# Lecture 3. Relativistic Dynamics

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```					     Lecture 3. Relativistic Dynamics

Outline:

Relativistic Momentum
Relativistic Kinetic Energy
Total Energy
Momentum and Energy in Relativistic Mechanics
Comment on 3- and 4-vectors
t  t
The length of a “3”                                        length      x2  x1    y2  y1    z2  z1                 
2                  2               2

vector is invariant                     x  x  Vt
y  y
     x2  x1    y2  y1    z2  z1 
                                                 Invariant
2                  2               2
under G.Tr.
z  z
Is there a combination of (x,y,z,t1) which remains invariant under L.Tr.? Indeed, such a
combination exists:
 x2  x1    y2  y1    z2  z1         c2  t2  t1    x '2  x '1    y '2  y '1    z '2  z '1   c 2 t '2  t '1  =Invariant
2             2                 2                  2                2                    2                2                    2

This is the square of the distance between two events
(ict1x1,y1,z1) and (,ict2,x2,y2,z2) in the 4-dimensional space.
i  1

This quantity (a.k.a. the interval) is invariant under L.Tr. (please show this explicitly at home
using L.Tr.), for two arbitrary events it might acquire any (zero, positive or negative) value,
unlike the distance in 3D space (Appendix II).
L.Tr. correspond to a rotation of a 4-dimensional
x '  x cos   y sin      RF through a fixed angle; these rotations preserve
           y '  y cos   x sin      the length of 4-vectors.
x'                                                      x   x   ct                                      x   x  i
  ict
x                                                  ct    ct   x                                     '    i x
1                                    i
y                                          cos  i                          sin  cos2   sin 2   1
y'                                       1  2                               1  2
V
tg  i  i
c
Comment on 3- and 4-vectors (cont’d)
Thus, if one can express some physical law in a form :
“4-vector A”=“4-vector B”, this would guarantee the invariance of this law under L.Tr!

Def.: A 4-vector is any set of four components which transforms
in the same manner as the space-time vector under L.Tr.
ict, x, y, z    x0 , x1, x2 , x3 

Examples of 4-vectors:     i, A , A , A 
x   y    z      A,  - the vector and scalar potentials in
Electrodynamics
 E                 
 i , px , p y , pz     p, E - the momentum and total energy
 c                                in relativistic mechanics

Note that both force and acceleration are 3-vectors. Thus, one should not expect that the
2nd Law, being expressed in terms of force and acceleration, remains invariant under
Lorentz Transformations. This is the reason why “force” and “acceleration” are not popular
in relativistic mechanics. We’ll discuss relativistic dynamics in terms of the “momentum”
and “total energy”.
Relativistic Momentum
p  mv           - the momentum of a particle in classical mechanics,
m is invariant (does not depend on the velocity)

dp    dv
Newton’s 2nd Law:       F           m  ma                 - expressed in terms of 3-vectors,
invariant under G.Tr. (but not L.Tr.!)
dt    dt
mv
Relativistic form of
the momentum
p                  mv      - definition of the momentum
2
v
(introduced by
Einstein):
1 2                      in relativistic mechanics
c
p   mv
Example: Calculate the momentum of an electron
moving with a speed of 0.98c.

me  0.98c 
p                   4.9mec
1  0.98 2

By ignoring relativistic effects, one would get
 v/c            1
p  0.98me c
Relativistic Kinetic Energy
Let’s calculate the kin. energy gained by an accelerated particle:                           2
v = vf
2
dp
2     2
dl
2         2
   mv       
K   Fdl   dl   dp   vd  mv    vd               
dt          dt 1                 1 v / c 
2   2
1       1         1                   1

mv
(integration by parts  xdy  xy   ydx, x  v y               )                        1
1 v / c
2   2
v=0

                                      
vf
mv 2                      vdv                                   1     vdv
 m                   d        1  v2 / c2                      
1  v2 / c2        0     1  v2 / c2                             c2 1  v2 / c2   
vf
 mv       2
     mv f 2
             mc 2 1  v 2 / c 2                   mc 2 1  v f 2 / c 2  mc 2 
 1 v / c                        0  1  v f 2 / c2
2    2

 v 2 / c2                              mc 2
2                             2 
 1  v f / c  mc                  mc 2
f                    2        2
mc
 1  v f 2 / c2                     1  v f 2 / c2
                               
mc2
K                        mc 2  mc 2   1         - kinetic energy of a particle of the mass
m moving with speed v
1  v2 / c2
Total and Rest Energies
We expect this result to be reduced to the classical KE at low speed:
mc 2                        2           v2
K         mc     1  mc 1 
2              2
 1  m
1  2
    2         2                 
K                                      Let’s rewrite the expression for K in the form:
mc 2

2                   E0  mc 2            the rest
mc                                               energy
K                          mc 2
1  v2 / c2                                             the total
 v/c         1                                             E  mc2  K               energy
Limit of                       v2
E  mc  m
2
- we must use Relativistic Mechanics when K and
small speed:                     2                    E0 become of the same order of magnitude.

E                                             The energy and momentum are conserved
mc 2                                        (the consequence of uniform and isotropic space).

For an isolated system of particles:
1                                        all particles                           all particles

 v/c        1
E                     Ei  const         p       
i 1
pi  const
i 1
Electron-Volt: convenient unit of energy
In relativistic mechanics, most of the time we deal with tiny particles like an electron or
proton (after all, it’s hard to accelerate a macroscopic body to v~c). In this case, the most
convenient unit of energy is an electron-volt, the kinetic energy acquired by an electron
accelerated through a potential difference of 1Volt.

K  qV  1.6 1019 C 1V  1.6 1019 J  1eV
1
Energy eV   Energy  J                 19
 Energy  J   6.2 1018 eV / J
1.6 10         J / eV
V  1V
For example, the rest energy of an electron:
8.2 1014 J
E0 e  9.110 kg   3 10 m / s   8.2 10
31                 2             14
8
J                    510, 000eV  0.5MeV
1.6 1019 J / eV
Thus, when the electrons are accelerated across a pot. difference ~ 10kV in a TV tube, they
still can be considered as non-relativistic particles (K<<m0c2)

E0 p  1.6726 1027 kg   3 108 m / s   938.3MeV
2
The rest energy of a proton:
E0 n  1.6749 1027 kg   3 108 m / s   939.6MeV
2
The rest energy of a neutron:
Problem
An electron whose speed relative to an observer in a lab RF is 0.8c is also being studied
by an observer moving in the same direction as the electron at a speed of 0.5c relative
to the lab RF. What is the kinetic energy (in MeV) of the electron to each observer?

K                                  The electron speed v’ as seen by the moving observer K’:
v                                               v V    0.8c  0.5c
V                   v'                            0.5c
vV 1   0.8 0.5
K’                                 1 2
c
                                                      
        1                               1             
In the lab IRF K:     K  mc 2                        1  0.5MeV                   1  0.34MeV
    1 v / c                     1   0.8         
2                               2
                                                      
                                                      
        1                                1            
In the moving IRF K’:   K '  mc 2                       1  0.5MeV                   1  0.08MeV
   1   v '/ c                   1   0.5         
2                               2
                                                      
Energy and Mass in Cl. M. and Rel. M.
Caution: some textbooks use the velocity-dependent mass m=m0 and the “rest” mass m0.
Einstein: “It is not good to introduce the concept of the mass m=m0 of a moving body for
which no clear definition can be given. It is better to introduce no other mass concept than
the “rest mass” m0. Instead of introducing m it is better to mention the expression for the
momentum and energy of a body in motion”.
“Relativistic” mass m=m0 – just another name for the energy (Occam’s Resor: “Entities
must not be multiplied beyond necessity”). See also Okun’s paper on our Webpage.

Important: for a system of many particles, the mass M includes
the potential energy of all interactions between the particles.

The potential energy U for the particles that attract each other is negative (for repelling
particles – positive). Thus, for the stability of a body, its mass should be smaller than the
sum of masses of all particles that constitute the body.
E         U          Ebind
M  2   mi  2   mi  2   mi
c    i     c    i      c    i

Mass and energy are different aspects of the same “thing”, they become interchangeable:
matter can be created or destroyed, but when it happens, an equivalent amount of energy
vanishes or comes into being.

Classical mechanics: conservation of mass and energy
Relativistic mechanics: conservation of energy
Mass Defect
Mass defect:       M  M   mi                  For the stability of a body, its mass
i                defect should be negative.

the mass of a composite body (system)        the sum of masses of its constituents

Binding energy:       Ebind  Mc 2      We’ll consider the binding energy in detail when
we consider nuclei and nuclear reactions.
Why was it difficult to notice in Cl. M.? Because in all processes of chemical
transformations (the most “violent” processes of the pre-20-century physics)
the energy release is tiny in comparison with the rest energy of reactants.

Example: dynamite explosion
When 1kg of the TNT explodes, the energy release is 5.4MJ. At the same time, the rest

mc  1kg   3 10 m / s   9 10 J
energy of 1kg is               2               8       2        16

m 5.4 106 J
            6 1011           - it’s very difficult to detect this mass change
m   9 1016 J
The mass loss for the Sun. The power of solar radiation P ~4 ·1026 J/s (the power per 1m2
on the Earth’s surface (~1400 W/m2) being multiplied by the area of a sphere with radius
1.5·1011 m (the Sun-Earth distance) .
dm P      4 1026 W
the mass loss per one second          2                  4.4 109 kg/s
 3 108 m 
2
dt c
Examples
An elementary particle (e.g. a free electron) cannot absorb/emit a photon. (Hint: use the
reference frame in which the electron was at rest before the collision/after absorption).

A composite particle can decay into two (or more) fragments if the mass of all fragments is
less than the particle mass.
Decay of a neutron: a free neutron is unstable (the lifetime ~ 15 min).
It decays into a proton, an electron, and an electron anti-neutrino:
n  p  e  e

m p  938.3MeV
The masses
mn  m p  me  m e 
involved:        mn  939.6MeV
me  0.5MeV
 939.6  938.3  0.5 MeV
m e  0                             0.8MeV

m p  me  mn     - otherwise, the neutron would be stable and most of the protons and
electrons in the early Universe would have combined to form neutrons,
leaving little hydrogen to fuel the stars.
However, neutrons are stable in nuclei:

mn  m  mn  Ebind per nuclon  m p        - the energy conservation prevents the
neutron from decaying in nuclei.

Without neutrons we would not have the heavier elements needed for building
complex systems such as life.
An important relationship between E and p
2
E       m2c 2
  
 c  1  v / c 
2   2
E
2
m 2 c 2 1  v 2 / c 2 
     p2                            m 2 c 2  Inv
m2v 2                 c             1  v 2 / c2 
p 
2

1  v2 / c2 
this combination of E and p
does not depend on the IRF!

(In fact, it is the length2 of a 4-vector formed by the components of p and i(E/c))

E  c p 2  m 2c 2            E 2  p 2c 2  m 2c 4

Def.: A 4-vector is any set of four components which transforms
in the same manner as the space-time vector under L.Tr.
ict, x, y, z    x0 , x1, x2 , x3 
                    
Examples of 4-vectors:        i , Ax , Ay , Az         A,  - the vector and scalar potentials in
 c                                        Electrodynamics
 E                    
 i , px , p y , pz       p, E - the momentum and total energy
 c                                   in relativistic mechanics
                   
  i , kx , k y , kz      k ,       - the wave vector and angular
 c                               frequency of a plane harmonic wave
Ultra-relativistic and massless particles
In the ultra-relativistic case (K>>m0c2)   E  cp
This equation “works” for the particles with zero rest mass (photons). They must move
with the speed of light: v=c (otherwise, both p and E are 0)
E  p c m c 0
2     2 2    2 4

Massless particles: the photon (carrier of the electromagnetic interaction), the gluon
(carrier of the strong interaction, never observed as a free particle), and, perhaps, the
graviton (carrier of gravitational interaction, remains to be discovered).
E ph  cp ph

External forces can bend the trajectories of massless particles (e.g., photons in
gravitational fields), but cannot accelerate (decelerate) them (v is always =c).
Problems
1. What is the momentum of an electron with K =mc2?                    E 2  p 2c 2  m2c 4
2
 mc  K 
2
E
2
p     m2 c 2            m c  4m c  m c  3mc
2 2    2 2   2 2

c               c    

2. How fast is a proton traveling if its kinetic energy is 2/3 of its total energy?

E   mc 2  K  E  3mc 2
2    2
K                                                                                   2
3    3                                              1                    V  1     8
mc 2                                           3 1     V    c
E                                  1  V / c               c 9     3
2

1  V / c 
2
Problem
An electron initially moving with momentum p=mc is passed through a retarding
potential difference of V volts which slows it down; it ends up with its final momentum
being mc/2. (a) Calculate V in volts. (b) What would V have to be in order to bring the
electron to rest?

(a)        p=mc:          E1  p c   mc
2 2

2 2
        mc    mc 
2 2       2 2
 2mc2
2
1 2
E2   mc    mc  
2 2   5 2
p=m0c/2:                                          mc
2               2
      5 2
E  E1  E2   2     mc  0.3mc  0.3  0.5 10 eV   1.5 10 eV
2               6              5
     2 
       
Thus, the retarding potential difference         V  1.5 105V

(b)    E1  2mc 2       E2  mc 2    E             
2  1 mc 2  2.1105 eV V  2.1105V
Problem (Relativistic Dynamics)
Beiser #38. A moving electron collides with a stationary electron and an electron-
positron pair comes into being as a result. When all four particles have the same velocity
after the collision, the kinetic energy required for this process is a minimum. Use a
relativistic calculation to show that Kmin=6mc2, where m is the electron mass.
p1                            4 p2
E1  mc 2  4 E2          energy conservation
before                       after                                  p1  4 p2                momentum conservation

                  2
 2 E1mc 2   mc 2   16  E2   16  mc 2    p1c  
1
 E1 
2                          2
E   mc          p c
2                                    2
2          2 2               2
1                        1                                                                          16       
E   mc          p c
2 2
 E1    p1c         2 E1mc   mc    
2 2
 16  mc    
2 2
2                             2                                            2           2           2
2                        2

E1  mc 2  4 E2                            mc  2 2

p1  4 p2                             E1  14  mc           
2 2
/ 2mc 2  7mc 2          K1  E1  mc 2  6mc 2

E1 '   ' mc2  2  mc 2    '  2
In the center-of-mass RF:
2 E1 '  4mc 2
p1 '              p1 '
 v '
2
1         3                                                 v'
1                     v'     c                  relative speed V 
before                                           c2              4        2                                                  2
after
v ' V             3      4 3   48                                             mc 2
v                                                                    K1                         mc 2  6mc 2
1   vv '/ c 2  1  3 / 4    7    49                                         1   48 / 49 
Problem (Relativistic Dynamics)
Find the minimum energy a proton must have to initiate the reaction
p p  p p p p
(production of anti-protons (Berkeley Bevatron 1954): the energy and, thus, the
cost of the accelerator, must be minimized)

The minimum energy – when the products of reaction are at rest in the “center of mass”
reference frame: all the incoming energy is transformed into the rest energy.

E  ( psys c)  Inv 
2
sys
2
 4m c 
p
2 2
in the rest frame if we want to minimize energy
The invariant has the same value in all RFs!
Let’s use the lab RF:
p                        p
p1  p           p2  0           E1  K  mc 2                E2  mc 2

 E1  E2    p1  p2  c   E1  mc                  2 2
     p1c    E1   2 E1mc   mc    
2 2
  p1c 
2                    2   2                                 2       2          2                            2

 2  mc   2 2
    2 E1mc  Inv  16  mc
2

2 2
For “colliding beam” accelerators (e.g.,
LHC), the center-of-mass frame and the
E1  7 mc 2               K  E1  mc 2  6mc 2                    lab frame are the same (each proton
should have E=2mc2)

Homework #2: Beiser Ch. 1, Problems 29, 30, 33, 34, 41, 42, 45, 49, 54, 56.

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