Physics 202Professor P. Q. Hung
311B, Physics Building
Physics 202 – p. 1/29
Relativity
Momentum in Special Relativity
Classically, the momentum is defined as~p = m~v = m∆~r
∆t .We also learned that momentum isconserved.We also learned that ~F = ∆~p
∆t . Momentumconservation is the consequence of zeroexternal force.
Requirement: The laws of physics must bethe same in all inertial frames.For instance, the total momentum should beconserved in a collision.
Physics 202 – p. 2/29
Relativity
Momentum in Special Relativity
Classically, the momentum is defined as~p = m~v = m∆~r
∆t .We also learned that momentum isconserved.We also learned that ~F = ∆~p
∆t . Momentumconservation is the consequence of zeroexternal force.
Requirement: The laws of physics must bethe same in all inertial frames.For instance, the total momentum should beconserved in a collision. Physics 202 – p. 2/29
Relativity
Momentum in Special Relativity
A detailed analysis reveals that, if we were touse ~p = m~v, the momentum might beconserved in one inertial frame but not inanother inertial frame. Should one give upmomentum conservation? NO. Redefine themomentum.
Physics 202 – p. 3/29
Relativity
Momentum in Special Relativity
Instead of ∆t, one should use the proper time
∆t0 =√
1 − v2
c2 ∆t.
The proper form for the momentum is~p = m~v
√
1− v2
c2
For v � c, one recovers the usual classical~p = m~v.
Physics 202 – p. 4/29
Relativity
Momentum in Special Relativity
Physics 202 – p. 5/29
Relativity
Momentum in Special Relativity: ExampleAn electron, which has a mass of 9.11 × 10−31 kg,moves with a speed of 0.750 c. Find its relativisticmomentum and compare this value with themomentum calculated from the classicalexpression.
Physics 202 – p. 6/29
Relativity
Momentum in Special Relativity: Example
p = mv√
1− v2
c2
= (9.11×10−31 kg)(0.750×3×108m/s)√1−0.7502
=
3.1 × 10−22 kg.m/s.
The classical result isp = mv = (9.11 × 10−31 kg)(0.750 × 3 ×108m/s) = 2.05 × 10−22 kg.m/s. A 50% smallerthan the relativistic result.
Physics 202 – p. 7/29
Relativity
Momentum in Special Relativity: Example
p = mv√
1− v2
c2
= (9.11×10−31 kg)(0.750×3×108m/s)√1−0.7502
=
3.1 × 10−22 kg.m/s.
The classical result isp = mv = (9.11 × 10−31 kg)(0.750 × 3 ×108m/s) = 2.05 × 10−22 kg.m/s. A 50% smallerthan the relativistic result.
Physics 202 – p. 7/29
Relativity
Relativistic EnergyWhat does the folkloric E = mc2 mean?Start with motion in blueone dimension forsimplicity. And also start the motion from rest.
Work done = Change in kinetic energy.W =
∫
F dx =∫
dpdt dx.
After some calculations, one findsW = mc2
√
1− v2
c2
− mc2
⇒K = mc2
√
1− v2
c2
− mc2
Physics 202 – p. 8/29
Relativity
Relativistic EnergyWhat does the folkloric E = mc2 mean?Start with motion in blueone dimension forsimplicity. And also start the motion from rest.
Work done = Change in kinetic energy.W =
∫
F dx =∫
dpdt dx.
After some calculations, one findsW = mc2
√
1− v2
c2
− mc2
⇒K = mc2
√
1− v2
c2
− mc2
Physics 202 – p. 8/29
Relativity
Momentum in Special Relativity
Physics 202 – p. 9/29
Relativity
Relativistic Energy
γ = 1√
1− v2
c2
.
Notice: For v � c, one has1
√
1− v2
c2
≈ 1 + 12
v2
c2
⇒ K ≈ mc2(1 + 12
v2
c2 ) − mc2 = 12mv2. The
classical result!
Physics 202 – p. 10/29
Relativity
Relativistic Energy
γ = 1√
1− v2
c2
.
Notice: For v � c, one has1
√
1− v2
c2
≈ 1 + 12
v2
c2
⇒ K ≈ mc2(1 + 12
v2
c2 ) − mc2 = 12mv2. The
classical result!
Physics 202 – p. 10/29
Relativity
Relativistic Energy
There is one term which does not depend onthe 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 andsubtracting E2 − p2c2):E2 = p2c2 + (mc2)2
For p2c2 � (mc2)2, one has E ≈ pc.
Physics 202 – p. 11/29
Relativity
Relativistic Energy
There is one term which does not depend onthe 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 andsubtracting E2 − p2c2):E2 = p2c2 + (mc2)2
For p2c2 � (mc2)2, one has E ≈ pc.
Physics 202 – p. 11/29
Relativity
Relativistic Energy
From Eq. (4), one also finds:vc = pc
E
Some units:1 eV = 1.602 × 10−19 joule.1 keV = 103 eV1 MeV = 106 eV1 GeV = 109 eV1 TeV = 1012 eV
Physics 202 – p. 12/29
Relativity
Relativistic Energy
From Eq. (4), one also finds:vc = pc
E
Some units:1 eV = 1.602 × 10−19 joule.1 keV = 103 eV1 MeV = 106 eV1 GeV = 109 eV1 TeV = 1012 eV
Physics 202 – p. 12/29
Relativity
Relativistic Energy: ExamplesExamples:1) The deuteron H2 consists of a neutron and aproton bound together. Its rest mass is1875.58 MeV . The rest masses of the proton andneutron are 938.26 MeV and 938.55 MeVrespectively, and whose sum is 1877.81 MeV >Rest mass of the deuteron. Therefore thedeuteron cannot spontaneously decay into aproton and a neutron. The difference betweenthe two: 1877.81 MeV − 1875.58 MeV = 2.23 MeVis the binding energy of the deuteron. 2.23 MeVmust be added in order to break up the deuteron.
Physics 202 – p. 13/29
Relativity
Relativistic Energy: Examples2) An electron and a proton are each acceleratedthrough a potential of 107V . Find the momentumand speed of each.
Physics 202 – p. 14/29
Relativity
a) For the electron:
Kinetic energy of both: K = 10 MeV
γ = 1 + Kmc2 = 1 + 10
0.51 = 20.6⇒ One cannot use the classicalnon-relativistic approximation here.
The rest mass of the electron is0.51 MeV � K. Thereforep ≈ E/c = (mc2 + K)/c = 10.51 MeV/c.
p = γmv = (γmc2)vc2 ⇒
vc = pc
γmc2 = 10.51 MeV20.6 0.51 MeV = 0.999
Physics 202 – p. 15/29
Relativity
a) For the electron:
Kinetic energy of both: K = 10 MeV
γ = 1 + Kmc2 = 1 + 10
0.51 = 20.6⇒ One cannot use the classicalnon-relativistic approximation here.
The rest mass of the electron is0.51 MeV � K. Thereforep ≈ E/c = (mc2 + K)/c = 10.51 MeV/c.
p = γmv = (γmc2)vc2 ⇒
vc = pc
γmc2 = 10.51 MeV20.6 0.51 MeV = 0.999
Physics 202 – p. 15/29
Relativity
a) For the electron:
Kinetic energy of both: K = 10 MeV
γ = 1 + Kmc2 = 1 + 10
0.51 = 20.6⇒ One cannot use the classicalnon-relativistic approximation here.
The rest mass of the electron is0.51 MeV � K. Thereforep ≈ E/c = (mc2 + K)/c = 10.51 MeV/c.
p = γmv = (γmc2)vc2 ⇒
vc = pc
γmc2 = 10.51 MeV20.6 0.51 MeV = 0.999
Physics 202 – p. 15/29
Relativity
a) For the electron:
Kinetic energy of both: K = 10 MeV
γ = 1 + Kmc2 = 1 + 10
0.51 = 20.6⇒ One cannot use the classicalnon-relativistic approximation here.
The rest mass of the electron is0.51 MeV � K. Thereforep ≈ E/c = (mc2 + K)/c = 10.51 MeV/c.
p = γmv = (γmc2)vc2 ⇒
vc = pc
γmc2 = 10.51 MeV20.6 0.51 MeV = 0.999
Physics 202 – p. 15/29
Relativity
a) For the proton:
γ = 1 + Kmc2 = 1 + 10
938 ≈ 1 ⇒ classical,non-relativistic approximation might be good.
12mv2 = 10 MeV ⇒ v
c ≈ 0.146.
Physics 202 – p. 16/29
Relativity
a) For the proton:
γ = 1 + Kmc2 = 1 + 10
938 ≈ 1 ⇒ classical,non-relativistic approximation might be good.12mv2 = 10 MeV ⇒ v
c ≈ 0.146.
Physics 202 – p. 16/29
Relativity
General relativityApplies to accelerated frame of references andprovides a theory of gravitation beyond that ofNewton.
Principle of equivalence: Experimentsconducted in a uniform gravitational field andin an accelerated frame of reference giveidentical results.Some consequences: A gravitational fieldbends light. The stronger the field is the morebend one gets. Observations: Bending oflight near the sun in 1919 by Eddington;Gravitational lensing, etc.... Physics 202 – p. 17/29
Relativity
General relativity
Physics 202 – p. 18/29
Relativity
General relativity
Physics 202 – p. 19/29
Relativity
General relativity
Physics 202 – p. 20/29
Relativity
General relativity
Physics 202 – p. 21/29
Relativity
General relativity
Physics 202 – p. 22/29
Relativity
General relativity
Black holes: We mentioned that lastsemester.Heuristic derivation of the Schwarschildradius:Escape veolcity: c =
√
2GMR ⇒ RS = 2GM
c2 .
Schwarschild radius of a black hole beyondwhich light cannot escape.
Physics 202 – p. 23/29
Relativity
General relativityExample:For a black hole with a mass comparable to thatof the Earth,RS = 2(6.67×10−11N.m2/kg2)(5.98×1024kg)
(3×108m/s)2 ≈ 9 mm
Physics 202 – p. 24/29
Relativity
General relativity
Physics 202 – p. 25/29
Relativity
General relativity
Physics 202 – p. 26/29
Relativity
General relativity
Physics 202 – p. 27/29
Relativity
General relativity
Physics 202 – p. 28/29
Relativity
General relativity
Physics 202 – p. 29/29