Theory of Relativity including the different consequences of Einstein's Special Theory of Relativity

1. THE
SPECIAL
THEORY OF
RELATIVITYClado — Orcullo — Salva

2. OBJECTIVES
To understand the different
consequences of special relativity
To familiarize different formulas and perform calculations
regarding the consequences of special relativity
To apply the lessons in real life

3. Did you know? If you could
travel at the speed of light
without breaking down to
your basic atoms, you’d be
able to go around the Earth
7 and a half times in a
single second
TRIVIA TRIVIATRIVIA
TRIVIATRIVIA
TRIVIA TRIVIATRIVIA
TRIVIA TRIVIATRIVIA TRIVIA TRIVIATRIVIA

4. BACKGROUND ON
SPECIAL RELATIVITY
Along with the nuclear physics, special
relativity is central to modern physics. It
describes the motion of particles with
speed ranging from zero to a value
close to the speed of light in vacuum.
In 1905, Albert Enstein published his special theory
of relativity. He based his theory on two postulates.

5. BACKGROUND ON
SPECIAL RELATIVITY
WHAT ARE THE TWO POSTULATES
OF SPECIAL RELATIVITY
Principle Of Relativity
Constancy Of The Speed Of Light
States that the laws of physics must be the same in all inertial frames
of reference. An inertial frame of reference is one where Newton’s
first law of motion is valid. It is a frame of reference where a particle is
observed to have no acceleration in the absence of an unbalanced
force, thus, an accelerating or rotating frame is not inertial.
States that the speed if light in a vacuum is constant in all inertial
frames of reference and is independent of the motion of the source.

6. BACKGROUND ON
SPECIAL RELATIVITY
LORENTZ
FACTOR
𝛾 =
1
1 −
𝑣
𝑐
2
This quantity is called the Lorentz factor
named after dutch physicist Hendrik
Lorentz. It is denoted by the symbol 𝛾
(read as gamma). The letter 𝑣 is the speed
of the object and 𝑐 is the speed of light
equal to 3𝑥108
𝑚/𝑠2

7. BACKGROUND ON
SPECIAL RELATIVITY
CONSEQUENCES FROM EINSTEIN’S
SPECIAL THEORY OF RELATIVITY
Relativistic Kinematics:
• Time dilation,
• Length contraction,
• Relativistic Doppler
effect, and
• Mass increase.
Relativistic Dynamics
• Relativistic mass and
momentum,
• mass-energy
equivalence, and
• Relativistic second law.

8. TIME DILATION
In 1971, physicist Joseph Hafele and
astronomer Richard Keating confirmed that
the time interval betweem two events
measured by a moving observer who views
these events as occuring in different places
is longer than the time interval measured by
an observer at rest, who views the events
as happening in the same place.

9. TIME DILATION
The first time interval is referred to as the dilated time
interval and the second time interval is the proper
time interval. The dilated time interval may be
computed using
𝛥𝑡 =
𝛥𝑡0
1 −
𝑣
𝑐
2
= 𝛾𝛥𝑡0
TIME
INTERVAL
𝛥𝑡
𝛥𝑡0
𝛾
𝑐
𝑣dilate time interval
proper time interval
speed of relative motion
speed of light in vacuum
Lorentz factor

10. TIME DILATION
TWIN
PARADOX
Imagine a pair of twins, one of whom is an
astronaut that travels to a distant star. The
stay-at-home twin will see his brother age
more slowly than him. You’d expect the space-
faring twin to see the same happen to his
Earth-bound counterpart since they’re moving
at the same speed relative to one another. But
if the twins are re-united, Einstein said that the
space-faring twin will have aged less than the
one on Earth, which is odd given that they’ve
both performed identical journeys relative to
each other.

11. TIME DILATION
TWIN
PARADOX
Sample Problem
Leo and Christian are twins. At the age of 30, Leo left for a round
trip to a distant star in a spaceship with a speed of 0.95𝑐 relative to
Earth. The whole trip took 20 years according to the shipboard
clock. Find their ages when Leo returns.
Given: 𝑣 = 0.95𝑐
Solution
The two events in this example are Leo’s departure from Earth and
his return to Earth. The proper time interval must be the time
recorded by the clock onboard the spaceship, that is 𝛥𝑡0=20 years
𝛥𝑡 =
𝛾𝛥𝑡0
1 −
𝑣
𝑐
2 𝛥𝑡 =
20𝑦𝑒𝑎𝑟𝑠
1 −
0.95𝑐
𝑐
2
𝛥𝑡 = 64.05𝑦𝑒𝑎𝑟𝑠 ≈ 64𝑦𝑒𝑎𝑟𝑠

12. LENGTH OF
CONTRACTION
Length of an object measured by an
observer in an inertial reference frame
that is moving with respect to the object
is less than its proper length
Happens only in dimensions along the
direction of motion

13. LENGTH OF
CONTRACTION
FORMULA FOR LENGTH
LENGTH OF CONTRACTION
𝐿′ = 𝐿 1 −
𝑣
𝑐
2
𝐿 =
𝐿
𝛾
where:
L = length if the object at rest
L’ = length of the object moving at speed 𝑣
𝛾 = the Lorentz Factor: 𝛾 =
1
1−
𝑣
𝑐
2

14. LENGTH OF
CONTRACTION
SAMPLE
PROBLEM
A spaceship traveling at 0.5c relative to earth is 45m long as
measured by its crew. How long is the spaceship as measured
by the mission control in Texas?
𝑣=0.5 𝑐
𝐿0=45m
Solution
𝐿′ = 𝐿 1 −
𝑣
𝑐
2
𝐿′ = 45𝑚 1 −
0.5𝑐
𝑐
2
𝐿′ = 38.97𝑚 ≈ 39𝑚

15. RELATIVISTIC
DOPPLER
EFFECT
Doppler effect for electromagnetic waves
Named after Austrian mathematician and
physicist Christian Johann Doppler
Doppler effect for electromagnetic waves
is given by:
- fobs is the observed frequency
- c is the speed of light
- v is the speed of source relative to the observer
- fs is frequency in the rest frame or source
- the upper sign is for observer moving toward the source
- the lower sign is for an observer moving away from the source
𝑓𝑜𝑏𝑠 = 𝑓𝑠
1 −
𝑣
𝑐
1 +
𝑣
𝑐

16. RELATIVISTIC
DOPPLER
EFFECT
DOPPLER EFFECT FOR LIGHT IS
GIVEN IN TERMS OF REDSHIFT OR
BLUESHIFT
RED SHIFT If the source of light is moving
away from the observer, the observed
wavelength is longer than the wavelength
emitted when the source is at rest
BLUE SHIFT if source
of light is moving
toward an observer,
the wavelength of light
is shorter than the
wavelength when the
source is at rest

17. RELATIVISTIC
DOPPLER
EFFECT
DOPPLER EFFECT FOR LIGHT IS
GIVEN IN TERMS OF REDSHIFT OR
BLUESHIFT
Doppler developed this theory in an
effort to explain the shift in frequency
light by moving atoms or astronomical
bodies
Edwin Hubble an American astronomer
used this theory to confirm that the
universe is expanding and that most
galaxies are receding from us.
Light emitted by
those galaxies
are redshifted

18. RELATIVISTIC
DOPPLER
EFFECT
Sample problem: suppose a space probe moves
away from earth at a speed of .350c. It sends a
radio-wave message back to earth at a frequency
of 1.50 GHz. At what frequency is message
received back on earth?
Solution
SAMPLE
PROBLEM
𝑓𝑜𝑏𝑠 = 𝑓𝑠
1 −
𝑣
𝑐
1 +
𝑣
𝑐
𝑓𝑜𝑏𝑠 = 150𝐺𝐻𝑧
1 −
0.35𝑐
𝑐
1 +
0.35𝑐
𝑐
𝑓𝑜𝑏𝑠 = 1.04𝐺𝐻𝑧

19. LORENTZ
TRANSFORMATIONS
Motion is relative and depends on the
frame of reference where motion is being
observed
Based from the figure: Galilean coordinate transformation
xB = xA – vt
yB = yA
zA = zB

20. LORENTZ
TRANSFORMATIONS
Based from the
figure: Galilean
coordinate
transformation
xB = xA – vt
yB = yA
zA = zB
differentiating the Galilean coordinate transformation with
respect to time
𝑑𝑥 𝐵
𝑑𝑡
=
𝑑𝑥 𝐴
𝑑𝑡
− 𝑣
𝑑𝑦 𝐵
𝑑𝑡
=
𝑑𝑦 𝐴
𝑑𝑡
𝑑𝑧 𝐵
𝑑𝑡
=
𝑑𝑧 𝐴
𝑑𝑡

21. LORENTZ
TRANSFORMATIONS
Let uA and uB be the velocity of the object as seen in frames A and
B, respectively
‣ Since frame B is just moving along the x-axis, the y- and z-
coordinates of the object viewed in the two frames are just the
same.
‣ Time tA in frame A is equal to time tb in frame B
If Speed of light measured in frame A is c, then, speed of light measured in frame B is:
uB = c – v
- this is contradictory in the speed of light (postulate 2) stating that the speed of light in
the vacuum is constant in all inertial frames of references and is independent of the
motion of the source
- postulate 2 requires that time dilation be considered in the coordinate and velocity
transformation

22. LORENTZ
TRANSFORMATIONS
LORENTZ COORDINATE
TRANSFORMATION
𝑥 𝐵 =
𝑥 𝐴 − 𝑣𝑡
1 −
𝑣2
𝑐2
𝑦 𝐵 = 𝑦 𝐴
𝑧 𝐵 = 𝑧 𝐴
𝑡 𝐵
=
𝑡 𝐴−
𝑣𝑥 𝐴
𝑐2

23. LORENTZ
TRANSFORMATIONS
LORENTZ VELOCITY
TRANSFORMATION
Velocity of (ux)B of the object along the x-axis as see in frame B:
𝑑𝑥 𝐵 =
𝑑𝑥 𝐴 − 𝑣𝑑𝑡 𝐴
1 −
𝑣2
𝑐2
𝑑𝑡 𝐵 =
𝑑𝑡 𝐴 −
𝑣𝑑𝑥 𝐴
𝑐2
1 −
𝑣2
𝑐2
Velocity of (ux)A of the object along the x-
axis as seen in frame A:
𝑢 𝑋 𝐴 =
𝑢 𝑥 𝐵 + 𝑣
1 +
𝑢 𝑥 𝐵 𝑣
𝑐2

24. LORENTZ
TRANSFORMATIONS
SAMPLE
PROBLEM
Sample problem: an electron is moving at 0l6c relative
to Lorentz. Einstein, who is inside a spaceship moving
at 0.8c relative to Lorentz, is also observing the
electron. What is the speed of the electron as
determined by Einstein?
Solution
𝑢 𝐵 =
𝑢 𝐴 − 𝑣
1 −
𝑢 𝐴 𝑣
𝑐2
Given: Lorentz is the fixed (stationary) frame and Einstein is in the moving frame.
uB = 0.6c – 0.8c / [1 – (0.6c)(0.8c)/ c2]
uB = -0.3846c ~-0.4c
- The negative sign means that relative to
Einstein, the electron is moving in the negative
direction.
v = 0.8c uA = 0.6c
solving for uB:

25. RELATIVISTIC
MASS
The mass of an object moving at a speed
v approaching that of light in vacuum (c) is
greater than its mass, called the rest
mass, when at rest relative to an observer.
The mass of a moving object, called the
relativistic mass is given by:
𝑚 =
𝑚0
1 −
𝑣
𝑐
2
= 𝛾𝑚0
Where m= relativistic mass and m0= rest mass. It follows that the relativistic
momentum p is given by:
𝑝 =
𝑚0 𝑣
1 −
𝑣
𝑐
2
= 𝛾𝑚0 𝑣

26. RELATIVISTIC
MASS
SAMPLE
PROBLEM
An electron is moving at 0.5c. What is the momentum
according to special relativity?
Given: v= 0.5c= 1.5 x 108 m/s, m0= 9.109 x 10-31kg
Solution
𝑝 =
𝑚0 𝑣
1 −
𝑣
𝑐
2
=
9.109x10−31
𝑘𝑔 (1.5x108 m
s
)
1 −
0.5𝑐
𝑐
2
= 1.5777 x 10-22 kg⋅m/s ≈ 1.6 x 10-22 kg⋅m/s

27. MASS-ENERGY
EQUIVALENCE
𝑬 = 𝒎0 𝒄 + 𝑲 = 𝒎𝒄2
Where E= Total energy, m0= rest
mass, c= speed of light in
vacuum, and K= relativistic
kinetic energy.
The equation shows that mass and energy are
equivalent. Therefore, a gain or loss in mass may
be considered a gain or loss in energy.

28. MASS-ENERGY
EQUIVALENCE
The relativistic kinetic energy is given as:
𝐾 =
𝑚0 𝑐2
1 −
𝑣
𝑐
2
− 𝑚0 𝑐2
=
1
1 −
𝑣
𝑐
2
− 1 𝑚0 𝑐2
= 𝛾 − 1 𝑚0 𝑐2
For the particle at rest, v= 0, and it
follows that K= 0. Therefore,
𝐸 = 𝐸0
𝑚0 𝑐2=
The energy 𝑬0is called rest energy

29. MASS-ENERGY
EQUIVALENCE
SAMPLE
PROBLEM
An object of mass 0.2 kg is moving at (2/3) c.
What is the kinetic energy in eV according to
special relativity?
Given: m= 0.2 kg, v= (2/3)c= 2 x 108 m/s
Solution
𝐾 = 𝛾 − 1 𝑚0 𝑐2
=
1
1 −
𝑣
𝑐
2
− 1 𝑚0 𝑐2
=
1
1 −
2
3
𝑐
𝑐
2
− 1 (0.2𝑘𝑔)(3x108
𝑚
𝑠
)
= 6.1495x1015
𝐽
1𝑒𝑉
1.602x10−19 𝐽
= 3.83866 x 1034 eV ≈ 3.8 x 1034 eV

30. RELATIVISTIC
SECOND LAW
In classical mechanics, force is
defined as the time rate of change
of momentum, that is
𝐹 =
𝑑𝑝
𝑑𝑡
This definition is still valid in relativistic
mechanics provided that the relativistic
momentum is used:
𝐹 =
𝑑
𝑑𝑡
𝑚𝑣
1 −
𝑣
𝑐
2

31. RELATIVISTIC
SECOND LAW
MORE
FORMULA
For force acting in the same direction,
𝐹 =
𝑚
1 −
𝑣
𝑐
2
3
2
𝑎
Solving for acceleration,
𝑎 =
𝐹
𝑚
1 −
𝑣
𝑐
2 3 2
This equation shows that is speed v is very small compared to the speed of light
c, the acceleration can be written as 𝑎 =
𝐹
𝑚
, which is Newton’s second law in
classical mechanics. As v approaches c, the acceleration approaches zero.
This implies that it is impossible to accelerate an object to a speed equal to or
greater than c (speed of light in vacuum).

32. RELATIVISTIC
SECOND LAW
SAMPLE
PROBLEM
A microsatellite has a mass of approximately 50 kg.
What is its acceleration when launched at a speed
of 0.6c by a force of 5000 N?
Given: m= 50kg, v= 0.6c, F= 5000N
Solution
𝑎 =
𝐹
𝑚
1 −
𝑣
𝑐
2 3 2
=
5000𝑁
50𝑘𝑔
1 −
0.6𝑐
𝑐
2 3 2
= 51.2 m/s2