Special Relativity - Relativistic Energy

A lump of clay whose rest mass is 4 kilograms is travelling at three-fifths the speed of light when it collides head-on with an identical lump going the opposite direction at the same speed. If the two lumps stick together and no energy is radiated away, what is the mass of the composite lump?

A. 4 kg
B. 6.4 kg
C. 8 kg
D. 10 kg
E. 13.3 kg

(GR9677 #36)

Solution:

No energy is radiated away = energy is conserved

E_rel(a) + E_rel(b)  = E_rest(a,b)

γ_am_ac² + γ_bm_bc²  = Mc²

Given:
m_a = m_b = m₀ = 4 kg
|v_a| = |v_b| = ³/₅ c

γ = 1/(1 − ^v2/_c²)^½ 
γ = 1/(1 − ⁹/₂₅)^½ = 1/(¹⁶/₂₅)^½ = 1/(⁴/5)^½ = ⁵/₄
γ_a = γ_b = γ = ⁵/₄  

γ_am_ac² + γ_bm_bc²  = Mc²
M = 2γm₀ = 2(⁵/₄ )(4) = 10 kg

Answer: D

Special Relativity - Relativistic Addition of Velocities

An atom moving at speed 0.3c emits an electron along the same direction with speed 0.6c in the internal rest frame of the atom. The speed of the electron in the lab frame equal to

A. 0.25c
B. 0.51c
C. 0.66c
D. 0.76c
E. 0.90c

(GR9677 #37)

Solution:

Relativistic Addition of Velocities:

v = speed of the moving observer
u' = speed of object in the moving observer
u = speed of object in the moving observer relative to the rest observer

Given:
v = speed of atom = 0.3c
u' = speed of electron in the rest frame of the atom = 0.6c

u = speed of electron in the lab frame
= (0.6c + 0.3c) / [1 + (0.6)(0.3)]
= 0.9c /1.18 
= 90/118 
= 0.762c

Special Relativity - Relativistic Energy

What is the speed of a particle having a momentum of 5 MeV/c and a total relativistic energy of 10 MeV?

A. c
B. 0.75 c
C. 1/√3 c
D. ½ c
E.  ¼ c

(GR9677 #38)

Solution:

p = γm₀v
E = γm₀c²

p/E = v/c²
v = pc²/E = (5 MeV/c) c²/10 MeV = ½ c

Answer: D

Special Relativity - Relativistic Speed

The half life of π⁺ meson at rest is 2.5 × 10⁻⁸ second. A beam of  π⁺ mesons is generated at a point 15 meters from a detector. Only ½ of the π⁺mesons live to reach the detector. The speed of the π⁺ mesons is

A.¹/₂ c
B. √²/₅ c
C. ²/_√5 c
D. c
E. 2c

(GR9677 #48)

Solution:

v = v₀/γ 

with γ = (1 − ^v2/_c²)^−1/2

Given:

t₀ =  2.5 × 10⁻⁸s
L₀ = 15 m (measured in lab)

v₀ = L₀/t₀ = 15/(2.5 × 10⁻⁸) = 3/5 × 10⁹ =  (10/5)(3 × 10⁸) = 2c 

v₀/c = 2

v = v₀(1 − ^v2/_c²)^1/2
v² = v₀²(1 − ^v2/_c²) = v₀² − v²v₀²/c² = v₀² − 4v²
v² +  4v² = v₀²
5v² = 4c²
v = ²/_√5 c

Answer: C

Special Relativity - Electric Field

The infinity xy-plane is a nonconducting surface, with surface charge density σ, as measured by an observer at rest on the surface. A second observer moves with velocity v relative to the surface, at height h above it. Which of the following expressions gives the electric measured by this second observer?

A.

B.

C.

D.

E. 

(GR9677 #49)

Solution:

E_z = γE_0z 
with γ = 1/(1 − ^v2/_c²)^½ 

E_z = E₀/(1 − ^v2/_c²)^½

For infinity plane: E = σ/2ɛ₀ (show)

For infinity xy-plane: E₀  = σ / 2ɛ₀

E_z =  σ / 2ɛ₀(1 − ^v2/_c²)^½

Answer: C

Special Relativity - Space-Time Interval

In inertial frame S, two events occur at the same instant in time and 3c·minutes apart in space. In inertial frame S’, the same events occur at 5c·minutes apart. What is the time interval between the events in S’?

A. 0 min
B. 2 min
C. 4 min
D. 8 min
E. 16 min

(GR9677 #50)

Solution:

dS²  =  (dx)² − (cdt)²

dS²  =  dS'²

(3c)²  =  (5c)² − (ct)²

9c²  =  25c² − (ct)² 

t²  =  25 − 9 = 16 

t = 4

Answer: C

Special Relativity - Doppler Effect

The Lyman alpha spectral line of Hydrogen (λ = 122 nanometers) differs by 1.8 × 10⁻¹² meter in spectra taken at opposite ends of the Sun’s equator. What is the speed of a particle on the equator due to the Sun’s rotation, in kilometers per second?

A. 0.22
B. 2.2
C. 22
D. 220
E. 2200

(GR9677 #60)

Solution:

The problem deals with v ≪ c since the  answer is in km/s.

For v ≪ c, Redshift parameter, z = Δλ/λ ≈ v/c

v =  cΔλ/λ

Given:
Δλ = 1.8 × 10⁻¹² m 
λ = 122 nm = 1.22 × 10⁻⁷ m
c = 3 × 10⁸ m/s

v = (3 × 10⁸)(1.8 × 10⁻¹²)/(1.22 × 10⁻⁷) 

= (5.4/1.22) × 10³ m/s  
= (5.4/1.22) km/s ≈ 2.2 km/s

Answer: B

General Relativity - Black Hole

A black hole is an object whose gravitational field is so strong that even light cannot escape. To what approximate radius would Earth (mass = 5.98 × 10²⁴ kilograms) have to be compressed in order to become a black hole?

A. 1 nm
B. 1 µm
C. 1 cm
D. 100 m
E. 10 km

(GR9677 #67)

Solution:

Schwarzschild radius: GMm/R = ½ mc²
→ R = 2GM/c² 

G = 6.67 × 10⁻¹¹ m³/(kg·s²)
M = 5.98 × 10²⁴ kg
c = 3 × 10⁸ m/s

R =  2 (6.67 × 10⁻¹¹)(5.98 × 10²⁴) / (3 × 10⁸)²
= (2·6.67·5.98 / 9) × 10⁻¹¹× 10²⁴ × 10⁻¹⁶
≈ (2·7·6 / 9) × 10⁻³ ≈ 1 cm

Answer: C

Special Relativity - Speed of Photon

A π⁰ meson (rest-mass energy 135 MeV) is moving with velocity 0.8c  in the laboratory rest frame when it decays into two photons γ₁ and γ₂. In the π⁰ rest frame, γ₁ is emitted forward and γ₂ is emitted backward relative to the direction of flight. The velocity of γ₂ in the laboratory rest frame is

A. −1.0c 
B. −0.2c 
C. +0.8c 
D. +1.0c 
E. +1.8c 

(GR9277 #37)

Solution:

(B), (C), (E) are FALSE
Photon moves with the speed of light (1.0c)

Since γ₂ is emitted backward, the velocity of γ₂ is −1.0c

Answer: A

Special Relativity - Time Dilation

Tau leptons are observed to have an average half-life of Δt₁ in the frame S₁ in which the leptons are at rest. In an inertial frame S₂, which is moving at a speed v₁₂ relative to S₁, the leptons are observed to have an average half-life of Δt₂. In another inertial reference frame S₃, which is moving at a speed v₁₃ relative to S₁ and v₂₃ relative to S₂, the leptons have an observed half-life of Δt₃. Which of the following is a correct relationship among two of the half lives, Δt₁, Δt₂, and Δt₃.

A.

B.

C.

D.

E.

(GR9277 #38)

Solution:

Time Dilation:

Δt_j =  γ_ijΔt_i



Δt_i = time in a rest frame
Δt_j = time in a moving frame

In this problem, Δt₁ = time in a rest frame
Time dilation only relates to time in a rest frame Δt₁ with time in a moving frame, Δt₂ or Δt₃

The only right answers:
Δt₂ =  γ₁₂Δt₁
Δt₃ =  γ₁₃Δt₁

(C) and (D) are FALSE (no Δt₁ in the equation)
(E) is FALSE (relates v₂₃ with Δt₁ and Δt₂)

(A) FALSE
Δt₂ = Δt₁ [1 − (v₁₂)²/c²]^½ = Δt₁ (1/γ₁₂) → Δt₁ = γ₁₂Δt₂ 

(B) TRUE
Δt₁ = Δt₃ [1 − (v₁₃)²/c²]^½ = Δt₃ (1/γ₁₃) → Δt₃ = γ₁₃Δt₁ 

Answer: B

Special Relativity - Momentum

A monoenergetic beam consists of unstable particles with total energies 100 times their rest energy. If the particles have rest mass m, their momentum is most nearly 

A. mc
B. 10 mc
C. 70 mc
D. 100 mc
E. 10⁴ mc

(GR9277 #70)

Solution:

E = γE₀ = 100E₀ → γ = 100
p = γm₀c = 100 mc

Answer: D

Special Relativity - Relativistic Energy

A free electron (rest mass m_e = 0.5 MeV/c²) has a total energy of 1.5 MeV. Its momentum p in units of MeV/c is about

A. 0.86
B. 1.0
C. 1.4
D. 1.5
E. 2.0

(GR9277 #85)

Solution:

E² = p²c² + m₀²c⁴

1.5² = p²c² + (0.5/c²)²c⁴

 p²c² = 1.5² − 0.5² = 2

p = ^√2/_c ≈ ^1.4/_c

Answer: C

Special Relativity - Velocity of Light

For blue light, a transparent material has a relative permittivity (dielectric constant) of 2.1 and a relative permeability of 1.0. If the speed of light in vacuum is c, the phase velocity of blue light in an unbounded medium of this material is

A. √3.1 c
B. √2.1 c
C. c/√1.1
D. c/√2.1
E. c/√3.1

(GR8677 #03)

Solution:

Permittivity, ϵ = 2.1 ϵ₀
Permeability, μ = μ₀
Speed of light in vacuum:
The velocity of light in medium:  

Answer: D

Special Relativity - Speed of Kaon

A positive kaon (K⁺) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly

A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

(GR8677 #20)

Solution:





Answer: E

Special Relativity - Space-Time Interval

Two observers O and O’ observe two events A and B. The observers have a constant relative speed of 0.8c. In units such that the speed of light is 1, observer O obtained the following coordinates:

Event A: x = 3, y = 3, z = 3, t = 3 

Event B: x = 5, y = 3, z = 1, t = 5

What is the length of the space-time interval between these two events, as measured by O’.

A. 1
B. √2
C. 2
D. 3
E. 2√3

(GR8677 #21)

Solution:

dS²  = dr² + (cdt)²  =  dx² + dy² + dz² − (cdt)² 
dS² = dS'² 

dx² = (5 − 3)² = 4
dy² = (3 − 3)² = 0
dz² = (1 − 3)² = 4
dt²  = (5 − 3)² = 4
c = 1

dS'² = 4 + 0 + 4 − 4 = 4
dS' = 2

Answer: C

Special Relativity - Electromagnetic Field

Which of the following statements most accurately describes how an electromagnetic field behaves under a Lorentz transformation?

1. The electric field transforms completely into a magnetic field.
2. If initially there is only an electric field, after the transformation there maybe both an electric and magnetic field.
3. The electric field is unaltered.
4. The magnetic field is unaltered.
5. It cannot be determined unless a gauge transformation is also specified.

(GR8677 #22)

Solution:

see analysis: Electromagnetism and Relativity

Answer: A

Special Relativity - Relativistic Properties

If a newly discovered particle X moves with a speed equal to the speed of light in vacuum, then which of the following must be true?

A. The rest mass of X is zero
B. The spin of X equals the spin of a photon
C. The charge of X is carried on its surface
D. X does not spin
E. X cannot be detected

(GR8677 #68)

Solution:

Relativistic Energy: E² = p²c² + m₀²c⁴

If the particle moves with v = c, de Broglie relation: E = hf = pc

E² = E² + m₀²c⁴ 
m₀²c⁴ = E² − E² = 0
m₀ = 0

Answer: A

Special Relativity - Length Contraction

Questions 69-71

A car of rest length 5 meters passes through a garage rest length 4 meters. Due to the relativistic Lorentz contraction, the car is only 3 meters long in the garage’s rest frame. There are doors on both ends of the garage, which open automatically when the front of the car reaches them and close automatically when the rear passes them. The opening or closing of each door requires a negligible amount of time.

The velocity of the car in the garage’s rest frame is

A. 0.4c
B. 0.6c
C. 0.8c
D. Greater than c
E. Not determinable from the data given

(GR8677 #69)

Solution:

Length contraction: L = L₀/γ
with γ = 1/(1 − ^v2/_c²)^½ 

Given for the car, L₀ = 5 m and L = 3 m

γ = L₀/L = 5/3 = 1/(1 − ^v2/_c²)^½ 
γ² = 25/9 = 1/(1 − ^v2/_c²)
1 − v²/c² = 9/25
v²/c² = 1 − 9/25 = 16/25
v = ⁴/₅ c = 0.8c

Answer: C

Special Relativity - Length Contraction

Questions 69-71

A car of rest length 5 meters passes through a garage rest length 4 meters. Due to the relativistic Lorentz contraction, the car is only 3 meters long in the garage’s rest frame. There are doors on both ends of the garage, which open automatically when the front of the car reaches them and close automatically when the rear passes them. The opening or closing of each door requires a negligible amount of time.

The length of the garage in the car’s rest frame is

A. 2.4 m
B. 4.0 m
C. 5.0 m
D. 8.3 m
E. Not determinable from the data given

(GR8677 #70)

Solution:

Length contraction: L = L₀/γ

Given for the garage, L₀ = 4 m
From problem GR8677 #69 → γ = 5/3

L = 4/(5/3) = 12/5 = 2.4 m

Answer: A

Special Relativity - Length Contraction

Questions 69-71

A car of rest length 5 meters passes through a garage rest length 4 meters. Due to the relativistic Lorentz contraction, the car is only 3 meters long in the garage’s rest frame. There are doors on both ends of the garage, which open automatically when the front of the car reaches them and close automatically when the rear passes them. The opening or closing of each door requires a negligible amount of time.

Which of the following statements is the best response to the question: “Was the car ever inside a closed garage?”

1. No, because the car is longer than the garage in all reference frames.
2. No, because the Lorentz contraction is not a “Real” effect.
3. Yes, because the car is shorter than the garage in all reference frames.
4. Yes, because the answer to the question in the garage’s rest frame must apply in all reference frames.
5. There is no unique answer to the question, as the order of door openings and closings depend on the reference frame.

(GR8677 #71)

Solution:

In the garage’s frame:
The length of the garage = 4 m
The length of the car = 3 m
L_car  L_garage → A. FALSE

In the car’s frame:
The length of the garage = 2.4 m (from problem GR8677 #70)
The length of the car = 5 m
L_car L_garage → C. FALSE

→ No unique answer, it depends on the reference frame

Answer: E