1 / 25

Relativity

Relativity. Lorentz transformation 4-vectors and 4-tensors Relativistic kinematics Electromagnetism . y. y’. = v/c in x direction. S. S’. x. x’. z. z’. Lorentz Transformation.

gwyn
Download Presentation

Relativity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


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

  2. 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

  3. 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

  4. 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

  5. More Examples • 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

  6. More Examples • Time Dilation: A clock in S’ is at the origin. It records a time T’ (from t’ = 0 to t’ = T’). What 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, 0,) (gcT 0 , bgcT 0, 0, 0) • NOTE – Stop event is at different place in S So elapsed time in S is gT0 Brian Meadows, U. Cincinnati

  7. 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

  8. 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

  9. 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

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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

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

  17. 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

  18. 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

  19. Relativistic Kinematics Brian Meadows, U. Cincinnati

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

More Related