Your (primed) frame frame

6 Special Relativity 6.1 Galiean Relativity 6.1.1 Spacetime Diagrams We keep seeing the word relativity appear in our discussion. To the person on the street, relativity (normally associated with Einstein) elicits the statement everything is relative, implying that we cannot know anything objectively. In fact, both Einstein and Galileo (who we will also discuss) would have emphatically disagreed with this assessment. We can say things objectively it s just that part of our objective statement needs to include information about how to translate what I see, to what you see. Our standard piece of equipment for this experiment will be a train moving to the right at some known speed, v. You will be on that train, while I will be on the platform. Prof. Goldberg s (unprimed) Your (primed) frame frame v v Note that the train is moving at all times. The Special in special relativity comes from the fact that there are no accelerations. Of course, you also know from your own experience that from your vantage, it appears that I am moving backwards at speed, v. In general, when we talk about different perspectives, we ll refer to the unprimed (meaning, in this case, my) or primed (meaning your) frame. For example, if I saw a puppy roll over 10 meters in front of me, I d say, x = 10m. However, if you also observed the same event, but it was behind you 5m, we d say, x = 5m, which is just a shorthand for saying that the position of the puppy rolling over is -5m from the perspective of the person in the train. I used the word event in the previous paragraph. For us, there are only events, and events have a time, and a position. Thus, The party begins at 9:00pm at the funky disco indicates an event. We can imagine a particularly simple 1-d universe full of events, and plot it up on a space-time diagram. 32

Time 3 "Events" in spacetime Stationary Observer Observer Traveling at 1/2 c Light Detected Space Light Detected and Reflected Light Beam Sent In general we tend to scale these things so that 1 unit in the x-direction corresponds to, say, 1 light-second, and 1 unit in the y-direction corresponds to a second, and thus, light travels at 45 degree angles (or -45 deg. if it s traveling to the left). Someone sitting still will move vertically through time (but not through space). And, of course, since no-one can travel faster than light, your world line can t ever be shallower than 45 degrees. On the space-time diagram, I ve drawn a light-beam fired from a ship moving half the speed of light toward an observer at rest (Event 1). The beam is detected by the stationary observer (event 2), and re-deflected. Finally, the the lightbeam is detected by the moving observer (event 3). For each observer, each event has a definite time and position. I have drawn things from the perspective of the stationary observer, since from your own perspective it always seems as though you are sitting still. 6.1.2 Galilean Relativity All of this is just notation leading us up to relativity. And when you think relativity, you no doubt think of Galileo. Well, probably not. But in reality, Galileo s relativity is the one in our everyday experience. Time: For Galileo, time was an absolute, and thus, all clocks ran at the same rate. So, even though it is legitimate (say if you and I are in different time-zones) for us to measure an event at different times say t 1 = 0, and t 1 = 3600s (if you measure a particular event as being one hour later than may because you never reset your watch from daylight savings), it is true that our watches run at the same rate. Thus, if we measure a second event, and I measure it at t 2 = 10s, you ll measure it as t = 3610s. According to Galileo, For both of us: Galileo : t = t 2 t 1 = t = t 2 t 1 Space: With space it s a different matter. Consider that for a moving observer, objects, and hence events, will look like they re in different positions at different times. Thus, if a particular event appears to me at 33

some position, x, you ll observe it as x = x vt which should accord with your intuition. If you don t believe me, think about how a stationary observer appears as you pull away from the station. As more time passes, his position looks more and more negative. Finally, we can relate the velocity of an observed object. Consider that velocity is simply defined as: u x t (34) where we simply observe a moving object at two different times. Using the relations: x = x v t (35) t = t (36) simple algebra yields: u = u v (37) In other words, if you were on a train going 100mph, and I fired a bullet parallel to the train at 150mph, you would observe the bullet to be moving at 50mph. Finally, note that these equations are completely symmetric. We could just as easily say: x = x + v t t = t u = u + v And thus, there s no way from Newton physics to tell who is moving and who is staying still. According to Newton, the only thing that matters are accelerations, and consider that a system which obeys Newton s laws in one frame will also satisfy them in any other inertial frame. No problem, right? Wrong. 34

6.2 Special Relativity: Space and time 6.2.1 A Simple Experiment with Light As I ve mentioned before, you are in the primed frame (the one inside the train). This is a very special train, one set up with mirrors and lasers, and sophisticated chronometers and the like. Light (in the form of lasers) is particularly important because of an experiment done in 1887 by Michelson and Morely, which showed that light travels at a constant speed for all observers. Using that fact, and the fact that at low speeds our picture of the universe should be the same as that suggested by Galileo, will be our basis of special relativity. So, consider the following apparatus within your train. 2 h 1 3 In this experiment there are 3 events: 1) A pulse is fired (at the speed of light) from the bottom of the train. 2) The pulse is reflected from the top of the train. 3) The pulse is detected at the bottom of the train again. View from the primed frame: Despite the small offset in the picture, events 1 and 3 take place at the same position (according to you), and thus: x =0 However, since the height of the train is h, the light must have traveled a distance of 2h between the beginning and end of the experiment, and thus: t = 2h c All we ve used here is the fact that light travels at the speed of light. View from the unprimed frame: Now, from the unprimed frame, the situation looks somewhat different. The detector has moved from when the laser was fired, and thus, from my perspective, the experiment looks like: 35

2 h v 1 3 The solid train represents the position of the train at the beginning of the experiment, and the dotted line represents the position of the train at the end of the experiment. The detector must have moved a distance: x = v t from event 1 to 3. This is the same result we would have gotten from Galilean relativity. Well, except for one thing. We are no longer assuming that t = t. In this case, we note that the total trip length is: (v t d tot =2 2 ) 2 + h 2 In case it s not obvious where I got that, each light beam can be written as an x-component (v t/2, where t is the time observed by me), and a y-component, h. I m just using the Pythagorean theorem to add them. We also note that the light travel distance can be related to the full time interval between 1 and 3 via the relation: where I ve used t =2h/c explicitly. t = d tot c (v t ) 2 = 2 + h2 2c c 2 (v t ) 2 ( ) 2 2h = + c c (v t ) 2 = + ( t c ) 2 Interesting... I now have t in terms of t. Squaring the expression, I get: t 2 = ( ) 2 v t + ( t ) 2 c 36

) (1 v2 t 2 = ( t ) 2 c 2 t 2 = t = ( t ) 2 1 v 2 /c 2 t 1 v2 /c 2 Thus, we ve found an interesting relation: where t = γ t + f(v) x (38) γ 1 1 v2 /c 2 (39) Note that I put a function f(v) in there. Why? Because we don t know what happens if x 0. We ll figure it out eventually. Also, by the same argument, since we found: then x = v t x = vγ t + g(v) x (40) These equations need to be symmetric. In other words, there shouldn t be anything which can distinguish between the moving (primed) frame and the stationary (unprimed frame). To you, I should just appear to be moving to the left. And thus, we should have the relations: and x = vγ t + g( v) x (41) t = γ t + f( v) x (42) To figure out what f(v) and g(v) are, we first note, that by definition, x = x. Thus, plugging equations 41 and 42 into 40 we get: x = vγ(γ t + f( v) x)+ g(v)( vγ t + g( v) x) = (vγf( v)+g(v)g( v)) x + ( vγ 2 vγg(v) ) t Since the left has to equal the right, the terms in front of x have to add to 1, and the terms in front of t have to add to 0. Thus: vγ 2 = vγg(v) vγf( v) = 1 g(v)g( v) The first of these yield: g(v) =g( v) =γ while plugging in gamma (after a bit of algebra) yields: We (finally!) have our Lorentz transforms: f(v) =γv/c 2 x = γ x vγ t (43) t = γ t v γ x c2 (44) x = γ x + vγ t (45) t = γ t + v c 2 γ x (46) 37

Note that if γ 1, then these are simply x = x v t, and t = t, just as with the Galilean transforms. Moreover, as with our Galilean transforms, we can combine these to figure out how velocity transforms: u = u v 1 uv c 2 (47) u = u + v 1+ u v c 2 (48) (49) 6.2.2 Some Examples A Lightbeam: Consider what happens if you shine a flashlight (u = c) on your train which is moving, say, at half the speed of light (v = c/2). The speed I observe the lightbeam to be traveling is: A Meter Stick: u = c + c/2 1+ c/2 c c 2 = 1.5c 1.5 = c Now, what happens if you have a meter stick in your ship (which is lying on the floor in the î direction)? You observe it as having a length, of course, of 1 m. I, however, notice something different. Since I m measuring the length at a particular moment, I measure t = 0 (the front and back are measured at the same time). Thus, according to equation 44, x = γ x. Or, if I am measuring the length of the rod in my frame: x = x γ = L γ Since γ is always greater than or equal to 1, I always measure moving meter sticks to be shorter than stationary ones. Not just meter sticks everything. You, your ship, your control panels, and so on. This is known as length contraction. Now here s the wacky thing because of the symmetry of the equations, you will find that a meter stick in my frame (again, oriented along the x-axis) looks short to you. A Clock: What about a clock? Well, imagine that a clock ticks on your ship t =1s. For you, the two ticks of the clock happen in the same place x = 0. Thus, according to equation 46, we have t = γ t. In other words your clock (and your heartbeat, and the oscillations of your molecules and everything else, indeed time itself) seems to run slow on your ship. But again to you, it appears that my clock (and heart, etc.) are running slow! This is known as time dilation. 6.2.3 A Spacetime Example Here s a snapshot of a stationary lab in which a light beam is bouncing back and forth as seen from within the lab, and also as seen from an observer moving at 1/2 the speed of light: 38

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6.3 The Relation of Energy to Momentum Thus far, we ve described how to translate between space and time intervals from one frame to another. And beyond a constant speed of light, we ve introduced very little actual physics into the mix. Where s momentum? Where s energy? I want to begin by reminding you of the fundamental definition of energy: E = v p (50) Remember, we derived this from the Work-Kinetic Energy theorem. Turning this into a derivative, we get: Remind you of anything? de dp = v How about: dx dt = v In other words, if we switch frames, momentum and energy chance. In the transforms, momentum plays the same role as time, and energy plays the same role as position. Thus: E = γe + vγp (51) p = γp + γ v c 2 E (52) But wait! There s more! Consider a particle to be stationary in the primed frame. There sits a particle with some (unknown) energy,e = E 0, and p = 0. Now, do a Lorentz transform into the unprimed frame with a velocity v << c. In this case, γ 1, and p = mv, since after all, we re in the Newtonian limit. Thus, according to equation 52: or, rearranging: mv = 0 + v c 2 E 0 Even the most stoic among you must admit that this is freakin awesome! E 0 = mc 2 (53) But now that we know the rest energy, we can solve for momentum in general. Plugging in the energy to equation 52: Exactly as we asserted in the first place! p = 0 + γ v c 2 mc2 = mvγ (54) (55) Or, using equation 51, we find: E = γmc 2 (56) 6.3.1 An example using an explosion Okay, so perhaps you have a little trouble buying into my math tricks. Who could blame you? To demonstrate how all of this fits together, we ll do a little example. 40

Now, the important thing to remember here is that we physicists love conserved quantities. Momentum and Energy should be conserved no matter who is looking at the experiment. If they re not conserved, we ve done something wrong. Before M After m m v v This is what we call a nuclear bomb. We start with a big mass (at rest), and it explodes into two equal parts (each of mass, m). Now, before getting into detail, the nice thing about this explosion is that it is symmetric. No matter whether you use the relativistic or non-relativistic version of momentum, the explosion seems reasonable since as much of the material flies to the left as to the right. Thus, momentum is conserved. However, this is not necessarily true if we look at the same explosion in a frame moving to the left at speed v: Before (primed frame) M v After (primed frame) m m v =2v/(1+v^2/c^2) Now, if Newton came along and looked at the explosion from this point of view, he d be baffled. The mass of each of the pieces of debris should just be half that of the original, and thus, the total momentum before the explosion is: p (Newton) i = Mv 41

while the total momentum afterwards is: p (Newton) f = M 2 2v 1+v 2 /c 2 which is not the same thing at all! In other words, our traditional interpretation of momentum does not work. We can get around this if we use the form of the momentum: p = mvγ (57) Well, with one more complication. In Einstein s view, the energy of a particle (kinetic+mass energy) is given by: E = mc 2 γ (58) where γ is based on the velocity that the particle is moving in a particular frame. Unsuprisingly, a particle which is at rest according to one observer, and moving according to another, will appear to have a higher energy according to the moving observer. Because the mass energy is so big, it is a huge reservoir for kinetic energy, and thus is no longer conserved. How much mass is lost? Looking at it in the unprimed frame, the initial energy is: (since the bomb is initially at rest γ = 1). The final energy is: E i = Mc 2 E f = 2(mc 2 γ) where the factor of 2 comes from the fact that there are two pieces of debris and the mc 2 γ includes all of the energy (mass energy+kinetic energy). Thus, conservation of energy gives us: 2mc 2 γ = Mc 2 2m M = 1 γ (59) (60) In other words, if the explosion is very mild (small v), then γ will be close to 1, and it looks like mass will be more or less conserved. However, if the pieces fly off at, say, v = c/2, then γ =1.15, and thus, about 13.4% of the mass will be converted into energy! 6.3.2 Putting some numbers in. The unprimed frame: Let s do a concrete example to show that our approach does work. Consider a bomb which has an initial mass of 1kg. Its rest energy is thus: E i =1kg c 2 =9 10 16 J It then explodes into two equal masses, each with a speed of 0.8c, giving us γ =1.67 (plug into the γ equation and verify for yourself!). Thus, according to equation 60, we have: 2m 1kg = 1 1.666 =0.6 In other words, each of the two particles will have a mass of 0.3kg. However, M =0.4kg were converted into energy. 42

Using this, energy (because that s how we derived the mass loss relation) and momentum (because it s zero before and after) are conserved in the rest (unprimed) frame. The primed frame: The primed frame is more complicated. Imagine that we re running to the left at a speed 0.8c. Before the explosion, the big mass is moving within that frame at a speed u M =0.8c (γ =1.67). One of the pieces of debris is at rest, while the other is moving at a speed: which puts γ =4.55. u m = 0.8c +0.8c 1 + 0.8c 0.8c/c 2 =0.976c So, before the explosion, the big mass has a momentum of: Likewise, the energy is: p M = mu M γ = 1kg (0.8 3 10 8 m/s) 1.67 = 4 10 8 kg m/s E M = Mc 2 γ = 1kg (3 10 8 m/s) 2 1.67 = 1.5 10 17 J Now, after the explosion, the stationary mass obviously has zero momentum, while the moving piece has a momentum of: p m = 0.3kg (0.976 3 108 m/s) 4.55 = 4 10 8 kg m/s And momentum is conserved! (As it must be) Likewise, the energy afterwards from both the stationary and the moving pieces are: Tada! E fin = mc 2 + mc 2 γ = 0.3kg (3 10 8 m/s) 2 +0.3kg (3 10 8 m/s) 2 4.55 = 1.5 10 17 J 6.3.3 A real world example You ll notice that the number of Joules, above, tend to be enormous, and thus particle physicists tend to work in units of energy called MeV (mega-electronvolts), and units of mass of MeV/c 2. It cuts down on the exponents. So, consider the following relation (the basic nuclear fusion relation we ve discussed earlier): 4H +2e + 4 He+2ν + γ (61) In this case, we find that the total masses on the left hand side of the equation are considerably less than the masses on the right. m p = 938.3MeV/c 2, m He = 3728.1MeV/c 2, and m e =0.511MeV/c 2. 43

So, E i = 3754.2, E f = 3728.1, so, E = 26.1MeV. Where does it go? It goes into the kinetic energy of the resulting helium atom, and the energy of the neutrinos and photons. 44