1. P780.02 Spring 2003 L2Richard Kass Relativistic Kinematics In HEP the particles (e.g. protons, pions, electrons) we are concerned with are usually moving at speeds close to the speed of light. The classical relationship for the kinetic energy of the particle in terms of its mass and velocity is not valid: Kinetic Energy 1/2 mv 2 Thus we must use special relativity to describe the energies and momentum of the particles. The total energy (E=rest +kinetic) of a particle with rest mass, m o, is: Here v is its speed, c = speed of light, and m is sometimes called the relativistic mass. The total momentum, p, of a particle with rest mass, m o, is: We can also relate the total energy, E, to a particle's total momentum, p: Read: Appendix A of M&S E=m 0 c 2 +T

2. P780.02 Spring 2003 L2Richard Kass Relativistic Kinematics 4-vectors It is sometimes convenient to describe a particle (or a collection of particles) by a 4-vector. The components of the momentum and energy 4-vector, p, are given by: The length of the 4-vector is given by: This relationship is true in ALL reference frames (lab, center of mass,…) because it is a Lorentz invariant. A 4-vector with length L 2 is classified as follows: Time like if L 2 >0 Space like if L 2 <0 Light like if L 2 =0 (think photon!) A particle is said to be on the mass shell if m 0 =rest mass

3. P780.02 Spring 2003 L2Richard Kass Lorentz Invariant Vs. Conserved quantity With a Lorentz Invariant you get the same number in two different reference systems (it is a scalar). Let E L and p L be energy and momentum measured in LAB frame Let E cm and p cm be energy and momentum measured in center of mass frame Then: E 2 cm -p 2 cm = E 2 L -p 2 L Since (E, p) is a Lorentz invariant (as long as both are measured in same system) With a conserved quantity you get the same number in the same reference system but at a different time. Let p iL =initial momentum in lab (before collision) Let p fL =final momentum in lab (after collision) Let p icm =initial momentum in CM (before collision) Let p fcm =final momentum in CM (after collision) Momentum conservation says: p iL = p fL AND p icm = p fcm BUT NOT p iL = p fcm

4. P780.02 Spring 2003 L2Richard Kass Relativistic Kinematics 4-vectors We can also manipulate 4-vectors using contravariant/covariant (up/down) notation: m 0 2 =p u p u =g uv p v p u In this notation g uv is a metric and is given (e.g.) by: Thus the (scalar) product of two 4-vectors (a, b) is given by: The sum of two 4-vectors is also a 4-vector. The length of the sum of the 4-vectors of two particles (1,2) is: =angle between particles, m 1, m 2 are rest masses m 12 is called the invariant mass or effective mass

5. P780.02 Spring 2003 L2Richard Kass Relativistic Kinematics 4-vectors Example: Consider a proton at rest in the lab frame and an antiproton with 10GeV/c of momentum also in the lab frame. What is the energy of the antiproton in the lab frame? Since the rest energy of a particle is a Lorentz invariant we can make us of: For an antiproton the rest mass, m 0, = 938 MeV/c 2. We can re-write the above as: c=1 Thus at high energies (E>>m 0 c 2 ) E |p|. How fast is the anti-proton moving in the lab frame ? We need to remember the energy/momentum relationship between the rest frame (particle at rest) of the anti-proton and the lab frame: Thus v = 0.996c (fast!)

6. P780.02 Spring 2003 L2Richard Kass More Relativistic Kinematics Colliding beam vs fixed target Collisions As we will see later in the course cross sections and the energy available for new particle production depend on the total energy in the center of mass (CM) frame. We define the CM where the total vector momentum of the collision is zero. By definition then, in the CM frame we have for two 4-vectors (p 1, p 2 ): (p 1 +p 2 )= (E 1 +E 2, p 1 +p 2 ) = (E 1 +E 2, 0) If the masses of the two particles are equal as in the case of proton anti-proton collisions then the above reduces to: (p 1 +p 2 )= (E 1 + E 2, p 1 + p 2 ) = (E 1 + E 2, 0)= (2E, 0) (twice energy of either particle) NOTE: the square of the energy in the CM is often called s. s= magnitude of (p 1 +p 2 ) How much energy is available in the CM from a 10 GeV/c anti-proton colliding with a proton at rest? Since (p 1 +p 2 ) is a Lorentz invariant we evaluate in any frame we please! We are given values in the lab frame: (p 1 + p 2 ) =(E 1 +m p, p 1 +0) The magnitude of this 4-vector is: s = (E 1 +m p ) 2 - p 1 2 = (10.044+0.938) 2 -10 2 = 20.6GeV 2 Thus the total energy in the CM is 4.54GeV We could have gotten the same CM energy with two beams = 2.27 GeV ! In general the energy available for new particle production increases as: (2m target E beam ) 1/2 for fixed target experiments 2E beam for colliding beam experiments colliding beams are more efficient for heavy particle production

7. P780.02 Spring 2003 L2Richard Kass Even More Relativistic Kinematics Our final example for this section concerns the discovery of the anti-proton. In the early 1950s many labs were trying to find evidence of the anti-proton. At Berkeley a new proton accelerator (BEVATRON) was being designed for this purpose. Assuming fixed target proton-proton collisions would be used to create the antiproton what energy proton beam (E b ) is necessary ? The simplest reaction that conserves all the necessary quantities (energy, momentum, electric charge, baryon number) is: The total energy in the CM is given by: (p b +p t ) 2 the sum of the 4-vectors of the beam and target proton (assumed to be at rest): (p b +p t ) 2 =(E b +m p, p b ) 2 = m b 2 +m t 2 +2m t E b =2m p 2 +2m p E b The trick now is to remember that (p b +p t ) 2 is Lorentz invariant and can be evaluated in any frame we choose. The most convenient frame is the one where all the final state particles are produced at rest. Here we have: (p b +p t ) 2 inirtial =(E b +m p, p b ) 2 = (p b +p t ) 2 final = (4m p ) 2 2m p 2 +2m p E b = (4m p ) 2 E b = 7m p = 6.6 GeV The anti-proton was discovered at Berkeley in 1955 (Nobel Prize 1959)

8. P780.02 Spring 2003 L2Richard Kass Time Dilation Most of the particles we are concerned with in HEP are not stable, i.e. they spontaneously decay into other particles after a certain amount of time. For example consider the lepton family: LeptonMean Lifetime (s) electron stable muon ( ) 2x10 -6 tau ( ) 3x10 -13 The above table gives the average lifetime of the leptons in their rest frame. However, often we need to know how long a particle will live (on average) in a frame where the particle is moving close to the speed of light (c). We need to use special relativity! Consider a particle moving with speed v as pictured below: The particle is moving in the lab frame with speed v along the x-axis. The relationship between time and distance measured in the lab and CM frame is: t lab = (t cm + /c x cm ) with = v/c. x lab = (x cm + c t cm ) In the lab frame the time between life and death of the particle is: lab = t 2lab - t 1lab = (t 2 cm + /c x 2cm )- (t 1 cm + /c x 1cm ) = (t 2 cm -t 1 cm )= Note: is called the proper lifetime and lab In CM: x 1cm =x 2cm

9. P780.02 Spring 2003 L2Richard Kass More Time Dilation Consider a muon (m 0 =0.106 GeV/c 2 ) with 1 GeV energy in the lab frame. On average how long does this particle live in the LAB? In the muons rest frame it only lives (on average) = 2 sec. But in the lab frame it lives (on average): lab =, and =E/(m 0 c 2 ) = 1/0.106 10 lab = =(10)(2 sec) = 20 sec On average how far does this particle travel in the LAB before decaying ? x lab = c with = p/(m 0 c) 1/0.106 10 x lab = c = 10c = (10)(3x10 8 m/s)(2x10 -6 s) = 6x10 3 m Big increase due to Special Relativity