Special Theory of Relativity

Transcription

1 Special Theory of Relativity In ~1895, used simple Galilean Transformations x = x - vt t = t But observed that the speed of light, c, is always measured to travel at the same speed even if seen from different, moving frames c = 3 x 10 8 m/s is finite and is the fastest speed at which information/energy/particles can travel Einstein postulated that the laws of physics are the same in all inertial frames. With c=constant he derived Lorentz Transformations light u sees v_l =c moving frame also sees v_l =c.not v_l = c-u P460 - Relativity 1

2 Derive Lorentz Transform Bounce light off a mirror. Observe in 2 frames: A vel=0 respect to light source A velocity = v observe speed of light = c in both frames -> c = distance/time = 2L / t A c 2 = 4(L 2 + (x /2) 2 ) / t 2 A Assume linear transform (guess) x = G(x + vt) let x = 0 t = G(t + Bx) so x =Gvt t =Gt some algebra (ct ) 2 = 4L 2 + (Gvt) 2 and L = ct /2 and t =Gt gives (Gt) 2 = t 2 (1 + (Gv/c) 2 ) or G 2 = 1/(1 - v 2 /c 2 ) A mirror L A x P460 - Relativity 2

3 Lorentz Transformations Define β = u/c and γ = 1/sqrt(1 - β 2 ) x = γ (x + ut) u = velocity of transform y = y between frames is in z = z x-direction. If do x ->x then t = γ (t + βx/c) + -> -. Use common sense can differentiate these to get velocity transforms v x = (v x - u) / (1 -u v x / c 2 ) v y = v y / γ / (1 - u v x / c 2 ) v z = v z / γ / (1 - u v x / c 2 ) usually for v < 0.1c non-relativistic (non-newtonian) expressions are OK. Note that 3D space point is now 4D space-time point (x,y,z,t) P460 - Relativity 3

4 Time Dilation Saw that t = γ t. The clock runs slower for an observer not in the rest frame muons in atmosphere. Lifetime = τ =2.2 x 10-6 sec cτ = 0.66 km decay path = βγτc β γ average in lab lifetime decay path µs 0.07 km µs 0.4 km µs 1.4 km µs 4.6 km µs 15 km P460 - Relativity 4

5 Time Dilation Short-lived particles like tau and B. Lifetime = sec cτ = 0.03 mm time dilation gives longer path lengths measure second vertex, determine proper time in rest frame If measure L=1.25 mm L and v =.995c t(proper)=l/vγ =.4 ps Twin Paradox. If travel to distant planet at v~c then age less on spaceship then in lab frame P460 - Relativity 5

6 Adding velocities Rocket A has v = 0.8c with respect to DS9. Rocket B have u = 0.9c with respect to Rocket A. What is velocity of B with respect to DS9? DS9 A B V =(v-u)/(1-vu/c 2 ) v = (.9+.8)/(1+.9*.8)=.988c Notes: use common sense on +/- if v = c and u = c v = (c+c)/2= c P460 - Relativity 6

7 Adding velocities Rocket A has v = 0.826c with respect to DS9. Rocket B have u = 0.635c with respect to DS9. What is velocity of A as observed from B? DS9 A Think of O as DSP and O as rocket B V =(v-u)/(1-vu/c B 2 ) v = ( )/(1-.826*.635)=.4c If did B from A get -.4c ( )/(1+.4*.635)=1.04/1.25 =.826 P460 - Relativity 7

8 Relativistic Kinematics E 2 = (pc) 2 + (mc 2 ) 2 E = total energy m= mass and p=momentum natural units E in ev, p in ev/c, m in ev/c 2 -> c = 1 effectively. E 2 =p 2 +m 2 kinetic energy K = T = E - m 1/2 mv 2 if v << c Can show: β = p/e and γ = E/m --> p = βγm if m.ne.0 or p=e if m=0 (many massless particles, photon, gluon and (almost massless) neutrinos) relativistic mass m = γm 0 a BAD concept P460 - Relativity 8

9 Derive Kinematics de = Fdx = dp/dt*dx = vdp = vd(γmv) assume p=γmv (need relativistic) d(γmv) = mγdv + mvdγ = mdv/γ v de = mvdv = = final 2 / γ mv 1 v 2 / c dv. v= 0 = γmc 2 - mc 2 =Total Energy - rest energy = Kinetic Energy P460 - Relativity 9

10 What are the momentum, kinetic, and total energies of a proton with v=.86c? V =.86c γ = 1/( ) 0.5 = 1.96 E = γ m = 1.96*938 MeV/c 2 = 1840 MeV T = E - mc 2 = = 900 MeV p = βe = γβm =.86*1840 MeV = 1580 MeV/c or p = (E 2 - m 2 ) 0.5 Note units: MeV, MeV/c and MeV/c 2. Usually never have to use c = 300,000 km/s P460 - Relativity 10

11 Accelerate electron to 0.99c and then to 0.999c. How much energy is added at each step? V =.99c γ = 7.1 v =.999c γ = 22.4 E = γ m = 7.1*0.511 MeV = 3.6 Mev = 22.4*0.511 MeV = 11.4 MeV step 1 adds 3.1 MeV and step 2 adds 7.8 MeV even though velocity change in step 2 is only 0.9% P460 - Relativity 11

12 Lorentz Transformations (px,py,px,e) are components of a 4-vector which has same Lorentz transformation px = γ (px + ue/c 2 ) py = py pz = pz u = velocity of transform between frames is in x-direction. If do px -> px E = γ (E + upx) + -> - Use common sense also let c = 1 Frame 1 Frame 2 (cm) Before and after scatter P460 - Relativity 12

13 center-of-momentum frame Σp = 0. Some quantities are invariant when going from one frame to another: py and pz are transverse momentum M total = Invariant mass of system dervived from Etotal and Ptotal as if just one particle How to get to CM system? Think as if 1 particle Etotal = E1 + E2 Pxtotal = px1 + px2 (etc) Mtotal 2 = Etotal 2 - Ptotal 2 γcm = Etotal/Mtotal and βcm = Ptotal/Etotal 1 2 at rest p1 = p 2 1 Lab 2 CM P460 - Relativity 13

14 Particle production convert kinetic energy into mass - new particles assume 2 particles 1 and 2 both mass = m Lab or fixed target Etotal = E1 + E2 = E1 + m2 Ptotal = p1 --> Mtotal 2 = Etotal 2 - Ptotal 2 Mtotal = (E1 2 +2E1*m + m*m - p1 2 ).5 Mtotal= (2m*m + 2E1*m) 0.5 ~ (2E1*m) 0.5 CM: E1 = E2 Etotal = E1+E2 and Ptotal = 0 Mtotal = 2E1 1 2 at rest p1 = p Lab CM P460 - Relativity 14

15 p + p --> p + p + p + pbar what is the minimum energy to make a protonantiproton pair? In all frames Mtotal (invariant mass) at threshold is equal to 4*mp (think of cm frame, all at rest) Lab Mtotal = (E1 2 +2E1*m + m*m - p1 2 ).5 Mtotal= (2m*m + 2E1*m) 0.5 = 4m E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp CM: Mtotal = 2E1 = 4mp or E1 and E2 each =2mp 1 2 at rest p1 = p Lab CM P460 - Relativity 15

16 Transform examples Trivial: at rest E = m p=0. boost velocity = v E = γ(e + βp) = γm p = γ(p + βe) = γβm moving with velocity v = p/e and then boost velocity = u (letting c=1) E = γ(e + pu) p = γ(p + Eu) calculate v = p /E = (p +Eu)/(E+pu) = (p/e + u)/(1+up/e) = (v+u)/(1+vu) prove velocity addition formula P460 - Relativity 16

17 A p=1 GeV proton hits an electron at rest. What is the maximum pt and E of the electron after the reaction? Elastic collision. In cm frame, the energy and momentum before/after collision are the same. Direction changes. 90 deg = max pt 180 deg = max energy βcm = Ptot/Etot = Pp/(Ep + me) Pcm = γcmβcm*me (transform electron to cm) Ecme = γcm*me ( easy as at rest in lab) pt max = Pcm as elastic scatter same pt in lab Emax = γcm(ecm + βcmpcm) 180 deg scatter = γcm(γcm*me + γββ*me) = γ γ*me(1 + β β) P460 - Relativity 17

18 p=1 GeV proton (or electron) hits a stationary electron (or proton) mp =.94 GeV me =.5 MeV incoming target βcm γcm Ptmax Emax p e MeV 1.7 MeV p p GeV 1.4 GeV e p GeV 1.7 GeV e e MeV 1 GeV Εmax is maximum energy transferred to stationary particle. Ptmax is maximum momentum of (either) outgoing particle transverse to beam. Ptmax gives you the maximum scattering angle a proton can t transfer much energy to the electron as need to conserve E and P. An electron scattering off another electron can t have much Pt as need to conserve E and P. P460 - Relativity 18