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f (b)) ≤ y is valid for P 1 . • For P 2 we combine −(x − (cid:5)b(cid:6)) ≤ −1 and x − y ≤ b with multipliers f (b) and 1: (x − (cid:5)b(cid:6))(1 − f (b)) ≤ y. By disjunction, (x − (cid:5)b(cid:6))(1 − f (b)) ≤ y is valid for conv(P 1 ∪ P 2) = conv(X). Slide 18 5 (cid:6) 3.5.3 Theorem (cid:7) (cid:7) (cid:5)n A...
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, j} ∈ E is called a clique. (cid:2) xi ≤ 1, i∈U for any clique U (∗) is valid. • A clique U is maximal if for all i ∈ V \ U , U ∪ {i} is not a clique. • (*) is facet defining if and only if U is a maximal clique. • U = {1, . . . , k}. Then, ei, i = 1, . . . , k satisfy (*) with equality. • For each i (cid:10)∈...
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x2 + x3 x1 x1 + x1 + x1 + x2 ≤ 1 ≤ 1 ≤ 1 x5 + x6 ≤ 1 + x6 ≤ 1. unique optimal solution x 0 = (1/2)(0, 1, 1, 1, 1, 1)(cid:2) . Do maximal clique inequalities describe convex hull? Slide 25 6 2 1 3 4 6 5 • x 0 does not satisfy x2 + x3 + x4 + x5 + x6 ≤ 2. (1) • Stable sets {2, 4}, {2, 5}, {3, 5}, {3, 6}...
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2, {2, 4}, {2, 5}, {3, 5}, {3, 6}, {4, 6}, and {1} satisfy it with equality. • 2x1 + x2 + x3 + x4 + x5 + x6 ≤ 2, is valid and defines a facet conv(S). 4.2.1 General principle (cid:11) Suppose S ⊂ {0, 1}n, Si = S∩ x ∈ {0, 1}n (cid:7) x1 = i , i = 0, 1, and is valid for S0 . (cid:12) (cid:7) (cid:5) n aj xj ≤ a0 (2...
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• P = conv {x ∈ {0, 1}6 | 5x1 + 5x2 + 5x3 + 5x4 + 3x5 + 8x6 ≤ 17}. • x1 + x2 + x3 + x4 ≤ 3 is valid for P ∩ {x5 = x6 = 0}. • Lifting on x5 and then on x6, yields x1 + x2 + x3 + x4 + x5 + x6 ≤ 3. • Lifting on x6 and then on x5, yields x1 + x2 + x3 + x4 + 2x6 ≤ 3. Slide 29 8 15.083J/6.859J Integer Optimization Le...
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, 1)(cid:2) , b3 = (1, −1)(cid:2) . Then, L(b1 , b2 ) = L(b2 , b3 ). Slide 3 x2 0011 00 4 11 1 0 11 00 00 11 3 0011 01 10 4 x1 01 01 01 b1 0011 2 00000 11 111 000111 b3 000 111 1 000111 01 01 00000 00 11 11111 b2 111 000 0011 11111 00000 111 000 0011 11111 00000 111 000 11 00011111111 00000 ...
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• For all x ∈ L: x = Bv with v ∈ Z d . • det(U ) = ±1, and det(U −1) = 1/ det(U ) = ±1. • x = BU U −1 v. • From Cramer’s rule, U −1 has integral coordinates, and thus w = U −1 v is integral. Slide 4 Slide 5 • B = BU . Then, x = Bw, with w ∈ Z d, which implies that B is a basis of L. d 1 , . . . , b ] be bas...
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if and only if −z ∈ A. Then A contains a non-zero lattice point. n n 2.5 Integer normal form • A ∈ Z m×n of full row rank is in integer normal form, if it is of the form [B, 0], where B ∈ Z m×m is invertible, has integral elements and is lower triangular. • Elementary operations: (a) Exchanging two columns; (...
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max{i : a1,i > 0}. If k = 1, then we have transformed A into a matrix of the form (1). Otherwise, k ≥ 2 and by applying k − 1 operations (c) we transform A to � � � � a1,1 A2, . . . , Ak−1 − a1,2 � a1,k−1 Ak, Ak, Ak+1, . . . , An . a1,k � A = A1 − • Repeat the process to A, and exchange two columns of ...
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(j, j) is −1. det(U 2) = −1. (iii) Adding f ∈ Z times column k to column j, corresponds to multiplying matrix A by a unimodular matrix U 3 = I + f I k,j . Since det(U 3) = 1, U 3 is unimodular. • Performing two elementary column operations corresponds to multiplying the corresponding unimodular matrices resulting...
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] = AU . Let b ∈ Z m and S = {x ∈ Z n | Ax = b}. (a) The set S is nonempty if and only if B −1b ∈ Z m (b) If S (cid:6)= ∅, every solution of S is of the form . Slide 10 x = U 1B −1b + U 2z, z ∈ Z n−m , where U 1, U 2: U = [U 1, U 2]. (c) L = {x ∈ Z n | Ax = 0} is a lattice and the column vectors of U 2 constitut...
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z ∈ Z n−m}. • Let L = {x ∈ Z n | Ax = 0}. By setting b = 0 in part (b) we obtain that L = {x ∈ Z n | x = U 2z, z ∈ Z n−m}. Thus, by definition, L is a lattice with basis U 2. 2.10 Example • Is S = {x ∈ Z 3 | Ax = b} is nonempty � � A = 3 4 6 5 1 5 Slide 12 and b = � � . 3 2 • Integer normal form: [B, ...
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and y(cid:2)b / ∈ Z. • The set S = ∅ if and only if there exists a y ∈ Qm, such that y ≥ 0, y(cid:2)A ∈ Z and y ∈ Z . (cid:2)b / m 2.12 Proof • Assume that S = ∅. If there exists y ∈ Qm, such that y (cid:2)A ∈ Z and y ∈ Z, (cid:2)b / m (cid:6) then y (cid:2)Ax = y (cid:2) b with y (cid:2)Ax ∈ Z and y (cid:2)...
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, z ∈ Z n−m}. • max c (cid:2)U 2z s.t U 2z ≥ −x 0 z ∈ Z n−m . • Different bases give rise to alternative reformulations max c (cid:2)Bz s.t. Bz ≥ −x 0 z ∈ Z n−m . 6 MIT OpenCourseWare http://ocw.mit.edu 15.083J / 6.859J Integer Programming and Combinatorial Optimization Fall 2009 For information about citi...
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4 SCETI LAGRANGIAN where Wn = (cid:88) (cid:88) −g)k ( k p erm k! (cid:32) n¯ · An(q1) · · · n¯ · An(qk) [n¯ · q1][n¯ · (q1 + q2)] · · · [n¯ · (cid:80) k i=1 qi] (cid:33) . (3.34) Here Wn is the momentum space version of a Wilson line built from collinear An gluon fields. In position space the corresponding Wilson li...
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ons, and momentum degrees of freedom. On the way to our final result we introduce the label operator which provide a simple method to separate large (label) momenta from small (residual) momenta. 4.1 SCET Quark Lagrangian Lets construct the leading order SCET collinear quark Lagrangian. This desired properties that t...
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⊥ and ¯n · k « n¯ · p corresponds to carrying out a multipole expansion for the interaction of the ultrasoft gluon with the collinear quark. The LO collinear quark propagator must be smart enough to give the correct leading order result without further expansions, irrespective of whether it later emits a collinear g...
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0 , n D⊥ ˆ D⊥ ˆ n ˆ n ϕn¯ i D/ ⊥ϕn = 0 , (4.4) since ξnPn = 0 and ϕ¯n¯Pn¯ = 0. These simiplifications leave us with the Lagrangian ¯ L = ξˆ n n/ 2 in · D ξˆn + ϕ¯iD/ ⊥ ξˆn + ξˆ n n iD/ ⊥ϕn¯ + ϕn¯ n/ 2 in¯ · Dϕn¯ . (4.5) So far this is just QCD written in terms of the ξˆn and ϕn¯ fields. However, the field ϕn¯ corr...
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0 (4.6) in¯ · Dϕn¯ + ϕn¯ = 1 in¯ · D n/¯ 2 ˆ iD/ ⊥ξn = 0 n/¯ ˆ iD/ ⊥ ξn , 2 where the second line is obtained by multiplying the first by /¯n/2 on the left, and the plus sign in the last = −i / ¯ Plugging this result back into our Lagrangian, two terms cancel, line comes from using /¯ and the other two terms g...
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inear and ultrasoft gauge fields, nor the corresponding momentum components. These remaining steps will be to 2. Separate the collinear and ultrasoft gauge fields. 3. Separate the collinear and usoft momentum components with a multipole expansion. We then can expand in the fields and momenta and keep the leading piece...
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µ = Aµ + Aµ + · · · . us n (4.9) Here the ellipsis stand for additional terms involving Wilson lines which only will become relevant when we formulate power corrections, and hence will be ignorded for our leading order analysis here (they are given below in Eq.()). The interpretation of Aµ n will also prove useful...
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us to formulate power corrections in a manner where operators give homogeneous contributions in λ order by order. For example, consider the denominator of the propagator of a quark with momentum pn + kus expanded to keep the leading and first subleading terms 1 (pn + kus)2 = − (pn 1 us)(pn + k− + ⊥ us) + (pn +...
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= (cid:90) (cid:90) (cid:90) dx dp1 dp2 dk e ip1x ik(0) e− ip2x ¯ e− ψ(p1)Aus(k)ψ(p2) dp1 dp2 dk δ(p1 − p2) ψ(p1)Aus(k)ψ(p2). ¯ (4.12) We see immediately that this corresponds to a 3-point Feynman rule where the small momentum k is ignored relative to the large momenta p1 and p2, and that total momentum is not conserv...
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to achieve this. It will allow us to 1) simply derive the corresponding momentum space Feynman rules, 2) simplify the formulation of gauge transformations in the effective theory, and 3) incorporate the multipole expansion into propagators rather than vertices. For the moment we only consider the quark part of the fie...
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so we have a grid as shown in Fig. 5 (for convenience we show only one of the pµ components). Note that any momentum p has a unique decomposition in terms of label and residual components. Since pc » pr the spacing between grid points is always much larger than the spacing between points in a box. This setup has the...
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= 0 does not define a collinear momentum. Indeed the pc = 0 bin corresponds to the ultrasoft modes. For an ultrasoft momentum p we simply have (cid:90) (cid:90) d4p → d4pr . With this momentum separation we can also label our fields by both components ˜ξn(p) → ˜ξn, p(cid:96)(pr) . We also have separate conservation l...
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these fields. Altogether, the above steps allow us to rewrite our hatted collinear field ξˆn(x) as (cid:90) ˆξn(x) = d4p −ip·x ˜ e (2π)4 ξn(p) = (cid:90) (cid:88) p (cid:96)(cid:54)=0 (cid:88) = p l=0(cid:54) e−ip(cid:96)·x ξn, p(cid:96)(x) . 4 d pr e −ip(cid:96)·x −ipr·x ˜ e ξn, p (pr) (cid:96) (4.21) We can identify ...
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, p£ (x). Recall that P µ and p only contains the components P ≡ n¯ · P ∼ p ∼ λ0 and P µ ∼ p ∼ λ. Therefore we have n · P = 0. Also (4.23) ⊥µ c µ c − c ⊥ µ c in¯ · ∂ « P , µ . i∂µ « P⊥ ⊥ (4.24) The main advantage of the label operator is that it provides a definite power counting for derivatives. It is als...
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.26) where the label operator acts on both fields. Consequently, conservation of label momenta is simply encoded by this phase factor and is manifest at the level of operators. Lastly, we must deal with anti-particles and gluons. For the anti-particles, we expand our Dirac field into two parts (cid:90) ψ(x) = d4p δ(...
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+ ξ− n, p£ n, −p£ (x) (4.29) where pc has either sign, but one picks out particles and one picks out antiparticles. Thus the action of the fields ξn,p£ and ξn,p£ is that for ¯ n¯ · pc > 0 : n¯ · pc < 0 : a particle is destroyed or created an antiparticle is created or destroyed The sign convention for the labe...
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destroys a gluon, while for q < 0 it creates a gluon. With our conventions the action of the label operator on a bunch of labelled fields is − c − c P µ(φ† φ† · · · q1 q2 φp1 φp2 µ µ 1 + p2 + · · · · ) = (p · · − q − q − · · · )(φ† φ† · · · q1 q2 µ 2 µ 1 φp1 φp2 · · · ). (4.34) Thus it gives a minus sign wh...
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result: (cid:18) L = ˆξn in · D + iD/ ⊥ 1 iD/ in¯ · D ⊥ (cid:19) ¯ ˆξn. n/ 2 Changing i∂µ → (Pµ + i∂µ) and ξˆ n → ξn and expanding our derivative operators, we have in · D = in · ∂ + gn · An + gn · An µ + gA ) n⊥ (cid:123)(cid:122) (cid:125) ∼λ P µ iD = ( ⊥ ⊥ (cid:124) + (i∂µ (cid:124) ⊥ + gAµ (cid:123)(cid:122) ∼λ2 ...
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+ gAµ n⊥ , ⊥ in¯ · Dn = P + gn¯ · An. Remarks: (4.39) (4.40) • Both terms with covariant derivatives in the (· · · ) in L are of order λ2 so the leading order La­ grangian is order λ4 (recalling that the fields scale as ξn ∼ λ). Since for a Lagrangian with collinear d4x ∼ λ−4 this gives us an action that is ∼ λ0 a...
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attitude that low energy locality is a desired property for the effective field theory. • If we are considering a situation without ultrasoft particles, and without hard interactions that do not couple to a particular component, then the interaction of collinear fermions alone could equally 29 4.2 Wilson Line Ide...
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· pc in/ 2 (¯n · pc)(n · pr) + (pc⊥)2 + i0 . (4.42) The leading order Lagragian is smart enough that it gives the correct propagator in different situations without having to make further expansions. This is important to ensure that the leading order Lagrangian strictly give O(λ0) terms, while subleading Lagrangian...
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= (P + gn¯ · An)Wn = 0 . (4.45) 30 k,uspplr)(,pplr)(+kus,pplr)(,pplr)(+,qqlr)(qr+ql(cid:54) 4.3 Collinear Gluon and Ultrasoft Lagrangians 4 SCETI LAGRANGIAN With this definition, the action of in¯ · Dn on a product of Wn and some arbitrary operator is P + gn¯ · An)WnO in¯ · Dn(WnO) = ( (cid:2) = (P + gn¯ · An)Wn ...
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P = Wn = Wn 1 1 W † . W † in¯ · D n . n in¯ · Dn n The first relation allows us to rewrite L as (0) nξ L(0) = e nξ −ix·P ¯ξn (cid:16) in · D + i /Dn⊥W † n 1 P Wni /Dn⊥ (cid:17) /¯n 2 ξn . (4.49) (4.49) (4.50) It is also useful to note that we can use the label operator to write a tidy expression for the Wilson line...
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Dµ, Dν ]. Expanding the covariant derivative as we did in the quark sector we keep only the leading order terms. For a covariant derivative acting on collinear fields the leading order terms are i g iDµ → iDµ = µ (P + gn¯ · An) + (P µ + gAµ n 2 Recall that the field Aµ varys much more slowly than Aµ n, so we can th...
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− /(cid:48) p/⊥ ⊥ p n¯ n¯·p n¯·p (cid:48) µ (cid:21) ¯ n/ 2 = ig2 T A T B n¯·(p−q) (cid:20) ⊥γν µ γ ⊥ − µ ⊥p/⊥ ¯n·p ¯nν − p (cid:48)/⊥γ⊥ ¯n·p (cid:48) ¯nµ + p (cid:48)/⊥p/⊥ n¯·p n¯·p (cid:48) µn¯ν n¯ ν (cid:21) ¯ n/ 2 + ig2 T B T A n¯·(q+p(cid:48)) ⊥γµ ν ⊥ − γ p/⊥ ¯n·p ¯nµ − p (cid:48)/⊥γ⊥ ¯n·p (cid:48) ¯nν + p (cid:48...
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collinear gluon Lagrangian is then L(0) ng = 1 2g2 (cid:8) Tr ([i µ, i D D µ (cid:9) ])2 + τ Tr ([i µ , An µ])2 + 2Tr cn[i us, [i µ, cn]] Dµ D Dus (cid:9) (cid:8) (cid:8) (4.54) (cid:9) . (4.55) For the Langrangian with only ultrasoft quarks and ultrasoft gluons, at lowest order we simply have the QCD actions. Using ...
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oft modes, L(0) = L (0) + L(0) + L(0) us ng nξ . (4.57) 32 μ , Appɂμ , Appɂμ , Aν , Bq 4.4 Feynman Rules for Collinear Quarks and Gluons 4 SCETI LAGRANGIAN = − i n¯ ·q n·k + q2 + i0 ⊥ (cid:18) µν − (1 − τ ) g µqν q n¯ ·q n·k + q2 ⊥ (cid:19) δ a,b = gf abcnµ (cid:8)n¯ · q1 gνλ − 1 (12 − 1 )[n¯τ λq1ν + n¯νq2λ](...
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that follow from the collinear quark and gluon Lagrangians. We do not show the purely ultrasoft interactions which are identical to those of QCD, nor do we show the purely collinear gluon interactions which are also identical to those of QCD. The Feynman rules that follow from the leading order collinear quark Lagra...
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size of momenta. 33 a, μb, ν(q, k)b, νc, λa, μq2q1a, μb, νc, λd, ρa, μb, νc, λd, ρ 4.5 Rules for Combining Label and Residual Momenta in Amplitudes 4 SCETI LAGRANGIAN Instead of this, we need to use a Continuum EFT picture where the EFT modes have propagators that extend over all momenta, but integrands which obta...
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the case with d-dimensions) is resolved, given a pair of momenta components (pc, pr) ∈ Rd−1 × Rd . The upshot is that in the simplest cases the residual momentum can simply be dropped or absorbed into a label momentum in the same direction (making it continuous), while in the most complicated cases the formalism lea...
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onshell condition k2 = 0. For the external lines that are collinear it suffices to take label momenta p = n¯ · pc − and pc⊥, and a single residual momentum p . This amounts to picking βµ above to contain the full pr and pµ components. The onshell condition for the collinear particles is then simply p − p − pp 2 = 0. All...
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particles or collinear loops, and residual momenta for external + ultrasoft particle, external collinear particles from p , and from collinear and ultrasoft loops. r First we note that if we integrate over all label momenta qc and residual momenta qr that this will be equal to an integration over all of the qµ momen...
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q+ r ) = (cid:96) (cid:88) (cid:90) q(cid:96) ddqr F (q− + q− q(cid:96) + q r , ⊥ (cid:96) (cid:90) ⊥, qr ) = + r d q F (q−, q⊥, q ) . (4.60) d + In the first step we use the fact that F is constant throughout each box in the grid picture of Fig. 5 so its the same with the first two arguments shifted by residual momenta...
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the momentum of a collinear propagator. These are referred to as 0-bin restrictions.4 We will discuss the change needed which handles this complication below. Often the results for collinear loop integrals are called “naive” if one uses Eq. (4.60). The result from this naive result will be correct if the added terms...
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for an ultrasoft loop integration. Both ultrasoft loop integrations and ultrasoft external particles introduce residual momenta into propagators that can not be absorbed by a rule like that in Eq. (4.59). If we consider a case with an ultrasoft loop integration, then there will be dependence on the residual momentum...
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(4.63) (4.63) where qc = 0 is simply a label to denote the fact that the label momentum qc must be large in order to correspond to a collinear particle carrying total momentum q. If qc = 0 then the particle would instead be ultrasoft, and we will often have included another diagram in SCET to account for the differen...
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− r , q⊥ (cid:96) + q⊥ r , q+ r ) − (cid:90) ddqr F 0(q− r , q⊥ r , q+ r ) q(cid:96) (cid:90) ddq F (q−, q⊥, q+) − (cid:90) ddqr F 0(q− r , q⊥ r , q+ r ) (cid:90) ddq (cid:2)F (q−, q⊥, q+) − F 0(q−, q⊥, q+)(cid:3) . (4.64) Here the integrand F 0 is derived from expanding the integrand for F by taking the label momenta ...
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scaling for the subtraction is shown pictorally in Fig. 8. The F 0 term subtracts singularities from F that come from the region where the collinear momentum behaves like an ultrasoft momentum. In general when the subtraction integration is non-trivial there will always exist a corresponding ultrasoft diagram where ...
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) (cid:54) (cid:54) (cid:54) (cid:54) (cid:54) (cid:54) (cid:54) MIT OpenCourseWare http://ocw.mit.edu 8.851 Effective Field Theory Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Introduction to Robotics, H. Harry Asada 1 Chapter 6 Statics Robots physically interact with the environment through mechanical contacts. Mating work pieces in a robotic assembly line, manipulating an object with a multi-fingered hand, and negotiating a rough terrain through leg locomotion are just a few examples ...
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i from link i+1 is then given by –fi,i+1. The gravity force acting at the mass centroid Ci is denoted mig, where mi is the mass of link i and g is the 3x1 vector representing the acceleration of gravity. The balance of linear forces is then given by f i ,1 i − f − + ii 1, + mi g = 0 , i (cid:34)= ,1 , n (6.1.1) No...
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1, + Link i C i iCi,r ,1−r i i Oi gim Link i+1 Actuator i iτ z O x y Figure 6.1.1 Free body diagram of the i-th link The force ,1−f i i and moment N i ,1− i are called the coupling force and moment between . These are the adjacent links i and i-1. For i=1, the coupling force and moment are interpreted...
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3x1 vectors. The number of coupling forces and moments involved is 2(n+1). Therefore two of the coupling forces and moments must be specified; otherwise the equations cannot be solved. The final coupling force and moment, environment. It is this pair of force and moment that the robot needs to accommodate in order ...
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forces and moments are so called “constraint forces and moments” merely joining adjacent links together. Therefore, constraint forces and moments do not participate in energy formulation. This significantly reduces the number of terms and, more importantly, will provide an explicit formula relating the joint torques...
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1 f i ,1 − i (6.2.2) Note that, although we use the same notation as that of a revolute joint, the scalar quantity iτ has the unit of a linear force for a prismatic joint. To unify the notation we use joints, and call it a joint torque regardless the type of joint. iτ for both types of iτ ,1−f i i ,1−N i i Revol...
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T q = q n Department of Mechanical Engineering Massachusetts Institute of Technology Introduction to Robotics, H. Harry Asada 5 The explicit relationship between the n joint torques and the endpoint force F is given by the following theorem: Theorem ...
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+nn exδ eδϕ Figure 6.2.2 Virtual displacements of the end effecter and individual joints (cid:34) We assume that joint torques the serial linkage system, while the joints and the end-effecter are moved in the directions geometrically admissible. Then, the virtual work done by the forces and moments is given by a...
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coordinates that are complete and independent. Therefore, for the above virtual work to vanish for arbitrary virtual displacements we must have: τ T= FJ This is eq.(6.2.4), and the theorem has been proven. The above theorem has broad applications in robot mechanics, design, and control. We will use it repeatedly i...
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1 ) (cid:65) 1 (cid:65) cos( cos + θ 1 2 (cid:65) cos( + θθ 2 + θθ 1 2 ) 2 1 ) ⎞ ⋅⎟⎟ ⎠ ⎛ ⎜ ⎜ ⎝ F x F y ⎞ ⎟ ⎟ ⎠ (6.2.8) Department of Mechanical Engineering Massachusetts Institute of Technology Introduction to Robotics, H. Harry Asada 7 y Endpoint Fo...
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. The null space N(J), on the other hand, represents the set of joint velocities that do not produce any velocity at the end-effecter. If the null space contains a non- zero element, the differential kinematic equation has an infinite number of solutions that cause the same end-effecter velocity. The lower half of F...
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(JT) mV∈F Figure 6.3.1 Duality of differential kinematics and statics The ranges and null spaces of J and JT are closely related. According to the rules of linear algebra, the null space N(J) is the orthogonal complement of the range space R(JT). Namely, if a non-zero n-vector x is in N(J) , it cannot also belong to...
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Introduction to Robotics, H. Harry Asada 9 6.4 Closed-Loop Kinematic Chains The relationship between joint torques and the endpoint force obtained in Theorem 6.1 can be extended to a class of parallel-link mechanisms with closed kinematic-chains. It can also be extended to multi-fingered hands, leg locomotion, and ...
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5 4 = 2 δ Work = ( ττ 3 1 T ) δθ ⎛ 1 ⎜⎜ δθ ⎝ 3 ⎞ −⎟⎟ ⎠ ( F x T ) F y x δ ⎛ e ⎜⎜ y δ ⎝ e ⎞ ⎟⎟ ⎠ . (6.4.2) For any given configuration of the robot, the virtual displacements of the end-effecter are uniquely determined by the virtual displacements of Joints 1 and 3. In fact, the former is related to the latter via th...
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reedom robot mechanism with n active joints. Assume that all the joints are frictionless, and that, for a given configuration of the robot mechanism, there exists a unique Jacobian matrix relating the virtual displacements of its end-effecter, , to the 1m ×ℜ∈δ p virtual displacements of the active joints, 1n ×ℜ∈δ ...
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by Department of Mechanical Engineering Massachusetts Institute of Technology Introduction to Robotics, H. Harry Asada δ Work = + δθτδθτδθτδθτ 4 + + 1 2 3 4 2 3 1 − xF δ x e − yF δ y e 11 (6.5.1) Note t...
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force applied to the object, as illustrated in the figure. This internal force is a grasp force that is needed for performing a task. 2θ y 1θ Finger 1 Grasped Object External Endpoint Force x , A y A Grasp Force Finger 2 4θ 3θ x Figure 6.5.1 Two-fingered hand manipulating a grasped object Department of ...
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Lecture 9 (On-Line Video) The Hydrogen Atom Today’s Program: 1. Angular momentum, classical and quantum mechanical. 2. The Hydrogen atom semi-classical approach. 3. The Hydrogen atom quantum mechanical approach. 4. Eigenfunctions and eigenvalues common to Hˆ , Lˆ2 and Lˆ z . 5. The radial dependence. 6. Energy...
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ivity of free space. II The Classical Hamiltonian   H r, p  2  p2 e  2 r Where:  m m e  mme p p  me is a reduced mass. Semiclassical Bohr model that yielded correct energy (but produced generally incorrect results) relied on the conservation of energy, angular momentum and the balance of forces....
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electron from the ground state). Also the Bohr radius also correctly described atomic dimensions. 2 III The Quantum mechanical Hamiltonian     Hˆ rˆ 1, pˆ 2, pˆ 1, rˆ 2   2 pˆ1 2m1  2 pˆ2 2m2  Vˆ   2  rˆ rˆ 1  Let’s perform the following coordinate transform – break our system ...
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 m2 , M  m1  m2 Since the Center of Mass Hamiltonian commutes with the relative motion Hamiltonian, it commutes with the whole system Hamiltonian, which means that energy of the center of mass is conserved and can be used to label states. Then the Hamiltonian above breaks into two independent problems – one for...
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�2  2 r r 2 r  1  2  r  2 2  2  1    tan  sin2   2   1 e2 r 3 IV Finding the eigenfunctions and eigenvalues of the Hamiltonian.  2  1     tan  sin2  2   1 e2 r   Hˆ rˆ, pˆ    r 2 1 2  2 r r 1 2 2 2 r r 2 r  2 2 1  ...
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�2  1  1 Then we need to solve the following eigenvalue problem for the time-independent Hamiltonian:  2 1 2     2 r r 2 r  Lˆ2 2r  e2  u r,,  Eu r,, r    2 Note that the angular momentum squared operator commutes with the Hamiltonian and recall that it also commutes with the z-co...
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� r m m z : Yl , we can look for the l – the number associated with the eigenvalues of the orbital angular momentum magnitude operator is called the orbital angular momentum quantum number. �m – the number associated with the eigenvalues of the z component of the orbital angular momentum called the magnetic qu...
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��3 n  l 1!  r l   n 2n n  l!  na0       2r   na0  2l1  Lnl1  , n  l, n  0 Where L2l1 is the generalized Laguerre polynomials. Substituting it into Mathematica, we can nl1 find the Bohr radius: 2 a0   e2  0.529 A Example of atomic orbitals – 1s: u100 r,,  R10 r  Y0  ...
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l 5 Conclusions: 1. The energy levels or the energy eigenvalues En of the hydrogen atom depend only�on n , which is called the principal quantum number. 2. Since the energy levels and radial decay rate depend only on the n number this�number is used to identify an electron shell. 3. For each energy ...
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probability of being near the origin. 9. The spectroscopic notation assigns a letter to each subshell: l  0  s l  1  p l  2  d l  3  f 6 Discussion Comparison of the energy level (energy eigenvalues) spacing in free particle, 1D box, 1D harmonic os...
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Introduction to Robotics, H. Harry Asada 1 Chapter 3 Robot Mechanisms A robot is a machine capable of physical motion for interacting with the environment. Physical interactions include manipulation, locomotion, and any other tasks changing the state of the environment or the state of the robot relative to the env...
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at some examples. Robot mechanisms analogous to coordinate systems One of the fundamental functional requirements for a robotic system is to locate its end- effecter, e.g. a hand, a leg, or any other part of the body performing a task, in three-dimensional space. If the kinematic structure of such a robot mechanism ...
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1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 i 3 3 3 3 3 3 3 3 3 3 3 3 x x x x x x x x x x x x z z z z z z z z z z z z 0 0 0 0 0 0 0 0 0 0 0 0 Figure 3.1.2 Cartesian coordinate robot Figure by MIT OCW. Photo removed for copyright reasons. GMF Robotics model M-100. Figure 3.1.3 Cylindrical coordinate robot Figure by MIT OCW. Departm...
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SCALAR type robot. Figure by MIT OCW. Department of Mechanical Engineering Massachusetts Institute of Technology Introduction to Robotics, H. Harry Asada 4 The second type, called an articulated robot or an elbow robot, consists of all three revolute joints, like a human arm. This type of robot has a great degre...
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formed by the five links and, thereby, the two serial link arms must conform to a certain geometric constraint. It is clear from the figure that the end-effecter position is determined if two of the five joint angles are given. For example, if angles 1? and 3? of joints 1 and 3 are determined, then all the link posi...
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to the base frame. The position and orientation of the moving platform are determined by the six independent actuators. The load acting on the moving platform is born by the six "arms". Therefore, the load capacity is generally large, and dynamic response is fast for this type of robot mechanisms. Note, however, tha...
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S S S S Moving Moving Moving Moving Moving Moving Moving Moving Moving Moving Platform Platform Platform Platform Platform Platform Platform Platform Platform Platform S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S ...
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Lecture 02 Voting classifiers, training error of boosting. 18.465 In this lecture we consider the classification problem, i.e. Y = {−1, +1}. Consider a family of weak classifiers H = {h : X → {−1, +1}}. Let the empirical minimizer be h0 = argmin 1 n n Xi=1 I(h(Xi) 6= Yi) and assume its expected error, 1 2 > ε = Error(h0),...
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final classifier: f = sign( αtht(x)). P Theorem 2.1. Let γt = 1/2 − εt (how much better ht is than tossing a coin). Then 1 n n Xi=1 T I(f (Xi) 6= Yi) ≤ 1 − 4γ2 t Yt=1 p Proof. I(f (Xi) 6= Yi) = I(Yif (Xi) = −1) = I(Yi αtht(Xi) ≤ 0) ≤ e−Yi P T t=1 αtht(Xi) T Xt=1 Consider how weight of example i changes: wT +1(i) = wT (i)...
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3.46 PHOTONIC MATERIALS AND DEVICES Lecture 14: Defects and Strain Lecture Notes Perfection • LRO (Long Range Order) • SRO (Short Range Order) Imperfection • Vibrating atom • Electronic change • Chemical impurity • Point defect • 1D defect • 2D defect • 3D defect (I, V) (dislocation) (grain boundary) ...
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A A + xBB U VA + xV + A A + BB B KS = [V ][VB ]x A Frenkel Pair A A → A I + VA KFP ( ) = [A ][VA ] A FP ( ) = [B ][VB ] K B I I Anti-Site Defect A A + BB U AB + B AS = [A ][BA ] K B A 3.46 Photonic Materials and Devices Prof. Lionel C. Kimerling Lecture 14: Defects and Strain Page 2 of 5 Lecture ...
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e + h U bond + ⎤ ⎤ −⎡BSi ⎥ + = ⎡ ⎥ + p ⎢LiI ⎦ n ⎢⎣ ⎦ ⎣ 3.46 Photonic Materials and Devices Prof. Lionel C. Kimerling Lecture 14: Defects and Strain Page 4 of 5 Notes Lecture T1 T 1 siB−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ n i B−⎡ ≡ ⎢ ⎣ si ⎤ ⎥ ⎦ ⎡ ILi + ⎤ ⎡ ⎢ ILi + ⎣ ⎥ ⎢ ⎦ ⎣ ⎤ ⎥ ⎦ Consequences: Lase...
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Copyright c Nancy Leveson, Sept. 1999 1980s: OO design: added inheritance, multiple inheritance, and polymorphism to ADT. In process added complexity and increased some types of connectivity. Lots of claimed advantages -- so far empirical evaluation is not supporting them well. 1990s: Architecture Patterns ...
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veson, Sept. 1999 3. Simplicity Emphasis on software that is clear, simple, and therefore easy to check, understand, and modify. 4. Restricted visibility Locality of information  Copyright c Nancy Leveson, Sept. 1999 General Software Design Concepts Implementations of the general principles Decomposition C...
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Concepts (4) Modularity Separation of concerns: 1. Deal with details of each module in isolation (ignoring details of other modules) 2. Deal with overall characteristics of all modules and their relationships in order to integrate them into a coherent system. Base on hierarchy and abstraction: Abstraction hand...
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make task of solving it more intractible, i.e., the design process is not neutral. 2. Expert designers engage in opportunistic behavior: As solution’s form emerges, problem solving strategy is adapted to meet new characteristics that are revealed, i.e., expert designers do not follow a single method. �  Copyr...
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real need for design skills, but does not create an objective forum for evaluation. Conclusions: Life belt has become waterlogged and acting more like a leg iron. Need to stop pretending that software design is largely a matter of following a set of well-defined activities. Recognize it as a creative process that ...
https://ocw.mit.edu/courses/16-355j-software-engineering-concepts-fall-2005/5c4493536832111845f291d3d62a6f38_cnotes5.pdf
Electricity and Magnetism • Review – Electric Charge and Coulomb’s Force – Electric Field and Field Lines – Superposition principle – E.S. Induction – Electric Dipole – Electric Flux and Gauss’ Law – Electric Potential Energy and Electric Potential – Conductors, Isolators and Semi-Conductors Feb 27 2002 Today • Fast s...
https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/5c48da61d3532b22af63ad3792b99f6b_2_27_2002_edited.pdf
12 Q1 F1,total Q2 • Note: – Total force is given by vector sum – Watch out for the charge signs – Use symmetry when possible Feb 27 2002 Superposition principle • If we have many, many charges – Approximate with continous distribution • Replace sum with integral! Feb 27 2002 Electric Field • New concept – Electric Fi...
https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/5c48da61d3532b22af63ad3792b99f6b_2_27_2002_edited.pdf
‘complicated’ surfaces and non-constant E: – Use integral • Often, ‘closed’ surfaces Feb 27 2002 Electric Flux • Example of closed surface: Box (no charge inside) dA E dA • Flux in (left) = -Flux out (right): ΦE = 0 Feb 27 2002 Gauss’ Law • How are flux and charge connected? • Charge Qencl as source of flux through c...
https://ocw.mit.edu/courses/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/5c48da61d3532b22af63ad3792b99f6b_2_27_2002_edited.pdf