Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 2. Special relativity – kinematic paradoxes • addition of
velocities • relativistic mass and momentum • relativistic energy •
energy-momentum 4-vector • decay and scattering in relativity •
nuclear energy
Sources: Feynman Lectures I, Chap. 15, Sects. 5-6 and Chap. 16;
Tipler and Llewellyn, Chap. 1, Sect. 6 and Chap. 2, Sects. 1-4
Kinematic paradoxes
The “Twin Paradox”
You go out into space for one year at 0.9 of the speed of
light. When you come back, how old will your twin brother
be? Is there a paradox here? How would you solve it?
The “Triplet Paradox”
Grumpy lives in Haifa; Dumpy and Jump live in Tel Aviv.
They are identical triplets with synchronized watches. One
day, Jump takes a high-speed (v = c/2) train to visit Grumpy.
Grumpy and Dumpy expect Jump’s watch to run slow during
the train ride. But Jump expects their watches to run slow!
He reasons, “When the train left the station, I saw that my
watch was still synchronized with Dumpy’s. So it was still
synchronized with Grumpy’s watch. Now I am sitting and
they are moving fast relative to me! So their watches must
run slower than mine!”
Who is right???
Event 1: Jump looks at his watch and at Dumpy’s watch.
He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be
x1′ = x1 = 0.
Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0.
Event 2: Grumpy looks at his watch and sees t2 = 0. Let his
location (Haifa) be x2 = 85 km.
Summary of Event 2 : t2 = 0, x2 = 85 km.
Event 1: Jump looks at his watch and at Dumpy’s watch.
He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be
x1′ = x1 = 0.
Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0.
Event 2: Grumpy looks at his watch and sees t2 = 0. Let his
location (Haifa) be x2 = 85 km.
Summary of Event 2 : t2 = 0, x2 = 85 km.
Hence
∆t = t2 – t1 = 0,
∆x = x2 – x1 = 85 km.
Event 1: Jump looks at his watch and at Dumpy’s watch.
He sees t1′ = 0 and t1 = 0. Let their location (Tel Aviv) be
x1′ = x1 = 0.
Summary of Event 1: t1′ = t1 = 0, x1′ = x1 = 0.
Event 2: Grumpy looks at his watch and sees t2 = 0. Let his
location (Haifa) be x2 = 85 km.
Summary of Event 2 : t2 = 0, x2 = 85 km.
Hence
∆t = t2 – t1 = 0,
∆x = x2 – x1 = 85 km.
What is ∆t′ ?
“The barn paradox”
The barn is 5 m long to the farmer.
The pole is 10 m long to the runner
and 5 m long to the farmer, i.e. we
have g = 2 and so b = v/c = 0.866.
According to the farmer, the pole fits into the barn at a
single instant. But the runner cannot agree! The pole is
10 m long and the barn is only 2.5 long!
What are the events?
Event 1: the rear end of the pole
reaches the entrance to the barn.
Event 2: the front tip of the pole
reaches the far end of the barn.
In the farmer’s reference frame, Event 1 occurs at time t1
and at place x1; Event 2 occurs at time t2 and at place x2 ;
Dt = t2 – t1 = 0 and Dx = x2 – x1 = 5 m.
In the runner’s reference frame:
Dt′ = g(Dt – vDx/c2) = (2)(-0.866)(5 m) / c = -28.9 ns,
Dx′ = g(Dx – vDt) = 10 m.
What are the events?
Event 1: the rear end of the pole
reaches the entrance to the barn.
Event 2: the front tip of the pole
reaches the far end of the barn.
For the runner, the two events are not simultaneous!
What does it mean that Dt′ is negative?
What are the events?
Event 1: the rear end of the pole
reaches the entrance to the barn.
Event 2: the front tip of the pole
reaches the far end of the barn.
Further applications of Lorentz transformations
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
How do we derive
time dilation from
the Lorentz
transformations?
If Alice (primed system) moves along the x-axis with
velocity v relative to Bob (unprimed system), then
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
How do we derive
time dilation from
the Lorentz
transformations?
cDt/2
L
cDt/2
vDt
☺
If Alice (primed system) moves along the x-axis with
velocity v relative to Bob (unprimed system), then
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
For Alice, the clock
stays in one place.
Hence Dx′ = 0.
cDt/2
L
cDt/2
vDt
☺
If Alice (primed system) moves along the x-axis with
velocity v relative to Bob (unprimed system), then
0 = Dx′ = g(Dx – vDt)
Hence Dx = vDt
Dt′ = g(Dt – vDx/c2) = g(Dt – v2Dt/c2) = g(Dt/g2 ) = Dt/g
For Alice, the clock
stays in one place.
Hence Dx′ = 0.
cDt/2
L
cDt/2
vDt
☺
Further applications of Lorentz transformations
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
How do we derive
length contraction
from the Lorentz
transformations?
If the runner (primed system) moves along the x-axis with
velocity v relative to the farmer (unprimed system), then
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
How do we derive
length contraction
from the Lorentz
transformations?
If the runner (primed system) moves along the x-axis with
velocity v relative to the farmer (unprimed system), then
Dx′ = g(Dx – vDt)
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
For the farmer, Dx = L and Dt = 0.
For the runner, Dx′ = L′.
If the runner (primed system) moves along the x-axis with
velocity v relative to the farmer (unprimed system), then
L′ = Dx′ = g(Dx – vDt) = gL
Dy′ = Dy
Dz′ = Dz
Dt′ = g(Dt – vDx/c2)
For the farmer, Dx = L and Dt = 0.
For the runner, Dx′ = L′.
Addition of velocities:
ux= dx/dt ; uy= dy/dt ; uz= dz/dt
ux′= dx′/dt′ ; uy′= dy′/dt′ ; uz′= dz′/dt′
For inertial reference frames moving with relative velocity v
along the x–axis,
ux′= (ux–v)/(1–vux/c2)
uy′= uy /g(1–vux/c2)
uz′= uz /g(1–vux/c2)
Thus you can never surpass the speed of light e.g. by shooting a
0.9c bullet from a 0.9c jet:
(0.9c+0.9c)/[1+(0.9c) · (0.9c)/c2] = (1.80/1.81)c = 0.9945c
Derivation: apply the Lorentz transformations to the differentials,
dx' = g dx - vg dt
dy' = dy
dz' = dz
dt ' = g dt - vg dx / c 2
and substitute them into ux′, uy′ and uz′ to obtain
ux - v
g dx - vg dt
dx - v dt
u x ' = dx' / dt ' =
=
=
2
2
g dt - vg dx / c
dt - v dx / c
1- v ux / c2
u y ' = dy' / dt ' =
u z ' = dz' / dt ' =
dy
g dt - vg dx / c
2
dz
g dt - vg dx / c
2
=
=
uy
g (1 - v u x / c 2 )
uz
g (1 - v u x / c 2 )
Relativistic mass and momentum
We now move from kinematics (pure description of motion) to
dynamics (description of forces).
We will discover that if momentum (mv) is conserved in one
inertial reference frame, it is not conserved in another!
But conservation of momentum is a law of physics. And Einstein
postulated that the laws of physics are the same in all reference
frames….
Two identical particles collide in their center-of-mass frame:
y
uy The blue particle has mass m and
speed v and so does the red particle.
x
In the center-of-mass frame, total
momentum is 0 before and after the
uy
collision.
ux
Two identical particles collide in their center-of-mass frame:
y
uy The blue particle has mass m and
speed v and so does the red particle.
x
In the center-of-mass frame, total
momentum is 0 before and after the
uy
collision.
ux
But now we go to a reference frame
moving to the right with speed ux:
Two identical particles collide in their center-of-mass frame:
y
uy The blue particle has mass m and
speed v and so does the red particle.
x
In the center-of-mass frame, total
momentum is 0 before and after the
uy
collision.
y
u
x
uy′
But now we go to a reference frame
moving to the right with speed ux:
(v' )ured
yy′ =
x
vured
yy
g [1 - u x2 c 2 ]
(v' )ublue
y y′ =
uy′
vublue
yy
g [1  u x2 c 2 ]
ux′ = 0
Since uy′ + uy′ ≠ 0, the y-component of momentum is not
conserved…
y
uy′
x
(v' )ured
yy′ =
vured
yy
g [1 - u x2 c 2 ]
(v' )ublue
y y′ =
uy′
vublue
yy
g [1  u x2 c 2 ]
ux′ = 0
Since uy′ + uy′ ≠ 0, the y-component of momentum is not
conserved…unless we redefine momentum!
y
uy′
x
(v' )ured
yy′ =
vured
yy
g [1 - u x2 c 2 ]
(v' )ublue
y y′ =
uy′
vublue
yy
g [1  u x2 c 2 ]
ux′ = 0
Since uy′ + uy′ ≠ 0, the y-component of momentum is not
conserved…unless we redefine momentum!
Let’s try p = m f(v) v with f(0)=1.
y
uy′
x
(v' )ured
yy′ =
vured
yy
g [1 - u x2 c 2 ]
(v' )ublue
y y′ =
uy′
vublue
yy
g [1  u x2 c 2 ]
ux′ = 0
y
Boost left ux
Boost right ux
x
ux
y
uy"
uy′
y
x
ux" = 0
x
Boost right v = ux"
uy′
uy"
ux′ = 0
2 2
g
=
1
/
1
v
/c :
Addition of velocities with v = ux" and
uuyy′ ' =
uy""
g [1 - vvu
u x"/
"/c ]
c2 2
=
uyy""
g [1 - v / c ]
2
2
= g uuyy"
y
uy" = – uy′
uy′
y
x
ux" = 0
x
uy" = – uy′
Boost right v = ux"
uy′
ux′ = 0
So now if we assume p = m f(v) v, then
mf (u ' ) u y ' = m f (u" ) u y "
= m f  (u x " ) 2  (u y " ) 2  u y "


f (u y ' ) u y ' / u y " = f  (u x " ) 2  (u y " ) 2 


g f (gu y " ) = f  (u x " ) 2  (u y " ) 2 


and in the limit uy"→ 0 we replace ux" with v to obtain
g = g f (0) = f (v)
and therefore
p = mγv
.
Relativistic momentum: p = gmv (or “relativistic mass”?)
gmv / mv
v/c .
v/c
Relativistic energy
We accelerate a body using a force F = dp/dt :



= vp -  pdv = v 2 mg -  vmg dv
dp
W = F  dx =
 vdt = vdp
dt
mv 2
m
=
1- v2 / c2 2
=
mv 2
1- v2 / c2

d (v 2 )
1- v2 / c2
 mc 2 1 - v 2 / c 2
If we accelerate the body from rest to a final speed v, we
find that the change in kinetic energy of the body is
W=
2
 mc 2  1 - v 2 / c 2 - 1


1- v2 / c2
mv
= mc 2 [g v 2 / c 2  g (1 - v 2 / c 2 ) - 1]
= gmc 2 - mc 2
In the limit c → ∞ the kinetic energy is mv2/2. But what
is the meaning of the huge “rest energy” mc2 ?
v
 
c
2
v
Energy-momentum 4-vector
E = mγc2, so E and pc form a 4-vector u = (E, pc) and
u 2 = (pc) 2 - ( E ) 2 = (mg v c) 2 - (mg c 2 ) 2
= -m 2 c 4
The energy-momentum 4-vector
transforms like any other 4-vector
under Lorentz transformations.
Hence both the energy E and the
momentum p are conserved in each
inertial reference frame.
Decay and scattering in relativity A fateful example
A heavy particle of mass M decays into two lighter particles of
mass m. What is the velocity of each of the lighter particles? How
much kinetic energy is released?
Let us treat this problem in the rest frame of the heavy particle.
Energy-momentum conservation:
(Mc2, 0) = (mγ1c2, mγ1v1c) + (mγ2c2, mγ2v2c) = (E1, p1c) + (E2, p2c)
Hence p1= – p2 and γ1= γ2 = γ from momentum conservation, and
Mc2 = 2mγc2 from energy conservation. Therefore γ = M/2m and
M 2 - 4m2
c.
since M / 2m = g =
we get v =
M
1 - v2 / c2
1
So in the rest frame of the heavy particle, the lighter particles
come out back to back with opposite velocities and equal speeds
M 2 - 4m2
v=
c
M
There is no kinetic energy initially. The final kinetic energy is
M

2
2
2mc (g - 1) = 2mc 
- 1 = Mc - 2mc
 2m 
2
2
More generally, if the heavy particle decays into several lighter
particles of masses mi, the kinetic energy released is
i
(  m )c
mi (g i - 1) c 2 = M -
i
i
2
Application: fission of U235 into Ba141 and Kr92 and also two
neutrons yields how much kinetic energy?
mass (a.m.u.)
U235
Ba141
Kr92
n
235.043923
140.914411
91.926156
1.008665
Kinetic energy yield
= (235.043923 – 140.914411 – 91.926156 – 2 × 1.008665) a.m.u.
× (1.660539 × 10–27 kg/a.m.u.) × (2.997925 × 108 m/s)2
= 2.77629 × 10–11 J
One gram of U235 contains 2.56 × 1021 atoms of U235, so fission
of one gram of pure U235 yields (2.56 × 1021 ) × (2.78 × 10–11 J)
= 7.12 × 1010 J of kinetic energy.
By way of comparison, consider two cars, each with 1500 kg
mass, at speed 90 km/hour = 25 m/s, in a head-on collision. The
total kinetic energy lost in the collision is (3000 kg) × (25 m/s)2
= 1.875 × 106 J. So fission of one gram of U235 has the
destructive power of about 40,000 such collisions.
The bomb dropped on
Hiroshima caused about
800 grams of U235 to
undergo fission…
...and about 80,000
deaths on that day, two
to three times as many
within five years.
Relativistic scattering: another fateful example
Two deuterium (1H2) nuclei collide and produce a helium-3 (2He3)
nucleus and a neutron. What are the velocities and energies of the
helium nucleus and the neutron?
The easiest way to solve this problem is to go to the “center-ofmomentum” frame, i.e. the inertial frame in which the total
momentum is zero:
The easiest way to solve this problem is to go to the “center-ofmomentum” frame, i.e. the inertial frame in which the total
momentum is zero:
In this case the problem reduces to the previous problem, with the
initial mass M of the decaying particle replaced by the initial
energy of the two deuterium nuclei divided by c2.
What is the minimum amount of kinetic energy produced by this
scattering process (in the center-of-momentum frame)?
What is the minimum amount of kinetic energy produced by this
scattering process (in the center-of-momentum frame)?
It is [2 × m(deuterium) – m(helium-3) – m(neutron)] × c2
Note: In fission of uranium-235, the fraction of mass converted
into energy is
235.043923 – 140.914411 – 91.926156 – 2  1.008665
= 0.000791
235.043923
Note: In fission of uranium-235, the fraction of mass converted
into energy is
235.043923 – 140.914411 – 91.926156 – 2  1.008665
= 0.000791
235.043923
In fusion of two deuterium nuclei to helium-4, it is
2  2.01410778 – 4.00260325
= 0.00636
2  2.01410778
Nuclear energy: the curve of binding energy
Average binding energy ∆E/A vs. mass number A
m/A
Nuclear energy: the curve of average nucleon mass
Average mass of nucleon m/A vs. mass number A