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String Pair Production in a Time–Dependent Gravitational Field Andrew J.Tolley 1,?and Daniel H.Wesley 1,?1Joseph Henry Laboratories,Princeton University,Princeton NJ,08544(Dated:September 29,2005)Abstract We study the pair creation of point–particles and strings in a time–dependent,weak gravitational ?eld.We ?nd that,for massive string states,there are surprising and signi?cant di?erences between the string and point–particle results.Central to our approach is the fact that a weakly curved spacetime can be represented by a coherent state of gravitons,and therefore we employ standard techniques in string perturbation theory.String and point–particle pairs are created through tree–level interactions between the background gravitons.In particular,we focus on the production of excited string states and perform explicit calculations of the production of a set of string states of arbitrary excitation level.The di?erences between the string and point–particle results may contain important lessons for the pair production of strings in the strong gravitational ?elds of interest in cosmology and black hole physics.
I.INTRODUCTION
The aim of this paper is a calculation the pair production of strings in a time–dependent background.This phenomenon is of paramount importance in cosmology,particularly during (p)reheating after in?ation,and near cosmological https://www.doczj.com/doc/149828857.html,ually one studies pair production using an e?ective?eld theory approach.Here we study a spacetime with weak gravitational?elds and compute the pair production of excited string states using worldsheet methods.This“stringy”calculation gives signi?cantly di?erent results than those found in the e?ective?eld theory approximation.Our results suggest speci?c signatures of the string pair creation phenomenon that may survive into the regime of strong gravitational?elds.
Recently[1,2,3]the pair creation of strings in a cosmological setting was estimated using e?ective?eld theory methods.In this work the excited states of the string are treated as massive particles for which the usual?eld theory calculation applies.While the production rate of a single massive state is small,the exponential Hagedorn growth of the number of string states can give rise to a signi?cant cumulative e?ect.Here we test this e?ective?eld theory approximation in a weak–?eld limit,where string pair creation may be computed exactly.Although our real interest is strong gravitational?elds,such as those that arise in cosmology,the formalism does not yet exist to calculate pair production of strings here(see [4]for an interesting approach to this and[5]for a calculation in a more speci?c background). Nevertheless interesting surprises can be seen already for weak?elds.Among them,we?nd that the enhancement of the production of excited string states due to the growing density of states is greatly suppressed due to the very mild hard scattering behavior of the string, which in turn arises from the fact that the string is an extended object.
To carry out our calculation of string pair production,we use the familiar idea that a curved geometry can be represented as a coherent state of gravitons[6,7].For weak gravitational?elds this gives a prescription relating S-matrix amplitudes on a weakly curved spacetime to a sum of those on Minkowski spacetime with multiple graviton vertex operators inserted.To lowest order in the closed string coupling,the pair production of strings is thus the four–point process gg→AB,with two gravitons g interacting to produce two strings in states A and B.This approach has the advantage of leading to a controlled calculation using standard string perturbation theory,but there are many other conceptual and technical obstacles to an understanding of strings in general time–dependent backgrounds.There are
subtleties involved in interpreting the S–matrix for strings on curved backgrounds(see,for example,[8,9]).More generally,in any quantum theory including gravity,the inevitable formation of singularities as predicted by the singularity theorems[10,11]implies the absence of asymptotically trivial“in”and“out”regions required to de?ne an S–matrix.We construct a speci?c geometry describing the collision of two small amplitude plane waves which we embed in an exact nonlinear solution of Einstein’s equations.An analysis of the full solution shows the formation of a spacelike singularity in the future of the collision of the plane waves,and also how by making the amplitude of both waves su?ciently small we may push the singularity arbitrarily far into the future of the initial collision region.Consequently perturbation theory is valid for an arbitrarily long period,and we believe the S–matrix formalism can be used to understand the physics in this region.This is as it should be,for the possible formation of a big crunch in our future should not prevent us from understanding physics today!
In this work we primarily focus on the pair production of long–wavelength string states. Since the theory of pair creation in?eld theory is well developed[12,13],and since string
√
theory should have an e?ective?eld theory description valid forλ?
α′,all momentum scales in the scattering process are well below the string scale.On the other hand,to produce excited string states,with masses √
m～1/
α′.This is beyond the validity of e?ective?eld theory,and indeed previous investigations have revealed uniquely“stringy”behavior in this regime[14,15,16].Thus, it is reasonable to expect the pair production of massive strings might be fundamentally di?erent from the pair production of massive particles.
Our calculation reveals two speci?c di?erences between the string and point–particle results.As we have mentioned above,and will discuss in more detail in Section IV,the pair production of strings is suppressed due to the mild hard scattering behavior of string amplitudes.Another intriguing di?erence relates to the production of identical string states. At tree level in?eld theory only identical particle pairs are produced.A surprising feature of the string case is that production of pairs of strings at the same excitation level are suppressed,going to zero asλ→∞.Instead,most created pairs are of di?erent mass levels.
Both of these features can be explained in the context of certain simple features of the string spectrum and scattering amplitudes,as we discuss in more detail below.
This paper is organised as follows:In Section II we discuss the perturbative approach to calculating pair production,using?eld theory as a guide to the string calculation.In Section III we discuss a speci?c example of a background,describing the collision of two plane waves,for which the perturbative calculation may be applied.
In Section IV we perform the explicit string calculation and discuss its properties.Finally, we present our conclusions in Section V.We also include two appendices with additional details on some elements of our work.The four–point amplitude describing pair produc-tion is discussed in Appendix A,and a more detailed account of perturbatively computing Bogolubov coe?cients is given in Appendix B.
II.PAIR PRODUCTION IN WEAK GRA VITATIONAL FIELDS
In this section we discuss how pair production may be studied perturbatively.Certainly there are examples where the production rate is nonperturbative;these include Schwinger’s classic calculation of e+e?production in an electric?eld[17]and more recent calculations with open strings[18,19,20].An example with a di?erent behavior is provided by the pair production of scalars of mass m in a?at FRW spacetime that undergoes a brief period of expansion[21].In this example we have the metric
ds2=a(η)2 ?dη2+dx23 ,a2(η)=1+b tanh(ρη),(1) and the number of particles produced in each mode k=(ω, k)is found to be
N
k =
sinh2[π(ω+∞?ω?∞)/2ρ]
A.Field Theory
In the case of?eld theory,studying pair production can be realized quite straightforwardly through the addition of new vertices in the Feynman diagrams of the theory.This point of view is developed in more detail in Appendix B,where it is shown that this technique yields precisely the same results as the more standard Bogolubov coe?cient calculation. The?eld theory case is useful for our present purposes as it provides a prototype for the string calculation.One expands the metric about Minkowski space
gμν=ημν+h(1)μν+h(2)μν+···,(4)
where h(1)μνis the?rst order metric perturbation,and h(2)μν+...,the corrections required at successively higher order so that the full metric gμνsatis?es the Einstein equations.If we consider a massive minimally coupled scalar?eld with action
Sφ=?1?g d D x,(5) Then the expansion(4)of the metric gives a corresponding perturbative expansion of the action,
Sφ=S(0)φ+S(1)φ+S(2)φ+ (6)
at n th order in the perturbation.
as well as a series of terms S(n)
φ
This allows a perturbative evaluation of the n–point functions in the background,de-noted by ··· b,in terms of the same quantities ··· 0in Minkowski space with additional insertions.For any combination of?elds O,the expansion of the action(6)inserted into the path integral yields,
O b= O 0+ T O kΣ(1)?k 0d D k(2π)D+ (7)
whereΣ(1)andΣ(2)are the new vertices that appear to?rst and second order in perturba-tions.These are determined by the expansion(6)of the action
,(8a)
Σ(1)=i S(1)
φ
1
Σ(2)=i S(2)
φ?
states.Therefore this prescription really only makes sense when the spacetime in question
is asymptotically Minkowski.In other cases it may be necessary to modify the de?nition of asymptotic states.
At this point,it is possible to see how even a weak gravitational background can lead to pair production.Pair production is the process with no“in”particles and two“out”particles.While forbidden by momentum conservation in Minkowski space,in the perturbed spacetime each term in the S–matrix carries three momenta;the two momenta of the“out”particles and the momentum carried by the new verticesΣ(n).Thus two positive energy
supplies the“missing”particles can appear in the“out”state if one of the termsΣ(n)
k
momentum.So,the amplitude to pair produce two particles with momenta k1,k2will be
nonzero provided that we haveΣ(n)
?k1?k2=0.More physically the missing momentum is coming from the background gravitons.
B.String Theory
Following the same procedure as the?eld theory calculation,one may insert the pertur-bative expansion of the metric gμν=ημν+h(1)μν+h(2)μν+···into the conformal gauge string worldsheet action
1
S=?
4πg c h(1)μν(k′) V A(k A)V B(k B)Vμνg(k′) 0d D k′
The?rst–order contribution to pair creation(10)vanishes when A and B are massive,due to momentum conservation when the graviton is on–shell(as it must be for a consistent string amplitude).Nevertheless,(10)is related to the phenomenon of“particle transmutation,”by which a string can change its mass and spin as it passes through a gravitational background [22,23,24,25].We will not investigate this phenomenon in the current work,but for weakly curved spacetimes it could in principle be studied within the S–matrix framework we apply herein.
Ideally we would be able to continue this expansion to higher order in the metric per-turbation,by analogy to the?eld theory case.However,at second order issues arise in the string theory that make the situation somewhat more complex.For one,consistency of the of the string amplitude(10)requires that each of the vertex operators appearing in be a conformal tensor of weight(h,ˉh)=(1,1),or in other words satisfy the physical state condi-tions.As is well known,this is equivalent to imposing the linearized Einstein equations on h(1)μν.At second order in the expansion of worldsheet amplitudes,a graviton vertex operator with polarization h(2)μνappears.By construction,the second–order perturbation h(2)μνsatis?es the second–order RG equations.In general,this means that h(2)μνwith not in fact satisfy the physical state conditions.Thus,the string amplitude with the corresponding vertex operator will not be consistent.
Another issue that arises is the presence of a nonzero“tadpole”for some?elds.Taking our prescription literally,to compute the S–matrix elements on our background to second order involves inserting two graviton vertex operators into the relevant correlation functions. For massive?elds A,the tadpole is zero at?rst order due to momentum conservation.At second order in the metric perturbation,the tadpole for some state A in the background is therefore given by
V A(k A) b= V A(k A)V g(k1)V g(k2) + (11)
with V g(k)=1
and states should change,which we have not taken into account.We expect that the failure of the tadpole to vanish is due to some combination of all of these e?ects.
Given these new complexities,it appears that problems are beginning to develop with the technique of inserting new vertex operators at second order.Nevertheless,by considering each of these new problems,we will argue in Sections II C that useful results can be obtained from the second order expansion.Therefore,for the purpose of studying pair production we will use the prescription
AB|S|vac =1(2π)D d D k′′
(?H)2?12m2χ2+λH2χ?1
2
The presence of theλcoupling implies that we cannot turn on a background?eld for H without also exciting a background value forχ,as a consequence of the equations of motion
? H=2λHχ+ (14)
? χ+m2χ=λH2+ (15)
Theμcoupling gives rise to pair creation ofχquanta in the presence of a nontrivial‘gravi-ton’background H.Before calculating this e?ect we must?rst correctly deal with the background.Assumingχ=0+O(H2)these equations may be solved perturbatively to second order in H as(here d?k= d d?1k/(2ω),ω=|k0|which is valid for massless and massive particles)
H(x)=H0(x)= d?k α?(k)e ik.x+α(k)e?ik.x +O(H30),(16)
χ(x)=λ d d yG ret(x,y)H20(y)+O(H30),(17) where G ret(x,y)is the retarded propagator for a scalar of mass m and we have chosen the following boundary conditions
χ(x)=0.(18)
lim
t→?∞
It is well known that in?eld theory a nontrivial background may be described by a coherent state.The initial state contains only gravitons and may be described by
|vac,t→?∞ =e?1
2 d?k|β|2e? d?k1
also determined by the condition that the amplitude to create a single particle must vanish (tadpole cancellation)
vac,t→+∞|√
√
2ω3?d(k3)
√
2ω3d(k3)
where H(u,X)is a harmonic function?2H(u,X)=0.These solutions may easily be extended to include nontrivial dilaton and NS2-form sources.However,for simplicitly of presentation we shall concentrate on pure vacuum solutions.We will focus on the class of “exact”plane waves for which H(u,X)=A ij(u)X i X j with i A ii(u)=0.All pp-wave spacetimes admit a null Killing vector?
This metric is a special case of a more general class of solutions given by
ds 2=2e
Γ(u,v )dudv +α(u,v ) i e 2βi (u,v )dx 2i ,(30)
where i βi =0.This is the form of an exact set of solutions to Einstein’s equations known as Einstein-Rosen waves.They satisfy ?u ?v α=0and ?u (α?v βi )+?u (α?v βi )=0.This form
is invariant under transformations u →f (ˉu ),v →g (ˉv )and exp(Γ)→exp(ˉΓ)/(f ,ˉu g ,ˉv ).We
may use this freedom to set α=(ˉu +ˉv )√
2then the βi satisfy 1
perturbed metric up to and including3rd order in perturbations is,
α=1+A2(α+(u)+α?(v))+O(A4),(32)
1
βi=A(βi+(u)+βi?(v))?
A2 iβi+(u)βi?(v)+O(A4),(34)
4
where,
1
α′′+(u)=?
4 iβi2?,v.(35) We are assuming thatβ+(u)andβ?(v)are localized functions,for instance with a Gaussian pro?leβ+=exp(?u2/L2)andβ?=exp(?u2/L2).From this it is clear that up to3rd order in A,βi andΓare both bounded functions localized near the interaction region u=v=0. However,as a consequence of(35),αis not bounded.We choose the solution
1
α+(u)=?
4 ∞?∞du iβi2+,u=0.(37) Trying to remove this divergence by adding the homogeneous solution?D+u causesα?(u) to diverge as u→?∞.This is the?rst indication that perturbation theory breaks down at late times.On dimensional grounds,D+～O(1)/L and so this coordinate system breaks down at u≈L/A2,v≈L/A2.Evidently by making A su?ciently small,we can push this region arbitrarily far into the causal future of the interaction region.
Remarkably,to3rd order in A,the growth ofαis a coordinate artifact and does not re?ect a genuine breakdown of perturbation theory.One may show this by removing the linear growth in u by means of the following coordinate transformation,
x i→x i(1+A2
D+ x2+O(A3),(39)
4
A2
v→v+
Similarly all components of the Riemann tensor are?nite and supported only near u=v=0. In fact we must go to4th order in perturbations to see the?rst signal that perturbation theory is breaking down.An explicit calculation shows that certain components of the Riemann tensor diverge linearly for large u and v in any orthonormal frame.However, unsuprisingly this only occurs when u≈L/A2,v≈L/A2.So again we may push the inevitable breakdown of perturbation theory o?to arbitrarily far in the future of the region of interest.
In what follows we shall only compute the string amplitudes to second order in the background perturbations.As we have seen,at second order the collision of two plane waves results in a localized interaction region and no pathologies occur.Thus there will be no problem interpreting the implications of our amplitudes.If we continue the amplitude calculations to higher order in perturbations,we may expect to see mildly pathological be-haviour associated with the eventual breakdown in perturbation theory.However,it should always be possible to separate this from the information which describes the interaction region.These above observations show the fundamental limitations of the application of the S-matrix formalism to a time-dependent geometry,these issues would not arise in a more Schrodinger like prescription where we consider the state at a given time,rather than the transition amplitude to a state at future in?nity.This suggests that string?eld theory or a similar formalism may be a more appropriate way to consider time-dependent spacetimes.
IV.PAIR PRODUCTION
The?rst nonvanishing contribution to string pair production arises from the four–point amplitude with two incoming gravitons.In this section we discuss this amplitude for an explicit set of string states,and obtain results in accord with previous investigations on the high–energy behavior of string scattering amplitudes[14,15,16].This suggests that the behavior we observe here may hold for more general excited string states as well.A detailed calculation of the relevant amplitude may be found in Appendix A.
We calculate the pair creation of“representative”string states given in oscillator notation by
|?;k =(N!)?1?μ1ˉμ1...μNˉμN N j=1αμj?1?αˉμj?1 |0;k ,(41)
where we assume that N>1.This set of states include a unique scalar state at each exciation level N,which provides the most direct comparison with?eld theory results.
Clearly the polarization?must be symmetric under the interchange of two holomorphic indices.The physical state conditions are satis?ed provided that
m2=
4
α′ N: N
j=1
?Xμjˉ?Xˉμj e ik·X(z,ˉz):(44)
where we have included a factor of the closed string coupling g c as is conventional.
As we are interested in comparing string results to those in?eld theory,we focus on the production of“long wavelength”string states,or equivalently those whose spatial momenta are small in comparison with1/
√
momenta as
k+=(m+,0,0)+(δω+,δk x,δ k T),(46a)
k?=(m?,0,0)+(δω?,?δk x,?δ k T),(46b)
k L=(?ω,?ω,0)+(?δωL,?δk′x,0),(46c)
k R=(?ω,+ω,0)+(?δωR,+δk′x,0),(46d)
Momentum conservation and the mass–shell conditions imply that that we may chooseδk x andδ k T to be the independent variables in the problem.
For our discussion,we will?nd it useful to de?ne
a=N+?α′t
4
,(47b) where t and u are the conventional Mandelstam variables.For the problem at hand,when the strings are produced with zero spatial momenta,these variables are given by
s=m2++m2?+2m+m?,(48a)
t=?m+m?,(48b)
u=?m+m?.(48c) It will be important for our argument below that in the case where the pair is produced with zero spatial momentum,these variables are all4/α′multiplied by integers.
With this parameterization of the momenta,standard techniques in string perturbation theory lead to the leading order term in the string S–matrix,given by
?+μ
1ˉμ1···μN+ˉμN+??ν
1ˉν1···νN?ˉνN?
?Lσˉσ?Rτˉτ(N+!)?1(N?!)?1×
16πi
2 N++N?N+ j=1kμj R kˉμj R N? k=1kνj R kˉνj Rηστηˉσˉτ×
Γ(a)Γ(b)sin[π(a+b)].(49)
This amplitude is then multiplied by the metric perturbations and integrated according to equation(12).There are two key features of this amplitude that di?er substantially from the?eld theory case.
The?rst remarkable feature of this amplitude is that the production of identical string pairs vanishes at zero spatial momentum.This follows directly from the behavior of the sine functions appearing in the amplitude(49).When N+=N?=N,and when the spatial momentum of the created pair vanishes,then using(47,48)and the string mass shell condition we?nd
a=b=2N?1,(50) and therefore the trigonometric factors vanish.For?eld theory,examination of the S–matrix element arising from(7)reveals that the pair production amplitude is independent of k x,T in the long–wavelength limit.For strings,by contrast,it appears that the pair production vanishes as a power law in the long–wavelength limit.Similar behavior obtains whenever the condition a,b∈Z is satis?ed,which occurs for pairs of excitation levels such as
(N+,N?)=(2,5),(2,10),(2,17),(3,9), (51)
in addition to the N+=N?cases corresponding to the creation of identical pairs.When N±are such that a/∈Z and b/∈Z,then the production of these string pairs are independent of spatial momentum in the long wavelength limit,similar to the?eld theory case.
This behavior is a consequence of some fundamental features of string theory.Recall that the open string Veneziano amplitude is essentially uniquely determined by the presence of poles corresponding to excited string states,as well as worldsheet duality.The closed string amplitudes are in a sense products of open string amplitudes,with additional sine factors to ensure that all of poles that occur in the product are simple ones.Again,the closed string amplitudes are essentially uniquely determined by basic features of the theory. Thus di?erent sine factors are zero when the s,t or u Mandelstam variables correspond to a physical string state and compensate for some of the poles arising from the gamma functions in the amplitude.In our case,when the string pairs are produced with zero spatial momentum,the s Mandelstam variable corresponds to an on–shell string state,but the t and u variables are the negative of on–shell values.Thus the sine functions have zeros, but now the gamma functions have no poles,and so the amplitude vanishes.Of course, this argument requires that we establish exactly some additional integers appearing in the arguments to the gamma functions,and this is done in Appendix A.The vanishing of this amplitude holds not only for the leading term in the S–matrix,but for terms of all orders in k appearing in this amplitude.
The second key di?erence between the string and?eld theory pair production amplitudes is an exponential suppression of the string pair production rate.To explore this feature we consider the case where excitation level of both strings is approximately N,in which case a～b～2N from(50).Applying Stirling’s approximation to the gamma function factors in (49)we?nd
Γ(2N)4
2?8N 1+1
N
V.CONCLUSIONS
In this work we have studied the pair production of excited strings in a time–dependent background.The speci?c background we consider consists of two colliding plane gravita-tional waves.Through studying the singularity structure of this background,as well as the nature of corrections needed to cancel“tadpole”diagrams,we have concluded that reliable results may be obtained through standard S–matrix methods at second order in the grav-itational wave amplitude.Our calculations revealed two essential di?erences between the ?eld and string theory results.The pair production of identical string states is suppressed for certain pairs of excitation numbers of the outgoing strings;most signi?cantly,the pro-duction of identical string states vanishes in the in?nite wavelength limit.In addition,the overall production of strings is suppressed relative to point–particle results,which may be viewed as a consequence of the mild hard scattering behavior of string amplitudes.
Our results are especially relevant to questions regarding the pair production of strings in cosmological spacetimes.In particular,these results may have implications for one mo-tivation for the current work,in which the pair production of strings is studied using an e?ective?eld theory approximation[1,2,3].The suppression of production of excited string states suggests that,despite the exponentially growing Hagedorn density of states,e?ective ?eld theory provides an overestimate of the production of individual string states.
Of course,we have established these results using techniques that are reliable only when spacetime is nearly Minkowski.Nevertheless,our?ndings suggest phenomena that may also be present in the strong?eld regime.It would be quite interesting to know if these e?ects persist in more highly curved,time–dependent backgrounds.If so,it would be an example of a uniquely“stringy”signature of relevance to cosmology,providing further clues to the role of quantum gravity in the early universe.
Acknowledgments
We would like to thank Steven Gubser,Justin Khoury,Paul Steinhardt and Herman Verlinde for useful comments and discussions.This work was supported in part by US Department of Energy Grant DE-FG02-91ER40671(AJT)and an NSF Graduate Research Fellowship(DHW).
APPENDIX A:gg→NM
Standard techniques in string perturbation theory give the form of the four–point string S–matrix amplitude as
M=8πi
2 κμj+kμj R2 κνk+kνk R
2 zκσ?kσ?
:: ?Xτ?iα′z?kτ?
α′ 2
从后面所接成分看 a p a i r o f的用法 The document was prepared on January 2, 2021
从后面所接成分看a pair of的用法 简单的来理解，a pair of不仅可以表示“一对”，还可以表示“一副”。那么a pair of在这两个不同的含义下，后面可以接怎样的成分呢?下面我们就来详细的看这部分a pair of的用法。 从后面所接成分看a pair of的用法： 1. a pair of后面可以接由两部分构成的单件事物名词，例如trousers，scissors，glasses，spectacles，jeans等等。 例句：My father has small eyes, wear a pair of glasses, looking more gently. 我的父亲有一双小小的眼睛，戴着一副眼镜，看上去很温和。 It was a pair of jeans from a denim shop. 那是一条从牛仔裤店偷来的牛仔裤。 注意事项：对于这样的由两部分构成的单件事物，我们不可以用one或者ones来替代它，而是要用pair或者pairs。 2. a pair of后面可以接包括两件东西且习惯上一起使用的名词，例如shoes，boots，gloves，socks，earrings，chopsticks等等。 例句：It's cold, I need a pair of gloves. 天气冷，我需要一副手套。 I brought my father a pair of socks on the Father's Day. 父亲节那天我为父亲买了一双短袜。 3. a pair of后面可以接表身体中两个相同的部位的名词。
PDI产品使用常见问题解答——PIT篇 PIT检测仪时间设置无法保存，什么原因？ PIT的时间是由主板上一个专门的电池负责供电的，电池寿命一般3-5年，当发现设备时间设置无法保存时，往往是这个电池没电了，您可联系相关区域的售后服务人员，更换电池即可解决，更换电池是收费的，具体收费需咨询我公司售后工程师。 PIT双通道采集仪的几种模式分别是什么意思？如何选用？ 目前，EPC在国内销售的PIT检测仪均为双通道，可进行常规单通道检测，也可进行瞬态阻抗法检测及双速度检测。检测时，共有四种模式可供选择，分别是：ACC、VEL、F+V 和V+V。ACC表示加速度模式，我们使用的传感器其实加速度传感器，ACC模式采集的就是原始的加速度信号，但通常我们不采用这种模式，因为不符合我们的分析习惯，如果不小心选了这个模式也没关系，后期在电脑软件中可以转换为速度信号分析；VEL表示速度模式，就是讲采集的加速度信号积分为速度信号显示，是我们最常用的模式；F+V 模式是瞬态阻抗法测试，需要用到内置传感器的力锤进行测试，对分析桩身浅部缺陷很有帮助；V+V模式为双速度模式，需用到两个侧置加速度传感器，主要用于测试既有基础下桩身完整性及确定桩长等。 PIT采集信号时，采样频率如何选择？ PIT-VV的采样频率为50kHz、100kHz、150kHz；PIT-XA及PIT-Q的采样频率范围为32~128kHz。 正常情况下，PIT-VV的采样频率我们选择50kHz，PIT-XA和PIT-Q的采样频率我们选择64kHz，即可满足要求。采样频率的不同，影响的是相邻采样点之间的距离，频率越高，相邻采样点之间的间隔就越小，信号精度就相应越高。但由于PIT的采样点数是固定的（1K），所以采样频率的改变，势必会影响到信号所能覆盖的桩长范围，频率越高，覆盖的桩长越短。因此，一般只有我们需要查看桩身浅部缺陷的时候，才会设置较高的采样频率，不过即使这样，精度的提高其实也是有限的。
play a part (in) 重要程度：★☆☆☆☆难易程度：★★☆☆☆ Have you realized the part computers have ___________ in the daily life? A. made B. given C. caused D. played 【参考答案】D 【拓展延伸】 1. play a part (in) 在……中扮演一个角色；参与；在……中起作用 2. play the role of 扮演……的角色 play an important role / part in...在……中起重要作用 play the leading role / part主演；起带头作用 3. take part in 参加 for the most part多半；在很大程度上 for one’s part就某人而言，对某人来说 1. Colors play an important ___________ in the way you look. A. part B. form C. effect D. pride 2. Mr. Huang will ___________ in the movement. A. play a leading part B. take parts C. play leading part D. take a part 3. __________ part that women ___________ in society is great. A. The; plays B. A; takes C. A; plays
Origin 使用问题集锦 1. 请教怎样反读出 origin 曲线上全部数据点？ 如，我用 10个数据点画出了一条 origin 曲线，并存为 project的.OPJ 格式。但，现在我想利用 OPJ 文件从这条曲线上均匀的取出 100个数据点的数值，该如何做？注：要一切都使用 origin 软件完成，不用其他曲线识别软件。 Answer： ORIGIN 中，在分析菜单（或统计菜单）中有插值命令，打开设置对话框，输入数据的起点和终点以及插值点的个数，OK！生成新的插值曲线和对应的数据表格。 2. 如何用origin 做出附件中的图： 其中标注的三角形、方块是怎么整上去的？ Answer: 选中左侧竖工具条中的 draw tool(显示是几个点，第七个工具），移动到你要标注的位置双击，就产生了一个点，依次标注完方块。再标注三角的第一个点，标注完后改成三角，以后标注的就都是三角了。改动点的类型的方法和正常画曲线方式一样。 3. 如何用origin 做出附件图中的坐标轴（带刻度）？
Answer: 你把刻度改成那样不就行了。 8.0 的具体方法是双击坐标轴，title & format --> 选左边那个 bottom，然后在右边把 axis 改为 at position=。同理，然后选左边的 left，把axis也改为 at position=。 4. origin能否读取导入曲线的坐标？ 一张 bmp 格式的图片，图片内容是坐标系和拟合曲线，但是不知道用什么软件绘制的。请问能否将该图片导入 origin，读出曲线上任意一点的数据？ Answer: (1). 1.ORIGIN 有一个图形数字化插件可完成该任务。 2.有许多专门的图形数字化软件也可完成此任务。个人感觉专门的比插件也用、便捷。推荐 WINDIG25 (2). origin下的数字化插件是digitizer,下载地 址：https://www.doczj.com/doc/149828857.html,/fileexchange/details.aspx?fid=8拖入origin即可，但使用不是很方便。比较方便的是un-scan-it。 5. 如何在origin7.5 中标峰值？ 用origin7.5 作的XRD图，怎样直接在峰上标数据？ Answer: Tools/Pick peaks 设置一下点击 Find Peaks 就 OK了。Positive和Negative 是标正负峰值的意思，其他数值改变一下就知道干吗用的了。 6. 关于origin 拟合曲线延长的问题？ 我想把拟合之后的直线向前或向后延长一段距离与坐标轴相交。但是不知道该怎么弄。是不是要改那个范围的最大值和最小值啊？可是怎么改？
【摘要】黄自先生是我国杰出的音乐家，他以艺术歌曲的创作最为代表。而黄自先生特别强调了钢琴伴奏对于艺术歌曲组成的重要性。本文是以黄自先生创作的具有爱国主义和人道主义的艺术歌曲《天伦歌》为研究对象，通过对作品分析，归纳钢琴伴奏的弹奏方法与特点，并总结黄自先生的艺术成就与贡献。 【关键词】艺术歌曲；和声；伴奏织体；弹奏技巧 一、黄自艺术歌曲《天伦歌》的分析 （一）《天伦歌》的人文及创作背景。黄自的艺术歌曲《天伦歌》是一首具有教育意义和人道主义精神的作品。同时，它也具有民族性的特点。这首作品是根据联华公司的影片《天伦》而创作的主题曲，也是我国近代音乐史上第一首为电影谱写的艺术歌曲。作品创作于我国政治动荡、经济不稳定的30年代，这个时期，这种文化思潮冲击着我国各个领域，连音乐艺术领域也未幸免――以《毛毛雨》为代表的黄色歌曲流传广泛，对人民大众，尤其是青少年的不良影响极其深刻，黄自为此担忧，创作了大量艺术修养和文化水平较高的艺术歌曲。《天伦歌》就是在这样的历史背景下创作的，作品以孤儿失去亲人的苦痛为起点，发展到人民的发愤图强，最后升华到博爱、奋起的民族志向，对青少年的爱国主义教育有着重要的影响。 （二）《天伦歌》曲式与和声。《天伦歌》是并列三部曲式，为a+b+c，最后扩充并达到全曲的高潮。作品中引子和coda所使用的音乐材料相同，前后呼应，合头合尾。这首艺术歌曲结构规整，乐句进行的较为清晰，所使用的节拍韵律符合歌词的特点，如三连音紧密连接，为突出歌词中号召的力量等。 和声上，充分体现了中西方作曲技法融合的创作特性。使用了很多七和弦。其中，一部分是西方的和声，一部分是将我国传统的五声调式中的五个音纵向的结合，构成五声性和弦。与前两首作品相比，《天伦歌》的民族性因素增强，这也与它本身的歌词内容和要弘扬的爱国主义精神相对应。 （三）《天伦歌》的伴奏织体分析。《天伦歌》的前奏使用了a段进唱的旋律发展而来的，具有五声调性特点，增添了民族性的色彩。在作品的第10小节转调入近关系调，调性的转换使歌曲增添抒情的情绪。这时的伴奏加强和弦力度，采用切分节奏，节拍重音突出，与a段形成强弱的明显对比，突出悲壮情绪。 c段的伴奏采用进行曲的风格，右手以和弦为主，表现铿锵有力的进行。右手为上行进行，把全曲推向最高潮。左手仍以柱式和弦为主，保持节奏稳定。在作品的扩展乐段，左手的节拍低音上行与右手的八度和弦与音程对应，推动音乐朝向宏伟、壮丽的方向进行。coda 处，与引子材料相同，首尾呼应。 二、《天伦歌》实践研究 《天伦歌》是具有很强民族性因素的作品。所谓民族性，体现在所使用的五声性和声、传统歌词韵律以及歌曲段落发展等方面上。 作品的整个发展过程可以用伤感――悲壮――兴奋――宏达四个过程来表述。在钢琴伴奏弹奏的时候，要以演唱者的歌唱状态为中心，选择合适的伴奏音量、音色和音质来配合，做到对演唱者的演唱同步，并起到连接、补充、修饰等辅助作用。 作品分为三段，即a+b+c+扩充段落。第一段以五声音阶的进行为主，表现儿童失去父母的悲伤和痛苦，前奏进入时要弹奏的使用稍凄楚的音色，左手低音重复进行，在弹奏完第一个低音后，要迅速的找到下一个跨音区的音符；右手弹奏的要有棱角，在前奏结束的时候第四小节的t方向的延音处，要给演唱者留有准备。演唱者进入后，左手整体的踏板使用的要连贯。随着作品发展，伴奏与旋律声部出现轮唱的形式，要弹奏的流动性强，稍突出一些。后以mf力度出现的具有转调性质的琶音奏法，要弹奏的如流水般连贯。在重复段落，即“小
从后面所接成分看i r o f 的用法 Newly compiled on November 23, 2020
从后面所接成分看a pair of的用法 简单的来理解，a pair of不仅可以表示“一对”，还可以表示“一副”。那么a pair of在这两个不同的含义下，后面可以接怎 样的成分呢下面我们就来详细的看这部分a pair of的用法。 从后面所接成分看a pair of的用法： 1. a pair of后面可以接由两部分构成的单件事物名词，例如trousers，scissors，glasses，spectacles，jeans等等。 例句：My father has small eyes, wear a pair of glasses, looking more gently. 我的父亲有一双小小的眼睛，戴着一副眼镜，看上去很温和。 It was a pair of jeans from a denim shop. 那是一条从牛仔裤店偷来的牛仔裤。 注意事项：对于这样的由两部分构成的单件事物，我们不可以 用one或者ones来替代它，而是要用pair或者pairs。 2. a pair of后面可以接包括两件东西且习惯上一起使用的名词，例如shoes，boots，gloves，socks，earrings，chopsticks 等等。 例句：It's cold, I need a pair of gloves. 天气冷，我需要一副手套。 I brought my father a pair of socks on the Father's Day. 父亲节那天我为父亲买了一双短袜。 3. a pair of后面可以接表身体中两个相同的部位的名词。
快速英语学习法（7）：“吃”的英文用法大全 小伙伴们都纷纷封自己为中国最合格的“小吃货”，当然，要把中国文化发扬光大，就要做最合格的国际吃货。 那么，一看到“吃”，你想到的第一个单词除了eat之外，还有别的？其实，这是应该根据不同的情景和意义给予恰当的表达。常见的表达法如下： 1．译作take或have。 The patient can't take food yet．病人还不能吃饭。 We are having dinner now．我们现在正在吃晚饭。 2．译作dine，feed，taste，touch，graze等词。 She invites me to dine with her tomorrow． 她邀请我明天和她一起吃饭。 Have you fed yet？你吃饭了吗？ He hasn't touched ／tasted food for two days．他两天没吃东西了。 The horses were grazing quietly in the field． 马群正在田野里静静地吃草。 3．译作like，love，prefer，enjoy，care for，be fond of等。 Would you like `ice-cream`？你想吃冰淇淋吗？ She loves bananas．她爱吃香蕉。 Southerners prefer rice while northerners prefer noodles． 南方人喜欢吃米饭而北方人喜欢吃面条。 He is enjoying his dinner．他正津津有味地吃晚饭。 Tom doesn't care for meat．汤姆不怎么喜欢吃肉。 Ants are fond of sweet food．蚂蚁喜欢吃甜食。
Population用法 population是一个集合名词（无复数形式)，它的用法有时较为特殊,所以很容易用错。 下面谈一下它的用法: 一、population常与定冠词the连用,作主语用时,谓语动词常用第三人称单数形式。 例如: The world\'s population is increasing faster and faster. 全世界的人口增长得越来越快。 At the beginning of the twentieth century, the world\'s population was about 1,700 million. 在二十世纪初,全世界的人口大约是十七亿。 二、当主语是表示\"人口的百分之几、几分之几\"时,谓语动词用复数形式。 例如: About seventy percent of the population in China are farmers. 中国大约有百分之七十的人口是农民。 三、有时population可用作可数名词,其前可用不定冠词。 例如: China has a population of about 1.3 billion. (=There is a population of about 1.3 billion in China.) 中国大约有十三亿人口。 New York is a big city with a population of over 10 million. 纽约是一个有一千多万人口的大城市。 在表示多个地区的人口时,population要用复数形式populations。 例如: Many parts of the world, which once had large populations and produced plenty of crops, have become deserts. 世界上很多地区一度人口众多,种植大量的农作物;现在,这些地区已经变成了沙漠。 四、表示人口的\"多\"或\"少\",不用\"much\"或\"little\",而要用\"large\"或\"small\"。 例如: India has a large population. 印度人口众多。 Singapore has a small population. 新加坡人口少。 五、询问某国、某地有多少人口时,不用\"How much...?\",而用\"How large...?\";在问具体人口时用\"What...?\" 例如: -How large is the population of your hometown? 你们家乡有多少人口? -The population of our hometown is nearly twice as large as that of yours. 我们家乡的人口是你们家乡人口的将近两倍。 -What is the population of Canada? 加拿大的人口有多少? -The population of Canada is about 29 million. 加拿大的人口大约有二千九百万。 六、population还表示\"某地、某类的动、植物或物品的总数\"。 例如: In India, however, the population of tigers has increased, from 2,000 in 1972 to about 5,000 in 1989. 然而在印度,老虎的总数已从1972年的2,000只增长到了1989年的大约5,000只。
人人网- 日志分享 1. 请教怎样反读出origin曲线上全部数据点？ 如，我用10个数据点画出了一条origin曲线，并存为project的.OPJ格式。 但，现在我想利用OPJ文件从这条曲线上均匀的取出100个数据点的数值，该如何做？ 注：要一切都使用origin软件完成，不用其他曲线识别软件。 https://www.doczj.com/doc/149828857.html,/bbs/viewthread.php?tid=1390313 [1] Answer： ORIGIN中，在分析菜单（或统计菜单）中有插值命令，打开设置对话框，输入数据的起点和终点以及插值点的个数，OK！生成新的插值曲线和对应的数据表格。 2. origin中非线性拟合中logistic模型的疑问？ origin 中非线性拟合中的logistic模型为 y = A2 + (A1-A2)/(1 + (x/x0)^p) 其初始参数设置为 sort(x_y_curve); //smooth(x_y_curve, 2); x0 = xaty50( x_y_curve ); p = 3.0;
A1 = max( y_data ); A2 = min( y_data ); A1 = min( y_data ); A2 = max( y_data ); 而据我看到的logistic的模型都是（自己origin中自定义的） y =A1/(1+(A1/A2-1)*exp(-k*x)) 也就是说 origin 中的logistic有4个数值需要确定，而自定义的有3个数值 从结果来看，没有太大区别，但为什么函数不一样呢？ 不是学数学，高人能否详细说明下。 https://www.doczj.com/doc/149828857.html,/bbs/viewthread.php?tid=1391522 [2] Answer: 你可以看一下这个文档，里面有数种不同形式的 logistic 模型： https://www.doczj.com/doc/149828857.html,/web/packages/drc/drc.pdf [3] 当然，这是一个 R (https://www.doczj.com/doc/149828857.html,) 包的文档，但不妨碍你看其中的公式。 R 是开源的啊，以 GPL 发布，可以从 https://www.doczj.com/doc/149828857.html, [4]上了解更多。I 3. 如何用origin做出附件中的图：其中标注的三角形、方块是怎么整上去的？https://www.doczj.com/doc/149828857.html,/bbs/viewthread.php?tid=1393739 [5] Answer:
我国艺术歌曲钢琴伴奏-精 2020-12-12 【关键字】传统、作风、整体、现代、快速、统一、发展、建立、了解、研究、特点、突出、关键、内涵、情绪、力量、地位、需要、氛围、重点、需求、特色、作用、结构、关系、增强、塑造、借鉴、把握、形成、丰富、满足、帮助、发挥、提高、内心 【摘要】艺术歌曲中，伴奏、旋律、诗歌三者是不可分割的重 要因素，它们三个共同构成一个统一体，伴奏声部与声乐演唱处于 同样的重要地位。形成了人声与器乐的巧妙的结合，即钢琴和歌唱 的二重奏。钢琴部分的音乐使歌曲紧密的联系起来，组成形象变化 丰富而且不中断的套曲，把音乐表达的淋漓尽致。 【关键词】艺术歌曲；钢琴伴奏；中国艺术歌曲 艺术歌曲中，钢琴伴奏不是简单、辅助的衬托，而是根据音乐 作品的内容为表现音乐形象的需要来进行创作的重要部分。准确了 解钢琴伴奏与艺术歌曲之间的关系,深层次地了解其钢琴伴奏的风 格特点,能帮助我们更为准确地把握钢琴伴奏在艺术歌曲中的作用 和地位，从而在演奏实践中为歌曲的演唱起到更好的烘托作用。 一、中国艺术歌曲与钢琴伴奏 “中西结合”是中国艺术歌曲中钢琴伴奏的主要特征之一，作 曲家们将西洋作曲技法同中国的传统文化相结合，从开始的借鉴古 典乐派和浪漫主义时期的创作风格，到尝试接近民族乐派及印象主 义乐派的风格，在融入中国风格的钢琴伴奏写作，都是对中国艺术 歌曲中钢琴写作技法的进一步尝试和提高。也为后来的艺术歌曲写 作提供了更多宝贵的经验，在长期发展中，我国艺术歌曲的钢琴伴 奏也逐渐呈现出多姿多彩的音乐风格和特色。中国艺术歌曲的钢琴
写作中，不可忽略的是钢琴伴奏织体的作用，因此作曲家们通常都以丰富的伴奏织体来烘托歌曲的意境，铺垫音乐背景，增强音乐感染力。和声织体，复调织体都在许多作品中使用，较为常见的是综合织体。这些不同的伴奏织体的歌曲，极大限度的发挥了钢琴的艺术表现力，起到了渲染歌曲氛围，揭示内心情感，塑造歌曲背景的重要作用。钢琴伴奏成为整体乐思不可缺少的部分。优秀的钢琴伴奏织体，对发掘歌曲内涵，表现音乐形象，构架诗词与音乐之间的桥梁等方面具有很大的意义。在不断发展和探索中，也将许多伴奏织体使用得非常娴熟精确。 二、青主艺术歌曲《我住长江头》中钢琴伴奏的特点 《我住长江头》原词模仿民歌风格，抒写一个女子怀念其爱人的深情。青主以清新悠远的音乐体现了原词的意境，而又别有寄寓。歌调悠长，但有别于民间的山歌小曲；句尾经常出现下行或向上的拖腔，听起来更接近于吟哦古诗的意味，却又比吟诗更具激情。钢琴伴奏以江水般流动的音型贯穿全曲，衬托着气息宽广的歌唱，象征着绵绵不断的情思。由于运用了自然调式的旋律与和声，显得自由舒畅，富于浪漫气息，并具有民族风味。最有新意的是，歌曲突破了“卜算子”词牌双调上、下两阕一般应取平行反复结构的惯例，而把下阕单独反复了三次，并且一次比一次激动，最后在全曲的高音区以ff结束。这样的处理突出了思念之情的真切和执著，并具有单纯的情歌所没有的昂奋力量。这是因为作者当年是大革命的参加者，正被反动派通缉，才不得不以破格的音乐处理，假借古代的
主谓一致用法总结 I.主谓一致定义 II.谓语受主语支配，和主语在人称和数上保持一致，这叫做主谓一致。 III.例: My favorite food is noodles. II. 主谓一致的重要原则 ?语法原则 ?意义原则 ?就近一致 （一）语法一致 IV.顾名思义，即谓语在语法角度上与主语保持一致，不考虑主语的意义。 V.以单数名词或代词，动词不定式短语作主语时，谓语动词要用单数;主语为复数时，谓语用复数。例如： 注意：不定式及动名词作主语时，谓语 动词单数。E.g. 2) 由and或both……and连接的并列成分作主语时，谓语动词用复数。例如： Both you and he are right. Mr. and Mrs. Brown have a son called Tom. 但并列主语如果指的是同一人，同一事物或同一概念，谓语动词用单数。例如：The poet and writer has written lots of books. The poet and the writer have come. 可通过名词前定冠词来判断。 3) 由and连接的并列单数主语之前如果分别由each, every修饰时，其谓语动词要用单数形式。例如： Now every man and every woman has the right to receive education. 4) 主语后面跟有 but ,except, besides, with ，together with 等介词短语时，谓语动词仍用与主语（即前面的词语）保持一致。例如： The teacher with his students is going to have a picnic in the park. The students with the teacher are going to have a picnic in the park. Nobody but two boys was late for class. 5) 集合名词作主语谓语动词要用复数。 如people, police, cattle, clothes等。 集合名词指可用来指称一群对象的词语，这些对象可以是人、动物、或是一群概念等事物。 常见集合名词：people, police, cattle, goods, clothes等。 e.g. The police are looking for him. ?有些集合名词如class, team, group, family 根据其表达意义不同，单复数用法也不同。
错误!不能读取或显示文件。特殊用法adjustable blade 可调整叶片 可调整刀片 变距桨叶, 活距桨叶 air cooled cascade blade 气冷叶栅桨 airscrew blade 螺旋桨叶 anchor blade 锚爪 articulated blade 【航空】铰接桨叶 back saw blade 锯条 backward curve blade 后曲叶片 beater blade 轮叶, 逐稿轮叶片, 击轮叶片 bent blade 弯刀, 弯形旋耕刀 blower blade 吹送机轮叶片, 鼓风机轮叶片 calling-on signal blade 引导信号臂板 cambered blade 曲面桨叶, 弯曲叶片 carbide blade (无心磨床的)硬质合金刀片, 硬质合金托板 cascade blade 叶栅的叶片 cast blade 铸造叶片 central blade
(多叶舵的)中间舵叶 centrifuge blade 离心叶片; 离心刀片 ceramic rotor blade 陶瓷动叶片 circular shear blade (圆剪床的)圆形刀片completeblade 整叶片 compressorblade 压缩机叶片 conduction-cooled turbine blade (导热)冷却式涡轮叶片 constant reaction blades 等反应度叶片 constant speed chopper blade 恒速调制盘叶片 counter blade 对刃 covering blade 覆土铲 Cuban blade 无柄刀(生产雪茄烟用) cutter blade (铣)刀片 切割器动刀片 Damascus blade 大马士革钢刀 detachable blade 可折桨叶 可换刀片[刀刃] diamond-impregnated blade 金刚砂刀片 die blade
play a part in的用法 Do you know this pretty girl Right! She is Audrey Hepburn who played many classic roles in a great many famous films. For example, < Roman Holiday> is the one of the films which earned Hepburn her first Academy Award for Best Actress. Audrey Hepburn played the part/role of Princess Ann in this film. play the part/role of…扮演……角色
Audrey Hepburn played a leading part in
Audrey Hepburn also played leading roles in < Funny Face>and < My Fair Lady〉.( My Fair Lady 窈窕淑女影片讲述下层卖花女被语言学教授改造成优雅贵妇的故事，从头至尾洋溢着幽默和雅趣 .Funny Face 甜姐儿) Play a role in…在……中扮演角色，在……中起作用 play a part/role in doing sth. 在做某事方面起作用,参与做某事 We can?all?play?a?role/part?in?reducing?our?dependence?on?plastic, if we started to?take some small?steps?in?our?everyday?lives?to?be
population 的用法 1.population 常以单数形式出现，意为“人口”“人数”。如果指世界不同地区的人口时，用复数。 （1）The population of this city is growing every year. (2) Many parts of the world, which once had large populations and produced plenty of crops, have become deserts. 2.做主语时，谓语动词的数的选用，population直接做主语，谓语动词一般用单数，前面有some, most 或者百分数时，谓语动词常用复数，前面有分数时，谓语动词单复数即可。 (1).The population is increasing faster and faster. (2).At the beginning of the twentieth century, the world\'s population was about 1,700 million. (3).About seventy percent of the population in China are farmers. (4).Most of the population of the city are workers. (5).One third of the population now smoke/smokes. 3.population 前的冠词的选用，population of 跟地名或者事物时，用定冠词the, population of 跟数词时，用不定冠词a, population 表示抽象意义时，用零冠词。 （1）.China has a population of about 1.3 billion. (2). New Y ork is a big city with a population of over 10 million. (3). The population of Canada is about 29 million. (4). The people’s living standard has risen, causing a rise in population. 4.询问某国、某地有多少人口时,不用\"How much...?\",而用\"How large...?\";在问具体人口时用\"What...?\"。表示人口的\"多\"或\"少\",不用\"much\"或\"little\",而要用\"large\"或\"small\" (1).How large is the population of your hometown? 你们家乡有多少人口? (2).The population of our hometown is nearly twice as large as that of yours. 我们家乡的人口是你们家乡人口的将近两倍。 (3).What is the population of Canada? 加拿大的人口有多少? 5. 在表示多个地区的人口时,population要用复数形式populations。例如: Many parts of the world, which once had large populations and produced plenty of crops, have become deserts. 世界上很多地区一度人口众多,种植大量的农作物;现在,这些地区已经变成了沙漠。
英语量词用法大全 常用语法 场景1：当和外国友人走在公园，突然看到一群鸽子。这时候你是不是就会说:There are many pigeons 或者a lot of呢？想再高大上一点吗？Sam, look! There is a flock of pigeons over there. 场景2:当说三张纸的时候，是不是很习惯说：“three papers”？错了哦。paper 是不可数名词。正确说法是：three pieces of paper。 ◆◆量词用法整理+讲解◆◆ 量词可以用于描述可数名词比如a herd of elephants(一群大象）, 也可以描述不可数名词；比如three pieces of paper（三张纸） 1、描述一群...量词+群+of+名词 一群人a/an crowd／group／army／team／of people； 一群牛、象、马、天鹅a herd of cattle／elephants／horses／swans
一群鸟、鹅、母鸡、羊、燕子a flock of birds／geese／hens／goats／swallows 一群猎狗、狼a pack of hounds／wolves 2、描述一丝／点／层 一丝怀疑a shadow of doubt 一线未来之光a glimpse of future 一缕月光a streak of moonlight 一层霜／雪／糖霜a layer of frost／snow／cream 3. piece块；片；段；项；件；篇；首；幅；张 a piece of bread／paper／wood／furniture／land／advice／news／meat ／cloth／music... 4、英译“一阵” 一阵哭泣／喝彩／炮击／ 雷声
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▲掌握小数，分数、百分数和时间、日期的表达法。 【复习要点】 （一）基数词num. 基数词用来表示数目，或者说表示数量的词叫基数词。最基本的基数词如下表所示。 1 one 11 eleven 100 a hundred 2 two 12 twelve 20 twenty 1000 a thousand 3 three 13 thirteen 30 thirty 1，000，000 a million 4 four 14 fourteen 40 forty 10，000，000 ten million 5 five 15 fifteen 50 fifty 100，000，000 a hundred million 6 six 16 sixteen 60 sixty 1，000，000，000 a billion 7 seven 17 seventeen 70 seventy 8 eight 18 eighteen 80 eighty 9 nine 19 nineteen 90 ninety 10 ten 说明： 1．13—19是由个位数加后缀-teen构成。注意其中13、15的拼写是thirteen 和fifteen。 2．20—90由个位数加后缀-ty构成，注意其中20—50的拼写分别是twenty, thirty, forty 和fifty;80的拼写是eighty。 3．其它非整十的两位数21—99是由整十位数加连字符“-”，再加个位数构成。如：81 eighty-one。 4．101—999的基数词先写百位数，后加and再写十位数和个位数。如：691 six hundred and ninety-one。5．1000以上的基数词先写千位数，后写百位数，再加and，最后写十位数和个位数。 如：5893 five thousand eight hundred and ninety-three。在基数词中只有表示“百”、“千”的单位词，没有单独表示“万”、“亿”的单位词，而是用thousand（千）和million（百万）来表达，其换算关系为：1万=10 thousand;1亿=100 million; 10亿=a thousand million=a billion。 Hundreds of Thousands of 6．多位数的读法： 1）1000以上的多位数，要使用计数间隔或逗号“，”。即从个位起，每隔三位加一个间隔或逗号。第一个间隔或逗号前是thousand（千），第二个间隔或逗号前是million（百万），第三个间隔或逗号前是a thousand million或a billion(十亿)。 2）每隔三位分段以后就都成了101—999。读的时候十位数（或个位数）的前面一般要加and。如： 888，000，000读作：eight hundred and eighty-eight million。 基数词的用法： 1. 基数词在句中的作用 基数词的作用相当于名词和形容词，在句中可作定语、主语、宾语（介宾）、表语、同位语等。 例如：Three and five is eight. 3+5=8 （作表语）How many oranges do you want?你要多少桔子？ I want eight. 我要八个。（作宾语）There are eight boats in the lake. 湖里有八条小船。（作定语） 2．Hundred, thousand, million, dozen, score这些词前面如有表示具体数字的词，它们不能加“s”,反之则须加“s”, 并要与of短语连用。例如：
The的用法归纳 1 表示特指的人或物 例：Please hand me the key on the desk. 请把桌上的钥匙递给我。 The girl in red is his sister. 穿红色衣服的女孩是他妹妹。 The building over there is the tallest in the town. 那边那幢大楼是这个城里最高的。 I like the music of the film. 我喜欢这部电影的音乐。 2 表示双方都知道的或心中明白的人或物 例：Shut the door, please. 请关门。 Has he returned the book? 那本书他还了吗？ Take the blue one, it is cheaper. 拿那个蓝的，它便宜些。 3 第二次提到某人或某物第一次提到时用不定冠词，第二次提到时要用定冠词。 例：He saw a house in the distance. Jim's parents lived in the house. 他看见远处有一所房子，吉姆的父母就住在那所房子里。 There was once an old fisherman. The old fisherman had a cat. The cat was white. 从前有一个老渔夫。这个老渔夫有一只猫。这只猫是只白猫。 4 用在世界上独一无二的名词前 这类词有： the sun太阳, the earth地球, the moon月亮, the sky天空, the world 世界 例：The moon goes round the earth. 月亮绕着地球转。 There is not any cloud in the sky. 天空中没有一丝云彩。 It was a fine day in spring. The sun shone brightly. 这是一个晴朗的春日，阳光灿烂。 He is the richest man in the world. 他是世界上最富的人。 5 用在表示方向、方位的名词前 这类词有： the east东方，the west西方，the south南方，the north北方，the right右边，the left左边 例：The birds are flying to the north. 这些鸟向北方飞去。 The moon rises in the east and sets in the west. 月亮从东方升起，在西方落下。 The wind was blowing from the south. 风从南方吹来。 She lived to the west of the Summer Palace. 她住在颐和园的西边。 Walk along the road and take the first turning on the right. 沿着这条路往前走，在第一个路口往右拐。He stood at the back of the door. 他站在门背后。 提示 方位词成对使用构成平行结构时，不用定冠词。 例：The river is two thousand kilometers long from west to east. 这条河自西向2000公里长。 They traveled through the country from south to north. 他们自南向北在这个国家旅行。 6 用在形容词最高级前 例：Summer is the hottest season of the year. 夏天是一年中最炎热的季节。 She is the best person for the job. 她是最适合这个工作的人。 Hangzhou is one of the most beautiful cities in the world. 杭州是世界上最美的城市之一。 The car is the most expensive of the four. 这部车是四部车中最贵的。 7 用在序数词等前 定冠词用在序数词前，也用在表示序列的next, last等前，还有在表示“同一”或“唯一”等的词前。 例：The first man to land on the moon is an American. 第一个登上月球的人是美国人。 She was the fifth to climb to the top of the mountain. 她是第五个到达山顶的人。 This may be the last chance. 这可能是最后一次机会。 If I miss this train I'll catch the next one. 如果赶不上这趟火车，我就赶下一趟。 He is the only person who knows the secret. 他是唯一一个知道这个秘密的人。 The two coats are of the same colour. 这两件外衣颜色相同。 This is the very book I want. 这正是我要的书。（用very表示强调）