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Relativity

Relativity. Lorentz transformation 4-vectors Relativistic kinematics. “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 • Relativistic kinematics 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: • Euclidean transformation (to a moving frame): 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 (to a moving frame): ct’ = g(ct - bx) x’ = g(x -bct) y’ = y z’ = z Where b = v/c and g = (1-b 2)-1/2 Linear transformation consistent with observed speed of e/m radiation = c in both frames Brian Meadows, U. Cincinnati

  5. y y’ • = v/c in -x direction S S’ x x’ z z’ Lorentz Transformation • This is equivalent to reversing the sign of ¯: • Lorentz transformation: ct’ = g(ct +bx) x’ = g(x +bct) y’ = y z’ = z Where b = v/c and g = (1-b 2)-1/2 Linear transformation consistent with observed speed of e/m radiation = c in both 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 Example: u’ = c and v=0.75c u = (c+0.75c)/(1+0.75)= c !! Brian Meadows, U. Cincinnati

  7. 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 • Message to take away: Time differs in moving framet° t’ NOTE that x=x’=0 So elapsed time in S is T = gT0 Brian Meadows, U. Cincinnati

  8. 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

  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 in space-time xm = (ct, x1, x2 , x3) = (ct, x) • Then the Lorentz transformation can be written as: x0’ =g(x0 - bx1) x1’ =g(x1- bx0) x2’ = y x3’ = z Where b = v/c and g = (1-b 2)-1/2 • Any quantity with 4 “components” that transforms this way is called a “relativistic 4-vector”. (1) Brian Meadows, U. Cincinnati

  11. The “4-Velocity” • A 4-vector must transform as given by (1) on previous page. • Easiest way to obtain a new 4-vector is to multiply an old one with a constant factor (or by an invariant). • One useful invariant is t= 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

  12. Relativistic Kinematics Brian Meadows, U. Cincinnati

  13. Two “4-Vectors” • Since dt (= dt/g) is invariant and dxmis a 4-vector, then: Um = dxm / dt = (gc, gdx1/dt, gdx2/dt, gdx3/dt) = (gc, gdx/dt ) = (gc, gv ) • The rest mass of particle m0 is also constant. So we can define another 4-vector: Pm = m0Um = (gm0c, gm0dx1/dt, gm0dx2/dt, gm0dx3/dt) = (gm0c, gm0dx/dt ) = (m0gc, m0gv ) • It is as if the “moving mass” is m = m0g and Pm = (mc, m0v) = (mc, mp ) Brian Meadows, U. Cincinnati

  14. Pm = (mc, mp) The 4-Momentum • The zeroth element is particularly interesting. We relate it to the energy E: P0 = E/c ie E = mc2 !! • We define three kinds of energy: • Total relativistic energy (when particle is moving) E = mc2 • The relativistic rest energy (when b = 0) E0 = m0 c2 • Kinetic energy (energy due to motion) T = E – m0 c2 Brian Meadows, U. Cincinnati

  15. Kinetic Energy • We find agreement with classical mechanics when b<<c: • Binomial expansion for b << c : g = (1-¯2)- ½ = (1 + ½b2 + 3/8b4 + …) • Therefore, for b << c : E = m0g c2 ¼ m0c2 (1 + ½b2 + 3/8b4 + …) ¼ m0c2 + ½ m0 v2 • So the kinetic energy T = E - m0c2 for b << c is: T¼ ½ m0v2 !! Brian Meadows, U. Cincinnati

  16. Invariants • These are quantities that have the same value in any frame of reference (S, S’, etc.) • For any 4-vector x¹, the quantity (x0)2 – (x1)2 – (x2)2 – (x3)2 is the same in S as it is in S’ie (ct’)2 – (x’)2 – (y’)2 – (z’)2 = (ct)2 – (x)2 – (y)2 – (z)2 etc. You can verify this for yourself ! How ?? Brian Meadows, U. Cincinnati

  17. Invariant (or “rest”) Mass • For a 4-momentum P¹, the invariant is called invariant mass: (moc)2 = (P0)2 – (P1)2 – (P2)2 – (P3)2 = (E/c)2 – (px)2 – (py)2 – (pz)2 so mo 2 = (E/c)2 – |p|2 This is the same in all reference frames, including the rest frame where it is equal to the “rest mass”. • This means that, for a particle with rest mass mo E2 = p2c2 + mo2c4 Brian Meadows, U. Cincinnati

  18. Invariant Mass • For two particles with 4-momenta (P¹)1 and (P¹)2 there is also an “invariant mass” m12 that is, in effect, their “rest mass” that is the same in all frames: (m12c)2 = [(E1+E2)/c]2 – |p1+p2|2 • For more than two particles the “invariant mass” m12…n (“rest mass”) is given by: (m12…nc)2 = [(E1+E2+…+En)/c]2 – |p1+p2+…+p2|2 Brian Meadows, U. Cincinnati

  19. Units and c • Factors of c are confusing so, in relativistic kinematics, it is common to use units in which c = 1. These are labelled as: • eV (energy) • eV/c (momentum) • eV/c2 (mass) • Then we can write • Pm = (E, p) • Um = (g, gu) … etc. • E2 = p2 + m02 • m12…n2 = (E1+E2+…+En)2 – |p1+p2+…+p2|2 • Kinetic energy T = E - m0 Brian Meadows, U. Cincinnati

  20. 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

  21. 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

  22. 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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