Lecture 5. The Lorentz Transformation

Transcription

1 Lecture 5 The Lorentz Transformation We have learned so far about how rates of time vary in different IRFs in motion with respect to each other and also how lengths appear shorter when in motion. What we want to do now is to develop a set of equations that will explicitly relate events in one IRF to a second IRF. This will allow us to quantify more complex events between uniformly moving frames. The Galilean Transformation Before proceeding to the relativistic transformation we review the classical, Galilean, transformations for two reference frames in motion with respect to each other. In both the following and the relativistic scenario we will deal with references frames that are in the standard configuration, that is, the origins coincide at time t = t = 0 and all axes are collinear (x is collinear with x, y with y, and z and z ). Two reference frames, S and S, are in relative motion at constant velocity. From the perspective of frame S, frame S moves in the positive x direction at speed v. (This same scenario will be used for the relativistic derivation). First, the perpendicular directions, y and z, are unaffected by the relative motion since the two frames are at rest with respect to each other along these directions. Thus, we have that y = y and z = z. In classical physics there is a universal time that ticks at the same rate for all observers. Thus in standard configuration we have t = t. For the remaining relation we consider an event, 1, in frame S given by the coordinates (x, y, z, t ). At the time t = t = 0, the position of x and x coincide and as the frame S moves off to the right the origin is measured to be at position x = vt when the event occurs and additional displacement is that observed at time t = 0, i.e. x. Thus when event 1 occurs the position as measured by S is x = x + vt. To complete the connection we note that t = t and this yields the Galilean transformation equations: x = x + vt y = y z = z t = t Thus given a point in the frame S these equations will give the position in S if the frames are in standard configuration (if they are not it is easy to translate the two frames). If we have the inverse scenario given a point in S and want to find it in frame S, we use the inverse Galilean transformation. This is easily found by setting v to v in the above equations and switching primed and unprimed coordinates: x = x - vt y = y z = z t = t Now that the classical result is understood we proceed to the relativistic case. The Lorentz Transformation The problem is the following, given a point in frame S, specified by the coordinates x and t, how does this point map into the frame S. That is, what are the coordinates; x(t,x ) and t(t,x )? (Again noting that perpendicular coordinates are unaffected, y = y, z = z ). To begin, we will examine a spacetime diagram representing the question at hand in frames S (on the left) and S (on the right). The goal is to express the point P, where we are given the x and ct coordinates, in the S frame. 1

2 2 EPGY Special and General Relativity

3 Combining these results we have the full Lorentz transformation, x = (x + vt ) y = y The Lorentz Transformation z = z (given point in S, these give the point in S) t = (t + vx /c 2 ) And just as before, the inverse transformation (if given a point in S) is found by setting v = -v. x = (x - vt) y = y The Inverse Lorentz Transformation z = z (given point in S, these give the point in S ) t = (t - vx/c 2 ) To Boldly Go Where No Object Can Ever Go. You are no doubt familiar with the old television and movie series Star Trek in which the starship Enterprise hurtles through space at speeds far exceeding that of light. We now want to show why this will always remain purely fiction unless we give up a basic philosophical tenet concerning cause and effect. There are several hints we ve already seen telling us that c is the highest speed that can be achieved by anything. Lets examine these results. We saw that time is observed to pass more slowly for moving objects, Explicitly t = t, and if an object travels at c with respect to frame S we see that time appears to stop for the moving object t = t =. It takes an infinite amount of time for the moving clock to tick. Now what happens if v > c? Well then, = 1 v2 c 2 1/ 2 = 1 v 2 c 1 2 1/2, v 2 c 2 > 1 the factor g becomes imaginary. Imaginary numbers can not represent real physical quantities so that t ~ i has no meaning. Similarly the observed length becomes imaginary as well. So we see that special relativity can not incorporate objects which move faster than light. This does not forbid objects traveling faster than c but just says that the theory of special relativity can not describe them. Such objects are called tachyons. In order to prove that objects can t move faster than light we will have to show that it would cause problems. To do this lets devise another gedankenexperiment, this one is a pretty farcical one but will prove the point. (See Box L-1 of your text to see an equivalent scenario involving a Star Trek theme). Our gedankenexperiment posits a alien race of beings who are very large and inhabit deep space. They also enjoy sports, especially American Football which they ve picked up on broadcasts. Owing to their immense size, they have developed their own version with some modifications. First, the size of the playing field is 4 light years long. They also require a large number of referees to establish reference frames to view the action. These creatures all travel 3

4 and throw the ball at speeds less than the speed of light. One year a young recruit is drafted for one of the teams as a quarterback. He is a very talented player who can throw the football at speeds faster than light, up to 3c. Thinking this will ensure an easy route to the Universe Bowl the team pays the young recruit lots of money. However, on his first play from scrimmage there is a problem, and the young recruit never manages to score a touchdown and becomes washed up within the first year. Let s see if we can determine why he is forced out. Let s examine one play where this team is at the 0 ly line. The young gun receives the hike of the ball, waits an appropriate amount of time, and then hurtles a bomb to the streaking receiver who runs at 0.6 c. The receiver catches the ball at the 3 ly line, runs into the end-zone and scores. The team celebrates until they see one referee throw a flag indicating a penalty. The referee states that more than one ball was used in the play, a 1 ly penalty, in this case, half the distance to the goal. To analyze this play let s plot the worldlines of the quarterback, the receiver, and the ball. Also, let s determine beforehand what the value of the dilation factor will be for the reference frames of the quarterback and the receiver, = 1 = = Let s plot the events in the quarterback s frame. Let s list the events in both frames via the Lorentz transformation equations, QB WR x(ly) t(y) x (ly) t (y) Start of play: Throw ball: Catch ball: With these values we can now plot the play in the wide receiver s frame. Notice the worldline of the football. It travels backwards in this frame! Notice at time t = -0.5 y there are 3 balls! For this is after the catch in the WR frame, before the QB throws the ball, AND the ball is en route to the receiver. 4

5 Notice that to make a call on this play, there needs to be a series of referees all running at the same speed as the receiver. Each referee sees a ball near him, and when they later confer they all say that the ball was in front of them at one particular time. This violates conservation of energy, (there are now 3 balls in existence in this frame as opposed to one in the QB frame). This also violates causality, the result (the WR catching the ball), occurs before the QB throws it. Imagine if this was some kind of gun which shot bullets which traveled faster than c. The WR would get shot before the gun was fired in one frame and after in another. Cause and effect no longer holds. Hence, to preserve causality, no thing (not even information), can travel faster than c. Spacetime Maps In a previous lecture we examined objects traveling faster than c by viewing events in different reference frames. Now we are going to examine spacetime maps in more detail. What we will find is a new, easy, way to examine events in many different reference frames. This method will also display the strange hyperbolic geometry of Lorentz spacetime. To begin, we go back to the example of the rocket ship moving at 0.4c with the light clock in it. Remember that in the rocket frame the emission and absorption of a light pulse occurs at the same spatial location. In the laboratory frame we concluded that the time must run slower in the rocket because of the path traced out by the light pulse is longer. We show the two spacetime diagrams here and remind you that the interval between the two events is the same. 5

6 Alice s frame s 2 = 1m 2 = (ct) 2 (x) 2 =(ct) 2 [1 (x)2 (ct) 2 ] Bob s frame s 2 = 1m 2 = (ct ') 2 = c 2 (3.3ns) 2 =(ct) 2 [1 v2 c ] = 2 (ct) (t) 2 =c 2 (3.6 ns) 2 Now consider a third frame, the super rocket frame, which is traveling in the same direction as the rocket but at 0.8c. In this frame the light pulse of the rocket and laboratory will move in the opposite direction. This being an IRF as well, the interval between these two events (emission and absorption) will be the same. We plot this frame as well. s 2 = 1m 2 = (ct '') 2 (x'') 2 =(ct '') 2 [1 v sr 2 c 2 ] = (ct '') (t '') 2 =c 2 (5.6 ns) 2 By placing these three diagrams within one diagram, since they have the same origin, we begin to notice a pattern, 6

7 We can also plot these events in any other IRF we can think of which moves along the x axis and shares the same origin. Again the interval is 1 m in all of these, By filling these in a line emerges, This line traces out all points which have the interval equaling 1 m with respect to the origin. This line is in the shape of, (the upper half), a hyperbola. The equation for a hyperbola is a 2 = x 2 y 2, where a is the point of closest approach to the origin. (To compare a 2 = x 2 + y 2 is the equation of a circle and 1 = x 2 /a 2 + y 2 /b 2 is the equation for an ellipse). 7

8 What we see is that to describe two events in any frame we can draw the invariant hyperbola to find the interval or event in any other IRF. This plot tells us is what the (x,t) event location will be in various frames. Note that the frame in which the two events occur in the same place, the proper frame, has the shortest time interval between two events. In all other frames, clocks appear to run slower, giving a larger time interval. Just to be complete, if all of these frames are coincident at the reception of the light pulse we could describe these events in all frames using the lower branch of the hyperbola. Now we want to discuss worldlines in more detail. We will stick to objects moving in one dimension for simplicity. Note that at no time does the particle go beyond 45 degrees of slope, that would correspond to faster than light travel. We examine now our light clock traveling in spacetime starting at the origin, heading out some distance, and then returning to the x = 0 point. This is similar to the discussion of the twin paradox before, but now we want to go into more detail about the proper time indicated on the light clock. Each mark indicates one roundtrip of a light pulse for each light clock. In the laboratory frame, the interval between events (two receptions) in the rocket can be found easily as s 2 = (ct ) 2 (x) 2. But since the two events occur at the same place in the rocket we see that this interval is the same as the proper time measured by the rocket s light clock, (or one tick of that clock). Note that the rocket is not always a free float frame, (IRF), since it accelerates at times, (its path is curved). We need to be careful with neighboring events not joined by a straight line; in this interval the frame is accelerating and the principles we ve set up do not need to hold. To get around this we will assume that all of the receptions are close enough together such that they are all connected by straight worldlines. We cut up the path into small segments. When the rocket returns home at event B the light clock will have read off the total proper time elapsed during its journey. To other reference frames, like the lab, this proper time will equal the cumulative interval of all the ticks along the path, i.e. taking the interval between 0-1, 1-2, 2-3, etc. and adding them up. All IRFs will agree on this cumulative interval which equals the elapsed proper time. The only way the lab frame finds the total proper time is to add up these segments separately. Finding the interval between O and B directly will not work. An analogy might suffice here. Consider a car traveling between two cities. The distance traveled by the car can easily be read off of the odometer. To an observer not in the car to measure the distance is not so easy. Simply measuring the distance between the 8

9 cities will not give the correct result since the road bends and twists. The only way for the observer to measure the distance the car has traveled is to cut up the path into small segments and add up the distance of the segments. Then the observer will get the same result as the car s odometer. We see that finding the interval between two events can vary if the path or worldline between the two events is curved. Before we only examined IRFs and we saw that there was only one IRF which can have the events occur at the same location. (Assuming that the events are timelike separated). But if we have a worldline which is not straight we will get a different result for the proper time. The proper time for the curved worldline will not equal the proper time for an IRF connecting the same two points. This is the same as in flat space (not spacetime), the shortest distance between two points is a straight line but there are infinitely many longer curved paths which can connect the two points. The same occurs here except the metric throws in a twist. What we see is that the straight worldline has the longest proper time. s 2 = (ct ) 2. Any other frame will require summing the end segments, s 2 = (ct i ) 2 (x i ) 2. Since we are subtracting off the i=start spatial part we see that it will be less than the straight worldline. Again, the straight worldline has the longest proper time. It is important to note that the total proper time along a worldline is an invariant. It has the same value in all IRFs, (since it is essentially the interval). What we see is that if we have a set of identical siblings and all but one sets off on a space journey and they all meet later, the one with the straight worldline, (the one who stays at home), will be the oldest. Consider a particle in deep space and picture its worldline between two events. If no outside forces act on this particle, (it s a free particle) it will not accelerate. Hence in special relativity it will have a straight worldline. Hence it will follow a worldline of maximal aging. This is raised to a principle, The Principle of Maximal Aging. This is not only true in deep space (in an IRF) but also in a region near a gravitating mass. In such regions we will need to use General Relativity to examine systems. This principle provides a bridge between SR and GR. The proper time (cumulative interval) is a fundamental method of comparing different worldlines between events. One last point to make is regarding the stretch factor and IRFs. Remember that we found that the time dilates between two uniform moving frames. t = t Now if one of these is the proper frame (indicating the proper time by ) we see we can define as follows, = t = 1 1 v2 c 2 9 = timeinframe proper time (Again this is for IRFs, if the worldline is curved we will need to cut up the worldline into segments where each segment is approximately an IRF). We can easily find the stretch factor if we know the proper time. Say a rocket goes by and you see 1 second go by on its clock while yours ticks off 3 seconds. Well then = 3s/1s = 3, from which you can invert to find the speed of the other clock. 1 3 = 1 2 1/ 9 = 1 2 v c = 8 = Hence can be used as a measure of speed. 9

10 Transforming Between Frames, Spacelike Separated Events We continue our discussion of spacetime maps and translating between reference frames. Examine the following plot for IRFs all observing the same interval between two events. This line gives all of the timelike separated frames. [The following refers back to the discussion in Box L-1 of your text, pg 108]. Remember back to our discussion of the object moving faster than the speed of light, discussed in Box L-1 of your text. The squared interval between the throw and reception is negative, Lab frame. x = 0 x = 3 ly t = 0 t = 1 ly from which s 2 = c 2 (1 y) 2 (3 ly) 2 = -8 ly 2. This was the same in the shuttle s frame (traveling at 0.6 c), x = 0 x = 3 ly t = 0 t = -1 y from which s 2 = - 8 ly 2. And now consider a frame traveling at 0.3c then, x = 0 x = 8 ly t = 0 t = 0 y from which s 2 = - 8 ly 2. The events occur simultaneously in this frame. We see that for spacelike separated events there is an invariant hyperbola lying outside the light cone along the x axis. And note now that transforming between these frame the temporal order between events can switch. (And for timelike separated events the spatial order can switch). For timelike and spacelike separated events there is a barrier, the speed of light, or the lightlike events. So that the type of separation between events is invariant. (timelike for one, timelike for all). Timelike (interval) 2 + Spacelike (interval) 2 - Lightlike (interval) 2 0 In three dimensions the lightcone creates a barrier, a natural partitioning of spacetime. s 2 = c 2 t 2 - x 2 -y 2 -z 2. Note that spacelike separated events can not be causally related, (no information can pass from one event to the other). 10