Special Relativity. Modern Physics 09/26, 09/28, 10/01-03/2018

Transcription

1 Special Relativity Modern Physics 09/26, 09/28, 10/01-03/2018 1

2 Introduction, to Chapter 2 This was the 1 st of the jewels of modern Physics In Physics 140, we studied motion, and said that it was in an inertial reference frame (e.g. K) Basically means that the reference frame in which we are viewing the motion has constant velocity ( v = 0 or in a straight line ) K moves relative to K If the two frames have no relative acceleration (i.e. relative constant velocity) then if Newton s laws are valid in one frame, then they are also valid in the other. If we observe an object s motion from another (K ) reference frame, we can do the transformations to get x, y, z in terms of x, y, z. Newtonian Relativity or Galilean Invariance 2

3 The Problem From Physics 140: x = x vt, y = y, z = z, t = t If vector v is along the x-axis Galilean transformation dx /dt = dx/dt v, speeds simply added up In Physics 150, you studied Electromagnetic waves, and it was shown that wave speed c = 1 / ( µ 0 ε 0 ) in vacuum In a medium wave speed = 1 / ( µ ε ) So, what would the wave speed look like when viewed from a different inertial frame? If we believe in the Galilean invariance, then the wave speed would change: c = c v or c = c + v properties of medium 3

4 The Oddity This felt a little odd: why should the wave speed be different in a different frame? A function of the fundamental properties of a medium People tried to resolve this by saying that waves in water seem to have different speeds when viewed from different frames Argument is that waves are IN water So, perhaps EM waves are also traveling in a medium, that is everywhere: called this aether. So it was resolved that the speed of an EM wave, c = 1 / ( µ 0 ε 0 ) was in a frame that was fixed with respect to this aether Great! Now what if we are in a frame that is moving relative to this aether? Can we measure the change in c? 4

5 Michelson-Morley Experiment Since c is very large (~3 x10^8 m/s), it was difficult to design an experiment (remember this was the late 1800 s) to see a shift in c v = speed another of another frame (v large enough) In this experiment, they used the speed of the earth as it orbits the sun This speed is ~10^-4c, so that experiment had to be sensitive to 1 part in 10,000 (doable) Assume that one arm of this device is moving relative to the aether. If you adjust the optical lengths, l 1 and l 2, then you should be able to observe interference fringes 5

6 Some Math What happens if you rotate the device 90? To observe fringes we need to satisfy 2(l 1 -l 2 ) = nλ, where n is an integer and λ is the wavelength of the light We expect the fringes will shift: let us see how speed is the same for A->D and D->A, however the aether drags the light and so one needs to account for the slight angle at which we send the light 6

7 Rotation Technology is called interferometer Author: [[User:Stigmatella aurantiaca]] Source: Own work The time difference in these 2 settings will manifest itself as a shift in the interference pattern. Since v ~10^-4 c, we can use an approximation to simplify this expression: 1 / (1-x) for small x is ~ 1+x and 1 / (1-x) ~ 1/(1-x/2) ~ 1+x/2 7

8 Fin Despite repeating the experiment many times, simultaneously refining it to become increasingly sensitive, they always came up with a null result i.e., no shift in fringes was observed. This put the aether theory in grave danger There were other ideas too (see aether drag on pg. 25) Physics was at an impasse: Galilean transformation did not appear to work for EM waves One could argue that laws of physics are different in different frames, but that is ugly! Then, in

9 Solution Einstein made an incredibly radical and bold move (recall he was working as a clerk in the Patent Office). He postulated 1. All laws of physics are the same in all inertial frames. There is no preferred frame. 2. The speed of light is a constant regardless of which frame you are in. In normal everyday life there is no effect because v << c Although these postulates seem quite tame, they completely overturned our concepts of space & time. Nothing was absolute anymore Everything was relative, even time! One casualty was the assumption that t = t -> according to these The Galilean transformations (x = x vt, t =t, y = y, z = z) for two frames moving relative to each other along the x-direction have to be replaced. postulates, each frame has its own time. -> Fig Simultaneous events in one frame are no longer simultaneous in a different frame. Simultaneity not real 9

10 Lorentz Transformation At t=t =0 the origins & axes of the two frames overlap Assume a flashbulb goes off. Since speed of light is a constant, we have x^2 + y^2 + z^2 = c^2 t^2 x ^2 + y ^2 + z ^2 = c^2 t ^2 distance d = speed * time We will not derive the equations, but I encourage you to re-read Section 2.4 Space and time have become enmeshed. That s why time is referred to as the 4 th dimension. When v/c << 1 then x = x vt and t = t 10

11 Consequences We can simplify these equations by writing β = v/c and γ = 1 / (1 β^2) = 1 / (1 v^2/c^2) β <= 1 and γ >= 1. And we get x = γ (x vt), y = y, z = z, t = γ (t v x / c^2) The earth orbits the sun at 30 km/s (i.e., ~20 miles/s or ~70,000 miles/hr.), which is pretty fast. However c = 3x10^5 km/s => β = 10^-4. and γ ~ !! So, relativistic effects are negligible Let s look at some implications of Lorentz transformations and relativity (Section 2.5) Time dilation: moving clocks run slower!! (Fig. 2.9) 11

12 Time Dilation In the frame K, time difference between sparkler being lit and going out = t 2 t 1 = T 0 T 0 is known as the proper time; in the moving frame Now in the fixed system x 2 = x 1 (and since γ >= 1 => T >= T 0 ) Consider Figure 2.10; DEMO Each person will say that the other s clock is slow!!! In Fig. 2.10, event happens in moving frame. To an observer in that frame light bounces up/down whereas in fixed frame light would travel in a different path 12

13 Analogy: Light Clock Imagine the following scenario: Person on a truck throws a ball up in the air What would the trajectories look like to K and K? Acc. to frame K, T 0 = t 2 -t 1 =2L/c. Acc. to frame K, watch?v=-o8lbichre0 YouTube video to review the Einstein clock (Brian Cox, UK). Especially useful if you missed our in-class demonstration K would say K says Next, we will consider the evidence for time dilation Re-Read Example 2.2 => Proper time is always the shortest time interval. 13

14 Copyright 2015 Hackaday, Hack A Day, and the Skull and Wrenches Logo are Trademarks of Hackaday.com 14

15 Experimental Evidence Muon decay (see 2.7): muons (elementary particles => ~ 200x heavier cousins of the e-) are produced when cosmic rays hit the Earth s atmosphere. Muons have a lifetime of about 2.2x10^-6 s N = N 0 e t/τ where τ = 2.2 x 10^-6 s => c * τ ~ 3x10^8 x 2.2x10^-6 ~ 660 m (assuming it is traveling very fast) So if relativity were not correct we (on earth) would observe that, after ~660 m, N = N 0 / e. However, we observe that many more muons survive the journey as they travel though the atmosphere This is explained by the fact that in our frame (Earth), the moving muon s clock slows down and τ is effectively larger by a factor γ 15 (books can define these things slightly differently)

16 Length Contraction x r : right end; x l : left end. Take a meter-stick in the two frames. L 0 = x r x l and L 0 = x r x l (as measured in each frame) The longest possible length ( at rest ) is called Proper Length Observer in K will see their own meter-stick at rest and measure its length as L0; similarly for observer in K and their own meter-stick L0 ( = L0). What happens when the observer in frame K observes the meter-stick in K as the latter moves along the x-axis? We require that the observer in K measure the two ends of the meter-stick in K at their same time, i.e., t r = t l = t 16

17 Lorentz Using the Lorentz transformation x = γ (x vt) for the x coordinate, we get that the observer in K sees x r x l = γ ( x r vt r ) γ (x l vt l ) = γ [ (x r x l ) v (t r t l ) ]. If we require that t r = t l => x r x l = γ (x r x l ) or (x r x l ) = (x r x l ) / γ = L 0 /γ That is, the observer in K sees that the meter-stick in K is shortened by factor γ. Remember that γ = 1 / (1 v^2/c^2) >= 1 Again we can turn this argument around and show that if the observer was in K, they will observe the meter-stick in K to be shorter by a factor γ. These two effects, time dilation and length contraction, are, of course, related. Let us now look at examples like 2.2/2.3 from the text 17

18 In-Class On-Board Examples What speed (relative to the Earth frame) would a starship have to achieve in order to reach the Andromeda Galaxy (2.5 million light-years away) during one human lifetime, as experienced by the occupants of that outbound spacecraft, despite Earth experiencing 2.5 million years approximately during the same frame of time? Note: You may quantitatively interpret how many years one human lifetime means in any reasonably justifiable way. Sun is ~8 light-minutes away (b.) Let s consider this from the point of view of the astronauts. They will claim that the distance between earth and Andromeda as measured in the earth s rest-frame is actually shorter! (c.) Assuming tech continues to improve, could this be really possible one day? 18

19 A muon has a mean lifetime of 2.2 µs, and in an experiment it s observed that it leaves a track 9.5 cm long. How fast is the µ travelling? Note - in Section 2.7, they talk about half-life and here it s mean lifetime. In my lecture notes I use mean lifetime. Assume that the µ decays after 2.2 µs (in its rest frame) To the observer, this would look like T = γ * 2.2 µs Since we say that µ s clock is running slow So, the distance the µ travels in the lab = v * T v * γ * 2.2 x 10^-6 = 9.5 cm = m Note that in reality, the mean lifetime (τ) of 2.2 µs implies that after 2.2 µs we are left with N = N 0 e t/τ = N 0 /e muons. Any given muon can decay whenever it decides to. When we do experiments to measure t, By measuring falloff we measure we need to collect lots of data. τ (Ideally, all µ have same speed) 19

20 Addition of Velocities In classical physics (Newton s Laws or Galilean transformation) the observer K will say that K is moving towards them at ( v 1 + v 2 ) On a highway, if two cars are driving side by side at the same speed, they will appear at rest to each other So if v 1 and v 2 are close to speed of light, c, this would imply that the relative speed would be > c => not allowed. Velocity addition another casualty of special relativity. Let s take an example based on your textbook 20

21 K star destroyer K x-wing v = 0.6 * c In normal space (not hyperspace) fires protons, with speed u = 0.99 * c To the x-wing fighter firing upon the TIE fighter, the proton torpedoes are moving with speed u. What is their speed relative to the Imperial star destroyer (stationary)? From Newton s mechanics, we will say u = u + v = ( ) = 1.59c, but that s > c -> A problem! We had in past x =γ(x-vt), y =y, z =z, t =γ(t-vx/c^2) In our case, we need to flip this around, i.e., what do these quantities look like as measured in frame K? x=γ(x +vt ), y=y, z=z, & t=γ(t +vx /c^2) (v sign change) Differentiate, dx = γ (dx + v dt ), dy = dy and dt = γ [ dt + (v/c^2) dx ], and dz = dz Using 1D always x for simplicity We can identify u x =dx/dt and u x as dx / dt 21

22 (re-read p. 45 to see experimental verification) divide RHS by dt in numerator and denominator (1) Similarly or (2) (3) 22

23 Answer So, in our original example u x = 0.99c and v = 0.6c This implies u x =(0.99c+0.6c)/[1+(0.6c/c^2)*0.99c] = 1.59c/1.594 = c (what if photon torpedoes?) Notice that we now also have expressions for v y & v z, which are different from v y and v z, not like y=y and z=z This just reflects that time is different in the two frames! In many cases, we can simplify things so that we put v y = v z = 0 -> i.e., we just choose a set of axes where everything is along x-axis Of course, if motion is truly 3-D, then we have to account for that. -> (see Example 2.4) Also, we can flip the equations and write u x = (u x v) / [ 1 (v/c^2) u x ], etc. -> (see Eqns. 2.24) 23

24 example (problem #32 in text book) A proton (p) and an anti-proton (p-bar) move toward each other in a head-on collision. If each has a speed of 0.8c with respect to the collision point, how fast are they moving with respect to each other? If we follow the arguments we just made, we now have three frames, (1) attached to p, (2) to p-bar, and (3) collision point: K 1, K 2, and K, respectively. Let s see things from the perspective of the collision point The proton is fixed in frame K 1, and the frame is moving in frame K. This is a tough one. Do in your groups. Correct answer for you to check against is 0.976c each 24

25 The Twin Paradox (Sect. 2.8) Advent of relativity was a boon for philosophers since it gave rise to apparent contradictions. One twin (fixed) remains on Earth and the other (moving) goes on a rocket ship to a star this is 8 light years away, and returns after many years. What is the situation? (1) According to the fixed twin, the moving twin s clock moves slower, so latter will be younger at the trip s end. (2) The moving twin says the same thing (since in their frame of reference, it is the fixed twin who is moving ) Now, both (1) and (2) cannot be simultaneously true, so perhaps the real answer is that they are the same age? 25

26 More Contradiction Rocket ship quickly reaches cruising speed, say, 0.8c, gets to the star, turns around and comes back at 0.8c. We can assume getting to cruising speed and time taken to turn back is negligible. According to the fixed twin, the moving twin takes 8 L-yr/0.8c = 10 years for each leg of the journey, i.e., a total of 20 years, so the fixed twin will be 20 yr older Because the moving twin s clock moves slower, the fixed twin would argue that the former would age less 10 yrs. / γ = 10* (1-v^2/c^2)=10 (1-0.8^2)=6 years, so they would only age by 12 years. What s going on? Be sure to review examples 2.2 & 2.3 yet again 26

27 Subtlety It is a subtle point (see slide 9 of notes as refresher) The 1 st postulate of relativity says that all laws of physics are the same in all inertial frames, i.e., frames which move with constant velocity with respect to each other. In our problems, the rocket ship is not in an inertial frame, when it has to turn around at the distant star, i.e., the rocket ship experiences acceleration or a change in speed (from 0.8c to -0.8c) Hence the symmetry of the problem is broken and there is NO paradox The moving twin is the younger one Actual analysis is more complex. For example, what if it is a one-way trip? What happens after the rocket reaches the star (and keeps going). No absolute t! 27

28 Space-Time (Section 2.9) I am just going to mention this concept/idea without going into any detail ( light cones ) Consider a situation where something (e.g., a spaceship) is moving with speed v (relative to a fixed frame) along one dimension. ct slope = 1. y, z x After time t, we will say that ship has traveled a distance = vt * If v is constant, slope of line in ct-x plane is c/v Light will travel along the line c, slope = 1. We can t have anything with v>c, so no line with slope <1 The future will always be within the conical region. Events outside of this region are not in causal contact with events inside (means no influence) 28

29 Space-Time Distances Although both x (again assuming 1-D motion for simplicity) and t change depending on the relative speed of two frames, it turns out that there is a quantity which is the same in both frames We have a relationship between x, x, t, & t If we write s^2 = x^2-(ct)^2; s ^2=x 2-(ct )^2 We discover that s^2 = s ^2 Just plug in x = γ (x - vt) and t = γ [ (t - vx/c^2) ] into formula for s ^2 In 3-D, this becomes s^2 = x^2 + y^2 + z^2 - (ct)^2 In classical physics distance is 3-dimensional. It = x^2 +y^2 +z^2 So, s^2 represents a distance in 4-d, i.e. in space-time. (We may revisit this later on). Invariant, while energy isn t P

30 Doppler Effect (Section 2.10) If you are standing by the side of the road, and an ambulance drives by, you notice that its first becomes shriller as it approaches you And then the reverse occurs as it moves away Don t confuse this with loudness/volume The frequency of the sound first increases (shriller) and then decreases. But now... We look at the frequency of light that is emitted by a distant star As you know, the universe is expanding, i.e., all distant objects are moving away from each other (quite rapidly, as it turns out) However, let s consider the case when the star is coming towards us (or, imagine an extremely fast-moving comet) 30 K c K v, star Earth

31 star Earth Imagine light waves In time T (in Earth s frame), light moves a distance ct; star moves a distance vt So the distance between the front and back of the train of light waves = ct vt Assume that there are n waves in this time Wavelength (λ) of emitted light = (ct-vt)/n Frequency = f = c/λ = cn / (ct-vt) = n/[t-(v/c)t] As observed by us, f = n / [ T (1 - v/c) ] Now in the star s rest frame, if the frequency of light is f 0, then in the proper time interval T0, we have n = f 0 T 0 [#waves/sec.*#seconds]. Now T = γ T 0. Then... 31

32 Since recall that the star is *approaching* us So, when the source is approaching us, then When the sources recedes from us, we just need to flip the sign of β relative to the previous case 32

33 Redshift and Blueshift This is the phenomenon known as red-shift In reality what we observe is a bunch of frequencies from any star These frequencies depend on what the star is made up of, what its stellar atmosphere is, etc. e.g., Na on earth will emit the same frequency on a star What we observe is the same bunch of frequencies all shifted by the same amount This helps us determine how fast the star is moving. If we can make an independent measurement of the distance to the star, we can derive Hubble s constant This tells us how fast the universe is expanding We will discuss other applications of the Doppler effect as we go further in the textbook R&T 33

34 example (problem #54 of text book) Setup: A spaceship moves with acceleration = 29.4 m/s^2 away from the Earth. They constantly observe Sodium vapor streetlamps as they move away (λ = 589 nm) Question: After how much time will the light from the lamps be so shifted that they are not observable in the visible (range is nm). Note: Frequency of red < frequency of blue light, hence red has the higher wavelength Hint: Since the earth is receding this light will get redshifted, i.e., frequency will keep decreasing until it becomes < f red (= 4.3 x 10^14 Hz). Use Equation 2.33! 34

35 Relativistic Kinematics (S 2.11 & 2.12) In Physics 140, you had learnt that p = momentum = mv (vector) KE = Kinetic Energy = (1/2) m v^2 Do these quantities change? Recall the example on slide 23 of these lecture notes where the proton and anti-proton approach each other We determined that to one, the other would appear to be moving with speed = 0.976c (and not 1.6c) We ll skip the derivations, but I encourage you to re-read the sections 2.11 and 2.12 For an object at rest, i.e., v=0, γ=1 thus p=0 and KE=0, as they should be When v/c much < 1 => p ~ mv (since v/c << 1 => v^2/c^2 is even smaller and can be neglected) Relativistic p = γmv Similarly, relativistic KE = mc^2 (γ-1) Let us understand the implications of this w/ 0.8c 35

36 Relativistic Energy We can write KE=γmc^2 mc^2 -> γmc^2=ke+mc^2 We identify total E = γmc^2 = KE +E 0 Another way to think of this is that E = (γm)c^2, i.e., object becomes heavier as v -> c So, how do we deal with the photon? It has v/c = 1 and m=0. -> Will come back to this It has energy and momentum, but no mass!! (rest energy, =mc^2) In this new scenario, there s some modification of the old and we call it conservation of mass-energy In classical physics, we had talked about conservation of energy and mass (i.e., total energy is conserved). 36

37 Collision Consider a collision of two objects One would think that M = 2m Initial total E = KE 1 + KE 2 + E 0,1 + E 0,2 = 2KE + 2mc^2 Final total E = Mc^2 since total Energy is conserved BUT Mc^2 = 2KE + 2mc^2, so ΔM = M-2m=2KE/c^2 Total Energy is what the text book calls mass-energy What happened was that some of the Kinetic energy got converted to mass In the collision, the two objects stick together by deforming each other and warming the surfaces, thereby raising internal E, which manifests itself as an increase in M(ass) Since mass and energy are equivalent, this is almost like book-keeping. There s a fractional increase in mass ΔM/2m a When v/c << 1, gamma is very close to 1, hence this increase is negligible. 37

38 Energy and Momentum We had said that p = γmv p = mv / (1 - v^2/c^2) = mv / (1 - β^2) = γmv (Multiply and divide by c^2 on RHS) where for a photon m = 0 => E^2 = p^2 * c^2 or E = p * c where rest mass 38

39 Change of Units for Calculations Section 2.13: Since the electron, proton, and neutrons have very small masses, and they move at very high speeds, we use a slightly different system of units for our calculations Electron (e - or e) charge -> x 10^-19 Coulombs mass -> x 10^-31 kg The rest energy of a proton = mc^2 which is (1.7x10^-27)(3x10^8)^2=1.5x10^-10 J. Using the conversion that 1 ev = 1.6 x 10^-19 J. we get that E 0 (proton) ~ 9.38 x10^8 ev or 938 MeV Proton and neutron (p + and n 0 or p and n) mass -> ~1.67 x 10^-27 kg (there is a small difference ~ x 10^-27 kg!) and charge equal and opposite of e If we accelerate an e- across a 1 Volt battery, it will have gain E = 1.6 x 10^-12 C * 1 V We simply call this 1 ev = 1 electron-volt=1.6x10-19 Joule. 39 p n

40 Mass-Energies and AMU Or, we simply say that the mass of the proton is 938 MeV/c^2; neutron mass ~ 939 MeV/c^2 Similarly, electron mass ~ MeV/c^2 or 511 kev In contrast, the Higgs boson has a mass of 125 GeV/ c^2, i.e., 129 x 10^9 ev/c^2 ~ 133 m proton!! And the top quark (a fundamental particle) has mass ~172 GeV/c^2, i.e., ~184 m proton -> almost as much as a gold atom!!! (Read part (a.) of Example 2.13) Atomic mass unit (u or amu) -> defined that mass ( 12 C atom) = 12 u isotope of hydrogen (1n + 1p) This gives us that u ~ 1.66 x kg ~ MeV/c^2. m proton = u, m neutron = , m deuteron =

41 Of Particular Interest Turns out that M deut M proton M neutron = -2.4x10-3 u That is, a deuteron weighs LESS than the sum of its constituents. The difference is called Binding Energy The energy required to hold the proton and neutron together in the nucleus, responsible for fission & fusion Will return to this later in semester. Now, last example Nuclear binding energy 41

42 Example (Problem #71) Note, that as KE increases, speed gets closer to c, but never equal to it. Only massless particles can travel AT speed of light. Do Homework 4, on relativity! Due this Mon. 42

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