a
r X i v :h e p -t h /0005085v 2 26 S e p 2000Preprint typeset in JHEP style.-HYPER VERSION
Justin R.David Department of Physics University of California Santa Barbara,CA 93106USA.justin@https://www.doczj.com/doc/e89439726.html, Abstract:The naive low energy e?ective action of the tachyon and the U (1)gauge ?eld obtained from string ?eld theory does not correspond to the world volume action of unstable branes in bosonic string theory.We show that there exists a ?eld rede?nition which relates the gauge ?eld and the tachyon of the string ?eld theory action to the ?elds in the world volume action of unstable branes.We identify a string gauge symmetry which corresponds to the U (1)gauge transformation.This is done to the ?rst non linear order in the ?elds.We examine the vector ?uctuations at the tachyon condensate till level (4,8).
Keywords:Bosonic Strings,D-branes,Conformal Field Models in String Theory.
1.Introduction
Recent works have shown that string?eld theory is useful to understand the physics of tachyon condensation[1].In bosonic string theory,the decay of the D25-brane has been shown to correspond to the rolling of the tachyon of open bosonic string
theory down to a new stationary point in the tachyon potential.The analysis of the tachyon potential was carried out using string?eld theory.It was also seen that level truncation is a good approximation scheme for the calculation of the value of the minimum of the tachyon potential and the physics of solitons in the potential [2,3,4,5,6,7,8].The level truncation seems to be a good approximation scheme in superstring?eld theory[9,10,11,12].
As the D25-brane has completely decayed at the new stationary point of the tachyon potential,there should be no dynamical open string degrees of freedom.In particular the U(1)gauge?eld should no longer be dynamical.It was conjectured in[1]that at this stationary point the e?ective action does not have a kinetic term for the U(1)gauge?eld.In fact the term multiplying the kinetic term was argued to be the tachyon potential[13].This would render this?eld auxiliary.It was also suggested it would be interesting to see this phenomenon directly from string?eld theory.The di?culty with the direct evaluation of the coe?cient of the kinetic term of the gauge?eld using string?eld theory is the problem of?eld rede?nition.
We will now state this problem.Consider eliminating all the massive modes and the auxiliary?elds of the string?eld using the equations of motion.We obtain the e?ective action S(?Aμ,?φ)as a function of the U(1)gauge?eld and the tachyon?φat the tree level.Can we identify the gauge?eld?Aμand the tachyon?φwith that of the gauge?eld of the D25-brane and its tachyon,and thus obtain the e?ective action of the D25-brane from string?eld theory?A string?eldΦhas the following gauge transformation in string?eld theory
δΦ=Q BΛ+Λ?Ψ?Ψ?Λ(1.1)
HereΛ,the in?nitesimal gauge parameter is a string?eld of ghost number zero and ‘?’is the star product in string?eld theory.Let us call the U(1)?eld present in the expansion ofΦas?Aμ.Then,it is clear that?Aμdoes not transform as
δ?Aμ=?μ?(1.2)
The transformation property of?A has extra terms arising from the non-Abelian like transformation property of the string?eldΦ.Therefore the gauge?eld that appears in the low energy e?ective Lagrangian Aμwhich has the conventional gauge
transformation property cannot be identi?ed with?Aμ.Thus the naive low energy e?ective action S(?Aμ,?φ)cannot be identi?ed with the world volume action of the unstable brane.
It must be possible to?nd a relationship between Aμand?Aμandφand?φ.This will involve a?eld rede?nition
?A
=?Aμ(Aμ,φ),?φ=?φ(Aμ,φ)(1.3)
μ
It also involves the identi?cation of the string?eldΛwhich corresponds to the conventional gauge transformation property of Aμ.In this paper we show that it is indeed possible to?nd such a relationship at the?rst non-linear level in the?elds. To do this we use methods developed to identify general coordinate transformation and anti-symmetric tensor gauge transformation as a set of string?eld theoretic symmetries developed for the non-polynomial closed string?eld theory by[15]1.
The fact that gauge?eld of the low energy e?ective action involves not only?Aμbut also other?elds is important in determining the coe?cient of the kinetic term of Aμfrom string?eld theory.Determining the naive kinetic term of?Aμis not enough. This?eld rede?nition involved in relating Aμand?Aμis also important in determining terms of the non-Abelian Dirac Born Infeld action from string?eld theory as done in[4].This would help in understanding the discrepancy in gauge invariance which arose in trying to determining the coe?cients of certain terms in the non-Abelian Dirac Born Infeld action from string?eld theory.
Thus it seems di?cult to compare the e?ective action S(?Aμ,?φ)obtained from string?eld theory with the world volume action of unstable branes and verify the claims of[13].It still must be possible to see if U(1)gauge?eld is no longer dynamical from string?eld theory.To this end,it is of interest to calculate the two point functions of all vector particles at the tachyon condensate using level truncation. It is of interest to compare with the situation in p-adic string theory[14].The kinetic terms for the translationaly zero modes of the soliton in p-adic string theory disappears after?eld rede?nition at the new stationary point2.
The organization of this paper is as follows.In section2we review Witten’s bosonic open string?eld theory[16]to set up our notations and conventions.We use the formulation of describing string?eld theory in arbitrary background?eld developed by[17]and[18].Then we show the existence of the?eld rede?nition(1.3). We also identify the set of string?eld theory transformations which correspond to the gauge transformation of the U(1)gauge?eld in the low energy e?ective action. These are shown to exist at the?rst non-linear order in the?elds.In section3we review of how the the free action for the tachyon and the gauge?eld arises from string ?eld theory.In section4we obtain the e?ective action S(?Aμ,?φ)by eliminating all other massive and auxiliary?elds.In section5we derive the constraint equations for?eld rede?nition.In section6we write down the consistency conditions which should be identically satis?ed.In section7we verify the consistency conditions are identically satis?ed and in section8we show the existence of the solutions for the consistency conditions and thus prove the existence of the?eld rede?nition up to the?rst nonlinear order in the?elds.This also identi?es the string?eld theory transformation which corresponds to the gauge transformation of the U(1)?eld of the low energy e?ective action.In section9we examine the two point function of the transverse photon.We show explictly that the kinetic term for the transverse photon does not decrease as fast as the tachyon potential approaches zero in the level truncation approximation.We show that at level(3,6)there is a physical vector excitation in the spectrum,but at level(4,8)this might be removed or pushed to a higher energy.
2.Review of string?eld theory
The open string?eld theory action is given by
S=?12 I?Φ(0)Q BΦ(0) +1
?eldΦcreates the state|Φ out of the SL(2,R)vacuum.
|Φ =Φ(0)|0 (2.2) AsΦhas ghost number one,the Hilbert space of the?rst quantized open string theory including the b and c ghost?elds is restricted to ghost number one.f?Φ(z) denotes the conformal transformation of the?eldΦby the map f(z).For example ifΦis a primary?eld of weight h then f?Φ(z)=(f′(z))hΦ(f(z)).The conformal transformations f1,f2,f3and I are are given below.
f1(z)=?e?iπ/3 1?iz1+iz 2/3+e iπ/3 (2.3) f2(z)=F(f1(z)),f3(z)=F(f2(z)),
F(z)=?1
z
It is convenient to work in a given basis of the Hilbert space of the?rst quantized string theory with ghost number one.Therefore we write
|Φ = rψr|Φ1;r (2.4) r denotes the various?elds as well as momenta.The subscript1in|Φ1;r stands for the ghost number.In terms of this basis the string?eld theory action is given by
S=?12?A(2)rsψrψs+1
where
A (2)rα= Φc 2;r |Q
B |Φ0;α
(2.9)A (3)rαs = f 1?Φc 2;r (0)f 2?Φ1;s (0)f 3?Φ0;α(0)
? f 1?Φc 2;r (0)f 2?Φ0;α(0)f 3?Φ1;s (0) (?1)n r +n g
r where n r and n g r denotes the fact that |Φ1;r is an n r th level secondary in the matter
sector,and n g r th level secondary in the ghost sector. Φc 2;r |is conjugate to the state |Φ1;r ,that is Φc 2;r |Φ1;s =δrs (2.10)
In addition to the gauge symmetries,the string ?eld theory action has a ‘trivial’symmetry of the form
δψr =K rs ({ψ})δS
g 2 dk ?φ
(?k )(2k 2?1)?φ(k )+?A μ(?k )(2k 2)?A μ(k )(3.2)?2F (?k )F (k )+4k μ?A μ(?k )F (k )
We are working in the unitsα′=2.The gauge transformation of these?elds can be determined by using(2.8).Restricting to only the linearized gauge transformation that is,only terms from A(2)rαin(2.8)we obtain
δ?φ(k)=0(3.3)
δ?Aμ(k)=?2kμλ(k)
δF(k)=2k2λ(k)
Hereλ(k)is the in?nitesimal gauge parameter.The quadratic piece of the action in(2.5)is invariant under linearized gauge transformation.It is easy to verify the action in(3.2)is invariant under the transformation of(3.3)
The auxiliary?eld F(k)can be eliminated using the equations of motion
F(k)=?kμ?Aμ(k)(3.4)
On eliminating F(k),the action reduces to that of the tachyon with the standard kinetic term for the U(1)gauge?eld.It is given by
S=?1
We divide the string?eldsψr into two kinds.ψr,ψs...stand for the tachyon?φ(k) and?Aμ(k)including their momentum index.The?eldsψa,ψb...stand for higher string modes,auxiliary?elds and their momenta.Solving forψa in terms ofψr up to the?rst non-linear order in the?elds we obtain
ψa=E(0)arψr+
1
2?A(2)
ab
E(1)brs+?A(3)ars+?A(3)abs E(0)br+?A(3)arb E(0)bs+?A(3)abc E(0)br E(0)cs=0
To solve these equations we need?A(2)ab to be invertible.The zero eigen modes of this operator arises from BRST exact states.So we restrict ourselves to states which are not BRST exact to ensure that the operator?A(2)ab is invertible.The linearized equation of motion forψr now becomes
?A(2)
sr
ψr+?A(2)sa E(0)arψr=0(4.3) From the gauge transformation ofψa at the linear order we obtain the relation
A(2)aα=E(0)ar A(2)rα(4.4) The gauge transformation ofψr up to the?rst nonlinear order can be written down as
δψr=A(2)rαλα+(A(3)rαs+A(3)rαa E(0)as)λαψs+O(ψ2)(4.5) It is clear from this gauge transformation,thatψr does not have the same gauge transformation property of the U(1)gauge?eld or that of the tachyon,because of the non-linear terms.The conventional low energy e?ective action of open bosonic string theory has the tachyon denoted byφ(k)and the gauge?eld,Aμ(k),The action should be invariant under the following transformation
δφ(k)=0δAμ=ikμη(k),(4.6) whereη(k)is the gauge transformation parameter.Let us call these low energy?elds asφi.The index i labels the tachyon,the gauge?eld and their momenta.Let the
gauge parameters of these?elds be labelled byηκ.It must be possible to relateψr to φi so thatφi has the gauge transformation given by(4.6).It also must be possible to relate the U(1)gauge invariance of the low energy action to a a string?eld theory gauge transformation.We show that this is possible in the next section.We use the method developed by[15]for?eld rede?nition in the closed non-polynomial string ?eld theory.We account for the elimination of all the massive and auxiliary?elds and the fact that we are dealing with open string?eld theory.We verify the consistency conditions for?eld rede?nitions in open string?eld theory.
5.Field rede?nition
In this section we?nd the conditions which need to be satis?ed to rede?ne the?elds that appear in the string?eld theory action to that which appear in the low energy e?ective action.Let us relateψr andφi by the following general formula up to the ?rst non linear order
ψr=C(0)riφi+
1
δψs +K ra
δS
Substituting(5.5)in the variation obtained from(5.1)we get
δψr=C(0)ri B(2)iκηκ+C(1)rijφi B(2)jκηκ(5.6) Now from(4.5),(2.8),(5.3),(5.4)we obtain
δψr=A(2)rαD(0)ακηκ+A(2)rαD(1)ακiηκφi+(A(3)rαs+A(3)rαa E(0)as)D(0)ακC(0)siηκφi(5.7) +K(1)rsκ(?A(2)st+?A(2)sa E(0)at)C(0)tiηκφi
From comparing(5.6)and(5.7)we get the following set of equations
C(0)ri B(2)iκη(ρ)κ=A(2)rαD(0)ακη(ρ)κ(5.8)
C(1)rijφ(m)
i B(2)jκη(ρ)κ=A(2)rαD(1)ακiη(ρ)κφ(m)
i
+(A(3)rαs+A(3)rαa E(0)as)D(0)ακC(0)siη(ρ)κφ(m)
i
(5.9)
+K(1)rsκ(?A(2)st+?A(2)sa E(0)at)C(0)tiη(ρ)κφ(m)
i
Where we have introduced a complete set of gauge transformations{η(ρ)κ}and a
complete set of?eld con?gurations{φ(m)
i}We have already seen in the previous section that we are able to?nd C(0)ri and D(0)ακsuch that(5.8)is satis?ed.It remains to be seen that if one can?nd C(1)rij,D(1)ακi and K(1)rsκsuch that(5.9)can be satis?ed. It is easy to extend these constraint equations to higher orders in?elds.There are obstructions to this.These obstructions can arise if under some conditions(5.9) reduces to equations involving known quantities which should be satis?ed identically. In the next section we will enumerate them.
6.Consistency conditions for?eld rede?nition
There are three consistency conditions in all for the(5.9).We will enumerate each of them in the following subsections.These are similar to the consistency conditions found in[15]for the closed string.The di?erence being that we have eliminated all the auxiliary and massive?elds using equations of motion and we are dealing with open string?eld theory.
6.1Condition A
Divide the complete set of?elds{φ(m)
i}to2sets{?φ(?m)i}and{ˉφ(ˉm)i}.The set{?φ(?m)i}
satis?es linearized equation of motion and the set{ˉφ(ˉm)
i}does not.
(?A(2)st+?A(2)sa E(0)at)C(0)ti?φ(?m)
i
=0for all s(6.1)
(?A(2)st+?A(2)sa E(0)at)C(0)tiˉφ(ˉm)
i=0for some s
Now divide the complete set of gauge transformations{η(ρ)κ}to sets{?η(?ρ)κ}and{ˉη(ˉρ)κ} such that
B(2)iκ?η(?ρ)κ=0for all i(6.2)
B(2)iκˉη(ˉρ)κ=0for some i
Thus{?η(?ρ)κ}are those gauge transformations for which there is no variation in the ?eldsφi.The index r is divided also to two sets into BRST invariant and non-invariant states given by
?Φc2;r|Q B=0 ˉΦc2;ˉr|Q B=0(6.3) Now restrict the indices r,m,ρin(5.9)to?r,?m and?ρ.The terms involving C(1) and K(1)drop out because of(6.1)and(6.2)respectively.The term involving D(1) drops o?because of
A(2)?rα=0(6.4) This follows from the de?nition of A(2)in(2.8)and(6.3).Thus we obtain the consistency condition
=0(6.5)
(A(3)rαs+A(3)rαa E(0)as)D(0)ακC(0)si?η(?ρ)κ?φ(?m)
i
This involves all known coe?cients and should be satis?ed identically.
6.2Condition B
The next obstruction arises because of C(1)rij is symmetric in i and j.Restrict the index r in(5.9)to be?r.The term involving D(1)drops out because of(6.4).Take φ(m)
to be of the form B(2)jκ′η(ρ′)κ′.Then term involving K(1)is given by
j
K(1)?r sκ(?A(2)st+?A(2)sa E(0)at)C(0)tiη(ρ)κB(2)jκ′η(ρ′)κ′=(6.6)
=K(1)?r sκ(?A(2)st+?A(2)sa E(0)at)η(ρ)κA(2)tαD(0)ακ′η(ρ′)κ′
=K(1)?r sκ(?A(2)st A(2)tα+?A(2)sa A(2)aα)η(ρ)κA(2)tαD(0)ακ′η(ρ′)κ′
=0
Where we have used(5.8)in the?rst step and(4.4)in the second step.We have
?A(2)
A(2)tα+?A(2)sa A(2)aα= I?Φ1;s Q2BΦ0;α =0(6.7)
st
To arrive at the above relation we have used completeness and Q2B=0.We also use the fact correlations vanish except when saturated by?elds with total ghost number three.The term involving K(1)also vanishes.Thus for this case from(5.9)we are left with
C(1)?r ij B(2)iκ′η(ρ′)κ′B(2)jκη(ρ)κ=(A(3)?rαs+A(3)?rαa E(0)as)D(0)ακC(0)si B(2)iκ′η(ρ)κη(ρ′)κ′(6.8) As C(1)?r ij is symmetric in i and j we have the following constraint
(A(3)?rαs+A(3)?rαa E(0)as)D(0)ακC(0)si B(2)iκ′(η(ρ)κη(ρ′)κ′?η(ρ′)κ′η(ρ)κ)=0(6.9) 6.3Condition C
We now?nd the third and?nal obstruction.This is due to the antisymmetry of K(1)rsαin r and s.In(5.8)choose the indexρto be?ρ.Then by(6.2)the term involving C(1)in(5.8)drops out.Now multiply the equation by(?A(2)rt′+E(0)ar?A(2)at′)C(0)t′jφ(m′)
j′
The term involving D(1)is then given by
A(2)rα(?A(2)rt′+E(0)ar?A(2)at′)C(0)t′j′φ(m′)
j′
D(1)ακi?η(?ρ)κφ(m)i(6.10)
=(A(2)rα?A(2)rt′+A(2)aα?A(2)at′)C(0)t′j′φ(m′)
j′
D(1)ακi?η(?ρ)κφ(m)i
Where we have used(4.4).Now using the fact that?A(2)is a symmetric matrix and (6.7),the term involving D(1)vanishes.Thus(5.9)becomes
(A(3)rαs+A(3)rαa E(0)as)(?A(2)rt′+E(0)ar?A(2)at′)C(0)t′j′D(0)ακC(0)si?η(?ρ)κφ(m′)
j′
φ(m)i+(6.11)
K(1)rsκ(?A(2)st+?A(2)sa E(0)at)C(0)ti?η(?ρ)κφ(m)
i (?A(2)rt′+?A(2)rb E(0)bt′)C(0)t′j?η(?ρ′)κ′φ(m′)
i′
=0
As K(1)rsαis antisymmetric in r and s,the symmetric component in r and s should be zero.Therefore we have the following constraint
(A(3)rαs+A(3)rαa E(0)as)(?A(2)rt′+E(0)ar?A(2)at′)C(0)t′j′D(0)ακC(0)si?η(?ρ)κ(φ(m′)
j′φ(m)i+φ(m)j′φ(m′)
i
)=0(6.12)
7.Veri?cation of the consistency conditions
In this section we show that each of the consistency conditions found in the previous section is satis?ed identically for the open superstring theory.
7.1Condition A
Let us?rst verify that equation(6.5)is satis?ed.De?ne
|?Ψ?(m)1 =C(0)si?φ(?m)i|Φ1;s +E(0)as C(0)si?φ(?m)i|Φ1;a (7.1) Now
Φ1;r|Q B|?Ψ?(m) =C(0)si?φ(?m)i(?A(2)rs+?A(2)ra E(0)as)=0(7.2) Where we have used(6.1).Also from(4.2)we have
Φ1;a|Q B|?Ψ?(m) =C(0)si?φ(?m)i(?A(2)as+?A(2)ab E(0)bs)=0(7.3) This implies that|?Ψ?(m) is a BRST invariant state.
Q B|?Ψ?(m) =0(7.4) Now de?ne
|?Λ?(ρ) =D(0)ακ?η(?ρ)κ|Φ0;α (7.5) We use(5.8)and(6.2)to show
Φc2;r|Q B|?Λ?(ρ) =D(0)ακ?η(?ρ)κ Φc2;r|Q B|Φ0α (7.6)
=D(0)ακ?η(?ρ)κA(2)rα
=C(0)ri B(2)iκ?η(?ρ)κ
=0
Also we have
Φc2;a|Q B|?Λ?(ρ) =D(0)ακ?η(?ρ)κ Φc2;a|Q B|Φ0α (7.7)
=D(0)ακ?η(?ρ)κA(2)aα
=C(0)ri B(2)iκ?η(?ρ)κE(0)ra
=0
Where we have used(4.4)and(6.2).From(7.6)and(7.7)we obtain that|?Λ?(ρ) is a BRST invariant state.
Q B|?Λ?(ρ) =0(7.8)
Using de?nitions(7.1),(7.5)and(2.9)the consistency condition(6.5)reduces to
f1??Φc?r f2??Λ?(ρ)f3??Ψ?(m) ? f1??Φc?r f2??Ψ?(m)f3??Λ?(ρ) =0(7.9) As all the?eld in(7.9)are BRST invariant?elds,if any of them is BRST exact,then the equation is automatically satis?ed.Therefore we have to look for BRST invariant but not BRST exact?elds.The linearized gauge transformation parameter for the tachyon?eld is zero.Thus by standard BRST analysis the only gauge transformation we have is that for the photon.This is given by|?Λ?(ρ) =ξ|0 The ghost number one
?elds in the formula(7.9)can by the on shell tachyon|?Ψ?(m) =c1|k with k2=1/2, the mass shell condition.Or it is the on shell gauge?eld given by|?Ψ?(m) =?μc1αμ?1|k with kμ?μ=0and k2=0.The conjugate?elds in(7.9)are the?elds conjugate to the on shell tachyon and the on shell gauge?eld.We can have Φc?r|= k|c?1c0with k2=1/2,or we have Φc?r|= k|c?1c0αμ?1aμwith aμkμ=0and k2=0.Evaluating the left hand side of(7.9)with any of these states it can be seen that it is zero.Thus the consistency condition(6.5)is identically satis?ed.
7.2Condition B
Now let us examine the obstruction(6.9).De?ne|Λ(ρ) =D(0)ακη(ρ)κ|Φ0;α .Then we have
C(0)si B(2)iκ′η(ρ′)κ′|Φ1;s +E(0)as C(0)si B(2)iκ′η(κ′)ρ′|Φ1;a (7.10) =A(2)sαD(0)ακ′η(ρ′)κ′|Φ1;s +E(0)as A(2)sαD(0)ακ′η(ρ′)κ′|Φ1;a
= Φc2;s|Q B|Φ0;α D(0)ακ′η(ρ′)κ′|Φ1;s + Φc2;a|Q B|Φ0α D(0)ακ′η(ρ′)κ′|Φ1;a
=D(0)ακ′η(ρ′)κ′Q B|Φ0;α
=Q B|Λ(ρ′)
Here we have used(5.8),(2.9)and completeness.Thus the consistency condition (6.9)can be written as
f1?Φc?r(0)f2?Λ(ρ)(0)f3?Q BΛ(ρ′)(0) ? f1?Φc?r(0)f2?Q BΛ(ρ′)(0)f3Λ(ρ)(0) ?(ρ?ρ′)
(7.11) Now,one can deform the contour of Q B in the?rst set of term so that it will act on Λρ.This is because|Φc?r is BRST invariant.Then,the?rst set of term cancel against the second set of terms.Thus the consistency condition(6.9)is satis?ed identically.
7.3Condition C
Finally we analyze the third consistency condition(6.12).De?ne
|Ψ(m) =C(0)tjφ(m)j|Φ1,t +E(0)at C(0)tjφ(m)j|Φ1,a =ψ(m)t|Ψ1;t +ψ(m)a|Ψ1;a (7.12) and|Λρ =D(0)ακη(ρ)κ|Φ0;α .Using these de?nitions(6.12)can be written as
A(3)rρs?A(2)rt′ψ(m′)
t′ψ(m)s+A(3)rρa?A(2)rt′ψ(m′)
t′
ψ(m)a+(7.13)
A(3)rρs?A(2)ra′ψ(m′)
a′ψ(m)s+A(3)rρa?A(2)ra′ψ(m′)
a′
ψ(m)a+(m?m′)=0
Where we have used the linearized equation of motion ofψ(m)
a
and(4.2).Now one can add to this equation the following equation
A(3)bρs?A(2)bt′ψ(m′)
t′ψ(m)s+A(3)bρa?A(2)bt′ψ(m′)
t′
ψ(m)a+(7.14)
A(3)bρs?A(2)ba′ψ(m′)
a′ψ(m)s+A(3)bρa?A(2)ba′ψ(m′)
a′
ψ(m)a+(m?m′)=0
This equation holds becauseψ(m)
a
satis?es linearized equation of motion.
?A(2) ar ψ(m)
r
+?A(2)abψ(m)
b
=0(7.15)
We are justi?ed in using the linear equations of motion because the terms in(6.12) are quadratic in the?elds.Adding(7.13)and(7.14)and using completeness,the constraint(6.12)becomes
f1?Q BΨ(m)(0)f2??Λ?(ρ)(0)f3?Ψ(m′)(0) ?(7.16) f1?Q BΨ(m)(0)f2?Ψ(m′)(0)f3??Λ?(ρ)(0) +(m?m′)=0 One can now deform the contour of Q B in the?rst set of terms so that it acts on|Ψ(m′) .It does not act on on|?Λ?(ρ) as it is BRST invariant.In deforming the contour one picks up minus sign,and another minus sign as Q B crossesΨ(m).Now these?rst set of terms cancel against the second set of terms.Thus(6.12)is satis?ed identically.
We have shown that all the consistency conditions found in the previous section for solving(5.9)are satis?ed identically.
8.Existence of?eld rede?nition
In this section we show that there exists C(1)and D(1)which solve(5.9).This establishes that there exists a?eld rede?nition such that in the new set of?elds the tachyon and the U(1)gauge?eld corresponds to the?elds seen in the low energy e?ective action to the?rst non-linear order in the?elds.This also establishes the existence of a string gauge transformations which corresponds to the U(1)gauge transformation of the low energy e?ective action.
In this section we will closely follow[15].We outline the steps involved for completeness.We divide the situations to three cases
8.1Case A:r∈ˉr
Let us consider the case when r belongs to the typeˉr.This implies that A(2)ˉrαhas a right inverse Mαˉs
A(2)ˉrαMαˉs=δˉrˉs(8.1) Therefore we can solve(5.9)for this case by choosing D(1)given by
D(1)ακiηκφi=Mαˉr C(1)ˉr ijφi B(2)jκηκ?A(2)ˉrβD(1)βκiηκφi(8.2)?(A(3)ˉrβs+A(3)rβa E(0)as)D(0)βκC(0)siηκφi+K(1)ˉr sκ(?A(2)st+?A(2)sa E(0)at)C(0)tiη(ρ)κφ(m)i 8.2Case B:r∈?r,ρ∈ˉr
In this case the term involving D(1)in(5.9)vanishes because of(6.3).Since the indexρis in the setˉρthe matrix S iˉρ=B(2)iκˉη(ˉρ)κhas a left inverse Nˉρj which satis?es
Nˉρi S iˉρ′=δρˉρ′(8.3)
We can now solve(5.9)by choosing C(1)given by
C(1)?r ij= (A(3)?rαs+A(3)?αa E(0)as)D(0)ακC(0)sj+K(1)?r sκ(?A(2)st+?A(2)sa E(0)at)C(0)tj ˉη(ˉρ)κNˉρi(8.4) It can be shown[15]that one can choose a basis for the?elds{φi}such that the C(1) obtained from the above equation is symmetric in i and j.
8.3Case C:r∈?r,ρ∈?ρ,m∈ˉm
The case r∈?r,ρ∈?ρand m∈?m is already covered in(6.5).We have shown in such a case the consistency condition is satis?ed identically.So,the case which remains is r∈?r,ρ∈?ρand m∈ˉm.In this case the terms involving C(1)and D(1)drop from (5.9)due to(6.2)and(6.3)respectively.The matrix T sˉm given by
(8.5)
T sˉm=(?A(2)st+?A(2)sa E(0)at)C(0)tjˉφ(ˉm)
i
Then the matrix T sˉm has a left inverse Uˉmr given by
Uˉmr T rˉn=δˉmˉn(8.6)
Then we can solve(5.9)by choosing
K(1)?r tκ?η(?ρ)κ=?(A(3)?rαs+A(3)?αa E(0)as)D(0)ακC(0)sj?η(?ρ)κˉφ(ˉm)
Uˉmt(8.7)
i
Notice that only the projection of K(1)?r tκonto the space spanned by?η(?ρ)κis determined. The other components can be chosen to be arbitrary.It can also be shown that the K(1)rsκcan be chosen in such a way that it is antisymmetric in r and s[15].
This completes the proof of the existence of?eld rede?nition of the tachyon and the gauge?eld so that they can be identi?ed with the?elds that appear in the low energy e?ective action.This also determines a string gauge transformation,λαwhich corresponds to the U(1)gauge transformation,ηκof the low energy?elds.
9.The transverse photon at the tachyon condensate
In this section we demonstrate explictly that?eld rede?nition is important in the relation of the Dirac-Born-Infeld action and the low energy e?ective action derived from Witten’s String?eld theory.To do this we evaluate the kinetic term for the transverse photon A Tμ,to level4in the tachyon condensate.We work in the Feynman-Siegel gauge and withα′=1.The quadriatic terms in the action involving the transverse photon is given by3
S= dkAμ(?k)Aμ(k) k23/4)R(k) (9.1)
Where R (k )is given by R (k )=3√4t ?(493?43k 2)v 13+113
u (9.2)+(15793?883k 2)A +(?5393+443
k 2)F +203D +193E ?203
C +(7853?5203k 2+643
k 4)B Here the variables t,u,v,A,B,C,D,E,F stand for the ?elds as de?ned in [3].It is now easy to extract the kinetic term.It is given by
S KE =
dk T A μ(?k )k 2A μ(k )(9.3)where
T =1
3312√12√162√324√27
√108√81√54√3√
81√81√81√(1,2)
.189937.9%(2,4)
.186137.2%(4,8).181436.3%
9.1Vector excitations at the tachyon condensate
To?nd the physcial excitations of the transverse photon we evaluate the physical poles in the two point function of the transverse photon.This is done by evaluting the zeros of the funtion
f(k0)=?k20
3/4)R(k0)(9.6)
Here we have substitued k2=?k20in the two point funtion of the transverse photon from(9.1).Poles in f(k0)represent masses of physical excitations.We?nd the following results
Level Zeros of the two point function
(9.7)
We note that the location of the pole in level4decreases as compared to level2. To?nd the physical excitations at the tachyon condensate it is not only su?cient to look at the poles in the transverse photon two-point function.We also need to ?nd the zeros in the determinant of the?uctuation matrix of all the vector particles. Above level(2,4),vector?elds at level3mix with the transverse photon.
We will now show that the pole found at level(2,4)exists even when we consider all the vector?elds at level(3,6).Consider the action including all the vector?elds till level3.We write the quadriatic terms in the action as
S= dkAμ(?k)Aμ(k) k23/4)R(k) (9.8)
+2Aμ(?k)αi(k)V iμ(k)+V iμ(?k)M ij(k)V jμ(k)
Where V iμare all the other vector?elds till level3.These are
V1μαμ?3c1|0 +V2μαμ?1b?1c?1c1|0 +V3μαμ?1L?2c1|0 (9.9) Evaluating the zeros of the determinant of the?uctation matrix is equivalent to elimating the?elds V iμusing their equations of motion and then evaulating the zeros of the two point funtion for the transverse photon.This is given by
S= dkAμ(?k)Aμ(k) k23/4)R(k)(9.10)
?αi(k)M?1ij(k)αj(k)
峰的高低说明了信号的强弱，宽窄表示的是分离效果，可以对比板胶电泳的亮度和条带的宽窄。重叠则说明了样品中有长度相同但是序列不同的片段。 Q-1.为什么提供新鲜的菌液？如何提供新鲜的菌液？返回顶端 A-1.首先，新鲜的菌液易于培养，可以获得更多的DNA，同时最大限度地保证菌种的纯度。如果您提供新鲜菌液，用封口膜封口以免泄露；也可以将培养好的 4~5ml菌液沉淀下来，倒去上清液以方便邮寄。同时邮寄时最好用盒子以免邮寄过程中压破。 Q-2.DNA测序样品用什么溶液溶解比较好?返回顶端 A-2.溶解DNA测序样品时，用灭菌蒸馏水溶解最好。DNA的测序反应也是Taq酶的聚合反应，需要一个最佳的酶反应条件。如果DNA用缓冲液溶解后，在进行测序反应时，DNA溶液中的缓冲液组份会影响测序反应的体系条件，造成Taq酶的聚合性能下降。 有很多客户在溶解DNA测序样品时使用TE Buffer。的确，TE Buffer能增加DNA样品保存期间的稳定性，并且TE Buffer对DNA测序反应的影响也较小，但根据我们的经验，我们还是推荐使用灭菌蒸馏水来溶解DNA测序样品。 Q-3.提供DNA测序样品时，提供何种形态的比较好?返回顶端 A-3.我们推荐客户提供菌体，由我们来提取质粒，这样DNA样品比较稳定。如果您可以提供DNA样品，我们也很欢迎，但一定要注意样品纯度和数量。如果提供的DNA量不够，我们就需要对质粒进行转化，此时需收取转化费。有些质粒提取法提取的DNA质量很好，象TaKaRa、Qiagen、Promega的质粒制备试剂盒等。提供的测序样品为PCR产物时，特别需要注意DNA的纯度和数量。PCR产物必须进行切胶回收，否则无法得到良好的测序效果。 有关DNA测序样品的详细情况请严格参照“测序样品的提供”部分的说明。 Q-4.提供的测序样品为菌体时，以什么形态提供为好?返回顶端 A-4.一般，菌体的形态有：平板培养菌、穿刺培养菌，甘油保存菌或新鲜菌液等。我们提倡寄送穿刺培养菌或新鲜菌液。 平板培养菌运送特别不方便，我们收到的一些平板培养菌的培养皿在运送过程中常常已经破碎，面目全非，需要用户重新寄样。这样既误时间，又浪费客户的样品。一旦是客户非常重要的样品时，其后果更不可设想。而甘油保存菌则容易污染。 制作穿刺菌时，可在1.5 ml的Tube管中加入琼脂培养基，把菌体用牙签穿刺于琼脂培养基（固体）中,37℃培养一个晚上后便可使用。穿刺培养菌在4℃下可保存数个月，并且不容易污染，便于运送。
学习 通常一份测序结果图由红、黑、绿和蓝色测序峰组成，代表不同的碱基序列。测序图的两端（本图原图的后半段被剪切掉了）大约50个碱基的测序图部分通常杂质的干扰较大，无法判读，这是正常现象。这也提醒我们在做引物设计时，要避免将所研究的位点离PCR序列的两端太近（通常要大于50个碱基距离），以免测序后难以分析比对。 我的课题是研究基因多态性的，因此下面要介绍的内容也主要以判读测序图中的等位基因突变位点为主。 实际上，要在一份测序图中找到真正确实的等位基因多态位点并不是一件容易的事情。由于临床专业的研究生，这些东西是没人带的，只好自己研究。开始时大概的知道等位基因位点在假如在测序图上出现像套叠的两个峰，就是杂合子位点。实际比对了数千份序列后才知道，情况并非那么简单，下面测序图中标出的两
个套峰均不是杂合子位点，如图并说明如下： 说明：第一组套峰，两峰的轴线并不在同一位置，左侧的T峰是干扰峰；第二组套峰，虽两峰轴线位置相同，但两峰的位置太靠近了，不是杂合子峰，蓝色的C峰是干扰峰通常的杂合子峰由一高一略低的两个轴线相同的峰组成，此处的序列被机器误判为“C”，实际的序列应为“A”，通常一个高大碱基峰的前面1～2个位点很容易产生一个相同碱基的干扰峰，峰的高度大约是高大碱基峰的1/2，离得越近受干扰越大。一个摸索出来的规律是：主峰通常在干扰峰的右侧，干扰峰并不一定比主峰低。最关键的一点是一定要拿疑似为杂合子峰的测序图位点与测序结果的文本序列和基因库中的比对结果相比较；一个位点的多个样本相比较；你得出的该位点的突变率与权威文献或数据库中的突变率相比较。通常，对于一个疑似突变位点来说，即使是国际上权威组织大样本的测序结果中都没有报道的话，那么单纯通过测序结果就判定它是突变点，是并不严谨的，因一份PCR产物中各个碱基的实际含量并不相同，很难避免不产生误差的。对于一个未知
各种单位换算及公式 长度单位面积单位 1 in = 25.4 mm 1 in 2 = 6.45 cm2 1 ft = 0.3048 m 1 ft2 = 0.09 3 m2 1 micron = 0.001 mm 体积单位 1 litre = 0.001 m3 1 cu.ft. = 0.0283 m3 1 cu.in. = 16.39 cm3 1 fluid oz.(imp) = 28.41 mL 1 fluid oz.(us) = 29.57 mL 1 gal(imp) = 4.546 L 1 gal(us) = 3.79 L 温度单位 (°F-32)X5/9=℃K-273.15 = ℃ 功及能量单位 1 Nm = 1 J 1 kgm = 9.807 J 1 kW/hr = 3.6 MJ 1 lbft = 1.356 J 功率单位 1 Nm/sec = 1 W 1 lbft/sec = 1.356 W 1 kgm/sec = 9.807 W 1 Joule/sec = 1 W 1 H.P.(imp) = 745.7 W 质量单位 1 lb = 453.6 g 1 tonne = 1000 kg 1 ton(imp) = 1016 kg 1 ton(us) = 907. 2 kg
流量计算公式 Q = Cv值X 984 = Kv值X 1100 Cv = So ÷ 18 力单位 1 kgf = 9.81 N 1 lbf = 4.45 N 1 kp(kilopound) = 9.81 N 1 poundal = 138.3 mN 1 ton force = 9.964 kM 力矩单位 1 kgm = 9.807 Nm 1 ft. poundal = 0.0421 Nm 1 in lb = 0.113 Nm 1 ft lb = 1.356 Nm 压力单位 1 psi = 6.89 kPa 1 kgf/cm 2 = 98.07 kPa 1 bar = 100 kPa 1 bar = 14.5 psi 1 mm mercury = 133.3 Pa 1 in mercury = 3.39 kPa 1 Torr = 133.3 Pa 1 ft water = 0.0298 bar 1 bar = 3.33 ft water 1 atmosphere = 101.3 kPa 1 cm water = 97.89 Pa 1 in water = 248.64 Pa 换算表 1psi=6.895kPa=0.07kg/cm2=0.06895bar=0.0703atm 1standard atmosphere=14.7psi=101.3kPa=1.01325bar 1kgf/cm2 = 98.07kPa=14.22psi = 28.96ins mercury 1m3 = 1000000cm3 1cu ft/min = 28.3 l/min
1.为什么TALEN和CAS9技术移码敲除项目需通过测序判定基因型？ TALEN和CAS9技术是通过特异识别和核酸酶切割，使DNA双链断开，机体启动DNA损伤修复机制，在非同源末端连接修复过程中，碱基的随机增减造成目标基因功能缺失。这一修复过程通常会产生1-30个左右的碱基缺失，PCR后凝胶电泳无法将碱基缺失链和wt链区分开，所以只能通过测序判断。 2.如何判断TALEN和CAS9技术移码敲除项目的基因型？ a. 野生型或基因敲除的纯合子基因型的判定： 1）如下图所示,只有单峰的为野生型或基因敲除的纯合子。 2）将只有单峰的序列文件与wt序列比对，可判定野生型或基因敲除的纯合子基因型（网站或生物软件比对均可）. 如下图的对比结果为野生型：
如下图的比对结果为缺失一个C的纯合子 b.杂合子基因型的判定： 1).如下图所示，测序图谱中出现叠峰的为杂合子 若将叠峰的序列文件直接与wt序列比对，叠峰后的序列会对应不上(因为叠峰的序列只显示了信号强一点的那个碱基，实际上每个叠峰对应两个碱基)，如下图所示：
所以不能通过直接比对来分析，需通过峰图分析，在峰图中峰的颜色与碱基对应关系如下：A：绿色 T：红色 C：蓝色 G：黑色 2)将wt的峰图与杂合子的峰图用软件同时打开，从出现叠峰的位置开始分析（杂合子的一条链是wt链，一条链是缺失链）： 上图第一个峰图是野生型的序列，与第二个峰图叠峰开始对应的序列为： Gttaact ccgagcagcaaagaaatgatgtccc 上图第二个峰图是杂合子的测序峰图，从叠峰位置开始的序列为： Gttaact ccgagcagcaaagaaatgatgtccc（wt链） Ccgagcagcaaagaaatgatgtcccaagcctt（缺失链） 由此可判定为该杂合子的基因型为缺失gttaact（-7）
DNA测序结果分析比对（实例） 关键词：dna测序结果2013-08-22 11:59来源：互联网点击次数：14423 从测序公司得到的一份DNA测序结果通常包含.seq格式的测序结果序列文本和.ab1格式的测序图两个文件，下面是一份测序结果的实例： CYP3A4-E1-1-1(E1B).ab1 CYP3A4-E1-1-1(E1B).seq .seq文件可以用系统自带的记事本程序打开，.ab1文件需要用专门的软件打开。软件名称：Chromas 软件Chromas下载 .seq文件打开后如下图： .ab1文件打开后如下图： 通常一份测序结果图由红、黑、绿和蓝色测序峰组成，代表不同的碱基序列。测序图的两端（下图原图的后半段被剪切掉了）大约50个碱
基的测序图部分通常杂质的干扰较大，无法判读，这是正常现象。这也提醒我们在做引物设计时，要避免将所研究的位点离PCR序列的两端太近（通常要大于50个碱基距离），以免测序后难以分析比对。 我的课题是研究基因多态性的，因此下面要介绍的内容也主要以判读测序图中的等位基因突变位点为主。 实际上，要在一份测序图中找到真正确实的等位基因多态位点并不是一件容易的事情。一般认为等位基因位点假如在测序图上出现像套叠的两个峰，就是杂合子位点。实际比对后才知道，情况并非那么简单，下面测序图中标出的两个套峰均不是杂合子位点，如图并说明如下：
说明： 第一组套峰，两峰的轴线并不在同一位置，左侧的T峰是干扰峰；第二组套峰，虽两峰轴线位置相同，但两峰的位置太靠近了，不是杂合子峰，蓝色的C峰是干扰峰通常的杂合子峰由一高一略低的两个轴线相同的峰组成，此处的序列被机器误判为“C”，实际的序列应为“A”，通常一个高大碱基峰的前面 1~2个位点很容易产生一个相同碱基的干扰峰，峰的高度大约是高大碱基峰的1/2，离得越近受干扰越大。 一个摸索出来的规律是：主峰通常在干扰峰的右侧，干扰峰并不一定比主峰低。最关键的一点是一定要拿疑似为杂合子峰的测序图位点与测序结果的文本序列和基因库中的比对结果相比较；一个位点的多个样本相比较；你得出的该位点的突变率与权威文献或数据库中的突变率相比较。 通常，对于一个疑似突变位点来说，即使是国际上权威组织大样本的测序结果中都没有报道的话，那么单纯通过测序结果就判定它是突变点，是并不严谨的，因一份 PCR产物中各个碱基的实际含量并不相同，很难避免不产生误差的。对于一个未知突变位点的发现，通常还需要用到更精确的酶切技术。 (责任编辑：大汉昆仑王)
测序结果的判读 测序结果为.abi格式，可用软件chrosmas打开，一种颜色的峰代表一个碱基，峰的高低表信号的强弱。一个正常的N表示机器没法判读是哪种碱基，原因是：杂峰的信号高于机器默认的值，机器会认为该处有两个峰，因此不能判断确定是哪个峰，需要人工判读。以下三种情况会出现N：有杂合子，有杂峰，反应已结束。
原因：测序产物纯化不够 注意：染料峰位于序列的前100 碱基以内;酒精峰位于序列的220 ~ 320 碱基之间
产生的原因是样品或毛细管内有灰尘等固体小颗粒 原因：测序反应失败。 解决办法：改进条件，重做反应。注意两个关键因素：引物与模板之间的比例：3.2 pmol: 200 ng。模板DNA 的纯度和用量：1.6 ~ 2.0
原因：残余的Dye 太多，纯化不够。有测序反应，但效率低下信号太弱 解决办法：纯化充分。避开引物峰，确定新的分析起点 1、PCR产物测序时出现重叠峰 问题图1(模板中有碱基缺失，往往是单一位点(1-1)或两个位点(1-2)碱基缺失导致测序结果移码) 解决方法：将PCR产物克隆到质粒(如T载体)中挑单克隆测序，或将PCR产物进行PAGE 纯化(至少琼脂糖充分电泳后切胶纯化)后再进行测序。 问题图2(PCR产物不纯，含部分序列一致的两种以上的片段，长度不一)
解决方法：主要原因是PCR产物没有纯化，含有部分序列一致的两种以上长度不一的片段，将PCR产物进行PAGE纯化(至少琼脂糖充分电泳后切胶纯化)后再进行测序，便可解决。 问题图3(测序引物有碱基缺失) 测序引物有碱基缺失(一般是引物的5'端缺失)，和模板的碱基缺失即图1有些类似，所不同的是模板碱基缺失一般是在一段正常测序序列后才出现移码，而引物碱基缺失的话，则从测序一开始就出现移码，表面在图形上便是一开始就是严重的峰形重叠。 解决方法：重新合成引物，或将引物进行PAGE纯化 2、克隆测序时出现峰形重叠
工程上常用的是兆帕（MPa）：1MPa=1000000Pa。 1个标准大气压力=1.00336×0.098MPa=0.10108MPa≈0.1Mpa。 1bar=0.1MPa 压力的法定单位是帕斯卡（Pa）：1Pa=1N/㎡（牛顿/平方米）。 压力单位换算： 1MPa=1000kPa 1kPa=10mbar=101.9716 mmH2O = 4.01463imH2O 10mWC＝1bar＝100kPa bar 巴= 0.987 大气压= 1.02 千克/平方厘米= 100 千帕= 14.5 磅/平方英寸 PSI英文全称为Pounds per square inch。P是磅pound，S是平方square，I是英寸inch。把所有的单位换成公制单位就可以算出：1bar≈14.5psi 1psi=6.895kPa=0.06895bar
1兆帕(MPa)=145磅/英寸2(psi)=10.2千克/厘米2(kg/cm2)=10巴(bar)=9.8大气压(atm) 1磅/英寸2(psi)=0.006895兆帕(MPa)=0.0703千克/厘米2(kg/cm2)=0.0689巴(bar)=0.068大气压(atm) 1巴(bar)=0.1兆帕(MPa)=14.503磅/英寸2(psi)=1.0197千克/厘米 2(kg/cm2)=0.987大气压(atm) 1大气压(atm)=0.101325兆帕(MPa)=14.696磅/英寸2(psi)=1.0333千克/厘米2(kg/cm2)=1.0133巴(bar) ------------------------------------------------------------------------------------- 压力单位换算方法 1. 1atm=0.1MPa=100KPa=1公斤=1bar=10米水柱＝14.5PSI 2.1KPa=0.01公斤 =0.01bar=10mbar=7.5mmHg=0.3inHg=7.5torr=100mmH2O=4inH2O 3. 1MPa=1N/mm2 14.5psi=0.1Mpa 1bar=0.1Mpa 30psi=0.21mpa,7bar=0.7mpa 现将单位的换算转摘如下： Bar---国际标准组织定义的压力单位。 1 bar=100,000Pa 1Pa=F/A, Pa: 压力单位, 1Pa=1 N/㎡ F : 力, 单位为牛顿(N) A: 面积, 单位为㎡ 1bar=100,000Pa=100Kpa 1 atm=101,325N/㎡=101,325Pa 所以,bar是一种表压力(gauge pressure)的称呼。
AWG 标准线径对照表 线径的粗细是以号数(xxAWG)来表示的，数目越小表示线径愈粗，所能承载的电流就越大，反之则线径越细，耐电流量越小。例如说：12号的耐电流量是20安培，最大承受功率是2200瓦，而18号线的耐电流量则是7安培，最大承受功率是770瓦。 为什么AWG号数越小直径反而越大？如这么解释你就会明白，固定的截面积下能塞相同的AWG线的数量，如11#AWG号数可塞11根而15#AWG号数可塞15根，自然的15#AWG的单位线径就较小。 美规线径值单一导体或群导体【各正值或负值】的线径值(Gauge)是以圆或平方厘米(mm2) 量测而得，平方厘米不常用在量测线径值，由于牵涉到不正确，因一般大部份的导体形体，包含长方形及其他怪异形状。因此我们拿全部的量测以圆平方厘米(c/m)为参考值 群导体计算的方法或公式： 加上单一导体的线径值总和，并比较上表求得。如果值落入两者之间，取比较少的值。 40股群导体线的线径值为，如每一芯为24 Guage = 40 x 405 c/m = 16，200 c/m = 9 AWG(得出值落入12960c/m和16440c/m之间) 快速求得线径值的方法： 两条(AWG)相加时，该单一线径值减3. ex. 2 x 18 AWG = (18-3=) 15 AWG 三条(AWG)相加时，该单一线径值减5. ex. 3 x 24 AWG = (24-5=) 19 AWG 四条(AWG)相加时，该单一线径值减6. ex. 4 x 10 AWG = (10-6=) 04 AWG 请记得“快速求得线径值的方法”一些案例也许边际会不正确，只采用此方式为大原则 AWG 标准线径规格对照表
各种单位换算及公式
各种单位换算及公式 长度单位面积单位 1 in = 25.4 mm 1 in 2 = 6.45 cm2 1 ft = 0.3048 m 1 ft 2 = 0.09 3 m2 1 micro n = 0.001 mm 体积单位 1 litre = 0.001 m3 1 cu.ft. = 0.0283 m3 1 cu.i n. = 16.39 cm3 1 fluid oz. (imp) = 28.41 mL 1 fluid oz. (us) = 29.57 mL 1 gal(imp) = 4.546 L 1 gal(us) = 3.79 L 温度单位 (°-32)X5/9= C K-273.15 = C 功及能量单位 1 Nm = 1 J 1 kgm = 9.807 J 1 kW/hr = 3.6 MJ 1 Ibft = 1.356 J 功率单位 1 Nm/sec = 1 W 1 lbft/sec = 1.356 W 1 kgm/sec = 9.807 W 1 Joule/sec = 1 W 1 H.P.(imp) = 745.7 W
质量单位 1 to nne = 1000 kg 1 lb = 453.6 g
流量计算公式 Q = Cv 值X 984 = Kv 值X 1100 Cv = So 48 力单位 1 kgf = 9.81 N 1 Ibf = 4.45 N 1 kp(kilopou nd) = 9.81 N 1 pou ndal = 138.3 mN 1 ton force = 9.964 kM 力矩单位 1 kgm = 9.807 Nm 1 ft. poun dal = 0.0421 Nm 1 in lb = 0.113 Nm 1 ft lb = 1.356 Nm 压力单位 1 psi = 6.89 kPa 1 kgf/cm 2 = 98.07 kPa 1 bar = 100 kPa 1 bar = 14.5 psi 1 mm mercury =133.3 Pa 1 in mercury =3.39 kPa 1 Torr = 133.3 Pa 1 ft water = 0.0298 bar 1 bar = 3.33 ft water 1 atmosphere = 101.3 kPa 1 cm water = 97.89 Pa 1 in water = 248.64 Pa 换算表 1psi=6.895kPa=0.07kg/cm2=0.06895bar=0.0703atm 1sta ndard atmosphere=14.7psi=101.3kPa=1.01325bar 1kgf/cm2 = 98.07kPa=14.22psi = 28.96i ns mercury 1m3 = 1000000cm3
常用线规号码与线径对照表
线规SWG BWG BG AWG 号码英寸毫米英寸毫米英寸毫米英寸毫米 7／0 6／0 5／0 4／0 3／0 2／0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 0．500 0．464 0．432 0．400 0．372 0．348 0．324 0．300 0．276 0．252 0．232 0．212 0．192 0．176 0．160 0．144 0．128 0．116 0．104 0．092 0．080 0．072 0．064 0．056 0．048 0．040 0．036 0．032 0．0280 0．0240 0．0220 0．0200 0．0180 12．700 11．786 10．973 10．160 9．449 8．839 8．230 7．620 7．010 6．401 5．893 5．385 4．877 4．470 4．046 3．658 3．251 2．946 2．642 2．337 2．032 1．829 1．626 1．422 1．219 1．016 0．914 0．813 0．711 0．610 0．559 0．508 0．457 -- -- 0．500 0．454 0．425 0．330 0．340 0．300 0．284 0．259 0．238 0．220 0．203 0．180 0．165 0．148 0．134 0．120 0．109 0．095 0．083 0．072 0．065 0．058 0．049 0．042 0．035 0．032 0．028 0．025 0．022 0．020 0．018 -- -- 12．700 11．532 10．795 9．652 8．639 7．620 7．214 6．579 6．045 5．588 5．156 4．572 4．191 3．759 3．404 3．048 2．769 2．413 2．108 1．829 1．651 1．473 1．245 1．067 0．839 0．813 0．711 0．635 0．559 0．508 0．457 0．6666 0．6250 0．5883 0．5416 0．5000 0．1152 0．3954 0．3532 0．3147 0．2804 0．2500 0．2225 0．1981 0．1764 0．1570 0．1398 0．1250 0．1313 0．0991 0．0882 0．0785 0．0699 0．0625 0．0556 0．0495 0．0440 0．0392 0．0349 0．03125 0．02782 0．02476 0．02204 0．01961 16．932 15．875 14．943 13．757 12．700 11．308 10．069 8．971 7．993 7．122 6．350 5．652 5．032 4．481 3．988 3．551 3．175 2．827 2．517 2．240 1．994 1．775 1．588 1．412 1．257 1．118 0．996 0．887 0．794 0．707 0．629 0．560 0．498 -- 0．5800 0．5165 0．4600 0．4096 0．3648 0．3249 0．2893 0．2576 0．2294 0．2043 0．1819 0．1620 0．1443 0．1285 0．1144 0．1019 0．0907 0。0808 0．0720 0．0648 0．0571 0．0508 0．0453 0．0403 0．0359 0．0320 0．0285 0．02535 0．02010 0．01790 0．01594 0．01420 -- 14．732 13．119 11．684 10．404 9．266 8．252 7．348 6．544 5．827 5．189 4．621 4．115 3．665 3．264 2．906 2．588 2．305 2．053 1．828 1．628 1．450 1．291 1．150 1．024 0．912 0．812 0．723 0．644 0．573 0．511 0．455 0．405 常用线规号码与线径对照表
American Wire Gauge AWG mm2 42 0.003 1/0.06 41 0.004 1/0.07 40 0.005 1/0.08 38 0.008 1/0.10 36 0.013 1/0.127 34 0.020 1/0.16 7/0.06 32 0.032 1/0.203 7/0.08 8/0.07 11/0.06 30 0.051 1/0.26 7/0.10 11/0.08 14/0.07 19/0.06 28 0.081 1/0.32 7/0.12 11/0.10 16/0.08 21/0.07 28/0.06 26 0.129 1/0.40 7/0.16 9/0.14 11/0.12 16/0.10 25/0.08 33/0.07 45/0.06 24 0.205 1/0.50 7/0.20 14/0.14 19/0.12 26/0.10 41/0.08 53/0.07 73/0.06 22 0.326 1/0.65 7/0.26 11/0.203 13/0.18 17/0.16 22/0.14 29/0.12 42/0.10 65/0.08 20 0.518 1/0.80 7/0.30 10/0.26 12/0.23 16/0.203 20/0.18 26/0.16 34/0.14 46/0.12 66/0.10 18 0.823 1/1.02 7/0.40 10/0.32 16/0.26 20/0.23 26/0.203 33/0.18 41/0.16 54/0.14 73/0.12 65/0.127 104/0.10 16 1.309 1/1.29 7/0.50 11/0.40 17/0.32 25/0.26 32/0.23 41/0.203 52/0.18 65/0.16 85/0.14 119/0.12 165/0.10 14 2.081 1/1.63 11/0.50 17/0.40 26/0.32 40/0.26 50/0.23 65/0.203 82/0.18 103/0.16 135/0.14 183/0.12 264/0.10 12 3.309 1/2.05 17/0.50 27/0.40 41/0.32 54/0.28 80/0.23 102/0.203 130/0.18 164/0.16 10 5.261 1/2.60 27/0.50 42/0.40 65/0.32 99/0.26 126/0.23 162/0.203 206/0.18 261/0.16 8 8.366 1/3.26 26/0.65 67/0.40 104/0.32 157/0.26 6 13.30 1/4.12 27/0.80 40/0.65 68/0.50 105/0.40 165/0.32 4 21.1 5 1/5.20 26/1.02 42/0.80 64/0.65 107/0.50 168/0.40 2 33.6 3 1/6.54 4 26/1.29 42/1.02 67/0.80 101/0.6 5 171/0.50 0 53.48 1/8.254 26/1.63 41/1.29 66/1.02 106/0.80 161/0.65 1. 基准线规直径：直径5 mil（0.005 inch）为36 AWG： 2. 相邻线号之间以几何级数计算：见右框图中公式。 例如：d18 = d36 × r (36-18) = 5 × 8.06053 = 40.3mils = 1.024mm d n = d 36× r (36 - n)( mil ) = 0.127 r(36 - n)( mm ) 其中，r = (460/5) 1/39 = 1.1229322
测序峰图 测序方法 现代DNA序列测序可分为一二三代。 一代测序 第一代测序技术包括链终止法和化学降解法，其主要特点是以待测DNA为模板，根据碱 基互补规则采用DNA聚合酶体外合成新链。由于新链中渗入了带有标记的碱基类似物，可用于制 备末端带有标记的DNA单链。这些末端带有标记的DNA单链是一个群体，在凝胶电泳时可以形 成彼此仅差一个碱基的梯形条带。根据末端碱基的特有标记可以读取待测DNA的序列组成。 链终止法（Sanger法） 反应分别在4个试管中进行，每一试管中都含有： 1.待测的DNA单链片段，作为模板； 2.DNA聚合酶； 3.序列已知且与模板3'端互补的引物； 4.四种脱氧核苷三磷酸(dATP、dGTP、dCTP和dTTP) ； 5.四种双脱氧核苷酸中的一种，作为DNA链合成的末端终止剂。 在适当条件下进行互补链的合成，由于新掺入的核苷酸只能同新生链3’端的-OH连接，而2’, 3’ -双脱氧核苷三磷酸核糖基的2’, 3’位-OH的氧已脱掉，失去了和后来的核苷酸连接的可能性，这时碱基的掺入就有两种可能性： 1.dNTP掺入新生链，使链继续延伸。 2.2’, 3’ -ddNTP掺入，使新生链的合成终止。 由于两者掺入的几率不同，就形成一套长度不一的DNA片段。最后通过A、T、G和C四组反应的凝胶电泳放射自显影图谱，即可读出所测样品互补链的核苷酸序列。 i ii 要求测序时使用的DNA聚合酶不能有5’-3’外切酶活性（可将新合成DNA链的5’-末端除去核苷酸而改变链的长度）、3’-5’外切酶活性（可将错配的碱基除去，再补上正确配对的核苷酸）。这两种活性会产生干扰条带。
Gauge 板材換算
钢材理论重量计算 钢材理论重量计算的计量单位为公斤（kg ）。其基本公式为： 钢的密度为：7.85g/cm3 ，各种钢材理论重量计算公式如下： 圆钢盘条（kg/m）W= 0.006165 ×d×d d = 直径mm 直径100 mm 的圆钢，求每m 重量。每m 重量= 0.006165 ×1002=61.65kg 螺纹钢（kg/m）W= 0.00617 ×d×d d= 断面直径mm 断面直径为12 mm 的螺纹钢，求每m 重量。每m 重量=0.00617 ×12 2=0.89kg 等边角钢（kg/m）= 0.00785 ×[d （2b – d ）+0.215 （R2 –2r 2 ）] b= 边宽 d= 边厚R= 内弧半径r= 端弧半径求20 mm ×4mm 等边角钢的每m 重量。从冶金产品目录中查出4mm ×20 mm 等边角钢的R 为3.5 ，r 为1.2 ，则每m 重量= 0.00785 ×[4 ×（2 ×20 – 4 ）+0.215 ×（3.52 – 2 ×1.2 2 ）]=1.15kg 不等边角钢（kg/m）W= 0.00785 ×[d （B+b –d ）+0.215 （R2 – 2 r 2 ）] B= 长边宽 b= 短边宽d= 边厚R= 内弧半径r= 端弧半径求30 mm ×20mm ×4mm 不等边角钢的每m 重量。从冶金产品目录中查出30 ×20 ×4 不等边角钢的R 为3.5 ，r 为1.2 ，则每m 重量= 0.00785 ×[4 ×（30+20 –4 ）+0.215 ×（3.52 –2 ×1.2 2 ）]=1.46kg 钢板（kg/m2）W= 7.85 ×d d= 厚厚度4mm 的钢板，求每m2 重量。每m2 重量=7.85 ×4=31.4kg 钢管（包括无缝钢管及焊接钢管（kg/m）W= 0.02466 ×S （D –S ）D= 外径 S= 壁厚外径为60 mm 壁厚4mm 的无缝钢管，求每m 重量。每m 重量= 0.02466 ×4 ×（60 –4 ）=5.52kg
线规SWG BWG BG AWG 号码英寸毫米英寸毫米英寸毫米英寸毫米 7／0 6／0 5／0 4／0 3／0 2／0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 0．500 0．464 0．432 0．400 0．372 0．348 0．324 0．300 0．276 0．252 0．232 0．212 0．192 0．176 0．160 0．144 0．128 0．116 0．104 0．092 0．080 0．072 0．064 0．056 0．048 0．040 0．036 0．032 0．0280 0．0240 0．0220 0．0200 0．0180 12．700 11．786 10．973 10．160 9．449 8．839 8．230 7．620 7．010 6．401 5．893 5．385 4．877 4．470 4．046 3．658 3．251 2．946 2．642 2．337 2．032 1．829 1．626 1．422 1．219 1．016 0．914 0．813 0．711 0．610 0．559 0．508 0．457 -- -- 0．500 0．454 0．425 0．330 0．340 0．300 0．284 0．259 0．238 0．220 0．203 0．180 0．165 0．148 0．134 0．120 0．109 0．095 0．083 0．072 0．065 0．058 0．049 0．042 0．035 0．032 0．028 0．025 0．022 0．020 0．018 -- -- 12．700 11．532 10．795 9．652 8．639 7．620 7．214 6．579 6．045 5．588 5．156 4．572 4．191 3．759 3．404 3．048 2．769 2．413 2．108 1．829 1．651 1．473 1．245 1．067 0．839 0．813 0．711 0．635 0．559 0．508 0．457 0．6666 0．6250 0．5883 0．5416 0．5000 0．1152 0．3954 0．3532 0．3147 0．2804 0．2500 0．2225 0．1981 0．1764 0．1570 0．1398 0．1250 0．1313 0．0991 0．0882 0．0785 0．0699 0．0625 0．0556 0．0495 0．0440 0．0392 0．0349 0．03125 0．02782 0．02476 0．02204 0．01961 16．932 15．875 14．943 13．757 12．700 11．308 10．069 8．971 7．993 7．122 6．350 5．652 5．032 4．481 3．988 3．551 3．175 2．827 2．517 2．240 1．994 1．775 1．588 1．412 1．257 1．118 0．996 0．887 0．794 0．707 0．629 0．560 0．498 -- 0．5800 0．5165 0．4600 0．4096 0．3648 0．3249 0．2893 0．2576 0．2294 0．2043 0．1819 0．1620 0．1443 0．1285 0．1144 0．1019 0．0907 0。0808 0．0720 0．0648 0．0571 0．0508 0．0453 0．0403 0．0359 0．0320 0．0285 0．02535 0．02010 0．01790 0．01594 0．01420 -- 14．732 13．119 11．684 10．404 9．266 8．252 7．348 6．544 5．827 5．189 4．621 4．115 3．665 3．264 2．906 2．588 2．305 2．053 1．828 1．628 1．450 1．291 1．150 1．024 0．912 0．812 0．723 0．644 0．573 0．511 0．455 0．405 常用线规号码与线径对照表
压力单位换算方法 1. 1atm=0.1MPa=100KPa=1公斤=1bar=10米水柱＝14.5PSI 2.1KPa=0.01公=0.01bar=10mbar=7.5mmHg=0.3inHg=7.5torr=100mmH2O=4inH2O 3. 1MPa=1N/mm2 14.5psi=0.1Mpa 1bar=0.1Mpa 30psi=0.21mpa,7bar=0.7mpa 现将单位的换算转摘如下： Bar---国际标准组织定义的压力单位。 1 bar=100,000Pa 1Pa=F/A, Pa: 压力单位, 1Pa=1 N/㎡ F : 力 , 单位为牛顿(N) A: 面积 , 单位为㎡ 1bar=100,000Pa=100Kpa 1 atm=101,325N/㎡=101,325Pa 所以,bar是一种表压力(gauge pressure)的称呼。 1Kg/c㎡=98.067KPa =0.9806bar 1bar=1.02Kg/ c㎡ 压力单位： 英制(IP) psi ,psf ,in.Hg ,inH2O 公制(metric) Kg/㎡, Kg/ c㎡ ,mH2O ISO公制(ISO metric) Pa , bar , 、兆帕（MPa）; 千帕（kPa）; 帕（Pa） 压力单位的兆帕符号为MPa,不要书写为Mpa、mpa ; 千帕符号kPa 不要书写为KPa、Kpa或kpa; 帕的符号Pa不要书写为pa； 1MPa=1000kPa=1000000Pa 2、磅力/英寸2（lbf/in2, psi） 压力单位的磅力/英寸2符号为lbf/in2, psi，不要书写为Ibf/ln2 Psi ; 1psi=0.00689474482MPa 3、毫米汞柱（mmHg） 压力单位的毫米汞柱符号为mmHg，不要书写为mmhg； 1mmHg=0.00013332231MPa 4、英寸汞柱（inHg） 压力单位的英寸汞柱符号为inHg，不要书写为 inhg； 1inHg=0.00338638672MPa 5、毫米水柱（mmH2O）
测序常见问题及解决策略 一、PCR常见问题 1.假阴性，不出现扩增条带 PCR出现假阴性结果，可从以下几个方面来寻找原因： 1）模板：①模板中有杂蛋白；②模板中有Taq酶抑制剂；③在提取制备模板时丢失过多；④模板核酸变性不彻底。 2）酶：酶失活或反应时忘了加酶。 3）Mg2+浓度：Mg2+浓度过高可降低PCR扩增的特异性，浓度过低则影响PCR 扩增产量甚至使PCR扩增失败而不出扩增条带。 4）反应条件：变性对PCR扩增来说相当重要，如变性温度低，变性时间短，极有可能出现假阴性;退火温度过低，可致非特异性扩增而降低特异性扩增效率退火温度过高影响引物与模板的结合而降低PCR扩增效率。 5）靶序列变异：靶序列发生突变或缺失，影响引物与模板特异性结合，或因靶序列某段缺失使引物与模板失去互补序列，其PCR扩增是不会成功的。 2.假阳性 假阳性：出现的PCR扩增条带与目的靶序列条带一致，有时其条带更整齐，亮度更高。常见原因有： 1）引物设计不合适：选择的扩增序列与非目的扩增序列有同源性，因而在进行PCR扩增时，扩增出的PCR产物为非目的性的序列。靶序列太短或引 物太短，容易出现假阳性。需重新设计引物。 2）靶序列或扩增产物的交叉污染：这种污染有两种原因：一是整个基因组或大片段的交叉污染，导致假阳性。这种假阳性可用以下方法解决：操作时应小心轻柔，防止将靶序列吸入加样枪内或溅出离心管外。二是空气中的 小片段核酸污染，这些小片段比靶序列短，但有一定的同源性。可互相拼接，与引物互补后，可扩增出PCR产物，而导致假阳性的产生，可用巢式PCR方法来减轻或消除。 3.出现非特异性扩增带 PCR扩增后出现的条带与预计的大小不一致，或大或小，或者同时出现特异性扩增带与非特异性扩增带。非特异性条带的出现，其原因：一是引物
压力单位换算方法1. 1atm= 0.1MPa=100KPa=1公斤=1bar=10米水柱＝14.5PSI 2.1KPa= 0.01公斤= 0.01bar=10mbar= 7.5mmHg= 0.3inHg= 7.5torr=100mmH2O=4inH2O 3. 1MPa=1N/mm 214.5psi= 0.1Mpa1bar= 0.1Mpa30psi= 0.21mpa,7bar= 0.7mpa现将单位的换算转摘如下： Bar---国际标准组织定义的压力单位。 1 bar=100,000Pa1Pa=F/A,Pa： 压力单位, 1Pa=1 N/㎡F ： 力,单位为牛顿(N)A：
面积,单位为㎡1bar=100,000Pa=100Kpa1 atm=101,325N/㎡=101,325Pa所以,bar是一种表压力(gauge pressure)的称呼。 1Kg/c㎡= 98.067KPa = 0.9806bar1bar= 1.02Kg/ c㎡压力单位： 英制(IP) psi ,psf ,in.Hg ,inH2O公制(metric) Kg/㎡, Kg/ c㎡,mH2OISO公制(ISO metric) Pa , bar ,N 1、兆帕（MPa）;千帕（kPa）;帕（Pa）压力单位的兆帕符号为MPa,不要书写为Mpa、mpa ;千帕符号kPa不要书写为KPa、Kpa或kpa;帕的符号Pa不要书写为pa；1MPa=1000kPa=1000Pa 2、磅力/英寸2（lbf/in2, psi）压力单位的磅力/英寸2符号为lbf/in2, psi，不要书写为Ibf/ln2 Psi ;1psi= 0.482MPa 3、毫米汞柱（mmHg）压力单位的毫米汞柱符号为mmHg，不要书写为mmhg；1mmHg= 0.231MPa 4、英寸汞柱（inHg）压力单位的英寸汞柱符号为inHg，不要书写为inhg；1inHg= 0.MPa 5、毫米水柱（mmH2O）压力单位的毫米水柱符号为mmH2O，不要书写为mmh2O；1mmH2O= 0.4MPa