# Special Relativity

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```							Special Relativity
5

1
Topics

   The Law of Motion

   Momentum & Energy

   E = mc2

   Relativistic Kinematics

   Summary
2
The Law of Motion
According to Principle 1, the laws of physics
should hold true in all inertial frames. In
t     t'       particular, this should be true
of the 2nd law of motion:
S’ frame
P                 F  ma
However, this law cannot
be valid at speeds
x' near that of light
Q
S frame                 3
O       B                 x
The Law of Motion
However, if Newton’s 2nd law of motion is
expressed the way Newton
t     t'                 did originally:

S’ frame             dp
F
P                   dt
then the 2nd law holds true
in special relativity
x'        provided
that…
Q
S frame                  4
O        B                x
Momentum & Energy
…momentum is defined as follows

t                    p   (u )mu
t'
where u is the speed of the
S’ frame   object relative to a
P      frame of reference.

WARNING: the -factor here
differs from the -factor
x' arising from the
relative motion of the frames
Q
S frame                 5
O            B            x
Momentum & Energy

Extra credit: show that   p   mu
implies
dp            3 u a
  ma        2
mu
dt               c
where        du
a
dt
due 8-Feb-08      6
Momentum & Energy
Consider the 2nd law of motion in
1-dimension
dp           3 ua
F      ma   2 mu
dt             c
du       3 u 
2
 m    2 
dt        c 
In Newtonian physics the            b
kinetic energy K arises from   K   Fdx
the work-energy theorem:            a      7
Momentum & Energy
We shall try the same definition of kinetic
energy in special relativity. Consider a particle
that starts from rest relative to some inertial
frame and is accelerated by a force F
du       u2 
b       b
K   Fdx   m     3 2  dx
a       a
dt       c 
b
      3 u 
2
  m     2  udu
a            c 
  mc  mc
2      2
8
Momentum & Energy
Einstein’s interpretation of the result
K   mc 2  mc 2
is that E =  mc2 is the total energy of the object
E =  mc2 = K + mc2
From this he arrived at the momentous
conclusion: the minimum energy of a particle is

E = mc2
9
E = mc2

An Example

10
How Long will the Sun Shine?
Power Output of Sun
 3.826 x 1026
Watts

Unit of Power
 1 Watt = 1
Joule/second

11
How Long will the Sun Shine?
Mass of H nucleus    (i.e, a proton)
1.007276 amu

Mass of He nucleus (i.e., an a-particle)
4.0015 amu
1 Atomic Mass Unit (amu) is

10-3 kg of Carbon-12 / NA = 1.66 x 10-27 kg

NA = 6.022 x 1023 is Avogadro’s Number
12
How Long will the Sun Shine?
Mass Deficit (of reaction   4 H -> He)
4 x 1.007276 amu =         4.0291   amu (4H )
-       4.0015   amu (He)
=       0.0276   amu

That is, 0.0276 / 4.0291 = 0.007 of the mass
of a proton disappears in this reaction!

The mass is converted to energy, in the form of
photons and neutrinos
13
How Long will the Sun Shine?
Available Hydrogen Fuel

   0.1 of the Sun’s mass is hot enough to
fuse hydrogen to helium

   0.007 of that mass is converted to energy

fuel     = (0.1) x (0.007) x (2 x 1030) kg
= 1.4 x 1027 kg
that is, about 233 times the Earth’s mass
14
How Long will the Sun Shine?
   Mass Destroyed Per Second
m     = E / c2
= 4 x 1026 J/s / (3 x 108 m/s )2

= 4.3 x 109 kg / s

   Mass Destroyed Per Year
m     = (3.15x107 s/yr) x (4.3 x 109 kg/s)

= 1.4 x 1017 kg / yr
15
How Long will the Sun Shine?

Available Fuel / Rate of Fuel Consumption

1.4 x 1027 kg / (1.4 x 1017 kg/yr)

10 billion years

16
Relativistic Kinematics

17
Spacetime Vectors

The values ct, x, y, z can be regarded as the
components of a vector in spacetime, a 4-vector

X  (ct , x, y, z )  (ct , r )
relative to some frame of reference. The
same vector can be expressed in a different
frame of reference
X  (ct ', x ', y ', z ')  (ct ', r ')
using the Lorentz transformation                 18
Spacetime Vectors
S frame       -> S’ frame
X  (ct , x, y, z)  (ct ', x ', y ', z ')

Lorentz Transformation

t '   (t  vx / c ) 2

x '   ( x  vt )
y'  y
z'  z                              19
Spacetime Vectors

The energy together with the momentum also
form a 4-vector

P  ( E , px c, p y c, pz c)  ( E , pc)

which transforms between reference frames
in a manner similar to the 4-vector X

20
Spacetime Vectors
S frame        -> S’ frame
P  ( E , px c, p y c, pz c )  ( E ', p ' x c, p ' y c, p ' z c )

Lorentz Transformation

E '   ( E  vpx )
p 'x   ( px  vE / c 2 )
p 'y  py
p ' z  pz
21
Example - Micrometeorite

micrometeorite m = 10-9 kg moves past Earth
at b = v/c = 0.01. What are the energy and
momentum as viewed by an observer in a starship
moving at b = 0.5 relative to the Earth?

micrometeorite          Starship

b = 0.01                b = 0.5

22
Example - Micrometeorite

P  ( E , px c, 0, 0)   Earth frame

E =(u) mc2 = (1.00005) x (10-9kg) c2   J
= 1.00005 x 10-9 c2         J
px= (u)mux = (1.00005) x (10-9kg) x (0.01 c) kg m/s
= 1.00005 x 10-11 c          kg m/s

b = 0.01                   b = 0.5

23
Example - Micrometeorite

P  ( E ', p 'x c, 0, 0)   Starship frame

E’ =  (E – v px) = 1.1547 x
[1.00005x10-9c2 - (0.5 c) (1.00005 x 10-11c) ]
= 1.14898 x 10-9 c2 J

p’x=  (px – vE/c2) = 1.1547 x
[1.00005 x 10-11c - (0.5 c) (1.00005x10-9c2)/c2 ]
=-56.6 x 10-11 c kg m/s

Starship measures energy 15 % greater and
momentum -57 times greater                   24
A Word on Units
   Energy – electron-Volt          (eV)
 1 eV       = 1.6 × 10-19 Joules
 1 kW•hr    = 3.6 × 106 Joules

   Mass – electron-Volt / c2    (eV/c2)
 1 eV/c2         = 1.78 × 10-36 kg
 Electron mass   = 0.511    MeV/c2
 Proton mass     = 938.3    MeV/c2
 Neutron mass    = 939.6    MeV/c2
 Top quark mass = 174.0     GeV/c2
25
Spacetime Vectors

Like vectors in space, one can add, scale
and take the dot product of 4-vectors.

However, the dot product of two 4-vectors
A and B is defined differently:

A  ( At , A), B  ( Bt , B)
AB  At Bt  A  B

The dot product is an invariant, that is, it
has the same value in all reference frames     26
Spacetime Vectors

Class Problem
Consider the momentum 4-vector:
P  ( E, pc)
Given that p   (u )mu and   E   (u )mc   2

compute the dot product of P with itself

The square root of this particular dot
product is called the invariant mass
27
Zero Mass Particles

The energy and momentum of a particle
are related as follows
E2 = (pc)2 + (mc2)2

This suggests that particles with m = 0 are
possible. For them, E = pc. Moreover, in vacuum
they move at the speed of light. Indeed, such
particles exist, for example:
the photon, the quantum of light, and
the gluon, the quantum of the strong force
28
But Zero + Zero ≠ Zero!

Consider two photons with 4-vectors
P  ( E1 , p1c) and P2  ( E2 , p2 c)
1

The 4-vector of the pair is just
P  P  P2  ( E1  E2 , p1c  p2 c)
1

So the squared mass of the pair is
P 2  ( E1  E2 ) 2  ( p1c  p2c) 2
which is not necessarily zero!
Review Example 2-12                           29
Conservation of Energy & Momentum
Energy and momentum are defined relative to
a frame of reference and, as we have seen, can
change as one goes from one frame to another.

However, as in Newtonian physics, within each
frame of reference, energy and momentum are
conserved.

30
Conservation of Energy & Momentum
Space is filled with photons, called the Cosmic
Microwave Background (CMB), with a typical
wavelength of 2.64 mm in our frame of reference.

If a cosmic proton is sufficiently energetic, the
reaction proton + photon -> delta can occur
in space. The delta is an unstable particle.

What proton energy (in our frame of reference)
is needed to trigger this reaction?

Extra Credit:    due 15 Feb             31
Summary
   Newton’s 2nd Law holds in relativity provided
one defines momentum as p = (u) m u

   Energy and mass are related by E = mc2. This
discovery solved the mystery of sun shine.

   Particles with zero mass exist, e.g., the
photon in vacuum.

   Energy and momentum can be combined into a
4-vector, whose dot products are invariant.
32

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