Boosts Along beam-axis Form a boost of velocity B along z axis e+ (E+Bp )+rp+ BE E→?(E+βp) y2E2+单)2+E Transform rapidity hn (E+p2)+B =y+h(+B) p)1-B y→y+yb Boosts along the beam axis with v=B will change y by a constant yb (prym)→(py+ybm)wthy→y+yb,y6≡lny(1+B) simple additive to rapidity Relationship between y, B, and e can be seen using p,= pcos(0)and p= BE 1+Bcos日 In 2 1-Bcos 6 or tanh y=B cose where p is the CMboost 13Jun2007 Tsinghua University
13-Jun-2007 Tsinghua University 11 Boosts Along beam-axis • Form a boost of velocity b along z axis – pz g(pz + bE) – E g(E+ bpz ) – Transform rapidity: • Boosts along the beam axis with v=b will change y by a constant yb – (pT ,y,f,m) (pT ,y+yb ,f,m) with y y+ yb , yb ln g(1+b) simple additive to rapidity – Relationship between y, b, and q can be seen using pz = pcos(q) and p = bE ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) b z z z z z z z z y y y y E p E p E p p E E p p E E p E p y + = + + - - + + = + - + + + + - + = g b b b g b g b g b g b ln 1 1 1 ln 2 1 ln 2 1 ln 2 1 b q b q 1 cos 1 cos ln 2 1 - + y = or where tanh y = b cosq b is the CM boost
dp, p, dp d≡ Phase Space (cont) E E Transform phase space element dt from(E, px pv pz to(p+ y, o, m) φd,=dpld&=(m2Em ay ay aE using hn E+p E p. p 2 E P dp. P: E-p Gives. dt 14d p E Basic quantum mechanics: do =Isi12dt If o 12 varies slowly with respect to rapidity do/dy will be -constant in y Origin of the rapidity plateau for the min bias and underlying event structure pply to jet fragmentation -particles should be uniform in rapidity wrt jet axis We expect jet fragmentation to be function of momentum perpendicular to jet axis This is tested in detectors that have a magnetic field used to measure tracks 13Jun2007 Tsinghua University
13-Jun-2007 Tsinghua University 12 Phase Space (cont) • Transform phase space element d from (E,px ,py ,pz ) to (pt , y, f, m) • Gives: • Basic quantum mechanics: ds = |M | 2d – If |M | 2 varies slowly with respect to rapidity, ds/dy will be ~constant in y – Origin of the “rapidity plateau” for the min bias and underlying event structure – Apply to jet fragmentation - particles should be uniform in rapidity wrt jet axis: • We expect jet fragmentation to be function of momentum perpendicular to jet axis • This is tested in detectors that have a magnetic field used to measure tracks E dp E p E p p E p E dp p E E y p y dy dp z z z z z z z z z = - - - = + = 2 2 2 2 dpx dpy dpT df 2 2 1 = d dp d dy T f 2 2 1 = E dp dp dp E d p d x y z = 3 & z z E p E p y - + ln 2 1 using
Transverse Energy and Momentum Defin Transverse-iommenttim: momentum perpendicular to beam direction pr=Rx+py or pr =psin 8 Transverse Energy defined as the energy if pz was identically 0: E-FE(p2=0) E7=p2+p2+m2=p7+m2=E-p How does E and p2 change with the boost along beam direction? Using tanh y=B cose and p:=pose gives P:=Etanh y theh ET=E-p=E-E tanh*y=Esech'y or E=Er cosh y which also means p=Er sinh y (remember boosts cause y>y+ yb) Note that the sometimes used formula ET ET= ESi 0 is not(strictly) correct But it's close-more later 13Jun2007 Tsinghua University
13-Jun-2007 Tsinghua University 13 Transverse Energy and Momentum Definitions • Transverse Momentum: momentum perpendicular to beam direction: • Transverse Energy defined as the energy if pz was identically 0: ETE(pz=0) • How does E and pz change with the boost along beam direction? – Using and gives – (remember boosts cause y → y + yb ) – Note that the sometimes used formula is not (strictly) correct! – But it’s close – more later…. 2 2 2 2 2 2 2 2 ET = px + py +m = pT +m = E - pz p E y z = tanh E E y T = cosh or 2 2 2 pT = px + py pT = psin q ET = Esin q tanh y = b cosq pz = pcosq E E p E E y E y T z 2 2 2 2 2 2 2 2 then = - = - tanh = sech or which also means p E y z T = sinh
Invariant Mass M1.2 of 2 particles p1, p Well defined: M2 12=(p1+p2)=m2+m2+2(EE2-p1 Switch to p=(pr,y, o, m)(and do some algebra.) P·P1=p、p+Pnh+PpEB2c0sA+ sinh y Sinh E= Er cosh y and月≡Pn/Er This gives Mi2=F+2+2E Er, cosh Ay-Pr Br, coS Ap Wihβ≡pEr Note For△y→>0and△φ→0, high momentum limit:M→>0: angles"generate"mass Forβ→1(m/p→0) M2=2EE2(ohy-cos△) This is a useful formula when analyzing data 13Jun2007 Tsinghua University
13-Jun-2007 Tsinghua University 14 Invariant Mass M1,2 of 2 particles p1 , p2 • Well defined: • Switch to pm=(pT ,y,f,m) (and do some algebra…) • This gives – With bT pT /ET – Note: • For Dy → 0 and Df → 0, high momentum limit: M → 0: angles “generate” mass • For b →1 (m/p → 0) This is a useful formula when analyzing data… = + + 2 (coshD - b b cosDf) 1 2 1 2 2 2 2 1 2 1,2 T T T T M m m E E y ( ) ( ) 1 2 1 2 2 2 2 1 2 1 2 2 M1,2 = p + p = m + m + 2 E E - p p E E y T with and = cosh bT pT ET = 2 (coshD -cosDf) 1 2 2 1,2 M E E y T T ( ) 1 1 1 2 cos sinh sinh 1 2 1 2 1 2 1 2 1 2 p p p p p p p p E E y y = x x + y y + z z = T T b T b T Df +