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