Physics 202
Professor P. Q. Hung
311B, Physics Building
Physics 202 – p. 1/2
Relativity
Momentum in Special Relativity
Classically, the momentum is defined as
r
p~ = m~v = m ∆~
∆t .
We also learned that momentum is
conserved.
p
We also learned that F~ = ∆~
∆t . Momentum
conservation is the consequence of zero
external force.
Physics 202 – p. 2/2
Relativity
Momentum in Special Relativity
Classically, the momentum is defined as
r
p~ = m~v = m ∆~
∆t .
We also learned that momentum is
conserved.
p
We also learned that F~ = ∆~
∆t . Momentum
conservation is the consequence of zero
external force.
Requirement: The laws of physics must be
the same in all inertial frames.
For instance, the total momentum should be
conserved in a collision.
Physics 202 – p. 2/2
Relativity
Momentum in Special Relativity
A detailed analysis reveals that, if we were to
use p~ = m~v , the momentum might be
conserved in one inertial frame but not in
another inertial frame. Should one give up
momentum conservation? NO. Redefine the
momentum.
Physics 202 – p. 3/2
Relativity
Momentum in Special Relativity
Insteadqof ∆t, one should use the proper time
v2
∆t0 = 1 − c2 ∆t.
The proper form for the momentum is
p~ = qm~v v2
1− c2
For v c, one recovers the usual classical
p~ = m~v .
Physics 202 – p. 4/2
Relativity
Momentum in Special Relativity
Physics 202 – p. 5/2
Relativity
Momentum in Special Relativity: Example
An electron, which has a mass of 9.11 × 10−31 kg,
moves with a speed of 0.750 c. Find its relativistic
momentum and compare this value with the
momentum calculated from the classical
expression.
Physics 202 – p. 6/2
Relativity
Momentum in Special Relativity: Example
p=
(9.11×10−31 kg)(0.750×3×108 m/s)
mv
q
√
=
2
v2
1−0.750
1− c2
−22
3.1 × 10
=
kg.m/s.
Physics 202 – p. 7/2
Relativity
Momentum in Special Relativity: Example
p=
(9.11×10−31 kg)(0.750×3×108 m/s)
mv
q
√
=
2
v2
1−0.750
1− c2
−22
3.1 × 10
=
kg.m/s.
The classical result is
p = mv = (9.11 × 10−31 kg)(0.750 × 3 ×
108 m/s) = 2.05 × 10−22 kg.m/s. A 50% smaller
than the relativistic result.
Physics 202 – p. 7/2
Relativity
Relativistic Energy
What does the folkloric E = mc2 mean?
Start with motion in blueone dimension for
simplicity. And also start the motion from rest.
Work done = Change in kinetic energy.
R
R dp
W = F dx = dt dx.
Physics 202 – p. 8/2
Relativity
Relativistic Energy
What does the folkloric E = mc2 mean?
Start with motion in blueone dimension for
simplicity. And also start the motion from rest.
Work done = Change in kinetic energy.
R
R dp
W = F dx = dt dx.
After some calculations, one finds
2
mc
W = q v2 − mc2
1− c2
⇒
K=
2
mc
q
2
1− vc2
− mc2
Physics 202 – p. 8/2
Relativity
Momentum in Special Relativity
Physics 202 – p. 9/2
Relativity
Relativistic Energy
γ=
q 1
.
v2
1− c2
Physics 202 – p. 10/2
Relativity
Relativistic Energy
γ=
q 1
.
v2
1− c2
Notice: For v c, one has
2
v
1
1
q
≈ 1 + 2 c2
v2
1− c2
2
⇒ K ≈ mc (1 +
classical result!
1 v2
2 c2 )
− mc2 = 12 mv 2 . The
Physics 202 – p. 10/2
Relativity
Relativistic Energy
There is one term which does not depend on
the speed: mc2 ⇒ Rest Energy of the particle.
Physics 202 – p. 11/2
Relativity
Relativistic Energy
There is one term which does not depend on
the speed: mc2 ⇒ Rest Energy of the particle.
Define the Total Energy of the particle as:
E = γ mc2 = K + mc2
Using p = γ mv, one finds (squaring both and
subtracting E 2 − p2 c2 ):
E 2 = p2 c2 + (mc2 )2
For p2 c2 (mc2 )2 , one has E ≈ pc.
Physics 202 – p. 11/2
Relativity
Relativistic Energy
From Eq. (4), one also finds:
pc
v
c = E
Physics 202 – p. 12/2
Relativity
Relativistic Energy
From Eq. (4), one also finds:
pc
v
c = E
Some units:
1 eV = 1.602 × 10−19 joule.
1 keV = 103 eV
1 M eV = 106 eV
1 GeV = 109 eV
1 T eV = 1012 eV
Physics 202 – p. 12/2
Relativity
Relativistic Energy: Examples
Examples:
1) The deuteron H 2 consists of a neutron and a
proton bound together. Its rest mass is
1875.58 M eV . The rest masses of the proton and
neutron are 938.26 M eV and 938.55 M eV
respectively, and whose sum is 1877.81 M eV >
Rest mass of the deuteron. Therefore the
deuteron cannot spontaneously decay into a
proton and a neutron. The difference between
the two: 1877.81 M eV − 1875.58 M eV = 2.23 M eV
is the binding energy of the deuteron. 2.23 M eV
must be added in order to break up the deuteron.
Physics 202 – p. 13/2
Relativity
Relativistic Energy: Examples
2) An electron and a proton are each accelerated
through a potential of 107 V . Find the momentum
and speed of each.
Physics 202 – p. 14/2
Relativity
a) For the electron:
Kinetic energy of both: K = 10 M eV
Physics 202 – p. 15/2
Relativity
a) For the electron:
Kinetic energy of both: K = 10 M eV
K
10
γ = 1 + mc
=
1
+
2
0.51 = 20.6
⇒ One cannot use the classical
non-relativistic approximation here.
Physics 202 – p. 15/2
Relativity
a) For the electron:
Kinetic energy of both: K = 10 M eV
K
10
γ = 1 + mc
=
1
+
2
0.51 = 20.6
⇒ One cannot use the classical
non-relativistic approximation here.
The rest mass of the electron is
0.51 M eV K. Therefore
p ≈ E/c = (mc2 + K)/c = 10.51 M eV /c.
Physics 202 – p. 15/2
Relativity
a) For the electron:
Kinetic energy of both: K = 10 M eV
K
10
γ = 1 + mc
=
1
+
2
0.51 = 20.6
⇒ One cannot use the classical
non-relativistic approximation here.
The rest mass of the electron is
0.51 M eV K. Therefore
p ≈ E/c = (mc2 + K)/c = 10.51 M eV /c.
p = γmv =
pc
v
c = γmc2 =
(γmc2 )v
⇒
c2
10.51 M eV
20.6 0.51 M eV
= 0.999
Physics 202 – p. 15/2
Relativity
a) For the proton:
K
10
γ = 1 + mc
=
1
+
2
938 ≈ 1 ⇒ classical,
non-relativistic approximation might be good.
Physics 202 – p. 16/2
Relativity
a) For the proton:
K
10
γ = 1 + mc
=
1
+
2
938 ≈ 1 ⇒ classical,
non-relativistic approximation might be good.
1
2
mv
2
= 10 M eV ⇒
v
c
≈ 0.146.
Physics 202 – p. 16/2
Relativity
General relativity
Applies to accelerated frame of references and
provides a theory of gravitation beyond that of
Newton.
Principle of equivalence: Experiments
conducted in a uniform gravitational field and
in an accelerated frame of reference give
identical results.
Some consequences: A gravitational field
bends light. The stronger the field is the more
bend one gets. Observations: Bending of
light near the sun in 1919 by Eddington;
Gravitational lensing, etc....
Physics 202 – p. 17/2
Relativity
General relativity
Physics 202 – p. 18/2
Relativity
General relativity
Physics 202 – p. 19/2
Relativity
General relativity
Physics 202 – p. 20/2
Relativity
General relativity
Physics 202 – p. 21/2
Relativity
General relativity
Physics 202 – p. 22/2
Relativity
General relativity
Black holes: We mentioned that last
semester.
Heuristic derivation of the Schwarschild
radius:
q
2GM
⇒
R
=
Escape veolcity: c = 2GM
S
R
c2 .
Schwarschild radius of a black hole beyond
which light cannot escape.
Physics 202 – p. 23/2
Relativity
General relativity
Example:
For a black hole with a mass comparable to that
of the Earth,
2(6.67×10−11 N.m2 /kg 2 )(5.98×1024 kg)
RS =
≈ 9 mm
(3×108 m/s)2
Physics 202 – p. 24/2
Relativity
General relativity
Physics 202 – p. 25/2
Relativity
General relativity
Physics 202 – p. 26/2
Relativity
General relativity
Physics 202 – p. 27/2
Relativity
General relativity
Physics 202 – p. 28/2
Relativity
General relativity
Physics 202 – p. 29/2