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Isospin quantum number of D_{s0}^+(2317)

Isospin quantum number of D_{s0}^+(2317)
Isospin quantum number of D_{s0}^+(2317)

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Isospin Quantum Number of D +s0(2317)

Arata Hayashigaki

Department of Physics,Kyoto University,Kyoto 606-8502,Japan

Kunihiko Terasaki

Yukawa Institute for Theoretical Physics,Kyoto University,Kyoto 606-8502,Japan

The D +s π0and D ?+

s γdecays of D +s 0(2317)are studied by assigning it to various charmed strange scalar mesons.As a result,it is found that its assignment to an iso-triplet four-quark meson is favored by the severest experimental constraint on the ratio of the rates for these decays,while assigning it to an I =0state (a four-quark or a conventional {c ˉs })is inconsistent with this constraint.

PACS numbers:PACS number(s):14.40.Lb

Recently,a narrow D +s π0

resonance of mass ?2317MeV has been observed [1,2],and various proposals regarding its assignments have been made [3,4,5,6],including the proposal that it is the ordinary scalar D ?+s 0~{c ˉs

}[7],which is the chiral partner of D +s [8].However,its isospin quantum number is not yet known de?nitely,although its neutral and doubly charged part-ners have not been observed [9].This is because the observation of di?erent charged states is closely related to their production rates,which are still not known.In

addition,if it is assigned to D ?+

s 0,it is not easy to satisfy the severest experimental constraint on the ratio [2],

R exp ≡

Γ(D +s 0(2317)→D ?+s γ)

2J A +1

q c

f π0

D +s |A π0|?F +I

(3)by using a hard pion technique with the PCAC (par-tially conserved axial-vector current)in the in?nite mo-mentum frame (IMF)[20,21].The asymptotic matrix

element of A π(i.e.,the matrix element of A πbetween single hadron states with in?nite momentum), B |A π|A ,gives the dimensionless ABπcoupling strength.Here,

we compare ?F +I

→D +s π0with ?δs +→ηπ+by us-ing asymptotic SU f (4)symmetry (roughly speaking,the SU f (4)symmetry of the asymptotic matrix elements).The above-mentioned hard pion technique,the asymp-totic ?avor symmetry and their important results are reviewed comprehensively in Ref.[21].The amplitude

for ?δs +→ηπ+is obtained by replacing (?F +I ,D +s ,π0)by (?δs +,η,π+)in Eq.(3).The usual η-η′mixing [16]

2 leads toη=cosΘ·ηn?sinΘ·ηs,whereηn andηs are

the{nˉn}and{sˉs}components ofη,respectively,and

Θ=χ+θP,with cosχ=1/

2 D+s|Aπ0|?F+I = ηs|Aπ?|?δs+ .This estima-

tion,however,is too naive,since the wave function over-

lapping in?F+

I→D+sπ0is small,as seen above,while this

is not the case in?δs+→ηπ+,because gluon exchanges in

the system,which are more often at a lower energy scale,

reshu?e color con?gurations and spin con?gurations and

cause con?guration mixings.As a result,the?δs+→ηπ+

decay would be nearly of the fall-apart type.Therefore,

we modify the previously employed asymptotic SU f(4)

relation with asymptotic matrix elements(including four-

quark mesons)by introducing the parameter

β=√

ηs|Aπ?|?δs+ .(4)

In the naive limit of the asymptotic SU f(4)symmetry, we haveβ=1,as seen above.We hereafter refer to the symmetry in theβ=1case as the modi?ed asymp-totic SU f(4)symmetry.For simplicity,we consider the limiting case in which there is no con?guration mixing for?F+

I

,while there is no suppression arising from the crossing matrices for?δs+(and there is maximum mixing of the color and spin con?gurations).In this case,we have|β|2=1

12has

been assumed.This result seems to be slightly larger than the measured width of D+s0(2317)[16]but still con-

sistent with the data if the SU f(4)symmetry breaking arising from the overlapping of spatial wavefunctions that was not considered above is taken into account.(Note

here that the asymptotic SU f(4)symmetry transform-ing q→c overestimates the size of the asymptotic ma-

trix elements involving a charmed meson state[19]by ~20?30%.)This suggests that the situation consid-ered above,in which the strong interactions are pertur-bative at the scale of charm meson mass while they are

non-perturbative at a scale<~1GeV,is approximately realized,at least in the system under consideration here.

In this case,the strange quark will be rather slim(with a smaller mass)at the scale of the charm meson mass,as will be brie?y discussed later.

When D+s0(2317)is assigned to an I=0state(the conventional scalar D?+s0or the four-quark?F+0),the decay D?+s0→D+sπ0or?F+0→D+sπ0does not con-serve isospin.As is well known,isospin violating phe-nomena(e.g.,the di?erence between d-u quark masses,?mπ=mπ±?mπ0,?m K=m K0?m K+,etc.)have been considered as second-order e?ects of QED[22],and the isospin non-conservingπ0-ηmixing[23,24]could be of the same origin.(However,this is not assumed in this article.)Therefore,the isospin non-conserving de-cay D?+s0→D+sπ0or?F+0→D+sπ0,which is assumed below to proceed throughπ0-ηmixing,is conjectured in-tuitively to be(much)weaker than the radiative decay, in contrast with the results obtained from speci?c mod-els[12,25,26,27].

Before studying isospin non-conserving decays of D?+s, D?+s0and?F+0,we?rst investigate their radiative de-cays under the vector meson dominance(VMD)hypoth-esis[28]with the?avor SU f(4)symmetry for the strong vertices and then compare the results with those for the D+sπ0decays.Because the VMD hypothesis with the idealω-φmixing and the?avor SU f(3)for the strong vertices works fairly well for the radiative decays of light vector mesons[29],we extend it to a system containing charm quarks below.The V→Pγamplitude can be written in the form

M(V→Pγ)=?μναβGμν(V)Fαβ(γ)A(V→Pγ),(6) where V,P and A(V→Pγ)denote a vector meson,a pseudoscalar meson,and the V Pγcoupling strength,re-spectively,and Gμν(V)and Fαβ(γ)are the?eld strengths of a vector meson(V)and a photon(γ),respectively. With the VMD hypothesis,A(V→Pγ)can be approxi-mated as

A(V→Pγ)? V′=ρ0,ω,φ,ψ X V′(0)

3

Also,the V P V ′coupling strength can be estimated from the measured rate for the ω→π0γdecay:Setting (V,P,V ′)=(

ω,π0,ρ0)in Eq.(7)and inserting the above value of X ρ(0),we obtain

|A (ω→π0ρ0)|?18GeV ?1

(8)

from Γ(ω→π0γ)exp =0.734±0.035MeV [16].

To estimate the uncertainties arising from the VMD and the SU f (4)symmetry for V P V ′vertices,we focus on the D ?→Dγdecays.The amplitudes A (D ?→Dγ)are given by the ρ0,ωand ψ-meson https://www.wendangku.net/doc/d08438866.html,ing the SU f (4)relations,2A (D ?0→D 0ρ0)=2A (D ?0→D 0ω)=√

2A (D ?+→D +ψ)=A (ω→π0ρ0),and Eq.(8),we obtain Γ(D ?+→D +γ)~2.4keV and Γ(D ?0→D 0γ)~19keV.The former is consistent with Γ(D ?+→D +γ)exp ?1.5keV (with ~30%uncertainty)if a possible reduction by ~40?60%arising from asymp-totic SU f (4)symmetry breaking [19]due to the overlap-ping of spatial wavefunctions is taken into account,as discussed above.This should be done because the VMD hypothesis with the SU f (4)symmetric V P V ′couplings again overestimates the rates for the radiative decays of charm mesons,here by ~40?60%.The latter is con-sistent with Γ(D ?0→D 0γ)exp <800keV [16].The large di?erence between the above two rates results from the fact that the contributions of the ρ0,ωand ψme-son poles interfere constructively in the latter decay and destructively in the former.

Now we are in a position to consider the D ?+s →D +

s γ

decay.We compare the result with that for D ?+s →D +s π0

later.The amplitude is given by φand ψ-meson https://www.wendangku.net/doc/d08438866.html,ing the SU f (4)symmetry relations,

√2A (D ?+s →D +

s ψ)=A (ω→π0ρ0),and

Eq.(8),we obtain a small rate,

Γ(D ?+s

D +s γ)

~0.8keV ,(9)

because of the destructive interference between the φand ψpole amplitudes.However,this result satis-?es the experimental constraint,Γ(D ?+s →D +

s γ)exp <1.8MeV [16].

Similarly,the amplitude for the radiative decay of the scalar meson S →V γcan be written in the form

M (S →V γ)=G μν(V )F μν(γ)A (S →V γ),

(10)

where S =?F +I

,?F +0or D ?+s 0.Under the VMD hypothesis,the above A (S →V γ)is given by

A (S →V γ)?

V ′=ρ0,ω,φ,ψ

X V ′(0)

4

from the crossing

matrices for color and spin [13].For (i)S =?F +I

,(ii)S =?F +0and (iii)S =D ?+s 0,the amplitude A (S →V γ)is dominated by (i)ρ0,(ii)ω,and (iii)φand ψpoles,respectively,because of the OZI rule.The input data

are |A (φ→?δ

s 0ρ0)|?0.020MeV ?1,obtained from Γ(φ→a 0(980)γ)exp =0.32±0.03keV [16],for (i)and (ii),and |A (χc 0→ψψ)|?0.0823MeV ?1,obtained from Γ(χc 0→ψγ)exp =119±15keV [16],for (iii).Inserting Eqs.(11)and (12)or Eqs.(11)and (13)into the ampli-tude,Eq.(10),and taking the values of the quantities X V (0)listed above,we estimate the rates for the radia-tive decays of the scalar mesons listed in Table I.The

rate for ?F +0

→D ?+s γis smaller by an order of magni-tude than that for ?F +I

→D ?+s γ,because of the relation 3X ω(0)?X ρ(0).

Table I.Estimated rates for the radiative de-cays of charmed scalar mesons.

?F +I D ?+s 0ρ0φ,ψ?45

~35

Γ(?F +I

→D +s π0)~0.005,(14)

which satis?es the constraint given in Eq.(1),as ex-pected.

In the above,we have studied the radiative decays of ?F +I

and ?F +0by taking only Γ(φ→a 0(980)γ)exp =0.32±0.03keV as the input data,although another ra-diative decay,φ→f 0(980)γ,has also been measured.The reason for this is that the latter is apparently more uncertain than the former.If the measured rate [16]Γ(φ→f 0(980)γ)exp =1.87±0.09keV were taken as the input data,the rate for the f 0(980)→γγunder the VMD hypothesis would be given by Γ(f 0(980)→γγ)V MD ?15keV,which is much larger than Γ(f 0(980)→γγ)exp =

0.39+0.10

?0.13keV [16].Also,the ratio R ?F I would be ob-tained as R ?F I ~0.2in the same way as in Eq.(14).This implies that the electromagnetic interaction would not be very much weaker than in the case of ordinary strong interactions,and thus it leads to an unnatural result.The above would imply that the input data Γ(φ→f 0(980)γ)exp is too large.By contrast,if we take Γ(φ→a 0(980)γ)exp ,as before,we can accurately repro-duce the above Γ(f 0(980)→γγ)exp .

4 Next,we discuss isospin non-conserving decays.The

amplitude for D?+s→D+sπ0is approximately given by

M(D?+s→D+sπ0)? m2D?s?m2D s

2 π+|Aπ+|ρ0 .Therefore,it follows that

2 D+s|Aπ0|D?+s =??sinΘ· π+|Aπ+|ρ0 .(16)

The size of π+|Aπ+|ρ0 can be estimated as

| π+|Aπ+|ρ0 |?1.0[20]from the measured rate,

Γ(ρ→ππ)exp?150MeV[16].Inserting Eq.(16)with

Θ?35?(θP??20?),as before,into Eq.(15)and using

the measured branching ratios for the D?+s→D+sπ0and

D?+s→D+sγdecays[16],we get

|?|~1.0×10?2,(17)

where we have used Eq.(9).This value of|?|is consistent

with the result given in Ref.[23]and with the second-

order e?ect of QED,as argued qualitatively above.

With the help of the above value of|?|,we?nally

consider two cases of the isospin non-conserving decay,

D+s0(2317)→D+sπ0:(a)assuming D+s0(2317)is the

scalar four-quark?F+0~[cn][ˉsˉn]I=0,and(b)assuming

D+s0(2317)is the conventional scalar D?+s0~{cˉs}.The

amplitude for the S+(=?F+0or D?+s0)→D+sπ0decay can

be approximated as

M(S+→D+sπ0)? m2S?m2D s

12and|

Γ(?F+0→D+sπ0)~7,(20)is still much larger than the experimental upper bound appearing in Eq.(1),although it is much smaller than that conjectured intuitively.This is because the radiative ?F+

0→D?+sγdecay is suppressed due to the smallγωcoupling.

In the case(b),the asymptotic SU f(4)symmetry leads to D+s|Aηs|D?+s0 = K+|Aπ+|K?00 .The size of the right-hand side is estimated to be| K+|Aπ+|K?00 |?0.29from the experimental resultΓ(K?00→K+π?)exp= 182±24MeV[16].Here we have assumed that K?0(1430) is the conventional3P0{dˉs}state[16].Using the above result for the asymptotic matrix elements,the value of |?|in Eq.(17),andθP=?20?,we have

Γ(D?+s0→D+sπ0)~0.6keV,(21) which is much smaller than the other estimates[11,12, 25,26,27,32].In this case,the ratio of the rate for the radiative decay(listed in Table I)to the isospin non-conserving rate is given by

R D?

s0≡

Γ(D?+s0→D?+sγ

)

5

For the radiative D ?+s 0→D ?+

s γdecay,the model [27]predicts Γ(D ?+

s 0→D ?+s γ)BEH ?1.74keV,which is not very far from the other predictions [11,12,25,26].In the model,however,the rate for the radiative decay is very sensitive to the assumed values of the constituent quark masses.The above small rate has been obtained by choosing m c ?3m s ,which is very far from the picture of the heavy charm quark.In addition,there is an ar-gument based on a semi-relativistic potential model [35]that m c ?1.4?1.5GeV and m s ?0.090?0.095GeV near the scale of the charm meson mass under consider-ation.Such a small value of m s might be supported by recent lattice QCD analyses [36].Moreover,the model predicts

Γ(D ?+s →D +s π0)

Γ(D ?+s

→D +

s γ)

exp

=0.062±0.026.(24)

The model studied in Ref.[26]predicts a small rate for

the D ?+s 0→D ?+

s γdecay that is consistent with the result discussed above.It also predicts a large rate,

Γ(D ?+s 0→D +s π0)?129keV ,

(25)

in the case that a large value is used for the isospin vio-lating parameter,?AG ?4.0×10?2,which has been es-timated by using the measured branching ratios for the

OZI violating decays [16],B (J/ψ(2S )→J/ψ(1S )π0)=(9.6±2.1)×10?4and B (J/ψ(2S )→J/ψ(1S )η)=(3.17±0.21)×10?2.As a consequence,a small ratio,R D ?s 0

≤0.015,which satis?es Eq.(1)has been obtained.However,the above value of ?AG is much larger than the value of |?|that we estimated from the measured ratio in Eq.(24).This seems to imply that the mechanism of the isospin violation discussed above is di?erent from the usual π0-ηmixing.For this reason,the model studied in Ref.[26]cannot reproduce Eq.(24).We conclude that it is di?cult to satisfy the experimental constraint given in Eq.(1)while simultaneously realizing consistency with

the well-known ratio Eq.(24)if D +

s 0(2317)is assigned to an I =0state.

In summary,we have studied the D +s 0(2317)→D ?+

s γand D +s 0(2317)→D +s π0decays using the VMD hy-pothesis,the hard pion technique in the IMF,and the (asymptotic)SU f (4).We used the measured rates of

Γ(χc 0→ψγ)exp and Γ(φ→a 0(980)γ)exp as the in-put data,although the assignment of a 0(980)to the iso-triplet four-quark meson,?δ

s ,has not yet been con?rmed;i.e.,it might contain some other con?guration,like a K ˉK

molecule [37].We have also checked ambiguities arising from the above approximations by comparing our results for D ?→Dγ(and D ?→Dπin Ref.[19])with exist-ing experimental data.In this way,we have observed

the followings.(i)The assignment of D +

s 0(2317)to the

iso-triplet four-quark meson,?F +I

~[cn ][ˉs ˉn ]I =1,is consis-tent with the experimental constraint,Eq.(1),as seen in Eq.(14).(ii)It is di?cult to reconcile the assign-ment of D +

s 0(2317)to an iso-singlet meson (the four-quark ?F +0or the conventional D ?+s 0

)with Eqs.(1)and (24)si-multaneously.This is quite natural,because the rates for the isospin non-conserving decays are proportional to |?|2=O (α2),while those for the radiative decays are proportional to α,the ?ne structure constant.Thus,it is

seen that the assignment of D +s 0(2317)to ?F +I is most rea-sonable.In addition,this assignment is compatible with

the observation that the estimated branching fraction for

B →ˉDD +sJ

would be consistent with the measured value if the D +

sJ are interpreted as four-quark mesons,while it would be much larger than the measured one if the D +sJ are interpreted as the conventional {c ˉs }[10].The remaining problem is that neutral and doubly charged

partners of ?F +I

have not yet been observed [9].To solve this problem,we need to know the production mechanism

of the ?F 0,+,++I

mesons.Determining this mechanism is left as a future project.(iii)Finally,the numerical re-sults obtained in this work should not be interpreted too strictly,because our theoretical framework may result in a total overestimate of ~40?60%in the rate.Even considering this ambiguity,however,the conclusion of the present work is not changed.

Acknowledgements

The authors would like to thank K.Abe and K.Hagi-wara of KEK for informing them of the status of experi-ments on the isospin of D +

s 0.One of the authors (K.T.)would like to thank T.Onogi of YITP for discussions and the members of the nuclear theory groups of the Tokyo Institute of Technology and Kyoto University for discus-sions and comments.The research of A.H.is supported by a research fellowship from the Department of Physics,Kyoto University.This work of K.T.is supported in part by a Grant-in-Aid for Science Research from the Ministry of Education,Culture,Sports,Science and Technology of Japan (Nos.13135101and 16540243).

[1]B.Aubert et al.(the BABAR Collaboration),Phys.Rev.Lett.90(2003),242001.P.Krokovny et al.(the BELLE Collaboration),Phys.Rev.Lett.91(2003),262002.

6

[2]D.Besson et al.(the CLEO Collaboration),Phys.

Rev.D68(2003),032002.

[3]T.Barnes,F.E.Close and H.J.Lipkin,Phys.Rev.

D68(2003),054006.

[4]H.-Y.Cheng and W.-S.Hou,Phys.Lett.B566

(2003),193.

[5]S.Nussinov,hep-ph/0306187.

A.P.Szczepaniak,Phys.Lett.B567(2003),23.

T.E.Browder,S.Pakvasa and A.A.Petrov,Phys.

Lett.B578(2004),365.

E.E.Kolomeitzev and M.

F.M.Lutz,Phys.Lett.B

582(2004),39.

E.van Beveren and G.Rupp,Phys.Rev.Lett.91

(2003),012003.

J.Hofmann and M.F.M.Lutz,Nucl.Phys.A733 (2004),142.

[6]K.Terasaki,Phys.Rev.D68(2003),011501(R).

[7]A.De R′u jura,H.Georgi and S.L.Glashow,Phys.

Rev.Lett.37(1976),785.

[8]M.A.Nowak,M.Rho and I.Zahed,Phys.Rev.D

48(1993),4370.

W.A.Bardeen and C.T.Hill,Phys.Rev.D49 (1994),409.

[9]B.Aubert et al.(the BABAR Collaboration),

hep-ex/0408067.

[10]J.Wang,hep-ex/0312039.

[11]P.Colangelo,F.De Fazio and R.Ferrandes,Mod.

Phys.Lett.A19(2004),2083.

[12]T.Mehen and R.P.Springer,Phys.Rev.D70

(2004),074014.

[13]R.L.Ja?e,Phys.Rev.D15(1977),267;ibid.15

(1977),281.

[14]N.N.Achasov and V.N.Ivanchenko,Nucl.Phys.B

315(1989),465.

[15]L.Maiani,F.Piccinini,A.D.Polosa and V.Riquer,

Phys.Rev.Lett.93(2004),212002.

[16]S.Eidelman et al.(the Particle Data Group),Phys.

Lett.B592(2004),1.

[17]E.M.Aitala et al.(the E791collaboration),Phys.

Rev.Lett.89(2002),121801.

[18]K.Terasaki,Soryushiron Kenkyu(Kyoto)108

(2004),F11;hep-ph/0309119;AIP Conf.Proc.717 (2004),556;hep-ph/0309279.

[19]K.Terasaki,hep-ph/0405146;K.Terasaki and Bruce

H J McKellar,Prog.Theor.Phys.114(2005),205.[20]K.Terasaki,S.Oneda and T.Tanuma,Phys.Rev.

D29(1984),456.

[21]S.Oneda and K.Terasaki,Prog.Theor.Phys.Suppl.

No.82(1985),1,and references therein.

[22]F.E.Close,An Introduction to Quarks and Partons

(Academic,New York,1979).

[23]R.H.Dalitz and F.von Hippel,Phys.Lett.10

(1964),153.

[24]P.Cho and M.B.Wise,Phys.Rev.D49(1994),

6228.

[25]S.Godfrey,Phys.Lett.B568(2003),254.

P.Colangelo and F.De Fazio,Phys.Lett.B570 (2003),180.

[26]Ya.I.Azimov and K.Goeke,Eur.Phys.J.A21

(2004),501;hep-ph/0403082.

[27]W.A.Bardeen,E.J.Eichten and C.T.Hill,Phys.

Rev.D68(2003),054024.

[28]See,for example,J.J.Sakurai,Currents and Mesons

(The University of Chicago Press,Chicago,1969), and references therein.

[29]K.Terasaki,Lett.Nuovo Cim.31(1981),457;Nuovo

Cim.66A(1981),475.The data used in these pa-pers now should be updated.It also should be noted that the SU f(3)for the V P V′couplings might over-estimate the rates for the K?→Kγdecays,by~20 per cent.

[30]D.W.G.S.Leith,Electromagnetic Interactions of

Hadrons,ed.A.Donnachie and G.Shaw(Plenum Press,New York,1978),p.345.

[31]S.D.Holmes,W.Lee and J.E.Wiss,Annu.Rev.

Nucl.Part.Sci.35(1985),397.

[32]M.Harada,M.Rho and C.Sasaki,Phys.Rev.D68

(2004),074002.

[33]K.Abe et al.(the BELLE Collaboration),Phys.

Rev.D69(2004),112002.

[34]E.W.Vaandering(the FOCUS Collaboration),

hep-ex/0406044.

[35]T.Matsuki and T.Morii,Phys.Rev.D56(1997),

5646.

T.Matsuki,K.Mawatari,T.Morii and K.Sudoh, Phys.Lett.B606(2005),329.

[36]R.Gupta,hep-lat/0502005.

[37]F.E.Close,N.Isgur and S.Kumano,Nucl.Phys.B

389(1993),513.

对数函数及其性质重点难点创新突破

对数函数及其性质重点难点创新 一、教学目标 课程标准对本节课的要求为:理解对数函数的概念及单调性,掌握对数函数的图象通过特殊点,依据学生的学习基础及自身特点结合课标要求,我确定了本节课的教学目标:知识目标:1、理解对数函数的定义,掌握对数函数的图象和性质; 2、会求和对数函数有关的函数的定义域; 3、会利用对数函数单调性比较两个对数的大小。 能力目标:1、通过对底数的讨论,使学生对分类讨论的思想有进一步的认识,体会由特殊到一般的数学思想; 2、通过例题、习题的解决,使学生领悟化归思想在解决问题中的作用。 情感目标:学生在参与中感受数学,探索数学,提高学习数学的兴趣,增强学好数学的自信心。 二、教学重难点: 教学重点:理解对数函数的定义,掌握对数函数图象和性质; 教学难点:底数a对函数值变化的影响及对数函数性质的应用。 三`教学方法: 通过让学生观察、思考、交流、讨论、发现对数函数的图象的特点 四、课堂结构设计: 本节课是概念、图象及性质的新授课,为了使学生更好的达成学习目标我设计了以学生活动为主体,以培养学生能力为中心,提高课堂教学质量为目标的课堂结构。这是我的课堂结构设计:

五、教学媒体设计: 根据本节课的教学任务和学生学习的需要,我设计了利用多媒体课件展示引例、例题、习题和练习……,增大教学的容量,也使学生易于接受,提高学生的学习兴趣和积极性;利用几何画板演示作图,展示图象的动态变化过程,有效地突出重点、突破难点、提高教学效率,增强直观性和准确性。这是我的教学媒体设计: 钟 15 分 钟 钟 钟 6 分 钟

六、教学过程设计 在对教材及学生全面深入了解的基础上,我设计了以下五个教学环节:

感官动词和使役动词

感官动词和使役动词 默认分类2010-05-28 23:14:26 阅读46 评论0 字号:大中小订阅 使役动词,比如let make have就是3个比较重要的 have sb to do 没有这个用法的 只有have sb doing.听凭某人做某事 have sb do 让某人做某事 have sth done 让某事被完成(就是让别人做) 另外: 使役动词 1.使役动词是表示使、令、让、帮、叫等意义的不完全及物动词,主要有make(使,令), let(让), help(帮助), have(叫)等。 2.使役动词后接受词,再接原形不定词作受词补语。 He made me laugh. 他使我发笑。 I let him go. 我让他走开。 I helped him repair the car. 我帮他修理汽车。 Please have him come here. 请叫他到这里来。 3.使役动词还可以接过去分词作受词补语。 I have my hair cut every month. 我每个月理发。 4.使役动词的被动语态的受词补语用不定词,不用原形不定词。 (主)He made me laugh. 他使我笑了。 (被)I was made to laugh by him. 我被他逗笑了。 使役动词有以下用法: a. have somebody do sth让某人去做某事 ??i had him arrange for a car. b. have somebody doing sth.让某人持续做某事。 ??he had us laughing all through lunch. 注意:用于否定名时,表示“允许” i won't have you running around in the house. 我不允许你在家里到处乱跑。 ******** 小议“使役动词”的用法 1. have sb do 让某人干某事 e.g:What would you have me do? have sb/sth doing 让某人或某事处于某种状态,听任 e.g: I won't have women working in our company. The two cheats had the light burning all night long. have sth done 让别人干某事,遭受到 e.g:you 'd better have your teeth pulled out. He had his pocket picked. notes: "done"这个动作不是主语发出来的。 2.make sb do sth 让某人干某事 e.g:They made me repeat the story. What makes the grass grow?

函数图象重难点分析

函数图象重难点分析 本资料为woRD文档,请点击下载地址下载全文下载地址www.5y kj.co m 用“五点法”画函数的简图,及函数,,的图像与正弦曲线的联系,参变数A,对图像的影响是本课的重点.弄清函数与图像的关系,特别是和对图形的影响是本课学生的一个难点. 克服难点的办法,是要让学生弄清: (1)在函数中,对函数性质所起的作用; (2)函数的图像是通过怎样的方法由正弦曲线变化而得到,三个参数在图像变换中起什么作用. 本节运用了对图像的三种变换: 振幅变换,是由A的变化引起的; 周期变换,是由的变化引起的; 相位变换(也叫沿x轴方向的平移变换):是由的变化引起的. 将函数图像与各点的横坐标不变,纵坐扩大到原来的2倍,得到的图像,将图像上各点的纵标不变,横坐标扩大到原来的2倍,得到的图像.在这里,学生往往弄不明白为什么沿y轴“扩大到2倍”是乘以2,沿x轴“扩大到2倍”

却是除以2?函数图像在横纵两个坐标轴上的拉伸为什么不一致.也弄不明白在横纵两轴的平移究竟是什么样子.其实这些问题在学生们学习了坐标轴的变换及曲线与方程的关系后很容易理解.我们可以通过“点变换”去认识“线变换”. (1)的图像与的图像上横坐标相同的相应两点与之间的关系要满足,可见,将图像点横坐标不变,纵坐标伸长(A>1)或压缩(A<1)到原来的A倍,变成了的图像.(2)的图像与的图像纵坐标相同的相应两点和,之间的关系要满足,即 ,因此,将图像上各点的纵坐标不变,横坐标压缩或伸长到原来的倍,就变成了的图像. 可以用类似的方法解释为什么时,把的图像向左平移个单位得到的图像,而时,要把的图像向右平移个单位得到的图像. 在图像变换的教学中,要教给学生利用观察、对比、分析找出变换的规律,弄清变换的原因,理解变换的过程,而不能死记变换的结论.特别要掌握“变换”中的辩证观点:由点变换认识线变换. “五点法”作图,是作函数的静态图,在学习初期对了解函数图像的形状有益,继续学习时,必须从国家的变换角度研究图像间的关系,也就是要教给学生在运动变化中,寻

The way常见用法

The way 的用法 Ⅰ常见用法: 1)the way+ that 2)the way + in which(最为正式的用法) 3)the way + 省略(最为自然的用法) 举例:I like the way in which he talks. I like the way that he talks. I like the way he talks. Ⅱ习惯用法: 在当代美国英语中,the way用作为副词的对格,“the way+ 从句”实际上相当于一个状语从句来修饰整个句子。 1)The way =as I am talking to you just the way I’d talk to my own child. He did not do it the way his friends did. Most fruits are naturally sweet and we can eat them just the way they are—all we have to do is to clean and peel them. 2)The way= according to the way/ judging from the way The way you answer the question, you are an excellent student. The way most people look at you, you’d think trash man is a monster. 3)The way =how/ how much No one can imagine the way he missed her. 4)The way =because

感官动词的用法

感官动词 1.see, hear, listen to, watch, notice等词,后接宾语,再接省略to的动词不定式或ing形式。前者表全过程,后者表正在进行。句中有频率词时,以上的词也常跟动词原形。 注释:省略to的动词不定式--to do是动词不定式,省略了to,剩下do,其形式和动词原形是一样的,但说法不同。 see sb do sth 看到某人做了某事 see sb doing sth 看到某人在做某事 hear sb do sth 听到某人做了某事 hear sb doing sth 听到某人在做某事 以此类推... I heard someone knocking at the door when I fell asleep. (我入睡时有人正敲门,强调当时正在敲门) I heard someone knock at the door three times. (听到有人敲门的全过程) I often watch my classmates play volleyball after school. (此处有频率词often) (了解)若以上词用于被动语态,须将省略的to还原: see sb do sth----sb be seen to do sth hear sb do sth----sb be seen to do sth 以此类推... We saw him go into the restaurant. → He was seen to go into the restaurant. I hear the boy cry every day. → The boy is heard to cry every day. 2.感官动词look, sound, smell, taste, feel可当系动词,后接形容词。 He looks angry. His explanation sounds reasonable. The cakes smell nice.

The way的用法及其含义(二)

The way的用法及其含义(二) 二、the way在句中的语法作用 the way在句中可以作主语、宾语或表语: 1.作主语 The way you are doing it is completely crazy.你这个干法简直发疯。 The way she puts on that accent really irritates me. 她故意操那种口音的样子实在令我恼火。The way she behaved towards him was utterly ruthless. 她对待他真是无情至极。 Words are important, but the way a person stands, folds his or her arms or moves his or her hands can also give us information about his or her feelings. 言语固然重要,但人的站姿,抱臂的方式和手势也回告诉我们他(她)的情感。 2.作宾语 I hate the way she stared at me.我讨厌她盯我看的样子。 We like the way that her hair hangs down.我们喜欢她的头发笔直地垂下来。 You could tell she was foreign by the way she was dressed. 从她的穿著就可以看出她是外国人。 She could not hide her amusement at the way he was dancing. 她见他跳舞的姿势,忍俊不禁。 3.作表语 This is the way the accident happened.这就是事故如何发生的。 Believe it or not, that's the way it is. 信不信由你, 反正事情就是这样。 That's the way I look at it, too. 我也是这么想。 That was the way minority nationalities were treated in old China. 那就是少数民族在旧中

对数函数 优秀教案

《对数函数》教学设计 一、教材分析 本小节选自《中等职业教育课程改革国家规划新教材-数学(基础模块上册)》第四章,主要内容是学习对数函数的定义、图象、性质及初步应用。对数函数是继指数函数之后的又一个重要初等函数,无论从知识或思想方法的角度对数函数与指数函数都有许多类似之处。与指数函数相比,对数函数所涉及的知识更丰富、方法更灵活,能力要求也更高。学习对数函数是对指数函数知识和方法的巩固、深化和提高,也为解决函数综合问题及其在实际上的应用奠定良好的基础。 二、学生学习情况分析 刚从初中升入高一的学生,仍保留着初中生许多学习特点,能力发展正处于形象思维向抽象思维转折阶段,但更注重形象思维。由于函数概念十分抽象,又以对数运算为基础,同时,初中函数教学要求降低,初中生运算能力有所下降,这双重问题增加了对数函数教学的难度。教师必须认识到这一点,教学中要控制要求的拔高,关注学习过程。 三、设计理念 本节课以建构主义基本理论为指导,以新课标基本理念为依据进行设计的,针对学生的学习背景,对数函数的教学首先要挖掘其知识背景贴近学生实际,其次,激发学生的学习热情,把学习的主动权交给学生,为他们提供自主探究、合作交流的机会,确实改变学生的学习方式。 四、教学目标 1.通过具体实例,直观了解对数函数模型所刻画的数量关系,初步理解对数函数的概念,体会对数函数是一类重要的函数模型; 2.能借助计算器或计算机画出具体对数函数的图象,探索并了解对数函数的单调性与特殊点; 3.通过比较、对照的方法,引导学生结合图象类比指数函数,探索研究对数函数的性质,培养学生运用函数的观点解决实际问题。 五、教学重点与难点 重点是掌握对数函数的图象和性质,难点是底数对对数函数值变化的影响. 六、教学过程设计 教学流程:背景材料→引出课题→函数图象→函数性质→问题解决→归纳小结 (一)熟悉背景、引入课题 1.让学生看材料: 如图1材料(多媒体):某种细胞分裂时,由1个分裂成2个,2个分裂成4个……,

英语中感官动词的用法

英语中感官动词的用法 一、感官动词 1、感官动词(及物动词)有:see/notice/look at/watch/observe/listen to/hear/feel(Vt)/taste(Vt)/smell(Vt) 2、连缀动词(含感官不及物动词) be/get/become/feel/look/sound/smell/taste/keep/stay/seem/ appear/grow/turn/prove/remain/go/run 二、具体用法: 1、see, hear, smell, taste, feel,这五个动词均可作连系动词,后面接形容词作表语,说明主语所处的状态。其意思分别为"看/听/闻/尝/摸起来……"。除look之外,其它几个动词的主语往往是物,而不是人。 例如:These flowers smell very sweet.这些花闻起来很香。 The tomatoes feel very soft.这些西红柿摸起来很软。 2、这些动词后面也可接介词like短语,like后面常用名词。 例如:Her idea sounds like fun.她的主意听起来很有趣。 3、这五个感官动词也可作实义动词,除look(当"看起来……"讲时)只能作不及物动词外,其余四个既可作及物动词也可作不及物动词,此时作为实义动词讲时其主语一般为人。 例如:She smelt the meat.她闻了闻那块肉。 I felt in my pocket for cigarettes.我用手在口袋里摸香烟。 4、taste, smell作不及物动词时,可用于"t aste / smell + of +名词"结构,意为"有……味道/气味"。 例如:The air in the room smells of earth.房间里的空气有股泥土味。 5、它们(sound除外)可以直接作名词,与have或take构成短语。 例如:May I have a taste of the mooncakes?我可以尝一口这月饼吗?taste有品位、味道的意思。 例如:I don’t like the taste of the garlic.我不喜欢大蒜的味道。 She dresses in poor taste.她穿着没有品位。 look有外观,特色的意思,例:The place has a European look.此地具有欧洲特色。 feel有感觉,感受的意思,watch有手表,观察的意思。例:My watch is expensive.我的手表很贵。 6、其中look, sound, feel还能构成"look / sound / feel + as if +从句"结构,意为"看起来/听起来/感觉好像……"。 例如:It looks as if our class is going to win.看来我们班好像要获胜了。 7、感官动词+do与+doing的区别: see, watch, observe, notice, look at, hear, listen to, smell, taste, feel + do表示动作的完整性,真实性;+doing 表示动作的连续性,进行性。 I saw him work in the garden yesterday.昨天我看见他在花园里干活了。(强调"我看见了"

(完整版)the的用法

定冠词the的用法: 定冠词the与指示代词this ,that同源,有“那(这)个”的意思,但较弱,可以和一个名词连用,来表示某个或某些特定的人或东西. (1)特指双方都明白的人或物 Take the medicine.把药吃了. (2)上文提到过的人或事 He bought a house.他买了幢房子. I've been to the house.我去过那幢房子. (3)指世界上独一无二的事物 the sun ,the sky ,the moon, the earth (4)单数名词连用表示一类事物 the dollar 美元 the fox 狐狸 或与形容词或分词连用,表示一类人 the rich 富人 the living 生者 (5)用在序数词和形容词最高级,及形容词等前面 Where do you live?你住在哪? I live on the second floor.我住在二楼. That's the very thing I've been looking for.那正是我要找的东西. (6)与复数名词连用,指整个群体 They are the teachers of this school.(指全体教师) They are teachers of this school.(指部分教师) (7)表示所有,相当于物主代词,用在表示身体部位的名词前 She caught me by the arm.她抓住了我的手臂. (8)用在某些有普通名词构成的国家名称,机关团体,阶级等专有名词前 the People's Republic of China 中华人民共和国 the United States 美国 (9)用在表示乐器的名词前 She plays the piano.她会弹钢琴. (10)用在姓氏的复数名词之前,表示一家人 the Greens 格林一家人(或格林夫妇) (11)用在惯用语中 in the day, in the morning... the day before yesterday, the next morning... in the sky... in the dark... in the end... on the whole, by the way...

指数、对数函数公式

指数函数和对数函数 重点、难点: 重点:指数函数和对数函数的概念、图象和性质。 难点:指数函数和对数函数的相互关系及性质的应用,以及逻辑划分思想讨论函数 y a y x x a ==,log 在a >1及01<≠01且叫指数函数。 定义域为R ,底数是常数,指数是自变量。 为什么要求函数y a x =中的a 必须a a >≠01且。 因为若a <0时,()y x =-4,当x =1 4 时,函数值不存在。 a =0,y x =0,当x ≤0,函数值不存在。 a =1时,y x =1对一切x 虽有意义,函数值恒为1, 但y x =1的反函数不存在,因为要求函数y a x =中的a a >≠01且。 1、对三个指数函数y y y x x x ==?? ? ? ?=21210,,的图 象的认识。 对图象的进一步认识,(通过三个函数相互关系的比较): ①所有指数函数的图象交叉相交于点(0,1),如y x =2和y x =10相交于()01,,当x >0 时,y x =10的图象在y x =2的图象的上方,当x <0,刚好相反,故有10222>及 10222--<。

②y x =2与y x =?? ?? ?12的图象关于y 轴对称。 ③通过y x =2,y x =10,y x =?? ?? ?12三个函数图象,可以画出任意一个函数y a x =(a a >≠01且)的示意图,如y x =3的图象,一定位于y x =2和y x =10两个图象的中 间,且过点()01,,从而y x =?? ???13也由关于y 轴的对称性,可得y x =?? ? ? ?13的示意图,即 通过有限个函数的图象进一步认识无限个函数的图象。 2、对数: 定义:如果a N a a b =>≠()01且,那么数b 就叫做以a 为底的对数,记作b N a =log (a 是底数,N 是真数,log a N 是对数式。) 由于N a b =>0故log a N 中N 必须大于0。 当N 为零的负数时对数不存在。 (1)对数式与指数式的互化。 (2)对数恒等式: 由a N b N b a ==()log ()12 将(2)代入(1)得a N a N log = 运用对数恒等式时要注意此式的特点,不能乱用,特别是注意转化时必须幂的底数和对数的底数相同。 计算: () 313 2 -log 解:原式==?? ?? ?-=3 131 2 222 13 1 3 log log 。 (3)对数的性质: ①负数和零没有对数; ②1的对数是零; ③底数的对数等于1。 (4)对数的运算法则: ①()()log log log a a a MN M N M N R =+∈+ , ②()log log log a a a M N M N M N R =-∈+ , ③()()log log a n a N n N N R =∈+ ④()log log a n a N n N N R =∈+ 1

感官动词的用法

1.感官动词用法之一:see, hear, listen to, watch, notice等词,后接宾语,再接动词原形或ing形式。前者表全过程,后者表正在进行。句中有频率词时,以上的词也常跟动词原形。 I heard someone knocking at the door when I fell asleep. (我入睡时有人正敲门) I heard someone knock at the door three times. (听的是全过程) I often watch my classmates play volleyball after school.(此处有频率词often) 若以上词用于被动语态,后面原有动词原形改为带to不定式: We saw him go into the restaurant. →He was seen to go into the restaurant. I hear the boy cry every day. →The boy is heard to cry every day. 2.感官动词用法之二:look, sound, smell, taste, feel可当系动词,后接形容词: He looks angry. It sounds good. The flowers smell beautiful. The sweets taste sweet. The silk feels soft. I felt tired. They all looked tired. 这些动词都不用于被动语态。如:The sweets are tasted sweet.是个病句。注意:如果加介词like,则后不可接形容词,而接名词或代词:

高一函数重难点突破

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不对。 Theway(that ,in which)you’re doingit is comple tely crazy.你这么个干法,简直发疯。 Weadmired him for theway inwhich he facesdifficulties. Wallace and Darwingreed on the way inwhi ch different forms of life had begun.华莱士和达尔文对不同类型的生物是如何起源的持相同的观点。 This is the way(that) hedid it. I likedthe way(that) sheorganized the meeting. 3.theway(that)有时可以与how(作“如何”解)通用。例如: That’s the way(that) shespoke. = That’s how shespoke.

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way 用法

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【例3】已知某一直线过点(1,-4)和点(4,-2), (1)求该直线所在的一次函数关系式; (2)求该直线与两坐标轴所围成的三角形的面积; (3)若函数图像上有两点(a,m+3)、(b,-2m+6)且a>b, 求m的取值范围. 【例4】一次函数y=kx+b的自变量的取值范围是-3≤x≤6,相应函数值的取值范围是-5≤y≤-2,求该一次函数的解析式. 【例5】如图,函数y=ax+b和y=kx的交于点P,则根据图象可得: (1)方程ax+b-kx=0的解是; (2)方程组y=ax+b, y=kx的解 是__________; (3)不等式ax+b

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对图象的进一步认识,(通过三个函数相互关系的比较): ①所有指数函数的图象交叉相交于点(0,1),如y x =2和y x =10相交于()01,,当x >0时,y x =10的图象在y x =2的图象的上方,当x <0,刚好相反,故有10222>及10222--<。 ②y x =2与y x =?? ? ? ?12的图象关于y 轴对称。 ③通过y x =2,y x =10,y x =?? ? ? ?12三个函数图象,可以画出任意一个函数y a x =(a a >≠01且)的 示意图,如y x =3的图象,一定位于y x =2和y x =10两个图象的中间,且过点()01,,从而y x =?? ? ? ? 13也由 关于y 轴的对称性,可得y x =?? ? ? ?13的示意图,即通过有限个函数的图象进一步认识无限个函数的图象。 2、对数: 定义:如果a N a a b =>≠()01且,那么数b 就叫做以a 为底的对数,记作b N a =l o g (a 是底数,N 是真数,log a N 是对数式。) 由于N a b =>0 故log a N 中N 必须大于0。 当N 为零的负数时对数不存在。 (1)对数式与指数式的互化。 由于对数是新学的,常常把不熟悉的对数式转化为指数式解决问题,如: 求lo g .032524?? ? ? ? 分析:对于初学者来说,对上述问题一般是束手无策,若将它写成log .032524?? ? ? ?=x ,再改写为指数式就比较好办。 解:设log .032524?? ? ? ?=x

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