Selected Topics in Physics a lecture course for 1st year students by W.B. von Schlippe Spring Semester 2007 Lecture 7 1. Relativistic Mechanics Charged particle in magnetic field 2. Relativistic Kinematics of Particle Collisions 1
Particle of mass m and charge q in a magnetic field B: The equation of motion is dp qv B dt = where v is the particle s velocity, p= γ mv is its momentum and F = qv B is the Lorentz force For simplicity of notation write B for qb and restore q at the end. Thus hence dp 1 v B p B dt = = γ m dp p 0, p ( = p B) = 0 dt 2
or dp dt 2 = 0 i.e. 2 p = const. and hence 2 2 2 2 4 E = p c + m c = const. and v p = = E const. Consider a solenoidal field in z direction: B= B Bz ( 0,0,1) ˆ 3
and then consider two cases: (i) p B, and (ii) p B. Case (i): p B: then hence dp 0 dt = p = const.; v = const. i.e. the particle continues to travel with its initial velocity. Case (ii): p B: p B= B p p ( y, x,0) hence dp dt z = 0, p = z const. and if initially p z = 0, then the particle continues to move in the (x,y) plane 4
Thus and since we get hence w.l.o.g. p = p ϕ ϕ p = pϕ ϕ ϕ ( cos,sin,0) ( sin,cos,0) 1 pb p = p B= γ m E pb = E ϕ = ( sin ϕ,cos ϕ,0) B E B ϕ = ϕ t E ( cos ϕ,sin ϕ, 0) ( 0, 0,1) ( t ) 0 0 ϕ = = 0, and t 0 0 0 5
hence ϕ = B t E Now since the vectors v and p are parallel, we have: hence t v = r = v ϕ ϕ ( cos,sin,0) E r = r + v dt = r + v B ( cos ϕ,sin ϕ, 0) ( sin ϕ, cos ϕ, 0) 0 0 0 and finally 2 2 p 0, r r = v= p E qb this is the equation of a circle of radius R = p/qb, remembering that we had absorbed the charge q in B. 6
Example of an application of our result: mass spectrometer. c) a) b) x rays Two spectrometers to measure the momenta of electrons, ejected from a metal foil illuminated by x-rays. The magnetic field is perpendicular to the plane of the drawing. a) narrow source, wide slit: the image of the source lies on the photographic emulsion. b) broad source, narrow slit: the image of the slit lies on the photographic emulsion. From R.W. Pohl, Optik und Atomphysik c) Spectrum of electrons ejected from silver atoms by x-rays of 60 kev. The different lines correspond to different atomic energy levels from which the electrons are ejected. 7
The next three pictures are photographs of bubble chamber events. A bubble chamber is a volume of superheated liquid, placed in a strong magnetic field. A charged particle passing through the liquid causes ionization along its track. The ions serve as seeds to begin boiling. When the bubbles have grown to a visible size, but before the entire volume of the chamber is filled with bubbles, a light flash is triggered and a photograph is taken From the measured curvature of the tracks the momenta of the particles are calculated. To do this we write the formula we have derived in the form of p = qbr The sense of the curvature gives the sign of the charge of the particle. The reconstruction of bubble chamber pictures has enabled particle physicists to discover many unknown particles, including neutral particles which leave no tracks but decay into pairs of charged particles. 8
e + e - pair recoil proton track of incident pion 9
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The discovery of the Ω - meson gave strong support to Gell-Mann s quark model. 11
The advantage of bubble chambers is the high quality of their pictures. The disadvantage is the slow rate at which pictures can be taken: there is a time of several seconds before the bubble chamber is ready for the next event. Therefore bubble chambers are today replaced by detectors consisting of an inner gas-filled tracking chamber surrounded by a calorimeter. The particle tracks are reconstructed by dedicated computers which determine the momenta of the charged particles from the curvature of the tracks. Calorimeters are devices which measure the energies of particles. Combining the information on energies and momenta one can calculate the particle masses. The following slides show an example of such a detector, the H1 detector at the DESY laboratory in Hamburg, Germany. 12
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superconducting coil 1.2 Tesla; diameter 5.2 m Central tracking chamber; length 2.6 m; diameter 2 m 14
Cross section of the H1 detector at the HERA electron-proton collider in DESY. superconducting coil 1.2 Tesla; diameter 5.2 m 15
Computer reconstruction of a collision event in the H1 detector. Momenta of particles are found from the curvature of the tracks; their energies are measured in the calorimeters. calorimeters 16
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CMS (Compact Muon Solenoid) under construction at CERN the superconducting coil gives a magnetic field of 4 Tesla. 18
January 2006: First cosmic rays seen in a complete sector of CMS muon chambers. The CMS solenoid at 100 Kelvin 19
14000 tons is an awful lot of weight, about the weight of a good-sized battle cruiser including 1000 sailors. And all this is hard stuff. Mostly iron. So you may be surprised that a particle, like a muon for instance, can pass through. But the even more surprising thing is that this huge bulk is mostly empty space to the muon. Let s assume for simplicity that the CMS detector is made purely of iron. Iron has atomic number A = 56, so one atom of iron weighs about 56 times the weight of a proton (or neutron, which is practically the same). But one proton weighs roughly 1.7 times 10-27 kg, so one iron atom weighs about 95 times 10-27 kg or, between friends, 10-25 kg Therefore the number of iron atoms in that CMS detector is N 14000 tons = = 1.4 10 25 10 kg 32 20
Now the volume of a nucleus is V = 4π R 3 3 and the nuclear radii are to a good approximation given by V R= A R ; R 1.4 10 1/3 15 0 0 m so the volume occupied by a single iron nucleus is 3 3 m 4AR = 4 56 1.4 10 0.6 10 3 3 45-42 0 m and hence the volume taken up by all iron nuclei of the CMS detector is V = N V 1.4 10 0.6 10 iron / CMS = 10 m 0.1 mm 10 3 3 the rest of the bulk of the detector is empty space. 32 42 3 m 21
A sector of the cross section of the CMS detector. The field of the SC solenoid is 4 T; outside the field is about 2 T in the return yoke. Simulated tracks of some charged particles are shown. 22
Particle collisions at relativistic energies Collision processes are of two kinds: (i) elastic collisions, (ii) inelastic collisions In relativistic kinematics such processes are studied within the framework of conservation laws: conservation of energy and of momentum. Relativistic kinematics is not concerned with the forces acting between the colliding particles. A typical problem of relativistic kinematics is the following: Consider two particles, a and b. Let b be initially at rest and a moving with 4-momentum p a. After their collision particle a has a different momentum and particle b has also acquired a momentum. Then one wants to know what the relation is between the momenta of a and b. 23
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We begin solving the problem by writing down the equation of 4-momentum conservation: p p + p = p + p a b a b To these we can add the four equations for the invariant squares of the 4-momenta: = = 2 2 ( E c) p ( m c) ( mc) ( mc) 2 2 a a a a 2 2 2 2 2 a = a b = b = b p, p p We can simplify our notation if we set c = 1. There is an additional advantage in doing that: now all equations must be homogenous in masses, energies and momenta, so it is easy to keep track of dimensionally correct calculations! Thus our invariant squares of 4-momenta will be of the general form of p = E p = m 2 2 2 2 25
and the 4-momenta of particles a and b in the case of elastic scattering are ( E p ) ( E p p p ) ( m ) ( E p p p ) p =,0,0,, p =,,,, a a a a a ax ay az p =,0,0,0, p =,,, b b b b bx by bz Given the momentum of the incident particle and the scattering angle we can now for example find the momentum of the scattered particle, and then also the recoil momentum and recoil angle. To do this we write down the energy conservation equation: W E + m = E + E a b a b 26
then express the energies in terms of the corresponding 3-momenta, hence W = m + p + m + p 2 2 2 2 a a b b and use momentum conservation to express the recoil momentum in terms of the momenta of the incident and the scattered particles: p = p + p a a b hence W = m + p + m + p p ( ) 2 2 2 2 a a b a a = m + p + m + p + p 2p p cosθ 2 2 2 2 2 a a b a a a a and solving for the momentum of the scattered particle we get p = a ( ± ) p a ( ) s+ m m cosθ ± 2W m m sin 2 2 2 2 2 a b b a 2 2 ( s+ pa θ ) 2 sin θ (A) 27
where ( p p ) 2 s = + a b This is an invariant, and we shall make use of its invariance to express s in different inertial frames. But at the moment we shall write it explicitly for the frame we have specified (i.e. with the target particle at rest): s = m + m + 2m E 2 2 a b b a There are two features in our result (A) that need discussing: (i) the expression under the square root (ii) the ± in front of the square root. (i) the expression under the square root ( radicand ) is m m sin θ 2 2 2 b a and this must be greater than or equal to zero for the momentum of the scattered particle to be real: 28
thus the reality condition demands m m sin θ 0 2 2 2 b a This will be satisfied for all scattering angles θ if the mass m b of the target particle is greater than the mass m a of the incident particle. However, if the mass m a is greater than m b, then the reality condition restricts the scattering angles: sinθ m m < 1 b a (ii) The ± sign of the square root: if m a < m b, then the lower sign gives a negative momentum and must therefore be rejected; if m a > m b, then both signs give a positive momentum, there are therefore two values of the momentum of the scattered particle for every value of θ 29
m a < m b all scattering angles θ give a real momentum; the lower sign must be rejected. m a > m b sinθ = m m < 1 max both signs of the square root must be kept. b a 30
Having found the momentum of the scattered particle we can get the recoil momentum and recoil angle from momentum conservation: hence ( ) 2 p = p + p a a b p = p p = p + p 2p p cosθ 2 2 2 b a a a a a a tanϕ = p a p a sinθ p cosθ a These results can be applied for instance in the design of collision experiments. Assume that we want to study elastic collisions at an energy at which some of the collisions are inelastic. Then we can select among all collisions the elastic ones by placing detectors at the appropriate angles. 31
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Compton scattering Consider the particular case when the incident particle is a photon, i.e. m a = 0. Then we have E a = p a etc. and the momentum of the scattered photon is p = a = p a mp b ( ) 2mpcosθ+ 2 p + m m b a a b b 2 2 2 ( mb + mbpa + pa θ ) 2 2 sin a ( ) ( ) p cosθ + p + m a a b 2 2 2 a + b a cos p m p θ where we have kept only the + sign because the incident particle (photon) has a smaller mass than the target particle (electron). Dividing by m b p a and taking the reciprocal we get 33
2 2 2 ( pa + mb) pa cos cosθ ( ) m p ( 1 cosθ ) mp θ b a = = ( p + m ) p p p + p + m a a a b = + b a a b a cosθ hence 1 1 = 1 1 cos p p m a a b or using the de Broglie relation and reinstating c we get ( θ ) p= h λ h λ λ = θ mc b ( 1 cos ) i.e. we have recovered the famous Compton effect formula. 34
Of course, we could have found the Compton effect formula much more easily in a few lines if we had started with the assumption of a zero-mass incident particle. Beginning with 4-momentum conservation, we have ( ) = + 2 p k k p or k+ p= k + p 2 2 ( ) ( ) = + + 2 p k k 2p k k ( k k ) 2 2 ( ) = + + 2 2 m m m k k hence ( ) = ( 1 cos ) m k k kk θ 1 1 cos m and finally λ λ = ( θ) 35
Laboratory and centre-of-mass frames The reference frame that has a target particle initially at rest is called laboratory frame (LAB). For many theoretical investigations it is more convenient to work in the centre-of-mass frame (CMS) which is defined by the momenta of the initial particles being equal in magnitude and opposite in direction. 36
I will label all CMS variables by an asterisk. Thus the 4-momenta for the case of elastic scattering a + b a + b are in the CMS and LAB: CMS = ( Ea p ) ( Eb p ) ( Ea p ) ( Eb p ) * * * p a,0,0,, * * * p b =,0,0,, * * * p a =,, * * * p b =,. LAB ( E p ) ( m ) p =,0,0,, a a a p =,0,0,0, b b ( E p p p ) ( E p p p ) p =,,,, a a ax ay az p =,,,. b b bx by bz I have previously defined the invariant s: ( p p ) 2 s = + and we can see that expressed in LAB variables it is given by a s = m + m + 2m E 2 2 a b b a b 37
and in the CMS it is * * ( ) 2 a b s = E + E i.e. s is the square of the total CMS energy. Now since E = m + p, E = m + p * 2 *2 * 2 *2 a a b b we can solve for p*: * 1 { ( ) 2 ( ) 2 } 12 a b a b p = s m m s m + m 2 s The velocity of the CMS relative to the LAB ( boost velocity ) is found by writing down the LT formula for the target particle and noting that by definition its LAB momentum is zero: * * ( ) p = γ p + VE = bz bz b 0, 38
hence V p = = E p * * bz az * * b Eb where in the last step I have used the definition of the CMS: p = p * * az bz The scattering angles in the LAB and CMS are related by sinθ tan θ ; tan γ where * = * θ = sinθ * * ( cosθ + V v ) γ ( cosθ V v) v = p E v= p E * * * 3 3 and 3 3 are the velocities of the scattered particle in the LAB and CMS, respectively. 39