11

Relativity (Option Relativity (Option A)A)A.4 Relativistic momentum and A.4 Relativistic momentum and energyenergy

22

Two basic physical quantities need to be Two basic physical quantities need to be modifiedmodified Length (contraction)Length (contraction) Time (dilation)Time (dilation)

Other physical quantities will need to be Other physical quantities will need to be modified as wellmodified as well

33

Relativistic momentumRelativistic momentum

To properly describe the motion of particles To properly describe the motion of particles within special relativity, Newton’s laws of within special relativity, Newton’s laws of motion and the definitions of momentum and motion and the definitions of momentum and energy need to be generalizedenergy need to be generalized

These generalized definitions reduce to the These generalized definitions reduce to the classical ones when the speed is much less classical ones when the speed is much less than c i.e. v << cthan c i.e. v << c

44

Relativistic momentumRelativistic momentum

To account for conservation of momentum in all To account for conservation of momentum in all inertial frames, the definition must be modifiedinertial frames, the definition must be modified

vv is the speed of the particle, is the speed of the particle, mm is its mass as measured by is its mass as measured by an an observer at restobserver at rest with respect to the mass with respect to the mass

When v << c, the denominator approaches 1 and soWhen v << c, the denominator approaches 1 and so pp approaches approaches mvmv

vmcv

vmp o

o

221

55

Problem: particle decayProblem: particle decay

An unstable particle at rest breaks up into An unstable particle at rest breaks up into two fragments of two fragments of unequal mass. unequal mass. The The mass of the lighter fragment is 2.50 × 10–mass of the lighter fragment is 2.50 × 10–28 kg, and that of the heavier fragment is 28 kg, and that of the heavier fragment is 1.67 × 10–27 kg. If the lighter fragment 1.67 × 10–27 kg. If the lighter fragment has a speed of 0.893has a speed of 0.893c c after the breakup, after the breakup, what is the speed of the heavier what is the speed of the heavier fragment?fragment?

66

An unstable particle at rest breaks up into two fragments of An unstable particle at rest breaks up into two fragments of unequal mass. unequal mass. The The mass of the lighter fragment is 2.50 × 10mass of the lighter fragment is 2.50 × 10–28–28 kg, and that of the heavier fragment is kg, and that of the heavier fragment is 1.67 × 101.67 × 10–27–27 kg. If the lighter fragment has a speed of 0.893 kg. If the lighter fragment has a speed of 0.893c c after the breakup, after the breakup, what is the speed of the heavier fragment?what is the speed of the heavier fragment?

Given:Given:

vv11 = 0.8 c = 0.8 c

mm11=2.50×10=2.50×10–28–28 kg kg

mm22=1.67×10=1.67×10–27–27 kg kg

Find:Find: vv22 = ? = ?

Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or

28

282 1 1 1 1 2

2.50 10 kg 0.8934.96 10 kg

1 0.893

cp p mv c

For the heavier fragment,

27

28

2

1.67 10 kg4.96 10 kg

1

vc

v c

which reduces to 23.37 1v c v c

and yields

77

Relativistic energyRelativistic energy

The definition of The definition of kinetic energykinetic energy requires requires modification in relativistic mechanicsmodification in relativistic mechanics

EEkk = = KE = KE = mmoocc22 – m – moo c c22 = ( = (γγ-1) -1) mmoocc22

The term The term mmoocc22 is called the is called the rest energyrest energy of the object of the object and is independent of its speedand is independent of its speed

The term The term mmoocc22 is the is the total energytotal energy, E, of the object , E, of the object and depends on its speed and its rest energyand depends on its speed and its rest energy

88

A particle has energy by virtue of its mass A particle has energy by virtue of its mass alonealone A stationary particle with zero kinetic energy has an A stationary particle with zero kinetic energy has an

energy proportional to its inertial massenergy proportional to its inertial mass

E = mE = moocc22

The mass of a particle may be completely The mass of a particle may be completely convertible to energy and pure energy may be convertible to energy and pure energy may be converted to particlesconverted to particles

99

Energy and Relativistic Energy and Relativistic MomentumMomentum

It is useful to have an expression relating total It is useful to have an expression relating total energy, E, to the relativistic momentum, penergy, E, to the relativistic momentum, p

EE22 = p = p22cc22 + (mc+ (mc22))22

When the particle is at rest, When the particle is at rest, p = 0p = 0 and and E = mcE = mc22 Massless particles (Massless particles (m = 0m = 0) have ) have E = pcE = pc See derivation in classSee derivation in class

This is also used to express masses in energy unitsThis is also used to express masses in energy units mass of an electron = 9.110 x 10mass of an electron = 9.110 x 10-31-31 kg = 0.511 MeVc kg = 0.511 MeVc-2-2

Mass of a proton = 1.673 x 10Mass of a proton = 1.673 x 10-27-27 kg = 938 MeVc kg = 938 MeVc-2-2

Conversion: 1 u = 931.5 MeVcConversion: 1 u = 931.5 MeVc-2-2

Practice ProblemsHL Physics 2nd ed Hamper

Calculate the momentum of an electron accelerated to a total energy of 2.0 MeV [1.9 MeVc-1]

Calculate the speed of an electron accelerated theough a potential difference (voltage) of 1 .0 MV [0.94c]

Calculate the potential difference (i.e. V) needed to accelerate an electron to a velocity of 0.80c [0.34 MV]

Calculate the speed of an electron with momentum of 2.0 MeVc-1

[v= 0.97c] Find the momentum of a particle of rest mass 100 MeVc-2 travelling at

0.80c [133MeVc-1] A particle of rest mass 200 MeV is accelerated to a Ek of 1.0 GeV.

Calculate its momentum and velocity [1183MeVc-1; 0.99c] A proton has momentum 150MeVc-1. Calculate its total energy and Ek [950

MeV; 12MeV]

Practice Problems Tsokos 5th ed p. 674

Find the momentum of a pion (rest mass 1.35 MeVc-2 ) whose speed is 0.80c [180 MeVc-1]

Find the speed of a muon (rest mass 105 MeVc-2) whose momentum is 228 MeVc-1 [0.91c]

Find the kinetic energy of an electron whose momentum is 1.5 MeVc-1 [1.07 MeV]

A particle at rest breaks apart into two pieces of masses 250 MeVc-2 and 125 MeVc-2 . The lighter fragment moves away at 0.85c. Using conservation of momentum and total energy find a) the speed of the other fragment and b) the rest mass of the original particle. [0.628c; 558.5MeV]