Equations of Einstein’s Special Relativity
E  mc 2
lv  l0
tv 
mv 
v2
1 2
c
t0
v2
1 2
c
m0
v2
1 2
c
E
energy of an object [joule J]
m
mass of object – relativistic mass – depends upon the velocity of the object w.r.t
an observer [kg]
v
speed of moving inertial frame of reference Fv w.r.t the stationary observer in
the inertial frame of reference Fs [m.s-1]
c
speed of light c = 3.0x108 m.s-1
lo
proper length – measured in the stationary inertial frame of reference Fs [m]
lv
relativistic length (contracted length) in moving inertial frame of reference Fv as
measured by the observer in the stationary inertial frame of reference Fs [m]
to
proper time interval – measured in the stationary inertial frame of reference Fs
[s]
tv
relativistic or dilated time interval in moving inertial frame of reference Fv as
measured by the observer in the stationary inertial frame of reference Fs [s]
Inertial frame of reference


Moves with a constant velocity or is stationary with rest to an observer in an inertial
frame of reference.
A non-accelerating frame of reference.
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In 1905, Albert Einstein (1879 – 1955) published his famous paper entitled: “On the
Electrodynamics of Moving Bodies”, in which he proposed his two postulates of relativity
and from these derived his Special Relativity Theory.
1.
The Principle of Relativity – All the laws of physics are the same in all inertial
reference frames – no preferred inertial frame exists.
2.
The Principle of the Constancy of the Speed of Light – the speed of light in free space
has the same value c, in all inertial frames, regardless of the velocity of the observer or
the velocity of the source emitting the light.
constant c  both space and time must be relative quantities
10
v/ c = 0.994
9
8
7
1
1
v2
c2
6


5
4
Newtonian
physics ok
3
2
1
0
0
0.2
0.4
0.6
0.8
1
v/ c
Einstein’s famous equation E  m c 2






Equivalence of mass and energy.
Energy can be converted into mass and vice versa
When a particle and its antiparticle collide, all the mass is converted into energy.
Mass is converted into energy in a nuclear fission reaction.
When a body gives off energy E in the form of radiation, its mass decreases by an
amount equal to E/c2.
In Special Relativity, the Law of Conservation of Energy and the Law of Conservation
of Mass have been replaced by the Law of Conservation of Mass-Energy.
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NUCLEAR BINDING ENERGY
Energy  Mass E = m c2
Why is the mass of a nucleus less than the combined mass of its nucleons?
The answer is that when nucleons combine to form a nucleus, the total energy of the
system (including rest mass energy) remains constant, although mass does not. The
difference in mass between the constituents of the nucleus and the nucleus itself appears as
additional kinetic energy of the nucleus beyond which the constituents initially possessed.
Correspondingly, this same mass deficiency must be made up by an addition of energy to
break up a nucleus. Consider a deuteron nucleus 2H1.
Mass of deuteron nucleus
2.014186 u
Mass of proton
1.007593 u
Mass of neutron
1.008982 u
Mass (proton + neutron)
2.016575 u
Mass of deuteron nucleus < mass (proton + neutron)
Mass of nucleus is less than the constituent nucleons. The difference is mass is called the
mass defect.
Mass defect m = (2.016575 - 2.014186) u = 0.002389 u
This mass defect is responsible for the energy that enables the nucleus to “stick together”.
The binding energy EB of the deuteron is
EB = m c2 = (0.002389)(1.660x10-27)(3.0x108)2 J = 3.57x10-13 J
EB = (3.57x10-13 / 1.602x10-19) eV = 2.22x106 eV = 2.22 MeV
1 amu = 1 u = 1.660x10-27 kg
c = 3.0x108 m.s-1
qe = e = 1.602x10-19 C
1 eV = 1.602x10-19 J
1 u = 931 MeV/c2
Alternatively EB =m c2 = (0.002389)(931)c2 MeV/c2 = 2.22 MeV
This number is confirmed by experiments that show that the minimum energy of a gamma
ray must be greater than 2.22 MeV to disrupt a deuteron nucleus.
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Lorentz-Fitzgerald Length Contraction
v2
L  L0 1  2
c
Contraction takes place in the direction of motion only
train at rest w.r.t. observer
train in motion w.r.t. observer
v
train is shorter in direction in
motion but just as high and wide as
it was at rest
Time dilation
tv 
t0
1
2
v
1 2
c
tv > t0
since
v2
1 2
c
1
vc
The moving clock is seen to “tick” more slowly than the identical clock at rest. The moving
clock has a greater time interval between its ticks than the clock that is at rets, a moving
clock runs slow. This is the time dilation effect.
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Example
Consider two trains with velocities v1 = 0.10c and v2 = 0.90c w.r.t. a stationary frame of
reference. In the stationary frame of reference, the duration of an event was 1.00 s. What
would be the duration of the event as measured by clocks on the moving frames of
reference?
Solution
How to approach the problem – Identify / Setup / Execute / Evaluate
Type of problem: special relativity / time dilation
Knowledge: moving clocks run slow
tv 
t0
1
v2
c2
Data: proper time t0 = 1.00 s
moving clock (1): v = v1 = 0.10c
moving clock (2): v = v2 = 0.90c
tv = t1 = ? s
tv = t2 = ? s
Execute: (substitute numbers into formula for time dilation)
t1 = 1.005 s
small increase in the time interval
t2 = 2.29 s
significance change increase in the time interval
For clock 2: To an observer on Earth, the time taken for the event is 2.29s. The Earth
observer sees that the train clock has slowed down. It is essential that you understand that
this is not an illusion. It makes no sense to ask which of these times is the “real” time. Since
no preferred reference frame exists all times are as real as each other. They are the real
times seen for the event by the respective observers.
Time dilation tells us that a moving clock runs slower than a clock at rest by a factor of
1/{1 – (v2/c2)}.
This result, however, can be generalised beyond clocks to include all physical, biological and
chemical processes. The Theory of Relativity predicts that all such processes occurring in a
moving frame will slow down relative to a stationary clock.
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RELATIVITY OF SIMULTANEITY
Thought experiment (Gedanken) to illustrate that time is relative. Imagine two observers
O1 and O2 standing at the midpoints of their respective trains (reference frames) F1 and F2.
F1 is moving at a constant speed v with respect to F. Just at the instant when the two
observers O1 and O2 are directly opposite each other, two lightning flashes (events) occur
simultaneously in the F2 frame. The question is: will these two events appear simultaneous
in the F1 frame?
From our F2 reference frame, it is clear that observer O1 in the F1 frame moves to the right
during the time the light is travelling to O1 from A1 and B1. At the instant that O2 receives
the light from A2 and B2, the light from B1 has already passed O1, whereas the light from A1
has not yet reached O1. O1 will thus observe the light coming from B1 before receiving the
light from A1. Since the speed of light along both paths O1 A1 and O1 B1 is c (according to
the second postulate), O1 must conclude that the event at B1 occurred before the event at
A1. The two events are not simultaneous for O1, even though they are for O2 – the two
events that are simultaneous to one observer are not necessarily simultaneous to a second
observer - there is no preferred reference frame, either description is equally valid simultaneity is not an absolute concept, but depends on the reference frame of the
observer.
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