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Recap: Relativistic definition of linear
momentum and moving mass
We have studied two concepts in earlier lecture: mass and
momentum in SR
Consider a mass moving with speed v in a rest frame
Classically, p = mv , m = mass is constant and not changing with its
state of motion
Relativistically, the mass of a moving object changes as its speed
changes: m = gm0. m is called the relativistic mass
m0 = rest mass = the mass measured in a frame where the object is
at rest. It’s value is a constant
Relativistically, p = mv = gm0v
I see the mass of M as
m = m0g;momentum of
M as p = mv=m0gv
O’
I see M is at rest. Its mass
is m0, momentum, p’ = 0
v
O
M
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Example
The rest mass of an electron is m0 =
m0
9.11 x 10-31kg.
If it moves with u = 0.75 c, what is its
relativistic momentum?
p = m0 g u
Compare it with that calculated with
classical definition.
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Solution
The Lorentz factor is g = [1-(u/c)2] -1/2
= [1-(0.75c/c)2] -1/2=1.51
Hence the relativistic momentum is simply
p = g x m0 x u = g x m0 x 0.75c
= 1.51 x 9.11 x 10-31kg x 0.75 x 3 x 108 m/s
= 3.1 x 10-22 kg m/s = Ns
In comparison, classical momentum gives
pclassical = m0 x 0.75c = 2.5 x 10-22 Ns – about 34% lesser
than the relativistic value
pclassical/pSR = g = [1-(u/c)2]-1/2
 1 when u<<c
pclassical/pSR = g = [1-(u/c)2]-1/2 >>1 when u 3 –>c
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Energy in SR
Recall the law of conservation of
mechanical energy you have studied in
classical mechanics:
Work done by external force on an object
(W) = the change in kinetic energy of the
object, (DK)
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DK = K2 - K1
K1
K2
F
F
s
Conservation of mechanical energy: W = DK
W=Fs
The total mechanical energy of the object, E = K + U. Ignoring
potential energy, E of the object is solely in the form of kinetic
energy. If K1 = 0, then E = K2. But in general, U also needs to be
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taken into account for E.
Force, work and kinetic energy
When a force is acting on an object, O, at rest
with rest mass m0, it will get accelerated (say
from rest) to some speed (say u) and increase
in kinetic energy from 0 to K
K as a function of u can be derived from first
principle based on the definition of:
Force,F = dp/dt,
work done, W =  F dx,
and conservation of mechanical energy, DK = 7W
In classical mechanics, mechanical energy (kinetic +
potential) of an object is closely related to its momentum
and mass
Since in SR we have redefined the classical mass and
momentum to that of relativistic version
mclass(cosnt) –> mSR = m0g
pclass = mclass u –> pSR = (m0g)u
we must also modify the relation btw work and energy so
that the law conservation of energy is consistent with SR
E.g, in classical mechanics, K = p2/2m = 2mu2/2. However,
this relationship has to be supplanted by the relativistic
version K = mu2/2 –> K = E – m0c2 = mc2 - m0c2
We will like to derive K in SR in the following slides
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Derivation of relativistic kinetic
Force = rate change of
energy
momentum
x2
x2
x2
dp
 dp dx 
W   F dx  
dx   
dx
dt
dx dt 
x1  0
x1  0
x1  0 
x2
Chain rule in
calculus
dp
dp
 dp du 
 
udx   
udx   udu
dx
du dx 
du
x1 0
0
0
dx
where, by definition, u 
is the velocity of the
dt object
u
u
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Explicitly, p = gm0u,
Hence, dp/du = d/du(gm0u)
= m0 [u (dg/du) + g ]
 m0 [g + (u2/c2) g3]  m0 (1-u2/c2)-3/2
in which we have inserted the relation
dg
d

du du
1
u
1
u 3
 2
 2g
3
/
2
c
u2 c  u2 
1- 2
1 - 2 
c
c 

u
integrate
2 -3 / 2
 u 
W  m0  u1 - 2  du
c 
0 
 K  W  m0gc 2 - m0c 2  mc2 - m0 c 2
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K  m0gc - m0 c  mc - m0c
2
2
2
2
The relativisitic kinetic energy of an object of
rest mass m0 travelling at speed u
E0 = m0c2 is called the rest energy of the
object. Its value is a constant for a given object
Any object has non-zero rest mass contains
energy as per E0 = m0c2
One can imagine that masses are ‘energies
frozen in the form of masses’ as per E0 = m0c2
E = mc2 is the total relativistic energy of an
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moving object
Or in other words, the total relativistic energy
of a moving object is the sum of its rest
energy and its relativistic kinetic energy
E  mc  m0 c  K
The mass of an moving object m is
larger than its rest mass m0 due to the
contribution from its relativistic kinetic
energy – this is a pure relativistic effect
not possible in classical mechanics
E = mc2 relates the mass of an object to
the total energy released when the
object is converted into pure energy
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2
2
Example, 10 kg of mass, if converted into pure
energy, it will be equivalent to
E = mc2 = 10 x (3 x108) 2 J = 9 x1017J
– equivalent to a few tons of TNT explosive
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Due to mass-energy equivalence, sometimes
we express the mass of an object in unit of
energy
Example
Electron has rest mass m0 = 9.1 x 10-31kg
The rest mass of the electron can be
expressed as energy equivalent, via
m0 c2 = 9.1 x 10-31kg x (3 x 108m/s)2
= 8.19 x 10-14 J
= 8.19 x 10-14 x (1.6x10-19)-1 eV
= 511.88 x 103 eV = 0.511 MeV
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Continue exploring the massenergy momentum formula…
In terms of relativistic momentum, the
relativistic total energy can be expressed as
followed
2
2
u
c p
E g m c ; p g m u  2  2
c
E
2
2
2 4
0
2
2
2
0
2
2
2 4 2

mc
m0 c E 
2
2 2 4
2 2

 E  g m0 c 
 m0 c
2
2 2 

u
E -c p 

1- 2
c
2 4
0
2
2 4 Conservation of
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energy-momentum
0
E  p c m c
2
2 2
Relativistic invariance
Note that, in general, E and p are framedependent (i.e they takes on different value
in different reference frame) but the quantity
E -p c
2
2 2
is an invariant – it’s the same in value (  m c )
in all reference frames. We call such quantity a
`relativistic invariant’
2 4
0
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