The Theory of Special Relativity
Learning Objectives

Einstein’s two postulates in his theory of special relativity:
The principle of relativity. (Same principle as in Newtonian physics)
The constancy of the speed of light. (Breaks from Newtonian physics)

Using Einstein’s two postulates, derive space and time transformations between
inertial reference frames (derived transformations are same as the Lorentz
transformations):
Learning Objectives

Einstein’s two postulates in his theory of special relativity:
The principle of relativity. (Same principle as in Newtonian physics)
The constancy of the speed of light. (Breaks from Newtonian physics)

Using Einstein’s two postulates, derive space and time transformations between
inertial reference frames (derived transformations are same as the Lorentz
transformations):
Einstein’s Postulates

The theory of special relativity was introduced in Einstein’s paper “On the
Electrodynamics of Moving Bodies” in 1905. In this paper, Einstein made two
postulates:
1. The Principle of Relativity. The laws of physics are the same for all
observers in uniform motion relative to one another (i.e., in all inertial
reference frames).
2.
The Constancy of the Speed of Light. Light
moves through a vacuum at a
constant speed c that is
independent of the motion of the light source relative
to
the observer.

The first principle is also postulated in Newtonian physics (as expressed in the
Galilean transformations). In Newtonian physics, however, the second postulate
contradicts the first, and therefore marks a break from Newtonian (classical)
physics.
Learning Objectives

Einstein’s two postulates in his theory of special relativity:
The principle of relativity. (Same principle as in Newtonian physics)
The constancy of the speed of light. (Breaks from Newtonian physics)

Using Einstein’s two postulates, derive space and time transformations between
inertial reference frames (derived transformations are same as the Lorentz
transformations):
Space and Time Transformations

To see how Einstein derived the Lorentz transformations from his two postulates,
let us write down the most general set of transformation equations between the
space and time coordinates (x, y, z, t) and (x´, y´, z´, t´) of the same event measured
from two inertial reference frames S and S´ in uniform relative motion
Space and Time Transformations

Say that a light bulb at rest in the S frame, located at (x, y, z), flashes at a time (t).
As measured in the S´ frame, what is the corresponding location (x´, y´, z´) and
time (t´) when the light bulb flashes?


In Newtonian physics, what would be the values of the coefficients?

Notice that the transformation equations are linear. Why would non-linear
transformations not make sense?
Space and Time Transformations

Say that a light bulb at rest in the S frame, located at (x, y, z), flashes at a time (t).
As measured in the S´ frame, what is the corresponding location (x´, y´, z´) and
time (t´) when the light bulb flashes?

If the transformations were non-linear, the displacement between two objects or
the time interval between two events would then depend on the choice of origin
for the S (and hence S´) frame. This is unacceptable as the laws of physics cannot
depend on the numerical coordinates of an arbitrarily chosen coordinate system.
Space and Time Transformations

Einstein’s first postulate, the Principle of Relativity, implies that lengths
perpendicular to u are unchanged; i.e., , y = y´ and z = z´.

To see this, imagine that frame S has a
meter stick oriented along the y-axis with
one end at the origin. Imagine also that
frame S´ has a meter stick oriented along
the y´-axis with one end at the origin.

Imagine further that paint brushes are
mounted perpendicular at both ends of each
meter stick, and that frames S and S´ are
separated by a sheet of glass.

If an observer in frame S sees the meter
stick to be shorter (or longer) in frame S´,
the line drawn by the paintbrush in frame S´
will be inside (outside) the line drawn by
the paintbrush in frame S.
Space and Time Transformations

Einstein’s first postulate, the Principle of Relativity, implies that lengths
perpendicular to u are unchanged; i.e., , y = y´ and z = z´.

To see this, imagine that frame S has a
meter stick oriented along the y-axis with
one end at the origin. Imagine also that
frame S´ has a meter stick oriented along
the y´-axis with one end at the origin.

Imagine further that paint brushes are
mounted perpendicular at both ends of each
meter stick, and that frames S and S´ are
separated by a sheet of glass.

Similarly, if an observer in frame S´ sees the
meter stick to be shorter (or longer) in
frame S, the line drawn by the paintbrush in
frame S will be inside (outside) the line
drawn by the paintbrush in frame S´.
Space and Time Transformations

Einstein’s first postulate, the Principle of Relativity, implies that lengths
perpendicular to u are unchanged; i.e., , y = y´ and z = z´.

To see this, imagine that frame S has a
meter stick oriented along the y-axis with
one end at the origin. Imagine also that
frame S´ has a meter stick oriented along
the y´-axis with one end at the origin.

Imagine further that paint brushes are
mounted perpendicular at both ends of each
meter stick, and that frames S and S´ are
separated by a sheet of glass.

Both sets of lines cannot lie inside (outside)
the other, so the only possible outcome is
that both lines must overlap; i.e., lengths
along y and y´ do not change.

The same arguments apply for lengths
along z and z´.
Space and Time Transformations

Einstein’s first postulate, the Principle of Relativity, implies that lengths
perpendicular to u are unchanged; i.e., , y = y´ and z = z´.

Thus, in the general set of transformations
the Principle of Relativity implies that
Space and Time Transformations

Einstein’s first postulate, the Principle of Relativity, implies that lengths
perpendicular to u are unchanged; i.e., , y = y´ and z = z´.

Thus, in the general set of transformations
the Principle of Relativity implies that
Space and Time Transformations

Measurement of time in Eq. (4.9) must give the same result if y is replaced by -y
or z is replaced by -z; i.e., time measurement cannot depend on the side of the
x-axis on which the event occurs.

Thus, in the general set of transformations
the above argument implies that
Space and Time Transformations

Measurement of time in Eq. (4.9) must give the same result if y is replaced by -y
or z is replaced by -z; i.e., time measurement cannot depend on the side of the
x-axis on which the event occurs.

Thus, in the general set of transformations
the above argument implies that
Space and Time Transformations

Consider the motion of the origin O´ of the
frame S´ relative to the frame S.

The clocks in both frames are synchronized
at t = t´ = 0 when the origins O and O´
coincide.

After a time t has elapsed as measured in
frame S, the origin O´ is located at x = ut.
Of course, in frame S´, the origin O´ is
always located at x´ = 0.

In this situation, Eq. (4.6)
becomes
which implies that
Space and Time Transformations

Consider the motion of the origin O´ of the
frame S´ relative to the frame S.

The clocks in both frames are synchronized
at t = t´ = 0 when the origins O and O´
coincide.

After a time t has elapsed as measured in
frame S, the origin O´ is located at x = ut.
Of course, in frame S´, the origin O´ is
always located at x´ = 0.

In this situation, Eq. (4.6)
becomes
which implies that
Space and Time Transformations

Collecting the results thus far, Eqs. (4.6-4.9) have been reduced to

In Newtonian physics, what would be the values of the coefficients?
Space and Time Transformations

Collecting the results thus far, Eqs. (4.6-4.9) have been reduced to

In Newtonian physics, what would be the values of the coefficients?
a11 = 1, a41 = 0, and a44 = 1, which reduces
the above equations to the Galilean transformation.
Space and Time Transformations

Collecting the results thus far, Eqs. (4.6-4.9) have been reduced to

So far we have only applied one of Einstein’s postulate, the Principle of Relativity,
which also is a basic principle of Newtonian physics.

As we shall see, when we apply Einstein’s 2nd postulate, Eqs. (4.10-4.13) do not
reduce to the Galilean transformations.
Space and Time Transformations

Einstein’s 2nd postulate, the constancy of the speed of light, implies that all
observers measure exactly the same value for the speed of light.

Suppose that a lightbulb, located at the origins O in the reference frames S, is
turned on at time t = t´ = 0 when the origins O and O´ coincide.

After a time t has elapsed in the S reference frame, an observer in this frame
would see a spherical wavefront with a radius ct moving away from the origin O
with speed c and satisfying
Equation for
sphere

In Newtonian physics, what would an
observer in the S´ frame see for the
same wavefront? What equation would
this wavefront satisfy?
Space and Time Transformations

Einstein’s 2nd postulate, the constancy of the speed of light, implies that all
observers measure exactly the same value for the speed of light.

Suppose that a lightbulb, located at the origins O in the reference frames S, is
turned on at time t = t´ = 0 when the origins O and O´ coincide.

After a time t has elapsed in the S reference frame, an observer in this frame
would see a spherical wavefront with a radius ct moving away from the origin O
with speed c and satisfying
Equation for
sphere

In Newtonian physics, what would an
observer in the S´ frame see for the
same wavefront? What equation would
this wavefront satisfy?
(x´+ut)2
+ y´2 + z´2 = (ct´)2
Space and Time Transformations

Einstein’s 2nd postulate, the constancy of the speed of light, implies that all
observers measure exactly the same value for the speed of light.

Suppose that a lightbulb, located at the origins O in the reference frames S, is
turned on at time t = t´ = 0 when the origins O and O´ coincide.

After a time t has elapsed in the S reference frame, an observer in this frame
would see a spherical wavefront with a radius ct moving away from the origin O
with speed c and satisfying
Equation for
sphere

Similarly, after a time t´ has elapsed in the S´ reference frame, an observer in this
frame would see a spherical wavefront with a radius ct moving away from the
origin O´ with speed c and satisfying
Equation for
sphere
Space and Time Transformations

Inserting Eqs. (4.10-4.13) into Eq. (4.15)
and comparing the result with Eq.
(4.14),
we find that
Assignment
question
Space and Time Transformations

Thus, if the speed of light is constant in all reference frames, we get
which are the same transformations as the Lorentz transformation equations. If
you differentiate x´ wrt to t´ (as we will do later), you will find that the speed of
light is constant in all reference frames.
Space and Time Transformations

Compare Lorentz with Galilean transformations:
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 0 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 0 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 0 – 0.1 c × 1 s = -0.1 c × 1 s
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 0 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 0 – 0.1 c × 1 s = -0.1 c × 1 s

According to the Lorentz transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 0 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 0 – 0.1 c × 1 s = -0.1 c × 1 s

According to the Lorentz transformation?
x´ = (0 – 0.1 c × 1 s) / √ (1 – 0.12)
= (-0.1 c × 1 s) / 0.995
< (-0.1 c × 1 s)
This reflects the contraction of space along
the x´ direction in frame S´ according to S.
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 10 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 10 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 10 – 0.1 c × 1 s
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 10 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 10 – 0.1 c × 1 s

According to the Lorentz transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Say that u = 0.1c. If an object is located at x = 10 in frame S, at what x´ will it be
located in frame S´ after t = 1 s according to the Galilean transformation?
x´ = 10 – 0.1 c × 1 s

According to the Lorentz transformation?
x´ = (10 –0.1 c × 1 s) / √ (1 – 0.12)
= (10 – 0.1 c × 1 s) / 0.995
> (10 –0.1 c × 1 s)
This reflects the contraction of space along
the x´ direction in frame S´ according to S.
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Why does t´ depend on x?
Space and Time Transformations

Compare Lorentz with Galilean transformations:
Implicit in this equation is that t´ = t at x = 0; i.e., where origins of S´ and S frames coincide.

Say that u = 0.1c. Suppose two flashbulbs, one located at x = 10 and the other at x
= 20, both flash at time t = 10 s in frame S. At what time t´ would they flash in
frame S´ according to the Galilean transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:
Implicit in this equation is that t´ = t at x = 0; i.e., where origins of S´ and S frames coincide.

Say that u = 0.1c. Suppose two flashbulbs, one located at x = 10 and the other at x
= 20, both flash at time t = 10 s in frame S. At what time t´ would they flash in
frame S´ according to the Galilean transformation? Both flash at t´ = 10 s

According to the Lorentz transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:
Implicit in this equation is that t´ = t at x = 0; i.e., where origins of S´ and S frames coincide.

Say that u = 0.1c. Suppose two flashbulbs, one located at x = 10 and the other at x
= 20, both flash at time t = 10 s in frame S. At what time t´ would they flash in
frame S´ according to the Galilean transformation? Both flash at t´ = 10 s

According to the Lorentz transformation?
t´1 = (10 − 0.1 c 10 / c2) / √ (1 – 0.12)
= (10 − 1 / c) / √ 0.995
t´2 = (10 − 0.1 c 20 / c2) / √ (1 – 0.12)
= (10 − 2 / c) / √ 0.995
frame S´.
of simultaneity.
They do not flash at the same time in
This is the concept of the loss
Space and Time Transformations

Compare Lorentz with Galilean transformations:
Implicit in this equation is that t´ = t at x = 0; i.e., where origins of S´ and S frames coincide.

Say that u = 0.1c. Suppose two flashbulbs, one located at x = 10 and the other at x
= 20, both flash at time t = 10 s in frame S. At what time t´ would they flash in
frame S´ according to the Galilean transformation? Both flash at t´ = 10 s

According to the Lorentz transformation?
t´1 = (10 − 0.1 c 10 / c2) / √ (1 – 0.12)
= (10 − 1 / c) / √ 0.995
t´2 = (10 − 0.1 c 20 / c2) / √ (1 – 0.12)
= (10 − 2 / c) / √ 0.995
The reason that t´1 ≠ t and t´2 ≠ t
reflects
the loss of synchronicity
because the clock
in frame S´ appears to
tick slower according to observer in frame S.
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Why does t´ depend on x? Reflects concept of the loss of simultaneity.
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Under what condition does the Lorentz transformation closely approach the
Galilean transformation?
Space and Time Transformations

Compare Lorentz with Galilean transformations:

Under what condition does the Lorentz transformation closely approach the
Galilean transformation? When u « c
The Lorentz Transformations

The Lorentz transformations were originally proposed by
the Dutch physicist Hendrik Antoon Lorentz and his
collaborators. They believed in the luminiferous ether
hypothesis, and were motivated purely by the desire to find
a mathematical transformation under which Maxwell’s
equations were invariant when transformed
from the ether to a moving reference frame
(i.e, so that light travels at the same speed no
matter the motion of the observer relative to the ether).
Hendrik A. Lorentz,
1853-1928
Hendrik A. Lorentz,
1853-1928 mirror
Half-silvered
Light source
Matter at rest in ether
Matter moving to right
with respect to ether
Direction of Earth’s motion through ether
The Lorentz Transformations

Lorentz believed that space is an absolute, and that objects
contract along the direction of motion such that the speed of
light is a constant no matter our speed relative to the ether.

Einstein discarded the notion that space (and time) is an
absolute, but instead has dimensions that depend on our
relative motion.
Hendrik A. Lorentz,
1853-1928
Albert Einstein, 18791955
Matter at rest in ether
Matter moving to right
with respect to ether
The γ factor

The Lorentz transformations:

The factor
is called the Lorentz factor, and is often used to estimate the importance of
relativistic effects. When u << c, γ ≈ 1 and the Lorentz transformation reduces to
the Galilean transformation. When u → c, γ > 1.

γ = 1.00001 when u = 1/222 c = 1341 km/s, ~170 times the speed of the space
shuttle (~28000 km/h = 7.8 km/s).

γ = 1.01 when u = 1/7 c.

γ = 1.1 when u = 5/12 c.
The γ factor
u/c
0.14
0.42
0.55
0.64
0.70
0.75
0.87
0.94
0.98
0.99

1.01
1.10
1.20
1.30
1.40
1.50
2.0
3.0
5.0
10.0
The inverse Lorentz Transformations

-
What if the reference frame is moving in the –x-direction relative to us? We then
use the inverse Lorentz transformations, which can be derived in a similar manner
as the (forward) Lorentz transformations, or using symmetry arguments by simply
switching primed and unprimed quantities and replacing u with –u.