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Relativity

Relativity. Lorentz transformation 4-vectors and 4-tensors Relativistic kinematics Electromagnetism. “Space-Time”. In relativity, we need to understand that space and time are connected.

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Relativity

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  1. Relativity • Lorentz transformation • 4-vectors and 4-tensors • Relativistic kinematics • Electromagnetism Brian Meadows, U. Cincinnati.

  2. “Space-Time” • In relativity, we need to understand that space and time are connected. • Before relativity, we believed that spatial position is defined by a vector s (with three components – x, y and z) and time (defined as a single quantity t) are independent of one another. • In relativity, s and t are related quantities that we use to define “space-time”. • The “where” and the “what” must BOTH be specified. Brian Meadows, U. Cincinnati

  3. y y’ • = v/c in x direction S S’ x x’ z z’ Euclidean Transformation • Relates two “standard frames” of reference S and S’ that coincide at time t=0: • Lorentz transformation: ct’ = ct x’ = x - bct y’ = y z’ = z Where b = v/c Linear transformation Brian Meadows, U. Cincinnati

  4. y y’ • = v/c in x direction S S’ x x’ z z’ Lorentz Transformation • Principal of special relativity relates two “standard frames” of reference S and S’ that coincide at time t=0: • Lorentz transformation: ct’ = g(ct - bx) x’ = g(x-bct) y’ = y z’ = z Where b = v/c and g = (1-b2)-1/2 Linear transformation consistent with observed speed of e/m radiation = c in both frames Brian Meadows, U. Cincinnati

  5. Lorentz Transformation • Inverse transformation (b –b) ct = g (ct’ + bx’) x = g (x’ + b ct’) y = y’ z = z’ • Can generalize: • Origin of S’ need not coincide with that of S when t’ = 0 • S’ may be differently oriented wrt S • However, many problems can be solved using the standard frames: Brian Meadows, U. Cincinnati

  6. Example • The velocity of a particle is u0 in the x0-direction in S0. What is its speed as seen in S? In S’ distance moved is Dx 0 =u 0Dt 0 in time. In S (use Lorentz transformation): Dx = g(Dx’ + bcDt 0) cDt = g(c D t ’ + bDx0) sou = Dx / Dt = g(Dx’ + bcDt ’) / {gc(c D t ’ + bDx’)} u = (u ’ + v) / (1 + u ’v/c2) • Note u < u’ + v Ltv!c (u) = c Brian Meadows, U. Cincinnati

  7. Length Contraction • A line of length L’ in S’ has one end at the origin, the other at x’=L’. What length is recorded in S? • Solution: • Both ends of the line must be observed at the same time in S, say at t=0. • These times will NOT be the same in S’ • “Other end” will be observed at ct’ = g(ct-bx) = - gbL • Define two events in space-time in each frame – one for each end: S’S One end of line: (0, 0, 0, 0) (0, 0, 0, 0) Other end: (-gbL, L’, 0, 0,) (0, L, 0, 0) So measured length in S is L = g(x’+bct’) = g(L’-b2 gL)  L = L’/g Brian Meadows, U. Cincinnati

  8. Time Dilation • A clock in S’ is at the origin. It records an elapsed time T’ (from t’ = 0 to t’ = T’). What elapsed time is recorded in S? • Solution: • Define two events in space-time in each frame: S 0S Start event: (0, 0, 0, 0) (0, 0, 0, 0) Stop event: (cT , 0, 0, 0,) (gcT 0 , bgcT 0, 0, 0) • NOTE – Stop event is at different place in S NOTE that x=x’=0 So elapsed time in S is T = gT0 Brian Meadows, U. Cincinnati

  9. More Examples • m§lifetime: If a m+ travels at speed v = 0.996 c towards Earth, how far can it travel in its lifetime t =1.6£ 10 -6s ? • Solution: Consider m+ to be the clock (at rest at origin of S’). Earth observer is at rest at origin of S. Then problem is as above with T’ =t since t is time measured by the m+. So distance moved in S is bgct where b = v/c and g=(1-b2)-1/2 In the present example: b = 0.9996, g = 35.36 so bg ct = 16.965 km NOTE – it only travels bct=475m (factor g less!) in its own rest frame Brian Meadows, U. Cincinnati

  10. 4-Vectors • Convenient to define coordinates by x0 = ct, x1 = x, x2 = y, x3 = z and use 4-vector notation for point xm in space-time: • Then the Lorentz transformation can be written as (NOTE - summation over repeated indices) Brian Meadows, U. Cincinnati

  11. Lorentz Transformations • Obvious properties of Lorentz transformations: • If axes of S’ are rotated wrt S with Euler angles (q, f, y) then generalized Lorentz transformation is: is a 4-dimensional representation of the rotation operator Brian Meadows, U. Cincinnati

  12. Generalized Lorentz Transformations • So • And then is the generalized Lorentz transformation. Brian Meadows, U. Cincinnati

  13. Invariant, Contra- and Co-variant • The space-time interval between two points evaluated in S is s = (Dx0)2 – (Dx1)2 – (Dx2)2 – (Dx3)2 • This is equal to the identical quantity evaluated in S’ s’ = (Dx0’)2 – (Dx1’)2 – (Dx2’)2 – (Dx3’)2 a requirement from the principle of special relativity. • A neat way to write this is s = gmnDxn Dxm(sum 03 over repeated indices m and n) This is an example of an “invariant” Brian Meadows, U. Cincinnati

  14. Invariant, Contra- and Co-variant • The signs of the terms requires that gmn be defined as • An even neater notation is to define Dxm = gmnDxn. Then s = DxmDxm • Vectors with low- (upp-)er indices are “contra-” (“co-”) variant Co-variant form: Contra-variant form: NOTE change of sign of 1-3 components Brian Meadows, U. Cincinnati

  15. Invariant, Contra- and Co-variant • Any 4-vector product is also an invariant • NOTE that the “g” are simply numbers (0, 1 or -1, depending on  and ). • Therefore: Brian Meadows, U. Cincinnati

  16. Other 4-Vectors • A 4-vector must transform as given by (1) • Easy way to obtain a new 4-vector is to multiply an old one with a scalar factor (invariant). Brian Meadows, U. Cincinnati

  17. The 4-Velocity • A 4-vector must transform as given by (1) • Easy way to obtain a new 4-vector is to multiply an old one with a constant factor (or by an invariant). • One useful constant is t the time elapsed in the rest frame of a particle at rest in S’. • An excellent example is the lifetime of a decaying particle (e.g. a ¹) in its own rest frame. This is an invariant, characterizing the particle. • In an arbitrary frame S, the time observed is t = gt, so dt = gdt • Since dt is invariant, then: Um = dxm / dt 4-velocity (gc, gdx1/dt, gdx2/dt, gdx3/dt) is a 4-vector • Other examples follow … Brian Meadows, U. Cincinnati

  18. D’Alembertian Operator • Derivative operator ( m´¶m = ¶ /¶xm ) is a 4-vector: • Summary Brian Meadows, U. Cincinnati

  19. Relativistic Tensors • 4-vectors transform according to • These are tensors of rank one (Scalars are rank zero) • They have 4 components • Tensors of rank 2 – each index must transform as above • These have 4x4=16 components Brian Meadows, U. Cincinnati

  20. Recall that the are just numbers (, , 0 or 1 that depend on  and ) Relativistic Tensors Rank 2 • Example – 4-dimensional “vector product” Fmn = AmBn – AnBm is a 4-tensor of rank 2. • Proof: Transform Fmn • Similarly, Fmn and Fmnare also 4-tensors of rank 2 Brian Meadows, U. Cincinnati

  21. Relativistic Kinematics Brian Meadows, U. Cincinnati

  22. 4-momentum (m0gc, m0g dx/dt) The 4-Momentum • The rest mass of particle m0 is constant. So we can define another 4-vector: Pm = m0 Um • Three kinds of energy: • Total relativistic energy E = m0g c2 i.e. P0 = E/c • The “rest energy” (when b = 0) is E0 = m0 c2 • Kinetic energy (due to motion) T = E – m0 c2 Brian Meadows, U. Cincinnati

  23. Kinetic Energy and Invariant Mass • In agreement with classical mechanics when b<<c: • Binomial expansion for b << c : E = m0g c2 ¼ m0c2 (1 + ½b2 + 3/8b4 + …) ¼ m0c2 + ½ m0 v2 T¼ ½ m0 v2 • Leads to concept of invariant mass: Relativistic mass m m = m0g ; Relativistic 3-momentum p p = m dx/dt (as in classical mechanics, but m  m0) “Invariant mass” m: m2c2 = pmpm (same in all frames)= (m0g c)2 – p2 = E2/c2 – p2 Brian Meadows, U. Cincinnati

  24. Units and c • Factors of c are tedious (or even confusing). • So, in relativistic kinematics, it is common to use units • eV (energy) • eV/c (momentum) • eV/c2 (mass) • Then we write • E2 = p2 + m2 • Pm = (E, p) • Um = (g, gu) … etc. • Kinetic energy T = E - m0 Brian Meadows, U. Cincinnati

  25. Collisions and Decays • Total energy and 3-momentum are conserved in all collisions • Relativistically, this means that all 4 components of Pm are separately conserved • Rest masses can change, so kinetic energy can : • decrease (endo-thermic) • increase (exo-thermic) OR • remain the same (elastic) Brian Meadows, U. Cincinnati

  26. Examples: • A beam of protons with momentum 3 GeV/c along the +x axis collides with a beam of 1 GeV/c anti-protons moving in the –x dierction to make two photons. Compute the maximum energy a photon can have. • Solution: The reaction is p+ + p- ga + gb Energy conservation: E=E- + E+ = (32 + 0.9382)½ + (12 + 0.93822)½ = 4.5145 (GeV) = pa + pb(for photons Ea,b = pa,b) Momentum conservation (only need 2 components): pa cos qa + pb cos qb = P = 2 (GeV/c) pa sin qa - pb sin qb = 0 (GeV/c) Solve pa = 0.5 (2EP cos qb – P2 – E2)/(P cos qb – E) Maximum energy of photon is for collinear collision (qa= 0, qb= p):  pa = 3.2573 (GeV) - this is the maximum energy  pb = 1.2573 (GeV) ga qa p p qb gb Brian Meadows, U. Cincinnati

  27. Examples: • A r0 resonance (meson) decays into two pions r0 p+p- • Compute the momentum pp of each pion in the r0 CMS. • Solution: Work in the r0 CMS where E = mr – the rest mass of the r0 and 3-momentum sum is zero: Momentum conservation: p+ = p- = pp (GeV/c) Energy conservation: mr = Ep+ + Ep- = 2 (pp2 +mp2) ½ So using mr = 0.752 (GeV/c2) andmp = 0.1396 (GeV/c2) pp2 = mr2 / 4 – mp2  pp = 0.349 (GeV/c) Brian Meadows, U. Cincinnati

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