Relativity

1
Relativity

• Lorentz transformation
• 4-vectors and 4-tensors
• Relativistic kinematics
• Electromagnetism
  

2
Lorentz Transformation

• Principal of special relativity relates two
  standard frames of reference S and S that
  coincide at time t0
• Lorentz transformation
• ct g(ct - b x)
• x g(x-b ct)
• 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
3
Lorentz Transformation

• Inverse transformation (b ? b)
• ct g (ct b x)
• 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

4
Example

• The velocity of a particle is u in the
  x-direction in S. What is its speed as seen in
  S?
• In S distance moved is D x 0 u D t
  in time.
• In S (use Lorentz transformation)
• D x g(D x b c D t )
• c D t g(c D t b D x )
• so u D x / D t
• g(D x b c D t ) / gc(c D t b D x
  )
• u (u v) / (1 u v/c2)
• Note
• u lt u u
• Ltv!c (u) c

5
More Examples

• Length contraction A line of length L in S
  has one end at the origin, the other at xL.
  What length is recorded in S?
• Solution
• Both ends of the line must be observed at the
  same time in S, say at t0.
• 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 (ct -gbL, L, 0, 0,) (0, L, 0, 0)
• So measured length in S is L g(xbct)
  g(L-b2g L)

? L L/g
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 0 S
• 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
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 -6 s ?
• 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
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)
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

10
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 gmn Dxn Dxm (sum 0?3 over
  repeated indices m and n)
• This is an example of an invariant

11
Invariant, Contra- and Co-variant

• The signs of the terms requires that gmn be
  defined as
• An even neater notation is to define Dxm gmn
  Dxn .
• Then
• s Dxm Dxm
• Vectors with lower (upper) indices are contra-
  (co-) variant
• Any 4-vector product
• k am bm am bm
• is also an invariant

12
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 dt the time elapsed in the
  rest frame of a particle at rest in S.
• In an arbitrary frame S, this is dt g dt
• Example of 4-vector
• Um dxm / dt 4-velocity (gc, gdx1/dt ,
  gdx2/dt , gdx3/dt)

13
DAlembertian Operator

• Derivative operator ( m \equiv m /xm )
  is a 4-vector
• Summary

14
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 4x416 components

15
Relativistic Tensors Rank 2

• Example vector product
• Fmn AmBn AnBm
• Is a 4-tensor of rank 2.
• Proof Transform Fmn
• Similarly, Fmn and Fmn are also 4-tensors of rank
  2

16
Relativistic Kinematics
17
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

18
The 4-Momentum

• In agreement with classical mechanics when bltltc
• Binomial expansion for b ltlt c
• E m0g c2
• ¼ m0c2 (1 ½ b2 3/8 b4 ) 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

19
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

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)

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 pa pb (NOTE Ea pa, Eb pb)
• (32 0.9382)½ (12 0.93822)½ 4.5145
  (GeV)
• Momentum conservation
• pa cosqa pb cosqb 2 (GeV/c)
• pa sinqa - pb sinqb 0 (GeV/c)
• 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)

22
Examples

• A r0 resonance (meson) decays into two pions
• r0 ? pp-
• 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) and mp 0.1396
  (GeV/c2)
• pp2 mr2 / 4 mp2 ? pp 0.349 (GeV/c)

23

• Electromagnetism

24
Maxwells Equations

• Aim to write all equations in co-variant form
  (same in all frames of reference)
• Maxwells equations
• Equation of continuity
• In the vacuum e m s 0

In homogenous , linear, isotropic, medium
with Conductivity ? Dielectric constant
? Permeability ?
25
4-Current

• In rest frame S0
• Charge density is ?0
• Charges moving at velocity u0 make current
  density
• Viewed from S
• Define 4-current (obviously a 4-vector)
• Equation of continuity becomes

Invariant is zero in all frames.
26
Potentials

• Introduce vector and scalar potentials (A, f)
• Substitute into Maxwells equations
• Auxiliary conditions

27
4-Potential

• It is tempting to define a new vector
• Then we would write Maxwells equations as
• Has form (scalar) x A? (4-vector)
• Lorentz condition is then

Therefore, A ? is a 4-vector
28
E/M Field Tensor

• The E and B fields are related to the potential
• This suggests that the fields are given by the
  anti-symmetric tensor
• For example

29
E/M Field Tensor

• The anti-symmetric field tensor is therefore
• An exercise for the student !

30
Lorentz Transformations

• Since f?? is a 4-tensor, we know how it
  transforms
• ( !) leads to
• Likewise J? 0 ??? J? leads to

31
Maxwells Equations (again!)

• In co-variant form
• Verify these !