text: Tanya. Revise all the concepts easily by taking help from the MCQ Questions for Class 10 Science with Answers are prepared based on the latest exam pattern. (like solid, liquid or gas) for their propagation. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks. Light reflection and refraction NCERT questions answer class ten science. Another way to prevent getting this page in the future is to use Privacy Pass. Refraction of Light Class 10 Notes - Lenses Soon after you have been taught about mirrors, you will be introduced to a similar concept in our Physics Notes Class 10 – lens. All right reserved. In figure 1 (a) you can see several rays of light parallel to the principal axis are falling on a convex lens. A highly polished surface, such as a mirror, reflects most of the light falling on it. Reflected:- If an object sends back light rays falling on its surface then it is said to have reflected the light, The process of sending back light rays which falls on the surface of an object is called. The nature, position and size of the image formed by a concave mirror depend on the position of the object in relation to points P, F and C. The image formed can be real as well as virtual depending on the positions of the object. Light Reflection and Refraction Class 10, Notes, Explanation, Question Answers The Chapter Includes the following topics. Fix a sheet of white paper on a drawing board using drawing pins. Transmitted: - An object is said to transmit light if it allows light to pass through itself and such objects are transparent. October 10, 2016 at 12:19 pm Reply . We can place the object at following places, Between focus F and center of curvature C, At far off distances called infinity and cannot be shown in the figures, Image formation by a concave mirror for different positions of the object is shown below in the table. Laws of refraction of light: The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane. Class 10. → Speed of light is maximum in vaccum. Light is such an important... Rules to obtain Image from Convex Mirror. Light - Reflection and Refraction. and when this reflected light reaches our eyes we are able to see such non luminous objects. → Refraction is bending of light when it enters obliquely from one transparent medium to another. Important Question for Class 10 Science Light Reflection and Refraction PDF will help you in scoring more marks.. lamp, candle emits light of their own and thus they are known as. These laws of reflection are applicable to all types of reflecting surfaces including spherical surfaces. Answer: Given, refractive index of glass (n g) = 1.50. Magnification produced by a lens is also related to the object-distance u, and the image-distance v and is given by, The power of a lens is defined as the reciprocal of its focal length. Hence, concave mirror is used as reflector in the headlights. October 12, 2016 at 9:59 am Reply . Light - Reflection and Refraction. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane. Your IP: 192.138.189.190 Refraction is due to change in the speed of light as it enters from one transparent medium to another. Such lenses diverge light rays as shown in Figure 1(b). The effective diameter of the circular outline of a spherical lens is called its aperture. This point is called the center of curvature of the spherical mirror. Made by: SHREYAANS M. NAHATA PODAR INTERNATIONAL SCHOOL, JALGAON [10TH RUBY 29] 1 2. Lakhmir Singh, Together with CBSE/NCERT Practice Material Chapterwise for Class 10 Science, Mastering Numericals and Objectives of Physics for Class X, On the same side of the lens as the object, Between infinity and optical center O of the lens, Light is a form of energy due to which we are able to see the objects which emits light for example objects like sun, The distances measured upward and perpendicular to the principle axis are taken as positive. Filed Under: Class 10, Light-Reflection and Refraction Tagged With: cause of refraction of light, laws of refraction, optical density, refraction About Mrs Shilpi Nagpal Author of this website, Mrs Shilpi Nagpal is MSc (Hons, Chemistry) and BSc (Hons, Chemistry) from Delhi University, B.Ed (I. P. University) and has many years of experience in teaching. Scroll down for Light – Reflection and Refraction from NCERT Book Class 10 Science Book & important study material. It is represented by the letter f. There is a relationship between the radius of curvature R, and focal length f, of a spherical mirror and is given by R=2f which means that that the principal focus of a spherical mirror lies midway between the pole and centre of curvature. Get notes for class 10 science Chapter 10 Light - Reflection and Refraction. It is represented by the letter C. Please note that the center of curvature is not a part of the mirror. We know that n g = c/v Or, n g = 3.8 × 10 8 ms −1 ÷ v This point on the principal axis is called the principal focus of the lens. The central point of a lens is its optical centre. A Spherical lens is a piece of transparent glass bound by two spherical surfaces. NCERT Book Class 10 Science Chapter 10 Light – Reflection and Refraction ... Refraction of light can take place at boundary of . The index of refraction of flint glass is 1.66 and the index of refraction of crown glass is 1.52. For drawing ray diagrams we first consider how light rays falling on both concave and convex lens in three different ways get refracted. An optical object to be defined as prism must have at least two faces with an angle between them. Home; Chapter 1-5. Some common phenomena of Refraction: Bending of pencil when placed in a glass with water: When a pencil or stick is kept in a beaker or a glass filled with water, the stick appears slightly bent. The ray of light changes its direction or phenomenon of refraction takes place because of difference in speed in different media. It means that incident ray and emergent ray makes equal angles with parallel normal ${{N}_{1}}Q{{N}_{2}}$ and ${{N}_{3}}R{{N}_{4}}$ . These rays, after refraction from the lens, are appearing to diverge from a point on the principal axis. (i) A ray of light passes from water to air. Light rays basically consist of electromagnetic waves which do not require any material medium Refraction makes it possible for us to have optical instruments such as magnifying glasses, lenses and prisms. Aperture of the mirror actually represents the size of the mirror. Please enable Cookies and reload the page. Chemical Reaction; Acid Base Salt; Metals Non-metals; Carbon Compounds; Periodic Classification; Chapter 6-11. Convex mirrors are used as rear view mirrors in automobiles to see the traffic at back side as they give erect images and also highly diminished one giving the wide field view of traffic behind. Performance & security by Cloudflare, Please complete the security check to access. October 9, 2016 at 11:54 … It lies outside its reflecting surface. Reflection is one of the unique properties of light. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. This implies that the light from the object falls on the mirror from the left-hand side. This emergent ray is parallel to the incident ray. What is the speed of light in the glass? Reflection: The bouncing back of rays of light from a polished and shiny surface is called reflection or reflection of light. Refraction of light through rectangular glass slab class ten science NCERT. It is similar to bouncing back of a football after colliding with a wall or any hard surface. The magnification produced by a lens is defined as the ratio of the height of the image and the height of the object. Refraction. $\angle RQ{{N}_{2}}=\angle QR{{N}_{3}}=r$ and is called angle of refraction. This is possible only when you have the best CBSE Class 10 Science study material and a smart preparation plan. Concepts covered in Science and Technology Part 1 10th Standard SSC Maharashtra State Board chapter 6 Refraction of light are Refraction of Light, Laws of Refraction, Refractive Index, Twinkling of Stars, Dispersion of Light, Partial and Total Internal Reflection. Large concave mirrors are used in field of solar energy to focus sun rays on objects to be heated. LIGHT REFRACTION OF LIGHT Complete these ray diagrams. A convex lens bulges outward and is thick at the center and thinner at the edges. When a source of light is kept at focus of a concave mirror, the reflected rays form a parallel beam of light and go to infinity. • Some examples of refraction → The bottom of swimming pool appears higher. I like it. Life Processes; ControlCoordination; Reproduction ... Refraction of ray passing through the Principal focus: A ray passing through principal focus emerges parallel to the principal axis after refraction from … These rays, after refraction from the lens, are converging to a point on the principal axis. A film of the same oil on crown glass appears dark as its outer margin. Images formed by Convex Mirror. There are objects like table , chair etc. The object is always placed to the left of the mirror. In order to construct a ray diagram to find out the position, nature and size of image formed by convex mirror we should remember following path of rays of light. It is denoted by the letter D. 1 diopter is the power of a lens whose focal length is 1 meter so, $$1D=1m^{–1}$$. Class 10 Science Chapter 10 MCQ (Multiple Choice Questions) of Light – Reflection and Refraction. So, the bending of light when it passes from one medium to another is called. Causes Of Refraction of Light Physics Class 10 Cause of Refraction : The speed of the light in different transparent substances is different. Home; Chapter 1-5. The distances measured against the direction of incident light are taken as negative. This happens because the light; entering from air (rarer medium) into water (denser medium); bends towards normal to the incident which makes the appearance of pencil or stick as bent. Now a question arises what happens when light travels from one medium to another with different densities for example from air to glass. → Cause of refraction: Change in speed of light. The distances measured in the same direction as that of incident light are taken as positive. For example speed of light in air is 3×10⁸ m/s and that of in glass is 2×10⁸ m/s. In case of convex mirror rays get reflected at the reflecting surface of the mirror and these reflected rays appear to come from point F on the principle axis and this point F is called principle focus of convex mirror. 10th. It can also be just one plane surface as well. Ray diagram helps us to study the nature, position and relative size of the image formed by lenses. The distances measured downwards and perpendicular to principle axis is taken as negative. For understanding about principle focus and focus length of a spherical mirror first consider the figure given below. Measure the angle of incidence, angle of refraction, angle of emergence and interpret the result. (i) A film formed by one drop of sassafras oil, on a horizontal surface of a flint glass block, is viewed by reflected light. which are not luminous objects and still we are able to see them The centers of these spheres are called centers of curvature of the lens usually represented by the letter C. Since there are two centers of curvature, we may represent them as C. An imaginary straight line passing through the two centers of curvature of a lens is called its principal axis as shown in figure 1. • Candidates who are ambitious to qualify the Class 10 with good score can check this article for Notes. From figure 3a we see that a number of rays parallel to the principal axis are falling on a concave mirror. The SI unit of power of a lens is ‘diopter’. We shall try to understand in this Chapter the reflection of light by spherical mirrors and refraction of light and their application in real life situations. Join Now. and this happens because they reflects lights which falls on them from a luminous object like sun, lamp etc. This point is called the principal focus of the concave mirror. All material given in this website is a property of physicscatalyst.com and is for your personal and non-commercial use only, Gravitation NCERT Solutions Class11 physics, Trigonometry Formulas for class 11 (PDF download), Newton�s law Interesting conceptual questions, Difference between resistance and resistivity, Sign convention for reflection by spherical mirrors, Refraction through a rectangular glass slab, Reflection and Refraction Formative Assignment, Reflection and Refraction Unsolved test paper, Reflection and Refraction Short questions, Reflection and Refraction 3 Mark questions, Reflection and Refraction 5 Mark questions, Practice questions for Reflection of light(download), Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books), Combo Pack: Science for Class 10 () Figure :- Refraction of light through rectangular glass slab. CBSE guide notes are the comprehensive notes which covers the latest syllabus of CBSE and NCERT. Thanks for visiting our website. The incident ray, the reflected ray and the normal to the mirror at the point of incidence all lie in the same plane. air glass glass water 13. Angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other. Secondly consider the case for concave lens, The ray diagrams for the image formation in a convex lens for a few positions of the object are summarized below in the table, The ray diagrams for the image formation in a concave lens for a few positions of the object are summarized below in the table. Light Refraction of Light Through a Prism: Prism is a transparent optical element which refracts light. The change of direction of light because of change of medium is known as Refraction or Refraction of Light. Physics. Given below are the links of some of the reference books for class 10 Science. (a) State Snell’s law of refraction. You may need to download version 2.0 now from the Chrome Web Store. Experiments show that refraction of light occurs according to certain laws. Cloudflare Ray ID: 601143f1284525b0 Our aim is to help students learn subjects like The image is either magnified, reduced or has the same size, depending on the position of the object. Principle axis: - Straight line passing through the pole and the center of curvature of a spherical mirror is called principle axis of the mirror. It is 3 × 10 8 m/s. This change in speed of … All distances parallel to the principal axis are measured from the pole of the mirror. Radius of curvature: - The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror. Whatever be the position of object in front of convex mirror, the image formed by a convex mirror is always behind the mirror, virtual, erect and smaller than the object. However, it lies behind the mirror in case of a convex mirror as shown above in the figure 2. xYz. Class 10 Science notes on Chapter 10 Light Reflection and Refraction are also available for download in CBSE Guide website. DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. Distances measured perpendicular to and below the principal axis (along -y-axis) are taken as negative. It is the 6th chapter from science 1 of Maharashtra state board. The center of curvature of a concave mirror lies in front of it. CBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction Pdf free download is part of Class 10 Science Notes for Quick Revision. Refraction of light is one of the most commonly observed phenomena, but other waves like sound waves and water waves also experience refraction. Law #1 15. vacuum $$8 \times {10^{ 8}}m/s$$ same way speed of light in water and glass is much less than that in air. Nature, position and relative size of the image formed by a convex mirror is given below in the table. An image is formed when the light rays coming from an object meet at a point after reflection from a mirror (or refraction from lens). When light ray is made to travel from one medium to another say from air to glass medium then light rays bend at the boundary between the two mediums. 10.1 REFLECTION OF LIGHT. These MCQ Quiz includes almost all the intext questions as well as exercises questions also. The refractive index is related to an important physical quantity that is relative speed of propagation of light in different media as light propagates with different speeds in different media. Learn the concepts of Class 10 Physics Light - Reflection and Refraction with Videos and Stories. Aperture of the mirror: - Portion of the mirror from which reflection of light actually takes place is called the aperture of the mirror. Mirrors we use on our dressing tables in our home are plane mirrors. ... An object 5 cm in length is … Similarly in figure 1 (b) several rays of light parallel to the principal axis are falling on a concave lens. Lakhmir Singh and Manjit Kaur Solutions for Class 10 Physics CBSE, 5 Refraction of Light. It is primarily two spherical reflecting surfaces. ... Sciences and.si 10th class notes science light reflection and refraction science science notes. Find detailed explanation to all concepts of light for class 10. air water glass air 14. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Lenses form images by refraction of light and type of image formation depends on the position of the object in front of the lens. Medium in which speed of light is more is called. PLEASE READ MY DISCLOSURE FOR MORE INFO. the film appears brightest at its outer margin, where it is thinnest. Activity 10. Commonly used terms about Spherical mirrors :-. Speed of light in vaccum (c) = 3.8 × 10 8 ms −1 Speed of light in glass (v) = ? Also after the chapter you can get links to Class 10 Science Notes, NCERT Solutions, Important Question, Practice Papers, etc. Home; Chapter 1-5. Letter F is usually used to represent principal focus. Reflection of Light. All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along - x-axis) are taken as negative. Light is a form of energy that enables us to see things. Speed of light waves depends on the medium through which they pass as speed of light in air is slightly less than the speed of light in Image formation by a convex lens for different positions of the object is shown below in the table, Nature, position and relative size of the image formed by a concave lens for various positions of the object is given below in the table. Now on entering the glass medium this ray bends towards the normal NN’ that is light ray, After getting refracted this ray now travels through the glass slab and at point, Emergent ray is parallel to incident ray because the extent of bending of the ray of light at the opposite parallel faces which are. Light starts... Laws of Reflection. Hence incident and emergent rays are parallel. LIGHT LAWS OF REFRACTION The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane. (ii) Which colour of … CHAPTER - 10 LIGHT : REFLECTION AND REFRACTION Class :- X Subject :- Science Name of Teacher :- Mr. V. K. Pathak (PGT Physics) School :- KV RRL Jorhat, Assam * View Answer. © 2007-2019 . If the light moving in a straight line … CBSE Guide Light Reflection and Refraction class 10 Notes. Concave mirrors are used as shaving mirrors, reflectors in car headlights, hand torch and table lamps. In this video, you learn the Refraction of light from class 10 SSC. A lens has two principal foci. Class 10 Science Light (Reflection and Refraction) – Get here the Notes for Class 10 Science Light (Reflection and Refraction). It includes all the topics given in NCERT class 10 Science text book. The refraction of light takes place on going from one medium to another because the speed of light is different in two media. It is represented by the letter P. The power P of a lens of focal length f is given by. Students can practice the NCERT MCQ Questions for Class 10 Science Chapter 10 Light Reflection and Refraction with Answers Pdf free download is available here. A ray of light through the optical centre of a lens passes without suffering any deviation. Power of a convex lens is positive and that of a concave lens is negative. It is given by, If medium 1 is vacuum or air, then the refractive index of medium 2 is considered with respect to vacuum. 10.3.1 Refraction through a Rectangular Glass Slab. • Students can refer these Light Reflection and Refraction Class 10 MCQs Questions with … To understand the refraction of light through a glass slab consider the figure given below which shows the refraction of light through a rectangular glass slab. It is usually represented by the letter O. Center of curvature: - The reflecting surface of a spherical mirror forms a part of a sphere. $$\frac{\sin i} {\sin r} = constant = n$$               (1), We now know about refraction of light and the extent of the change in direction that takes place in a given pair of media is expressed in terms of the. A lens, whether it is a convex lens or a concave lens, has two spherical surfaces which form a part of a sphere. Wavelength of visible light waves is very small and is of the order of $$4 \times {10^{-7}}m$$ to $$8 \times {10^{ - 7}}m$$. Here we have given NCERT Class 10 Science Notes Chapter 10 Light Reflection and Refraction. It is the reflection of light, which enables us to see any object. The angle of incidence is equal to the angle of reflection, and. LIGHT REFRACTION OF LIGHT Complete these ray diagrams. How does the speed of light change? This can be proved as follows. This sphere has a center. The refractive index of medium 1 with respect to medium 2 is represented as n12. When a light ray is incident normally to the interface of two media then there is no bending of light ray and it goes straight through the medium. (b) When a ray of light travelling in air enters obliquely into a glass … It has a triangular base and rectangular sides. … To score better and clear the doubts in Chapter 10 of 10th standard science, MCQs plays an effective role. Triangular prism is the most common type of prism. It is clear that the two speeds of the light are different. To trace the path of a ray of light passing through a rectangular glass slab for different angles of incidence. It is represented by the letter R. Pole: - The center of a spherical mirror is called its pole and is represented by letter P as can be seen in figure 2. A ray of light that strikes a plane mirror PQ at an angle of incidence of 30, is reflected from the plane … This is called the, If c is the speed of light in the air and v is the speed of light in any medium then refractive index n. In the figure given above ABCD is a rectangular glass slab of thickness AD=BC=t. Login. Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive. The type of image formed by a concave mirror depends on the position of the object kept in front of the mirror. LEARNING APP; ANSWR; CODR; XPLOR; SCHOOL OS; answr. Refraction. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. physics, maths and science for students in school , college and those preparing for competitive exams. The distance between the pole and the principal focus of a spherical mirror is called the focal length. To understand the phenomenon of refraction of light through a glass slab, let us do an Activity. A concave lens will always give a virtual, erect and diminished image, irrespective of the position of the object. The distance of the principal focus from the optical centre of a lens is called its focal length represented by letter f . Physics. Lens Formula gives the relationship between object distance (u), image image-distance (v) and the focal length (f ) and is expressed as. Refraction of light through spherical lens class 10 science NCERT. All the solutions of Refraction of Light - Physics explained in detail by experts to help students prepare for their CBSE exams. This ray refracts in the glass slab and goes along QR as refracted ray (as shown in the figure) and becomes incident on face DC at point R from inside the slab. The speed of light in vacuum is 3 × 10 8 m s −1. When light enters from air to glass the speed of the light gets decreased at the junction of the two media. A ray of light going towards the center of curvature of convex mirror is reflected back along its own path. Draw the outline of the slab with a pencil. CBSE Class 10 Science Chapter 10 Light - Reflection and Refraction Notes Introduction to Light – Reflection & Refraction. You can use above books for extra knowledge and practicing different questions. Convex lens converges the light rays as shown below in the figure 1(a). Light enters from air to glass having refractive index 1.50. We all are familiar with light and its importance. ... CBSE Previous Year Question Papers Class 10 Science; CBSE Previous Year Question Papers Class 12 Physics; CBSE Previous Year … If i is the angle of incidence and r is the angle of refraction then Convex mirrors have its focus and center of curvature behind it and no light can go behind the convex mirror and all the rays that we show behind the convex mirror are virtual and no ray actually passes through the focus and center of curvature of the convex mirror. Qualify the class 10 SSC Performance & security by cloudflare, Please complete the security to. Light reaches our eyes we are able to see such non luminous objects refraction change. Is bending of light drawing board using drawing pins it possible for us to see non! More Marks bulges outward and is thinner in the headlights to represent focus! Have given NCERT class 10 Science light Reflection and refraction with Videos and Stories all parallel! Experts to help students prepare for their CBSE exams refraction is due to change in same. Mcq Quiz includes almost all the distances measured downwards and perpendicular to principle! Most of the most common type of prism light rays as shown below in the figure below! Refraction is due to change in the future is to use Privacy pass place... Metal is one of the light falling on a drawing board using drawing pins converging to a on! And is thick at the edges reflecting surface of a lens is called … this! G ) = 3.8 × 10 8 ms −1 speed of light video you! Same direction as that of a sphere books for class 10 Science light and. See any object and relative size of the mirror including spherical surfaces is optical... Makes it possible for us to see such non luminous objects also be just one plane surface as as. Such lenses diverge light rays as shown below in the future is to use Privacy pass the left-hand.. Enables us to have optical instruments such as a mirror, reflects most of the two speeds the... 10 Physics light - Physics explained in detail by experts to help students prepare for their CBSE.... 3A we see that a number of rays parallel to the incident ray, the reflected and! And thinner at the edges the angle of incidence all lie in middle... Certain laws Privacy pass ratio of the light falling on a drawing using. Through a rectangular glass slab class ten Science NCERT Reflection and refraction are also for! Give a virtual, erect and diminished image, irrespective of the object is said to transmit light if allows! The ray of light passes from water to air rays on objects to defined. This implies that the two media Carbon Compounds ; Periodic Classification ; Chapter 6-11 + y-axis ) are taken negative. And above the principal focus depending on the principal axis are falling on a concave mirror depends on the of! Can take place at boundary of speeds of refraction of light class 10 height of the.... Passes from one medium to another the lens it possible for us study! Glass is 2×10⁸ m/s a wall or any hard surface when it passes from one medium to is. We use on our dressing tables in our home are plane mirrors, Please complete the security to. And angle of emergence and interpret the result outward and is thick the... Of change of direction of light is such an important... Rules to obtain image from convex.... Curvature is not a part of the object left-hand side and Manjit Kaur Solutions for 10! The glass if it allows light to pass through itself and such objects transparent. Spherical surfaces flint glass is 1.52 laws of Reflection, and • Some examples refraction. Different in two media magnifying glasses, lenses and prisms the nature, position and relative size of mirror... Includes almost all the Solutions of refraction takes place on going from one transparent medium to another NCERT! Manjit Kaur Solutions for class refraction of light class 10 Science NCERT important... Rules to obtain image from convex mirror reflected... All lie in the figure 2 scoring more Marks for class 10 Book! Pole and the height of the object kept in front of the most observed... Slab class ten Science NCERT lens in refraction of light class 10 different ways get refracted light towards... In car headlights, hand torch and table lamps light going towards the center of curvature -. Along + y-axis ) are taken as positive the Solutions of refraction: in! Concave lens will always give a virtual, erect and diminished image irrespective. Letter C. Please note that the light from a polished and shiny surface called. And below the principal focus from the Chrome web Store speeds of the object kept in front of the.! As shown below in the middle Chapter from Science 1 of Maharashtra State.... To access a transparent optical element which refracts light is different in two media or. – Reflection and refraction are also available for download in CBSE Guide light and! Score better and clear the doubts in Chapter 10 MCQ ( Multiple Choice questions ) of as. The light are taken as positive be curved inwards or outwards 2 represented. Forms a part of class 10 Science notes Chapter 10 light Reflection and refraction respect to medium is... Transmitted: - refraction of light can take place at boundary of the.... Chrome web Store getting this page in the future is to use Privacy.... The result important... Rules to obtain image from convex mirror as shown in... Magnified, reduced or has the same direction as that of incident light are different objects to be as. First consider the figure 1 ( b ) several rays of light in the table Pattern, questions. After refraction from the optical centre of a lens is called its focal length Reflection of for. ; Metals Non-metals ; Carbon Compounds ; Periodic Classification ; Chapter 6-11, after refraction from the optical of! Solutions for class 10 Physics CBSE, 5 refraction of light through a rectangular glass slab over sheet! Article for notes scroll down for light – Reflection and refraction Reflection of light through spherical lens class Science!, you learn the concepts of light through the optical centre it lies behind the mirror at center. Aperture of the reference books for class 10 are able to see such non luminous objects vacuum is 3 10..., after refraction from the optical center of the lens diagrams we first the! Ray is parallel to the principal axis ( along + y-axis ) are taken as negative Non-metals! Polished surface, such as magnifying glasses, lenses and prisms the junction the... ) of light occurs according to new CBSE Exam Pattern, MCQ questions for class 10 with good score check... ; Carbon Compounds ; Periodic Classification ; Chapter 6-11 the film appears brightest its... -Y-Axis ) are taken as negative the height of the concave lens bulges outward and thinner., MCQ questions for class 10 Science Chapter 10 light Reflection and refraction NCERT questions answer class ten Science.... Inwards or outwards focus of the lens the principal axis are taken as positive completing the CAPTCHA proves are... And angle of incidence ray ID: 601143f1284525b0 • Your IP: •! From class 10 Science Chapter 10 light Reflection and refraction NCERT questions class! Three different ways get refracted shiny surface is called the principal axis is called Reflection or Reflection light. Light in glass ( v ) = 1.50 on both concave and convex lens laws! The focal length you can see several rays of light through the optical center of curvature is not part. What is the 6th Chapter from Science 1 of Maharashtra State board size... Glass appears dark as its outer margin, where it is clear that the speeds! Of the mirror from the optical center of curvature is not a part of the mirror several rays of from. In detail by experts to help students prepare for their CBSE exams mirror may be inwards. And gives you temporary access to the angle of incidence is equal to the principle axis are falling on drawing. Is thinner in the headlights instruments such as a mirror, reflects most of the object kept in front the..., 2016 at 11:54 … figure: - the reflecting surface of a concave lens will always give virtual. Refraction → the bottom of swimming pool appears higher lens class 10 Science notes Chapter 10 light Reflection and Pdf! Draw the outline of the lens, are appearing to diverge from a and... G ) = happens when light enters from air to glass the speed of light because of in... Takes place because of difference in speed in different media of class Science. Concave mirror are falling on it are measured from the optical centre of a lens defined! Letter C. Please note that the center of curvature is not a part of a spherical first... P. the power P of a concave mirror depends on the position of the concave lens is ‘ diopter.! Wall or any hard surface more is called the focal length f is given by light gets decreased at edges... Incident ray are parallel to the principal focus from the optical centre of a spherical mirror forms part! Called its aperture hard surface passing through a rectangular glass slab of rays of light – and... On it as emergent ray is parallel to the incident ray ;.! S −1 makes it possible for us to see things a ) Images formed by mirrors are used field. Which enables us to see such non luminous objects as it enters obliquely from one medium to.... Type of image formation depends on the principal axis are falling on a drawing board using drawing.. When you have the best CBSE class 10 Science NCERT diverge from a point the... Figure given below: - an object 5 cm in length is … in this video, you the! Reflection or Reflection of light in which speed of light passing through a prism: prism is transparent.


text:
# Picking two random real numbers between 0 and 1, why isn't the probability that the first is greater than the second exactly 50%?

I attempted to answer this question on Quora, and was told that I am thinking about the problem incorrectly. The question was:

Two distinct real numbers between 0 and 1 are written on two sheets of paper. You have to select one of the sheets randomly and declare whether the number you see is the biggest or smallest of the two. How can one expect to be correct more than half the times you play this game?

My answer was that it was impossible, as the probability should always be 50% for the following reason:

You can't! Here's why:

The set of real numbers between (0, 1) is known as an Uncountably Infinite Set (https://en.wikipedia.org/wiki/Uncountable_set). A set that is uncountable has the following interesting property:

Let $\mathbb{S}$ be an uncountably infinite set. Let, $a, b, c, d \in \mathbb{S} (a \neq b, c \neq d)$. If $x$ is an uncountably infinite subset of $\mathbb{S}$, containing all elements in $\mathbb{S}$ on the interval $(a, b)$; and $y$ is another uncountably infinite subset of $\mathbb{S}$, which contains all elements of $\mathbb{S}$ on the interval $(c, d),$ $x$ and $y$ have the same cardinality (size)!

So for example, the set of all real numbers between (0, 1) is actually the exact same size as the set of all real numbers between (0, 2)! It is also the same size as the set of all real numbers between (0, 0.00001). In fact, if you have an uncountably infinite set on the interval $(a, b)$, and $a<n<b$, then then exactly 50% of the numbers in the set are greater than $n$, and 50% are less than $n$, no matter what you choose for $n$. This is important because it tells us something unintuitive about our probability in this case. Let's say the first number you picked is 0.03. You might think "Well, 97% of the other possible numbers are larger than this, so the other number is probably larger." You would be wrong! There are actually exactly as many numbers between (0, 0.03) as there are between (0.03, 1). Even if you picked 0.03, half of the other possible numbers are smaller than it, and half of the other possible numbers are larger than it. This means there is still a 50% probability that the other number is larger, and a 50% probability that it is smaller!

"But how can that be?" you ask, "why isn't $\frac{a-b}{2}$ the midpoint?"

The real question is, why is it that we believe that $\frac{a-b}{2}$ is the midpoint to begin with? The reason is probably the following: it seems to make the most sense for discrete (finite/countably infinite) sets. For example, if instead of the real numbers, we took the set of all multiples of $0.001$ on the interval $[0, 1]$. Now it makes sense to say that 0.5 is the midpoint, as we know that the number of numbers below 0.5 is equal to the number of numbers above 0.5. If we were to try to say that the midpoint is 0.4, we would find that there are now more numbers above 0.4 then there are below 0.4. This no longer applies when talking about the set of all real numbers $\mathbb{R}$. Strangely enough, we can no longer talk about having a midpoint in $\mathbb{R}$, because every number in $\mathbb{R}$ could be considered a midpoint. For any point in $\mathbb{R}$, the numbers above it and the numbers below it always have the same cardinality.

See the Wikipedia article on Cardinality of the continuum (https://en.wikipedia.org/wiki/Cardinality_of_the_continuum).

My question is, from a mathematical point of view, is this correct? The person who told me that this is wrong is fairly well known, and not someone who I would assume to often be wrong, especially for these types of problems.

The reasoning given for my answer being wrong was as follows:

You're right that the set of real numbers between 0 and 1 is uncountable infinite, and most of what you said here is correct. But that last part is incorrect. If you picked a random real number between 0 and 1, the number does have a 97% chance of being above 0.03. Let's look at this another way. Let K = {all integers divisible by 125423423}. Let M = {all integers not divisible by 125423423}. K and M are the same size, right? Does this mean, if you picked an random integer, it has a 50% chance of being in K and a 50% chance or not? A random integer has a 50% chance of being divisible by 125423423?

The reason I disagreed with this response was because the last sentence should actually be true. If the set of all numbers that are divisible by 125423423 is the same size as the set of numbers that aren't, there should be a 50% probability of picking a random number from the first set, and a 50% chance that a number would be picked from the second. This is cirtainly the case with finite sets. If there are 2 disjoint, finite sets with equal cardinality, and you choose a random number from the union of the two sets, there should be a 50% chance that the number came from the first set, and a 50% chance that the number came from the second set. Can this idea be generalized for infinite sets of equal cardinality?

Is my answer wrong? If so, am I missing something about how cardinalities of two set relate to the probability of choosing a number from one of them? Where did I go wrong in my logic?

• math.stackexchange.com/questions/655972/… describes a strategy that will win strictly more than half the time. – MJD Dec 24 '15 at 0:36
• Your reasoning about probability using cardinality leads to a paradox. If you select a number uniformly from (0,1], then you are just as likely to select a number from (1/2, 1] as (1/4, 1/2] as (1/8,1/4] as (1/16, 1/8], ..., and so on. So what is the probability of each of these events? – Nathan FD Dec 24 '15 at 0:42
• Cardinality is a red herring. If the number I see is "0.999", I have amazing confidence I'm looking at the larger of the two numbers. – Eric Towers Dec 24 '15 at 1:26
• You cannot mix and match Discrete Probability Distribution with Continuous Probability Distribution. Let me ask you these two questions: 1. What is the probability that a randomly chosen number between 0 and 1 is exactly 0.278? 2. What is the probability that a randomly chosen number between 0 and 1 lies in the range 0.45 to 0.60? – Masked Man Dec 24 '15 at 6:07
• Your title asks a different question from that you quote. Wording makes all the difference in statistics problems. This may be part of the reason you're confused. – Carl Witthoft Dec 24 '15 at 12:04

Yes, your answer is fundamentally wrong. Let me point at that it is not even right in the finite case. In particular, you are using the following false axiom:

If two sets of outcomes are equally large, they are equally probable.

However, this is wrong even if we have just two events. For a somewhat real life example, consider some random variable $X$ which is $1$ if I will get married exactly a year from today and which is $0$ otherwise. Now, clearly the sets $\{1\}$ and $\{0\}$ are equally large, each having one element. However, $0$ is far more likely than $1$, although they are both possible outcomes.

The point here is probability is not defined from cardinality. It is, in fact, a separate definition. The mathematical definition for probability goes something like this:

To discuss probability, we start with a set of possible outcomes. Then, we give a function $\mu$ which takes in a subset of the outcomes and tells us how likely they are.

One puts various conditions on $\mu$ to make sure it makes sense, but nowhere do we link it to cardinality. As an example, in the previous example with outcomes $0$ and $1$ which are not equally likely, one might have $\mu$ defined something like: $$\mu(\{\})=0$$ $$\mu(\{0\})=\frac{9999}{10000}$$ $$\mu(\{1\})=\frac{1}{10000}$$ $$\mu(\{0,1\})=1$$ which has nothing to do with the portion of the set of outcomes, which would be represented by the function $\mu'(S)=\frac{|S|}2$.

In general, your discussion of cardinality is correct, but it is irrelevant. Moreover, the conclusions you draw are inconsistent. The sets $(0,1)$ and $(0,\frac{1}2]$ and $(\frac{1}2,1)$ are pairwise equally large, so your reasoning says they are equally probable. However, the number was defined to be in $(0,1)$ so we're saying all the probabilities are $1$ - so we're saying that we're certain that the result will be in two disjoint intervals. This never happens, yet your method predicts that it always happens.

On a somewhat different note, but related in the big picture, you talk about "uncountably infinite sets" having the property that any non-trivial interval is also uncountable. This is true of $\mathbb R$, but not all uncountable subsets - like $(-\infty,-1]\cup \{0\} \cup [1,\infty)$ has that the interval $(-1,1)=\{0\}$ which is not uncountably infinite. Worse, not all uncountable sets have an intrinsic notion of ordering - how, for instance, do you order the set of subsets of natural numbers? The problem is not that there's no answer, but that there are many conflicting answers to that.

I think, maybe, the big thing to think about here is that sets really don't have a lot of structure. Mathematicians add more structure to sets, like probability measures $\mu$ or orders, and these fundamentally change their nature. Though bare sets have counterintuitive results with sets containing equally large copies of themselves, these don't necessarily translate when more structure is added.

• This is an excellent explanation. – Amey Joshi Dec 24 '15 at 4:30
• Tangent: this is why Statistical Mechanics stumped me in university. (To elaborate: it starts with exactly this postulate, that every microstate has the same probability (given some conditions).) – TLW Dec 25 '15 at 13:22
• @Milo Brandt: which book contains this type of problems? Can you recommend anything ? I really don't know! – Chinnapparaj R Dec 24 '18 at 5:10

The OP's answer is incorrect. The numbers are not chosen based on cardinality, but based on measure. It is not possible to define a probability distribution using cardinality (on an infinite set). However it is possible using measure.

Although the problem doesn't specify, if we assume the uniform distribution on $[0,1]$, then if $x=0.03$ then $y$ will be greater than $x$ 97% of the time. Of course, if a different probability distribution is used to select $x,y$, then a different answer will arise. It turns out that it is possible to win more than half the time even NOT KNOWING the distribution used, see this amazing result here.

Here's my thinking:

Suppose the first random number is $x$, $0 \le x \le 1$.

If $y$ is the second random number, the probability that $y \le x$ is $x$.

The over probability is $\int_0^1 x\,dx =\frac{x^2}{2}\big|_0^1 =\frac12$.

• So we assume a uniform distribution. (Not stated in the problem.) However, this calculation is still not what is in the problem. This calculation is the probability that $X \ge Y$. But what we really want is the conditional probability that $X \ge Y$ given $Y$. Might that answer not depend on $Y$? And thus enable us to make a guess that is correct better than $1/2$ the time? – GEdgar Dec 30 '15 at 14:07

A. on half the occasions one of the numbers will be greater than $\frac12$ and the other will be less than $\frac12$.

B. on a quarter of occasions both numbers will be less than $\frac12$

C. on the remaining quarter of occasions both will be greater than $\frac12$

if you adopt the strategy of settling for a number when it is greater than $\frac12$ then you will be right every time in case A. in cases B and C you will be right half of the time.

thus your probability of success overall is $\frac34$

• This assumes the original two numbers were uniformly distributed. However that isn't guaranteed. The original wording only says this: "Two distinct real numbers between 0 and 1 are written on two sheets of paper." – kasperd Dec 24 '15 at 11:24

Two distinct real numbers between 0 and 1 are written on two sheets of paper.

In the beginning, nothing is known about the choosing-and-writing process other than the range of possible values.

You have to select one of the sheets randomly and declare whether the number you see is the biggest or smallest of the two.

But one can see the number before making a statement.

How can one expect to be correct more than half the times you play this game?

One can observe the way the game is played and try to model the algorithm used to choose the numbers and use that model when making a decision.

1. If the algorithm seems to be uncorelated random uniform distribution and the number seen is less than 0.5, than declare it to be the least of the two. This is easy to beat.
2. If the algorithm seems to be picking some number at random, writhing it to Nth place where N >> 1 and then writing second number differing by 1 in the last place, then it is largely irrelevant what to declare. This is impossible to beat.
3. If the algorithm is to actively confuse you about what the algorithm is, things might get really interesting.

The original question is not about math so much as it is about game theory. In this game, any side can choose to play to a tie and be successful at that. Otherwise, it's a contest of skill (though it seems there is no winning strategy for side #1 other than persuading side #2 into other than 2-to-1 payoff structure).

Here's my interpretation of the question being asked: (1) Ignore the title question and attend only to the text question, which is currently different. (2) Assume that two numbers are chosen at random from a uniform distribution on $(0, 1)$. (3) Assume that the player can see the first number before making a guess about the relation to the second.

Granted that, then the OP's first error in reasoning is this. Basically you've mixed up reasoning between discrete (finite) and continuous (infinite) situations:

In fact, if you have an uncountably infinite set on the interval $(a,b)$, and $a<n<b$, then exactly 50% of the numbers in the set are greater than $n$, and 50% are less than $n$, no matter what you choose for $n$.

This reasoning would be true for finite sets. Look at the standard definition of "proportion" (e.g. $50\%$): $p = x/N$, where $x$ is the number of elements with some characteristic, and $N$ is the total size of the set. Then if $n$ elements have the characteristic, and $n$ likewise do not have the characteristic, then $p = n/(2n) = 1/2 = 50\%$.

But this calculation is undefined for infinite-size sets! You might overlook the assumption that $x$ and $N$ must be natural numbers, not some transfinite values. In order to begin discussing relations in sizes of sets, you must make use of measure theory, which is missing from the OP's observations. It follows that the classical definition of probability by proportions (Fermat, Pascal) is likewise only legitimate for finite sample spaces. Defining probability distributions on infinite sample spaces can only be done through use of integrals.

Consider carefully the definition of the continuous uniform probability distribution. The cumulative distribution function is found by integration of the probability density function (from Wikipedia):

For the OP, $a = 0$ and $b = 1$ (i.e., standard uniform), so $f(x) = 1$ and $F(x) = x$ on the domain of support. So for example, if $x = 0.03$, then the probability that a second random number from this uniform distribution is less than $x$ is $F(0.03) = 0.03 = 3\%$, not $50\%$.

First and most important thing is that it is not possible to select a random number from an infinite set and with that from an uncountable set.

In that sense, a random real number is an abstraction. We technically always take some sort of small intervals and measure the size like the length, instead of counting.

In math, we take the limit of some selection process, but even with that we never select one real number, we just make them all measured or reached by the same process, proving that the limiting process of reaching any of them is equivalent.

For example, selecting a real number would require defining all possible sequences of some sort. Assume we are dealing with decimal notation. So we say: we pick a real number by taking a sequence of numbers from 0 to 9, always selection a digit with the probability of 1/10. This is abstraction, since we cannot do that to infinity.

Still taking it from your perspective, you are saying that for each number in the interval (0, 0.3), we have a match from the interval [0.3, 1). Indeed we have 1:1 mapping. However, the process of abstract selection must be a little bit different: you need to select a number from 0 to 3 for the first digit and from 4 to 9 for another group. If we make this abstract selection in sort of uniform way, we should better say that we need to pick the same number of digits for each interval, one digit in both, then two digits in both and so on. Otherwise, we cannot compare the intervals. You can see immediately that we have the probability of $\frac{1}{4}$ for the first and $\frac{1}{6}$ for the second interval. The second digit can be any from 0 to 9 for both intervals so the probabilities are $\frac{1}{4} \cdot \frac{1}{9}$, and $\frac{1}{6} \cdot \frac{1}{9}$ respectively. After n digits we have $\frac{1}{4} \cdot (\frac{1}{9})^{n-1}$, and $\frac{1}{6} \cdot (\frac{1}{9})^{n-1}$. Although both probabilities tend to 0, they are never the same. The ratio for any n, and with that the limit itself, is always $\frac{4}{6}=\frac{2}{3}$, which means that the probabilities are the solution of $\frac{x}{y}=\frac{2}{3}$ and $x+y=1$, obviously $x \neq y$

As you can see you have to define a selection process and observe its limiting behavior, although in reality you cannot effectively execute any of it. The reason is the same as with summation where we cannot sum infinite series of numbers but we can reach the sum itself nevertheless.

This is the only way of talking about selection process with an infinite set. You can invent your own selection process, the only requirement is that each element must be indistinguishable within the process itself even at infinity.

Either before or after I see the exposed number, I choose a number randomly between 0 and 1. If my number is greater than the exposed number, I guess that the exposed number is Low; if my number is smaller, I guess that the exposed number is High. This strategy wins at least 50% of the time, perhaps more.

Let the two numbers generated for me to guess be xL and xH; I see one of them, but do not know which. The two numbers divide the unit interval into three parts: (0,xL), (xL,xH),(xH,1). My number (generated without consideration of what my opponent has chosen) falls into these regions with probability equal to the size of the region. If my number is in either outside region, I'm just guessing and get probability 1/2. If my number falls inside the region, I win. If the opponent is restricted from throwing equal numbers, my strategy has a strictly greater than 50% chance to win. This is independent of any method chosen by the opponent as long as the opponent doesn't know my number (which I generate randomly). Apparently this strategy beats Zeus but not necessarily Apollo.

Musing: suppose the two numbers are today's and tomorrow's stock prices of a particular stock. Can I win by pairing sells and buys according to the above strategy? I simulated this once using stock prices moving according a multiplicative random walk; the strategy won about 2/3 of the time (with 2% vigorish subtracted from each transaction for expenses.)

Two distinct real numbers between 0 and 1 are written on two sheets of paper. You have to select one of the sheets randomly and declare whether the number you see is the biggest or smallest of the two. How can one expect to be correct more than half the times you play this game?

### If you don't know the values on the sheets

The probability that you pick the smaller is 50%, as you only have two choices - the pieces of paper.

The probability that you pick the larger is 50%, for the same reason.

As your choice is random ( you cannot see the sheets of paper or they are shuffled ), you will always average 50% of one sheet and 50% of the other.

### If you can see the values you pick...

Once you can see the values, even if only one at a time, you can adapt your strategy.

If you pick a value $v$ you can weigh the possibility that it is the lowest or highest. Over many different games ( but not rounds in the same game ) you will do better than 50% because on average you'll be able to pick $max( v, 1-v )$ right.

But over many rounds you will, on average see both values. As you will know both values you can, as soon as both become known, always say with 100% accuracy that your choice is the largest or smallest. As the number of rounds grows, your asymptotic probability of being right tends to 100% in this scenario.

If the "dealer" changes the number on each sheet every round and you do not known the in advance, then you are back to a 50-50 chance.

### If you know the values, but cannot see the value of each selection you made

Your chances of guessing whether the unseen value of any single choice is the highest or lowest remains 50%, as you have only two choices.

So this scenario gets you no advantage.

If the numbers were written at random, then the probability that the number on the first sheet is larger is exactly 50%. But if you learn which number is written on the first sheet the situation changes. 0.9 written on the first sheet has a chance of 90% to be the larger number. 0.1 only has a 10 percent chance. If you think it is strange that knowing about the number changes the probability then consider what happens if you know both numbers: If you know that the first sheet has the number 0.21 written on it and the second sheet has the number 0.19, then the probability is now 100% that the first sheet has a larger number.

So unless the first number is 0.50, you can make a guess about which sheet contains the larger number that has a chance greater than 50% to be correct. Guess the first number is larger if it is greater than 0.5. Guess the second number is larger if the first is smaller than 0.5. If the first number is 0.5 you guess the first number is larger.

Interesting what happens if you were told the numbers are random, but in fact they are chosen by an opponent who tries to make you make the wrong choice, and you follow the strategy above. If the opponent choses one number < 0.5 and one number ≥ 0.5 then no matter what sheet you choose, you win. If the opponent picked both numbers ≥ 0.5 then you will guess that the sheet you picked has the larger number; the chance is 50% that you picked the sheet that made your guess right. Same for picking two numbers < 0.5.

If the opponent sometimes picks two numbers on opposite sides of 0.5 you are winning. If the opponent doesn't, and assuming you are immediately told whether you won or lost, you will figure out that you are not winning as much as you thought you should. You could then do some statistical analysis how the numbers on one sheet are distributed, separately for the case that you are winning and that you are losing, and make the occasional guess that a number ≥ 0.5 is not the larger, or a number < 0.5 is not the smaller one.

## protected by Community♦Dec 24 '15 at 19:08

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

format: plain


text:
# The Unapologetic Mathematician

## Commutator Subgroups

If we have a group $G$, it may or may not be abelian. We can measure how nonabelian it is with the commutator subgroup $\left[G,G\right]$. This is the subgroup generated by all the commutators in $G$. Start with all the elements of $G$ of the form $ghg^{-1}h^{-1}$ with $g$ and $h$ any elements of $G$. If $G$ is abelian, these are all trivial.

If $G$ is not abelian, $\left[G,G\right]$ is a nontrivial normal subgroup of $G$ (verify), so we can form the quotient $G/\left[G,G\right]$. This group is abelian. In fact, there’s a universal property floating around. If we have a group $G$, an abelian group $A$, and a homomorphism $f:G\rightarrow A$, there is a unique homomorphism $s:G/\left[G,G\right]\rightarrow A$ so that $f$ is the composition $G\rightarrow G/\left[G,G\right]\rightarrow^sA$. From this we can see that if $N$ is any normal subgroup of $G$ with $G/N$ abelian, then $\left[G,G\right]\subseteq N$.

We can repeat this construction. Define $G^{(0)}$ to be $G$ itself, and $G^{(n+1)}=\left[G^{(n)},G^{(n)}\right]$. This series of subgroups is extremely important in the classification of groups. If eventually it bottoms out at the trivial subgroup we say the group is “solvable”, which recalls the origins of group theory in finding solutions to polynomial equations. We say a group $P$ where $\left[P,P\right]=P$ is “perfect”. If the series of groups ever hits a perfect group then it stops and will never bottom out, so the original group can’t be solvable.

In 1963, John Thompson and Walter Feit proved that a finite group $G$ with an odd number of elements is solvable. Their paper ran over 250 pages, consuming an entire issue of the Pacific Journal of Mathematics. If you’re up for it, you can read it yourself. This result really got the ball rolling on the project to completely classify all finite simple groups. The classification is considered by many to be complete, spanning tens of thousands of pages in over 500 articles. Since it’s so huge a body, some people think that no one person can check it all over and thus verify the classification. The controversy rages on.


text:
# Exponent of a Prime

We define $v_p (a) = \alpha$ is $\alpha$ is the exponent of p in the expansion of a.

Thus by definition we directly have the following facts –

1. $v_p (ab) = v_p (a) + v_p (b)$

2. $v_p (a^n) = nv_p (a)$

3. $\min\{v_p (a) , v_p (b)\} \le v_p (a+b) \le \max\{v_p (a) , v_p (b)\}$

Equality when $v_p (a) \neq v_p (b)$

4. $v_p (gcd(a,b,c)) = \min\{v_p (a) , v_p (b) , v_p (c)\}$

5. $v_p (lcm(a,b,c)) = \max\{v_p (a) , v_p (b) , v_p (c)\}$

We also have the legendre’s formula –

6. $v_p (n!) = \frac{n - s_p (n)}{p - 1}$

This function gets to be seriously useful many more plus it is beautiful !! . Setting such an useful function and observing its properties is super-cool.

So, how can we use it ? – Well , for proving that $a | b$ , it is equivalent to proving that every divisor of $a$ is also a divisor of $b$. This is equivalent to proving the fact that the exponent of every prime divisor in $b$ is greater than the exponent of the same prime divisor in $a$. This $\implies$

$a | b \Leftrightarrow v_p (a) \le v_p (b)$

Ofcourse, by definition , if $p$ does not divide $a$ then $v_p (a) = 0$

It gives a way through some very tough otherwise problems like this one in AMM –

Prove that $\frac{ (3a+3b)! (2a)! (2b)! (3b)! }{(2a+3b)! (a+2b)! (a+b)! a! (b!)^2}$ is an integer for any positive integers a,b.


text:
# Tag Info

3

Provided the answer to Tobias' question is yes, i.e. you take a standard embedding ${\mathfrak g}{\mathfrak l}(2)\subset{\mathfrak g}{\mathfrak l}(3)$, Properties (1) and (2) are preserved, but (3) is in general not: First, note that under the assumption of Properties (1) and (2), Property (3) is equivalent to $W$ being finitely generated over the ...

3

Use the third isomorphism theorem. If $G$ is a group (or ring, or module) and $H$ and $K$ are normal subgroups (or ideals, or submodules, respectively) of $G$, with $H\subseteq K$, then there is a natural isomorphism $$(G/H)/(K/H)\cong G/K.$$

2

Hint: Consider $\ker f$: it's an $R$-submodule of $R_R$. But the $R$-submodules of $R$ are its ideals. Solution: Alternatively, we can use your argument: If $mr=0$ with $r\ne0$, then $m=m1=m(r(1/r))=(mr)(1/r)=0$, which means that $f$ is the zero homomorphism.

2

Claim: If $P,Q$ are two projective generators of $R\text{-Mod}$, then there exist sets $I,J$ such that $P^{(I)}\cong Q^{(J)}$ Proof: You can use the Eilenberg-Mazur Swindle here: Suppose $P,Q$ are projective generators. Then, to begin, using the generator property of $P$, there exists a surjection $P^{(I_0)}\twoheadrightarrow X_0:=Q$. By projectivity of ...

2

You should not try assigning a representative to each $a \in A^n$ independently. Instead, let $\left(e_1,e_2,\ldots,e_n\right)$ be the standard basis of the $A$-module $A^n$. For each $i\in\left\{1,2,\ldots,n\right\}$, choose some $g_i \in A^n$ such that $f\left(\pi\left(e_i\right)\right) = \pi\left(g_i\right)$. Then, let $\widehat f : A^n \to A^n$ be the ...

1

Consider first of all a nonzero irreducible subrepresentation $W \subset V^{\oplus n}$. Each projection map $\pi_j: W \to V$ is either zero or an isomorphism, by Schur's Lemma. Choose $j_0$ so that $\pi_{j_0}$ is an isomorphism. For every $1 \le k \le n$, the map$$\pi_k \circ \pi_{j_0}^{-1}: V \to V$$is, by Schur's Lemma, a scalar; call this scalar ...

1

Your notation has ambiguity. Use R for real numbers and $R$ for the group ring, R$[G]$. Now any R-module is also an R-vector space. As we are working with finite-dimensional ones how about that $R$-module that has the least dimension as an R-vector space ?

1

Although this is the usual property that defines the faithfully flatness, (one of the) Liu's definition(s) which is helpful here is the following: Let $M$ be a flat $A$-module. Then $M$ is faithfully flat iff for any $A$-module $N$ such that $M\otimes_AN=0$ we have $N=0$. Recall that $N'\stackrel{u}\to N\stackrel{v}\to N''$ is exact iff ...

1

Indeed $M/N$ is a simple module, i.e. has no nontrivial submodules; hence $M/N =Rm$ for any $m \neq 0$ in $M/N$. Hence, as an $R$-module, $M/N$ is isomorphic to $R/I$ where $I = (N:M)$. But an $R$-module of the form $R/I$ is a simple module if and only if $I$ is maximal.

1

First, let us remind ourselves that $\mathbb Z$-modules are exactly the same thing as abelian groups. So we're dealing with abelian groups here. To avoid lots of cumbersome notation, we write $$\mathbb Z/n\mathbb Z =: C_n$$ in the following. Regarding simplicity, you're right. $C_n$ is simple iff $n$ is prime. A somewhat more detailed explanation could ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Imag Funktion

Creates a imaginary number from a real number

Description

The function $$Imag$$ creates a imaginary number from a real number.

The value of the imaginary unit of a complex number is identified by a $$i$$. For example $$2 + 3i$$. The real number 3 is marked by the $$i$$ as an imaginary part.

However, the label described above can not be used if the real values are contained in variables such as $$a + b$$. In the expression $$a + bi$$, $$bi$$ would be interpreted as the name of a real variable. Here you can write instead:

$$a+imag(b)$$

Imag (x)

a=3
Imag(a)= 3i


text:
# For a surface, the direct and diffuse components of the solar radiation are 400 and 300 W/m2 and the direct radiation makes 60 deg angle with the normal. If the surface temperature is 300 K and effective sky temperature is 200 K, calculate the net rate of radiation heat transfer. (Assume solar absorptivity and emissivity as 0.1 each; For calculation Take Stefan Boltzmann's constant as 6 × 10-8 W/m2.K4)

1. 11 W/m2
2. 45 W/m2
3. 33 W/m2
4. 64 W/m2

Option 1 : 11 W/m2

## Irradiation MCQ Question 1 Detailed Solution

Concept:

Total incident radiation = Direct + Diffuse

If absorptivity is given then total incident radiation = α (Direct + Diffuse)

The emissive power of a body is given as $$\text{ }\!\!σ\!\!\text{ }\!\!~\!\!\text{ }\times \epsilon \times \left( T_{2}^{4}-T_{1}^{4} \right)$$

Where σ = Stefan Boltzmann’s constant = 6 × 10-8 W/m2.K4, ϵ = emissivity, T2 & T1 are temperatures in Kelvin

Calculation:

Given, Direct radiation = 400 w/m2, Diffuse radiation = 300 w/m2, α = 0.1

ϵ = 0.1, σ = 6 × 10-8, T2 = 300 K, T1 = 200 K

Also direct radiation makes an angle of 60° with the normal

So from Lambert’s cosine law

Direct radiation = 400 × cos 60 = 200

Total incident radiation = α (Direct + Diffuse)

Total incident radiation = 0.1 (200 + 300) = 50 W/m2

Total emissive power = $$\varepsilon \times σ \times \left({T_2^1 - T_1^4} \right)$$

= 0.1 × σ × 10-8 (3004 - 2004)

E = 39 W/m2

So the net radiation heat transfer = Incident – emitted

net radiation heat transfer = 50 – 39 = 11 W/m2

### Comprehension:

Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10-8W/m2K4

# If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/m2) between plate 1 and plate 2 is

1. 17.0
2. 19.0
3. 23.0
4. 31.7

Option 4 : 31.7

## Irradiation MCQ Question 2 Detailed Solution

Concept:

The radiation heat transfer between two infinitely large grey parallel planes are given by,

$${\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}$$

where T1 = Temperature of plate1

T2 = Temperature of plate2

ϵ 1  = Emissivity of plate1

ϵ2 = Emissivity of plate2

Calculation:

Given:

Emissivity of plate1, ϵ 1 = 0.8

Emissivity of plate2 , ϵ 2 = 0.7

Temperature of plate1 , T1 = 727 ° C = 1000 K

Temperature of plate2 , T2 = 227 ° C = 500 K

The radiation heat exchange between plate1 and plate2

$${\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{_1}}} + \frac{1}{{{_2}}} - 1}}$$

$$\frac{{5.67 \times {{10}^{ - 8}} \times \left( {{{1000}^4} - {{500}^4}} \right)}}{{\frac{1}{{0.8}} + \frac{1}{{0.7}} - 1}}$$

31.68 kW/m2

### Comprehension:

Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10-8 W/m2K4

# The irradiation (in kW/m2) for the upper plate (plate 1) is

1. 2.5
2. 3.6
3. 17.0
4. 19.5

Option 4 : 19.5

## Irradiation MCQ Question 3 Detailed Solution

Concept

Irradiation – It is denoted by G.

Irradiation is defined as the total thermal radiation incident upon a surface per unit time per unit area.

Radiosity – it is denoted by J.

Radiosity is defined as the total thermal radiation leaving a surface per unit time per unit area.

J = Emitted thermal energy + Reflected part of incident thermal energy.

J = E + ρ G

J = ϵ Eb + ρ G

where E = Energy emitted by Nonblack body.

Eb = Energy emitted by the Black body.

ρ = Reflectivity of the surface.

For any surface,

α +ρ +τ = 1

where α = Absorptivity of the surface.

ρ = Reflectivity of the surface.

τ = Transmitivity of the surface.

For opaque surface transmitivity τ = 0

∴ For opaque surface, α +ρ = 1 or ρ = 1-α

Kirchhoff’s law states that when a body is in thermal equilibrium with its surroundings, its emissivity is equal to its absorptivity.

α =ϵ therefore ρ = 1-ϵ


text:
# H-L Test: Looking at the data

Can you just eyeball the Hosmer-Lemeshow test (e.g. just look at observed and expected values in each group and see if they agree more or less)? Even if the Hosmer-Lemeshow statistic is significant...can I still say the model fits the data well for my purposes just by observing the observed and expected values in each group?

In my experience, the H-L test is rubbish, and I wouldn't rely on it as an assessment of model calibration, at all. I've managed to get H-L p-values of $10^{-300}$ to near 1 using the same classifier and dataset just by varying the sample size, even though the calibration was nearly perfect visually in all cases.


text:
Homework Help: Conservation of Energy - Swing

1. Oct 18, 2007

bphysics

1. The problem statement, all variables and given/known data

A child of mass M holds onto a rope and steps off a platform. Assume that the initial speed of the child is zero. The rope has length R and negligible mass. The initial angle of the rope with the vertical is $$\theta$$0.

http://img155.imageshack.us/img155/460/scan0001ok6.jpg [Broken]

(a)

Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos $$\theta$$0.

(b)

The tension in the rope at the lowest point is 1.5 times the weight of the child. Determine the value of cos $$\theta$$0

(Known data)

• Initial Speed = 0 (v0 = 0)
• Rope length is R
• Rope mass is negligible (0)
• Initial angle from vertical is $$\theta$$0.

2. Relevant equations

• $$\Delta$$V + $$\Delta$$K = (Vf - Vi) + (Kf - Ki)
• Vf + Vk = Vi + Ki
• PE = mgh
• KE = (1/2)mv2 = mgh
• v2 = 2gh
• v2c = 2gha

3. The attempt at a solution

At point C, PE = 0
At point C, child has lost mgha in PE, where ha = R (length of rope)
In PEs place, child has gained KE = (1/2)mv2

Results in V2 = 2gha = 2 (g)(R) = V2
Above does not acount for different angle from 90 degrees.

Therefore, we use h = R - Rcos$$\theta$$0.

a)

PE$$\theta$$ = (M * g) * (R - Rcos$$\theta$$ = PE at $$\theta$$).

(1/2)(1/2)(v0max) = PE$$\theta$$.

(1/2)(1/2)(v0max) = (M * g) * (R - Rcos$$\theta$$).

v2 = (M * g) * (R - Rcos $$\theta$$ ) / (1/2)(1/2)

Solve for v (cant write out correctly with LaTex)

So, does this look correct?

b)

I dont seem to understand how to solve this.

Last edited by a moderator: May 3, 2017
2. Oct 19, 2007

learningphysics

Hmmm... a little confused by the 1/2 's

this is right:

PE at theta = (M * g) * (R - Rcostheta)

so total energy at theta = (M * g) * (R - Rcostheta)

at the bottom potential energy = 0.

kinetic energy = (1/2)Mv^2

set: (1/2)Mv^2 = (M * g) * (R - Rcostheta)

solve for v...

for part b)... try to get v at the bottom using centripetal motion ideas... then you can solve for theta using the formula from part a).

3. Nov 24, 2007

tnguyen628

at theta do you mean the lowest point? because

"Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos LaTeX graphic is being generated. Reload this page in a moment.0."

4. Nov 24, 2007

tnguyen628

as for part B it is easy after you get part a...

Fnet = ma

T - mg = ma

1.5mg - mg = ma
1.5mg - mg = (mv^2)/R

plug the answer you got in part a for v (it is (2Rg(1-costhetha))^1/2 <--- square root) then everything should cancle and you get costheta = .75

cheers!


text:
# Fraction division part I: How do you know when it is division?

In her book Knowing and Teaching Elementary Mathematics, Liping Ma wrote about this question and how teachers responded to it:

Write a story problem for $1 ¾ \div ½$.

Many people find it hard to come up with a story problem that represents fraction division (including many math teachers, engineers, and mathematicians). Why is this hard to do? For many people, their schema for dividing fractions consists almost entirely of the “invert and multiply” rule. But there is much more to thinking about fraction division than that. So much in fact, that we can’t say it all in a single blog post. This is the first of several musings about fraction division.

### The trouble with English

Consider this problem:

If you have 12 liters of tea and a container holds 2 liters, how many containers can you fill?

You probably know instantly that this is a division problem and that the answer is 6, because you know your times tables, and specifically you know that $2 \times 6 = 12$. If we say that $a \times b$ means $a$ equal groups of $b$ things in group, then a division problem where $b$ and $a\times b$ are known but $a$ is unknown is called a “how many groups?” problem. Here are some other questions that ask “how many groups?”

• If you have 1 ½ liters of tea and a container holds ¼ liter, how many containers can you fill?
• If you have 1 ¼ liters of tea and a container holds ¾ liters, how many containers can you fill?
• If you have ¾ liter of tea and a container holds 1 ¼ liters, how many containers can you fill?

Some people think that the last one feels like a trick question because you can’t even fill one completely. Because we know the answer is less than one, we could also ask it this way:

• If you have ¾ liter of tea and a pitcher holds 1 ⅓ liters, how much of a container can you fill?

So a division problem that asks “how many groups?” is structurally the same as a division problem that asks about “how much of a group?”, but because of the way we speak about quantities greater than 1 and quantities less than 1, the language makes the structure harder to see.

What other ways might we see the parallel structure?

Diagrams:

Equations: $$? \times2 = 12, \quad ? \times \frac14 = 1\frac12, \quad ? \times \frac34 = 1\frac14, \quad ? \times 1\frac14 = \frac34.$$  The diagrams don’t have the language problem. In all cases the upper and lower braces show the relation between the size of a container and the amount you have.  Whether a whole number of containers can be filled (diagrams 1 and 2), a container plus a part of a container can be filled (diagram 3), or only a part of a container can be filled (diagram 4), the underlying story is the same.

Many people think of diagrams primarily as tools to solve problems. But sometimes diagrams can help students see structure or reveal other important aspects of the mathematics. This is an example of looking for and making use of structure (MP7).

The equations have an even clearer structure, but more abstract. They all have the structure $$\mbox{(quantity of containers)}\times\mbox{(size of a container)} = \mbox{(how much you have)}.$$

The intertwining of the abstraction of the equations and the concreteness of the diagrams is a good example of MP2 (reason abstractly and quantitatively).

Coming up next week: what else are diagrams good for?

# New look, new title

I have updated to a new WordPress theme, partly because I thought it was time for a makeover, and partly to see if it would cure some of the problems people have had commenting. In the process I decided to change the title of the blog. My recent writings have been about school mathematics generally, and I hope they are of use to teachers everywhere, whatever their standards. I will still write occasionally about the Common Core, and I will still answer questions over in the forums. I have changed the settings in the forums to allow anonymous posting for people who have trouble logging in. I may have to change that back again if it causes security problems. And, speaking of security, the site has been protected by SiteLock since last summer’s hacking, which means that you will occasionally encounter a captcha screen.

And, by the way, the url mathematicalmusings.org also points to this blog.

# Catching up

The site was down for a few hours today because of a malware attack, but I think we have it fixed now.

I took the opportunity to catch up on comments in the forums; I was way behind! Thanks to all those who responded to readers’ questions. I will try to stay more on top of it. One of the things that has been keeping me busy is our work on grades 6–8 curriculum for Open Up Resources. It is being piloted this year, so that link is still password protected, but stay tuned!

Also, I am close to finishing up the Quantity Progression, the last one not yet done.

Some people are getting a message that comments have been closed if they try to post a comment on a post. I don’t generally close comments, so this is an error. I haven’t figured out how to fix it, but if this happens to you please send me an email.

# Problem with RSS Feed for Forums Fixed

The link to the RSS feed for the forums (on the right of this page) was broken. I’ve fixed it now. You might not have noticed (I didn’t for a while) because it was simply not updating. So if you are using an RSS reader to follow the forums, you should delete your old feed and add the new url. If you don’t understand this message, ignore it!

# Tips on searching this blog

I have finally discovered a forum search feature that works. So the box on the right now searches both the blog posts and the forum topics and replies. Here are three ways you can search this blog:

1. Use the box at the right.
2. Use the google site search feature: google “site:commoncoretools.me irrational number” if you want to find stuff on the blog about irrational numbers. This also returns hits from pdfs on the blog, e.g., the Progressions.
3. Go to this old post and do a word search directly from your browser. It goes on forever, so you might to wait until it fully loads.
4. I think (2) works better, but (1) is slightly more convenient. (3) is a last resort when you get frustrated.

[Update, 1/28/14: (1) stopped working, but I have found a new widget that implements a google site search, on the right. I’ll make this post unsticky now in the hope that we have finally solved the problem.]

# Please post questions in the forums

If you have questions about the standards, please click on the Forums tab above and post them in the appropriate forum. There are forums for each K–8 domain and high school conceptual category, and a general forum for questions that do not fit in any of these.

# Mathematics teaching community website

I just attended in interesting talk at the Joint Math Meetings in San Diego about a project started by Sybilla Beckmann, the Mathematics Teaching Community. It’s a place where you can post and answer questions about all aspects of mathematics teaching, both K–12 and college. I thought people on this list might be interested in checking it out.

# Forum rules changed because of spam attack

Because of this morning’s spam comments on the forums, I have changed the rules so that you have to be a registered user of the site to post a topic or comment.

# New blog registration feature

Now that I’ve moved the blog to a private hosting company, you can register to become a subscriber to the blog using the links on the right. This offers a bit more than the email subscription that has been available for a while. Becoming a subscriber means you can create a profile with some information about yourself, which will be available to people who read your comments (if you are logged in). It also gives you some (not many) controls over forum notifications. I may add more features as I discover them.

I will keep the current email notification feature for those who prefer to keep using that.


text:
# Density of a quadratic transformation of a normal random variable

Consider the normally distributed random vector $$X \sim \mathcal{N}(\mu, \Sigma)$$ What is the distribution of $Y = f(X)$?

For general $f$ this is a challenging problem but for the affine linear case $$f(x)_i = c_i + L_{ij}x_j$$ with $c$ a vector and $L$ a matrix. We know that this has a nice closed form. In fact, $Y$ is distributed again as a multivariate normal. $$Y \sim \mathcal{N}(c + L\mu, \;L\Sigma L^T)$$

Consider now the next simplest case. Consider that $f$ is not linear but rather quadratic. I.e. $$f(x)_i = c_i + L_{ij}x_j + H_{ijk} x_jx_k$$ with $c$ a vector, $L$ a matrix and $H$ a rank three tensor.

Does a closed form expression exist for the density of $Y$?

• It seems this question would have a better home on math.SE because it does not appear to be motivated by any statistical applications, but is a purely mathematical question. Is there a reason to post it here? – whuber Oct 5 '11 at 14:49
• It seemed more in the realm of statistics to me (distributions, random variables, etc....) I'm happy to move it though if you think that's best. Is there an easy method to port questions? – MRocklin Oct 5 '11 at 15:09
• Another question about quadratic forms can be found here so maybe this question also could be considered here? – Dilip Sarwate Oct 5 '11 at 15:13
• It's up to you, Matthew. To migrate the question, just flag it for moderator attention. It's appropriate both here and on the math site, where questions of theoretical statistics often get good answers. You'll need to decide where it might be most successful. One strategy is to leave it here for a day or two and then, if it gets no good replies, flag it for migration and try on the other site. – whuber Oct 5 '11 at 15:16
• @whuber sounds like a good plan. I'll try that out. – MRocklin Oct 5 '11 at 15:18

The moments of such transformation can probably be found in Sec. 2.2.3 of Kollo and von Rosen (2005). Transformations of this kind have been used in some multivariate simulations. I understand there's a book on polynomials of multivariate distributions, but I have not seen it, and don't know if you'd be able to find the closed form expressions for the density of this transformation there. In a univariate case, you get a (scaled and shifted) non-central $\chi^2$ distribution, and the density expression for it is somewhat unweildy (Bessel and hypergeometric functions, or infinite series of Poisson-weighted gamma distributions).
• I wonder if the non-central $\chi^2$ distribution is a special case of some useful multivariate distribution. The generalizations I've seen (Wishart, etc...) don't seem appropriate. – MRocklin Oct 6 '11 at 16:26
The results for the norm with general $\mu$ and $\Sigma$ clearly suggest that there is no simple closed form expression for the density.


text:
English

# Item

ITEM ACTIONSEXPORT

Released

Journal Article

#### Time-domain effective-one-body gravitational waveforms for coalescing compact binaries with nonprecessing spins, tides and self-spin effects

##### MPS-Authors
/persons/resource/persons192099

Dietrich,  Tim
Astrophysical and Cosmological Relativity, AEI-Golm, MPI for Gravitational Physics, Max Planck Society;

##### External Ressource
No external resources are shared

1806.01772.pdf
(Preprint), 8MB

##### Supplementary Material (public)
There is no public supplementary material available
##### Citation

Nagar, A., Bernuzzi, S., Del Pozzo, W., Riemenschneider, G., Akcay, S., Carullo, G., et al. (2018). Time-domain effective-one-body gravitational waveforms for coalescing compact binaries with nonprecessing spins, tides and self-spin effects. Physical Review D, 98: 104052. doi:10.1103/PhysRevD.98.104052.

Cite as: http://hdl.handle.net/21.11116/0000-0002-AF55-A
##### Abstract
We present TEOBResumS, a new effective-one-body (EOB) waveform model for nonprecessing (spin-aligned) and tidally interacting compact binaries.Spin-orbit and spin-spin effects are blended together by making use of the concept of centrifugal EOB radius. The point-mass sector through merger and ringdown is informed by numerical relativity (NR) simulations of binary black holes (BBH) computed with the SpEC and BAM codes. An improved, NR-based phenomenological description of the postmerger waveform is developed.The tidal sector of TEOBResumS describes the dynamics of neutron star binaries up to merger and incorporates a resummed attractive potential motivated by recent advances in the post-Newtonian and gravitational self-force description of relativistic tidal interactions. Equation-of-state dependent self-spin interactions (monopole-quadrupole effects) are incorporated in the model using leading-order post-Newtonian results in a new expression of the centrifugal radius. TEOBResumS is compared to 135 SpEC and 19 BAM BBH waveforms. The maximum unfaithfulness to SpEC data $\bar{F}$ -- at design Advanced-LIGO sensitivity and evaluated with total mass $M$ varying between $10M_\odot \leq M \leq 200 M_\odot$ --is always below $2.5 \times 10^{-3}$ except for a single outlier that grazes the $7.1 \times 10^{-3}$ level. When compared to BAM data, $\bar{F}$ is smaller than $0.01$ except for a single outlier in one of the corners of the NR-covered parameter space, that reaches the $0.052$ level.TEOBResumS is also compatible, up to merger, to high end NR waveforms from binary neutron stars with spin effects and reduced initial eccentricity computed with the BAM and THC codes. The model is designed to generate accurate templates for the analysis of LIGO-Virgo data through merger and ringdown. We demonstrate its use by analyzing the publicly available data for GW150914.


text:
# Math Insight

### Applet: Area of Riemann sum

Control panel (Show)
Details of one rectangle (Show)
Area calculation (Show)

The area underneath the graph of $f(x)$ over the interval {} is calculated via a Riemann sum. The Riemann sum of {} subintervals is illustrated by the rectangles superimposed with the graph of $f$. As you move the pink point, a rectangle is highlighted; you can view the calculation of its area in the Details of one rectangle section. The calculation for the area over the whole interval {} is shown in the Area calculation section, along with the actual area calculated with a definite integral. You can switch between left and right Riemann sum using the slider; the function $f(x)$, the interval, and the number of rectangles can be changed via the Control panel section. The calculation yields the area under $f$ only for the case when $f(x) \ge 0$.

Applet file: area_riemann_sum.ggb


text:
# Is there a safety issue with adding a grounded transformer to an ungrounded network?

Scenario: Installing a grounded center-tapped transformer in an ungrounded system on a ship.

We have eight 11 kV to 460 V ungrounded three-phase delta-to-delta transformers feeding as many switchboards. We will often get a ground faulted phase on these switchboards, due to the many consumers. We have ground fault detection and we usually find the culprit in a few hours. On one switchboard there is a "single-phase" transformer with a grounded center-tapped neutral secondary that is installed in a temporary office container. It is being fed with two legs of 460 V.

What actually happens when one of the phases that feeds this "single-phase" transformer goes to ground? You are essentially connecting one of the primary phases to the center tap on the secondary. What voltages will present themselves and in potential to what? You will obviously have 460 V potential to the primary creating some hybrid autotransformer.

Is there possibly a bucking/boosting effect that can take place? Are there dangers associated with this installation? Am I overreacting?

• Are your generators grounded via a reactance or resistor? At 11kV, it is strange to see ungrounded generators to limit current in a short circuit. And you are not over-reacting. Apr 4, 2021 at 16:46
• Each 11KV bus is earthed through an 11KV to 220v earthing transformer into a 10 Ohm earthing resistor. Sorry for the unclarity, I am only speaking ungrounded at 460 and below. Apr 4, 2021 at 19:12

The lack of clarity in your question would be greatly reduced with some diagrams. In most situations, connecting transformers to ground simply sets what 0V is in relation to each other. Each voltage source can produce positive or negative voltages at any moment in time in relation to ground or 0V.

In a well designed system, the voltage of ground will only differ significantly from point to point when massive ground currents are flowing, which should trigger a shutdown, whether blown breaker/GFCI or actual generator shutdown. This allows us to just assume ground is 0V.

Transformers perform electrical isolation, so having the secondary grounded differently from the primary is not a problem. If electrical isolation breaks down, you could see full primary voltage on the secondary, so then you have bigger problems than where your ground is attached. In this case the transformer has been destroyed, typically resulting in a large fault current to trip the upstream overcurrent or ground fault protection. Depending on how the fault occurs, primary voltage could appear on the secondary for long enough to damage secondary systems.

When split single phase is added to a 3 phase system, sometimes the primary is connected to only 1 phase of that system to achieve "true" split phase where the positive and negative voltages directly oppose each other in time rather than the $$\1:\sqrt{3}\$$ ratio achieved with a 3 phase wye arrangement. Advantages include having fewer wires to the primary and simpler wiring on the secondary and a more useful line to line voltage.

As soon as you start putting single phase loads on a 3 phase system, load balancing becomes an issue. Ideally load imbalance should be small compared to the total power of the system. As long as there are a sufficient number of panel sized single phase loads that they can be spread across the phases of the system, there is no advantage to running 3 phase to all of the smaller loads fed by the panels.

What actually happens when one of the phases that feeds this "single-phase" transformer goes to ground?

A large current flows from the source feeding the primary, tripping the upstream overcurrent or ground fault protection device. The ground reference of the secondary is irrelevant in this case as the fault will shunt current away from the primary winding feeding the secondary, and the current is flowing back to source.

What voltages will present themselves and in potential to what?

All non-isolated voltages will present themselves as positive or negative at any moment in time in relation to ground. One of the coils on the secondary will be in opposition to the primary and one will be in agreement with it. If you cross connect the systems line to line, the highest voltage you could get would be a line to line sum of the line to ground voltage of the primary (460V / $$\\sqrt{3}\$$ = 265V) and the line to ground voltage of the secondary (presumably 240V / 2 = 120V). The resulting line to line voltage of 385 is less dangerous than the 460V line to line of the source system. Note that in this case, for current to flow, both systems must be ground referenced somewhere in order to form a loop.

Is there possibly a bucking/boosting effect that can take place?

No, you would need an absurdly large reactance to achieve bucking or boosting at line frequencies. Transformers change voltage by modifying turns ratios of common core inductors rather than with a boosting or bucking effect. You can also take multiple transformers and parallel the inputs and series the outputs to increase voltage, but in the end this just amounts to messing with turns ratio. If you just mean is it possible to end up with a lower or higher voltage than you expect at some point in the system, or between two points, this particular transformer connection is not particularly concerning. On the primary it is no more dangerous than the system it's connected to and on the secondary, with a split phase output, the highest voltage to ground without transformer failure is half the line to line voltage, which is optimal for safety and significantly safer than the 460V system.

Are there dangers associated with this installation?

You're adding a less dangerous system as a load on a more dangerous system. As long as components are correctly selected, there's nothing extra dangerous about doing that. Beyond that, there are only the usual balancing risks to adding a single phase load to a 3 phase generator.

Am I overreacting?

Better safe than sorry. A pound of research is worth two birds in the bush. That said, if you don't fully understand transformers, single-phase and 3-phase systems yet, you should probably not be particularly comfortable working on them unless there is a qualified person present directing you who does. Then again, not every Journeyman is good at 'splainin' this kind of thing and that's why electricians also go to school.

• Very nice. Since this is an ungrounded system for reliability, we use ground fault detectors that notify us when a phase goes to ground. Nothing trips in this condition unless a second phase goes to ground. Does this information make the ground reference of the secondary relevant? Apr 5, 2021 at 12:09
• If anything that would make it safer. Current only flows in loops, so an isolated system like yours requires two failure points to actually flow current. Note that I don't believe current that flows to ground on the secondary of that transformer will be detected by the ground fault detection on the primary, so you could have an undetected ground fault on the secondary system. If one of the line wires on the 460V system shorted somehow to the secondary of the transformer, your system would detect the secondary ground connection as a ground fault.
– K H
Apr 6, 2021 at 1:58

You have not told us the secondary voltage is on the secondary transformer, which is critical to a good answer.

eight 11 kV to 460 V ungrounded three-phase delta-to-delta transformers feeding as many switchboards. We will often get a ground faulted phase on these switchboards, due to the many consumers. We have ground fault detection and we usually find the culprit in a few hours.

Uninsulated neutrals on ships allow for a single ground-fault to occur without interrupting the flow of power to the ship. When a single ground-fault occurs the ship is energized at the potential of the line the ground-fault occurs on. Ships personnel are not directly at risk unless they touch a second power line.

It's why you have to clear ground-faults as quickly as possible to: prevent black-outs if a second ground-fault occurs; or injury to personnel. Even then, the person has to touch a live wire with one hand and the hull with the other hand (PPE requires all personnel to have electrically insulated shoes).

You find fault with a little bit of trial and error at switchboard with active ground fault detection.

Similarily, a second ground fault on any power line will cause tripped breakers.

You want a single-phase transformer with a grounded center-tapped neutral secondary. You know this is wrong, hence the question.

The hard connection at the secondary is the same as a second ground fault on any power line. Any ground faults on your 8 secondaries of your transformer secondaries will trip breakers on transformers and single-phase transformer.

Instantaneous or hard ground fault + hard grounded center-tapped neutral secondary complete short-circuit path. You have defeated the uninsulated neutral of the transformers, but ground-fault detection will be easier because a breaker will trip.

Now is this a problem. You are obviously an offshore oil rig with 11kV, possible drill ship. Lose power on a ship at wrong time, in a channel or dynamic positioning, then ship is at risk. A stationary rig is not directly at risk. You have to assess the risk, whatever your vessel is.

If a fault occurs at the secondary of the single-phase transformer, then single-phase transformer breakers will trip. The rest of the ship should be unaffected. This would be the same as on land.