text:
## Latex Rotate Page

 LaTeX : Fit a Table Into One Page. pdf file as your image! You could also produce geometric diagrams within LaTeX as shown in the Asymptote article. Rotate and wrap text within node using Tikz First post but I am hoping someone can help. Above Figure shows how to insert figure in LaTeX. Should I Flip or Rotate My Natural Latex Mattress? Normally there is no need to flip or rotate your Savvy Rest or Sleeptek natural latex mattress. pdf with the PDF rotated by 1. It is free software, you can change its source code and distribute your changes. I have tried that several times with several tweaks of the job option settings and I have managed to produce a landscape PDF, but once again, when viewing those PDFs in illustrator, or trying to upload to my vendor, they either automatically rotate to portrait, or appear to be falling off the page showing only a quarter of what should actually. Rmd in RStudio (press Knit PDF button) results in a pdf with all pages in landscape layout:. Note that the data is released under the GPL 2 licence. Probably not the solution you are after but a barely workable workaround. LaTeX will obviously take care of typesetting the footnote at the bottom of the page. A text area appears and the package can be put there. You can change the view so that two pages are displayed side-by-side, just like a book. You can upload your own PDF file, paste the URL to an online PDF, and make your own PDF from scratch. When I did this "O" appeared on the landscape page. They have been extracted from the reference manuals. For example, two pictures side by side, two tables next to a text or a picture or vice versa. This is a guide to making tables in Latex. Using Word cannot be the solution in a LaTeX based environment. 2645 contributors have contributed to it. Preparing Figures in Matlab and LATEX for Quality Publications Azad Ghaﬀari Cymer Center for Control Systems and Dynamics - UC San Diego Second Edition, January 2014. Ask Question Asked 9 years, 7 months ago. Export (png, jpg, gif, svg, pdf) and save & share. I use LaTeX like this to combine pdfs this way and rotate their pages as necessary all the time — super convenient and I find it easier (and cheaper!) than Adobe Pro. All you need is \usepackage{rotating} in the preamble, and \begin{turn}{45} \end{turn} around the text you wish to rotate (in this case, by an angle of 45 degrees). csv) to LaTeX There are two disadvantages of writing tables by hand as described in this tutorial. To include pages as landscape you can use the following: \usepackage{pdfpages} \includepdf[landscape=true]{circuit. Jun 30, 2012 · which latex editor are u using, wju dont you try mathType and then convert the equation already to latex, its very easy i write equation in word and then convert to tex file. There is a little update to the epslatex terminal introduction page. Think of a wall full of clocks the axis of revolution for all is the same and is normal to the wall. First, rendering this. Design it in InDesign, where you can go to Pages -> Rotate Page View. Here are some features of the pdfpages package: Several logical pages can be arranged onto each sheet of paper and the layout can be changed individually. I don't know if/how it will work with multiple pages, but you can extract one page of interest with pdftk: pdftk A=myfile. The Simpsons Mr Burns Latex Mask Montgomery Burns Monty Burns is satisfactory. I know that to correctly assign R,S priorities I have to rotate the above figure until the H is projecting into the page (away from me). The > pages of the book are typeset for A5 paper with LaTeX, so producing a > PostScript or PDF file is not a problem. This interface provides facilities that go far beyond what the graphics package offers on its own. To rotate a line, select the Pointer Tool on the Home tab in the Tools group, and then click the line and then drag one of the end points. You can then resize the photo frame to fit your photo by selecting any of the handles and dragging to change the dimensions. See the article about headers and footers for more concise information and examples about fancyhdr. This comment has been minimized. Rotate an image. Alt click with line tool to get the angle. PDF Rotate Pages 1. The way my project is structured (a party flyer), the front page must be rotated 180° when printing, and the back page must stay as-is (or the oth Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Jun 25, 2018 · Some PCs have hotkeys that quickly rotate the screen when pressed. However, this does not rotate my caption, as it has to be within the table environment. , you don't have to pay for it). When I asked the contractor, he asked his flooring guy who had been doing tile for 25 years and "never puts thinset under cement board". To put a blank page into your Word document, place the cursor where you want the new page to begin and then click Insert > Blank Page. There are multiple ways to rotate in inkscape, and this FAQ will show you the basics for four of them. However, if you have a large table that will take up the whole page, you might want to rotate the page instead of the table. The Control key no longer offers the rotation icon. This rotates the entire page and its contents, so that you can design something upside down (bellybands that wrap around a box or 2-sided packages folded on the short end, for example) and then rotate it back to design the right-side-up part. Lines and paragraphs reflow automatically, or you can click and drag to resize elements. Delete Select the page that needs to be deleted then: (a) click the Delete Pages(s) button, or (b) right-click then choose Delete Page(s) from the menu. = path d=3D"M27. Photoshop supports a two finger rotate gesture, as well as two finger pinch gesture for zooming, that users accidentally invoke while using the trackpad on their Macbook or Magic Trackpad. With PDF Rotator, users are able to rotate PDF document pages properly, and the program also allows users to rotate pages in a specified page range. Rotating Text, Tabulars and Images∗ A L T E X Documentation HerbertVoß† April22,2003 Abstract To rotate objects like words, tables or images seems to be very easy with LATEX but it is sometimes diﬃcult together whith a ﬂoating envi-. Huge selection of memory foam, latex, and hybrid mattresses. 2645 contributors have contributed to it. Product Code: 34-1570: Cutsheet: Download Download. header and footer lines, page formats, page numbers). Advanced Batch Converter does exactly what its title suggests, allowing you to easily convert multiple graphics files at the click of a button. you can't include additional packages. Including Landscape Figures and Tables in a LATEX Document To insert landscape ﬁgures or tables, you'll need two packages, graphicx and rotating, which are included with most TEX/LATEX distributions. How do I make a single page *portrait*? I've got a fully landscape document and want one page to be in landscape. How do i increase a figure's width/height only in latex? Increasing a figure's width/height only in latex. It is recommended to use more than one positioning parameter to prevent unexpected results. As an example, let $y=x^2$. Extract pages from multiple PDF files. Writing a thesis with LATEX Lapo F. DONOTEDITTHISFILE!!!!! !!!!!!!!!!///// !!!"!&!&!+!+!S!T![!^!`!k!p!y! !!!"""'" !!!&& !!!'/'notfoundin"%s" !!!) !!!5" !!!9" !!!EOFinsymboltable !!!NOTICE. 1 day ago · Overleaf rotate table. To rotate a line by an exact number of degrees, use the Size & Position window. They have been extracted from the reference manuals. To my knowledge macOS does not have any one native command line Unix executable that can rotate all pages in a PDF (while keeping text based ones text based). The Comprehensive T e X Archive Network (CTAN) is the central place for all kinds of material around T e X. csv files in LaTeX. Use your phone, laptop or a printed calendar to remind yourself to periodically rotate your mattress through the year. Die landscape-Umgebung produziert standardmäßig fehlerhafte Ergebnisse in Dokumenten der Klasse twoside, d. If rotate by 5#5degrees6#6 is given, conforming terminals will try to write the text at the specified angle; non-conforming terminals will treat this as vertical text. pdf cat 1W output out. GlowScript is an easy-to-use, powerful environment for creating 3D animations and publishing them on the web. I don't know if/how it will work with multiple pages, but you can extract one page of interest with pdftk: pdftk A=myfile. Aug 17, 2010 · The Swedish summer fades into a rainy autumn, which means it is time again to prepare lectures for the coming months. Select a file from your computer or drag-and-drop it in the box. As an example, let $y=x^2$. The package automatically takes care of the rotation direction for twoside documents. The first page of hits will almost always solve your problem. the application is this, we need to do a print out of a data table with the. The value of a counter can be changed with a command of the type \setcounter{equation}{0}. The Swedish summer fades into a rainy autumn, which means it is time again to prepare lectures for the coming months. There is a directive called minsize which according to the logrotate man page is the only size directive that can be used in conjunction with the time ones. LaTeX has rules about how much text it requires a page to have; it will push figures back until it meets those rules. Rotate your grid to that angle. In the meanwhile, the attached file contains the rotated. com Rotate all pages. Default value is FALSE. As a long-time LaTeX user, I use the LaTeX Beamer package for all my slides. Both within LaTeX's \twocolumn and within the multicols environment defined by the multicol package (often preferrable to \twocolumn, since it balances the columns on the last page and lets you change between two- and one-column modes without starting a new page), you can insert full-width floats by using the starred form of the float. LaTeX: Changing the Font Size Latex provides 10 different font sizes. how do i rotate an image 90 degrees using html? yahoo. Then, create a new. If you are having a problem with graphic placement, search for " LaTeX graphic placement". x - Scientific WorkPlace & Scientific Word. Adobe Acrobat is the software you need to create a Portable Document Format (PDF) document from scratch or convert other document formats to PDF. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ly/puthesis Abstract. The package pgf/tikz can be used to create beautiful graphics, especially diagrams in LaTeX. Büchern, bei denen gerade und ungerade Seiten für beidseitigen Druck unterschiedlich gesetzt werden. You may want to have a look at the pdfpages package which "simplifies the inclusion of external multi-page PDF documents in LATEX documents". LaTeX examples: How to reference a figure or table | alvinalexander. I use LaTeX like this to combine pdfs this way and rotate their pages as necessary all the time — super convenient and I find it easier (and cheaper!) than Adobe Pro. The Control key no longer offers the rotation icon. When you click on the image or the selection with this tool a Rotation adjustment dialog is opened. But what I'd like to do is rotate the entire layout view so that my Portrait template is presented to me in Landscape on screen, so I can look at and manipulate the contents of the data frame in the same. Anyone who works with LaTeX knows how time-consuming it can be to find a symbol in symbols-a4. Inplace rotate square matrix by 90 degrees | Set 1. The pdfpages package makes it easy to insert pages of external PDF documents without worrying about the print space. SAi FlexiPRINT HP Basic Edition brings you an easy-to-use design-to-print workflow with closely integrated HP media management. Then the automatic pages rotation works fine with both packages lscape and pdflscape. 9, released on 10 August 2008. For script authors, the primary resource is Paul A. Recently I found out that I can't use Adobe Acrobat to rotate a multi-page pdf file by. If you save the pdf as an image (jpg) and use Microsoft Word you can rotate. I believe the time complexity is O(n2), but I'd like to know for sure, as well as. Know better sleep & comfortable adjustability with Sleep Number. However, I don't know how to rotate the H and the rest of the molecule when the H is in the plane of the page. In the secondary toolbar, click Crop Pages. Generate the entire document in PDF from LyX, load it into your favorite PDF viewer, print it to the PDF driver and extract just the pages you want. I am using TeXStudio to write my PhD thesis, and I am trying to format thumb indices using this source. 515 0 0 0 3. Besides 90 degrees rotation, the package also provides a command and environment to rotate content at an arbitrary angle. PDFView is an Universal Binary, and requires Mac OS X 10. When I asked the contractor, he asked his flooring guy who had been doing tile for 25 years and "never puts thinset under cement board". Open an example in ShareLaTeX Further reading. The package automatically takes care of the rotation direction for twoside documents. You can upload your own PDF file, paste the URL to an online PDF, and make your own PDF from scratch. What I like most is the reference manager and the ability to choose between latex, markdown and WYSIWYG per paragraph. You must convert your DVI file to a PDF document and view it in a PDF viewer to see the rotation in effect. There are various available sophisticated computer programs in the market that claim to do so, but PDFelement Pro PDFelement Pro remains unrivaled due to its efficiency and performance. Manual improvements - Pepper the manual with screen shots and graphics This will be of interest for the HTML and TeX renditions of the texinfo manual. So I have a regular PDF file (flat scan - no text) that I need to rotate. Some packages come with the LATEX 2" base distribution (See Table 1. Should I Rotate my Leesa or Leesa Hybrid Mattress? When it comes to rotation of your mattress, many Leesa sleepers like to periodically rotate their foam or hybrid mattress to promote even wear. This is not a derivative of Ghostview. ly/puthesis Abstract. Plots in LaTeX - Visualize data with pgfplots The pgfplots package from tikz/pgf enabled you to plot data directly from. Sale ends soon. Each volume of the Code contains amendments published in the Federal Register since the last revision of that volume of the Code. I have a scanned PDF document on an iPad 2 (iOS 6) that I need to rotate but cannot find any easy way to do this. Location: CTAN Comprehensive T e X Archive Network The Comprehensive T e X Archive Network. Next comes the actual table. With it, you can quickly switch data from columns to rows, or vice versa. The document viewer then rotates this page when viewing. Many applications that produce PostScript files are too lazy to calculate the true bounding box, or they think they are generating a whole page with a figure somewhere in the middle. There are instructions on how to save your vector images in a PDF format understood by LaTeX and have LaTeX manage the text styles and sizes in the image automatically. Can I do this with LaTex?. There, you can set the rotation axis, marked with a point, and the rotation angle. Poppler remained as alternative engine for a time, and from version 0. edu Address Mechanical Engineering Department Northwestern University 2145 Sheridan Road Evanston IL 60208 USA Abstract This article provides useful tools to write a thesis with LATEX. It would be too tedious given that I will have multi-page tables. The > pages of the book are typeset for A5 paper with LaTeX, so producing a > PostScript or PDF file is not a problem. For example, if your data looks like this, with Sales Regions in the column headings and and Quarters along the left side:. The main difference between the various types of matrix is the kind of delimeters that surround them. Note 2: The resulting image using this procedure will be a raster. Graphics in LATEX using TikZ Zo a Walczak Faculty of Mathematics and Computer Science, University of Lodz zofiawal (at) math dot uni dot lodz dot pl Abstract In this paper we explain some of the basic and also more advanced features of. I don't know if/how it will work with multiple pages, but you can extract one page of interest with pdftk: pdftk A=myfile. Zoom and rotate pages, use bookmarks to navigate in documents, Windows 8, follow these steps to launch PDF Reader if you do not have a touch screen: Rotate A Page documents | PDFs Download. 1958 Simmons is the first company to introduce the Queen and King bed sizes. I have a scanned PDF document on an iPad 2 (iOS 6) that I need to rotate but cannot find any easy way to do this. Dec 17, 2018 · RotaTeq (rotavirus oral vaccine) contains up to five strains of rotavirus. While it isn't mentioned on the ViSpring mattress care page, I've always found it useful to rotate any "active" boxspring/torsion system at least once a year. Alt click with line tool to get the angle. It will simply draw a horizontal line between the paragraphs that proceed and follow it when the document is rendered. See the article about headers and footers for more concise information and examples about fancyhdr. Donna Baker – November 14, 2013 Open the document with pages you want to resize by cropping, and click the Page Thumbnails tab. As a long-time LaTeX user, I use the LaTeX Beamer package for all my slides. you rotate, you rotate about the point on the left-hand edge of the box that meets the baseline. PDF Rotate Pages 1. I'd like to describe the y-axis with some text. ps should be used, means latex->dvips->ps2pdf. Steps to rotate a vector:-Calculate the Perpendicular component, multiply it by the cosine of the angle you are trying to rotate through-Calculate the cross product between the vector you are trying to rotate about and the vector you are rotating, multiply it by the sine of the angle-Add the results together. Download; PDF Rotate Pages is a high performance server tool from Traction Software for Windows PC, Unix Aix, Linux, Macintosh OSX & SUN Sparc Solaris NOTE: This Software does NOT require Acrobat, PDF Rotate Pages is designed for server usage. I'm not really sure if this is a well-defined and widely accepted difference, but to me, "Rotate about" implies that the object is spinning in place (i. The idea behind the minipage command is that within an existing page "built in" an additional page. Also other projects work more and more on the wiki. , \rotateleft. The environment rotate and command turnbox are equivalent: they rotate but in a box of zero size, so that the contents overlap the surroundings. For submission to The PracTEX Journal. If used with latex->dvipdf, the pages in pdf are not rotated (the same happans to lscape). May 11, 2017 · Open Inkscape, click on the text tool (capital A) and type in your text in your document. com by David Winterbottom #:2# # commandlinefu. It allows you to specify the column headings such that it prints on each page. I used to utilize the landscape environment to rotate tables, but I don´t like that this also rotates the section titles, as this is quite unhandy when flipping through the hardcopy version (of course with the digital PDF this is no problem). Export (png, jpg, gif, svg, pdf) and save & share. Delete Select the page that needs to be deleted then: (a) click the Delete Pages(s) button, or (b) right-click then choose Delete Page(s) from the menu. To rotate around multiple local origins, change the pivot point to "individual origins. How to rotate an image, table or paragraph in LaTeX by. How to Import/Place an Image in a PDF Just a bit over three years ago, I sat in the audience of my first legal technology conference. OK, assuming that you can now easily get your images converted in to EPS, it is time to see the mechanisms for importing them in LaTeX. In GNUPlot, the text is rotated. Cheers, Tom. Shows how to determine both the figure number and the page number a figure is on, using the \ref and \pageref commands. You begin figures with \begin{figure}[loc] where loc is a sequence of 0 to 4 letters, each one specifying a location where the figure or table may be placed, as follows:. And typeset documents with LaTeX with just the click of a button. LaTeXDraw is developed in Java and thus runs on top of Linux, Windows, and Mac OS X. I had originally posted a comment on this blogger's post asking for help because I didn't get what to do with everything and where to put it. Header = 1” Footer = 1” • Depending on the computer system being used, 0. Each page has two buttons to rotate that single page: or. This will automatically rotate the table for us, so it can be read when flipping the page sideways: Tables from Excel (. In Excel, we can rotate a cell text by 45 or 90 degrees as we need, but have you ever tried to rotate the footer or header text by 90 degrees? Supposing you want to print a sheet in landscape mode, and you want to rotate the header or footer for viewing the page number conveniently as below screenshot show. During the last few months, I have developed some new helper commands and environments for LaTeX slides, which I want to present here. pdf You can do much more in pdftk, changing page order, rotation, merging various PDF files. Quand je di. Hellmund (Leipzig) PGF/TikZ 1 / 22. Then the automatic pages rotation works fine with both packages lscape and pdflscape. All support on the internet seems to be the other way around. Mix PDF files where a number of PDF files are merged, taking pages alternately from them. pdf cat A1 output page1. SAi FlexiPRINT HP Basic Edition brings you an easy-to-use design-to-print workflow with closely integrated HP media management. pdf with the PDF rotated by 1. From open source projects to private team repositories, we’re your all-in-one platform for collaborative development. Je veux que l'orientation de la page sur laquelle le tableau est tracé soit en pysage (et non en portrait comme pour l'emplacement conventionel des pages). This is handy for those of you that have particularly large figures or tables, that would fit better in landscape. Yes I have auto-rotate checked, I did a G sensor cal both with it checked and not checked. Drag the Rotate wheel, or enter a degree value in the adjacent field to specify the angle at which you want to rotate the object. = path d=3D"M27. Jun 25, 2008 · How can I rotate a table 90 degrees in microsoft word? I need to rotate the entire table, including text by 90 degrees, so I can make the columns wider. It is recommended to use more than one positioning parameter to prevent unexpected results. I'd like to describe the y-axis with some text. To rotate your screen with hotkeys, press Ctrl+Alt+Arrow. As a long-time LaTeX user, I use the LaTeX Beamer package for all my slides. There is a maximum of 9 footnotes with symbols can be counted. Header = 1” Footer = 1” • Depending on the computer system being used, 0. Oct 01, 2013 · Our real-time collaborative LaTeX editor automatically produces & displays your typeset document. txt) or read online for free. In either case, the bounding box corresponds to the full page even though the printed figure occupies only a portion of it. I used to utilize the landscape environment to rotate tables, but I don´t like that this also rotates the section titles, as this is quite unhandy when flipping through the hardcopy version (of course with the digital PDF this is no problem). \begin{rotate}{angle} text \end{rotate} \turnbox{angle}{text} where sideways rotates text by 90 , turn by an arbitrary angle. A blank page opens for you to add content. Design by Made By Argon. Kind regards. com by David Winterbottom #:2# # commandlinefu. Animation 1: Screen capture of procedure above. 63 m / 64 in) 6 Based on a comparison of HP Latex Ink technology to competitors with leading market share as of December, 2013 and analysis of published MSDS/SDSs and/or internal evaluation. Here you can see a diagram showing all the variables defining the page. In the fourth video in this series we look at adding images into LaTeX documents. How to rotate an image in gs. Extract pages from multiple PDF files. If you want to rotate all pages, click on one, then press Ctrl+a to select all. It will allow you add multiple files at once and rearrange them in the order you prefer. This post will tell you which one is the best. How to Track and Rotate Text in Inkscape. 2645 contributors have contributed to it. There is a directive called minsize which according to the logrotate man page is the only size directive that can be used in conjunction with the time ones. To start a new section in your document, heed these steps: Position the toothpick cursor where you want the new section to start. For example, a spreadsheet could print horizontally, while a newsletter could print vertically. Yes I have auto-rotate checked, I did a G sensor cal both with it checked and not checked. Beside the fact is made with latex, which makes it even more awesome. tex with the content above and then I run pdflatex rotated. It is a Java application that reads a formatting object (FO) tree and renders the resulting pages to a specified output. 1 day ago · Overleaf rotate table. The service enables you to easily rotate, manage, and retrieve database credentials, API keys, and other secrets throughout their lifecycle. Use a PDF printer driver (such as PDFCreator for the Windows platform or cups-pdf on Linux) that can output selected pages. Writing a thesis with LATEX Lapo F. May 11, 2017 · Open Inkscape, click on the text tool (capital A) and type in your text in your document. 75” may need to set for the Header and Footer. It enables you to create vector graphics from within your document, without the need of external tools such as Inkscape or Adobe Illustrator and thus is much more flexible. Please try again later. You mentioned previously that Urban Mattress was coming out three times a year to rotate your mattress, so see if they will rotate your box spring the next time. The easiest solution is to put a minipage environment inside a node. Each provides a table for expressions, aligned in rows and columns. CTAN has currently 5768 packages. A menu appears, listing several items. Rotate [g, θ] is equivalent to Rotate [g, θ, {Center, Center}]. Probably not the solution you are after but a barely workable workaround. 1995 The iconic bowling ball commercial shows not all spring mattresses are the same when it comes to motion transfer. Page size is automatically determined from the Document Structuring Comments. I don't know if/how it will work with multiple pages, but you can extract one page of interest with pdftk: pdftk A=myfile. For a more in depth. pdf), Text File (. I want to rotate 180 degrees every odd page using pdfjam. PSfrag is a LaTeX package that allows one to overlay Encapsulated PostScript (EPS) figures with arbitrary LaTeX constructions, properly aligned, scaled, and rotated. Sumatra PDF Viewer is a lightweight free Windows program lets you view PDF documents without a lot of extra features that slow things down. The tabular environment is the default L a T e X method to create tables. A minimal example of rotated text in LaTeX. to be done with the mouse instead of left/right arrow and Page Up/ Page Down. Ask Question How to create a single page PDF file. I use LaTeX like this to combine pdfs this way and rotate their pages as necessary all the time — super convenient and I find it easier (and cheaper!) than Adobe Pro. It is built on top of the Donald Knuth's TeX typesetting system. To rotate your screen with hotkeys, press Ctrl+Alt+Arrow. This will automatically rotate the table for us, so it can be read when flipping the page sideways: Tables from Excel (. I'm looking for easy-to-use desktop software that shows thumbnails of the pages in a PDF document, and the pages can be sorted using click-and-drag, right-click rotate, or right-click delete. 055 0 0 1-5. It is made from both human and animal sources. This package is not part of the graphics bundle and must be obtained. 2645 contributors have contributed to it. This document describes how to use the puthesis (Purdue University thesis) LaTeX typesetting system documentclass to format Purdue University master's bypass reports, master's theses, PhD dissertations, and PhD preliminary reports. However, this does not rotate my caption, as it has to be within the table environment. The Edit Text & Images tool outlines each text box, so it is clear what text is affected. png's--you can even use a 1 page. U+2190 – U+21FF (8592–8703) Characters 8592–8597 and 8616 are present in Microsoft’s WGL4 character set. Note 2: The resulting image using this procedure will be a raster. These packages can be defined by clicking on the button: at the bottom of the user interface. centering”, but about the plugin that do you use for write latex code in your page. Single pages or page ranges can be selected to create a new PDF file containing only the pages wanted. Rotate and wrap text within node using Tikz First post but I am hoping someone can help. Here you can see a diagram showing all the variables defining the page. A minimal example of rotated text in LaTeX. If TRUE, the column names are displayed vertically in LaTeX. They have been extracted from the reference manuals. You may need to move the objects a little to line them up to the page after doing all. download overleaf rotate table free and unlimited. The pdfpages Package Andreas MATTHIAS [email protected] When I imagine tilting the molecule back I get a see-saw shaped molecule obviously incorrect. Plots in LaTeX - Visualize data with pgfplots The pgfplots package from tikz/pgf enabled you to plot data directly from. You cannot drag a text block to another page, or move or rotate individual characters or words within a text box. Infection with rotavirus can affect the digestive system of babies and young children, causing severe stomach or intestinal illness. The different font sizes are listed below. In any case, most of the brands recommend that you should rotate or flip the bed anywhere from once every three months to once a year. ly/puthesis Abstract. The \fancyplain command takes two arguments: one for pages LaTeX forces into plain (first pages of chapters, etc), and one for the others. Alt click with line tool to get the angle. at 2002/12/14 Abstract This package makes it easy to insert pages of external PDF documents. Click the Layout tab on the Ribbon. Export (png, jpg, gif, svg, pdf) and save & share. I have been reading on here about floating text boxes and placing them over etc but that isn't working for me 😟 I can rotate the text but not place it. More advanced and powerful features are therefore not working, e. pdf \includegraphics[page=1]{Test} \end{document} put more then one page on a page By using option nup from usepackage pdfpages, it is also possible to put more then one page from the origin pdf file on one page of the new document. How to Rotate Linked or Embedded Objects in MS Word 2010 - Duration: 8:07. Placement for page number in the upper right 1” margin → 1. Online LaTeX editor with autocompletion, highlighting and 400 math symbols. What happens when we rotate the vector\vec j=\langle 0,1 \rangle\$ in the same way. Here you can see a diagram showing all the variables defining the page. Just in case someone else stumbles upon this answer in search of a rotation of 180 degrees (making the orientation 270 degrees): don't forget to account for the extent of the label. Hey there! I have a couple of tables that are both too wide for portrait and too long to fit on one single page. These packages can be defined by clicking on the button: at the bottom of the user interface. A coding challenge that rotates an image 90 degrees counterclockwise. This post will tell you which one is the best.


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# 12.3 million of concurrent WebSockets

One thing about WebSockets is that you need a lot of resources on the client's side to generate high enough load for the server to actually eat up all the CPU resources.

There are several challenges you have to overcome because the WebSockets protocol is more CPU demanding on the client's side than on the server's side. At the same time you need a lot of RAM to store information about open connections if you have millions of them.

I've been lucky enough to get a couple of new servers for a limited period of time at my disposal for the hardware "burnout" tests. So I decided to use my Lua Application Server — LAppS to do both jobs: test the hardware and perform the LAppS high load tests.

Let's see the details

# Challenges

First of all WebSockets on the client's side require a pretty fast RNG for traffic masking unless you want to fake it. I wanted the tests to be realistic, so I discarded the idea of faking the RNG by replacing it with a constant. This leaves the only option — a lot of CPU power. As I said, I have two servers: one with dual Intel Gold 6148 CPUs (40 cores in total) 256GB RAM (DDR4-2666) and one with quad Intel Platinum 8180 CPUs (112 cores in total) 384GB RAM (DDR4-2666). I'll be referring to them as Server A and Server B hereafter.

The second challenge — there are no libraries or test suites for WebSockets that are fast enough. Which is why I was forced to develop a WebSockets client module for LAppS (cws).

# The tests setup

Both servers have dual port 10GbE cards. Port 1 of each card is aggregated to a bonding interface and these ports on either cards are connected to each other in RR mode. Port 2 of each card is aggregated to a bonding interface in an Active-Backup mode, connected over an Ethernet switch. Each hardware server was running the RHEL 7.6. I had to install gcc-8.3 from the sources to have a modern compiler instead of the default gcc-4.8.5. Not the best setup for the performance tests, yet beggars can't be choosers.

CPUs on Server A (2xGold 6148) have higher frequency than on Server B (4xPlatinum 8180). At the same time Server B has more cores, so I decided to run an echo server on Server A and the echo clients on Server B.

I wanted to see how LAppS behaves under high load with millions of connections and what maximum amount of echo requests per second it can serve. These are actually two different sets of tests because with the connections amount growth the server and clients are doing an increasingly complex job.

Before this I have made all the tests on my home PC which I use for development. The results of these tests will be used for comparison. My home PC has an Intel Core i7-7700 CPU 3.6GHz (4.0GHz Turbo) with 4 cores and 32GB of DDR4-2400 RAM. This PC runs Gentoo with 4.14.72 kernel.

Specter and Meltdown patch levels
• Home PC
/sys/devices/system/cpu/vulnerabilities/l1tf:Mitigation: PTE Inversion; VMX: conditional cache flushes, SMT vulnerable
/sys/devices/system/cpu/vulnerabilities/meltdown:Mitigation: PTI
/sys/devices/system/cpu/vulnerabilities/spec_store_bypass:Mitigation: Speculative Store Bypass disabled via prctl and seccomp
/sys/devices/system/cpu/vulnerabilities/spectre_v1:Mitigation: __user pointer sanitization
/sys/devices/system/cpu/vulnerabilities/spectre_v2:Vulnerable: Minimal generic ASM retpoline, IBPB, IBRS_FW
• Servers A and B
/sys/kernel/debug/x86/ibpb_enabled : 1
/sys/kernel/debug/x86/pti_enabled : 1
/sys/kernel/debug/x86/ssbd_enabled : 1
/sys/kernel/debug/x86/ibrs_enabled : 1
/sys/kernel/debug/x86/retp_enabled : 3

I'll make an extra note with the results of the tests of Servers A and B whether these patches are enabled or not.

On my Home PC patch level is never changed.

## Installing LAppS 0.8.1

Installing prerequisites and the LAppS

LAppS depends on the luajit-2.0.5 or higher, the libcrypto++8.2 and the wolfSSL-3.15.7 libraries and they have to be installed from sources in RHEL 7.6 and likely in any other linux distribution.

The prefix for installation is /usr/local. Here is the pretty much self-explanatory Dockerfile part for wolfSSL installation

ADD https://github.com/wolfSSL/wolfssl/archive/v3.15.7-stable.tar.gz ${WORKSPACE} RUN tar xzvf v3.15.7-stable.tar.gz WORKDIR${WORKSPACE}/wolfssl-3.15.7-stable

RUN ./autogen.sh

RUN ./configure CFLAGS="-pipe -O2 -march=native -mtune=native -fomit-frame-pointer -fstack-check -fstack-protector-strong -mfpmath=sse -msse2avx -mavx2 -ftree-vectorize -funroll-loops -DTFM_TIMING_RESISTANT -DECC_TIMING_RESISTANT -DWC_RSA_BLINDING" --prefix=/usr/local --enable-tls13 --enable-openssh --enable-aesni --enable-intelasm --enable-keygen --enable-certgen --enable-certreq --enable-curve25519 --enable-ed25519 --enable-intelasm --enable-harden

RUN make -j40 all

RUN make install

And here is the part for the libcrypto++ installation

ADD https://github.com/weidai11/cryptopp/archive/CRYPTOPP_8_2_0.tar.gz ${WORKSPACE} RUN rm -rf${WORKSPACE}/cryptopp-CRYPTOPP_8_2_0

RUN tar xzvf ${WORKSPACE}/CRYPTOPP_8_2_0.tar.gz WORKDIR${WORKSPACE}/cryptopp-CRYPTOPP_8_2_0

RUN make CFLAGS="-pipe -O2 -march=native -mtune=native -fPIC -fomit-frame-pointer -fstack-check -fstack-protector-strong -mfpmath=sse -msse2avx -mavx2 -ftree-vectorize -funroll-loops" CXXFLAGS="-pipe -O2 -march=native -mtune=native -fPIC -fomit-frame-pointer -fstack-check -fstack-protector-strong -mfpmath=sse -msse2avx -mavx2 -ftree-vectorize -funroll-loops" -j40 libcryptopp.a libcryptopp.so

RUN make install

And the luajit

ADD http://luajit.org/download/LuaJIT-2.0.5.tar.gz ${WORKSPACE} WORKDIR${WORKSPACE}

RUN tar xzvf LuaJIT-2.0.5.tar.gz

WORKDIR ${WORKSPACE}/LuaJIT-2.0.5 RUN env CFLAGS="-pipe -Wall -pthread -O2 -fPIC -march=native -mtune=native -mfpmath=sse -msse2avx -mavx2 -ftree-vectorize -funroll-loops -fstack-check -fstack-protector-strong -fno-omit-frame-pointer" make -j40 all RUN make install One optional dependency of the LAppS, which may be ignored, is the new library from Microsoft mimalloc. This library makes significant performance improvement (about 1%) but requires cmake-3.4 or higher. Given the limited amount of time for the tests, I decided to sacrifice mentioned performance improvement. On my home PC, I'll not disable mimalloc during the tests. Lets checkout LAppS from the repository: WORKDIR${WORKSPACE}

RUN rm -rf ITCLib ITCFramework lar utils LAppS

RUN git clone https://github.com/ITpC/ITCLib.git

RUN git clone https://github.com/ITpC/utils.git

RUN git clone https://github.com/ITpC/ITCFramework.git

RUN git clone https://github.com/ITpC/LAppS.git

RUN git clone https://github.com/ITpC/lar.git

WORKDIR ${WORKSPACE}/LAppS Now we need to remove "-lmimalloc" from all the Makefiles in nbproject subdirectory, so we can build LAppS (assuming our current directory is${WORKSPACE}/LAppS)

# find ./nbproject -type f -name "*.mk" -exec sed -i.bak -e 's/-lmimalloc//g' {} \;
# find ./nbproject -type f -name "*.bak" -exec rm {} \;

And now we can build LAppS. LAppS provides several build configurations, which may or may not exclude some features on the server side:

• with SSL support and with statistics gathering support
• with SSL and without statistics gathering (though minimal statistics gathering will continue as it is used for dynamic LAppS tuning in runtime)
• without SSL and without statistic gathering.

Before next step make sure you are the owner of the /opt/lapps directory (or run make installs with sudo)

sed -e "s/IPADDR/$(cat IP1)/g" /opt/lapps/apps/benchmark/benchmark.lua > /opt/lapps/apps/benchmark${i}/benchmark${i}.lua done Now we modify the /opt/lapps/etc/conf/lapps.json on Server B to use these two benchmark services: Server B lapps.json { "directories": { "app_conf_dir": "etc", "applications": "apps", "deploy": "deploy", "tmp": "tmp", "workdir": "workdir" }, "services": { "benchmark1": { "auto_start": true, "instances" : 112, "internal": true, "preload": [ "nap", "cws", "time" ] }, "benchmark2": { "auto_start": true, "instances" : 112, "internal": true, "preload": [ "nap", "cws", "time" ] } } } Are we ready? No we are not. Because we are intent to generate 2 240 000 outgoing sockets to only two addresses and we need more ports on the server side. It is impossible to create more then 64k connections to the same ip:port pair (actually a little less then 64k). On the Server A in the file LAppS/include/wsServer.h there is a function void startListeners(). In this function we will replace line 126 LAppSConfig::getInstance()->getWSConfig()["port"], with this: static_cast<int>(LAppSConfig::getInstance()->getWSConfig()["port"])+i, Rebuild LAppS: make CONF=Release.AVX2 install # Running the tests for 2 240 000 client WebSockets. Start the LAppS on Server A, then start the LAppS on Server B and redirect the output to a file like this: /opt/lapps/bin/lapps.avx2 > log Edit, NB: if you want to run several LAppS instances, then create separate directory for each instnace (like run1,run2, etc) and run each instance from within these directories. This is required for lapps.log file and of course for resulting standard output, for not to be overlapped/overwritten by concurrent LAppS instances.  This may take a while till all the connections are established. Let's monitor the progress. Do not use netstat to watch established connections, it is pointless while it runs indefinitely after like 150k connections which are established in some seconds. Look at the lapps.log on the Server A, in the directory which you were in when started LAppS. You may use following one-liner to see how the connections are established and how are they distributed among IOWorkers: date;awk '/ will be added to connection pool of worker/{a[$22]++}END{for(i in a){ print i, a[i]; sum+=a[i]} print sum}' lapps.log | sort -n;date

Here is an idea on how rapidly these connections are established:

375874
Sun Jul 21 20:33:07 +06 2019

650874
Sun Jul 21 20:34:42 +06 2019   2894 connections per second

1001874
Sun Jul 21 20:36:45 +06 2019   2974 connections per second

1182874
Sun Jul 21 20:37:50 +06 2019   2784 connections per second

1843874
Sun Jul 21 20:41:44 +06 2019   2824 connections per second

2207874
Sun Jul 21 20:45:43 +06 2019   3058 connections per second

On the Server B we may check amount of benchmarks finished establishing their connections:

# grep Sockets log | wc -l
224

After you see that the number is 224 let the servers work for a while, monitor the CPU and memory usage with top. You might see something like this (Server B at left and Server A at the right side):

Or like this (Server B at left and Server A at the right side):

It is clearly the Server B, where the benchmark client services are running, is under heavy load. The Server A is under heavy load too but not always. Some times it chills while the Server B struggles with amount of tasks to work on. The traffic ciphering on TLS is very CPU intensive.

Let's stop the servers (press Ctrl-C several times) and modify the log as before but with respect to changed amount of benchmark services (224):

echo -e ':%s/ms/^V^M/g\n:%g/^$/d\nGdd1G224/Sockets\n224\ndggZZ' | vi$i
awk -v avg=0 '{rps=($1/$5);avg+=rps;}END{print ((avg*1000)/NR)*224}' log

You might want to delete several last lines as well, to account for print outs from stopping benchmark services. You've already got an idea on how to do this. So I'll post the results for different test scenarios and proceed with problems I've faced.

# Results for all the other tests (4.9 million and beyond)

test #5 balanced for equal CPU sharing

test #6

test #8

test #12

test #13

# Problems.

Marked in above table with red.

Running more then 224 benchmark instances on the Server B proved to be the bad idea, as the server on client side was incapable of evenly distribute the CPU time among processes/threads. First 224 benchmark instances which have established all their connections took most of the CPU resources and the rest of benchmark instances are lagging behind. Even using renice -20 does not help a lot(448 benchmark instances, 2 LAppS instances):

448 benchmark instances

The Server B (left side) is under very heavy load and the Server A still has free CPU resources.

So I doubled the benchmark.max_connections instead of starting more of the separate LAppS instances.

Still for 12.3 million of WebSockets to run, I have started 5th LAppS instance (tests 5 and 6) without stopping four already running ones. And played the role of CFQ by manually suspending and restarting prioritized processes with kill -STOP/-CONT or/and renice their priorities. You can use following template script for this:

while [ 1 ];
do
kill -STOP <4 fast-processes pids>
sleep 10
kill -CONT <4 fast-processes pids>
sleep 5
done

Welcome to 2019 RHEL 7.6! Honestly, I used the renice command first time since 2009. What is worst, — I used it almost unsuccessfully this time.

I had a problem with scatter-gather engine of the NICs. So I disabled it for some tests, not actually marking this event into the table.

I had partial link disruptions under heavy load and the NIC driver bug, so I had to discard related test results.

# End of story

Honestly, the tests went much smoother than I've anticipated.

I'm convinced that I have not managed to load LAppS on Server A to its full potential (without SSL), because I had not enough CPU resources for the client side. Though with TLS 1.3 enabled, the LAppS on the Server A was utilizing almost all available CPU resources.

I'm still convinced that LAppS is most scalable and fastest WebSockets open source server out there and the cws WebSockets client module is the only of it's kind, providing the framework for high load testing.

Please verify the results on your own hardware.

Note of advice: Never use nginx or apache as the load balancer or as a proxy-pass for WebSockets, or you'll end up cutting the performance on order of magnitude. They are not build for WebSockets.

format: plain


text:
Sqrt(x)

Implement integer square root function. There are two ways to solve this problem. The simplier one is using binary search.

public class Sqrt {
public int mySqrt(int x) {
if (x <= 0) return 0;
if (x <= 3) return 1;

long a = 1;
long b = x >> 1;
long mid;

while (a <= b) {
mid = a + ((b - a) >> 1);
if (mid * mid > x) {
b = mid - 1;
} else if (mid * mid < x) {
a = mid + 1;
} else {
return (int)mid;
}
}

return (int)b;
}
}

A more advanced way is Newton's method. To apply Newton's method we have to take notice of:

1. Newton's method is possible to converge to a negative root. On LeetCode, a positive root is required.
2. Intermediate result may exceed int's range
3. We take a max difference of 1 as end condition. Require may lead to infinite looping.
public class Sqrt {
public int mySqrt(int x) {
if (x <= 0) return 0;
if (x <= 3) return 1;

long x0, x1 = x << 1;

do {
x0 = x1;
x1 = (x0 * x0 + x) / (2 * x0);
} while (Math.abs(x0 - x1) > 1);

if (x1 * x1 > x) return (int)Math.abs(x1) - 1;
else return (int)Math.abs(x1);
}
}

Yet another way is to use bit manipulation. We start from the highest bit position and gradually test if a bit can be 1 to the lowest bit position. As for a 32bit number, the highest 1 in binary of its square root is at most at the 16 bit position, so we start from there.

public class Sqrt {
public int mySqrt(int x) {
if (x <= 0) return 0;

long X = x;
long res = 0;
long marker = (1 << 16);

while (marker > 0) {
long tempr = res | marker;
while (tempr * tempr > X) {
marker >>= 1;
tempr = res | marker;
}
res |= marker;
marker >>= 1;
}

return (int)res;
}
}


text:
# Math Help - Show me where I went wrong, trig equation

1. ## Show me where I went wrong, trig equation

$cos\theta = cos(\theta+60)$

$cos\theta = cos\theta cos60 - sin\theta sin60$

$cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta$

than multiply eachterm by '2', so I get;

$2cos\theta = cos\theta - \sqrt{3}sin\theta$

$2cos\theta - cos\theta = - \sqrt{3}sin\theta$

$cos\theta = - \sqrt{3}sin\theta$

$\frac{cos\theta}{sin\theta} = -\sqrt{3}$

$tan\theta = -\sqrt{3}$

my book says the answer is $tan\theta = -\frac{1}{\sqrt{3}}$

Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

Thank you

2. Originally Posted by Tweety
$cos\theta = cos(\theta+60)$

$cos\theta = cos\theta cos60 - sin\theta sin60$

$cos\theta = \frac{1}{2}cos\theta-\frac{\sqrt{3}}{2} sin\theta$

than multiply eachterm by '2', so I get;

$2cos\theta = cos\theta - \sqrt{3}sin\theta$

$2cos\theta - cos\theta = - \sqrt{3}sin\theta$

$cos\theta = - \sqrt{3}sin\theta$

$\frac{cos\theta}{sin\theta} = -\sqrt{3}$

$tan\theta = -\sqrt{3}$

my book says the answer is $tan\theta = -\frac{1}{\sqrt{3}}$

Can someone please show me what I have done wrong, cause I have been through my working quite a few times still get the same answer!

Thank you
Only your last step is wrong

$\frac{cos(x)}{sin(x)} \neq tan(x)$

$\frac{cos(x)}{sin(x)} = cot(x) = \frac{1}{tan(x)}$

If you take the reciprocal of the answer you have it will be right

3. yes the last step cos / sin isnot tan ,, sin / cos is tan so it will be ( cos / sin ) = - square root ( 3 )
=> 1/tan = -square root (3) => tan = 1/-squr root 3
got it ?

4. AAH thanks, can't actually believe that I did not spot that myself!

I know that cos/sin does not equal tan, but was not looking at it properly.

5. it happens tweety


text:
# Cos -3pi 2

cos x - cos y = -2 sen( (x-y)/2 ) sen( (x + y)/2 ) Tabla Trig de Ángulos Ordinarios; ángulo 0 30 45 60 90; sen ^2 (a) 0/4 : 1/4 : 2/4 : 3/4 : 4/4 : cos ^2 (a) 4/4

Note: Chapter BC 2 was renumbered ch where sin 2 θ means (sin θ) 2 and cos 2 θ means (cos θ) 2. This can be viewed as a version of the Pythagorean theorem, and follows from the equation x 2 + y 2 = 1 for the unit circle. This equation can be solved for either the sine or the cosine: Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history cos x - cos y = -2 sin( (x-y)/2 ) sin( (x + y)/2 ) Trig Table of Common Angles; angle 0 30 45 60 90; sin 2 (a) 0/4 1/4 : 2/4 : 3/4 : 4/4 cos 2 (a) 4/4 : 3/4 2/4 : 1/4 College of the Sequoias is a two-year California community college offering educational & career technical programs for the residents of its district in Tulare & Kings counties. Cos(A + B) = Cos A cos B – Sin A sin B. Let’s equate B to A, i.e A = B. And then, the first of these formulae becomes: Cos(t + t) = Cos t cos t – Sin t sin t. so that Cos 2t = Cos 2 t – Sin 2 t. And this is how we get second double-angle formula, which is so called because you are doubling the angle (as in 2A).

sin (theta) = a / c. csc (theta) = 1 / sin (theta) = c / a. cos (theta) = b / c. sec (theta) = 1 / cos (theta) = c / b. tan (theta) = sin (theta) / cos (theta) = a / b. cot (theta) = 1/ tan (theta) = b / a. sin (-x) = -sin (x) Cos is the cosine function, which is one of the basic functions encountered in trigonometry.

## The easiest way to calculate this integral is to use a simple trick. First, we write \cos^2 (x) = \cos (x)\cos (x) and apply integration by parts: If we apply integration by parts to the rightmost expression again, we will get ∫\cos^2 (x)dx = ∫\cos^2 (x)dx, which is not very useful.

In other words, these two functions differ by 1. Log in or register to reply now! Forums. Homework Help.

### Solve for x cos(x)=2/5. Take the inverse cosine of both sides of the equation to extract from inside the cosine. Evaluate. The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from to find the solution in the fourth quadrant. Simplify .

sin (theta) = a / c. csc (theta) = 1 / sin (theta) = c / a. cos (theta) = b / c. sec (theta) = 1 / cos (theta) = c / b.

Layer 2 to Layer 3 mapping can also be configured by matching on the CoS value, since switches already have the capability to match and set CoS values.

See full list on trigidentities.net See full list on surgere.com cos 2 2 x cos 2 2 x sin 2 x dx cos 2 2 x cos 4 2 x sin 2 x dx 1 6 cos 3 2 x 1 from MATH 279 at University of California, Berkeley sin^2(x) + cos^2(x) = 1, so combining these we get the equation. cos(2x) = 2cos^2(x) -1. Now we can rearrange this to give: cos^2(x) = (1+cos(2x))/2. So we have an equation that gives cos^2(x) in a nicer form which we can easily integrate using the reverse chain rule.

cot ^2 (x) + 1 = csc ^2 (x). sen(x y) = sen x cos y cos x sen y. cos(x y) = cos x cosy sen x sen y If the inequality sin 2 x + a cos x + a 2 > 1 + cos x holds for any x ϵ R, then the largest negative integral value of a is View solution For which interval for θ , the inequation ( 2 sin 2 θ − 5 sin θ + 2 ) > 0 . 2 cos(A−A)+ 1 2 cos(A+A) cos2 A = 1 2 cos0+ 1 2 cos2A. Simplifying this and doing the same with equation (11) we ﬁnd the two identities cos2 A = 1 2 (1+cos2A), (14) sin2 A = 1 2 (1−cos2A). (15) 7 Identities involving tangent Finally, from equations (2) and (3) we can obtain an identity for tan(A+B): tan(A+B) = sin(A+B) cos(A+B) = sinAcosB 12/25/2017 1/24/2018 where sin 2 θ means (sin θ) 2 and cos 2 θ means (cos θ) 2.

4/12/2007 Solve for x cos(x)=2/5. Take the inverse cosine of both sides of the equation to extract from inside the cosine. Evaluate. The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from to find the solution in the fourth quadrant. Simplify .

It is a virtual document, similar to a database record.

csgo 20 ročná minca
trhový kôš sushi kalórie
btc cena najnovšie správy
5 000 pakistanských rupií v amerických dolároch
grafy top 200

### The following (particularly the first of the three below) are called "Pythagorean" identities. sin2(t) + cos2(t) = 1. tan2(t) + 1 = sec2(t). 1 + cot2(t) = csc2(t).

Cos 2.06 Responsibilities of owners. Cos 2.07 Responsibilities of the cosmetology manager. Cos 2.08 Responsibilities of licensees. Note: Chapter BC 2 was renumbered ch where sin 2 θ means (sin θ) 2 and cos 2 θ means (cos θ) 2. This can be viewed as a version of the Pythagorean theorem, and follows from the equation x 2 + y 2 = 1 for the unit circle.


text:
# Spin representation

In mathematics, the spin representations are particular projective representations of the orthogonal or special orthogonal groups in arbitrary dimension and signature (i.e., including indefinite orthogonal groups). More precisely, they are representations of the spin groups, which are double covers of the special orthogonal groups. They are usually studied over the real or complex numbers, but they can be defined over other fields.

Elements of a spin representation are called spinors. They play an important role in the physical description of fermions such as the electron.

The spin representations may be constructed in several ways, but typically the construction involves (perhaps only implicitly) the choice of a maximal isotropic subspace in the vector representation of the group. Over the real numbers, this usually requires using a complexification of the vector representation. For this reason, it is convenient to define the spin representations over the complex numbers first, and derive real representations by introducing real structures.

The properties of the spin representations depend, in a subtle way, on the dimension and signature of the orthogonal group. In particular, spin representations often admit invariant bilinear forms, which can be used to embed the spin groups into classical Lie groups. In low dimensions, these embeddings are surjective and determine special isomorphisms between the spin groups and more familiar Lie groups; this elucidates the properties of spinors in these dimensions.

## Set-up

Let V be a finite-dimensional real or complex vector space with a nondegenerate quadratic form Q. The (real or complex) linear maps preserving Q form the orthogonal group O(V, Q). The identity component of the group is called the special orthogonal group SO(V, Q). (For V real with an indefinite quadratic form, this terminology is not standard: the special orthogonal group is usually defined to be a subgroup with two components in this case.) Up to group isomorphism, SO(V, Q) has a unique connected double cover, the spin group Spin(V, Q). There is thus a group homomorphism h: Spin(V, Q) → SO(V, Q) whose kernel has two elements denoted {1, −1}, where 1 is the identity element. Thus, the group elements g and −g of Spin(V, Q) are equivalent after the homomorphism to SO(V, Q); that is, h(g) = h(−g) for any g in SO(V, Q).

The groups O(V, Q), SO(V, Q) and Spin(V, Q) are all Lie groups, and for fixed (V, Q) they have the same Lie algebra, so(V, Q). If V is real, then V is a real vector subspace of its complexification VC = VR C, and the quadratic form Q extends naturally to a quadratic form QC on VC. This embeds SO(V, Q) as a subgroup of SO(VC, QC), and hence we may realise Spin(V, Q) as a subgroup of Spin(VC, QC). Furthermore, so(VC, QC) is the complexification of so(V, Q).

In the complex case, quadratic forms are determined uniquely up to isomorphism by the dimension n of V. Concretely, we may assume V = Cn and

${\displaystyle Q(z_{1},\ldots ,z_{n})=z_{1}^{2}+z_{2}^{2}+\cdots +z_{n}^{2}.}$

The corresponding Lie groups are denoted O(n, C), SO(n, C), Spin(n, C) and their Lie algebra as so(n, C).

In the real case, quadratic forms are determined up to isomorphism by a pair of nonnegative integers (p, q) where n = p + q is the dimension of V, and pq is the signature. Concretely, we may assume V = Rn and

${\displaystyle Q(x_{1},\ldots ,x_{n})=x_{1}^{2}+x_{2}^{2}+\cdots +x_{p}^{2}-(x_{p+1}^{2}+\cdots +x_{p+q}^{2}).}$

The corresponding Lie groups and Lie algebra are denoted O(p, q), SO(p, q), Spin(p, q) and so(p, q). We write Rp,q in place of Rn to make the signature explicit.

The spin representations are, in a sense, the simplest representations of Spin(n, C) and Spin(p, q) that do not come from representations of SO(n, C) and SO(p, q). A spin representation is, therefore, a real or complex vector space S together with a group homomorphism ρ from Spin(n, C) or Spin(p, q) to the general linear group GL(S) such that the element −1 is not in the kernel of ρ.

If S is such a representation, then according to the relation between Lie groups and Lie algebras, it induces a Lie algebra representation, i.e., a Lie algebra homomorphism from so(n, C) or so(p, q) to the Lie algebra gl(S) of endomorphisms of S with the commutator bracket.

Spin representations can be analysed according to the following strategy: if S is a real spin representation of Spin(p, q), then its complexification is a complex spin representation of Spin(p, q); as a representation of so(p, q), it therefore extends to a complex representation of so(n, C). Proceeding in reverse, we therefore first construct complex spin representations of Spin(n, C) and so(n, C), then restrict them to complex spin representations of so(p, q) and Spin(p, q), then finally analyse possible reductions to real spin representations.

## Complex spin representations

Let V = Cn with the standard quadratic form Q so that

${\displaystyle {\mathfrak {so}}(V,Q)={\mathfrak {so}}(n,\mathbb {C} ).}$

The symmetric bilinear form on V associated to Q by polarization is denoted .,..

### Isotropic subspaces and root systems

A standard construction of the spin representations of so(n, C) begins with a choice of a pair (W, W) of maximal totally isotropic subspaces (with respect to Q) of V with WW = 0. Let us make such a choice. If n = 2m or n = 2m + 1, then W and W both have dimension m. If n = 2m, then V = WW, whereas if n = 2m + 1, then V = WUW, where U is the 1-dimensional orthogonal complement to WW. The bilinear form .,. associated to Q induces a pairing between W and W, which must be nondegenerate, because W and W are totally isotropic subspaces and Q is nondegenerate. Hence W and W are dual vector spaces.

More concretely, let a1, am be a basis for W. Then there is a unique basis α1, ... αm of W such that

${\displaystyle \langle \alpha _{i},a_{j}\rangle =\delta _{ij}.}$

If A is an m × m matrix, then A induces an endomorphism of W with respect to this basis and the transpose AT induces a transformation of W with

${\displaystyle \langle Aw,w^{*}\rangle =\langle w,A^{\mathrm {T} }w^{*}\rangle }$

for all w in W and w in W. It follows that the endomorphism ρA of V, equal to A on W, AT on W and zero on U (if n is odd), is skew,

${\displaystyle \langle \rho _{A}u,v\rangle =-\langle u,\rho _{A}v\rangle }$

for all u, v in V, and hence (see classical group) an element of so(n, C) ⊂ End(V).

Using the diagonal matrices in this construction defines a Cartan subalgebra h of so(n, C): the rank of so(n, C) is m, and the diagonal n × n matrices determine an m-dimensional abelian subalgebra.

Let ε1, εm be the basis of h such that, for a diagonal matrix A, εk(ρA) is the kth diagonal entry of A. Clearly this is a basis for h. Since the bilinear form identifies so(n, C) with ${\displaystyle \wedge ^{2}V}$, explicitly,

${\displaystyle x\wedge y\mapsto \varphi _{x\wedge y},\quad \varphi _{x\wedge y}(v)=2(\langle y,v\rangle x-\langle x,v\rangle y),\quad x\wedge y\in \wedge ^{2}V,\quad x,y,v\in V,\quad \varphi _{x\wedge y}\in {\mathfrak {so}}(n,\mathbb {C} ),}$[1]

it is now easy to construct the root system associated to h. The root spaces (simultaneous eigenspaces for the action of h) are spanned by the following elements:

${\displaystyle a_{i}\wedge a_{j},\;i\neq j,}$ with root (simultaneous eigenvalue) ${\displaystyle \varepsilon _{i}+\varepsilon _{j}}$
${\displaystyle a_{i}\wedge \alpha _{j}}$ (which is in h if i = j) with root ${\displaystyle \varepsilon _{i}-\varepsilon _{j}}$
${\displaystyle \alpha _{i}\wedge \alpha _{j},\;i\neq j,}$ with root ${\displaystyle -\varepsilon _{i}-\varepsilon _{j},}$

and, if n is odd, and u is a nonzero element of U,

${\displaystyle a_{i}\wedge u,}$ with root ${\displaystyle \varepsilon _{i}}$
${\displaystyle \alpha _{i}\wedge u,}$ with root ${\displaystyle -\varepsilon _{i}.}$

Thus, with respect to the basis ε1, εm, the roots are the vectors in h that are permutations of

${\displaystyle (\pm 1,\pm 1,0,0,\dots ,0)}$

together with the permutations of

${\displaystyle (\pm 1,0,0,\dots ,0)}$

if n = 2m + 1 is odd.

A system of positive roots is given by εi + εj (ij), εi εj (i < j) and (for n odd) εi. The corresponding simple roots are

${\displaystyle \varepsilon _{1}-\varepsilon _{2},\varepsilon _{2}-\varepsilon _{3},\ldots ,\varepsilon _{m-1}-\varepsilon _{m},\left\{{\begin{matrix}\varepsilon _{m-1}+\varepsilon _{m}&n=2m\\\varepsilon _{m}&n=2m+1.\end{matrix}}\right.}$

The positive roots are nonnegative integer linear combinations of the simple roots.

### Spin representations and their weights

One construction of the spin representations of so(n, C) uses the exterior algebra(s)

${\displaystyle S=\wedge ^{\bullet }W}$ and/or ${\displaystyle S'=\wedge ^{\bullet }W^{*}.}$

There is an action of V on S such that for any element v = w + w in WW and any ψ in S the action is given by:

${\displaystyle v\cdot \psi =2^{\frac {1}{2}}(w\wedge \psi +\iota (w^{*})\psi ),}$

where the second term is a contraction (interior multiplication) defined using the bilinear form, which pairs W and W. This action respects the Clifford relations v2 = Q(v)1, and so induces a homomorphism from the Clifford algebra ClnC of V to End(S). A similar action can be defined on S, so that both S and S are Clifford modules.

The Lie algebra so(n, C) is isomorphic to the complexified Lie algebra spinnC in ClnC via the mapping induced by the covering Spin(n) → SO(n)

${\displaystyle v\wedge w\mapsto {\tfrac {1}{4}}[v,w].}$

It follows that both S and S are representations of so(n, C). They are actually equivalent representations, so we focus on S.

The explicit description shows that the elements αiai of the Cartan subalgebra h act on S by

${\displaystyle (\alpha _{i}\wedge a_{i})\cdot \psi ={\tfrac {1}{4}}(2^{\tfrac {1}{2}})^{2}(\iota (\alpha _{i})(a_{i}\wedge \psi )-a_{i}\wedge (\iota (\alpha _{i})\psi ))={\tfrac {1}{2}}\psi -a_{i}\wedge (\iota (\alpha _{i})\psi ).}$

A basis for S is given by elements of the form

${\displaystyle a_{i_{1}}\wedge a_{i_{2}}\wedge \cdots \wedge a_{i_{k}}}$

for 0 ≤ km and i1 < ... < ik. These clearly span weight spaces for the action of h: αiai has eigenvalue −1/2 on the given basis vector if i = ij for some j, and has eigenvalue 1/2 otherwise.

It follows that the weights of S are all possible combinations of

${\displaystyle {\bigl (}\pm {\tfrac {1}{2}},\pm {\tfrac {1}{2}},\ldots \pm {\tfrac {1}{2}}{\bigr )}}$

and each weight space is one-dimensional. Elements of S are called Dirac spinors.

When n is even, S is not an irreducible representation: ${\displaystyle S_{+}=\wedge ^{\mathrm {even} }W}$ and ${\displaystyle S_{-}=\wedge ^{\mathrm {odd} }W}$ are invariant subspaces. The weights divide into those with an even number of minus signs, and those with an odd number of minus signs. Both S+ and S are irreducible representations of dimension 2m1 whose elements are called Weyl spinors. They are also known as chiral spin representations or half-spin representations. With respect to the positive root system above, the highest weights of S+ and S are

${\displaystyle {\bigl (}{\tfrac {1}{2}},{\tfrac {1}{2}},\ldots {\tfrac {1}{2}},{\tfrac {1}{2}}{\bigr )}}$ and ${\displaystyle {\bigl (}{\tfrac {1}{2}},{\tfrac {1}{2}},\ldots {\tfrac {1}{2}},-{\tfrac {1}{2}}{\bigr )}}$

respectively. The Clifford action identifies ClnC with End(S) and the even subalgebra is identified with the endomorphisms preserving S+ and S. The other Clifford module S is isomorphic to S in this case.

When n is odd, S is an irreducible representation of so(n,C) of dimension 2m: the Clifford action of a unit vector uU is given by

${\displaystyle u\cdot \psi =\left\{{\begin{matrix}\psi &{\hbox{if }}\psi \in \wedge ^{\mathrm {even} }W\\-\psi &{\hbox{if }}\psi \in \wedge ^{\mathrm {odd} }W\end{matrix}}\right.}$

and so elements of so(n,C) of the form uw or uw do not preserve the even and odd parts of the exterior algebra of W. The highest weight of S is

${\displaystyle {\bigl (}{\tfrac {1}{2}},{\tfrac {1}{2}},\ldots {\tfrac {1}{2}}{\bigr )}.}$

The Clifford action is not faithful on S: ClnC can be identified with End(S) ⊕ End(S), where u acts with the opposite sign on S. More precisely, the two representations are related by the parity involution α of ClnC (also known as the principal automorphism), which is the identity on the even subalgebra, and minus the identity on the odd part of ClnC. In other words, there is a linear isomorphism from S to S, which identifies the action of A in ClnC on S with the action of α(A) on S.

### Bilinear forms

if λ is a weight of S, so is λ. It follows that S is isomorphic to the dual representation S.

When n = 2m + 1 is odd, the isomorphism B: SS is unique up to scale by Schur's lemma, since S is irreducible, and it defines a nondegenerate invariant bilinear form β on S via

${\displaystyle \beta (\varphi ,\psi )=B(\varphi )(\psi ).}$

Here invariance means that

${\displaystyle \beta (\xi \cdot \varphi ,\psi )+\beta (\varphi ,\xi \cdot \psi )=0}$

for all ξ in so(n,C) and φ, ψ in S in other words the action of ξ is skew with respect to β. In fact, more is true: S is a representation of the opposite Clifford algebra, and therefore, since ClnC only has two nontrivial simple modules S and S, related by the parity involution α, there is an antiautomorphism τ of ClnC such that

${\displaystyle \quad \beta (A\cdot \varphi ,\psi )=\beta (\varphi ,\tau (A)\cdot \psi )\qquad (1)}$

for any A in ClnC. In fact τ is reversion (the antiautomorphism induced by the identity on V) for m even, and conjugation (the antiautomorphism induced by minus the identity on V) for m odd. These two antiautomorphisms are related by parity involution α, which is the automorphism induced by minus the identity on V. Both satisfy τ(ξ) = ξ for ξ in so(n,C).

When n = 2m, the situation depends more sensitively upon the parity of m. For m even, a weight λ has an even number of minus signs if and only if λ does; it follows that there are separate isomorphisms B±: S±S± of each half-spin representation with its dual, each determined uniquely up to scale. These may be combined into an isomorphism B: SS. For m odd, λ is a weight of S+ if and only if λ is a weight of S; thus there is an isomorphism from S+ to S, again unique up to scale, and its transpose provides an isomorphism from S to S+. These may again be combined into an isomorphism B: SS.

For both m even and m odd, the freedom in the choice of B may be restricted to an overall scale by insisting that the bilinear form β corresponding to B satisfies (1), where τ is a fixed antiautomorphism (either reversion or conjugation).

### Symmetry and the tensor square

The symmetry properties of β: SSC can be determined using Clifford algebras or representation theory. In fact much more can be said: the tensor square SS must decompose into a direct sum of k-forms on V for various k, because its weights are all elements in h whose components belong to {1,0,1}. Now equivariant linear maps SS → ∧kV correspond bijectively to invariant maps ∧kVSSC and nonzero such maps can be constructed via the inclusion of ∧kV into the Clifford algebra. Furthermore, if β(φ,ψ) = ε β(ψ,φ) and τ has sign εk on ∧kV then

${\displaystyle \beta (A\cdot \varphi ,\psi )=\varepsilon \varepsilon _{k}\beta (A\cdot \psi ,\varphi )}$

for A in ∧kV.

If n = 2m+1 is odd then it follows from Schur's Lemma that

${\displaystyle S\otimes S\cong \bigoplus _{j=0}^{m}\wedge ^{2j}V^{*}}$

(both sides have dimension 22m and the representations on the right are inequivalent). Because the symmetries are governed by an involution τ that is either conjugation or reversion, the symmetry of the ∧2jV component alternates with j. Elementary combinatorics gives

${\displaystyle \sum _{j=0}^{m}(-1)^{j}\dim \wedge ^{2j}\mathbb {C} ^{2m+1}=(-1)^{{\frac {1}{2}}m(m+1)}2^{m}=(-1)^{{\frac {1}{2}}m(m+1)}(\dim \mathrm {S} ^{2}S-\dim \wedge ^{2}S)}$

and the sign determines which representations occur in S2S and which occur in ∧2S.[2] In particular

${\displaystyle \beta (\phi ,\psi )=(-1)^{{\frac {1}{2}}m(m+1)}\beta (\psi ,\phi ),}$ and
${\displaystyle \beta (v\cdot \phi ,\psi )=(-1)^{m}(-1)^{{\frac {1}{2}}m(m+1)}\beta (v\cdot \psi ,\phi )=(-1)^{m}\beta (\phi ,v\cdot \psi )}$

for vV (which is isomorphic to ∧2mV), confirming that τ is reversion for m even, and conjugation for m odd.

If n = 2m is even, then the analysis is more involved, but the result is a more refined decomposition: S2S±, 2S± and S+S can each be decomposed as a direct sum of k-forms (where for k = m there is a further decomposition into selfdual and antiselfdual m-forms).

The main outcome is a realisation of so(n,C) as a subalgebra of a classical Lie algebra on S, depending upon n modulo 8, according to the following table:

 n mod 8 Spinor algebra 0 1 2 3 4 5 6 7 ${\displaystyle {\mathfrak {so}}(S_{+})\oplus {\mathfrak {so}}(S_{-})}$ ${\displaystyle {\mathfrak {so}}(S)}$ ${\displaystyle {\mathfrak {gl}}(S_{\pm })}$ ${\displaystyle {\mathfrak {sp}}(S)}$ ${\displaystyle {\mathfrak {sp}}(S_{+})\oplus {\mathfrak {sp}}(S_{-})}$ ${\displaystyle {\mathfrak {sp}}(S)}$ ${\displaystyle {\mathfrak {gl}}(S_{\pm })}$ ${\displaystyle {\mathfrak {so}}(S)}$

For n ≤ 6, these embeddings are isomorphisms (onto sl rather than gl for n = 6):

${\displaystyle {\mathfrak {so}}(2,\mathbb {C} )\cong {\mathfrak {gl}}(1,\mathbb {C} )\qquad (=\mathbb {C} )}$
${\displaystyle {\mathfrak {so}}(3,\mathbb {C} )\cong {\mathfrak {sp}}(2,\mathbb {C} )\qquad (={\mathfrak {sl}}(2,\mathbb {C} ))}$
${\displaystyle {\mathfrak {so}}(4,\mathbb {C} )\cong {\mathfrak {sp}}(2,\mathbb {C} )\oplus {\mathfrak {sp}}(2,\mathbb {C} )}$
${\displaystyle {\mathfrak {so}}(5,\mathbb {C} )\cong {\mathfrak {sp}}(4,\mathbb {C} )}$
${\displaystyle {\mathfrak {so}}(6,\mathbb {C} )\cong {\mathfrak {sl}}(4,\mathbb {C} ).}$

## Real representations

The complex spin representations of so(n,C) yield real representations S of so(p,q) by restricting the action to the real subalgebras. However, there are additional "reality" structures that are invariant under the action of the real Lie algebras. These come in three types.

1. There is an invariant complex antilinear map r: SS with r2 = idS. The fixed point set of r is then a real vector subspace SR of S with SRC = S. This is called a real structure.
2. There is an invariant complex antilinear map j: SS with j2 = idS. It follows that the triple i, j and k:=ij make S into a quaternionic vector space SH. This is called a quaternionic structure.
3. There is an invariant complex antilinear map b: SS that is invertible. This defines a pseudohermitian bilinear form on S and is called a hermitian structure.

The type of structure invariant under so(p,q) depends only on the signature p q modulo 8, and is given by the following table.

 p−q mod 8 Structure 0 1 2 3 4 5 6 7 R + R R C H H + H H C R

Here R, C and H denote real, hermitian and quaternionic structures respectively, and R + R and H + H indicate that the half-spin representations both admit real or quaternionic structures respectively.

### Description and tables

To complete the description of real representation, we must describe how these structures interact with the invariant bilinear forms. Since n = p + qpq mod 2, there are two cases: the dimension and signature are both even, and the dimension and signature are both odd.

The odd case is simpler, there is only one complex spin representation S, and hermitian structures do not occur. Apart from the trivial case n = 1, S is always even-dimensional, say dim S = 2N. The real forms of so(2N,C) are so(K,L) with K + L = 2N and so(N,H), while the real forms of sp(2N,C) are sp(2N,R) and sp(K,L) with K + L = N. The presence of a Clifford action of V on S forces K = L in both cases unless pq = 0, in which case KL=0, which is denoted simply so(2N) or sp(N). Hence the odd spin representations may be summarized in the following table.

n mod 8 1, 7 3, 5
p-q mod 8 so(2N,C) sp(2N,C)
1, 7 R so(N,N) or so(2N) sp(2N,R)
3, 5 H so(N,H) sp(N/2,N/2) or sp(N)

(†) N is even for n > 3 and for n = 3, this is sp(1).

The even-dimensional case is similar. For n > 2, the complex half-spin representations are even-dimensional. We have additionally to deal with hermitian structures and the real forms of sl(2N, C), which are sl(2N, R), su(K, L) with K + L = 2N, and sl(N, H). The resulting even spin representations are summarized as follows.

n mod 8 0 2, 6 4
p-q mod 8 so(2N,C)+so(2N,C) sl(2N,C) sp(2N,C)+sp(2N,C)
0 R+R so(N,N)+so(N,N) sl(2N,R) sp(2N,R)+sp(2N,R)
2, 6 C so(2N,C) su(N,N) sp(2N,C)
4 H+H so(N,H)+so(N,H) sl(N,H) sp(N/2,N/2)+sp(N/2,N/2)

(*) For pq = 0, we have instead so(2N) + so(2N)

(†) N is even for n > 4 and for pq = 0 (which includes n = 4 with N = 1), we have instead sp(N) + sp(N)

The low-dimensional isomorphisms in the complex case have the following real forms.

 Euclidean signature Minkowskian signature Other signatures ${\displaystyle {\mathfrak {so}}(2)\cong {\mathfrak {u}}(1)}$ ${\displaystyle {\mathfrak {so}}(1,1)\cong \mathbb {R} }$ ${\displaystyle {\mathfrak {so}}(3)\cong {\mathfrak {sp}}(1)}$ ${\displaystyle {\mathfrak {so}}(2,1)\cong {\mathfrak {sl}}(2,\mathbb {R} )}$ ${\displaystyle {\mathfrak {so}}(4)\cong {\mathfrak {sp}}(1)\oplus {\mathfrak {sp}}(1)}$ ${\displaystyle {\mathfrak {so}}(3,1)\cong {\mathfrak {sl}}(2,\mathbb {C} )}$ ${\displaystyle {\mathfrak {so}}(2,2)\cong {\mathfrak {sl}}(2,\mathbb {R} )\oplus {\mathfrak {sl}}(2,\mathbb {R} )}$ ${\displaystyle {\mathfrak {so}}(5)\cong {\mathfrak {sp}}(2)}$ ${\displaystyle {\mathfrak {so}}(4,1)\cong {\mathfrak {sp}}(1,1)}$ ${\displaystyle {\mathfrak {so}}(3,2)\cong {\mathfrak {sp}}(4,\mathbb {R} )}$ ${\displaystyle {\mathfrak {so}}(6)\cong {\mathfrak {su}}(4)}$ ${\displaystyle {\mathfrak {so}}(5,1)\cong {\mathfrak {sl}}(2,\mathbb {H} )}$ ${\displaystyle {\mathfrak {so}}(4,2)\cong {\mathfrak {su}}(2,2)}$ ${\displaystyle {\mathfrak {so}}(3,3)\cong {\mathfrak {sl}}(4,\mathbb {R} )}$

The only special isomorphisms of real Lie algebras missing from this table are ${\displaystyle {\mathfrak {so}}^{*}(3,\mathbb {H} )\cong {\mathfrak {su}}(3,1)}$ and ${\displaystyle {\mathfrak {so}}^{*}(4,\mathbb {H} )\cong {\mathfrak {so}}(6,2).}$

## Notes

1. Fulton & Harris 1991 Chapter 20, p.303. The factor 2 is not important, it is there to agree with the Clifford algebra construction.
2. This sign can also be determined from the observation that if φ is a highest weight vector for S then φφ is a highest weight vector for mV m+1V, so this summand must occur in S2S.

## References

• Brauer, Richard; Weyl, Hermann (1935), "Spinors in n dimensions", American Journal of Mathematics, American Journal of Mathematics, Vol. 57, No. 2, 57 (2): 425–449, doi:10.2307/2371218, JSTOR 2371218.
• Cartan, Élie (1966), The theory of spinors, Paris, Hermann (reprinted 1981, Dover Publications), ISBN 978-0-486-64070-9.
• Chevalley, Claude (1954), The algebraic theory of spinors and Clifford algebras, Columbia University Press (reprinted 1996, Springer), ISBN 978-3-540-57063-9.
• Deligne, Pierre (1999), "Notes on spinors", in P. Deligne; P. Etingof; D. S. Freed; L. C. Jeffrey; D. Kazhdan; J. W. Morgan; D. R. Morrison; E. Witten (eds.), Quantum Fields and Strings: A Course for Mathematicians, Providence: American Mathematical Society, pp. 99–135. See also the programme website for a preliminary version.
• Fulton, William; Harris, Joe (1991), Representation theory. A first course, Graduate Texts in Mathematics, Readings in Mathematics, 129, New York: Springer-Verlag, ISBN 0-387-97495-4, MR 1153249.
• Harvey, F. Reese (1990), Spinors and Calibrations, Academic Press, ISBN 978-0-12-329650-4.
• Lawson, H. Blaine; Michelsohn, Marie-Louise (1989), Spin Geometry, Princeton University Press, ISBN 0-691-08542-0.
• Weyl, Hermann (1946), The Classical Groups: Their Invariants and Representations (2nd ed.), Princeton University Press (reprinted 1997), ISBN 978-0-691-05756-9.


text:
# What are dihedral groups?

I have trouble understanding what exactly a dihedral group is. I read about how they are rotations and reflections along the faces of a polygon. But then what does that mean? Whatever you do to a polygon you get the same polygon right? Could someone give a dihedral group for dummies explanation?

And also what does it mean for a shape to have a certain dihedral symmetry?

• Think of it as taking a $n$-sided polygon where the sides are numbered $1, 2, \ldots , n$. The "new" polygons are different because $1$ will be somewhere else or $7$ will be somewhere else. The trick is that the group itself is comprised of the transformations. – AJY Sep 23 '16 at 14:55
• Youconsider all movements of the polygon that do not change the polygon as such. But the $n$ vertices of it are permuted. – Hagen von Eitzen Sep 23 '16 at 14:57

Cut out a regular 5-gon from a paper sheet with a blade, and get 5-gon and paper sheet with the 5-gonal hole. Write your name on the 5-gon with blue pen on the one side, and with red pen on the other side. Then try to fit it back in the hole. How many different ways to do it? Two ways are different if your name have different color or different position. The answer to your question: each "way" is the element of the dihedral group $D_5$.
The final note is that $5$ may be any natural number.


text:
# What speed do subatomic particles travel around the nucleus?

1. Dec 30, 2004

### Gamish

What speed do subatomic particles travel around the nucleus? Do the speeds vary with different paricles, IE Electron, Proton, Neutron? Are the speeds differnet with its idenical positron? I do know that electrons and protons travel around the nuclueus at differnt speeds, this we call "heat", in which photons are sometimes released at the according frequency.

Last edited: Dec 30, 2004
2. Dec 30, 2004

### Gokul43201

Staff Emeritus
Gamish,

Protons make up the nucleus and so, do not revolve around it.

Older models of the atom (Rutherfold, Bohr) had the electrons revolving around the nucleus, but we know that this is only a very crude representation. The behavior of an electron (or any particle, for that matter) is determined from something known as its wavefunction. The wavefunction of an electron in a stable, non-interacting atom does not evolve in time, so we can discard the picture of revolving electrons when we treat the atom quantum mechanically.

Heat has nothing to do with the "speeds of revolution". Heat, in a material medium comes from the kinetic energy of the atoms/molecules themselves. And when electrons are involved, it is the free electrons that contribute to the heat capacity. These are the electrons that are not considered to be bound strongly to any one nucleus.

Yes, electronic transitions do result in an emission or absroption of photons and vice versa. However, excited electronic states are very short lived, and they end up transfering energy into the free electrons or phonons (vibrational modes of the atoms). It is through these that heat propagates along a medium.

3. Dec 31, 2004

### Gamish

OK, you did not answer my question, I am assuming that you do not know the answer. If you did not know, electrons do travel around a nuclues, I am asking at what speed. Lets say the speed of the electron on one atom of hydrogen, at 50 degreas C.

4. Jan 1, 2005

### elas

Electrons form a shell, would you ask 'what speed does the earth's crust travel around the core'? The question cannot be answered in the form you desire because there is no fixed orbital speed, all speeds up to the speed of light are possible depending on the degree of activity.
Gokul43201 has the answer in proffessional form, I hope my amatuer approach will add some enlightenment.

5. Jan 1, 2005

### Gamish

. Actually, no, according to special relativity, no object with rest mass can reach the speed of light (c). As an objects mass increases near speeds of c, more energy will be put towered the moving of the object rather than the actual speeding up of the object.

Also, can you please tell me what variables are involved that determine the speed at which electrons travel around the nucleus of it's atom?

6. Jan 1, 2005

### dextercioby

$$\langle\hat{v}\rangle_{|n,l,m\rangle}=\frac{1}{m}\langle\hat{p}\rangle_{|n,l,m\rangle}=\frac{1}{m}\langle n,l,m|\hat{p}|n,l,m\rangle$$

So it suffices to know the quantum state of the electron in one certain representation and in the same representation the form of the momentum operator.

Daniel.

7. Jan 2, 2005

### elas

Also, can you please tell me what variables are involved that determine the speed at which electrons travel around the nucleus of it's atom?

Spin (and therefore rotation) are slowed down in experiments that show the wave nature of electrons, as this is done by cooling; we know that temperature affects speed of rotation. Distance from the centre also plays a part in determining rotation speeds when electron is viewed as a particle. Variations in shell density probably cause variations in shell rotation even within a single atom, leading to (three dimewnsional) wave rotation.

You are of course, correct about c.

8. Jan 3, 2005

### tozhan

i think i heard a while ago that an electron takes about 150 attoseconds (150 quintillionths ($$10^{-18}$$) of a second) to 'orbit' a nucleus. if we take the diameter of an atom to be $$10^{-10}$$ metres we can VERY roughly make a guess at its speed.

$$\frac{\pi10^{-10}}{150^{-18}}=2.1*10^{8}ms^{-1}$$

does this seem reasonable?? It would be about about 1.4 times its rest mass if you apply special relativity to it.

i only have very patchy idea of classical physics so all you quantum guys can tell me what it really is!

also, the Pauli exclusion principle says electrons can orbit in pairs when the electrons are of opisite spin, but i though all naturally occuring electrons were spin-left? i also dont understand lone pairs, is this the same thing?

9. Jan 3, 2005

### dextercioby

1.The result u obtained for the speed of 'orbit' is absurd and it comes from the fact u used a wrong value for the period of orbit.I suggest u take a good look into the Bohr's theory of atom and then u can come up with a decent value for the period.The velocity for the first Bohr orbit should be 10 times less than u found and then u can conclude that relativistic effects are not essential.

2.Pauli's exclusion principle applies to all fermions without discrimination.Electrons have 2 possbile 'orientations' for their spin:up (m_{s}=+1/2) and down (m_{s}=-1/2),so there are two types of electrons judging after their spin projection on Oz eigenvalues.

Daniel.

10. Jan 3, 2005

### tozhan

being out by a factor of 10 isnt bad for me ;-) my sketchy mathes is always a little wrong! can anyone come up with a real answer? or shall i search google?

Tom

11. Jan 3, 2005

### dextercioby

The first orbital speed for the H atom in the theory of Bohr is
$$v(n=1)=\alpha c\sim\frac{c}{137}$$
,where $\alpha$ is the famous fine structure constant of Sommerfeld and it has the approximate value of 1/137,number which can be obtained if u plug in the constants in its definition
$$\alpha=:\frac{e^{2}}{4\pi\epsilon_{0}\hbar c}$$

Daniel.

12. Jan 4, 2005

### Gokul43201

Staff Emeritus
Thanks Dexter, for the clarification.

In my hurry, I did a terribly shoddy job. I started out with the hydrogen atom, but eventually decided to just write down how to find the velocity operator in a general 1-d case, but forgot to remove the subscripts that I had started out with.

What I simply meant to suggest was that

$$<\hat {v}_x> = \frac {\hbar}{im} \int_{- \infty}^{\infty} \psi }^* \frac {\partial \psi }{\partial x }~dx$$

What I was hoping to show was that information about "velocity" can be got from the wavefunction, and that systems such as those being discussed can not be treated classically. I suspected that Gamish may be looking for a classical answer to his question (since he brought up "heat" and so on). But now, I think, by "heat" he merely meant radiation.

Last edited: Jan 4, 2005
13. Jan 4, 2005

### Gamish

Example

Can you give me an example using that equasion, im not good with understanding tex formatting. And can you please explaine what all those variables are?

14. Jan 4, 2005

### dextercioby

Gokul gave a particular realization of the formula i posted in the post no.6 of this thread.

$<v_{n,l,m}>$ should have been put under the form
$\langle \hat{v}_{x}\rangle _{|n,l,m\rangle}$ and represents the average of the operator v_{x} computed with the Q system in the pure state $|n,l,m\rangle$.\psi and \psi star represent the wave functions of the Q system (probably H atom he,meant,but in that case,the formula is totally wrong;let's assume he didn't consider the H atom,though the indices "n","l","m" make us think that way) in the coordinate representation.

The partial derivative wrt to 'x' is a part of the formula giving the 'x' component of the momentum operator in the position representation.As i said before (when spaeking about the H atom),if the wave functions depend on all 3 variables,then Gokul's formula is incomplete,as the integral must be evaluated over all three coordinates.But let's stick to this simple unidimensional case.

U virtually have to find and use a wavefunction of the system psi(x) and with it compute 2 quantities:its derivative wrt to "x" and its complex conjugate,multiply the 2 results and integrate over the real axis.Then simply multiply with the reduced Planck's constant and divide through the product 'im'.U'll find the average of the velocity operator in the Q state u've chosen.

Daniel.

Last edited: Jan 4, 2005
15. Jan 5, 2005

### da_willem

In Quantum mechanics (QM) all you can know about a partice is it's state. Where in Classical mechanics you're solving the equation of motion (Newtons F=ma eg) to find the position of a particle as a function of time, in QM you solve (in the nonrelativistic case) the Schrödinger equation. This equation:

$$i\hbar\frac{\partial \Psi}{\partial t}=\hat{H}\Psi$$

gives you, when you know the phsyical circumstances ($H$), the state $\hat{\psi}$. This state does not directly give you the position of a particle as a function of time. You can only use it to find a 'probabality density' for the particle's position, from wich an 'expectation value' (mean) for the particles position can be found. In QM this is all we can know about a particles position, the probablility that upon measurement you find it here or there!

Now for a molecule this Schrödinger equation can only be a analytically solved for a hydrogen atom (and some similar atoms), a system of a proton and an electron. The relevant part of the state you need to say something about the electrons position or velocity is called the wavefunction. It's 'square' (actually its modulus) gives you the aforementioned probability distribution. The result in the form of a wavefuncion you can find in almost any textbook on QM. It is denoted by $$\psi_{n,l,m}$$ and is actually a whole collection of wavefunction depending on what numbers of n, l and m you are interested in. They represent the different orbitals, or excited states, of the hydrogen molecule. Note that n is the same variable as in Bohrs theorem of the hydrogen atom and is called the 'principle quantum number'.

As an example I will try to answer your question for the ground state (n=1, l=m=0) of a hydrogen atom. The wavefunction as a function of the radial coordinate r is

$$\psi_{1,0,0}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$ with $$a=\frac{4 \pi \epsilon_0 \hbar^2}{me^2}=0,529E-10 m$$ (Bohr radius in meters)

As said before the 'square' gives you the probability density:

$$|\psi_{1,0,0}|^2=\frac{1}{\pi a^3}e^{-2r/a}$$

And tells you the elektron has the greatest probability to be found at r=o wch is at the nucleus! Furtermore the wavefunction is spherically symmetric. But all this is not really relevant for finding out the velocity, so let's move on.

As said before you cannot find the position of the electron, simply because i has no position. You can find the expectation value for the position though. Suppose you know a particle can be found with probability at x=1, probability (1/4) at x=2 and probability (1/4) x=4 you can find the expectation value by calculating the following sum:

$$<x>=\sum_i P(x_i)x_i = (1/2)*1+(1/4)*2+(1/4)*4=2.$$

But now we don't deal with a discrete spectrum of probabilities ((1/2), (1/4) and (1/4)) but a continuum of probabilities. So instead of summing over the probabilities you will have to integrate:

$$<x>=\int |\psi_{1,0,0}|^2 x dx$$

Where the probabilities P are replaced by the probability density $|\psi_{1,0,0}|^2$. This integral has to be evaluated over all space.

Also the electron has no definite speed, but only a probability to be measured that speed, or that speed. So I will have to dissapoint you, again all we can find out about the velocity is it's expectation value <v>. Now:

$$<v>=\frac{d<x>}{dt}=\int \frac{\partial |\psi_{1,0,0}|^2}{\partial t} x dx = \frac{-i \hbar}{m} \int \psi_{1,0,0}^* \frac{\partial \psi_{1,0,0}}{\partial x} dx$$

Where the last step follows after some juggling with the integral and using the Scrödinger equation. [I don't know have far your mathematics skills reach but $|\psi_{1,0,0}|^2 = \psi_{1,0,0}^* \psi_{1,0,0}$ is called the 'modulus' and a star denotes 'complex conjugation'. In case your wavefunction is real and this means just respectively squaring and doing nothing.]

This is (for the groundstate of hydrogen) the result dextercioby and gokul pointed out and hopefully you now know more or less know where it came from.

16. Jan 5, 2005

### da_willem

But anyway. Using the equation to find the velocity of an electron in a hydrogen atom will not work. Because the atom as a whole stands still. <x> of the electron is zero. And the expectation value of the velocity will be zero as well, because it moves with the same speed but opposite sign in the x and -x direction. With a real wavefunction like the of the electron in a hydrogen atom this can also be seen in a different (mathematical) way. The formula for <v> involves an 'i'. But as the wavefunction is real and ofcourse <v> is real thi leaves only one possibility: the integral vanishes and <v>=0.

17. Jan 12, 2005

### vluth

In the Bohr model of the Hydrogen atom (which gives some right answers, but is known to be essentially incorrect) electrons _do_ spin around the nucleus. In the simplest case of a hydrogen atom with a single electron spinning around a single proton, the electron moves at about 1/137 of the speed of light, which is MUCH faster than sound. Sound travels at about 1100 feet per second, while light travels at 186,000 miles per second.. You do the math.

As a side note, with the Bohr model, the 'inner' electrons in atoms with greater atomic numbers would be moving faster...

18. Jan 12, 2005

### dextercioby

Neat trick u used... But hopefully u're aware of the tedious proof that
$$\langle \hat{v}_{x}\rangle _{|\psi\rangle} =\frac{d\langle \hat{x}\rangle _{|\psi\rangle}}{dt}$$

,independent of the description...

As in Gokul's case,please remove the indices (1,0,0).They remind of the H atom and it's not the case with your example/formula.

Daniel.

PS.I didn't trust your formula,so i did the proof by my own.It wasn't nice at all. :yuck: Juggling with the Heisenberg and Schroedinger descriptions...Anyways,i'm a theorist,so i shouldn't complain.

19. Jan 13, 2005

### da_willem

I didn't know the proof was that tedious. But Griffiths only 'postulated' it, so that should say something. And about the (1,0,0), My post was about the hydrogen atom, wich has a fairly easy ground state wavefunction, so it's okay they remind you about it.


text:
Substitute function for json

Discussion created by JaredHague on Jul 29, 2015
Latest reply on Jul 30, 2015 by beverly

Ok its been a long day so my brain may be just not thinking about this logically but I just can't figure out why..

I want to replace all my " in my text with \" and any \ with \\ . So in the case where myText is aa\bb"cc i need it to be aa\\bb\"cc

so at first I tried this

Substitute ( myText ; ["\\" ; "\\\\"];[ "\"" ; "\\\"" ]  )

and the output was aa\\\bb\\\"cc" not what I was expecting.  It appears the replacement value doesn't use the escapes so after experimenting this worked.

Substitute ( myText ; [ "\"" ; "\"" ];["\\" ; "\\"] )

what I am trying to accomplish is to have myText a value that would be safe to use as a value in a JSON object so if anybody knows a cf or something that world be great.

Thanks


text:
This site is supported by donations to The OEIS Foundation.

Given an integer ${\displaystyle a}$ and an odd prime ${\displaystyle p}$, the former is a quadratic nonresidue of the latter if the congruence ${\displaystyle x^{2}\equiv a\mod p}$ does not have a solution. For example, 5 is a quadratic nonresidue of 7 because ${\displaystyle x^{2}\equiv 5\mod 7}$ has no solutions.


text:
Publication
Title
Subcutaneous trastuzumab (Herceptin) versus intravenous trastuzumab for the treatment of patients with HER2-positive breast cancer: A time, motion and cost assessment study in a lean operating day care oncology unit
Author
Abstract
 Objective: The subcutaneous (SC) formulation of trastuzumab represents an alternative to the intravenous (IV) infusion in the treatment of patients with HER2-positive metastatic and early breast cancer. We compared the two formulations in terms of time and cost differential. Study desing: We conducted a time, motion and cost assessment study in a lean operating day care oncology unit to determine and compare the time and costs of trastuzumab SC versus IV administration in patients with HER2-positive breast cancer. Outcomes were the mean costs and the mean dedicated time of the health care professional (HCP) and patient chair time. Direct observation methodology was applied to collect data and statistical analysis was performed. Results: The total preparation and administration time for trastuzumab IV was 4.07 times longer than the total time required for the trastuzumab SC administration. The total patient time spent in the day care oncology unit (in minutes) was 71% shorter with using SC administration. IV administration costs (sic)50.4 ($54,89) more in HCP time and consumable supplies and (sic)162.53 ($177.00) of drug wastage. SC administration was associated with a total time saving of 53.7 min for the HCPs and 122.5 min for the patients. The administration of trastuzumab SC was translated in a cost saving of (sic)212.93 ($231.73) per patient episode compared to trastuzumab IV, which could lead to a total potential saving of (sic)3,832.74 ($4,171.06) over a full course of treatment (18 cycles) Conclusion: SC administration of trastuzumab was associated with a substantial reduction in active HCP time, patient chair time, unit time, and overall cost. These time and cost could be used to increase capacity within existing resources in a lean operating day dare oncology unit. (C) 2017 Published by Elsevier Ireland Ltd.
Language
English
Source (journal)
European journal of obstetrics and gynecology and reproductive biology. - Amsterdam, 1973, currens
Publication
Amsterdam : Elsevier , 2018
ISSN
0301-2115 [print]
1872-7654 [online]
Volume/pages
221 (2018) , p. 46-51
ISI
000429185900009
Pubmed ID
29245056
Full text (Publisher's DOI)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
# Topological Quantum Field Theory Seminar

### Stress-Energy-Momentum Tensors

J. Marsden and I present a new method of constructing a stress-energy-momentum tensor for a classical field theory based on covariance considerations and Noether theory. Our stress-energy-momentum tensor ${T}^{\mu }{}_{\nu }$ is defined using the (multi)momentum map associated to the spacetime diffeomorphism group. The tensor ${T}^{\mu }{}_{\nu }$ is uniquely determined as well as gauge-covariant, and depends only upon the divergence equivalence class of the Lagrangian. It satisfies a generalized version of the classical Belinfante-Rosenfeld formula, and hence naturally incorporates both the canonical stress-energy-momentum tensor and the "correction terms" that are necessary to make the latter well behaved. Furthermore, in the presence of a metric on spacetime, our ${T}^{\mu \nu }$ coincides with the Hilbert tensor and hence is automatically symmetric.

References:
1. Gotay, M. J. and J. E. Marsden [1992], Stress-energy-momentum tensors and the Belinfante-Rosenfeld formula, Contemp. Math. 132, 367-391.
2. Forger, M. and H. Römer [2004], Currents and the energy-momentum tensor in classical field theory: A fresh look at an old problem, Ann. Phys. 309, 306-389.