Copyright © 2012 Pearson Education Inc. PowerPoint ® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Chapter 37 Relativity

Copyright © 2012 Pearson Education Inc. Goals for Chapter 37 To understand the two postulates (axioms) of relativity and their motivation To see why two observers can disagree on simultaneity To learn why moving clocks run slow To see how motion affects length To understand why velocity depends on the frame of reference To calculate relativistic momentum and kinetic energy To explore the key concepts of general relativity

Copyright © 2012 Pearson Education Inc. Einstein’s first postulate Einstein’s first postulate, known as the principle of relativity, states that the laws of physics are the same in every inertial reference frame. Einstein’s second postulate is that the speed of light in vacuum is the same in all inertial frames of reference and is independent of the motion of the source. The second postulate implies that it is impossible for an inertial observer to travel at the speed of light in vacuum.

Copyright © 2012 Pearson Education Inc. The Galilean transformation The Galilean transformation is a transformation between two inertial frames of reference. For a frame S′ that is moving with speed u in the +x direction with respect to a stationary frame S, with the two frames aligned, we have The Galilean velocity transformation is: The Galilean transformation would suggest that beam of light moving at speed c in frame S′ would have a speed in frame S, which violates Einstein’s 2 nd postulate. We will find that in order to get out of this conundrum, we have to gain a new appreciation of simultaneity.

Copyright © 2012 Pearson Education Inc. A thought experiment in simultaneity Consider this thought experiment in simultaneity. Bolts of electricity simultaneously mark spots A and B in frame S, and A′ and B′ in moving frame S′ (a railway car). Light from the flashes start moving toward an observer at O in S (Stanley), and an observer at O′ in S′ (Mavis). Because Mavis is moving, she catches up to the light from B′ first, then sees the light from A′ later. She says the bolts did not hit simultaneously. Stanley, who is not moving, sees both simultaneously. So Stanley and Mavis do not agree on the simultaneity of the two events. Because light travels at the same speed c in Mavis’s frame, it is not just a speed issue.

Copyright © 2012 Pearson Education Inc. Relativity of time intervals The two observers (Mavis and Stanley) measure different time intervals due to their relative motion. Consider a clock made of a light and mirror in Mavis’s frame. A light pulse goes from the floor to ceiling a distance d, and back to the floor, where it is detected after a time Stanley, observing from a frame outside, sees the light travel along the paths l, which have length, which takes time Eliminating d and solving for  t, we have Stanley sees Mavis’s clock tick more slowly => time is slower (dilated) in moving frames.

Copyright © 2012 Pearson Education Inc. Proper time It is important to realize that none of this has to do with the travel time of light, which could be a factor for particular measurements. The thought experiments we are setting up eliminate this variable. To eliminate this, a frame of reference can be pictured as a coordinate system with a grid of synchronized clocks, as in the figure on the right. If two observers disagree on simultaneity, you might ask what happens when two balls collide as viewed from different frames. Could they disagree on the two balls hitting? No, they only disagree on simultaneity of two events at different places. They will always agree if the two events occur at the same place. Proper time is the time interval between two events that occur at the same point.

Copyright © 2012 Pearson Education Inc. Examples of time dilation Example 37.1: Time dilation at c. A muon at rest has a lifetime of 2.20  s. What is its lifetime if the muon is traveling at 99% of the speed of light? Example 37.2: Time dilation at airliner speeds: An airplane travels from San Francisco to NY (4800 km) at a steady speed of 300 m/s (670 mph). How much time does the trip take for an observer on the ground? For an observer on the plane? Use binomial theorem (Appendix B): Most calculators will calculate this as 1.

Copyright © 2012 Pearson Education Inc. Example 37.2, cont’d Binomial theorem: Put into form: i.e. if a > b, factor out a: Then use: Keep as many leading terms as necessary. In our case: Time on plane is More interesting is the difference in the two clocks:

Copyright © 2012 Pearson Education Inc. Examples of time dilation Example 37.3: Proper time. Mavis boards a spaceship and zips past Stanley on Earth at a relative speed of 0.600c. At the instant she passes him, they both start timers. (a) A short time later, Stanley observes that Mavis passes a space station 9.00  10 7 m beyond him. What time does Stanley’s timer read? What time does Mavis’s timer read? (b) Stanley starts to blink as Mavis passes him, and Mavis measures the blink to take s. According to Stanley, what is the duration of the blink? (a) In this first case, the two events are “Mavis passes Stanley” and “Mavis passes space station.” The two occur at different places to Stanley, but the same place for Mavis. Mavis will measure “proper time,”  t 0, and Stanley’s time will be  t. The speed, u, is measured by Stanley as 0.600c, and the distance is measured also by Stanley, so Stanley’s time is  t = 9.00  10 7 m / 0.6c = 0.5 s. Also, for u = 0.6 c,  = Mavis’s time is (b) In the second case, the two events are beginning and end of Stanley’s blink. This time Stanley’s is the proper time  t 0, and Mavis measures  t, so

Copyright © 2012 Pearson Education Inc. Twin Paradox On Earth, one twin boards a spaceship and travels at near the speed of light to a distant star, while the other twin stays on Earth. The traveling twin then reverses course and comes back to Earth. The Earth twin sees the clock of the traveling twin as ticking more slowly (time dilation) during the trip, so the traveling twin arrives back at Earth much younger that the one who stays home. But also during the trip, the traveling twin sees the clock of the Earth twin also ticking more slowly (by symmetry of the situation). How can both be true? The answer is that the traveling twin changed speeds, and her frame is not an inertial frame, while the Earth-bound twin’s frame is (approximately) inertial the whole time. This fact breaks the symmetry. In fact, it is true that the traveling twin arrives back at Earth younger than the one who did not travel.

Copyright © 2012 Pearson Education Inc. Relativity of length If the foregoing seems confusing, we will now see that part of the reason for the confusion is that we have only half of the story. We also have to examine what happens to lengths, after which we will derive some equations that take both into account (and things will be a little clearer). We will now discuss length contraction. In the figure below, the length of the ruler is measured by two observers moving relative to each other. The proper length is the length measured in the frame at which a body is at rest. Mavis measures the length of a ruler in her frame, using a source and mirror as before, l 0 = c  t 0 /2, or Stanley sees the light go a distance d 1 = l + u  t 1, which is longer than the ruler (because of the movement of the far end), bounce off the mirror, then go a shorter distance d 2 = l – u  t 2 coming back. But d 1 = c  t 1, and d 2 = c  t 2. Eliminating d 1 and d 2, we have But so

Copyright © 2012 Pearson Education Inc. Length Contraction The result means that lengths in moving frames (as viewed from a stationary frame) appear shorter, or contracted, along the direction of motion. There is no length contraction for lengths perpendicular to the direction of relative motion. Stanley and Mavis agree on the length of this ruler. A consequence of this is that if a ruler is at some angle  from the direction of motion in a moving frame, it will appear at a steeper angle from the stationary frame.

Copyright © 2012 Pearson Education Inc. Examples of length contraction Example 37.4: A spaceship flies past Earth at a speed of 0.990c. A crew member on board the spaceship measures its length, obtaining the value 400 m. What length do observers on Earth measure? Example 37.5: The observers at O 1 and O 2 are 56.4 m apart on Earth. How far apart does the spaceship crew measure them? Now the situation is reversed, so l 0 is in the Earth frame and we use the same expression: Note that this does not mean the spaceship crew measures their own length as 7.96 m. They would claim that the Earth observers did not measure the two ends of the spaceship simultaneously.

Copyright © 2012 Pearson Education Inc. The Lorentz transformations The Lorentz transformations relate the coordinates and velocities in two inertial reference frames. They are more general than the Galilean transformations and are consistent with the principle of relativity. Consider a point P at position x,y in frame S, and the same point at coordinates x′,y′ in moving frame S′, whose origins coincide at time t = t′ = 0. The Galilean transformation had but now we know that in S the coordinate x′ will be Lorentz contracted, i.e. The principle of relativity says that the situation of S viewed from S′ must be symmetrical to that of S′ viewed from S, except for the sign of u. Thus and solving for t′ we have

Copyright © 2012 Pearson Education Inc. The Lorentz Transformations The final statement of the Lorentz transformations is: An aside: These expressions can be made to look more symmetrical if we consider time as a length, by multiplying by c., i.e.  = ct. Then the equations become It is also typical for working in special relativity to use  = u/c, which makes them appear even simpler: Note: these define the motion of the moving frame (the velocity u) to be in the +x direction.

Copyright © 2012 Pearson Education Inc. Examples using the Lorentz transformations Example 37.6: Mavis wins an interstellar race by crossing the finish line at a speed of 0.600c relative to the line. A “hooray” message is sent from the back of her 300-m long ship (event 2) at the instant the nose crosses the line (event 1). Stanley is at the finish line and at rest relative to it. When and where does he measure these two events? In this problem, u = 0.6c (  = 1.25), the proper length is in Mavis’s frame, which is the primed (moving) frame since we are asked about events in Stanley’s stationary frame, so x′ =  300 m (the back of her ship). The time between events 1 and 2 in Mavis’s frame is t′ = 0 (they happen simultaneously in her frame). The primed coordinates in Mavis’s frame are (x′, t′) = (0,0) and (  300,0). We have everything we need to calculate x and t for the two events. Event 1: Event 2: This means Stanley observes the “hooray” message happen 0.75  s before the nose of her ship crossed the finish line.

Copyright © 2012 Pearson Education Inc. Lorentz Transformations and Velocities To learn how to transform velocities, we just take the differentials of the Lorentz transformation equations i.e., Then which can be written Again, to get the symmetrical form for v x, just swap primes and the sign of u.

Copyright © 2012 Pearson Education Inc. Examples using the Lorentz transformations Example 37.7 using the figure below: (a) A spaceship moving away from Earth at 0.900c fires a robot space probe in the same direction at 0.700c relative to the spaceship. What is the probe’s velocity relative to Earth? (b) A scoutship is sent to catch up with the spaceship by traveling at 0.95c relative to Earth. What is the velocity of the scoutship relative to the spaceship? (a) In Earth’s frame, the spaceship is the primed frame moving at speed u, and the probe has a velocity v x ′ in the primed frame (and is positive). We use (b) In this part we need to solve for the speed in the moving frame, so we use

Copyright © 2012 Pearson Education Inc. Doppler effect for electromagnetic waves In the figure below, the distance between wave crests is = (c – u) T. Then the frequency In the stationary frame, we have a frequency f 0 = 1/T 0, where T 0 is the proper time period of the wave. In the moving frame, the period just due to time dilation is T =  T 0, which can be written Inserting f 0 = 1/T 0, and solving for f, we have for a source moving toward the observer, or, changing the sign of u, for a source moving away form the observer. For small u/c

Copyright © 2012 Pearson Education Inc. Relativistic momentum We will state without proof how momentum changes with speed in a moving frame: So we see that faster objects not only get more momentum proportional to their speed, but the momentum skyrockets near the speed of light. Figure at the right shows a graph of the momentum of a particle as a function of its speed. Notice that the momentum approaches infinity as the speed approaches the speed of light. The expression for momentum is sometimes used to talk about “relativistic mass,” meaning the momentum acts as if objects get more massive with speed. (not a good concept)

Copyright © 2012 Pearson Education Inc. Newton’s second law and relativistic mass Once we have an expression for relativistic momentum, we can use Newton’s second law to relate to force, then work and energy. But in a synchrotron accelerator, for example, the force in a magnetic field is due to the Lorentz force, where the charged particles move in uniform circular motion. In that case, F and v are perpendicular and the speed is a constant, i.e. Work (or kinetic energy) is: (case of F and v along x)

Copyright © 2012 Pearson Education Inc. Relativistic work and energy The relativistic kinetic energy is K = (  – 1)mc 2. The graph in Figure (right) shows the kinetic energy as a function of speed. Note that the kinetic energy approaches infinity as the speed approaches the speed of light. The rest energy is mc 2. The total energy of a particle is E = K + mc 2 =  mc 2. The total energy, rest energy, and momentum are related by E 2 = (mc 2 ) 2 + (pc) 2. Method: Subtract (2) from (1) and rearrange

Copyright © 2012 Pearson Education Inc. Examples of relativistic energy Example 37.10: Energetic Electrons: (1) Find the rest energy of an electron (m =  10  31 kg, q =  e =   10  19 C) in joules and in eV. (b) Find the speed of an electron that has been accelerated by an electric field from rest, through a electric potential of 20.0 kV or of 5.00 MV (typical of a high- voltage X-ray machine) (a) Rest energy is E = mc 2 = (9.109  10  31 kg)(3.00  10  m/s) 2 =  10  14 J. Since 1 eV =  10  19 J, this is 5.11  10  5 eV = 511 keV. (b) An electron accelerated through 20 kV gains (by definition) 20 keV of kinetic energy. K = (  – 1)mc 2 = 20 keV, but mc 2 = 511 keV, so (  – 1) = 20/511 = and  = For the higher energy, (  – 1) = 5000/511 = 9.78, so  = Rearrange For the two cases:

Copyright © 2012 Pearson Education Inc. Examples of relativistic energy Fusion reaction in the Sun The energy from fusion in the Sun comes from combining 4 protons (H nuclei) to create a He nucleus. The mass of the proton is  10  kg, while the mass of a He nucleus is  10  kg. How much energy is released in the reaction? The total mass before the reaction is (4)(1.673  10  kg) =  10  kg, so a He nucleus actually has less mass by  m =  10  kg –  10  kg = 4.7  10  kg This “missing mass” is converted to energy by E = mc 2, and releases Huge numbers of these reactions are going on every second, converting the Sun’s mass into energy. E = mc 2 = (4.7  10  29 kg)(3.00  10  m/s) 2 = 4.23  10  12 J = 26.4 MeV.

Copyright © 2012 Pearson Education Inc. General relativity Einstein realized that his theory of relativity only described the situation in inertial frames, so was “special” in that sense. It took him 10 years of very hard work to generalize the idea to non-inertial frames (frames that are accelerating). Frames anywhere near a gravitating body are such non-inertial frames. To deal with gravitation, Einstein used a similar axiom as before—that one cannot do any experiment to tell the difference between a constant acceleration and a region of uniform gravity. In Figure at the right, the astronaut cannot distinguish the effects of uniform gravity from the effects of uniform acceleration. This leads to many surprising results such as the effect of gravity on light (gravitational lensing), the warping of space-time, etc.