Energy Dependence of Particle Production in nucleus-nucleus collisions at the CERN SPS
夸克的英语作文300字
夸克的英语作文300字Title: Exploring the Enigmatic World of Quarks。
Quarks are the fundamental building blocks of matter, combining to form protons and neutrons, which in turn constitute the nucleus of atoms. Despite their significance, quarks remain mysterious entities, fascinating physicists and enthusiasts alike with their elusive properties and behavior.Firstly, let's delve into the peculiar nature of quarks. These elementary particles possess fractional electric charges, either -1/3 or +2/3 times the elementary charge. This fractional charge makes quarks distinct from familiar particles like electrons and protons. Furthermore, quarks exhibit a property known as color charge, which is analogous to electric charge but involves three types: red, green, and blue. Quarks must combine in such a way that the resulting particle has a neutral color charge, leading to the formation of composite particles called hadrons.Quarks come in six flavors: up, down, charm, strange, top, and bottom. Each flavor corresponds to different mass and properties, contributing to the diversity of matter in the universe. For instance, up and down quarks are the most common, found in everyday matter, while the heavier flavors are produced in high-energy collisions, such as those occurring in particle accelerators.The study of quarks falls under the realm of quantum chromodynamics (QCD), a branch of theoretical physics that describes the strong interaction between quarks and gluons, the carriers of the color force. QCD has provided invaluable insights into the behavior of quarks within the confines of atomic nuclei, shedding light on phenomena such as confinement and asymptotic freedom.One of the most remarkable aspects of quarks is their confinement within hadrons. Despite their fractional charges, quarks have never been observed in isolation due to the strong force binding them together. This phenomenon, known as quark confinement, remains a central puzzle inparticle physics. The confinement of quarks within hadrons has significant implications for our understanding of the structure of matter and the behavior of the universe at the smallest scales.Moreover, quarks play a crucial role in the grand scheme of cosmic evolution. In the early universe, during the epoch of quark-gluon plasma, quarks and gluons existed freely before cooling and condensing into hadrons. Studying the properties of quark-gluon plasma provides insights into the conditions prevailing microseconds after the Big Bang, enriching our understanding of the universe's origins and evolution.In conclusion, quarks represent a fascinating frontier in particle physics, offering profound insights into the nature of matter and the fundamental forces governing the universe. Through experimental investigations and theoretical frameworks like quantum chromodynamics, scientists continue to unravel the mysteries surrounding these enigmatic particles, furthering our comprehension of the cosmos at its most fundamental level.。
AP物理1和AP物理2考试必备知识点
2014年5月，AP物理B考试完成了它的使命，就此退出历史舞台。
2015年，这门考试将被两门全新的考试所取代，分别为AP物理1和AP物理2。
虽然作为最后一次AP物理B考试，2014年5月的这次考题仍能给我们带来很多启发，并对之后的新考试、新题型给出非常重要的参考。
首先，本次考试的解答题部分题目数量较之前有所改变，在之前历年的考试中，AP物理B的考试一般包括6道大题，需要学生在90分钟的时间内完成，而在2014年的考试中，较少见的出了7道大题，时间仍是90分钟，这意味着两件事：1.每道题目内的小问数减少，同时题目的难度降低了；2.在解答题中考察的知识点范围更宽泛了。
在2015年即将到来的两门新考试中，在解答题的考察上会有比较明显的改变，我们来比较一下三门考试的解答题部分的异同：AP物理B：解答题共6道大题，时间90分钟，分值占50%。
对各道解答题的考试形式没有明确说明。
AP物理1：解答题共5道大题，时间90分钟，分值占50%。
明确将考察一道实验设计题，一道计算题，三道短问答题（其中一道需要学生进行辩证分析和叙述）。
AP物理2：解答题共4道大题，时间90分钟，分值占50%。
明确将考察一道实验设计题，一道计算题，两道短问答题（其中一道需要学生进行辩证分析和叙述）。
通过比较，我们可以发现，在2015年的两门新考试中，解答题部分的题目数减少了，对学生的辩证分析能力和叙述能力的要求进一步提高，解答题部分也将不再局限于对学生计算能力的考察，更将考察学生的文字叙述和书面表达能力。
其次，在考察内容上，本次考试7道大题分别考察了：单摆，流体力学，热学，静电力学，电磁感应，光电效应，光的折射这几大知识点，同时这些内容也是历年AP物理B考察的核心知识点，在每次考试中都是重点考察的对象，那么在2015年的新考试中，重点考察的内容会有什么变化呢？我们来比较一下三门考试的重点考察知识点：AP物理B：牛顿力学，流体力学，热学，电磁学，波动学和光学，现代物理。
PRL-2008-100-063002
Initial State Dependence in Multielectron Threshold Ionization of AtomsAgapi Emmanouilidou,1Peijie Wang,2,*and Jan M.Rost21ITS,University of Oregon,Eugene,Oregon97403-5203,USA2Max Planck Institute for the Physics of Complex Systems,No¨thnitzer Strasse38,D-01187Dresden,Germany (Received9October2007;published13February2008)It is shown that the geometry of multielectron threshold ionization in atoms depends on the initial configuration of bound electrons.The reason for this behavior is found in the stability properties of the classicalfixed point of the equations of motion for multiple threshold fragmentation.Specifically for three-electron breakup,apart from the symmetric triangular configuration also a breakup of lower symmetry in the form of a T shape can occur,as we demonstrate by calculating triple photoionization for the lithium ground andfirst excited states.We predict the electron breakup geometry for threshold fragmentation experiments.DOI:10.1103/PhysRevLett.100.063002PACS numbers:32.80.Fb,05.45.ÿa,34.80.DpThree-body Coulomb dynamics,in particular,two-electron atoms,are very well studied in the energy regime of single as well as double ionization[1–4].Much less is known about correlated dynamics in four-body Coulomb systems,more precisely on differential observables for fragmentation of a three-electron atom in its nucleus and all electrons[5–8].A recent experiment provides for the first time detailed information in terms of differential cross sections on the angular and energetic breakup parameters of three electrons following impact double ionization of Helium[9].For small excess energies E(each continuum electron carries away about9eV energy),it was found that the electrons form an equilateral triangle upon breaking away from the nucleus.This is expected in accordance with Wannier’s theory[10],quantified for three electrons in [11].There,it is shown that thefixed point of classical dynamics,through which full fragmentation near threshold E 0should proceed,is given for a three-electron atom by an equilateral triangle with the nucleus in the center and the electrons at the corners.In two-electron atoms the correspondingfixed point implies a collinear escape of the electrons in opposite directions[10,12].The normal mode vibration about this collinear configuration is stable.This is in marked contrast to the three-electron case,where the triangular configura-tion is linked to two unstable,degenerate normal modes [11,13].We will show that the latter property has the conse-quence that the preferredfinal geometry of the three escap-ing electrons becomes initial state dependent and can change between an equilateral triangle and a less symmet-ric T-shaped escape.While the former is realized,e.g.,in electron impact double ionization of helium,the latter should be seen in triple photoionization of lithium.These are only two prominent examples.The general pattern and the reason for it will be detailed below.Because of the scaling of the Coulomb potential,states offinite total angular momentum L will all behave like the L 0state close to threshold which is therefore sufficient to consider[14].In hyperspherical coordinates with the radial variable w instead of the hyperradius r w2,the Hamiltonian for a three-electron atom with total angular momentum L 0readshp2w8w222w4Cw2;(1) where 1; 2; 1; 2 y contains all angular variables describing the positions of the electrons on the hypersphere of radius r and ,the so-called grand angular momentum operator[15],is a function of and all conjugate mo-menta.The total Coulomb interaction V C=r acquires in this form simply an angular dependent charge C .In terms of the familiar vectors r1,r2,r3pointing from the nucleus to each electron,the hyperspherical coordinates are given byr r21r22r231=2;(2a)1 arctan r1=r2 ;(2b)2 arctan r2e=r3 ;(2c)1 arctan r1 r2= r1r2 ;(2d)2 arctan r1 r3= r1r3 ;(2e) where r2e r21 r22 1=2.Threshold dynamics is governed by motion along the normal modes about thefixed point of the Hamiltonian H hÿE w2[16].The special form of H ensures that the dynamics remains regular approaching thefixed point radially,i.e.,w!w 0,while at thefixed point is defined through r C j 0.The equations of motion can be expressed as a system offirst order differ-ential equations(ODE)_ÿ G rÿH for the phase space vectorÿ p w;P ;w; y,withG0ÿ1f1f0(3)a block matrix composed from0and unity matrices of dimension f f[17],where f is the number of degrees of freedom,here f 5.Since the differential equations are still singular at thefixed point(w , )a change of themomentum variables P conjugate to is neededp!j P!j=w(4)as well as a new time variable related to the original time t conjugate to the Hamiltonian Eq.(1)through dt w3d .Finally,the normal modes can be obtained from the modified ODE,d =d G r ~H by diagonalizing the matrix@2G~H= @ @ j ,where refers to the new phase space variables with the(noncanonical)momenta from Eq.(4).The eigenvalues are the Liapunov expo-nents j and in the normal mode basis f^u j g threshold dynamics assumes an oscillatorlike form of u j exp j u j 0 with the unit vectors^u j^u j u j 0 =j u j 0 j(5) defining the normal mode basis.The ÿ are excursions of from theirfixed point values and are expressed as a linear combination of the u j .We recall briefly the familiar three-body breakup in a two-electron atom with hyperradius r2e and the angles 1, 1defined as in Eq.(2).The charge corresponding to C in Eq.(2)is for the two-electron problemC2e 1; 1 ÿZsin 1ÿZcos 111ÿsin 2 1 cos 1 1=2:(6)Thefixed point analysis reveals a pair of unstable 1=2 ÿ 0 and stable 3=4 ÿ!0 i!Liapunov expo-nents( 0,!0, ,!>0)with a shift(ÿ 0andÿ!0, respectively),compared to standard symplectic dynamics. The shift formally arises through the noncanonical trans-formation of the momentum variables Eq.(4)necessary to obtain normal mode motion about the singularfixed point w 0.The resulting eigenvectors^u i reveal orthogonal motionalong 1and 1,i.e.,any phase space vector1describing linearized motion in the subspace spanned byp1, 1,can be expressed as a linear combination of twoeigenvectors1 aexp 1 ^u1 bexp 2 ^u2.An analogous relation holds for linearized motion in thesubspace p1, 1,realized through two different eigenvec-tors,1 aexp 3 ^u3 bexp 4 ^u4.Hence,110at all times .The coincidence ofthe eigenspaces with the respective dynamics of 1and 1 has the important consequence that thefixed point value 1 is preserved through its relation to the stable eigenmode while all energy sharings occur through their relation to the unstable eigenmode along 1.This coincidence of normal modes with subspaces of observables is special to two-electron atoms and does not hold for more electrons.Moreover,for three-electron atoms,a new feature emerges in threshold dynamics, namely,the existence of two degenerate pairs of unstable normal modes with Liapunov exponents 1=2 3=4 ÿ 0 .Note that the fragmentation dynamics close to thefixed point will take place in the phase space of the two unstable normal modes with an arbitrary linear combina-tion of the two eigenvectors^u1and^u3belonging to the two (equal)positive Liapunov exponents 1 3 ,exp c1^u1 c3^u3 ;(7) which allows forflexibility in the four-body breakup,as we will see.One sees directly from Table I that unlike the two-electron case the normal mode dynamics about thefixed point for geometrical angles i and hyperangles i is not separated,i.e.,iiÞ0.Even more impor-tantly,all phase space variables are linked to the unstable normal modes.Hence,thefixed point geometry does not provide necessarily a preference for thefinal angles of the electrons.On the other hand,this opens the way for the initial state to have an influence on thefinal observables, even close to threshold.Relevant for threshold ionization is the spatial electron distribution at the time(we label it 0)when all elec-trons to be ionized have received enough energy(through collisions)to leave the atom.We call this distribution the transient threshold configuration(TTC).In a two-electron atom,the necessary energy transfer between the two electrons leading to three-body fragmen-tation happens through a single collision.At this time 0both electrons are naturally close together,so that 1 0 0holds and defines the TTC,independent of the initial bound electron configuration.In a three-electron atom,the situation is more compli-cated since at least two collisions are necessary to distrib-ute the energy among the electrons so that all of them can escape.For triple photoionization of lithium we know from classical calculations that the1s photoelectron(3),which has absorbed the photon initially,collides immediately with the other1s electron(2)and subsequently(about60 attoseconds later)either electron2or the photoelectron itself collides with the2s electron(1).This can be ex-pressed with the two collision sequences s1 32;21 and s2 32;31 [18].The time delay of the second collision,TABLE I.The two eigenvectors^u i belonging to the positive Liapunov exponent >0in the basis of the phase space variables from Eq.(2).Basis^u1^u31ÿ8:22 10ÿ32:67 10ÿ1 p1ÿ2:97 10ÿ29:64 10ÿ1 2ÿ1:81 10ÿ10p2ÿ9:83 10ÿ1017:12 10ÿ30p16:57 10ÿ3ÿ4:46 10ÿ3 23:33 10ÿ37:40 10ÿ3 p2ÿ2:75 10ÿ48:92 10ÿ3respectively,is due to the ‘‘distance’’of the 2s shell from the 1s shell.It leaves an asymmetric situation after the second collision when the transient threshold configuration is reached at 0.While the two electrons participating in the last collision are close to each other,the third one is further away.Concentrating on the collision sequence (32,21),this implies in terms of distances to the nucleusr 1 r 2Þr 3:(8)Hence, 1 0 0while 2 0 Þ0.This can be easily accommodated with a suitable linear combination choos-ing,e.g.,c 1 1=a and c 2 ÿ1=b in Eq.(7),where a andb are the coefficients of 1with respect to ^u1and ^u 3(see first row of Table I ).The scenario described is still within the overall picture of threshold breakup in the spirit of Wannier and therefore in accordance with the power law / E=E 0 for the dependence of the total breakup cross section on the excess energy E [20–22].The exponent 1 3 = 2 r 2:162,where r 2:5088is the radial Liapunov exponent and 1 3 5:4240.Yet,the TTC with unequal dis-tances of the electrons to the nucleus breaks the complete symmetry among the electrons.This,in turn,leads to a preferred final geometry of the three electrons since essen-tially with the third electron being far away,the geometry determination reduces to that of two electrons escaping in an equivalent way as in the two-electron breakup,i.e.,back-to-back with an angle of 1 .The question re-mains if there is a preferred angle between this electron pair escaping along a line and the third electron.Assuming for simplicity that r 3 r 1 r 2and therefore sin 2 0holds (the opposite case would lead to the same result),we can expand C in Eq.(1)in powers of 2Cÿ1X 3n 0c n n :(9)One can show that to lowest order in ,the problem to find a stable configuration is that of the two-electron system (here with nuclear charge Z 3)with the well-knownsolution 1 and 1 =4[10].These values mini-mize c 2for any value 2.Its value 2 =2is deter-mined from @c 3=@ 2 0which is a stable solution.[Theis used for the fixed point solution of C when r 3 r 1 r 2to distinguish it from the global fixed point solution of C mentioned previously in the Letter.]This constitutes the T shape of the three electrons as a preferred asymptotically ( !0)stable geometrical configuration within the globally unstable two-dimensional subspacespanned by ^u1,^u 3.The fragmentation dynamics in the subspace is completely degenerate—hence,we have looked on a ‘‘higher order’’correction,which would give a preference within the degenerate subspace:that is the asymptotically stable T shape.Indeed,this configuration was found numerically close to threshold in triple photo-ionization [19].One can double check this new insight into the role of the TTC of the electrons in the presence of degenerate Liapunov exponents by considering initial configurations which lead to different transient configurations.For an initial excited state Li 1s 2s 2 the two collisions in which energy is transferred from the (1s )photoelectron to the two 2s electrons happen close in time (and therefore in space).Hence,the transient threshold configuration after the second collision at 0is r 1 r 2 r 3close to the fixed point and therefore giving preference to the symmet-ric breakup with a dominant angle 120 between the electrons,similar to electron impact double ionization of helium.Numerical results confirm this prediction as seen from Fig.1where the probability to find an angle between two electrons in triple photoionization of the initial Li 1s 2s 2 is shown for different excess energies.Clearly,the most likely angle is cos ÿ1=2corresponding to 120 ,in-−11−11−101cos θC (c o s θ)/ P3+FIG.1.Probability to find the angle between two electrons for triple ioniza-tion from the initial Li1s 2s 2 state for excess energies above the triple ioniza-tion threshold as indicated.The preferred breakup geometry is the equilateral tri-angle.0246cos θC (c o s θ)/ P3+FIG.2.Same as Fig.1,but for triple ionization from the initial Li 1s 22s ground state for excess energies above the triple ionization threshold as indi-cated.The preferred breakup geometry is the T shape.dicated by the thin vertical line.The method used to ob-tain the results in Figs.1–3is the classical trajectory Monte Carlo method detailed in [19].On the other hand,for the ground state as initial state the TTC is asymmetric as previously described and we expect a final T shape geometry with peaks at 90 and 180 .This is indeed the case,as can be seen in Fig.2.As a final test we may use again the excited initial state Li 1s 2s 2 but take the 2s electron as the photoelectron.This process is due to the smaller dipole coupling by more than an order of magnitude suppressed compared to the ionization with the 1s electron absorbing the photon.However,here we use this only as an illustration for the initial state depen-dence of threshold ionization.According to our reasoning we have in this case (although it is the same initial state as before)a different TTC since the first collision of the photoelectron 1happens with the 1s electron 2while later on the collision of the 1s electron 2with the 2s electron 3will take place.Consequently,for the TTC r 3 r 2Þr 1holds which is structurally identical to Eq.(8),the situation when a final T shape geometry appears.As can be seen in Fig.3the T shape indeed also emerges if the photoelectron comes from the 2s shell.To summarize,we predict that for three-electron breakup near threshold two preferred geometrical patterns for the electrons exist,namely,an equilateral triangle and a T shape.Which of them is realized depends on the transient threshold configuration,that is the spatial distribution of the electrons at the time when the energy among them is distributed such that all of them can escape.The transientthreshold configuration is strongly influenced by the initial state of the electrons,as has been discussed in detail for the case of lithium.A complete overview of the three-electron breakup pattern for three-electron systems is provided in Table II .Whenever the two electrons to which energy is transferred in the course of the fragmentation of the atom are in the same shell,a symmetric triangular geometry is expected.If the two electrons are from different shells,we expect a T shape.Recent experimental results on electron impact double ionization of helium [9]are consistent with our predictions but do not provide sufficient information to consider our prediction as experimentally already con-firmed.Based on the present consideration it is also pos-sible to predict the breakup geometries and their realization dependent on the initial state for more than three electrons.*Present address:The Beijing Key Laboratory for Nano-Photonics and Nano-Structure,Capital Normal University,Beijing 100037People’s Republic of China.[1]G.Tanner,K.Richter,and J.M.Rost,Rev.Mod.Phys.72,497(2000).[2]J.S.Briggs and V .Schmidt,J.Phys.B 33,R1(2000).[3]T.N.Rescigno,M.Baertschy,W.A.Isaacs,and C.W.McCurdy,Science 286,2474(1999).[4] C.W.Byun,N.N.Choi,M.-H.Lee,and G.Tanner,Phys.Rev.Lett.98,113001(2007).[5]J.Colgan and M.Pindzola,J.Phys.B 39,1879(2006).[6]J.R.Go ¨tz,M.Walter,and J.S.Briggs,J.Phys.B 39,4365(2006).[7] A.Emmanouilidou,Phys.Rev.A 75,042702(2007).[8] A.W.Malcherek and J.S.Briggs,J.Phys.B 30,4419(1997).[9]M.Du¨rr,A.Dorn,J.Ullrich,S.P.Cao,A.Czasch,A.S.Kheifets,J.R.Go¨tz,and J.S.Briggs,Phys.Rev.Lett.98,193201(2007).[10]G.H.Wannier,Phys.Rev.90,817(1953).[11]H.Klar and W.Schlecht,J.Phys.B 9,1699(1976).[12]J.M.Rost,Phys.Rev.Lett.72,1998(1994).[13]M.Y .Kuchiev and V .N.Ostrovsky,Phys.Rev.A 58,321(1998).[14]J.M.Rost,Phys.Rep.297,271(1998).[15] F.T.Smith,Phys.Rev.120,1058(1960).[16]see Sec.3.4in [14].[17] A.M.Ozorio de Almeida,Hamiltonian Systems:Chaosand Quantization (Cambridge University Press,Cambridge,U.K.,1988),p.11.[18]Note that for the present purpose we have relabeled theelectrons compared to [19].[19] A.Emmanouilidou and J.M.Rost,J.Phys.B 39,4037(2006).[20]J.A.R.Samson and G.C.Angel,Phys.Rev.Lett.61,1584(1988).[21]R.Wehlitz,T.Pattard,M.-T.Huang,I.A.Sellin,J.Burgdo¨rfer,and Y .Azuma,Phys.Rev.A 61,030704(R)(2000).[22]J.Bluett,D.Lukic,and R.Wehlitz,Phys.Rev.A 69,042717(2004).TABLE II.The relation of excitation process,initial configu-ration,and geometry of the electrons for various three-electron breakup processes close to threshold.Impact by Initial configurationBreakup geometryElectron 1s 2(He) Electron 1s 2s (He)?Photon 1s 22s (Li)?Photon 1s 2s 2(Li) Photon1s 2s 3s (Li)?C (c o s θ)/ P 3+cos θFIG.3.Same as Fig.1,but with a 2s electron as photoelectron.。
waiting times在物理中的意思
waiting times在物理中的意思英文版Waiting times in physics refer to the amount of time a particle or system of particles spends in a particular state or position before transitioning to another state or position. This concept is crucial in understanding the behavior of various physical systems, such as radioactive decay, particle collisions, and chemical reactions.In the context of radioactive decay, waiting times refer to the time it takes for a radioactive nucleus to decay and emit radiation. This waiting time is determined by the half-life of the nucleus, which is the time it takes for half of the radioactive nuclei in a sample to decay. By studying waiting times in radioactive decay, scientists can predict the rate at which a radioactive substance will decay and the amount of radiation it will emit.In particle collisions, waiting times play a key role in determining the likelihood of certain outcomes. For example, in high-energy particle collisions, the time it takes for particles to interact and produce new particles can vary depending on the energy of the collision and the types of particles involved. By analyzing waiting times in particle collisions, physicists can gain insights into the fundamental forces and interactions that govern the behavior of particles at the subatomic level.Similarly, in chemical reactions, waiting times refer to the time it takes for reactant molecules to come into contact and form products. This waiting time is influenced by factors such as the concentration of reactants, the temperature of the reaction, and the presence of catalysts. By studying waiting times in chemical reactions, chemists can optimize reaction conditions to maximize the yield of products and improve the efficiency of chemical processes.In conclusion, waiting times in physics are a fundamental concept that underpins our understanding of the behavior of particles and systems at the atomic and subatomic levels. By analyzing waiting times in various physical processes, scientists can uncover theunderlying mechanisms that govern the interactions and transformations of matter in the universe.中文版在物理学中，等待时间指的是粒子或粒子系统在转换到另一个状态或位置之前在特定状态或位置停留的时间。
Chapter5-Radiation+and+Materials
5.4 GAMMA RAY INTERACTIONS WITH MATTER (γ 射线与物质的相互作用 γ 射线与物质的相互作用) We now turn to a group of three related processes involving (使…陷于) gamma ray photons produced by nuclear reactions. These have energies as high as a few MeV. The interactions include simple scattering of the photon, ionization by it, and a special nuclear reaction known as pair production.
When we put radiation and materials together, a great variety of possible situations must be considered. Bombarding particles may have low or high energy, they may be charged, uncharged, or photons; they may be heavy or light in the scale of masses. The targets may be similarly (同样地) distinguished, but they may also exhibit (展示) degrees of binding that range from none (“free” particles), to weak (atoms in molecules and electrons in atoms), to strong (nucleons in nuclei).
第2章 夸克与轻子 (2)
第二章夸克与轻子Quarks and leptons2.1 粒子园The particle zoo学习目标Learning objectives：我们怎样发现新粒子？能否预言新粒子？什么是奇异粒子？大纲参考：3.1.1￣太空入侵者宇宙射线是由包括太阳在内的恒星发射而在宇宙空间传播的高能粒子。
如果宇宙射线粒子进入地球大气层，就会产生寿命短暂的新粒子和反粒子以及光子。
所以，就有“太空入侵者”这种戏称。
发现宇宙射线之初，大多数物理学家都认为这种射线不是来自太空，而是来自地球本身的放射性物质。
当时物理学家兼业余气球旅行者维克托·赫斯（Victor Hess）就发现，在5000m高空处宇宙射线的离子效应要比地面显著得多，从而证明这种理论无法成立。
经过进一步研究，表明大多数宇宙射线都是高速运动的质子或较小原子核。
这类粒子与大气中气体原子发生碰撞，产生粒子和反粒子簇射，数量之大在地面都能探测到。
通过云室和其他探测仪，人类发现了寿命短暂的新粒子与其反粒子。
μ介子（muon）或“重电子”（符号μ）。
这是一种带负电的粒子，静止质量是电子的200多倍。
π介子（pion）。
这可以是一种带正电的粒子（π＋）、带负电的粒子（π－）或中性不带电粒子（π0），静止质量大于μ介子但小于质子。
K介子（kaon）。
这可以是一种带正电的粒子（K＋）、带负电的粒子（K－）或中性不带电粒子（K0），静止质量大于π介子但小于质子。
科学探索How Science Works不同寻常的预言An unusual prediction在发现上述三种粒子之前，日本物理学家汤川秀树（Hideki Yukawa）就预言，核子间的强核力存在交换粒子。
他认为交换粒子的作用范围不超过10－15m，并推断其质量在电子与质子之间。
由于这种离子的质量介于电子与质子之间，所以汤川就将这种粒子称为“介子”（mesons）。
一年后，卡尔·安德森拍摄的云室照片显示一条异常轨迹可能就是这类粒子所产生。
Nparticle
maya教程之nDynamics-nParticleNucleu粒子（内核粒子系统）是Maya2009新增加的一个粒子特效系统，支持粒子间碰撞，可实现真正意义上的粒子堆叠效果，并且粒子生成的方式相比以前要简单的多。
因为n粒子部分属性与常规粒子的基本属性作用是相同的，在这就不重复说明此类属性（常规粒子的参数说明有专门的内容）。
Nucleus内核节点(nCloth内核布料系统也受该解算器控制)此节点是用于Maya的n布料系统和n粒子系统的常规解算器。
它包含控制力场的设置（Gravity重力和Wind风力）、地平面属性，以及时间和缩放属性，这些属性应用于所有链接到特定内核解算器的n布料节点（或n粒子节点）。
Enable启用开启后，当前内核解算器将对内核系统的内核物体进行模拟数据的计算。
关闭后，该内核系统下的任何内核物体都不会被进行动力学模拟的解算。
Gravity and Wind重力和风力Gravity重力定义Maya内核解算器的重力值.数值为0意味着没有重力，默认值是9.8米/秒平方－模拟地球的重力。
Gravity Direction重力方向定义重力的方向轴。
默认的XYZ是0,-1,0，表示重力朝着Y轴负方向向下。
Air Density空气密度定义Maya内核解算器的空气密度。
空气密度与场景中的大气质量有关，它会影响对物体的相对拖拽。
默认值为1。
低空气密度值会降低作用于物体上的空气阻力，同时也就减少了气流拖拽的效果。
例如，设置空气密度为0，创建一个模拟真空的环境，将不会有空气分子与物体进行碰撞。
高数值的空气密度会增加空气阻力和对物体的拖拽效果。
例如，高数值可用于模拟一个物体的水下移动。
注意：空气密度影响n布料和n粒子物体的Drag（拖拽）和Lift（抬升）属性。
数值为0的空气密度，实际上是将这些属性关闭了。
Wind Speed风速定义Maya内核解算器的风速。
风速决定了风力和风强。
高数值意味着迅捷的风速，其对内核物体的影响也会显著。
晨光杯青年优秀论文评选活动
发表期刊
Phys.Rev.Lett. 99,142002(2007)
高能物理研究所 华中师范大学 中国科技大学
PQCD calculation for b→+ in the standard model Examining the Crossover from the hadronic to Partonic Phase in QCD Identified Baryon and Meson Distributions at Large Transverse Momenta from Au+Au Collisions at sqrt(sNN)=200 GeV
三等奖
姓 名 Iftikhar Ahmad 吕 龑 单 位 中科院研究生院 论文名称
The spectrum of curvature perturbation for multi-field inflation with a small-field potential
发表期刊
Journal of Cosmology and Astroparticle Physics（14757516/08/02002） Phys.Review C 75, 054906 (2007)
a resonance X(1835) in J/γ+- 95:262001,2005Leabharlann 获奖名单一等奖房双世
二等奖 王小龙 王玉明 许明梅 刘海东
高能物理研究所
高能物理研究所 高能物理研究所 华中师范大学 中国科技大学
三等奖 Iftikhar Ahmad 中科院研究生院 吕 龑 中国科技大学 沈成平 高能物理研究所
Phys.Rev.D 74:034026,2006 Phys.Rev.Lett. 100,092301(2008) Phys.Rev.Lett. 97, 152301 (2006)
LINEARMOMENTUMLinearMomentum,Lawof…：线性动量的线性动量，法…
8. LINEAR MOMENTUM.Key words: Linear Momentum, Law of Conservation ofMomentum, Collisions, Elastic Collisions, Inelastic Collisions,Completely Inelastic Collision, Impulse, Impulse – MomentumTheorem.The law of conservation of energy is one of the of several great conservation laws in physics. Among the other quantities found to be conserved are the physical quantities studied in Mechanics Linear Momentum and Angular Momentum. The conservation laws are maybe most important ideas in science. In this course, we will pay intention to the law of conservation of linear momentum. It is called linear momentum to distinguish it from angular momentum, which plays an important role in rotational motion. We will call linear momentum just momentum, except places where there is danger of confusion with angular momentum.Energy is scalar quantity and its conservation helps us to predict magnitude quantity (for example, speed – magnitude of the velocity of an object). The linear momentum is vector quantity. Its conservation will help us to predict not only magnitude but also the direction of an object’s motion. The simultaneous application two laws of conservation – energy and linear momentum is the very useful tool to analyze collisions that we will do in this Chapter.8.1. Momentum and its Relation to the Force.When particle with mass m moves with velocity v, we define its Linear Momentum p as product of its mass m and its velocity v:p = m v (8.1) Unit of linear momentum is kg m / s. There is no special name for this unit. The importance of momentum can be traced to its close relation to Newton’s 2nd law. Actually, Newton originally stated his 2nd law in terms of momentum (he called it the quantity of motion). If the mass of a particle is constant, we can write the 2nd law for the case of constant forces as follows:ΣF = m a = m (Δv / Δt) = Δ (m v)/ Δt = Δp /Δt Newton’s 2nd law in terms of momentum.The vector sum of forces acting on a particle equals the rate of changeof momentum of the particle with respect to time:ΣF = Δp /Δt (8-2) The advantageous of this form for the 2nd Newton’s law with respect offormula that we used in previous chapters is that (8-2) is more generalexpression. It is hold for all cases including the object that are moving withvariable mass (for example, a rocket).The importance of momentum follows also from the fact that we canintroduce the total momentum of a system of two or more particles. TheTotal Momentum P of any number of particles (a system of particles) is equal to the vector sum of the momenta of the individual particles:P = p A + p B + p C +…….. (8-3) Where p A is the momentum of the particle A, p B is the momentum of the particle B, p C is the momentum of the particle C, and so on.But most important feature of momentum is existence of the law ofconservation of momentum.8. 2. The Law of Conservation of Momentum.Consider a definite collection of particles that we will call a system. For anysystem, the various particles exert forces on each other. We’ll call these forces internal forces F int. In addition, forces may be exerted on any part ofthe system by objects outside the system. We will call these forces external forces F ext. All forces acting on the system can be represented now as follows: ΣF = ΣF int + ΣF ext. The Newton’ s 2nd Law for the System ofParticles can be written for the case of constant forces by analogy to the forone particle (8-2) as follows:ΣF int + ΣF ext = ΔP /Δt (8-4) The sum of the internal forces according to the Newton’s 3rd law must be equal zero: ΣF int= 0 because in the interaction between any to particles are equal in magnitude and opposite in direction so cancel each other.A system that is acted upon by no external forces is called an Isolated System. So for an isolated system ΣF ext= 0. Thus, we can formulate The Law ofThe Law of Conservation of Momentum:The total momentum of a system is constant whenever the vector sum of the external forces on the system is zero. In particular, the total momentum of an isolated system is constant.If ΣF ext= 0, then ΔP = 0, so P = constant, and in any instant of timeP1 = P2 (8.5) EXAMPLE 8.1. Two objects initially at rest. Recoil problem. An atomic nucleus at rest decays radioactively into two-piece: alpha particle (α – particle) and a smaller nucleus. The new nucleus and the alpha particle will recoil in opposite directions. What will be the speed of this recoiling nucleus V if the speed of the α – particle vα is 2.5•10^5 m/s? Nucleus has a mass M 57 times greater than the mass mα of the α – particle.vα = 2.5•10^5 m/sM = 57 mαV ?When two parts of the system initially at rest and then one part begin to move, other part, because of the law of conservation of momentum will be recoiled. We will use the law of conservation of momentum (8-5) to consider this situation. Initially two particles are at rest, therefore the P1 =0. After decay particles are moving in opposite directions. If we choosethe direction of the motion of nucleus as positive, then P2 = MV --mαvα . Then, the momentum conservation gives us Æ 0 = MV – mαvα , 0 = (57mα)V – mα(2.5 × 105 m/s),ÆV= 4.4 m/sIn the next example only one object initially at rest. After the interaction both of them are moving.EXAMPLE 8.2. A bullet with mass m = 0.019 kg traveling with speed v1 = 190 m/s penetrates a block of wood with mass M = 2.0 kg and emergesgoing with speed v ′1 = 150 m/s. If the block is stationary on a frictionless surface when hit, what will be its speed after the bullet emerges?m = 0.019 kgv 1 = 190 m/sv2 = 0M = 2.0 kg Surface is frictionlessv ′2 ?M M v 2 ′ v 1′v 2 =Fig. 8.1 Example 8.2.The surface is frictionless it means that there are no external force (in this problem frictional force would be external problem, but it is absent. The block is initially at rest. It means that its initial speed is zero: v2 = 0.From the fact that the bullet emerged with the speed smaller than initial, we can figure out that some amount of its initial momentum was lost during the penetration through the wooden block. Because of the law of conservation of momentum the wooden block acquired this amount. As a result, after the collision with bullet the block begins to move in the direction of the bullet’s motion. Momentum conservation in general case could be written for system composed of two objects as follows:mv 1 + Mv 2 = mv ′1 + Mv ′2, (8-6)Substituting data, we can write:(0.012 kg)(190 m/s) + 0 = (0.012 kg)(150 m/s) + (2.0 kg)v 2′,which gives v ′2 = 0.24 m/s.Why the total momentum of an isolated system is conserved? It wasmentioned before that the laws of conservation in Physics result from thebasic properties of space and time. Particularly, the law of conservation of momentum stems from uniformity of space. In some ways, the law of conservation of momentum is more general than the law of conservation of mechanical energy. For example, it is valid even when the internal forces are not conservative; by contrast, mechanical energy is conserved only when the internal forces are conservative (otherwise the mechanical energy is transformed into other forms of energy). Below, we’ll analyze some situations in which both momentum and mechanical energy are conserved (these are collisions that are called elastic collisions) and others in which only momentum are conserved (inelastic collisions). The laws of conservation of energy and momentum play a fundamental role in all areas of physics, and we will encounter them throughout study of other branches of physics.8.3. Collisions.Collisions are defined in physics as any strong interaction between two objects that lasts a relatively short time. So this definition includes macroscopical collisions like impact of a meteor on the Arizona desert, car accidents and balls hitting on a billiard table as well as microscopic collisions between atoms elementary particles. At the subatomic level, physicists learn about the structure of nuclei and their constituents, and about the nature of the forces involved, by careful study of collisions between nuclei and / or elementary particles. It should be stressed that actually most of our knowledge about properties of elementary particles were derived from physical experiments of collisions. If the interaction forces are much larger than any external forces, we can treat the system as an isolated system, neglecting the external forces entirely and applying the law of conservation of momentum.If there are no conservative forces or we can neglect them, the application of the laws of conservation of energy and momentum together to the collisions allowed getting important information. Usually we can neglect the changes of the potential energy during collision, so we will take into account only situation with kinetic energy. If the interaction forces between the objects are conservative, the total kinetic energy of the system is the same after the collision as before. Such a collisions called an Elastic Collisions. Collisions in which the total kinetic energy after collisions is less than that before collisions are called Inelastic Collisions.8.4. Inelastic Collisions.In inelastic collisions we have situation when total momentum is conserved but kinetic energy is not conserved. This situation can be described as follows:P1 = P2K1≠ K2We have here actually one equation and can determine only one variable. We could not find speeds of two objects after collision, if we are given initial speeds of these two objects and their masses. But there is some important exclusion – Completely Inelastic Collisions. In those types of collisions, the colliding objects stick together and move as one object after the collisions. If these are 1D head on collisions, we can write the law of conservation of momentum as follows:m1 v1 + m2 v2 = (m2 + m2) v f (8-7)where m1 and m2 are the masses of objects 1 and 2, v1 and v2 are their velocities before collision, v f is their final velocity. From (8-5) we can derive expression for final velocityv f = (m1 v1 + m2 v2)/ (m1 + m2) (8-8) EXAMPLE 8.3. Completely inelastic collision of two train cars. During assembling a freight train car 1 (mass m1 = 6.5 • 10^4 kg, initial speed v1 = 0.80 m/s) collides with a car 2 (mass m2 = 9.2• 10^4 kg, initial speed v2 = 0). Two cars become coupled. (a) What is the common velocity v f of the two cars after they become coupled? (b) Show that the total kinetic energy after the completely inelastic collisions less than before the collision by finding the ratio K tot1 /K tot2.m1 = 6.5 • 10^4 kgv1 = 0.80 m/sm2 = 9.2• 10^4 kgv2 = 0(a) v f ?(b) K tot1 / K tot2?(a) Because two cars become coupled, this is the completely inelastic collision. We will use relationship (8-8) to find v f.v f = (m1 v1 + m2 v2)/ (m1 + m2)= [(6.5• 10^4 kg) (0.80 m/s) + 0] / [6.5 • 10^4 kg + 9.2• 10^4 kg]v f = 0.331 m/s(b) K tot1 = ½ m1v1² + ½ m2v2² = 0 + ½ m2v2² (8-9)Calculating K tot2, we will use expression (8-8).K tot2 = ½ (m1 + m2) v f² = ½ [m1 + m2][m1/ (m1 +m2)]² (v f)² (8-10)Dividing (8-10) by the (8-9), we find that the ratio of final to initialkinetic energy in the case of the completely inelastic collision:K tot2 / K tot1 = m1 / (m1 + m2) (8-11)Because always (m1 + m2) > m1, always, in the completely inelasticcollision between initially moving object and another object at rest, final kinetic energy always is smaller than initial total kinetic energy of the system. It could be right even if another object initially is also in motion.In our problem: K tot2 / K tot1 = 0.414.8.5. Elastic Collisions.In elastic collisions we have situation when total momentum and the total kinetic energy of the system are conserved. This situation can be described as follows:P1 = P2 (8-12) K1 = K2 (8-13) We have here actually two equations and can determine two variables.We will consider the simplest case: an elastic collision between object 1 and object 2. The collision is supposed to be head-on collision, in which all velocities lie along the same line. It is convenient to choose this line as X-axis. The equations (8-12) and (8-13) now can be written as follows:m1v1 + m2v2 = m1v′1 + m2v′2 (8-14) ½ m1 v1² + ½ m2 v2² = ½ m1 v′1² + ½ m2 v′2² (8-15)where: m1 and m2 are masses of objects 1 and 2 correspondingly; v1,v2 their initial and v′1, v′2 final velocities. We will concentrate on the particular case in which object 1 is initially moving in the direction of object 2 (initially at rest). The object 1 is called usually the projectile, the object 2 – the target. In this case, the pair of simultaneously equations can be solved for the two final velocities in terms of the masses and initial velocity v1. It could be found thatv′1 = [(m1 -- m2) /(m1 + m2)] v1 (8-16)v′2 = [2 m1 / (m1 + m2)] v1 (8-17) EXAMPLE 8.4. Consider elastic head-on collisions between 2 objects in which one initially is moving in the direction of the another that is at rest in the following cases: (a) object 1 has much greater mass than object 2 (m1 >> m2); (b) object 1 has much smaller mass than object 2 (m1 << m2); (c) two objects have equal masses (m1 =m2 =m).(a)Because m1 >> m2, we can neglect the m2 in the equations (8-16), (8-17). Than we will have v′1 ≈v1, v′2 = 2 v1. So the object 1 willcontinue to move with approximately the same velocity, the object 2will move with the twice-greater velocity in the same direction.(b)Because m1 << m2, we can neglect the m in the equations (8-16), (8-17). Then we will have v′1 ≈ -- v1, v′2 ≈ 0. So the object 1 willbounced and move back with approximately the same magnitude ofthe velocity, the object 2 will continue to be at rest.(c) Because m1 = m2 = m, the masses of the objects will disappear fromequations (8-16), (8-17). Then we will have v′1 = 0, v′2 = v1. So theobject 1 will stop, the object 2 will move with the initial velocity ifobject 1. The object 1 gives all its momentum and kinetic energy to the object that was initially at rest. This behavior is familiar to pool players. It is possible that this type of elastic collision gave hint for scientist of XVII century to discover the law of conservation ofmomentum.There is one more very helpful relationship that is working in the case of elastic collisions. The relative velocity has the same magnitude, but opposite direction, before and after the collision.v1 – v2 = -- (v′1 -- v′2) Whenever this condition is satisfied, the total kinetic energy is also conserved. Therefore it can be used together with the equation (8-14) instead of the equation (8-15). Equation (8-15) is quadratic equation and its solution could involve some difficulties.8.6. Impulse.In a collision of two macroscopic objects, both objects are deformed, often considerably, because of the large forces involved. The time interval of interaction Δt is usually very distinct and very small. Studying collisions in previous sections, we were interesting to learn how to find final motions of the colliding objects form initial their motion. We assumed that we know nothing about the forces acting during the collision and the interval of time of interaction. To analyze these physical quantities, we will pay attention only to the one of colliding objects (usually the object that has much smaller mass) and consider again the2nd Newton’s law of motion in terms of momentum. For the constant force we will have:F = Δp /Δt ÆΔp = F Δt (8-18)The right side of equation (8-18) consists of product of two quantities F and Δt that we try to study in collisions. Therefore it is useful to introduce a new physical quantity called Impulse and will be defined as follows: When constant force F acts on as object, the impulse of the force, designated by J, is the product of force and the time interval during which it acts. Impulse is the vector quantity: it has the same direction as that of the force. Unit is N s, or the same unit as the unit of momentum kg m/s.J = FΔt (8-19) Combining this definition (8-19) and equation (8-18) we get thefollowing relation, called the Impulse – Momentum Theorem:When a constant force F acts on an object during a time interval Δt = t2 – t1, the change in the object momentum is equal to the impulse of the force acting on the object:Δp = p2 –- p1 = F(t2 -– t1) = FΔt = J (8-20) To simplify situation we introduce an average force F av = const that has the same effect as real variable force. In this case, the (8-20) can bewritten as follows:Δp = p2 –- p1 = F av(t2 -– t1) = FΔt = J (8-21) EXAMPLE 8.5. A baseball with the mass m pitched at a speed of v1. The average force exerted by the bat is F av. (a) What is the initial momentum of the ball? (b) What is the final momentum of the ball? (c) What is the impulse of the ball? (d) What interval of time the ball is in the contact with the ball?mv1v2Fav(a)p1 ?(b)p2 ?(c) J ?(d)Δt ?Solving problems related to the Impulse – Momentum theorem, it isimportant to remember the vector nature of all involved quantities. We should choose the direction of coordinate axis. For example, in thisproblem it is convenient to choose it along the direction of velocity of object after the collision. Then the direction of the initial velocity will be in the negative direction of the X-axis.(a)p1 = -- mv1(b)p2 = + mv2(c)J = Δp = p2 – p1 = mv2 – (--mv1) = m (v2 + v1)(d)J = F avΔt ÆΔt = J / F av。
AtmosphericChemistry.ppt
( Rare Gases do not react readily )
Argon has three isotopes
(36Ar 0.337) (38Ar 0.063) (40Ar 99.60)
EC Decay 40K
40Ar
( t1/2 =精1选.文2档8 x 109y )
Isotopes of Xe
Aminonitrile
NH2CH2CN + 2 H2O NH2CH2COOH + NH3
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Murchison Meteor
A number of the compounds discovered in the discharge fractions are precursors to life.
indicated in the (Red Layers) of the Grand Canyon. O2 is believed to have entered the atmosphere about 1.8 Billion years ago Fe2+ and oxygen reactions may have delayed entry of oxygen into the atmosphere.
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Oxygen Rich Planet
Oxygen Rich Planet The build up of Oxygen in the atmosphere
led to the formation of the Ozone Layer at 15 to 60 km above the
earth. Ozone O3 absorbs harmful UV light and