Why do no particles have a 1/3 spin? Why are all particles' spin either a half-integer or integer? How would a particle with such a spin behave, as a fermion, boson, or neither?
It seems to me that we could change all the current spin values of particles by multiplying them by two. Then we could describe Bosons as even spin particles and Fermions as odd spin particles. Is there some consequence or reason why we can't simply change 1/2 as the smallest unit of spin into 1?  It just seems prettier this way. In fact, there's almost an interesting relation to even and odd functions in that if you switch any two bosons around the wave function stays the same while if you switch any two fermions around the wave function becomes negative.
I have an Ubuntu server and I've installed MySQL.  How would I use my two physical static IP addresses in my.cnf bind-address -- can I add two IP addresses? Hydrogen radial wave function infinity at $r=0$ When trying to solve the Schrödinger equation for hydrogen, one usually splits up the wave function into two parts: $$\psi(r,\phi,\theta)= R(r)Y_{\ell,m}(\phi,\theta).$$ I understand that the radial part usually has a singularity for the $1s$ state at $r=0$ and this is why you remove it by writing: $$R(r) = \frac{u(r)}{r}$$ But what is the physical meaning of $$R(r=0) = \infty~?$$ Wouldn't this mean that the electron cloud is only at the centre of the atomic nucleus? My processor is 64-bit - does that mean I need the amd64 image? Class moderncv moving footer I am using documentclass moderncv to style my resume, and I'd like to put the information that normally goes at the bottom of the document, such as phone number and email, to the space next to the name aligned on the left of the document. I'm a beginnner user, so I'd love to get some enlightenment on this.  \documentclass[11pt,letterpaper,sans]{moderncv}  % moderncv themes \moderncvstyle{casual} \moderncvcolor{blue} \usepackage[scale=0.75]{geometry}  % personal data \firstname{John} \familyname{Doe} \mobile{+1~(234)~567~890}  % content  \begin{document} \makecvtitle \pagestyle{empty}  \section{Education} \cventry{1998-2007}{Degree}{Institution}{City}{\textit{Grade}}{Description}   \section{Computer skills}  \thispagestyle{plain} \end{document}    Simply put, I want to move \mobile from the bottom to the top left. Convert object string to JSON Bizarre spacing with 4 $\star$s I'm new to Blender and this site, so I apologize in advance if I'm not following any of the proper etiquette.  I'm no artist, but I'm trying to learn Blender so that I have some programmer art for me to work around in Unity.    I followed a tutorial for making a basic human model, I rigged it with rigify, and now I'm starting to export it following yet another tutorial.  The thing is, when I was trying to rig the model, I couldn't get the model to go transparent so that I could see the bones inside the model.  Eventually, I managed to get it into wireframe mode, but I can't get it back into solid mode.    It's probably just a display thing and not a property of the model itself, but I want to be sure.  I'm looking at the display options under the "Figure" pane in Blender, and none of the Display check boxes are checked, including the one that says "Wire".  I don't want to import a plain wireframe model into Unity, and even if that doesn't happen, this is functionality of Blender that I should really know anyway, so I'd appreciate any help that anyone has to offer.  If necessary, I can upload the .blend file, but I'm not sure what site people would prefer that I upload it to. 1.7.4 Problem with mobs spawning How do people at Hogwarts usually tell the time? Classes and lessons at Hogwarts are presumably scheduled at fixed times of day. The first class might start, for example at nine in the morning.  To be on time for one's lessons, it is therefore crucial to know the current time but we know that electric watches of course do not work at Hogwarts because of all the magical interference. Are there magic watches? Is there maybe a spell to tell the time?  How do people at Hogwarts usually tell the time? which statistic should I use for inter-rater agreement on nominal data, multiple responses I have over 400 open-ended responses. The total has been coded by one rater, and a 10% identical sample of the total has been coded by 2 raters. There is a choice of 14 nominal codes, of which the rater can use any number per case, though most cases have between 1-4 codes. Which statistical test is the most appropriate to determine if I have inter-rater agreement on the 10% sample?   Also am I right in assuming that I will need to treat each code as a variable and look for the level of agreement for each variable?   I'm using SPSS 21.  Much thanks, Kellie Many LaTeX “hacks” begin with \makeatletter and end with \makeatother. What do these commands do? Transitive subgroup of symmetric group $S_n$ containing an $(n-1)$-cycle and a transposition How to backup settings and list of installed packages If I want to make a backup of everything I have done since the fresh install of Ubuntu, what are the possible options? What all should I backup? I want to get all the settings that I changed, all the packages I installed, etc. Why can you see virtual images? Check if convergent in the end points of the convergence interval I have to find the radius of convergence for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$, for x$\in$R. I think it is $r^{-1}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{ \frac{n+1}{n+2} }{ \frac{n}{n+1}}=1$ so r=1. Then I have to find out if $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$ is convergent in  the end points of the convergence interval. I think I have to check for $|r|$. But we can't use ratio test (L=1) or integral test(not decreasing) for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}$. How can I then check if it's convergent in the end points of the convergence interval? I have used the code shown below to draw a static CMOS inverter. I have tried to use the TikZ library calc to calculate the point at which drains of the PMOS and the NMOS devices meet; and the mid-point of the line which connects the gates of the two devices. Apparently calc gets the y coordinates of the points right. However, calc seems to get the x coordinates of the points wrong. What could I be doing wrong?  Here is the TikZ code:  \documentclass{standalone} \usepackage[americaninductors]{circuitikz} \usetikzlibrary{calc} \begin{document} \begin{circuitikz} \draw % Draw MOSFETs: (0,0) node [pigfete,yscale=-1,anchor=S] (M1) {} (M1.D) node[nigfete,anchor=D] (M2) {} (M1.G)--(M2.G) % The line connecting the gates of the MOSFETs. ; % Draw the power rails: \draw ($(M1.S)-(2.0,0)$) -- (M1.S) -- ($(M1.S)+(1.2,0)$); \draw ($(M2.S)-(2.0,0)$) -- (M2.S) -- ($(M2.S)+(1.2,0)$); % Input and output nodes: \node (I) at ($0.5*(M2.G)+0.5*(M1.G)$) {}; % The mid-point between the gates. \node (O) at (M1.D) {}; % The point at which the drains of the MOSFETs are conneted. % The Input and output lines. \draw ($(I)-(.8,0)$) -- (I); \draw (O) -- ($(O)+(0.8,0)$); \end{circuitikz} \end{document}   Here is the figure that I get: Note that that input and the output lines are not connected to the rest of the circuit. Can we attribute the problem of the lines not being connected, to the rest of the circuit, to the way I am using calc? Where did newtonian size setting go in 2.79a I have two checkboxes in a lightning component, when we click on the first checkbox it marks true to the other checkbox, it works well only with the UI tags but doesn't work when I convert it to lightning tag  &lt;ui:inputCheckbox aura:id="box3" change="{!c.selectAll}"/&gt;  &lt;ui:inputCheckbox aura:id="boxPack" value=""                                                          change="{!c.checkboxSelect}"/&gt;   below are the lightning tags  &lt;lightning:input type="checkbox" aura:id="box3" onclick="{!c.selectAll}"/&gt;  &lt;lightning:input type="checkbox" aura:id="boxPack" value=""                                                          onclick="{!c.checkboxSelect}"/&gt;   controller  checkboxSelect: function(component, event, helper) {         // get the selected checkbox value           var selectedRec = event.getSource().get("v.value");         // get the selectedCount attrbute value(default is 0) for add/less numbers.          var getSelectedNumber = component.get("v.selectedCount");         // check, if selected checkbox value is true then increment getSelectedNumber with 1          // else Decrement the getSelectedNumber with 1              if (selectedRec == true) {             getSelectedNumber++;         } else {             getSelectedNumber--;         }         // set the actual value on selectedCount attribute to show on header part.          component.set("v.selectedCount", getSelectedNumber);     },      selectAll: function(component, event, helper) {         //get the header checkbox value           var selectedHeaderCheck = event.getSource().get("v.value");         // get all checkbox on table with "boxPack" aura id (all iterate value have same Id)         // return the List of all checkboxs element          var getAllId = component.find("boxPack");         // If the local ID is unique[in single record case], find() returns the component. not array            if(! Array.isArray(getAllId)){             if(selectedHeaderCheck == true){                  console.log('true');                 component.find("boxPack").set("v.value", true);                 component.set("v.selectedCount", 1);             }else{                 component.find("boxPack").set("v.value", false);                 component.set("v.selectedCount", 0);             }         }else{             // check if select all (header checkbox) is true then true all checkboxes on table in a for loop               // and set the all selected checkbox length in selectedCount attribute.             // if value is false then make all checkboxes false in else part with play for loop              // and select count as 0              if (selectedHeaderCheck == true) {                 for (var i = 0; i &lt; getAllId.length; i++) {                     component.find("boxPack")[i].set("v.value", true);                     component.set("v.selectedCount", getAllId.length);                 }             } else {                 for (var i = 0; i &lt; getAllId.length; i++) {                     component.find("boxPack")[i].set("v.value", false);                     component.set("v.selectedCount", 0);                 }             }          }        }, Processing time/speed for Mosaic to New Raster I have a question on speeding up “Mosaic to New Raster” or finding an alternative.  After completing a suitability analysis we are trying to define movement corridors.  We used a modified CircuitScape that left 4 folders, each containing 1080 raster tiles across our study area.  We now need combine all 1080 rasters in each folder then combine these 4 combined rasters into a final raster which will be analyzed for paths.  I tried a python script that just wouldn’t work so I went on to Model Builder.  It iterates through the grids in the folder, collects the values (raster names), then mosaics all to a new raster.  It has been running for 5 days now and the results window indicates it is working.  I’m hoping it will conclude in the next day or 2.  But, we still have 3 more folders, each with 1080 rasters, that need to be mosaicked.  So….is there a way to speed up Mosaic to New Raster or is there an alternative I could use to produce the final grid?    The cell size is 270 meters, 540 x 540 columns and rows, 1 band, 32 bit floating. What do you call a person who "shuts down" ideas?