PS10_relativity_solutions

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Physics 341: Problem Set #10 due December 6 by 2:00 PM in PDF format on Canvas [40 pts] You are encouraged to work in groups on these problems, but each student must write up the solutions individually. You must also list your collaborators on your solutions, and cite any external sources you used (other than the course notes or textbook). Using online solutions to these or similar problems is not allowed. Show your work and explain your reasoning. The number of points for each problem is listed; partial credit will be given. Be careful with units. Questions asking about concepts require at most a few sentences to answer. Please make your answers clear and concise . 1. Each part of this question covers a key concept. Each requires at most a few sentences to answer; some are much shorter. Please be concise . [10 pts] (a) Suppose you see a friend moving past you at a constant speed. Explain why your friend can equally well say that she is stationary and you are moving past her at a constant speed. Reference frames moving at constant speed are inertial and cannot be distin- guished from stationary by any experiment. Hence you and your friend can both feel stationary and argue that the other is moving at a constant relative speed. (b) If your friend is moving past you at a high constant speed, you will notice that her time appears to run slowly and her length is contracted in the direction of motion. How will she perceive her own time and length? How will she perceive your time and length? She will perceive her own time and length as normal, because the laws of physics work the same in all inertial reference frames. She will perceive your time to be dilated and your length to be contracted due to the relative speed. (c) Decide whether the following statements make sense or not. Explain your reason- ing. i. You and a friend are standing at opposite sides of a room, and you each eat a peanut at the same instant. According to the theory of relativity, it is possible for a person moving past you at a constant speed to observe that you ate your peanut before your friend ate his. This makes sense. Simultaneous events in one reference frame are not gener- ally simultaneous in another frame with a non-zero relative velocity. 1

ii. An object moving by you at very high speed will appear to have a higher density than it has at rest. This makes sense; the transverse dimensions are unaffected, but length ap- pears contracted in the direction of motion, making the volume smaller. This will make the density appear higher. iii. If you could travel away from the Earth at a speed close to the speed of light, you would find yourself feeling uncomfortably heavy because of your increased mass. This does not make sense. In your own reference frame, your mass will not change; your reference frame is still inertial unless acceleration occurs. If you do accelerate, you will feel weight in the direction of acceleration due to your inertia, but your mass will not change. iv. If you had a sufficiently fast spaceship, you could leave today, make a round trip to a star 500 light-years away, and return home to Earth in the year 2050. This does not make sense. Nothing can travel faster than the speed of light, so you cannot make this round trip in less than 1000 years. However, if you accelerated sufficiently throughout your trip, the “gravitational” time dilation associated with that acceleration would allow you to age far less than 1000 years during the trip, so you might only appear ∼ 38 years older upon your return in Earth year 3023, in which case your own clock would read “2061”. 2. SN-1987A was a supernova which occurred in the Large Magellanic Cloud, and was detected on Earth in 1987. When a supernova occurs, 100 times as much energy is emitted in neutrinos as in photons. SN-1987A was the first (and so far only) supernova to have occurred nearby to Earth after we built neutrino detectors capable of seeing the supernova neutrino burst. For SN-1987A, 24 neutrino events were detected by a number of detectors all over the Earth. [10 pts] (a) The neutrinos from SN-1987A had an energy of approximately 20 MeV. The rest mass of a neutrino is as yet unknown, but must be less than 1 eV /c 2 . Let us assume that the mass of a neutrino is 1 eV /c 2 . How much slower than light (in m/s) were the neutrinos from SN-1987A moving? If the neutrinos were lighter, how would this answer change? Hint: for very large γ ≫ 1 , you may use the Taylor series β = 1 − 1 2 γ 2 A particle with rest mass m has energy given by E = γmc 2 . 2

Therefore, for our neutrinos with E = 20 MeV = 2 × 10 7 eV, the Lorentz factor γ is γ = 2 × 10 7 eV (1 eV /c 2 ) c 2 = 2 × 10 7 . This corresponds to a speed β = v/c of γ = 1 p 1 − β 2 β = r 1 − 1 γ 2 β = r 1 − 1 4 × 10 14 ≈ 1 − 1 2 × 4 × 10 14 = 1 − 1 . 25 × 10 − 15 v = c − 1 . 25 × 10 − 15 c = c − 3 . 75 × 10 − 7 m/s . Therefore, the photons are moving 3 . 75 × 10 − 7 m/s slower than light (or 0.375 micrometers per second slower). If the mass of the neutrino was smaller, than it would have a larger gamma factor, and so it would be moving faster. (b) SN-1987A occurred at a distance of 51.474 kpc from Earth. Assuming a mass of 1 eV /c 2 , what was the elapsed time between the emission of a neutrino from SN-1987A and our detection of it here on Earth, according to the neutrino? The neutrinos experience their emission from the supernova and their arrival on Earth as occurring at the same location in their coordinate system, separated by some time. Therefore, the elapsed time they experience is the proper time ∆ τ . The time that we see the neutrinos taking to arrive is: ∆ t = ∆ x v , where ∆ x is the distance to the supernova (51.474 kpc or 1 . 6788 × 10 5 ly) and v is the speed of the neutrinos. As we calculated in (a), this speed is very close to c , so ∆ t = 1 . 6788 × 10 5 ly c = 1 . 6788 × 10 5 years = 5 . 2981 × 10 12 s . Our measured elapsed time is related to the proper time by ∆ t = γ ∆ τ ∆ t = ∆ t γ = 5 . 2981 × 10 12 s 2 × 10 7 = 2 . 6491 × 10 5 s This is about 3 days of elapsed time. 3

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(c) Again assuming a mass of 1 eV /c 2 , what was the distance between the Earth and SN-1987A according to the neutrino? We see a distance to the supernova of 51.474 kpc or 1 . 6788 × 10 5 ly, so using the length contraction formula, the distance seen by the neutrino is L = ℓ γ = 1 . 6788 × 10 5 ly 2 × 10 7 = 8 . 394 × 10 − 3 ly = 7 . 94 × 10 13 m . Note that this is the distance that an object moving very close to the speed of light would cover in about 3 days (our answer for part (b)). 3. Imagine you have built a spaceship capable of traveling at relativistic speeds. You set off in this ship traveling at v = 0 . 999 c relative to Earth. [10 pts] (a) What is your Lorentz gamma factor? The normalized speed β is β = v/c = 0 . 999 . The Lorentz gamma factor is γ = 1 p 1 − β 2 = 1 √ 1 − 0 . 999 2 = 22 . 37 . (b) For each gram of mass on your ship, how much energy (in joules) must have been used accelerating the mass to v = 0 . 999 c ? What is this energy in units of kilotons of TNT (1 kton = 4 . 184 × 10 9 J)? The total energy of a particle of mass m with Lorentz factor γ is given by E = γmc 2 . However, we are interested in the kinetic energy added, so we must subtract off the rest mass: K = ( γ − 1) mc 2 . For m = 1 g and γ = 22 . 37, the kinetic energy in joules is K = (22 . 37 − 1)(10 − 3 kg)(3 × 10 8 m/s) 2 = 1 . 92 × 10 15 J. 4

or, in ktons of TNT, this is K = 1 . 92 × 10 15 J 4 . 184 × 10 9 J/kton = 4 . 60 × 10 5 kton , That is, each gram of ship-mass requires 460 megatons of TNT-energy equivalent to be accelerated to these speeds. The largest nuclear weapon ever detonated, the Soviet Tsar Bomba, was 50 megatons of TNT-equivalent. (c) As you travel, you will see starlight (from stars that are at rest relative to the Earth’s frame of reference) hitting your ship from all directions. Consider photons emitted which are emitted by their stars at 600 nm. If this photon reaches the ship from the direction of travel, what wavelength will you perceive it to be? If this photon reaches the ship from the direction exactly opposite the direction of travel, what wavelength will you see? Are stars in front of you blueshifted or redshifted? What about stars behind you? The relativistic Doppler shift is λ obs λ emit = s 1 + β 1 − β . Here β is positive for objects moving away from me, and negative for objects moving towards me. For stars in front of the ship, I see them approaching me, β = − 0 . 999. For stars behind the ship, I am seeing them recede at β = +0 . 999. Therefore, the 600 nm light from stars in front of the ship appears as: λ obs , front = r 1 − 0 . 999 1 + 0 . 999 (600 nm) = (0 . 0224)(600 nm) = 13 . 4 nm This is the far ultraviolet, so the light has been blueshifted. For stars who’s light reaches me from the direction opposite the one I am traveling, λ obs , back = r 1 + 0 . 999 1 − 0 . 999 (600 nm) = (44 . 7)(600 nm) = 26800 nm This is the far infrared so the light has been redshifted. Thus, when traveling at relativistic velocities, starlight in front of you appears bluer (to the point of being invisible in the UV, if you are traveling fast enough), and stars behind you appear redder (possibly to the point of being invisibly in the IR). 5

4. In the movie Interstellar, the crew visits planets in orbit around a supermassive black hole named Gargantua. Gargantua has a mass of 1 billion M ⊙ . At the first planet, the time dilation is so strong that only one hour passes on the planet compared to 7 years for an observer far away. Hint: Be careful with precision here. Throughout this problem you will find cases where you need to use a Taylor Expansion and compare how your results differ from 1 in order to not lose small numbers. [10 pts] (a) What is the Schwarzschild radius (event horizon) of Gargantua? The Schwarzschild radius is R S = 2 GM/c 2 . R S = 2 × 6 . 67 × 10 − 11 kg − 1 m 3 s − 2 × 10 9 M ⊙ × 2 × 10 30 kg / M ⊙ (3 × 10 8 m / s) 2 = 2 . 96 × 10 12 m = 2 . 96 × 10 9 km (b) Given the gravitational time dilation at the first planet, how far away must it be from the event horizon of Gargantua? The gravitational time dilation factor is given by ∆ t ( r ) ∆ t ( ∞ ) = r 1 − R S r Rearranging, 1 hr × 3600 s 7 × 3 . 15 × 10 7 s 2 = 1 − R S r R S r = 1 − ( 1 . 63 × 10 − 5 ) 2 r = R S 1 − (1 . 63 × 10 − 5 ) 2 ≈ R S (1 + ( 1 . 63 × 10 − 5 ) 2 ) Where in the last step I used a Taylor Expansion, 1 1 − x 2 = 1 + x 2 . r = R S + (2 . 96 × 10 9 km) × ( 1 . 63 × 10 − 5 ) 2 ) = R S + 0 . 786 km So the planet is less than 1 km from the event horizon of the supermassive black hole! (To be fair, a spinning black hole has a smaller event horizon than a non- rotating black hole, and Gargantua is supposed to be spinning. However, the math is well beyond what we will do in this class, so we will stick to the non-rotating black hole.) 6

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(c) What velocity would be required to leave the planet near Gargantua? How much smaller than the speed of light is this? Hint: Consider the fact that the time dilation is a measurement of the depth of the gravitational potential well. A ship leaving the potential well of Gargantua at some v esc would have a kinetic energy γmc 2 . The general relativistic gravitational time dilation is ∆ t ( ∞ ) ∆ t ( r ) = 1 − 2 GM c 2 r − 1 / 2 If we think of the time dilation as a measurement of the gravitational potential well depth, then energy required to overcome the potential is 1 − 2 GM c 2 r − 1 / 2 mc 2 Equating the two 1 − 2 GM c 2 r − 1 / 2 mc 2 = γmc 2 1 − 2 GM c 2 r − 1 / 2 = γ Hence, we can use γ to solve for the minimum velocity necessary to escape that potential well. γ = s 1 1 − v 2 esc /c 2 = ∆ t ( ∞ ) ∆ t ( r ) = 7 × 3 . 15 × 10 7 s 1 hr × 3600 s = 61250 1 γ 2 = 1 − v 2 esc /c 2 = 1 61250 2 v esc c = p 1 − 1 /γ 2 ≈ 1 − 1 2 γ 2 ≈ 1 − 1 . 33 × 10 − 10 which is about 4 cm per second slower than the speed of light. 7

PS10_relativity_solutions

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