Hints for solving Relativity problems1. Draw a picture and define the reference frames, showing their motion

relative to each other.

2. Identify objects and their motion in the reference frames and events thathappen at a specific place and time.

3. Identify any proper time intervals and proper lengths. These are measuredin an object’s rest frame.

4. Identify all other given information about coordinates/velocities withrespect to moving frames.

5. Apply the principle of relativity and the Lorentz transformations

6. Check if the results are consistent with Galilean relativity when velocity issmall.

7. Resolving paradoxes: Remember the laws of physics must be invariant ineach observer’s frame.

Relativistic Momentum

Classical definition of momentum:

Conservation of momentum: : When two or more isolated particlescollide (no net forces acting on them), the total momentum is conserved.

According to the principle of relativity, momentum must be conserved inall reference frames.

Using the Lorentz transformations, does the classical definition ofmomentum obey conservation of momentum?

p = m

dx

dt= mu

Relativistic MomentumConsider an inelastic collision of two particles of equal mass approaching eachother with speed v in the lab frame S.

In frame S, the classical definition of momentum is conserved:

Total momentum before collision:

Total momentum after collision:

In frame S’ moving at velocity v relative to frame S, using the Lorentztransformations,

Before collision After collision

v v1 2 1 2V=0

mu

1+ mu

2= mv ! mv = 0

V = 0

!u1=

u1" v

1" u1v c

2= 0, !u

2=

u2" v

1" u2v c

2=

"2v

1+ v2

c2

, !V =V " v

1"Vv c2= "v

Relativistic MomentumConsider an inelastic collision of two particles of equal mass approaching eachother with speed v in the lab frame S.

In frame S’, the classical definition of momentum is not conserved:

Total momentum before collision:

Total momentum after collision:

Before collision After collision

v v1 2 1 2V=0

m !u1+ m !u

2=

"2mv

1+ v2

c2

2m !V = "2mv

Relativistic MomentumWe must modify our definition of momentum so that conservation of momentumholds in all inertial frames. Try using

Using this definition one can verify that conservation of momentum holds.

In the low velocity limit the relativistic momentum, the denominatorapproaches 1, so that we recover the classical momentum as expected:

p =mu

1! u2

c2

Relativistic form of Newton’s 2nd LawThe modified definition of momentum results in a relativistic form of Newton’s 2ndlaw:

Consider a constant force along x. Then

F =dp

dt=

d

dt

mu

1! u2

c2

"

#$$

%

&''

Fx=

dpx

dt=

d

dt

mux

1! ux

2 c2

"

#$$

%

&''= m 1!

ux

2

c2

"

#$

%

&'

!3/ 2

dux

dt= m 1!

ux

2

c2

"

#$

%

&'

!3/ 2

ax

ax=

Fx

m1!

ux

2

c2

"

#$

%

&'

3/ 2

Relativistic form of Newton’s 2nd Law

ax=

Fx

m1!

ux

2

c2

"

#$

%

&'

3/ 2

• As ux approaches c, the acceleration ax approaches 0.• When ux=c, the acceleration is 0, so the speed cannot increase any more.• Thus a particle cannot be accelerated to a speed beyond c.

Momentum of subatomic particlesIn a particle accelerator, when electrons accelerated to 0.999c collide with atarget, the collision produces a muon which moves in the direction of theelectron with a speed of 0.95c. What is the muon’s momentum in the lab frameand in the frame of the electron beam?

Lab frame: SElectron’s frame: S’Relative velocity of the two frames: v = velocity of electron = 0.999c

In frame S:

Velocity of muon: 0.95cMomentum of muon:

p =mu

1! u2 c2

=1.9 "10

!28 kg( ) 0.95c( )

1! 0.952

= 1.73"10!19 kgm / s

Momentum of subatomic particlesIn a particle accelerator, when electrons accelerated to 0.999c collide with atarget, the collision produces a muon which moves in the direction of theelectron with a speed of 0.95c. What is the muon’s momentum in the lab frameand in the frame of the electron beam?

Lab frame: SElectron’s frame: S’Relative velocity of the two frames: v = velocity of electron = 0.999c

In frame S’:Velocity of muon:

Momentum of muon:

!p =m !u

1" !u 2 c2

=1.9 #10

"28 kg( ) "0.962c( )

1" 0.9622

= "2.01#10"19 kgm / s

!u =u " v

1" uv c2=

0.95c " 0.999c

1" (0.95)(0.999)= "0.962c

Relativistic EnergyRecall the work done by a force F to move a particle from x1 to x2 is

W = Fdxx

1

x2

! =dp

dtdx

x1

x2

!

p =mu

1! u2

c2

dp

dt=

d

dt

mu

1! ux

2 c2

"

#$$

%

&''= m 1!

u2

c2

"

#$%

&'

!3/ 2

du

dt

Given the relativistic momentum

W = Fdxx

1

x2

! =dp

dtdx

x1

x2

! = m 1"u2

c2

#

$%&

'(

"3/ 2

du

dtdx

x1

x2

! = m 1"u2

c2

#

$%&

'(

"3/ 2

du

dtudt

0

uf

!

W =mc

2

1! u2

c2

! mc2

Relativistic Energy

The work-energy theorem: Work done by all forces on a particle =change in kinetic energy of the particle.If the initial kinetic energy is 0, then work = final kinetic energy:

W =mc

2

1! u2

c2

! mc2

K =W =mc

2

1! u2

c2

! mc2

Define

!u=

1

1" u2

c2

K = !

umc

2" mc

2

Relativistic Energy

In the low velocity limit we recover the classical expression for K: K = !

umc

2" mc

2

!u=

1

1" u2

c2

# 1+1

2

u2

c2+ ...

K = !umc

2" mc

2# mc

2+

1

2

u2

c2

mc2" mc

2=

1

2mu

2

Mass-Energy Equivalence

Rearranging the kinetic energy equation, K = !

umc

2" mc

2

!

umc

2= K + mc

2

The total energy of a particle is (mass-energy equivalence)

It is equal to the sum of the kinetic energy and the rest mass mc2.

E = !

umc

2