EPGY Special and General Relativity. Lecture 4B
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1 Lecture 4B In the previous lecture we found that the proper description of the universe is one consisting of a four-dimensional manifold (space) endowed with a Lorentzian metric, (of course we are restricting discussions to IRFs only and far away from gravitating bodies). The distance between two events in spacetime is called the interval and is determined by the metric equation, s 2 = c 2 t 2 r 2, this is the analog of using the Pythagorean theorem to find distances in space. This type of geometry is called Lorentzian geometry. In this lecture we want to discuss a little more about the properties of spacetime, how to measure events and intervals, and examine some simple examples. When we discuss situations within spacetime we must now put observations in terms of events, and not just places or times. It should be clear that we can not describe a physical system by only saying that something happened at position x. This no longer is a complete description of the state. (Classically, consider a projectile motion problem where we only state how high the object reaches and want to know exactly where it will land. Well we have not specified where it starts from; there is only partial information and the answer can not be determined). Hence, the fundamental measurements that are made are now events, not times and places separately. Along with the invariant interval we have enough to describe object s motions through spacetime. How do we actually measure intervals? (Distances seem obvious, we lay down a ruler and measure them). Obviously you will need something to measure distance, say a meter stick, and another device to measure time intervals, a clock. Consider yourself sitting in your chair, at rest, in your own IRF. All you need to measure intervals along your worldline is a clock. This is always true, no matter how you move within your own reference frame, (inertial or not), you will only need a clock to measure intervals along your own worldline. This is obvious since in your reference frame you always sit at the spatial origin, x = 0 m for your position always. To quote Buckeroo Bonzai Where ever you go, there you are. In order to determine the interval between two points just look at your watch! (Of course you may need to multiply by c to get the units correct). The squared interval between two events which occur at the origin is simply s 2 = c 2 t 2. Now consider two events, which do not occur at the same spatial point, in a particular reference frame. How do you measure the interval between these two events if you always sit at the origin. Classically, this question is never considered to much extent. For if you sit at the origin you must look out and observe events happening at a distance. Since it takes a finite amount of time for light to reach you, (how else are you going to observe the event?), your observations may be skewed in relation to others. To correct for this, there are two possible methods to make sure that you measure the interval accurately: 1) Set yourself in an IRF moving at such a velocity that the two events occur right under your nose. That is, change IRFs to one for which the two events occur at the origin. This is not always convenient. 1
2 2) Get a friend with a synchronized clock to stand at the spatial location where the second event will occur, while you are at the spatial location of the first. Wait. When the first event occurs, you note your time. When the second event occurs, your friend notes the time. You meet later and compare notes to find the interval between the two events to be s = c 2 (t f t you ) 2 (x f x you ) 2. Although all inertial reference frames are equivalent, there is one frame which does deserve special recognition (though it is on no way universally special). This is the frame where the two events occur exactly at the same spatial location in one particular frame. (Like we described above). The time between two events in this one special frame is given a special name, the proper time. Of course, in this frame the proper time is equal to the interval divided by c between the two events. Such a clock used to measure between the two events is called the proper clock. This reference frame is only special for these two events and is equivalent to all other IRFs. Note well what this clock reads for the interval is the same in all other IRFs. [In the example with Alice and Bob, Bob is in a proper frame]. We now examine some specific intervals and introduce some more terminology. Consider two events occurring at exactly at the same instance of time in your reference frame. (An example might be you and a friend setting off firecrackers. You are in San Francisco and your friend is in New York. You have previously synchronized your clocks and set your firecrackers off at the same time, (say New Years in Times Square). When both of you record the events and compare notes you will observe the following data, 2
3 c t = 0 m, x = D (where D is the distance between SF and NY) and s 2 = 0 m - x 2 = - D 2. This may seem strange to you. The squared distance (interval) between two events is negative. You have always been told that distances are always positive, no matter how you measure them. That is the classical, Euclidean space view. In Lorentzian geometry we can have negative squared intervals. Of course this means that s is imaginary, s = D 2 = 1 D id. To answer why this occurs, imagine what reference frame would be required to have these two events occur at the same position, the proper frame. As these two events occur at the same time in this frame, the speed of the proper frame would need to be infinite! More importantly, the frame would be moving greater than the speed of light. Examine the squared interval, s 2 = c 2 t 2 - x 2. By examining the metric equation it is clear that the elapsed time in meters is less than the spatial difference between the two events. Whenever two events are separated by a distance in which a ray of light could not make it in time to reach the second event, the squared interval will always be negative. When two events are separated such that a proper frame would need to travel faster than c, the two events are called spacelike separated. For events for which a proper frame connecting the two needs to travel less than c, the term used for such events is timelike separated. For timelike separated frames the interval is always positive. There exist events in which a light ray can exactly connect the two events - the proper frame travels at the speed of light. Such frames are termed either light-like separated or null-like separated. For light-like separated events the interval is zero. This is since c t = x. These types of intervals segment spacetime into two regions and a dividing line. This division has profound consequences that we will return to later. If event 1 occurs at the origin then the interval between event 1 and 2 will have the following forms shown in the figure depending upon which section event 2 lies in. 3
4 Let s look at an example to see how we apply this invariance. Consider a spaceship that is heading off to Proxima Centauri from Earth. It passes Earth at a time ct = 0 m in both the Earth s reference frame and that of the spaceship. Consider twin astronauts, Alice and Bob, both qualified astronauts but Bob is chosen to join the mission to Proxima Centauri, the nearest star to Earth at a distance of 4.3 light years from Earth, ((ly) that is the distance it takes light to travel in one year). Since Alice and Bob are twins, they are both 20 years old at the time when the ship passes Earth. The ship travels at 90% the speed of light, (2.7 x 10 8 m/s). Let s examine what both Alice and Bob observe during this historic mission. First, Alice calculates the interval that occurs between the two events of Bob leaving Earth and returning. In Alice s frame the distance to P.C. is x = 4.3 ly. Alice calculates it should take Bob t = 4.3 ly/v = 4.3 ly/(0.9 c) = ly/c to reach P.C. A word about units, if it takes light to travel 1 ly in one year, then clearly 1 ly/c = 1 year. Hence, Alice figures it takes Bob 4.78 years to reach Proxima Centauri. Let s change scenes and view what is happening on the ship. Bob is eagerly watching his watch awaiting arrival at Proxima Centauri. When he arrives he notes his watch and dispatches an electronic postcard from Proxima Centauri to Earth that he has arrived at P.C. the two Suns are pretty, wish you were here, and that it only took him 2.08 years to get there. 4.3 years later, Alice receives the postcard and reads that it only took 2.08 years to get there. Alice, always in competition with Bob, is outraged that he made it in 2.08 years though by her own calculations it should take 4.78 years. She thinks, as usual, Bob is trying to upstage her by saying that he can get there quicker, until she recalls her basic NASA relativity training. For then Alice recalls that the interval in her own frame is s 2 = (4.77 ly) 2 (4.3 ly) 2 = 4.34 ly 2. Hence the interval between the two events is 2.08 ly and this is the same in Bob s frame. Alice soon realizes that Bob says it took 2.08 years to reach Proxima Centauri in Bob s frame, since x =0 m in that own frame, and in her own it takes 4.78 years. This result may seem strange but it stems directly from what we have discussed about the invariance of the interval. However now imagine the return trip. After dispatching the postcard, Bob s research team immediately returns back to Earth at the same speed, 0.9c. As this situation is completely symmetric, Alice will calculate that the interval between Bob leaving P.C. and him returning to Earth will again be 2.08 ly. When Bob returns and meets Alice, he once again brags how it only took 2.08 years to get back. Alice begins to explain how, since they agree on the interval, that the time differences will be different, but then they look at each other and notice that Alice looks slightly older, a few wrinkles, a few gray hairs. It then dawns on Alice that she is in fact actually older than Bob by 2x(4.78 y) 2x(2.08 y) = 5.4 years older than Bob. Although they started out as identical twins, now Alice is 5.4 years older. What has happened? Nothing really, but we will come back to this later. The example just described should make it clear that time differences and spatial differences may change in different inertial reference frames. This is a bizarre result that never appears in classical physics. Let s examine this more closely. We stated that the interval is invariant in all IRFs and now we will observe the consequences in a few examples. Time now to examine what is the same in two IRFs and what can differ between two events. To begin, we stated that the interval between two events is the same in all IRFs. Also, by Einstein s first postulate, the laws of physics are the same in all IRFs. In addition, the second 4
5 postulate stipulates that the value of the speed of light is the same in all IRFs. Lastly, the fundamental constants of nature are the same, c, e, G, h (respectively, the speed of light, the fundamental charge of the electron, the universal constant of gravitation, and Planck s constant (relevant for quantum phenomena)) The more interesting question is, what is not the same in different reference frames. What may differ in different IRFs. (Will differ if the frames are moving with respect to each other). - time intervals - spatial distances - velocities - accelerations - forces - electric and magnetic fields - energy - momentum The last items we will discuss in a future lecture. Lack of Rigidity in Special Relativity. The fundamental nature of objects is not as it appears in classical mechanics when considering relativity. Consider for example a very long rod that is made of the strongest steel. Classically we treat this as a solid object that we can manipulate. In order to fully understand the nature of this object we must consider what is holding it together. Electric forces hold this object together. We have been learning here that light, an electromagnetic wave, travels at c. It is also true that the electric force between charged atoms travels at c. Thus the force which holds this object together travels at (or less than) c. The implication is that we can no longer have an absolutely rigid object. If one side of this rod is bumped, pushed, or moved, the effect of that is not known to the other side of the rod until the force makes it down there. Thus the object can warp if sufficient force is applied. To repeat; there is no such thing as a completely rigid object. Twin Paradox In our discussion of time dilation we presented the gedankenexperiment with Alice and Beth traveling to Proxima Centauri. Now we will fast forward 100 years and consider two other identical twins, Bob and Bill. Bob gets in a new super fast spaceship which travels at 75% the speed of light and heads off to Pollux in the constellation of Gemini, about 30 ly away. You probably think that all of these effects we ve been discussing are strange but they ve all been observed in the laboratory. You may have thought that observers moving relative to each may see each other s clocks run slow but that when one stops, and they are at rest with each other, the clocks match up again, i.e. the observed variation in clock rates is just that, something that is observed but is not really occurring. The answer to that is no; it is a real effect and time does pass more slowly in a moving frame. This leads us to a famous paradox (in quotes since there is a resolution), known as the twin paradox. This leads us back to Bob and Bill. Bill is an accountant who spends his entire life on Earth. Bob is an astronaut and is slated for the flight to Pollux. Bob s flight begins when they are 5
6 both 20 years old. The flight plan calls for Bob to fly directly to Pollux at 0.75c spend a few days and return at the same rate. The calculations to find the change in the clocks proceeds exactly as for the previous case. In Bill s frame it takes Bob t = D/v = 30 ly/0.75 c = 40 y to reach Pollux. So Bill has to wait 80 years for his brother to return. From time dilation, we see that Bill observes that Bob s clock runs slow and during the trip only, t ' = t 1 v2 c 2 = 80 y 1 (0.75)2 = 80 y(0.66) = 53y passes. So when Bob arrives the identical twins will no longer be identical, Bill will be 100 y old and Bob will be 73 y old. This is a strange result but there is no apparent paradox until we consider the problem from Bob s frame of reference. In Bob s frame, Bill and the Earth zoom off at 75%c and he returns back to Bob at 75%c. Again to Bob, it takes Bill 40 y to go off and 40 y to return. 80 y will have passed on Bob s clock and he applies the time dilation formula to see that only 53 y has passed on Bill s clock. So when they arrive back together, from Bob s perspective, Bob will be 100 y old and Bill will be 73 y old. How can this be? It is not possible for them to have two different ages standing next to each other. This is the statement of the twin paradox. Resolution of the twin paradox. So what is the answer? Whose perspective is correct? Is Bob 100 y old or Bill? The resolution is simple but the actual movements of the clocks are not so simple to spell out. The answer is that Bill s frame is the correct one and that after they return, Bill is 100 y old and Bob is 73 y old. The reason is that Bob s frame is an IRF during the entire scenario. Bob is not always in the same IRF. Depending on how he turns around, there will be a time when he is not in an inertial frame so these results do not apply. Bob s frame is the proper frame for these two events, (departure and arrival). We will explain this resolution further when we discuss spacetime maps. 6
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