Special Theory of Relativity

1
Special Theory of Relativity

• Up to 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 cconstant
  he derived Lorentz Transformations

light
u
moving frame also sees vl c NOT vl c-u
sees vl c
2
Derive Lorentz Transform

• Bounce light off a mirror. Observe in 2 frames
• A velocity0 with 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) / t2 A
• Assume linear transform (guess)
• x G(x vt) let x 0
• t G(t Bx) so xGvt
  tGt
• some algebra
• (ct)2 4L2 (Gvt)2 and L ct /2 and
  tGt
• gives
• (Gt)2 t2(1 (Gv/c)2) or G2 1/(1 - v2/c2)

mirror
L
A
A x
3
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 lt 0.1c non-relativistic
  (non-Newtonian) expressions are OK. Note that 3D
  space point is now 4D space-time point (x,y,z,t)

4
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

5
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 L1.25 mm and v .995c t(proper)L/vg
.4 ps
L
Twin Paradox. If travel to distant planet at vc
then age less on spaceship then in lab frame
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/c2) v (.9.8)/(1.9.8) .988c
Notes use common sense on /- if v c and u
c v (cc)/2 c
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 DS9 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)/(1.4.6
35)1.04/1.25 .826
8
Relativistic Kinematics

• E2 (pc)2 (mc2)2 E total energy m mass
  and pmomentum
• natural units E in eV, p in eV/c, m in eV/c2 ? c
  1 effectively. E2p2m2
• kinetic energy K T E - m
  _at_ 1/2 mv2 if v ltlt c
• Can show b p/E and g E/m
• ? p bgm if m.ne.0 or pE if m0
• (many massless particles, photon, gluon and
  (almost massless) neutrinos)
• relativistic mass m gm0 a BAD concept

9
Derive Kinematics

• dE -Fdx -dp/dtdx -vdp -vd(gmv)
• assume pgmv (need relativistic for p)
• d(gmv) mgdv mvdg mdvg3

dv
gmc2 - mc2 Total Energy - rest energy
Kinetic Energy
10

• 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 1.96938 MeV/c2 1840 MeV
• T E - mc2 1840 - 938 900 MeV
• p bE gbm .861840 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 in calculation

11

• 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 7.10.511 MeV 3.6 Mev
• 22.40.511 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

12
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
13
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 derived from E(total)
  and P(total) as if just one particle
• How to get to C.M. system? Think as if 1 particle
• E(total) E1 E2 Px(total) px1 px2
  (etc) M(total)2 E(total)2 -
  P(total)2
• g(c.m.) E(total)/M(total) and
  b(c.m.) P(total)/E(total)
  
  

1 2 at rest
p1 p2
1 2
Lab C.M.
14
Particle production

• convert kinetic energy into mass - new particles
  assume 2 particles 1 and 2 both mass m
• Lab or fixed target E(total) E1 E2 E1
  m2
• P(total) p1 ? M(total)2 E(total)2 -
  P(total)2
• M(total) (E122E1m mm - p12).5
• M(total) (2mm 2E1m)0.5
  (2E1m)0.5
• CM E1 E2 E(total) E1E2 and P(total)
  0
• M(total) 2E1
  

1 2 at rest
p1 p 2
1 2
Lab CM
15
p p ? p p p pbar

• what is the minimum energy to make a
  proton-antiproton pair?
• In all frames M(total) (invariant mass) at
  threshold is equal to 4mp (think of cm frame,
  all at rest)
• Lab M(total) (E122E1m mm - p12).5
• M(total) (2mm 2E1m)0.5 4m
• E1 (16mpmp - 2mpmp)/2mp 7mp
• CM M(total) 2E1 4mp or E1 and E2 each 2mp
  
  

at threshold all at rest in c.m. after reaction
1 2 at rest
p1 p 2
1 2
Lab CM
16
Transform examples

• Trivial at rest E m p0. 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 c1)
  E g(E pu)
  p g(p Eu)
  calculate v p/E (p Eu)/(Epu) (p/E
  u)/(1up/E) (vu)/(1vu)
• prove velocity addition formula

17

• A p1 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 gcmbcmme (transform electron to cm)
• Ecme gcmme (easy as at rest in lab)
• pt max Pcm as elastic scatter same pt in lab
• Emax gcm(Ecm bcmPcm) 180 deg scatter
  gcm(gcmme gbbme)
  ggme(1 bb)

18

• p1 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 cant transfer much energy to the
  electron as need to conserve E and P. An electron
  scattering off another electron cant have much
  Pt as need to conserve E and P.