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# choice of unit in physics

Course at a Glance. Unit Guides. If the same weight is suspended from another wire of the same material … It only takes a minute to sign up. For any measurement, the unit being used must be stated to give an understanding of the scale of the measurement. Spell. Ltd. What are you looking for? Key Concepts: Terms in this set (10) A baseball is thrown vertically upward and feels no air resistance. An object moves according to the function x = t7/2 where x is the distance traveled and t is the time. Explanation: 1 coulomb is the charge carried by 6.25*10^18 electrons. Unit 1 - Introduction. Match. AP Physics Practice Test Motion In One Dimension. is a unit of electrical voltage is not true. 7 College Course Equivalent 7 Prerequisites 8 Lab Requirement. 12 . UNIT 1: Electrostatics 43. Answer: A. Contact. Find Tutor. Become our. PHYSICS CXC Education. About the AP Physics C: Electricity and Magnetism Course. D The property of a conductor that restricts its internal flow of electrons is resistance. Video transcript - [Teacher] Not only are there many different kinds of energies, but both objects, and systems of objects, can have energy. A. coulomb. The theoretical basis for the development of laser was provided by Albert Einstein in 1917. Galaxies and atoms may seem far removed from your daily life, but as you begin to explore this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice. Download latest questions with answers for Physics Units and Measurements in pdf free or read online in online reader free. COURSE FRAMEWORK. Unit 1 - Safety and Review. Physics Quiz : Measurement Units The standard system of units used by scientists worldwide is called SI (Systeme Internationale d'Unites). 22. D. watt. C. joule. Need assistance? NCERT Solutions Class 9 Science Chapter 5 The Fundamental Unit of Life – Here are all the NCERT solutions for Class 9 Science Chapter 5. SI units. Test. CBSE Class 11 Physics Important Questions Chapter 1 Units and Measurement are provided here for students. its momentum is not conserved, but its mechanical energy is conserved. This Practice Book Contains PHYSICS TEST. Unit –I LASER Engineering Physics Introduction LASER stands for light Amplification by Stimulated Emission of Radiation. Explanation are given for understanding. In SI, there is a base unit for the current, the ampere [A], derived from it the unit of charge, coulomb [C]. Section 2 – Current electricity multiple choice questions. Also, they boost their exam preparation for the annual exam. Physics Questions for both High school and Senior secondary school level examinations, suitable for CBSE,ICSE ICSE and IGCSE . 0 m/s^2 B. Capacitance and Resistivity. CBSE Class 11 Physics Units And Measurements MCQs. PLAY. Multiple Choice Questions If the atoms or molecules in a solid are periodical at regular intervals of distances in three dimensions, then that solid is known as: ( ) crystalline … - Selection from Engineering Physics … Course Framework Components. Test. Resources. The SI unit of temperature is Kelvin. Flashcards. By practising these questions, students get thorough with the important topics of this chapter. This solution contains questions, answers, images, step by step explanations of the complete Chapter 5 titled The Fundamental Unit of Life of Science taught in class 9. Learn. Physics is a branch of science which studies the properties of matter, energy and their mutual relationship. CBSE; JEE; NEET; NTSE; BITSAT; UGC NET; CTET ; ICAI; Important Topics; CBSE. Flashcards. UNIT 2: Conductors, Capacitors, Dielectrics 57. However, the questions often have two answers that could, at first glance, be correct. Home; CBSE; ICSE; IGCSE; Wednesday, 1 October 2014. Physics MCQ with detailed explanation for interview, entrance and competitive exams. Physics 132 Practice Final Exam Multiple Choice Questions. October 16, 2020 – by admin. It is exactly $$\frac {1}{273.16}$$ of the thermodynamic temperature of the triple point of water. AP Physics 1 review of Waves and Harmonic motion. Education Franchise × Contact Us. ELASTICITY MULTIPLE CHOICE QUESTIONS 1. 21. Practice the multiple choice questions to test understanding of important topics in the chapters. If you have applied for Physics Teachers jobs test here you will learn objective type MCQs online quiz exams tests for upcoming jobs preparations. Felixthenetherderpy. Physics 12 Unit 1 Test … There are seven base SI units (for mass, length, time, electric current, temperature, luminous intensity, and amount of substance) from which the other units may be derived. AP Physics 1 Review of Charge and Circuits. 1800-212-7858 / 9372462318. 11 . Physics Unit 1 Multiple Choice Cape Chemistry Unit 2 P1 Answer Sheet Scribd. 9. These unit systems also show that the choice of base units is arbitrary. 8. "Grade 10 Physics Quiz" PDF … It was well known at the time that a single pony could lift 220 pounds of coal up a 100-foot mine shaft in one minute. Unit 7 - Review . For Study plan details. Physics is about an applied mathematics. Its kinetic energy is proportional to: (A) t5/2 (B) t7/2 (C) t5 (D) t3/2 (E) t7 2. 10. Unit 5 - Atoms and Molecules. That translates to 22,000 ft-lb/min. B. volt. These fundamental and derived physical quantities are the integral part of physics as nothing can be defined in physics without referring to the relevant unit of that quantity. As it is rising. Physics Multiple Choice. Pages. Physics 11. Next lesson. 1.Whatsup group: https://chat.whatsapp.com/JwrnnET3JWDLCtlfZDHWgU 2. Download CBSE Class 10 Physics Electricity MCQs in pdf, Physics chapter wise Multiple Choice Questions free. "O Level Physics MCQ" book helps with fundamental concepts for self-assessment with theoretical, analytical, and distance learning. Unit 3 - Stoichiometry. Created by. A. Gravity. Chemistry 11. Which of the following best describes the relationship between force and potential energy? In this text, you learn about the laws of physics. Spell. O Level Physics Multiple Choice Questions and Answers (MCQs): Quizzes & Practice Tests with Answer Key (O Level Physics Quick Study Guide & Course Review Book 1) contains course review tests for competitive exams to solve 896 MCQs. Introduction . Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. PDF | On Nov 2, 2002, K.Sathyanarayana Reddy published CHOICE OF A TELETHERAPY UNIT: COBALT 60 UNIT vs LINEAR ACCELERATOR | Find, read and cite all the research you need on ResearchGate Write. What is the magnitude of her acceleration? STUDY. He then incorrectly assumed that a regular horse could do 50% more work and arbitrarily defined the horsepower as 33,000 ft-lb/min, which is the same as 550 ft-lb/s. 6 - Electromagnetism. Itlrc Com PHYS 2002 Multiple Choice 1 And Answers On. "Grade 10 Physics MCQ" with answers includes fundamental concepts for theoretical and analytical assessment tests. Mainmann. Ask your interesting physics related questions for free. Created by. This developmental approach will also provide the students with a substantial foundation on which to build their understanding of key principles and laws of physics. An ohm (?) or own an. 13 Science Practices 15 Course Content 18 . 22 Using the Unit Guides 25 . The AP Physics 1 Exam consists of two sections: a multiple-choice section and a free-response section. He created a unit based on how much work a single pit pony could do. Basic Physics Quiz, Multiple Choice Questions With Answers. 4 m/s^2 C. 9.8 m/s^2 D. you need to know her mass to be able to calculate her acceleration. B Applect Learning Systems Pvt. 5 - Atomic and Nuclear Physics. Write. Tricia_Glenn. Grade 10 Physics Multiple Choice Questions and Answers (MCQs): Quizzes & Practice Tests with Answer Key (10th Grade Physics Quick Study Guide & Course Review) covers subject tests for competitive exams to solve 1150 MCQs. PSI AP Physics C – Work and Energy (With Calculus) Multiple Choice Questions 1. This is a section 2 of MCQs on current electricity (Physics) with solutions. STUDY. The unit of electrical charge is the. The band diagram of the primitive unit cell is on the left and the doubled unit cell is on the right. Center of Mass. Formats: Info Page: Quiz: Review: Multiple choice. The initial Physics unit could be followed by the unit on Mechanics and then the Thermal Properties and Heat unit. Learn and free practice physics general knowledge (gk) objective type questions & answers with explanation for all competitve exams and quiz. Learn. Unit 2 - The Mole. Your Performance 1. Test and improve your knowledge of Units & Measurement in Physics with fun multiple choice exams you can take online with Study.com The multiple-choice section consists of two question types. A physical quantity is something that can be measured. Match. Key Concepts: Terms in this set (34) Little Nellie Nelson is in free-fall. C In physics, friction is the force that opposes the relative motion of two bodies in contact. The Brillouin zone for the doubled unit cell is half the Brillouin zone of the original, so is it just cutting the original band diagram in half and reflecting it (or is this horribly wrong)? A small car has a head-on collision with a large truck. 1. PLAY. Unit 4 - Organic Chemistry. Unit 2 - Kinematics in 1D. AP Physics 1 free response questions. Academic Partner. Multiple Choice Questions (MCQ) for CBSE Class 12-science Physics chapters on Topperlearning. Here are some examples of Basic or fundamental and derived units. Contact us on below numbers. Download MCQs for NEET Physics Units and Measurements, Get MCQs for Units and Measurements Physics for important topics for all chapters based on 2021 syllabus and pattern. Courses. Search Tutors Tutor Registration Tutor Login. The Gaussian system does without a dedicated unit for electricity. Unit 6 - Solution Chemistry. Unit 2 Physics Test. Basic Physics Quiz, Multiple Choice Questions With Answers. Gravity. Single-select questions are each followed by four possible responses, only one of which is correct. AP Physics 1 review of Torque and Angular momentum. These MCQ's are extremely critical for all CBSE students to score better marks. In 1960, the first laser device was developed by T.H. 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Authors

Stefano Sarao Mannelli, Eric Vanden-Eijnden, Lenka Zdeborová

Abstract

We study the dynamics of optimization and the generalization properties of one-hidden layer neural networks with quadratic activation function in the overparametrized regime where the layer width m is larger than the input dimension d.

We consider a teacher-student scenario where the teacher has the same structure as the student with a hidden layer of smaller width m*<=m.

We describe how the empirical loss landscape is affected by the number n of data samples and the width m* of the teacher network. In particular we determine how the probability that there be no spurious minima on the empirical loss depends on n, d, and m*, thereby establishing conditions under which the neural network can in principle recover the teacher.

We also show that under the same conditions gradient descent dynamics on the empirical loss converges and leads to small generalization error, i.e. it enables recovery in practice.

Finally we characterize the time-convergence rate of gradient descent in the limit of a large number of samples.

These results are confirmed by numerical experiments.

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# Cartesian Product of Countable Sets is Countable/Corollary/Proof 1

## Corollary to Cartesian Product of Countable Sets is Countable

Let $k > 1$.

Then the cartesian product of $k$ countable sets is countable.

## Proof

Let $S_1, S_2, \ldots, S_k$ be countable sets.

By the same argument, there exists an injection $g: S_1 \times S_2 \times \cdots \times S_k \to \N^k$.

Now let $p_1, p_2, \ldots, p_k$ be the first $k$ prime numbers.

From the Fundamental Theorem of Arithmetic, $f: \N^k \to \N$ defined as:

$f \left({n_1, n_2, \ldots, n_k}\right) = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}$

is an injection.

The result follows from Composite of Injections is Injection and Domain of Injection to Countable Set is Countable, as above.

$\blacksquare$


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# Tag Info

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I will just sketch this out. The Killing equation $\xi_{a;b} + \xi_{b;a} = 0$ tells you that $\nabla^i\xi^k$ is antisymmetric in indices. I am going to write this as the commutator between $U$ the vector along which we differentiate $\xi$ with $\nabla_U$ so in compact notation we have $$Q = -\frac{1}{\kappa}\int[U, \xi]d\sigma$$ I now use Stokes' rule to ...

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Your first equation looks a bit like GR with a dilaton. IIRC, the analogous equations in supergravity will naturally have spinors coupling to the dilaton.

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Detecting the emitted photons from a charge accelerating in Earth's gravity looks like a hopeless task, given that the radiated power is tiny (see formula 9 in Anna's answer). But we can try to exploit the fact that the number of emitted photons always diverges no matter how small the acceleration. While very low energy photons (so-called "soft photons") ...

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Let me answer this question systematically. 1)The first is how to think of causality and locality. Locality (sometimes people use locality to talk about microcausality but that's not very important) is the statement that two events cannot communicate with each other if they are separated by spacelike distances. So what does this mean? Suppose you have a ...

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The true meaning of all the trouble is that coordinates are just tools for calculations and don't have any intrinsic meaning in GR. You should never interpret the coordinates alone. These can be very misleading and in particular become singular in some regions of spacetime (because coordinates are local). And never trust the labels of coordinates: Fact that ...

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In special relativity the transition from one frame to another is given by the Lorentz boosts. This is not quite the same as an acceleration, but a transformation that relates observations on one frame with another. We can think of an acceleration as being a succession of infinitesimal Lorentz boosts that map one frame to another. The infinitesimal distance ...

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There is some misunderstanding here. whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate if you mean a charge in free fall. In this calculation:, from the conclusion It is found that the "naive" conclusion from the principle of equivalence - that a freely falling ...

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The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ...

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Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ...

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All motion is relative because - motion is phenomena of change of position of an object. Now if you think about it - change of position has always to be relative to something, otherwise, how do you know that the position changed? Thus all detected motion is relative. There can be absolute motion, but that can not be detected. For example, suppose you are ...

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The conformal transformation $g'_{\mu\nu} = e^{-2\sigma}g_{\mu\nu}$, $sigma = sigma(x)$ leads to the transformation of the Ricci scalar $$R' = e^{2\sigma}R - 12e^{2\sigma}(2\sigma_{,\mu}^{,\nu} - 2\sigma_{,\mu}\sigma^{,\mu}$$ Since $\phi' = e^{\sigma}$ then $$\frac{1}{12}\phi'^2R' = \frac{1}{12}\phi^2 R - \phi^2(2\square\sigma - ... 3 First of all we need the equation: $$\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT}$$ We have:$$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} ...

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Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of $v_0$, the initial velocity in terms of Schwarzschild coordinates $v_i$ would then be $$\dot{\theta}(0)\cdot r(0) = \frac{{v\perp}_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal component, ...

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There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is $3 r_s$. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, $(3/2) ... 3 Indeed,$f$is a symmetric form, since$\omega$and$\omega '$are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes$\omega$, the dimension of the space is$6$. A basis is given ... 2 Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ... 0 Easy way Let me first state the straight-forward way to do this computation. $$\langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d$$ First equality follows from compatibility, second equality uses ... 2 I will answer this with a simple example. Let us consider the metric for weak gravity, $$ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2.$$ The$g_{tt}$metric element is largest by a factor of$c^2$and we have $$\Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}.$$ Now let us work with the geodesic equation that is ... 4 As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ... 0 A spacetime is a manifold with two different, and technically independant, structures : a metric, which describes the distances between two points, and a connection, which describes how to transport a vector "straight". In general relativity, there is a formula to link the two, but at their core, the two notions are independant. A free particle is one that ... 0 Since I'm in a car right now I can't double check my answer, but I'll try my best not to lie to you. To begin with I am a little unclear as to what you are asking. In most circumstances (i.e if the region under consideration is geodesically complete, which in physics is the case if there is no singularity present) any two points can be connected by a ... 2 The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula$S=-k\sum_n p_n log(p_n)$would give a ... 0 The quasinormal mode frequencies are complex-valued numbers. For a given type of field (for example, scalar, vector, gravitational, fermionic, etc) there are a discrete infinity of these frequencies which are of course independent of the metric used to describe the background. If the background possess some symmetry, for example spherical symmetry, then the ... 0 A slightly different and, perhaps, more simple proof follows.$\def\lie{\mathit{£}}$For$K^a$a Killing vector, we have (Kostant formula), $$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$ Using this, we may prove that the covariant and the Lie derivatives commute: $$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary$X^a$. We have: ... 0 The spatial part of a metric is just a 3-sphere: $$d\Omega_3^2 = d\chi^2 + \sin^2\chi d\Omega_2^2 \, ,$$ and in fact the metric on a$n$-dimensional sphere can always be written in a resursive way using the metric on an$(n-1)$-dimensional sphere in the same manner: $$d\Omega_n^2 = d\chi^2 + \sin^2\chi d\Omega_{n-1}^2 \, .$$ In these coordinates the ... 4 The Einstein field equations $$R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu}$$ for zero stress energy means that the Ricci Curvature$R_{\mu\nu}$is proportional to the metric with$R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar$R~=~R_{\mu\nu}g^{\mu\nu}$this is a spacetime of constant ... 1 I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique ... 0 Let$v^\mu \equiv \frac{d}{dt} (x^\mu \circ \gamma)$, where$x^\mu$is a chart, and a tangent vector$u^a$, such that$v^a u_a = 0$,$u^a u_a \neq 0$for certain$t$. Such a vector always exists for$\dim X > 1$. For$\gamma$a geodesic, we parallel-transport$u^a$, hence$v^a \nabla_a (v^b u_b) = 0$for every$t$in the definition of the curve; ... 4 You can easily see this isn't the case by considering the special case of the stress-energy tensor equal to zero i.e. the vacuum solutions. These include the Minkowski metric, which is flat, but also the Schwarzschild and Kerr metrics and of course gravitational waves. 1 Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards ... 2 Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will ... 0 If the event horizons overlap you get one big horizon. EM forces can not counteract gravity if the curvature is too large since the force required to counteract gravity becomes infinite at the horizon. You can see this in the equation $$F=\frac{G\cdot M\cdot m}{r^2\cdot\sqrt{1-r_s/r}}$$ which becomes infinite at the horizon$r_s$. Since from the outside ... 1 The answer is that we don't know. Why? Because the theory of gravity which we have and use, GR, has a singularity. Things which should be finite in a physical theory, like the density, become infinite. And theories with a singularity are simply wrong, they need a modification, and this modification is necessary not only at the singularity itself, but already ... 4 The "theory" describing the black hole interior (in the classical approximation) is the same theory that implies the existence of the black holes, namely the general theory of relativity. As the OP correctly said, the singularity at the event horizon is a coordinate singularity – one that is an artifact of a bad choice of coordinates. When a coordinate ... 0 An orbiting observer is a bit problematic because there are no stable orbits for$r \le 3r_s$, so let's instead consider an observer hovering at some distance$r$from the black hole. In that case as$r \rightarrow r_s$the blue shift does indeed$\rightarrow\infty$and the observer would indeed be roasted. But this shouldn't be surprising. The acceleration ... 1 The reason to consider such metrics was not a particular interpretation, but that for this ansatz one could hope to find some exact solutions. It is known as the Kerr-Schild metric, and its role in the process of finding exact solutions has been described by Kerr in Wiltshire, Visser, Scott (eds.), The Kerr Spacetime. One can, of course, try to develop ... 2 Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ... 0 Not sure about the etiquette of this but I think I can now answer my own question. Please post if there is a better answer. The problem is that I misunderstood the meaning of the statement "the geodesic is the path the optimises the proper time". What this means is that given two endpoints, say$x_0^\alpha$and$x_1^\alpha$a geodesic is a path ... 1 our aging is directly proportional to metabolism of body and division,growth and death of cells of body and if gravity has some effect on the rate of above aspects astronauts will be definitely younger 0 Actually, only a gravitational wave and not a uniform gravitational field is composed of gravitons. I think that just like en electromagnetic wave, a gravitational wave is a sum of sinusodial waves with gravitons of different energy for different wavelengths in that sum. An isolated blackhole doesn't emit gravitons but a pair of orbiting blackholes does. I ... 4 We can always find such a vector field when$n=2m$and$M$is orientable. Proof. Let$\gamma:[0,a]\to M$be the geodesic in question, with$\gamma(0)=\gamma(a)=p$. Let$P_t:T_{p}M\to T_{\gamma(t)}M$be the parallel transport along this geodesic. From Picard-Lindelöf, it is clear that$P_a\gamma'(0)=\gamma'(a)=\gamma'(0)$. Thus$P_a$is an isomorphism of ... 4 Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ... 2 Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor$g_{\mu\nu}$in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action$S_{\rm m}[\Phi, g]$should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ... 2 The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer. It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you ... 4 This is anwered in Gravity on the International Space Station - General Relativity perspective, where we learn that time dilation in the ISS with respect to Earth equator is 1.00000000028655. So after 17 years for us, the astronauts would come back younger by about 0.15 seconds than if they have stayed on the ground. Note that a full GR treatment is ... -1 The most Natural way where to base units on is to base them on the Planck length and Planck time wich are universal lengths. So you can set the Planck length and Planck time equal to one. They are not based on human measures or alien measures. Every scientific thinking being in the universe would agree on how big these units are. And because all other units ... 2 The acceleration should be $$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$ with$r$as the height above the center of mass and the Schwarzschildradius $$r_s = \frac{2\cdot G\cdot M}{c^2}$$ The force to hold the ball at rest is $$F=m\cdot a$$ As one can see it now takes an infinite force and energy to keep a body at a fixed height when$r=r_s$. 0 A very peculiar fact is that in a compact space THERE IS a preferred inertial system. Indeed even if locally there is no way to single out a preferred inertial system, globally you can do it. Is the topology that tells you that an observer doing a loop around a torus is topologically different from an observer moving around simply connected loops. So for ... 2 The commutator of two vector fields$n^a$and$X^b$is$[n,X]^a = n^b \nabla_b X^a - X^b \nabla_b n^a$. Since this vanishes, it follows that$n^b \nabla_b X^a = X^b \nabla_b n^a$, and the second step follows from there. I don't have my copy of Wald in front of me, but I'm 99% sure that the commutator of two vector fields is defined in terms of the ... 6 The metric reads, restoring$c$, $$\mathrm{ds}^2 = -(1+gz/c^2)^2 c^2\mathrm{dt}^2 + \mathrm{dz}^2 + \mathrm{dx}^2\:.$$ The metric can be used to determine the geodesics by means of the associated quadratic Lagrangian $$L = -(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2\tag{0}$$ where the dot denotes the derivative with respect to the affine parameter$s\$. ...

Top 50 recent answers are included


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Plot the distribution of a sample as bars and add a histogram line for visualizing the sample characteristics.

 Source code
##' Plot distribution of a sample with bars, lines and rug.##' @title ##' @param x ##' @return ##' @author Dimtiris Kavroudakis, [email protected], www.dimitrisk.grplotDistribution = function (x) {  N = length(x)  hist( x,col = "light blue",        probability = TRUE)  lines(density(x), col = "red", lwd = 3)  rug(x)} x = c(3,4,4,4,7,5,5,5,6,8,4,2)plotDistribution(x)


text:
In ΔABC, AD is median on side BC. E, G and F, H are points on AD and AC, respectively such that AE : EG : GD = 1 : 1 : 1 and EF // GH // DC. If the area of trapezium EFHG is 12√7 cm2, then what is the area of the ΔABC?

1. 72√7 sq.cm
2. 64√7 sq.cm
3. 76√7 sq.cm
4. 75√7 sq.cm

Answer (Detailed Solution Below)

Option 1 : 72√7 sq.cm

Detailed Solution

Given:

AD is the median of side BC of ΔABC

E, G and F, H are points on AD and AC, respectively

AE : EG : GD = 1 : 1 : 1 and EF // GH // DC

Area of trapezium EFHG = $$12\sqrt 7 {\rm{\;c}}{{\rm{m}}^2}$$

Formula used:

In two similar triangles ratio of sides is equal to the square of ratio of area of two triangles

Calculation:

From the given information the diagram can be drawn as

In ΔABC

AD is median of ΔABC

⇒ Area of ΔADC = Area of ΔADB = 1/2 × Area of ΔABC

AE : EG : GD = 1 : 1 : 1

⇒ AE : AG : AD = 1 : 2 : 3

ΔAEF ~ ΔAGH ~ ΔADC

⇒ Ratio of Area of ΔAEF : Area of ΔAGH : Area of ΔADC = 1 : 4 : 9

Now,

Area of trapezium EFHG = (4 – 1) units = 12√7

⇒ 3 units = 12√7

⇒ 1 unit = 4√7

⇒ 9 units = 36√7

Area of ΔABC = 2 × Area of ΔADC

⇒ Area of ΔABC = 2 × 9 units = 2 × 36√7 = 72√7

∴ Area of ΔABC = 72√7 cm2


text:
Monotonic alpha-divergence minimisation

2 S2A - Signal, Statistique et Apprentissage
LTCI - Laboratoire Traitement et Communication de l'Information
5 TIPIC-SAMOVAR - Traitement de l'Information Pour Images et Communications
SAMOVAR - Services répartis, Architectures, MOdélisation, Validation, Administration des Réseaux
Abstract : In this paper, we introduce a novel iterative algorithm which carries out $\alpha$-divergence minimisation by ensuring a systematic decrease in the $\alpha$-divergence at each step. In its most general form, our framework allows us to simultaneously optimise the weights and components parameters of a given mixture model. Notably, our approach permits to build on various methods previously proposed for $\alpha$-divergence minimisation such as gradient or power descent schemes. Furthermore, we shed a new light on an integrated Expectation Maximization algorithm. We provide empirical evidence that our methodology yields improved results, all the while illustrating the numerical benefits of having introduced some flexibility through the parameter $\alpha$ of the $\alpha$-divergence.
Keywords :
Document type :
Preprints, Working Papers, ...
Domain :

https://hal.telecom-paris.fr/hal-03164338
Contributor : Kamélia Daudel Connect in order to contact the contributor
Submitted on : Tuesday, March 9, 2021 - 8:55:47 PM
Last modification on : Tuesday, September 21, 2021 - 2:16:05 PM
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Kamélia Daudel, Randal Douc, François Roueff. Monotonic alpha-divergence minimisation. 2021. ⟨hal-03164338⟩

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text:
# Escape velocity and gravitation

1. Aug 28, 2004

### Cyrus

Hi, me again. I have another question concerning the gravitation equation. My physics book shows through a proof that the escape velocity of an earth satelite is just $$sqrt(2)$$ greated than the orbital velocity. But I am having trouble seeing how this is so. My instinct tells me that if I increase the velocity by $$sqrt(2)$$, I will start to move out farther in space, but as a consequence I will also increase my radius, until which point I have achived the right speed/distance to become a satelite again, but this time at some further distance from the earth. I dont see how increasing my speed by the square root of two will allow me to drift off into space and no longer stay in orbit.

Part 2.

Lets say I have a delta V rocket that can go the escape velocity of 25kmi/hr. Now, lets also say I have an estees model rocket that can go 50mi/hr. Clearly the delta V has the required velocity to go off into space. What happens to the model rocket at 50mi/hr. If fuel is not an issue, it will always be going at 50mi/hr. Wouldent this small model rocket go off into infinity also. Is there a conflict with the equation to find the escape veloctiy then?, since I could send a delta V at 25kmi/hr or a meager model rocket at 50mi/hr off into infinity?
Cyrus Abdollahi.

Last edited: Aug 28, 2004
2. Aug 28, 2004

### Janus

Staff Emeritus
The greater the radius of the Orbit the smaller the orbital velocity needed So, if you increase velocity you start to pull away from the Earth. The further you are form the Earth, the slower you have to be moving to maintain orbit. Now as you pull away from the Earth you will also lose some velocity, but if you are moving fast enough you never lose all of it.

For instance, you are orbiting at a velocity of V, you increase your velocity by x and start increasing your distance from the Earth. When you reach a radius of Y, you would need to have a velocity of less than V by a certain amount to maintain orbit. You have lost some velocity climbing to that new radius, but the your new velocity is still more than you need to maintain orbit, so you climb further to a higher orbit with even a lower orbital speed, but you still haven't lost enough velocity and have some left over and climb to a even higher orbit. If your value of x was large enough, the required orbital velocity will continue to drop faster than you lose speed, and you will continue to keep climbing to an "higher orbit" forever.
The 50mi/hr velocity for the Estees rocket is the velocity it will reach after burning up all its fuel. The rocket continues to accelerate until it uses up all the propellant. So its top speed is dependant on its fuel.

But the question I think that you are asking is what happens if you were to climb at a constant 50 mph, and how do you escape Earth's gravity without exceeding the escape velocity?

The further you are from the Earth, the smaller the escape velocity. For instance, at the distance of the moon it is down to 1.4 km/sec. If you keep climbing at 50 mph, you eventually reach a height where the escape velocity is less than 50 mph. At which point you could shut your engine off and continue away from the Earth forever without ever having the reach the 11 km/s escape velocity needed from the surface of the Earth. Of course, you will burn a lot more fuel reaching that distance at 50 mph than you would if you just accelerated up to 11 km/sec right at the start.

3. Aug 29, 2004

### Cyrus

WOW! that is the best and clearest response i have gotten from a person yet! thanks soooooooooooooooooo much ;-) ! :-). Yippie im happy now. hehehe. If you could, would you please explain how come it would require alot more fuel to go the steady 50mph than the 11km/sec from the start. Why is it alot easier and use less fuel if you do it all at the start than over some period of time? I think my first problem was that I was forgetting that the velocity will decerease with distance. I was assuming you would have to go faster to orbit further away, which is opposite what is true. Your example made things very clear. THANKS! So I guess things like comets which 'seem' to go off forever, but really dont since they return back to the earth, are just in very very large orbits and have very small speeds, (elliptical orbits) i take it, and so they seem to come and go, but in actuality they go off into an ellipse which is too big to see them when there gone, and then they return again. Would the thing with the escape velocity leave in the earth in a straght line, or in a parobolic arch or what? If it were launched straight up and just continued on straight from there, it would seem to go straight forever. But if it were in orbit, it would seem to me based on your analogy, it would orbit bigger and bigger forever, so it would go off into infinity in some arched space curve. Or perhaps, would it move off into space in some striaght line tangent to the orbit, at that instant when the rocket was fired? Now my brain hurts, Ill leave it to the experts like you :-).

Cheers,

Cyrus

Last edited: Aug 29, 2004
4. Aug 29, 2004

### enigma

Staff Emeritus
If you're in a bounded orbit (less than escape velocity), then you're in an elliptical orbit. e<1

If you've got escape velocity, you're in a parabolic orbit. e=1

If you're going faster than escape velocity, you're in a hyperbolic orbit. e>1

5. Aug 29, 2004

### Janus

Staff Emeritus
If you burn all your fuel early on you don't have to lift that fuel. If you burn your fuel slowly, you have to lift at the early part of the trip the fuel you are going to burn later, which means you need more fuel to lift that fuel, and even more fuel to lift that fuel, etc.
An object traveling at exactly escape velocity would follow a parabola. If it was traveling at greater than escape velocity, it would travel in a hyperbola.

6. Aug 29, 2004

### Cyrus

Another reply, sorry :-). You said at first that you will loose some velocity to go to a higher orbit. I was wondering how is this so. Since there is nothing in space to retard your motion, if you go further from the earth and increase your radius, how would your velocity around the earth decease slightly? Would you not keep that velocity you gained when you fired your engine and just keep gaining distance forever with that same constant velocity?

7. Aug 29, 2004

### Janus

Staff Emeritus
Earth's gravity slows you down, for the same reason that when you throw a ball up into the air gravity slows it to a stop and pulls it back down. Just because it is moving too fast for gravity to ever completely bring it to a stop doesn't mean thart gravity still can't slow it down as it climbs away. The object still exchanges kinetic energy for potential energy.

8. Aug 29, 2004

### Cyrus

Last edited: Aug 30, 2004
9. Aug 30, 2004

### BobG

If you're going exactly the escape velocity, the spaceship traces out a parabola. In other words, if you took the distance the object was at launch, doubled it, and drew a line perpendicular to the (doubled) launch radius, the object would always be the same distance from the line and the center of the Earth. (If the object eventually came to rest, it would be pulled back into the strongest source of gravity).

If you launched straight up relative to the ground, you would have a lateral velocity, as well, since the launch site is moving with the Earth's rotation. The only exception would be if you launched from the North or South pole. If you launched in a direction that took you directly away from the Earth, it would not be a parabolic orbit. It would be a line. It is somewhat similar to a parabolic orbit. Parabolas are planes that intersect a cone parallel to the slant height. A line would be a plane that both intersects the cone parallel to the slant height and passes through the vertex of the cone. The object would continue on a straight line forever (excepting the influence from the Moon, Sun, planets, etc.)

If your booster failed and only got you halfway to escape velocity, your rocket would crash into the Earth. Technically, it would put you into a closed orbit (elliptical or circular). But, at only half the speed, the specific energy would be -46.87. That means the semi-major axis of your orbit would only be 4252 km, which is less than the radius of the Earth.

A velocity just barely below escape velocity would get you a really large orbit, though.

10. Aug 30, 2004

### Cyrus

Yes, But what if i launched from the north pole as you said, and my engine failed. Now I would be going off in a straight line into space, but not have enough speed. Would I still return in an elliptical orbit, because would I not just fall straight back down like a stone thrown up will fall straight back down? (at the north pole there would be no rotational effects so the stone would just come right back down and not be turned sideways a small amount due to the rotation, same as a rocket launched at the north pole.) I would still like an anwser to my other question as well please. If I have the right escape velocity to leave earth, does that mean that when I reach the point infinity, will my velocity now be zero, and I can stay at infinity without being pulled back in. Likewise, If I only have half the escape velocity, and I get halfway to infinity, will I eventually get to a velocity of zero, and then start drifting back down towards earth slowly going faster and faster? Just like a stone thrown upwards.

-Cyrus Abdollahi

Last edited: Aug 30, 2004
11. Aug 30, 2004

### BobG

Yes, if your path were directly away from the Earth, and you failed reach escape velocity, the return path would be a straight line back toward the Earth. In a theoretical world, where the Earth wouldn't stop the object, the object would oscillate back and forth in a straight line reaching the same distance away from Earth on opposite sides. In the real world.....

The only problem with the idea of reaching a zero velocity at 'infinity' is that you can never reach 'infinity'. Infinity is just away of saying an infinite never ending distance - and it takes an infinite never-ending amount of time to reach it. In a parabolic orbit or hyperbolic orbit, an object would continue getting further and further away from the Earth forever (ignoring the effects of other objects inside and outside the solar system).

But, infinity is the start point for measuring the potential energy of the Earth's gravitational pull and, even though non-existant, it is the point where a zero velocity would result in an object remain stationary instead of being pulled towards the Earth.

12. Aug 30, 2004

### Cyrus

Thanks BobG and Janus I am, you dont know how greatful I am to the both of you for anwsering all my questions. These questions were troubling me for a bit of time, as I could not find a direct anwser to them, but thanks to the both of you for helping me! Hopefully, however doubtfully, I can repay the favor. ( You guys are just too smart for me :-) ).

13. Aug 30, 2004

### Janus

Staff Emeritus

If you only attain 1/2 escape velocity you would reach a maxmum altitude of 2126 km above the surface of the Earth before you start to fall back. And ignoring air resistance, when you hit you would be going at 1/2 escape velocity.

14. Aug 30, 2004

### Cyrus

thats amazing, you would think that with half of the escape velocity you would be able to go much further into space. I also have a side question. In doing my physics I did the proof for keplers law and how sector area is constant with time, which proves that the magnitue and direction of the angular momentum is conserved, and the book says this helps explain why the planets orbit along a common plane. But I somehow dont buy into this explination, becuase from the start, we assumed that the planets were traveling along a plane when we calculated sector area. Suppose the planets moved around some funny zizaged circle, then it would not be a sector area, but some line integral (or is it surface integral) since as it goes around in a circle it might also go up or down in the same processs, the area of the shape it traces would no longer be planar, which is what we assumed was true when we did keplers law.

15. Aug 31, 2004

### Tide

But the planets don't just zig zag along. Their orbits are essentially elliptical.

To see how they go coplanar think about this example. Suppose two planets orbit a star in nearly circular orbits and further suppose that these orbits are perpendicular to each other. What influence do these planets have on each other's orbits? They will, in fact, tend to pull each other toward a common plane!

16. Aug 31, 2004

### Rogerio

As you can see at thread https://www.physicsforums.com/showthread.php?t=39745 , even 99% escape velocity is not enough to reach 384000km (aboute the Earth-Moon distance) .

17. Aug 31, 2004

### BobG

Good point. The tendency for geosynchronous satellites to want to re-orient their orbital plane with the Moon/Sun is one of the major perturbations satellite operators have to compensate for.

Except, this is an external force acting to change angular momentum (at least when considering individual planets) instead of a demonstration of conservation of angular momentum.

Unless, you want to look at the whole system and how a change in angular momentum of one object causes a change to another object in order to keep the system's overall angular momentum constant. The end result is the same regardless.

As a demonstration of conservation of momentum of the individual planets, they are relying on the idea of the Sun and planets all being created from the same cloud of rotating dust. Even after coalescing into big balls of matter, the overall angular momentum of the dust particles remain constant, hence the planets all starting out in the same plane from the get-go.

Obviously not a perfect demonstration, since there are very small differences in each planet's and each moon's orbital plane. And Pluto's orbit and the orientation of Uranus's rotational axis record some very significant events (external interference) in the solar system's history.


text:
Article

# Rainbow Connections of Graphs: A Survey

(Impact Factor: 0.39). 01/2013; 29(1):1--38. DOI: 10.1007/s00373-012-1243-2
Source: arXiv

ABSTRACT The concept of rainbow connection was introduced by Chartrand et al. in 2008.
It is fairly interesting and recently quite a lot papers have been published
about it. In this survey we attempt to bring together most of the results and
papers that dealt with it. We begin with an introduction, and then try to
organize the work into five categories, including (strong) rainbow connection
number, rainbow $k$-connectivity, $k$-rainbow index, rainbow vertex-connection
number, algorithms and computational complexity. This survey also contains some
conjectures, open problems or questions.

### Full-text

Available from: Xueliang Li, Aug 21, 2014
4 Followers
·
• Source
• "The minimum number of colors required to rainbow color a graph G is called the rainbow connection number, denoted by rc(G). For more results on the rainbow connection, we refer to the survey paper [21] of Li, Shi and Sun and a new book [22] of Li and Sun. "
##### Article: Monochromatic connectivity and graph products
[Hide abstract]
ABSTRACT: The concept of monochromatic connectivity was introduced by Caro and Yuster. A path in an edge-colored graph is called a \emph{monochromatic path} if all the edges on the path are colored the same. An edge-coloring of $G$ is a \emph{monochromatic connection coloring} ($MC$-coloring, for short) if there is a monochromatic path joining any two vertices in $G$. The \emph{monochromatic connection number}, denoted by $mc(G)$, is defined to be the maximum number of colors used in an $MC$-coloring of a graph $G$. In this paper, we study the monochromatic connection number on the lexicographical, strong, Cartesian and direct product and present several upper and lower bounds for these products of graphs.
• Source
• "See for example, Caro et al. [2], Chartrand et al. [6], and Krivelevich and Yuster [12]. Recently, Li et al. [14], and Li and Sun [17], published a survey and a book on the current status of rainbow connection. "
##### Article: Rainbow k-Connection in Dense Graphs
[Hide abstract]
ABSTRACT: An edge-coloured path is rainbow if the colours of its edges are distinct. For a positive integer k, an edge-colouring of a graph G is rainbow k-connected if any two vertices of G are connected by k internally vertex-disjoint rainbow paths. The rainbow k-connection number rc k (G) is defined to be the minimum integer t such that there exists an edge-colouring of G with t colours which is rainbow k-connected. We consider rc2(G) when G has fixed vertex-connectivity. We also consider rc k (G) for large complete bipartite and multipartite graphs G with equipartitions. Finally, we determine sharp threshold functions for the properties rc k (G) = 2 and rc k (G) = 3, where G is a random graph. Related open problems are posed.
Journal of Combinatorial Mathematics and Combinatorial Computing 05/2015; 93.
• Source
• "Chartrand, Johns, McKeon and Zhang introduced in [3] the concept of rainbow connection in graphs, and since then such topic has been broadly studied, see [5] for a survey on the matter. The rainbow connection for digraphs was first presented by Dorbec, Schiermeyer, Sidorowicz and Sopena in [4]. "
##### Article: A Note on the Rainbow Connectivity of Tournaments
[Hide abstract]
ABSTRACT: An arc-coloured digraph $D$ is said to be \emph{rainbow connected} if for every two vertices $u$ and $v$ there is an $uv$-path all whose arcs have different colours. The minimun number of colours required to make the digraph rainbow connected is called the \emph{rainbow connection number} of $D$, denoted $\stackrel{\rightarrow}{rc}(D)$. In \cite{Dorbec} it was showed that if $T$ is a strong tournament with $n\geq 5$ vertices, then $2\leq \stackrel{\rightarrow}{rc}(T)\leq n-1$; and that for every $n$ and $k$ such that $3\leq k\leq n-1$, there exists a tournament $T$ on $n$ vertices such that $\stackrel{\rightarrow}{rc}(T)=k$. In this note it is showed that for any $n\ge6$, there is a tournament $T$ of $n$ vertices such that $\stackrel{\rightarrow}{rc}(T)=2$.


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# 2: The Mathematics of Chemistry

Mathematics is used widely in chemistry and are absolutely necessary to explore important concepts in chemistry. Without some basic mathematics skills, these calculations, and therefore chemistry itself, will be extremely difficult. However, with a basic knowledge of some of the mathematics that will be used in your chemistry course, you will be well prepared to deal with the concepts and theories of chemistry.

• 2.1: Measuring Matter
Scientists use all kinds of equipment to measure matter.  Balances are used to measure mass while pipettes are used to measure volume. Errors in measurements can be made if the scientist does not know how to properly use the equipment or if the equipment is damaged. It is important that you know the types of equipment a scientists uses and the basic units that are found on the equipment. In addition, memorize and be able to apply the necessary conversion factors for this course.
• 2.2: Dimensional Analysis
Units can be converted to other units using the proper conversion factors. Conversion factors are constructed from equalities that relate two different units. Conversions can be a single step or multistep. Unit conversion is a powerful mathematical technique in chemistry that must be mastered.
• 2.3: Density
Density is a physical property that is defined as a substance’s mass divided by its volume. Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity.
• 2.4: Temperature
One of the fundamental quantities in science is temperature. Temperature is a measure of the average amount of energy of motion, or kinetic energy, a system contains. Temperatures are expressed using scales that use units called degrees, and there are several temperature scales in use. In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as “degrees Fahrenheit”).
• 2.E: The Mathematics of Chemistry (Exercises)
These are homework exercises to accompany Chapter 2 of the Furman University's LibreText for CHE 101 - Chemistry and Global Awareness.

This page titled 2: The Mathematics of Chemistry is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Elizabeth Gordon.


text:
## Hello $$\LaTeX$$

A learning resource for mathematics teachers.

High school mathematics teachers use a wide variety of digital platforms and technologies to communicate with their students, colleagues, and the broader learning community.

LaTeX (pronounced lay-tech) is a long-established technology that allows you to write portable, professional-quality mathematics and share it with everyone. It can be used to integrate mathematical symbols and notation into a wide variety of documents across many digital platforms.

These pages are structured as a self-paced online workshop that introduces high school mathematics teachers to LaTeX and the many platforms that support its use.

## Module 0: What is LaTeX?

LaTeX is a digital typesetting technology used by scientists and mathematicians to create publication-quality documents and precisely formatted mathematics.

## Module 1: Getting Started

Begin writing mathematics with LaTeX using online LaTeX equation editors. Learn how to create nicely formatted mathematics that can be saved as images and inserted into documents that do not have native LaTeX support.

## Module 2: Essential LaTeX Commands

Continue to build your repertoire of LaTeX commands and understanding of LaTeX syntax.

## Module 3: LaTeX on Various Platforms

LaTeX in Google Docs? In Brighspace? On any HTML page? You bet.

## Module 4: Cloud-Based LaTeX Environments

Learn some of the essential LaTeX commands that will help you communicate effectively with anyone across many platforms.

## Module 5: LaTeX as a Digital Literacy

Who can benefit from LaTeX? Well, maybe everybody.

## LaTeX Commands Quiz

Review the commands that you have learned.

## Simple Command Tester

A simple editor for LaTeX commands.


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sage: ex=atan(sin(1/5*pi)/cos(1/5*pi, hold=True))

The problem is that cos(pi/5) gets immediately simplified to (sqrt(5)+1)/4 in Sage before Maxima can be called. This can be prevented by saying cos(pi/5,hold=True).


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Dynamical Systems Seminar

Richard RandCornell University
A remarkable singularity in a delay-differential equation
Monday, September 11, 2017 - 3:30pm
Malott 224

Based on a paper coauthored with grad students Matt Davidow and B. Shayak.

The differential equation $x'' + x + x^3 =0$ is conservative and admits no limit cycles. If the linear term $x(t)$ is replaced by a delayed term $x(t-T)$, where $T$ is the delay, the resulting delay-differential equation exhibits an infinite number of limit cycles for positive values of $T$ in the neighborhood of $T=0$, their amplitudes going to infinity in the limit as $T$ approaches zero.

This bifurcation will be illustrated through the use of a variety of tools from nonlinear dynamics: harmonic balance, Melnikov's integral and numerical integration.


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What would happen if some signal could move faster than light?

The two postulates of STR doesn't say that any signal cannot move faster than light. It also doesn't assert that any signal except light cannot have velocity equals to that of light. So at the very beginning of analysis of STR we cannot say anything about the argument of "light having the ultimate(fastest) speed". Infact STR shows that we cannot accelerate a material particle to achieve and exceed the speed of light but we cannot just deny the existence of particles which could be generated similar to photons which at their creation can have speed equals to or greater than $c$ because they have achieved this speed at their birth they are not been accelerated to achieve this speed.

So Suppose there exists a hypothetical signal named schite having velocity=infinity but the speed of light is still same for all the inertial observers "would the events A and B be dissimultaneous in train's frame to validate constancy of speed of light in the Einstein's rail thought experiment"?

Moreover could we use this schite signal to synchronize clocks?

In other words would time dilation and length contraction exist?

I have heard little bit about tachyons. It is said using tachyons we can transmit signals at higher speed than $c$ if (assuming hypothetically) such particles exists can we use them to synchronize clocks?

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Showing that trans-luminal signalling is equivalent to sending messaged backward in time (and putting causality in question) in a standard exercise. And sure you could "synchronize" clocks but it wouldn't mean anything reliable because time dilation still means that the proper time of various paths between two time-like separated events still varies. –  dmckee Feb 23 '14 at 15:11
@dmckee If we synchronize two clocks seperated in space using schite(infinite speed signal) signal would they remain synchronized in all inertial frames moving at a certain speed? –  user31782 Feb 23 '14 at 15:16
–  John Rennie Feb 23 '14 at 16:09
I have learnt somewhere from the Lorentz transformation of the tachyons, that when tachyon interacts with a normal particle and it creates another tachyon and a particle without disturbing the momentum of the particle (by virtue of itself, tachyon can never be detected). If we look at this way, I think tachyons can't transfer information to normal particles. This is just my idea, could be wrong. I suggest you look for how tachyons (as particles not field) interact with normal particles. –  user35952 Feb 26 '14 at 5:37
@user35952 techyons are explain at the last of my book but i am struck at the beginning. I haven't reached even at Lorentz transformation. What to say there are enumerable questions in my head. –  user31782 Feb 26 '14 at 11:57

The two postulates of STR doesn't say that any signal cannot move faster than light.

This superficial appearance is deceiving. After all, the postulates involve notions such as inertial frame and speed; therefore they require and presume definitions of

Correspondingly, here c is of course understood as (magnitude of) signal front velocity (instead of "phase velocity", or "group velocity").

In summary: the detailed definitions of the notions and quantities appearing in the famous formulation of the postulates of STR imply that any "speed" is determined in reference to unambiguous "signal speed".

And applied to the question title asked above:
it is of course an absurd proposition that some signal should "move faster" (from some particular sender, to some particular receiver) than any signal exchanged between this particular pair of participants.

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we cant track a signal moving with a velocity more than C.we can say that it will not interfere with the material matter.it can pass through any matter with out making a interaction with it as it will pass through the matter as of its momentum hf/c2*infinity in this instance which is very high. even though it passes through it the body will be perfectly elastic and will retain its orginal shape with no deformation.

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Frankly you are not answering my question. It appears you are new here. Please use latex and be to the point to the question. You are going to have downvotes so this is my advice. –  user31782 Feb 23 '14 at 16:03


text: 한국농림기상학회지, 제 3권 제1호(2001) (pISSN 1229-5671, eISSN 2288-1859) Korean Journal of Agricultural and Forest Meteorology, Vol. 3, No. 1, (2001), pp. 37~43 ⓒ Author(s) 2014. CC Attribution 3.0 License. 기상자료 공간내삽과 작물 생육모의기법에 의한전국의 읍면 단위 쌀 생산량 예측 윤진일(1), 조경숙(2) (1)경희대학교 생명자원과학연구원/농학과 449-701 경기도 용인시 기흥읍 서천리 (2)기상연구소 응용기상연구실 156-720 서울시 동작구 신대방동 460-18 (2001년 02월 16일 접수) Yield and Production Forecasting of Paddy Rice at a Sub-countyScale Resolution by Using Crop Simulation andWeather Interpolation Techniques JinI Yun(1), KyungSook Cho(2) (1)Institute of Life Science and Natural Resources/Department of Agronomy Kyung Hee University, Suwon 449-701, Korea (2)Applied Meteorology Lab., Meteorological Research Institute 460-18 Sindaebang-dong, Dongjak-gu, Seoul 156-720, Korea (Manuscript received 16 February 2001) ABSTRACT Crop status monitoring and yield prediction at higher spatial resolution is a valuable tool in various decision making processes including agricultural policy making by the national and local governments. A prototype crop forecasting system was developed to project the size of rice crop across geographic areas nationwide, based on daily weather pattern. The system consists of crop models and the input data for 1,455 cultivation zone units (the smallest administrative unit of local government in South Korea called “Myun”) making up the coterminous South Korea. CERES-rice, a rice crop growth simulation model, was tuned to have genetic characteristics pertinent to domestic cultivars. Daily maximum/minimum temperature, solar radiation, and precipitation surface on 1km by 1km grid spacing were prepared by a spatial interpolation of 63 point observations from the Korea Meteorological Administration network. Spatial mean weather data were derived for each Myun and transformed to the model input format. Soil characteristics and management information at each Myun were available from the Rural Development Administration. The system was applied to the forecasting of national rice production for the recent 3 years (1997 to 1999). The model was run with the past weather data as of September 15 each year, which is about a month earlier than the actual harvest date. Simulated yields of 1,455 Myuns were grouped into 162 counties by acreage-weighted summation to enable the validation, since the official production statistics from the Ministry of Agriculture and Forestry is on the county basis. Forecast yields were less sensitive to the changes in annual climate than the reported yields and there was a relatively weak correlation between the forecast and the reported yields. However, the projected size of rice crop at each county, which was obtained by multiplication of the mean yield with the acreage, was close to the reported production with the $\small{r^2}$ values higher than 0.97 in all three years. Keyword: CERES-rice, GIS, spatial interpolation, crop models, yield forecasting, rice crop MAIN REFERENCES


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# Gas station without pumps

## 2014 June 17

### Segmenting noisy signals revisited

This is the fourth in a series—the three previous are

A student (well, junior specialist this year, but student again in the fall) and I were trying to write up a journal paper describing the method and implementation, and I realized that the presentation in this blog was not clear enough for him to rewrite the math, and that I had left in some bits of fluff that were confusing. In this blog post I’ll try to merge the three previous posts into a single, coherent presentation using a Bayesian approach.

The segmenter uses a divide-and-conquer approach:

segment( time interval=[s,t) )
if the time t-s is too short, return no breakpoints.
Select the best breakpoint, i, to divide into two intervals [s,i) [i,t)
If the breakpoint is poor, reject it and return no breakpoints
otherwise return segment(s,i) + [i] + segment(i,t)

There are two essential parts to this algorithm: selecting the breakpoint and deciding whether to accept it.  Both rely on a Bayesian scoring scheme.  For the score, we’ll need two likelihood models, that is computable functions that compute the probability of seeing the observed data if it is generated by the corresponding model.  To score a potential breakpoint, i, we compare a model with the breakpoint (so that [s,i) and [i,t) are modeled independently) to one where the entire interval is modeled as a unit.

The simplest model, and one that has worked well for DNA data from the nanopore, is that each segment consists of a constant current with independent Gaussian noise added to each sample.  The likelihood for the data $x_{s}, \ldots , x_{t-1}$ is just $\prod_{s\leq j < t} P_{\mu,\sigma}(x_{j})$, where µ and σ are the mean and standard deviation of the Gaussian generating the samples in [s,t). I’ll abbreviate this product as $P_{\mu,\sigma}(s:t)$.

The log of the likelihood ratio for a breakpoint at position is $L_{i}=\ln \frac{P_{\mu_{1},\sigma_{1}}(s:i) P_{\mu_{2},\sigma_{2}}(i:t)}{P_{\mu_{0},\sigma_{0}}(s:t)} = \ln P_{\mu_{1},\sigma_{1}}(s:i) +\ln P_{\mu_{2},\sigma_{2}}(i:t)- \ln P_{\mu_{0},\sigma_{0}}(s:t)$,

where µ0 and σ0 are the mean and standard deviation for the whole window from s to t, µ1 and σ1 are the mean and standard deviation for s to i, and µ2 and σ3 are the mean and standard deviation for i to t.  Because we are using maximum likelihood estimates for all the µ and σ values, we are guaranteed that the log-likelihood Li is non-negative.

The log-probability for a single sample with a Gaussian distribution is
$\ln P_{\mu,\sigma}(x) = \ln \left( \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \right) = -0.5 \ln(2\pi) -\ln(\sigma) - \frac{(x-\mu)^2}{2\sigma^2}$
When we chose the optimal μ and σ for a segment, we did it so that σ2 is the variance, that is the average value of $(x-\mu)^2$. So when we sum the log probabilities over a segment of length n, using μ and σ optimized for that segment, we get $n( -0.5 \ln(2\pi) -\ln(\sigma) - 0.5)$.

That means that the log likelihood ratio we want simplifies to $L_{i} = (s-i)\ln \sigma_{1} + (i-t)\ln \sigma_{2} - (s-t)\ln \sigma_{0}$. This simplification works for any modeling that results in independent Gaussian residuals, so we can do stepwise sloping segments by doing linear regression almost as easily as doing stepwise constant segments.

Choosing the best breakpoint is now easy—we just pick the value for i that maximizes Li.

Deciding whether to keep or reject the breakpoint requires looking not just at the likelihood ratio, but at the posterior-odds ratio, which is the likelihood ratio times the prior odds ratio. The prior odds ratio is a parameter we can set independent of the data. For example, if we have an estimate of the expected number of segments per second (sps) for a particular experimental setup and our sampling frequency is fs, then the prior probability of a step at each sample is $\mbox{\it sps}/f_{s}$, the prior odds ratio is $\frac{ \mbox{\it sps} }{f_{s}-\mbox{\it sps}}$, and the score we want to use (log posterior odds ratio) is $L_{i} + \ln \mbox{\it sps} - \ln (f_{s}-\mbox{\it sps})$.

If we are willing to accept no more than FP false positives per second, then we want the probability of a score arising by chance at a sample to be less than $\mbox{FP}/ f_{s}$

Although I have no proof for it, if the samples are all drawn independently from a Gaussian distribution, empirically the log-likelihood scores seem to be perfectly distributed on an exponential distribution $P(L_{i} > x) = e^{-x}$. Note that this distribution does not depend on the length of the window, μ, or σ, though I did avoid looking at the first and last few positions in the window, in case there are edge effects.

However, if we low-pass filter the Gaussian noise, then the samples are no longer independent, and the log-likelihood scores go up. The distribution with a cutoff frequency of k times the Nyquist frequency is empirically $P(L_{i} > x) = e^{-k x}$. (The Nyquist frequency is half the sampling frequency and is the highest frequency that can be represented by the samples.)

If these empirically observed distributions are correct, then we have
$\frac{\mbox{FP}}{f_{s}} > P(\mbox{score} > x) = P\left(L_{i} > x \right) -\ln \mbox{\it sps} + \ln (f_{s}-\mbox{\it sps})= e^{-k x} \left(\frac{\mbox{\it sps} }{f_{s}-\mbox{\it sps}}\right)^{k}$
that is
$x > \ln \frac{\mbox{\it sps}}{f_{s}-\mbox{\it sps}} - \frac{1}{k}\ln\left(\frac{\mbox{FP}}{f_{s}}\right)$.

This is the same result as in More on segmenting noisy signals and Segmenting filtered signals, but there I made the mistake of changing parameterization from FP (false positives per second) to $\mbox{\it over} = \frac{\mbox{FP}}{\mbox{\it sps}}$ (false positives per expected segment), which makes the parameterization more confusing, and I was not clear about where the scaling went for the filtering.

On the example I’ve been checking, I get good results with sps=10 segments/sec and FP=1e-4 false positives/sec, but neither parameter is very critical—changing them by a factor of 1000 does not change the results much.  Dropping FP to 1e-200 does cause segments to be obviously missed.