1st year Relativity - Notes on Lectures 6, 7 & 8

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1 1st year Relativity - Notes on Lectures 6, 7 & 8 Lecture Six 1. Let us consider momentum Both Galilean and relativistic mechanics define momentum to be: p = mv and p i = P = a constant i i.e. Total momentum in a system of interacting particles is conserved - has its origin in Newton s third law and it is a principle that holds both relativistically and non-relativistically. Therefore we have conservation of momentum in special relativity. However, as p classically v - not allowed in special relativity! Therefore: p = mf(v)v where f(v) 1 as v 0 and must be dimensionless, f(v) can only depend upon the magnitude of v and thus can only depend upon terms such as v (and higher even powers). Because it is dimensionless then it is reasonable to say f(v) v /c. Therefore f(v) = f(v /c ) or f(v) = f(γ) In fact f(v) = γ = 1 1 v /c To get the identity f(v) = γ requires an analysis of collisions between equal mass particles (see later). So we have: mv p = mγv = 1 v /c m is the REST MASS and I will always use it as such to avoid confusion. Thus Newton s second law becomes: F = dp = m d (γv) N.B. we now see that F and dv. Force: no longer have to point in the same direction! F = dp = m d (γv) F = γma + mv dγ where a = dv so if a leads to a change of speed, dγ is finite. Therefore force equals mass times acceleration is NOT valid in relativistic mechanics.

2 3. Motion of a particle in a field The Lorentz force law for EM fields: f elect = qe f mag = qv B F total = d (γmv) = d [ ] mv = q(e + v B) (1 v /c ) 1 q must not depend upon v otherwise the hydrogen atom would not be neutral! (confidence in this statement is at the 1 : 10 0 level experimentally). Shown by King (1960). 4. Motion in an electric field Diagram is on viewgraph (p4 of Rosser fig.). Form above [ ] d v = qe (1 v /c ) 1 m Then integrate with respect to time to give: [ ] v v = qe (1 v /c ) 1 m [t]t 0 0 v = (1 v /c ) q E t v = v = multiply top and bottom by mc/qet to get: v = m q E t ( ) m 1 + q E t m c qet m ( ) 1 + q E t 1 m c c ( ) 1 + m c 1 q E t If t is very big (long acceleration time in the field) then m c q E t (although always less than c). In Newtonian mechanics, 1 and v c a = f m = qe m v = at = qet m v as t

3 from above if qet mc 1 then v qet m as required. This was confirmed experimentally by Bertozzi (1964). A van der Graaf accelerator was used to accelerate electrons to 1.5MeV and then to 15MeV using a linac [French p7 - p9]. Times of flight over 8.4m were measured. For the speed of light, t = 8ns. t pulse = 3ns, therefore time of flight is long compared to the pulse wih [see viewgraph]. 5. What about x? If at x = 0, t = 0 then, v = dx = x 0 dx = t 0 qet m ( ) 1 + q E t 1 m c qet m ( ) 1 + q E t 1 m c this can be integrated by the substitution w = 1 + q E t /m c giving (do this as an exercise) ( x = mc 1 + q E t ) 1 1 qe m c From Newton x = 1 at = 1 qet m theorem: x mc qe 1 qet m expand the above equation for x using the Binomial ( q E t ) m c The above approximation is OK for v < 3000kmh What about B fields? Here v is a constant because: Work done = W = f.dl then dw = f. dl = f.v f = qv B dw = qv.v B 0 Therefore the kinetic energy is constant, v is constant, thus f = d (γmv) = γmdv = qv B acceleration a = qvb γm but a = v r = qvb γm

4 and the period is given by r = γmv qb p = γmv = qbr T = πr v = πγm qb then we have an expression for the Larmor frequency ω L : ω L = qb γm 7. Transformation of Velocity ****THIS IS NOT REQUIRED FOR PRELIMS**** Consider two inertial frames S and S, where S is moving at a constant relative speed v with respect to S along their mutual x axes. Then we have In S, u = dx In S, u = dx 8. Then applying the Lorentz transformations gives x = γ(x βct) y = y ct = γ(ct βx) z = z u x = dx = dx dx ( ) dx = γ v = γ(u x v) ( = γ 1 dx ) v c division gives u x = u x v (1 u x v/c ) if we now look at the y components we get u y = u y γ(1 u x v/c ) and similarly for u z. Thus even for relative motion of the inertial frames along x alone, u x, u y and u z are ALL changed. 9. If u y and u z are not simple transformations then is there some set of quantities which does transform more like x and ct? 10. Momentum from above if we stick to mv as p then conservation of momentum becomes FRAME DEPENDENT.

5 11. As an example consider the viewgraph showing the collision of two particles. Consider an elastic collision between two equal mass particles from the point of view of two inertial frames S and S. S is the centre of mass frame and thus both particles approach each other with equal and opposite velocities. Before After v 1 = ( v x, v y ) v 1 = ( v x, v y ) v = (v x, v y ) v = (v x, v y ) the y component of velocity is reversed in the collision. In S, which has a velocity v x along O x with respect to S, particle has no component of velocity v x. Before After v 1 = ( v 1x, v 1y) v 1 = ( v 1x, v 1y) v = (0, v y) v = (0, v y) We can now use the transformation of velocities to relate v y and v y. v 1y = v y = v y γ(1 + vx/c ) v y γ(1 vx/c ) but the masses are the same, so the momentum in the y direction is not conserved in S if it is in S. The problem is that time is NOT an invariant, we need PROPER TIME: τ = t/γ or t = γτ where τ is the proper time. In the particle s rest frame γ = 1 so t = τ. This leads us to a definition of the relativistic velocity of the form: V = dx dτ = dx dτ = γ(v)dx we then have a definition for relativistic momentum of the form: P = mv = γ(v)mv Does this fix the conservation of momentum problem above? - YES. In S: p y = m dy dτ In S : p y = m dy dτ But m is constant and τ an invariant so if y = y then p y = p y. Therefore if momentum is conserved in y direction in S it will remain so in S. 1. What happens to p x? p x = m dx but x = γ(x βct) dτ ( ) dx p x = mγ(v) β(v)c dτ dτ p x = γ(v) (p x mβ(v)cγ (u))

6 This is nearer our goal of a Lorentz transformation but the second term above is still a problem - we need the energy! 13. Relativistic Energy The relativistic force is given by: Then the work done W is: F rel = dp W = F.dx To bring a particle at rest at t = 0 to a speed v at time t we have W = t F dx = F dx t 0 = F v 0 F v = v dp = vm d (γv) = mv dγ dv + mγv The first term contains dγ W = = d ( 1 v /c ) 1 = 1 ( ( 1 v /c ) 3 c v dv ) = γ3 c v dv [ ( ) ] t v mγ 3 + mγ v dv 0 c (a) The factor in square brackets in this expression is: [( ) ] [ v mγ γ v /c ] [ ] + 1 = mγ c 1 (v /c ) = mγ = mγ 3 1 v /c Thus the integrand is mγ 3 v dv, but from (a), mγ 3 v dv = mc dγ t W = mc dγ = mc 0 [ W = mc (γ 1) = mc γ=γ γ=1 dγ 1 (1 v /c ) 1 1 ] This is the expression for the Relativistic Kinetic Energy. 14. What happens as v/c 0? For small v /c, (1 v /c ) 1 1 v /c W = mc. v c = 1 mv!

7 15. An Invariant p = γmv pc = γmvc Now DEFINE E = W + mc = γmc. Then consider E (pc) : E (pc) = (mc ) γ (mc) γ v = (mc) γ (c v ) But γ = 1 1 v /c = c c v E (pc) = m c 4 INVARIANT Lecture Seven 16. Let s recap the results of the last lecture: E p c = m c 4 INVARIANT p c E = m c 4 c.f. x c t = c τ INVARIANT When v = 0, γ = 1 and W = 0, therefore E = mc This is the REST ENERGY of the particle. 17. Transformation of E and p From above p x = γ(v) (p x mβ(v)cγ (u)) The extra bit is now recognisable as E/c. Therefore p x = γ(p x βe/c) and E = γ(e βp x c) Thus we can now write the transformation of relativistic momentum and energy in special relativity as: p x = γ(p x βe/c) p y = p y p z = p z E /c = γ(e/c βp x ) This is just the same Lorentz transformation as for (x, t, x, t ). So in general we can write: X µ = L µν X ν

8 where X is a 4-vector and L µν is the Lorentz transformation matrix. Thus quite generally: x µ = (x, y, z, ict) p µ = (p x, p y, p z, ie/c) γ 0 0 iβγ L µν = iβγ 0 0 γ 18. What about velocity? If you remember the transformation of velocity discussion in lecture 6 (NOT ON SYLLABUS!) then you will, I hope, see that we can now define a 4-velocity as: U µ = dx µ dτ = γ (u) (u, ic) Then U µ = L µν U ν or in the full, expanded form (NOT ON SYLLABUS!), γ(u ) u x u y u z ic = γ 0 0 iβγ iβγ 0 0 γ u x u y u z ic γ(u) which will lead to the transformation of velocity derived earlier - as an exercise for the brave expand the above to recover the transformation equations derived in lecture 6, note that the ic term lets you eliminate γ(u ). 19. Therefore we have for momentum p µ = mu µ = mγ(v)(v, ic) imc p 4 = imcγ(v) = (1 v /c ) 1 expanding gives or p 4 c i p 4 imc ( ) 1 + v c +... ( ) mc + mv +... = ie 0. I hope that from the above you can now see that it is the ENERGY and the MOMENTUM and not the velocity that are the natural quantities to use in relativistic kinematics. Conservation of energy and momentum holds so that in a closed system (no external forces) E i and p i are CONSERVED.

9 1. The relevance of E rest = mc This is an amazing result, starting from a re-examination of ideas of space, time and simultaneity we now find that mass and energy are in some sense INTER- CHANGEABLE. Because c is so big, small mass changes release huge amounts of energy e.g. fission and fusion.. Zero mass particles If m = 0 then E = pc. Thus radiation (photons) have an associated momentum - already proven by Maxwell (whose equations are, of course, correct relativistically). Thus all massless particles must travel at the speed of light. 3. Relativistic Kinematics Particle physicists don t like c! They use units more suited to collisions involving sub-atomic particles. Everything is measured in terms of energy in MeV or GeV. 1eV = J 1MeV = 10 6 ev, 1GeV = 10 9 ev, 1TeV = 10 1 ev We have E = p c + m c 4 thus we say that mass is measured in units of MeV/c, and that momentum is measured in units of MeV/c. This gives E = p + m 4. There is an important, and useful, limit to the above equations. In particle physics β 1 in many cases. If this is the case then γ becomes the important quantity and the relations for E and p are: E = γmc p = γmcβ γ = E β = p c mc E m always refers to the rest mass quantity. Thus in particle physics units, γ = E m β = p E 5. Centre of Mass and Centre of Momentum Frame and t part = γτ part = Eτ part /m The C of M frame is convenient in both non-relativistic and relativistic mechanics. To transform to the C of M of a system of particles we need to find the velocity of the C of M in terms of E i and p i for the particle in an arbitrary frame. For a single particle we know that E p c is an invariant. Because the Lorentz transformation is linear it is straightforward to show that ( ) ( ) A = E i p i c i is also invariant for a group of particles. In the C of M frame, by definition i p i = 0. So in C of M frame ( ) A = = (E cm) i E i i

10 where Ei is the energy of the i th particle in that frame. Then, by analogy with the single particle case i E i γ CM = E cm and β CM = i p i c i E i where β CM is the velocity of the centre of mass of the particles in the frame in which the momenta and energies are p i, E i, and E cm is the total energy of those particles in the C of M frame. 6. Threshold Energies Because E and m are interchangeable we can create particles out of energy. This energy is usually provided by colliding other, stable, particles such as protons, electrons, etc. Suppose I have a proton beam and a hydrogen target and I want to make a p p pair - what is the minimum energy that the proton beam must have? p + p p + p + (p + p) In the C of M frame, the question is easy, E cm must be just big enough to make 4 proton masses E cm = 4m p c But A = E cm is an invariant and therefore we must have same value in the LAB frame in which there are only two protons, one at rest. Thus in the LAB frame: a E al, p b m p c and A = (E L a + m p c ) (pc) = (E L a ) (pc) + m p c E L a + (m p c ) but (E L a ) (pc) = (m p c ) (invariants again) A = m p c (m p c + E L a ) but A threshold = (E cm ) = (4m p c ) 16(m p c ) = m p c (m p c + Ea L ) 8(m p c ) = (m p c + Ea L ) Ea L = 7m p c

11 or, the kinetic energy of the proton beam must be T L a = E L a m p c = 6m p c Note by the use of invariants we have NOT needed Lorentz transformations in this problem. 7. General point about body collisions A particle of mass m 1 collides with a particle of mass m at rest, m 1 has energy and momentum given by p 1, E 1. What is E cm, v cm? β cm = p 1 E 1 + m A = (E cm ) = (E 1 + m ) p 1 E cm = (m 1 + m + m E 1 ) 1 Therefore the available energy for creating particles increases only as E 1 when working with a stationary target. This is the reason that colliding beam accelerators are used. 8. Two Body Decay (a) Decay of a stationary particle: a E a * E b * b -q X q In the rest frame of the parent particle (equivalent to the C of M frame of the decay products) X a + b Decay products must have equal and opposite momenta q. What are the energies E a,b? Conservation of Energy and m X c = E a + E b (1) m ac 4 = Ea (qc) () m bc 4 = Eb (qc) (3) subtract equation 3 from equation to give ( ) m a m b c 4 = Ea Eb (4)

12 Divide 4 by 1 to give (m a m b) c finally, add or subtract 5 to/from 1 and get m X = E a E b (5) E a = (m X + m a m b)c m X E b = (m X + m b m a)c m X from we can then recover qc (exercise for the reader!) to give the final result (in particle physics units) q = m4 X + m 4 a + m 4 b m Xm a m Xm b m am b m X = q + m a + q + m b 4m X Thus everything apart from the direction of q is fixed in terms of the particle masses. Lecture Eight 9. Two Body Decay (b) Decay of moving particle: In the LAB system we need to use the Lorentz transformations explicitly. Suppose particle X has energy and momentum E X and p X, take p X along O X. p a E a * q* p X, E X E a q a q* X E b q b p b - q* E b * LAB C of M N.B. the β and γ for the Lorentz transformations are those of particle X, so γ X = E X m X c β X = p X c E X Since we want a Lorentz transformation from the rest frame of X to the LAB frame, we use the inverse transform. In this way we can use results derived in the rest frame. The energy is given by E a = γ X (E a + β X q c cos θ )

13 and similarly for E b. From this relation we see that E a varies between two limits, θ = 0 (max) and θ = π (min). Therefore γ X (E a β X q c) E a γ X (E a + β X q c) These cases correspond to a along the line of flight of X. Now lets look at the angles θ a and θ b. The component of p along O X is: p a cos θ a = γ X (q cos θ + β X Ea/c) p a sin θ a = q sin θ The solution is now straightforward, if messy, remembering that we also have p b cos θ b = γ X ( q cos θ + β X Eb /c) p b sin θ b = q sin θ If you want to follow this derivation in detail see Lectures on Special Relativity by M.G. Bowler. I ll take a special example at this point, π 0 γ. Since γ is massless, E = q c, we can use this and take the ratio of the angular equations above: tan θ a = tan θ b = taking these two together we get tan(θ a + θ b ) = sin θ γ X (cos θ + β X ) sin θ γ X ( cos θ + β X ) β X γ X sin θ (γ X 1) sin θ 1 Thus in this decay in the LAB frame we get a range of opening angles which can be obtained by varying θ between π/ and π (one γ must go backwards) 30. Photons E = hν = hc (m rest = 0) λ p c = E or p = E/c as we have seen before, photons are the ultimate relativistic particle. 31. Compton Scattering This is the scattering of photons by electrons. Apart from being an important process itself, it was also historically important in confirming the quantum theory of light and providing very accurate tests of the relativistic velocity - momentum relation. Let s work in the LAB frame in which an incoming photon of frequency ν scatters from a stationary electron. A scattered photon is produced of frequency ν at an angle θ relative to the incoming photon direction γ + e γ + e

14 ν ν θ φ Ε E = hν, E = hν, E = (p c) + (mc ) Now we conserve energy and momentum to give E + mc = E + E (6) q = q + p p = q q (p c) = (qc) + (q c) q.q c However, E photon = p photon c and so (p c) = E + E EE cos θ = E (mc ) (7) We can use 6 to eliminate E, E = E E + mc (E E + mc ) (mc ) = E + E EE cos θ (E E ) + (E E )mc = EE cos θ + E + E EE + (E E )mc = EE cos θ mc (E E ) = EE (1 cos θ) (8) But we know that E = hν = hc/λ and that E = hc/λ, therefore ν ν = hνν (1 cos θ) mc λ λ = h (1 cos θ) mc This is the expression for the change in wavelength of the scattered photon. 3. We can also find the energy and angle of the recoil electron. Rewriting 6, E mc = E E

15 the left hand side is the kinetic energy of the electron. Rearranging 8 to give E in terms of E and θ: E [E(1 cos θ) + mc ] = Emc E = We can therefore write the kinetic energy T e as T e = E E Emc = E E(1 cos θ) + mc T e = E (1 cos θ) E(1 cos θ) + mc Emc E(1 cos θ) + mc (9) This is a maximum when θ = π (photon is back-scattered) then: T e = E E + mc We can get the angle of scattering by considering the conservation of momentum q = E/c q = q cos θ + p cos φ E = E cos θ + p c cos φ (p x ) (10) 0 = E sin θ p c sin φ (p y ) (11) thus from 11 we have p c = E sin θ sin φ If we now substitute this into 10 we have using equation 9 we can write E mc E = E (cos θ + sin θ cot φ) E = 1 = E E (1 cos θ) + 1 mc cos θ + sin θ cot φ E mc (1 cos θ) + 1 (1 cos θ) + 1 = cos θ + sin θ cot φ ( ) E mc + 1 = cos θ ( E mc + 1 ) + sin θ cot φ

16 at this point I ll introduce a trick, let a = (E/mc + 1) and t = tan θ/ so that (using a trig. formula from the SMP tables) sin θ = t 1 + t cos θ = 1 t 1 + t a = a (1 t ) t cot φ + (1 + t ) (1 + t ) a(1 + t ) = a at + t cot φ at = t cot φ cot φ = at ( ) E cot φ = mc + 1 tan θ/ 33. Recoil shifts in emission and absorption of radiation For an atom or nucleus, let the excitation energy be E 0 and the ground state rest mass energy be mc. Define m c = E 0 + mc (a) Emission Atom/nucleus in an excited state decays at rest A A + γ Since A recoils, E 0 is not the energy of the emitted γ! What is the difference? As in a -body decay Energy m c = E A + E γ (1) The momenta are the same, therefore Eliminate E A using 1 13 to give (mc ) = E A (pc) and 0 = E γ (pc) E γ = (pc) E A = E γ + (mc ) (13) (m c E γ ) = E γ + (mc ) (m c ) m c E γ = (mc ) E γ = (m c ) (mc ) m c

17 But, from energy level diagram, mc = m c E 0 E γ = (m c ) (m c E 0 ) m c E γ = E 0 E 0 m c (b) Absorption Ground state A at rest, absorbs photon of energy E γ to produce and excite state A which recoils γ γ E γ + mc = E E 0 + mc = m c The momentum of A equals that of the γ so eliminate E to give E = E γ + (m c ) (E γ + mc ) = E γ + (E 0 + mc ) E γmc = E 0 + E 0 mc E γ = E 0 + E 0 mc For most nuclei and atoms the difference between m and m can be ignored. This is not so for sub-nuclear particles. The shifts for atoms and nuclei are small and are generally unimportant for atoms, but are relatively more important for nuclei. The criterion is the size of the shift (E 0/mc ) compared to the natural line wih obtained from quantum mechanical considerations: Γ h τ τ = lifetime 34. Mössbauer recoilless spectroscopy Consider the radioactive decay detailed below ν ν 57 7Co 57 6Fe (excited state) + e + + ν e 57 6Fe 57 6Fe + γ (14.4 kev) ( 57 6Fe ) = (Pound and Rebeka, PRL 3, (1959))

18 The linewih of the emitted γ is extremely small. Therefore in this special case we have the opportunity to measure the relativistic correction. From above, for emission: ( E γ = E 0 1 E ) 0 E E = E 0 m c = Thus the effect is easily seen. m c = In 1958 R.L. Mössbauer, aged 9, showed that when radioactive nuclei embedded in a solid emit γ-rays, many fail to recoil as free atoms. Instead they behave as if locked rigidly to the rest of the solid. The recoil is then communicated to the solid as a whole, thus giving an effective m many orders of magnitude bigger than that of the nucleus alone (mass of the crystal). Thus E 0 and we have a so called recoilless process. For example assume that the mass of the crystal is 1g, then E E = E 0 m c = = The Mössbauer effect has been used to test relativity as it gives an extremely precise tool with which to measure any energy shifts. Some of these are detailed below (a) Doppler effect test using resonant scattering (one moving Fe source and one stationary Fe absorber). (b) Tests of the gravitational red shift - Pound and Snider, Phys. Rev. 140, B (1965), PRL 4, (1960). (c) Test of the twins paradox using the thermal motion of Fe nuclei at different temperatures, PRL 4, (1960).