FQ IEH EOHHFGTKWH AHHH

---

#4 Firetruck
TDEGYJ NHGKW AX NZZ FZXAWHN UGXP

---

#5 Gorilla
UJEKQD UOGWN UE UOT UXDDFJI UUGQ

---

#6 Tree
NZNX CK GJZGWJ NZJ HIJWJWL HQOW

---

#7 APPLE
HE JA YIU FQSJDIGQB JMA MVDTDJE QDP YIYHFD OYDQ

---

#8 Dog
CDB EGL EI ALT DVDUK ESERL

---

What do these codes mean? What do they say? Where should you go next?

Good Luck

---

Note I will be posting Part 4 in one to two days. I will select the correct answer for Part 3 before, but I will post the answer for Part 3 if no one gets it

Answer

Before trying to break the other eight codes, it would be a good idea to figure out what the message written in marker says first.

Every ⁴⁄₁₆

WDFPSSDOQWXLVMKYSGFACRVLKJJPDDDHMNNAZXZBQWSEXNXTKJDIXXXC

If we take every fourth letter in the large string of letters, we get "POLYALPHABETIC", which most likely means that the next eight codes are encoded using some sort of polyalphabetic cipher.

If we then apply the Vigenère cipher (the most common form of polyalphabetic cipher) to each of the codes on the sticky notes with their respective keys, we get the following:

1. CODE NUMBER FIVE IS CORRECT
2. CODE NUMBER FIVE ISNT CORRECT

3. CHECK IN THE PRESIDENTS DESK
4. NUMBER SEVEN IS THE CORRECT CLUE
5. NUMBER THREE IS THE CORRECT CLUE
6. THIS IS BEFORE THE CORRECT CLUE
7. GO TO THE PENINSULA TWO HUNDRED AND THIRTY NINE
8. YOU ARE AT THE WRONG PLACE

A bunch of mutually contradictory statements! What do we do?

As always, look to the Bible. Recall in my answer to part 2, that the key to the puzzle was the end of Romans 7:6:

⁶ But now, by dying to what once bound us, we have been released from the law so that we serve in the new way of the Spirit, and not in the old way of the written code.

Five of these codes were handwritten. If we throw these out, then we are left with:

1. CODE NUMBER FIVE IS CORRECT
5. NUMBER THREE IS THE CORRECT CLUE
8. YOU ARE AT THE WRONG PLACE

Which tells us that we are at the wrong place, and we need to check inside the President's desk. Back to the scene of the crime we go!

acoustics - speed of sound relative to density of medium through which sound travels

I know that sound travels faster in water compared to air and say faster in steel than in what're so What would the density have to be to cause sound to approach the speed of light

Answer

Have a look at http://en.wikipedia.org/wiki/Speed_of_sound#Basic_formula for info on how the speed of sound depends on the medium it's passing through.

Generally the important factors are the stiffness of the medium and it's density. To get sound to travel faster you need a stiffer lighter medium. For ordinary matter you'll never get speeds at anything like the speed of light because the electromagnetic forces that hold matter together are far too weak to permit this. However in neutron stars the stiffness of the degenerate matter they're made from is determined by nuclear forces and the speed of sound can approach the speed of light.

reflection - What properties make a good barrier for microwave (oven) radiation?

Suppose I have plenty of food I want to heat (which will provide load) in the microwave, and one item I don't want to heat. What properties would make a material a a good shield, to reduce or prevent heating of that item? I know metal can work as a barrier in the form of a Faraday cage, but that there are also potential issues with arcing. Presumably some kind of smoothly flexible metal mesh would be a good candidate - what properties would it need to be effective? (For example, we might consider mesh spacing, wire diameter, and the choice of metal.)

_{(Note: this is based on this question, which was recently migrated over to Seasoned Advice (food and cooking), but I think that it could really use physics expertise, so this is a rephrased version, addressing the complaints I saw leading to that question's migration.)}

soft question - Why is the harmonic oscillator so important?

I've been wondering what makes the harmonic oscillator such an important model. What I came up with:

• 
  

  It is a (relatively) simple system, making it a perfect example for physics students to learn principles of classical and quantum mechanics.

  
  


• 
  
  

  The harmonic oscillator potential can be used as a model to approximate many physical phenomena quite well.

  
  

The first point is sort of meaningless though, I think the real reason is my second point. I'm looking for some materials to read about the different applications of the HO in different areas of physics.

Answer

To begin, note that there is more than one incarnation of "the" harmonic oscillator in physics, so before investigating its significance, it's probably beneficial to clarify what it is.

What is the harmonic oscillator?

There are at least two fundamental incarnations of "the" harmonic oscillator in physics: the classical harmonic oscillator and the quantum harmonic oscillator. Each of these is a mathematical thing that can be used to model part or all of certain physical systems in either an exact or approximate sense depending on the context.

The classical version is encapsulated in the following ordinary differential equation (ODE) for an unknown real-valued function $f$ of a real variable: \begin{align} f'' = -\omega^2 f \end{align} where primes here denote derivatives, and $\omega$ is a real number. The quantum version is encapsulated by the following commutation relation between an operator $a$ on a Hilbert space and its adjoint $a^\dagger$: \begin{align} [a, a^\dagger] = I. \end{align} It may not be obvious that these have anything to do with one another at this point, but they do, and instead of spoiling your fun, I invite you to investigate further if you are unfamiliar with the quantum harmonic oscillator. Often, as mentioned in the comments, $a$ and $a^\dagger$ are called ladder operators for reasons which we don't address here.

Every incarnation of harmonic oscillation that I can think of in physics boils down to understanding how one of these two mathematical things is relevant to a particular physical system, whether in an exact or approximate sense.

Why are these mathematical models important?

In short, the significance of both the classical and quantum harmonic oscillator comes from their ubiquity -- they are absolutely everywhere in physics. We could spend an enormous amount of time trying to understand why this is so, but I think it's more productive to just see the pervasiveness of these models with some examples. I'd like to remark that although it's certainly true that the harmonic oscillator is a simple an elegant model, I think that answering your question by saying that it's important because of this fact is kind of begging the question. Simplicity is not a sufficient condition for usefulness, but in this case, we're fortunate that the universe seems to really "like" this system.

Where do we find the classical harmonic oscillator?

(this is by no means an exhaustive list, and suggestions for additions are more than welcome!)

1. Mass on a Hooke's Law spring (the classic!). In this case, the classical harmonic oscillator equation describes the exact equation of motion of the system.


2. Many (but not all) classical situations in which a particle is moving near a local minimum of a potential (as rob writes in his answer). In these cases, the classical harmonic oscillator equation describes the approximate dynamics of the system provided its motion doesn't appreciably deviate from the local minimum of the potential.


3. Classical systems of coupled oscillators. In this case, if the couplings are linear (like when a bunch of masses are connected by Hooke's Law springs) one can use linear algebra magic (eigenvalues and eigenvectors) to determine normal modes of the system, each of which acts like a single classical harmonic oscillator. These normal modes can then be used to solve the general dynamics of the system. If the couplings are non-linear, then the harmonic oscillator becomes an approximation for small deviations from equilibrium.


4. Fourier analysis and PDEs. Recall that Fourier Series, which represent either periodic functions on the entire real line, or functions on a finite interval, and Fourier transforms are constructed using sines and cosines, and the set $\{\sin, \cos\}$ forms a basis for the solution space of the classical harmonic oscillator equation. In this sense, any time you are using Fourier analysis for signal processing or to solve a PDE, you are just using the classical harmonic oscillator on massively powerful steroids.


5. Classical electrodynamics. This actually falls under the last point since electromagnetic waves come from solving Maxwell's equations which in certain cases yields the wave equation which can be solved using Fourier analysis.

Where do we find the quantum harmonic oscillator?

1. Take any of the physical systems above, consider a quantum mechanical version of that system, and the resulting system will be governed by the quantum harmonic oscillator. For example, imagine a small system in which a particle is trapped in a quadratic potential. If the system is sufficiently small, then quantum effects will dominate, and the quantum harmonic oscillator will be needed to accurately describe its dynamics.


2. Lattice vibrations and phonons. (An example of what I assert in point 1 when applied to large systems of coupled oscillators.


3. Quantum fields. This is perhaps the most fundamental and important item on either of these two lists. It turns out that the most fundamental physical model we currently have, namely the Standard Model of particle physics, is ultimately based on quantizing classical fields (like electromagnetic fields) and realizing that particles basically just emerge from excitations of these fields, and these excitations are mathematically modeled as an infinite system of coupled, quantum harmonic oscillators.

electromagnetism - Nuclear Fusion: Why is spherical magnetic confinement not used instead of tokamaks in nuclear fusion?

In nuclear fusion, the goal is to create and sustain (usually with magnetic fields) a high-temperature and high-pressure environment enough to output more energy than put in.

Tokamaks (donut shape) have been the topology of choice for many years. However, it is very difficult to keep the plasma confined within the walls because of its high surface area (especially in the inner rings).

Why hasn't anyone used spherical magnetic confinement instead (to mimic a star's topology due to gravity)? - Apart from General Fusion

E.g. injecting Hydrogen into a magnetically confined spherical space and letting out the fused energy once a critical stage has been reached?

Answer

Not an expert, but I believe the answer lies in the hairy ball theorem.

You see, for a magnetic field to turn charged particles back from a surface, the field must be parallel to the surface, which means that to have a fully confining geometry you must have a smooth, everywhere non-zero, and continuous vector field mapped onto a surface.

But the theorem say that you can't do that on a sphere (or topologically equivalent shape).

mathematics - A closed path on the Rubik's cube

Is it possible to draw a closed path on the surface of a standard $3\times3\times3$ Rubik's cube

• such that the path traverses each of the $54$ little squares exactly once, and


• such that the path does not go through any corner of these little squares?

Answer

An observation is..

that you can traverse a cube's face from one corner and end up diagonally opposite or in same column (or row).

Using this observation, one of many possible solutions is (left image is front of cube, right image is inside view):

newtonian mechanics - Why do rocket engines have a throat?

Diagrams of rocket engines like this one,

^(source)

always seem to show a combustion chamber with a throat, followed by a nozzle.

Why is there a throat? Wouldn't the thrust be the same if the whole engine was a U-shaped combustion chamber with a nozzle?

Answer

The whole point to the throat is to increase the exhaust velocity. But not just increase it a little bit -- a rocket nozzle is designed so that the nozzle chokes. This is another way of saying that the flow accelerates so much that it reaches sonic conditions at the throat. This choking is important. Because it means the flow is sonic at the throat, no information can travel upstream from the throat into the chamber. So the outside pressure no longer has an effect on the combustion chamber properties.

Once it is sonic at the throat, and assuming the nozzle is properly designed, some interesting things happen. When we look at subsonic flow, the gas speeds up as the area decreases and slows down as the area increases. This is the traditional Venturi effect. However, when the flow is supersonic, the opposite happens. The flow accelerates as the area increases and slows as it decreases.

So, once the flow is sonic at the throat, the flow then continues to accelerate through the expanding nozzle. This all works together to increase the exhaust velocity to very high values.

From a nomenclature standpoint, the throat of a nozzle is the location where the area is the smallest. So a "U-shaped chamber with a nozzle" will still have a throat -- it's defined as wherever the area is the smallest. If the nozzle is a straight pipe then there is no throat to speak of.

english - Longest Word With *Only* Repeating Character Pairs

Inspired by and Opposite of: Longest Word without repeating character-pairs

The challenge is to find an English word in which every character pair appears at least twice. ("Word" does not include acronyms or proper names. Hyphenated words are included but the hyphen doesn't count as part of a pair.)

Bad word: mama contains the pair ma twice but am only once
Bad word: aa seems clever but still only has the pair once.
Good "word": aaa is not a word but it does contain its only pair aa twice.

Preference will be given to words found on Dictionary.com

Here's a check method in Excel: Input the word in the cell A1 and then input this formula somewhere else as an array formula using Ctrl+Shift+Enter: (TRUE = valid word, FALSE = invalid word)

=(SUM(LEN($A$1)-LEN(SUBSTITUTE(UPPER($A$1),MID(UPPER($A$1),ROW(OFFSET($A$1,0,0,LEN($A$1)-1)),2),"")))/4)+1>=LEN($A$1)

I've been told this javascript function will perform the same check although I haven't used the language in years and can't verify:

function test(e){for(l=[],i=1;i

NOTE: The answers thus far and the scratches I tried to come up with were all of the form [string][same string][first letter of string]. I'd be interested in finding one of a different form although the longest of any form will be accepted as the answer.

Answer

I've found a possible word, though it seems... invalid

11 letters

kinnikinnik
It's on Dictionary.com as kinnikinnick, though it includes the above form in the "Also" section.

electrostatics - Why can two (or more) electric field lines never cross?

The the title is self explanatory, I guess.

Why can two (or more) electric field lines never cross?

Answer

Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field.

At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two directions at once in the same location.

Field lines do cross, or at least intersect, in the sense that they converge on charge. If there is a location with charge, the field lines will converge on that point. However we typically say the field lines terminate on the charge rather than crossing there.

energy - Exertion from swinging on a playground swing

I've read about how by tilting one's body one changes one's center of mass while swinging on a playground swing, and thereby increases the energy of the swing.

But I would like to get a better sense of why one can get into a swing of ~1.5 meters above the ground on the back and front peaks of the swing (elevating an adult body weight that height each time), and do that repeatedly for, say, 10 minutes, and yet not at all feel fatigued--whereas if you were trying to jump up from the ground to that height you would be fatigued after just a few jumps.

I get the sense that the answer is related to the idea that you are "loading" the swing gradually with potential energy on each swing, and that each iteration of that requires only a little effort, but I'm still sort of surprised at how much sustained energy you can generate with so little feeling of exertion (it feels like one can keep swinging high with little effort for an hour, easily), and I thought there might be some subtleties to this that I don't understand.

Answer

Consider an idealized setting where there is essentially no friction or air resistance. Suppose someone gave you a quick push while you are on the swing. You will keep on swinging without having to expend any energy yourself. A rock sitting on the swing would behave in exactly the same way; it wouldn't have to "pump" to keep this idealized swing going up & down and back & forth once it was already moving. (As for how this phenomenon can occur, gravitational potential energy and conservation of mechanical energy would be a good starting search terms.)

Now, starting from rest at the bottom of this idealized swing, you would indeed need to do a bit of pumping to get started on your own. But you would only need to exert an amount of energy equal to your maximum change in potential energy desired (i.e., roughly $mgh$), and you would only need to exert this amount of energy once (well, a little at a time until you reach your desired height, but you wouldn't have to keep exerting this amount of energy each period).

Okay, the actual real wold. There are dissipative forces at play (air resistance, friction). Once you are already moving at your desired height/amplitude, the pumping that you do is only to counteract these dissipative forces. For, without these pesky forces, you wouldn't need to pump at all.

So, you aren't pumping/expending energy to make the swing up/down or back/forth, but rather to overcome the work done by dissipative forces.

everyday life - Infrared remote flashes blue light in camera

I know that if you held an infrared remote in front of a digital camera, it'll flash a blue/purplish light when you press the buttons. Why?

(Image by Zombie Rider.)

mathematics - Mathematical Rebus

Mathematical Rebus II

Mathematical Rebus III

---

$$ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\\ (-\infty,...,-1,0,1,...,\infty)\times(-\infty,...,-1,0,1,...,\infty)\\ \forall\begin{bmatrix}{-1}&{0}\\ {0}&{-1} \end{bmatrix} $$

Answer

The first line is

equal to Pi.

The second line is

the integers, Z, multiplied by itself, making Z².

The third line is

multiplying some matrix ∀ by the negative identity matrix, negating its value. The letter ∀ negated is A.

All together,

Pi + Z² + A = Pizza.

thermodynamics - How does temperature affect the elasticity and spring constant of a rubber band

Rubber bands expand when chilled and contract when heated (no stretching, just when heated/chilled and at rest) to my knowledge. Why are hot rubber bands able to be stretched longer than cold rubber bands when acted upon by the same force? For example, a hot rubber band will stretch 10 cm when a weight it pulling it down, but a cold rubber band will stretch 5 cm when the same weight is pulling it down. Is it because of entropy? Or potential energy? Is it Hooke's Law, where K-constant decreases if temperature increases?

general relativity - Does acceleration warp space?

I know that mass warps spacetime and gravity and acceleration are equivalent so does acceleration warp spacetime too?

Answer

Sort of. You are correct in saying (with some caveats) that gravity and acceleration are equivalent. According to general relativity, gravity is manifested as curvature of spacetime. As we know from special relativity and Einstein's famous equation $E = mc^2$, energy and mass are equivalent. As a result, any type of energy contributes to gravity (i.e. to the curvature of spacetime). This relationship can be seen directly from Einstein's Field Equations of General Relativity:

\begin{equation} G_{\mu\nu} = 8\pi T_{\mu\nu}, \end{equation}

where the left hand side of the equation (called the Einstein tensor) contains information about the curvature of spacetime and the right hand side (called the stress-energy tensor) contains information about the mass and energy contained in that spacetime.

Recall that Minkowski spacetime is the spacetime of special relativity. That is, it has no curvature (no gravity) and is the shape of spacetime when you are in an inertial (non-accelerating) reference frame. So, let's ask the question: what happens when you accelerate in Minkowski space?

The answer is that spacetime no longer looks flat to accelerated observers. This is precisely the equivalence principle; locally we cannot tell if we are in a gravitational field or accelerating. Thus, when we are in fact accelerating in a flat spacetime, everything will locally appear as though we are in a spacetime that is curved due to gravity.

There are other interesting similarities between accelerated observers in flat spacetime and observers in gravitational fields. For example, accelerated motion leads to horizons similar to the event horizon of a black hole because if you accelerate at a constant rate for long enough then there will be portions of the spacetime to which you can never send or recieve light signals. There is also an analog of Hawking radiation that occurs for accelerated observers in Minkowski space, called the Unruh effect.

Coriolis force and conservation of angular momentum

I'm trying to understand the relations between the existance of Coriolis force and the conservation of angular momentum. I found this example on Morin, which confuses me.

A carousel rotates counterclockwise with constant angular speed $ω$. Consider someone walking radially inward on the carousel (imagine a radial line painted on the carousel; the person walks along this line), at speed $v$ with respect to the carousel, at radius $r$. [...]

Take $d/dt $ of $L = mr^2ω$, where $ω$ is the person’s angular speed with respect to the lab frame, which is also the carousel’s angular speed. Using $dr/dt =− v$, we have $$dL/dt =− 2mrωv+mr^2(dω/dt)\tag{1}$$

What if the person doesn’t apply a tangential friction force at his feet?

Then the Coriolis force of $2mωv$ produces a tangential acceleration of $2ωv$ in the rotating frame, and hence also in the lab frame (initially, before the direction of the motion in the rotating frame has a chance to change), because the frames are related by a constant $ω$. This acceleration exists essentially to keep the person’s angular momentum (with respect to the lab frame) constant. [...] To see that this tangential acceleration is consistent with conservation of angular momentum, set $dL/dt = 0$ in Eq. (1) to obtain $2ωv = r(dω/dt)$ (this is the person’s $ω$ here, which is changing). The right-hand side of this is by definition the tangential acceleration. Therefore, saying that $L$ is conserved is the same as saying that $2ωv$ is the tangential acceleration (for this situation where the inward radial speed is $v$).

There is no friction force acting here, nor any other real force therefore the angular momentum of the person does not change.

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $\omega r$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the motion is on a straight line (as far as I can see).

The two highlighted parts in text are the most confusing to me.

How is acceleration produced in the lab frame initially? (Coriolis does not act there)

It seems that Coriolis force is there to keep the angular momentum of the person constant in the lab frame. But this cannot be true since Coriolis force is a fictitious force, existing only in the rotating frame. I don't see clearly the link between Coriolis force and conservation of angular momentum in this example.

---

So firstly will the angular momentum of the person (who will move on a straight line in the lab) be conserved in the lab frame?

Secondly can anyone give some further explanations about the links between Coriolis force and conservation of angular momentum in this example?

Answer

This is indeed confusing. The confusion comes from this very peculiar hypothesis:

What if the person doesn’t apply a tangential friction force at his feet?

It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay immobile in the carousel, they need to at least counterbalance centrifugal acceleration.

So let's imagine how the person could be "frictionless" tangentially yet "frictionful" radially: suppose there are slippery concentric rails all over the carousel, on which the person can lean to move radially, but which prevent them to control rotational speed.

Suppose the person starts immobile with respect to the rail of radius $r$ on which they stand. When the person steps inwards, they undergo the said tangential Coriolis acceleration, which makes them start to glide counterclockwise along the inner rail of radius $r-δr$ on which they now stand, at $δω$ with respect to the carousel. Their rotational speed with respect to the lab is now $ω+δω$, and $δω$ is such that their angular moment has not changed: $rω=(r-δr)(ω+δω)$.

electromagnetism - Faraday's paradox

I have studied that Faraday's law of induction and motional emf are two different lines of thinking but are essentially same.

But then, how can Faraday's paradox be explained by Faraday's law of induction? Particularly in paradoxes in which Faraday's law of induction predicts zero EMF but there is a non-zero . The wikipedia article is more confusing rather than enlightening.

I need an explanation regarding the article regarding this picture

EDIT : I wanted to know how is flux changing if faraday's law is valid .

enigmatic puzzle - Locate the hacker: Part 2

This is a continuation of Locate the hacker. Be sure to read through part 1 and the accepted answer first so you can get an idea of how to solve this. There are no tricks here. There’s no breaking of the “fourth wall”, no corrupt partners, no hacking of emails or falsified information.

Note: Reading through part 1 is not required to solve this puzzle.

---

You arrive at the address in Indiana and find an abandoned warehouse. There's no way he could have known you were coming! But alas, Mr. Jones is nowhere to be found. The warehouse is rather small and you wonder why he chose this location.

You look around but there's not much to see. There’s a small empty desk by the far wall. The only thing on the desk is a thermometer and a notebook. The thermometer reads 43 degrees C.

You open the notebook to find only one page filled with text.

Contents:

I know I can’t run,
No end ’til it’s done,
But why not play,
Before I pay?
HEX4

It’s not for Private P,
More for a never ending,
Sorry it’s missing 80D,
It’s a kind of wet bending.
61 70 70 65 6e 64 0d 0a
F T2 R F2
F T2 F2 TFT
F T2 FT F2 T
F T2 F T3 F
HEX4

56 Monument damns the square. 2000 prostitutes. 92 kills 43 from 1862.
HEX4
89 a plane falls down,
170 total with 18 crown,
Where I rest lays a clue,
Not part one but two,
Drop the trailers to find,
And add the arrow if you don’t mind

This guy is losing his mind. None of this makes sense. What’s that about prostitutes? Right as you finish reading the notebook you receive an email from your partner.

While on your flight here, your team picked up a transmission on a message board by someone under the pseudonym DrBit. Could it be him? Only three messages were written by this user. The user login time is exactly the same as the first message post time.

Transcript:

DrBit [23:22:10]: http://www.globalclassroom.org/minus80.html
DrBit [23:51:38]: 5642 doesn’t sink prepend_temp
DrBit [00:23:14]: 8328 doesn’t sink

Find Mr. Jones. You're looking for an exact address.

Hint:

Figure out the transcript first. Remember Mr. Jones’ M.O. from part one.

Hint 2:

The answer is not a U.S. address.

Hint 3:

false, true, repeat

Hint 4:

Pay very close attention to the timestamps

Answer

Notebook

It’s not for Private P,
More for a never ending,
Sorry it’s missing 80D,

It’s a kind of wet bending.

P_(ASCII) = 80_(Decimal)
The clue from the transcript (Waterloo) doesn't quite get us all the letters that this piece of the puzzle does -- it is missing the letter P. This tries to make up for that by providing the clues loop (never ending) and, by extension, waterloop (wet bending).

F T2 R F2
F T2 F2 TFT
F T2 FT F2 T
F T2 F T3 F

-- I did figure this one out, but realize now that ericm301 did it first --

This is binary notation, F = false = 0, T = true = 1, and R = repeat (from beginning). When properly written, you get 01101100 01100101 01101001 01101110, which translates to lein.

56 Monument damns the square. 2000 prostitutes. 92 kills 43 from 1862.

Theme: Amsterdam, Netherlands

The WWII memorial "National Monument" was opened in Amsterdam's Dam Square in 1956 and is the site of a Remembrance of the Dead ceremony held every May 4. [source]

Although it has been habitually tolerated for many years, in the year 2000, prostitution (specifically, operation of brothels) was officially legalized in Amsterdam. [source]

A cargo plane (El Al Flight 1862) crashed into an Amsterdam neighborhood in 1992, killing 43 people (all 4 aboard and 39 on the ground) according to the official record. [source]

89 a plane falls down,
170 total with 18 crown,
Where I rest lays a clue,
Not part one but two,

Drop the trailers to find,
And add the arrow if you don’t mind

-- Credit to LeppyR64 for pinpointing the event --

In 1989, international passenger airliner UTA Flight 772 (a DC-10) exploded in midair, killing all 170 people (from 18 different countries) onboard. [source]

The UTA 772 Memorial is located at 16°51′53.748″N 11°57′13.362″E, about 10 km away from the crash site. My understanding of the language is still not quite matching the intent... The endgame here is 1011 PG, so working backwards from the conclusion, "where I lie" may refer to the distance, and "second part" may refer to the longitude. Dropping the trailers (decimals) gets us 1011, but I'm still lost on the arrow.

Transcript

DrBit [23:22:10]: http://www.globalclassroom.org/minus80.html
DrBit [23:51:38]: 5642 doesn’t sink prepend_temp
DrBit [00:23:14]: 8328 doesn’t sink

We've been told to pay close attention to the timestamps. The amount of time between the first and second is 29 minutes and 28 seconds. Time between second and third is 31 minutes and 36 seconds.

By following the "instructions" given in the chat log, we find ourselves at 43°29'28.5642", -80°31'36.8328", presumably in the Manulife Financial building located in Waterloo, Ontario, Canada.

We are looking for DrBit at Waterlooplein! Also known as Waterloo Square.

(left on the record since it is interesting)

56 Monument damns the square.

- The Stalin monument in Budapest's City Park was famously torn down during the Hungarian Revolution of 1956 (possibly erected within the entrance area known as Heroes' Square). [source]
- There is also now a "Memorial to the 1956 Hungarian Revolution and War of Independence" which juts into "the Square of Silence". [source]

2000 prostitutes
It is estimated that, on average, 2000 prostitutes engage in 'window' prostitution every day in the Netherlands. (Women stand in front of a window which prospective clients must knock on to gain entry) [source]

89 a plane falls down, (file under close but no cigar)
Despite the deaths of 111 people onboard, the United Airlines Flight 232 crash of 1989 has been held as a prime example of successful emergency handling by a flight crew due to the number of lives saved (185 survived) and ability to land without proper controls. [source]

170 total with 18 crown, (file under close but no cigar)
- There are 172 total Nobel Laureates for the Prize in Chemistry, which Donald J. Cram, Jean-Marie Lehn, and Charles J. Pedersen were awarded in 1987 [source] for their synthesis of the 18-Crown-6 compound (C12H24O6), which is used in the lab as a phase transfer catalyst. [source]
- The 11th generation of the Toyota Crown was known as the S170 model. [source]
- The Neoplan Jumbocruiser measures 18m in length and holds the Guinness World Record for largest bus with a 170-passenger capacity. [source]

My best guess is that the language means to ignore the latitude, take the longitude, drop the numbers after the decimal, and include the direction, so 11E, but having read up on Amsterdam postal codes, they consist of 4 digits and two letters, so not sure where to go from here, though 11 as a start is legit for the Amsterdam area. [source]

Netherlands postal code finder tool!

$-80°F = -62.\overline{2}°C$, $-76°F = -60°C$, $43°C = 109.4°F$ Using the degrees in the format given, using the coordinates -80°29'28"S, 43°31'36"E puts us in the middle of Antarctica!

geometry - Two difficult "Seventeen right isosceles triangles into a square" tilings

Similar to: Unlucky tiling: Arrange thirteen right isosceles triangles into a square

Five graded difficulty isosceles right triangle into square tilings

V.hard problem, 20 right isosceles triangles into a square

Each tiling has only one solution, these might be possible by hand, computers allowed.

The two challenges are to arrange $17$ right isosceles triangles of the listed areas into a square of area $968$ with no gaps or overlaps. The square has a diagonal of length $44$.

$1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 49, 50, 64, 81, 162, 169, 242$ $1, 2, 4, 8, 9, 16, 25, 32, 36, 49, 50, 64, 98, 100, 121, 128, 225$

The answer tick will be given to whomever posts the solution to the second puzzle first, or the first if nobody gets the second. Only because the second looks slightly harder to me.

By way of illustration/clarification, here are the right isosceles triangles of area

$1, 2, 4, 9, 16, 18, 50$

arranged into a $10\times 10$ square:

Answer

Here are the solutions to both questions:

HAISU (Room Count): An original grid-logic challenge

HAISU is a portmanteau of three Japanese words - 'hairu', to enter, 'su', number, and 'hausu', an English borrow word meaning house, of course.
Together, we get a meaning of 'enter number house', which I have roughly translated to English as 'Room Count'.

The rules are simple - draw a path from the O to the X, passing through every cell in the grid exactly once. The grid is divided into several rooms. When your path passes over a cell with the big number N, it must be the Nth time you have entered the room. If a room has a small number m in the top left corner, you must enter that room a total of m times. An example Haisu puzzle and its unique solution are shown below.

Hopefully this example puzzle clarifies the rules. Your actual challenge is this!

No guessing, no handwavy steps, just pure logic required to solve this puzzle!

Edit: A long awaited fix.

Answer

Here is what I came up with by adding lines 1 by 1 where I could prove it was 100%. Please forgive the atrocious paint display after all my redo and deletions.

bottom right : only way to hit the 2 on the second entrance while keeping an exit open to go up after.

bottom right : only way to hit every tiles without entering more than 3 times.

right : only way to hit the 1 on first entrance and the 2 on second entrance. Cannot loop any higher without blocking the final loop for the exit(takes quite a bit of thinking to see it; could have been left for later when it is more obvious)

Only way to hit the 1 on first entrance without dooming the tile to its right to eternal solitude.



Must touch the 1 on first time in and must only enter 2 times.

Follow the only possible path to allow the correct amount of entrances.

More logical deductions based on required number of entrances.