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Solved Examples on Relativity Question 1:- Your starship passes Earth with a relative speed of 0.9990c. After travelling 10.0 y (your time), you stop at lookout post LP13, turn, and then travel back to Earth with the same relative speed. The trip back takes another 10.0 y (your time). How long does the round trip take according to measurements made on Earth? (Neglect any effects due to the accelerations involved with stopping and turning.) Solution:- On the journey out, the start and end of the journey occur at the same location in your reference frame, namely on your ship. Hence, you measure proper time Δt0 for the trip, which is the given 10.0 y. Equation Δt = Δt_{0}/√[1 – (v/c)^{2}] gives us the corresponding time Δt as measured in the Earth reference frame: Δt = Δt_{0}/√[1 – (v/c)^{2}] = 10.0 y/√[1 – (0.9990c/c)^{2}] = (22.37) (10.0 y) = 224 y On the journey back, we have the same situation and the same data. Thus the round trip requires 20 y of your time but Δt_{total} = (2) (224 y) = 448 y of Earth time. In other words, you have aged 20 y while the Earth has aged 448 y. Although you cannot travel into the past (as far as we know), you can travel into the future of, say, Earth, by using high-speed relative motion to adjust the rate at which time passes. ___________________________________________________________________________________________________ Question 2 (JEE Advanced):- Caught by surprise near a supernova, you race away from the explosion in your spaceship, hoping to outrun the high-speed material ejected toward you. Your Lorentz factor relative to the inertial reference frame of the local stars is 22.4. (a) To reach a safe distance, you figure you need to cover 9.00×10^{16} m as measured in the reference frame of the local stars is 22.4. (b) How long does that trip take according to you (in your reference frame)? Solution:- (a) The length L_{0} = 9.00×10^{16} m is a proper length in the reference frame of the local stars, because its two ends are at rest in that frame. For large Lorentz factor, your speed relative to the local stars is v ≈ c. So, with that approximation, to move through length L_{0} requires the time Δt = L_{0}/v _{ = }L_{0}/c = [9.00×10^{16} m] / [3.00×10^{8} m/s] = 3.00×10^{8} s = 9.49 y (b) From your reference frame, the distance you cover is a contracted length L that races past you at relative speed v ≈ c. Equation L = L_{0} √[1 – (β)^{2}] = L_{0}/γ tells us that L = L_{0}/γ. So, the time you measure for the passage of that contracted length is Δt = L/v = (L_{0}/γ) / v = L_{0}/cγ = [9×10^{16} m] / [3.00×10^{8} m/s] [22.4] = 1.339×10^{8} s = 0.424 y This is a proper time, because the start and end of the passage occur at the same point in your reference frame (at your ship). ____________________________________________________________________________________________________ Question 3:- (a) What is the total energy E of a 2.53 MeV electron? (b) What is its momentum p? (c) What is the Lorentz factor γ for the electron? Solution:- (a) From equation E = mc^{2} + K The value of mc^{2} for an electron is 0.511 MeV; so E = 0.511 MeV + 2.53 MeV = 3.04 MeV Therefore, the total energy E of a 2.53 MeV electron would be 3.04 MeV. (b) From equation E^{2} = (pc)^{2} + (mc^{2})^{2}, we can write (3.04 MeV)^{2} = (pc)^{2} + (0.511 MeV)^{2} Then pc = √ (3.04 MeV)^{2} – (0.511 MeV)^{2} = 3.00 MeV and, giving the momentum in units of energy divided by c, we have p = 3.00 MeV/c Thus the momentum will be 3.00 MeV/c. (c) From equation E = mc^{2} + K, we have E = γmc^{2} With E = 3.04 MeV and m = 9.11×10^{-31} kg, we then have γ = E/mc^{2} = [(3.04×10^{6} eV) (1.6×10^{-19} J/eV)] / [(9.11×10^{-31} kg) (3.00×10^{8} m/s)^{2}] = 5.93 From the above observation we conclude that, the Lorentz factor γ for the electron would be 5.93.
Your starship passes Earth with a relative speed of 0.9990c. After travelling 10.0 y (your time), you stop at lookout post LP13, turn, and then travel back to Earth with the same relative speed. The trip back takes another 10.0 y (your time). How long does the round trip take according to measurements made on Earth? (Neglect any effects due to the accelerations involved with stopping and turning.)
On the journey out, the start and end of the journey occur at the same location in your reference frame, namely on your ship. Hence, you measure proper time Δt0 for the trip, which is the given 10.0 y. Equation
Δt = Δt_{0}/√[1 – (v/c)^{2}] gives us the corresponding time Δt as measured in the Earth reference frame:
Δt = Δt_{0}/√[1 – (v/c)^{2}]
= 10.0 y/√[1 – (0.9990c/c)^{2}]
= (22.37) (10.0 y)
= 224 y
On the journey back, we have the same situation and the same data. Thus the round trip requires 20 y of your time but
Δt_{total} = (2) (224 y) = 448 y
of Earth time. In other words, you have aged 20 y while the Earth has aged 448 y. Although you cannot travel into the past (as far as we know), you can travel into the future of, say, Earth, by using high-speed relative motion to adjust the rate at which time passes.
___________________________________________________________________________________________________
Caught by surprise near a supernova, you race away from the explosion in your spaceship, hoping to outrun the high-speed material ejected toward you. Your Lorentz factor relative to the inertial reference frame of the local stars is 22.4.
(a) To reach a safe distance, you figure you need to cover 9.00×10^{16} m as measured in the reference frame of the local stars is 22.4.
(b) How long does that trip take according to you (in your reference frame)?
(a) The length L_{0} = 9.00×10^{16} m is a proper length in the reference frame of the local stars, because its two ends are at rest in that frame. For large Lorentz factor, your speed relative to the local stars is v ≈ c. So, with that approximation, to move through length
L_{0} requires the time
Δt = L_{0}/v
_{ = }L_{0}/c
= [9.00×10^{16} m] / [3.00×10^{8} m/s]
= 3.00×10^{8} s
= 9.49 y
(b) From your reference frame, the distance you cover is a contracted length L that races past you at relative speed v ≈ c. Equation L = L_{0} √[1 – (β)^{2}] = L_{0}/γ tells us that L = L_{0}/γ. So, the time you measure for the passage of that contracted length is
Δt = L/v
= (L_{0}/γ) / v
= L_{0}/cγ
= [9×10^{16} m] / [3.00×10^{8} m/s] [22.4]
= 1.339×10^{8} s
= 0.424 y
This is a proper time, because the start and end of the passage occur at the same point in your reference frame (at your ship).
____________________________________________________________________________________________________
(a) What is the total energy E of a 2.53 MeV electron?
(b) What is its momentum p?
(c) What is the Lorentz factor γ for the electron?
(a) From equation E = mc^{2} + K
The value of mc^{2} for an electron is 0.511 MeV; so
E = 0.511 MeV + 2.53 MeV = 3.04 MeV
Therefore, the total energy E of a 2.53 MeV electron would be 3.04 MeV.
(b) From equation E^{2} = (pc)^{2} + (mc^{2})^{2},
we can write
(3.04 MeV)^{2} = (pc)^{2} + (0.511 MeV)^{2}
Then
pc = √ (3.04 MeV)^{2} – (0.511 MeV)^{2}
= 3.00 MeV
and, giving the momentum in units of energy divided by c, we have
p = 3.00 MeV/c
Thus the momentum will be 3.00 MeV/c.
(c) From equation E = mc^{2} + K, we have
E = γmc^{2}
With E = 3.04 MeV and m = 9.11×10^{-31} kg, we then have
γ = E/mc^{2}
= [(3.04×10^{6} eV) (1.6×10^{-19} J/eV)] / [(9.11×10^{-31} kg) (3.00×10^{8} m/s)^{2}]
= 5.93
From the above observation we conclude that, the Lorentz factor γ for the electron would be 5.93.
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