Special Theory of Relativity - PowerPoint - PowerPoint by theoryman

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									 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 108 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
 “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
    c2 = 4(L2 + (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 = 4L2 + (Gvt)2 and L = ct /2 and t’=Gt
 gives
  (Gt)2 = t2(1 + (Gv/c)2) or G2 = 1/(1 - v2/c2)

                 mirror
            A                                     L
                                          A’ x’
                      P460 - Relativity               2
    Lorentz Transformations
Define b = u/c and g = 1/sqrt(1 - b2)

    x’ = g (x + ut)    u = velocity of transform
    y’ = y            between frames is in
   z’ = z            x-direction. If do x’ ->x then
   t’ = g (t + bx/c) + -> -. Use common sense

can differentiate these to get velocity transforms
  vx’ = (vx - u) / (1 -u vx/ c2)
  vy’ = vy / g / (1 - u vx / c2)
  vz’ = vz / g / (1 - u vx / c2)

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
           Time Dilation
• Saw that t’ = g t. The “clock” runs
  slower for an observer not in the
  “rest” frame
• muons in atmosphere. Lifetime = t
  =2.2 x 10-6 sec ct = 0.66 km
  decay path = bgtc
  b        g            average in lab
                    lifetime          decay path
  .1      1.005         2.2 ms          0.07 km
  .5      1.15          2.5 ms          0.4 km
  .9       2.29         5.0 ms          1.4 km
  .99      7.09          16 ms          4.6 km
  .999    22.4           49 ms          15 km
                  P460 - Relativity                4
           Time Dilation
• Short-lived particles like tau and B.
  Lifetime = 10-12 sec ct = 0.03 mm
• time dilation gives longer path lengths
• measure “second” vertex, determine
  “proper time” in rest frame

                          If measure L=1.25 mm
                                     and v = .995c
          L               t(proper)=L/vg = .4 ps

Twin Paradox. If travel to distant planet
at v~c then age less on spaceship then in
              “lab” frame
                 P460 - Relativity                   5
       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/c2)
      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
       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

                B
          V’ =(v-u)/(1-vu/c2)
  v’ = (.826-.635)/(1-.826*.635)= .4c
        If did B from A get -.4c
(.4+.635)/(1+.4*.635)=1.04/1.25 = .826

                  P460 - Relativity         7
     Relativistic Kinematics
• E2 = (pc)2 + (mc2)2 E = total energy
  m= mass and p=momentum


• natural units E in eV, p in eV/c, m in
  eV/c2 -> c = 1 effectively. E2=p2+m2
• kinetic energy K = T = E - m
  @ 1/2 mv2 if v << c
• Can show: b = p/E and g = E/m
  --> p = bgm if m.ne.0 or p=E if m=0
(many massless particles, photon, gluon and
  (almost massless) neutrinos)
• relativistic mass m = gm0 a BAD concept
                  P460 - Relativity      8
       “Derive” Kinematics
• dE = Fdx = dp/dt*dx = vdp = vd(gmv)
• assume p=gmv (need relativistic)
• d(gmv) = mgdv + mvdg = mdv/g


                        v = final
 dE =  mvdv / g = v =0
                                    mv 1  v 2 / c 2 . dv

             = gmc2 - mc2
    =Total Energy - “rest” energy
           = Kinetic Energy



                  P460 - Relativity                 9
• What are the momentum, kinetic, and
  total energies of a proton with v=.86c?

· V = .86c g = 1/(1.86*.86)0.5 = 1.96


• E = g m = 1.96*938 MeV/c2 = 1840 MeV
• T = E - mc2 = 1840 - 938 = 900 MeV

•   p = bE = gbm = .86*1840 MeV = 1580 MeV/c
    or p = (E2 - m2)0.5

Note units: MeV, MeV/c and MeV/c2. Usually never
  have to use c = 300,000 km/s




                     P460 - Relativity      10
• Accelerate electron to 0.99c and then
  to 0.999c. How much energy is added
  at each step?

· V = .99c g = 7.1    v = .999c          g = 22.4


• E = g 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
    Lorentz Transformations
(px,py,px,E) are components of a 4-vector which has
   same Lorentz transformation

   px’ = g (px + uE/c2)      u = velocity of transform
   py’ = py             between frames is in
  pz’ = pz            x-direction. If do px’ -> px
  E’ = g (E + upx) + -> - Use common sense
                         also let c = 1
  Frame 1                        Frame 2 (cm)




   Before and after scatter
                     P460 - Relativity           12
  center-of-momentum frame
Sp = 0. Some quantities are invariant when going
  from one frame to another:
  py and pz are “transverse” momentum
  Mtotal = 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)
  Mtotal2 = Etotal2 - Ptotal2
  gcm = Etotal/Mtotal and bcm = Ptotal/Etotal



    1    2 at rest               p1 = p 2

                             1
                             2
            Lab                          CM
                     P460 - Relativity         13
         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 --> Mtotal2 = Etotal2 - Ptotal2
  Mtotal = (E12+2E1*m + m*m - p12).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 2

                             1
                             2
            Lab                          CM
                     P460 - Relativity         14
   p + p --> p + p + p + pbar
 what is the minimum energy to make a proton-
  antiproton pair?
• In all frames Mtotal (invariant mass) at threshold is
  equal to 4*mp (think of cm frame, all at rest)
Lab Mtotal = (E12+2E1*m + m*m - p12).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 2

                              1
                              2
            Lab                           CM
                      P460 - Relativity          15
      Transform examples
• Trivial: at rest E = m p=0. “boost”
  velocity = v
  E’ = g(E + bp) = gm
  p’ = g(p + bE) = gbm


• moving with velocity v = p/E and then
  boost velocity = u (letting c=1)
  E’ = g(E + pu)
  p’ = g(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
• 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
 bcm = Ptot/Etot = Pp/(Ep + me)
· Pcm = gcmbcm*me         (transform electron to cm)
· Ecme = gcm*me           (“easy” as at rest in lab)
·
• pt max = Pcm as elastic scatter same pt in lab

•   Emax = gcm(Ecm + bcmPcm)              180 deg scatter
    = gcm(gcm*me + gbb*me)
    = g*g*me(1 + b*b)
                      P460 - Relativity                17
• p=1 GeV proton (or electron) hits a
  stationary electron (or proton)
  mp = .94 GeV me = .5 MeV
    incoming target bcm gcm                Ptmax   Emax
       p       e       .7       1.5 .4 MeV 1.7 MeV
       p       p       .4        1.2 .4 GeV 1.4 GeV
       e       p       .5       1.2 .5 GeV 1.7 GeV
       e      e       .9995 30 15 MeV 1 GeV
 Emax 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

								
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