Option H: Relativity
H4 Consequences of special relativity
The twin paradox
H.4.1 Describe how the concept of time dilation
leads to the twin paradox.
H.4.2 Discuss the Hafele-Keating experiment.
Option H: Relativity
H4 Consequences of special relativity
The twin paradox
Describe how time dilation leads to the paradox.
This has already been done in Topic H3.
EXAMPLE: The twin paradox: Suppose Einstein has a
twin brother who stays on Earth while Einstein
travels at great speed in a spaceship. When he
returns to Earth Einstein finds that his twin has
aged more than himself! Explain why this is so.
SOLUTION: Since Einstein is in the moving
spaceship, his clock ticks more slowly. But his
twin’s ticks at its Earth rate. The twin is thus
older than Einstein on his return!
By the way, this
is NOT the
paradox. The
paradox is on
the next slide…
Option H: Relativity
H4 Consequences of special relativity
The twin paradox
Describe how time dilation leads to the paradox.
This has already been done in Topic H3.
EXAMPLE: The twin paradox: From Einstein’s
perspective Einstein is standing still, but his
twin is moving (with Earth) in the opposite
direction. Thus Einstein’s twin should be the one
to age more slowly. Why doesn’t he?
SOLUTION: The “paradox” is resolved by general
relativity (which is the relativity of noninertial reference frames). It turns out that
because Einstein’s spaceship is the reference
frame that actually accelerates, his is the one
that “ages” more slowly.
Option H: Relativity
H4 Consequences of special relativity
The twin paradox
Discuss the Hafele-Keating experiment.
From Wikipedia: “The Hafele–Keating experiment
was a test of the theory of relativity. In 1971,
Hafele, a physicist, and Keating, an astronomer,
took four atomic clocks aboard
This experiment
commercial airliners.”
obviously tested
“They flew twice around the world, time dilation.
first eastward, then westward, and
compared the clocks against others
that remained at the United States
Naval Observatory.”
“When reunited, the three sets of
clocks were found to disagree with
one another, and their differences
were consistent with the predictions
of special and general relativity.”
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
H.4.3 Solve one-dimensional problems involving
the relativistic addition of velocities.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
Suppose you are in Car B, shown below.
Suppose car A is moving at ux = +20 m s-1 and your
car B is moving at v = +40 m s-1 as shown.
As far as you are concerned, the velocity ux’ of
car A relative to you is -20 m s-1 , because A is
moving backwards relative to you.
ux’ = ux - v
velocity of A relative to B
FYI
This is the Galilean transformation which we
learned way back in Topic 2.1.
B
A
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
Now consider two
spaceships leaving Earth
in opposite directions:
Suppose ux = 0.75c and
v = 0.50c (to the left).
Then according to the Galilean transformation,
the ux’ (the velocity of A relative to B) would be
ux’ = ux – v = 0.80c - -0.50c = 1.3c.
This is in contradiction to the tenet that c is
the maximum speed anything can have.
FYI
The Galilean transformation for addition of
velocities fails at relativistic speeds.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
At relativistic speeds we will use:
relativistic
u - v
ux’ = x
Where
u
and
v
are
x
u v
velocity
1 - x2
signed
quantities.
c
addition
EXAMPLE: Spaceship B is traveling leftward at 0.50c wrt Earth.
Spaceship A is traveling rightward at 0.80c wrt Earth. What
is the speed of A relative to B?
SOLUTION: v = -0.50c, ux = +0.80c:
● ux’ = (ux - v) / (1 - uxv/c2)
= (0.8c – -0.5c) / (1 - -0.5c  0.8c / c2)
= 1.3c / (1 + 0.4) = 0.93c.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
●Use ux’ = (ux – v)/(1 – uxv/c2), where ux = 0.8c
and v = -0.8c (since they are traveling in
opposite directions.
●Then ux’ = (0.8c – -0.8c)/[1 – (0.8c)(-0.8c)/c2].
ux’ = 1.6c/[1 + 0.64].
ux’ = 0.98c.
FYI
Make sure you are consistent with your signs and
that you get an answer that is LESS THAN c!
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
●A Galilean transformation assumes that time
and length are absolute (the same in all
reference frames) and fails under conditions
of relativistic speeds.
●A Galilean transformation simply adds
velocities in a straight-forward manner
according to “intuition.”
●EXAMPLE: ux’ = ux – v.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
●Use ux’ = ux – v, where ux = 0.9800c and
v = -0.9800c (since they are traveling
in opposite directions).
●Then ux’ = 0.98c – -0.98c = 1.960c.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
●Use ux’ = (ux – v)/(1 – uxv/c2), where ux =
0.98c and v = -0.98c (since they are
traveling in opposite directions.
ux’ = (0.98c – -0.98c)/[1 – (0.98c)(-0.98c)/c2]
= 1.96c/[1 + 0.982]
= 0.9998c.
Option H: Relativity
H4 Consequences of special relativity
Velocity addition
Solve one-dimensional problems involving the
relativistic addition of velocities.
●In (b)(i) v > c which is not possible.
●Thus the Galilean transformation is not
applicable to this problem.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
H.4.4 State the formula representing the
equivalence of mass and energy.
H.4.5 Define rest mass.
H.4.6 Distinguish between the energy of a body at
rest and its total energy when it is moving.
H.4.7 Explain why no object can ever attain the
speed of light in a vacuum
H.4.8 Determine the total energy of an
accelerated particle.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
State the formula representing the equivalence of
mass and energy.
●Recall E = mc2. It has been slightly changed:
E0 = m0c2
Equivalence of
mass and energy
●We used this formula when we looked at mass
defect in nuclear energy problems. It is the
energy of a mass m0 in its rest frame.
PRACTICE:
A nuclear power plant converts about 30 kg of
matter into energy each year. How many joules is
this? How many watts?
Much of this energy is
SOLUTION: wasted in conversion to electrical power.
●E0 = m0c2 = 30(3108)2 = 2.71018 J.
●P = E0/t = 2.71018/[365243600] = 8.61010 W.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Define rest mass.
●It is beyond the scope of this course, but not
only do time and length change with speed, but so
does mass!
1
m = m0
relativistic
where  =
mass
1 - v2/c2
●We call m the relativistic mass. Recall that  is
the Lorentz factor, and it is instrumental in
solving special relativity problems.
●We call m0 the rest mass. This is the mass of the
object as measured in a reference frame in which
it is at rest.
FYI
Note that as v  c that   .
●Thus as v  c we see that m  .
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Define rest mass.
m = m0
where  =
1
1 - v2/c2
relativistic
mass
PRACTICE: At CERN a proton can be accelerated to
a speed such that its relativistic mass is that
of U238. How fast is it going?
Make sure you can
SOLUTION:
reproduce both of
●First, find the Lorentz factor:
these graphs.
m = m0
238mp = mp
 = 238.
●Then solve for v:
(1 – v2/c2)1/2 = 1/238
1 – v2/c2 = 1/2382
1 – 1/2382 = v2/c2
0.9999823 = v2/c2  v = 0.9999912c.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Distinguish between the energy of a body at rest
and its total energy when it is moving.
●Since mass increases with velocity according to
m = m0, clearly the total energy of a moving mass
is E = m0c2.
1
E0 = m0c2
relativistic
where  =
E = m0c2
energy
1 - v2/c2
PRACTICE: Show
reduces to the
SOLUTION:
●If v = 0 then
●Then E = m0c2
that the relativistic energy E
rest energy E0 when v = 0.
 = 1/(1 - 02/c2)1/2 = 1/11/2 = 1.
= 1m0c2 = m0c2 = E0.
FYI
Thus E = m0c2 is the total energy of the object,
whether the object is moving or not.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Explain why no object can ever attain the speed
of light in a vacuum
●Various arguments can be presented to show that
no object with a non-zero rest mass can attain
the speed of light.
EXAMPLE:
Argument 1: m   as v  c.
●If v = c then  = 1/(1 - c2/c2) = 1/0 = .
●Since m = m0 then m   also.
●But there is not even an infinite amount of mass
in the universe. (Reductio ad absurdum).
Argument 2: E   as v  c.
●Since E = m0c2 then as    so does E.
●But there is not even an infinite amount of
energy in the universe. (Reductio ad absurdum).
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Recall that the acceleration of a charge e
through a potential difference V produces a
kinetic energy given by EK = eV.
●The total energy E of a particle of rest mass m0
is just the sum of its rest energy E0 and its
kinetic energy EK = eV.
E = E0 + EK
total energy of
mc2 = m0c2 + eV
an accelerated
particle
where m = m0
FYI
We can not use (1/2)mv2 = eV at relativistic
speeds to find v because it assumes all of the
energy eV is going into the velocity. But the
mass also changes at large speeds.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Use E = m0c2 = E0. Then
3E0 = E0   = 3.
●Since  = 1/(1 – v2/c2)1/2
(1 – v2/c2)1/2
1 – v2/c2
v2/c2
v
=
=
=
=
=
3, then
1/3
1/9
8/9
0.94c.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●The
with
●The
rest
mass as measured by an observer at rest
respect to the mass.
mass as measured by an observer in the
frame of the mass.
●From the formula we see that v  V.
●Thus if V is large enough, v > c, which
cannot happen.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Use E = E0 + EK.
●Then
mc2 =
mc2 - m0c2 =
mc2 =
m =
m =
m =
●Alternate method…
m =
m =
m0c2 + eV.
eV.
eV
eV/c2
e(5.0106 V)/c2
5.0 MeV c-2
(1.610-19)(5106)/(3108)2
8.910-30 kg.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Use E = E0 + eV. Then
mc2 = m0c2 + eV.
m0c2 = m0c2 + eV.
 = 1 + eV/(m0c2)
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Use
 =
 =
 =
 =
 = 1 + eV/(m0c2).
1 + e(500 MV)/(938 MeVc-2c2)
1 + 500 MeV/(938 MeV)
1 + 500/938 = 1.53
1/(1 – v2/c2)-1/2 = 1.53
1 – v2/c2 = 1/1.532 = 0.427
1 - 0.427 = v2/c2
v2 = 0.573c2
v = 0.76c.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●As v  c,   .
●Since E = m0c2 we see that
●As   , E  .
●Since there is not an infinite amount of
energy in the universe, you cannot
accelerate an object with a rest mass to
the speed of light.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
EK = eV = e(6.00106) V = 6.00 MeV.
E = E0 + EK = 0.51 MeV + 6.00 MeV = 6.51 MeV.
(For an electron, E0 = m0c2 = 0.51 MeV.)
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●E = m0c2 = (0.51 MeV) = 6.51 MeV
● = 6.51 MeV / 0.51 MeV = 12.17.

1 – v2/c2
1 - 0.0067
v2
v
=
=
=
=
=
1/(1 – v2/c2)-1/2 = 12.17
1/12.172 = 1/12.172
v2/c2
0.9933c2
0.997c.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●Rest mass energy is E0 = m0c2 and is the
energy that a particle has in its rest
frame.
●Total energy is E = m0c2 + EK and is the
sum of the rest mass energy and the kinetic
energy EK = eV.
●For these problems we always assume there
is no potential energy.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
●E0 = m0c2 = (938 MeV c-2)c2 = 938 MeV.
Option H: Relativity
H4 Consequences of special relativity
Mass and energy
Determine the total energy of an accelerated
particle.
● = 1/(1 – v2/c2)-1/2
= 1/(1 – 0.9802c2/c2)-1/2 = 5.03.
●E = m0c2 = m0c2 + eV
eV
V
V
V
=
=
=
=
( - 1)m0c2
( - 1)m0c2/e
(5.03 - 1)(938 MeV)/e
3780 MeV.