Lecture 3 - Lorentz Transformations

Transcription

1 Leture - Lorentz Transformations A Puzzle... Example A ruler is positioned perpendiular to a wall. A stik of length L flies by at speed v. It travels in front of the ruler, so that it obsures part of the ruler from your view. When the stik hits the wall it stops. In your referene frame, the stik is shorter than L. Therefore, right before it hits the wall, you will be able to see a mark on the ruler that is less than L units from the wall. But in the stik s frame, the marks on the ruler are loser together. Therefore, when the wall hits the stik, the losest mark to the wall that you an see on the ruler is greater than L units. Whih view is orret (and what is wrong with the inorret one)? Solution The first reasoning is orret. You will be able to see a mark on the ruler that is less than L units from the wall. (In fat, you will atually be able to see a mark even loser to the wall than L, as we ll show below). The main point of this problem (and many other ones) is that signals do not travel instantaneously. The bak of the stik does not know that the front of the stik has ome into ontat with the wall until a finite time has passed. Let s be quantitative about this and alulate (in both frames) the losest mark to the wall that you an see. Consider the lab referene frame. The stik has length L. Therefore, when the stik hits the wall, you an see mark a distane L from the wall. You will, however, be able to see a mark even loser to the wall, beause the bak end of the stik will keep moving forward, sine it doesn't yet know that the front end has hit the wall. The stopping signal (shok wave, et.) takes time to travel. Let's assume that the stopping signal travels along the stik at speed. (You an work with a general speed u. But the speed is simpler, and it yields an upper bound on the losest mark you an see.) Where will the signal reah the bak end? Starting from the time the stik hits the wall, the signal travels bakwards from the wall at speed, and the bak end of the stik travels forwards at speed v (from a point L away from the wall). So the relative speed (as viewed by you) of the signal and the bak end is + v. Therefore, the signal hits the bak end after a time L/. +v During this time, the signal has traveled a distane L/ from the wall. The losest point to the wall that you an +v see is therefore the

2 2 Leture nb L + v = mark on the ruler where we have used the notation L (+β) = L -β +β /2 () β v (2) Now onsider the stik's referene frame. The wall is moving to the left towards it at speed v. After the wall hits the right end of the stik, the signal moves to the left with speed, and the wall keeps moving to the left with speed v. Where is the wall when the signal reahes the left end? The wall travels v as fast as the signal, so it travels a distane L v in the time that the signal travels the distane L. This means that it is L - v away from the left end of the stik. In the stik's frame, this orresponds to a distane L - v on the ruler (beause the ruler is length-ontrated). So the left end of the stik is at the mark on the ruler, in agreement with the result we found above. Lorentz Transformation Introdution L ( - β) = L -β +β /2 () Consider a oordinate system, S', moving relative to another system, S. Let the onstant relative speed of the frames be v. Let the orresponding axes of S and S' point in the same diretion, and let the origin of S' move along the x-axis of S, in the positive diretion. Nothing exiting happens in the y and z diretions (sine there is no transverse length ontration), so we ll mostly ignore them. Our goal is to look at two events (an event is simply anything that has spae and time oordinates) in spae-time and relate the Δx and Δt of the oordinates in one frame to the Δx' and Δt' of the oordinates in another. We therefore want to find the onstants A, B, C, and D in the relations, Δx = A Δx' + B Δt' Δt = C Δt' + D Δx' The four onstants here will end up depending on v (whih is onstant, given the two inertial frames). We have assumed in the above equation that Δx and Δt are linear funtions of Δx' and Δt'. And we have also assumed that A, B, C, and D are onstants (that is, dependent only on v, and not on x, t, x', t'). The first of these assumptions is justified by the fat that any finite interval an be built up from a series of many infinitesimal ones. But for an infinitesimal interval, any terms suh as, for example, (Δt) 2, are negligible ompared to the linear terms. Therefore, if we add up all the infinitesimal intervals to obtain a finite one, we will be left with only the linear terms. Equivalently, suppose everyone deided that instead of meter stiks and seonds, we should all use half-meter stiks and half-seonds. In suh a world, although all lengths and times would be doubled, speeds would remain the same and we would expet that all of the relativity results we have seen so far will be the same (beause of the postulate that spae is homogeneous). However, if there were any non-linear terms in the Lorentz transformations, we would be able to distinguish these two senarios. For example, Δx = A Δx' + B Δt' + E Δx' Δt' using meters and ()

3 Leture nb distinguish example, using seonds would beome 2 Δx = 2 A Δx ' + 2 B Δt ' + E Δx ' Δt ' in half-seonds (Δt' = 2 Δt ', Δt = 2 Δt ) and halfmeter stiks (Δx' = 2 Δx ', Δx = 2 Δx ) and therefore the Lorentz transformation equation would be written as Δx = A Δx ' + B Δt ' + 2 E Δx ' Δt ' whih violates the postulate that spae is homogeneous. (The astute reader may be asking, "What if we use full seonds and half-meter stiks (i.e. Δt' = Δt ', Δt = Δt, Δx' = 2 Δx ', Δx = 2 Δx )?" In suh a ase, all veloities would also double, so upon substituting the four relations together with v = 2 v and = 2 we one again regain the same form of the Lorentz transformations.) The seond assumption - that A, B, C, and D are onstants - an be justified in various ways. One is that all inertial frames should agree on what non-aelerating motion is. That is, if Δx' = u' Δt', then we should also have Δx = u Δt, for some onstant u. This is true only if the above oeffiients are onstants. Another justifiation omes from the seond of our two relativity postulates, whih says that all points in (empty) spae are indistinguishable. With this in mind, let us assume that we have a transformation of the form, say, Δx = A Δx' + B Δt' + E x' Δx'. The x' in the last term implies that the absolute loation in spae-time (and not just the relative position) is important. Therefore, this last term annot exist. Before embarking on atually finding the oeffiients in the Lorentz transformation, onsider what they would be for the usual Galilean transformation (whih are the ones that hold for everyday relative speeds, v). Then we would have Δx = Δx' + v Δt and Δt = Δt' (that is, A = C =, B = v, D = 0). We will find, however, under the assumptions of Speial Relativity, that this is not the ase. The Galilean transformation is not the orret transformation. But we will show below that the orret transformation does indeed redue to the Galilean transformation in the limit of slow speeds, as it must. Finding the Lorentz Transformation The four onstants A, B, C, D are four unknowns, and we an solve for them by using the following four effets: Effet Condition Result Time Dilation Δx'=0 Δt = Δt' Length Contration Δt'=0 Δx' = Δx/ Relative v of Frames Δx=0 Δx' = -v Δt Head Start Effet Δt=0 Δt' = -v Δx'/ 2 Most of the time you will see these equations (and all equations dealing with Lorentz transformations) without the Δ symbols in front. That makes the equations look leaner, but in suh ases it must be understood that x really means Δx and so on. We are always onerned with the differene between oordinates of two events in spaetime. The atual value of any oordinate is irrelevant, beause there is no preferred origin in any frame. Now would be a great plae to pause and make sure that you understand the above four effets and why they lead to their orresponding results. For example the negative sign in the Head Start Effet orresponds to the fat that the rear lok on a train is ahead (i.e. the front lok shows less time). Thus the lok with the higher x' value shows the lower t' value. (In other words, suppose a train is moving to the right with speed v in frame S. Let the first event be "you read the front lok" and the seond event be "you read the rear lok." Although in frame S you do these simultaneously (Δt = 0), in frame S' your ations would be seen as having Δt' < 0 whih, supposing

4 Leture nb you simultaneously ( ), your having supposing that the loks are synhronized in frame S', signifies that the person in S first read the front lok and then read the rear lok.) Substituting in the four onditions and their orresponding results into the Lorentz transformation formula Δx = A Δx' + B Δt' Δt = C Δt' + D Δx' yields our four onstants. Time Dilation yields C =, Length ontration yields A =, Relative v of Frames yields B = v and hene B = v, and the Head Start Effet yields D = v and hene D = v. Therefore, the Lorentz A C 2 2 transformation, in all its glory, is Δx = (Δx' + v Δt') Δt = Δt' + v Δx' 2 (6) Δy = Δy' Δz = Δz' where =. We have taked on the trivial transformations for Δy and Δz, but we won t bother writing these - v2 2 /2 in the future. We ould solve the above equations for the inverse relationships of Δx' and Δt' in terms of Δx and Δt Δx' = (Δx - v Δt) Δt' = Δt - v Δx 2 Of ourse, these "inverse relationships" simply depend on your point of view. But it s intuitively lear that the only differene between the two sets of equations is the sign of v, beause S is simply moving bakwards with respet to S'. The Fundamental Effets Let us now see how the Lorentz transformations imply the three fundamental effets (namely, the loss of simultaneity, time dilation, and length ontration). Of ourse, we just used these effets to derive the Lorentz transformation, so we know everything will work out. We ll just be going in irles. But sine these fundamental effets are, well, fundamental, let s belabor the point and disuss them one more time, with the starting point being the Lorentz transformations. Loss of Simultaneity Let two events our simultaneously in frame S'. Then the separation between them, as measured by S', is (Δx', Δt') = (Δx', 0). Using the seond of Equation (6), we see that the time between the events, as measured by S, is Δt = v Δx' 2. This is not equal to zero (unless x' = 0). Therefore, the events do not our simultaneously in frame S. Time Dilation Consider two events that our in the same plae in S'. Then the separation between them is (Δx', Δt') = (0, Δt'). Using the seond of Equation (6), we see that the time between the events, as measured by S, is Δt = Δt' (if Δx' = 0) (8) The fator is greater than or equal to, so Δt Δt'. The passing of one seond on S' s lok takes more than one seond on S s lok. S sees S' drinking his offee very slowly. The same strategy works if we interhange S and S'. Consider two events that our in the same plae in S. The separation between them is (Δx, Δt) = (0, Δt). Using the seond of Equation (7), we see that the time between the events, as measured by S', is (5) (7)

5 Leture nb 5 Therefore, Δt' Δt. Δt' = Δt (if Δx = 0) (9) Note: If we write down the two above equations by themselves, Δt = Δt' and Δt' = Δt, they appear to ontradit eah other. This apparent ontradition arises beause we usually omit the onditions they are based on. The former equation is based on the assumption that Δx' = 0. The latter equation is based on the assumption that Δx = 0. Length Contration This proeeds just like the time dilation above, exept that now we want to set ertain time intervals equal to zero, instead of ertain spae intervals. We want to do this beause to measure a length, we simply measure the distane between two points whose positions are measured simultaneously. That s what a length is. Consider a stik at rest in S', where it has length l'. We want to find the length l in S. Simultaneous measurements of the oordinates of the ends of the stik in S yield a separation of (Δx, Δt) = (Δx, 0). Using the first of Equation (7), we have Δx' = Δx (if Δt = 0) (0) But Δx is by definition the length in S. And Δx' is the length in S', beause the stik is not moving in S'. Therefore, l = l'. Sine, we have l l', so S sees the stik shorter than S' sees it. Now interhange S and S'. Consider a stik at rest in S, where it has length l. We want to find the length in S'. Measurements of the oordinates of the ends of the stik in S' yield a separation of (Δx', Δt') = (Δx', 0). Using the first of Equation (6), we have Δx = Δx' (if Δt' = 0) () But Δx' is by definition the length in S'. And Δx is the length in S, beause the stik is not moving in S. Therefore, l' = l, so l' l. Note: As with time dilation, if we write down the two above equations by themselves, l = l' and l' = l, they appear to ontradit eah other. But as before, this apparent ontradition arises from the omission of the onditions they are based on. The former equation is based on the assumptions that Δt = 0 and that the stik is at rest in S'. The latter equation is based on the assumptions that Δt' = 0 and that the stik is at rest in S. They have nothing to do with eah other. A Further Note: Here is a wrong way to try and pull out length ontration from the Lorentz transformation. Suppose a stik is at rest in S', so that you measure its length to be (Δx', Δt') = (Δx', 0). What is the length in S? The first of Equation (6) suggests that it is Δx = Δx', but that is the wrong relation (it predits that the stik will expand rather than ontrat). What went wrong? Using the seond of Equation (6), we find Δt = v Δx', whih 2 shows that the measurement in frame S was not done at the same time, whih is why the orret length of the stik was not measured. This highlights the subtlety required when using the Lorentz transformation - you must ensure that you use the right assumptions for your setup. Rather than using the Lorentz transformation, it is often easier to double hek your answer by merely onsidering Length Contration, Time Dilation, and the Head Start Effet. Comments on the Lorentz Transformation. The plus or minus sign in the Lorentz transformations orresponds to how the oordinate system on the lefthand side sees the oordinate system on the right-hand side. If you ever get onfused about how they are related for two frames A and B, simply write down Δx A = (Δx B ± v Δt B ) and then imagine sitting in system A and looking at a fixed point in system B. This point satisfies Δx B = 0 so that Δx A = ± v Δt B. If the point moves to the right (i.e. if Δx A inreases with time) then pik the "+" sign; if it moves to the left, pik the "-" sign.

6 6 Leture nb 2. In the limiting ase v, the Lorentz transformation equations redue to Δx = Δx' + v Δt + O v 2 Δt = Δt' + O v 2 (2) whih as disussed above is the Galilean transformation. This must be the ase, beause we know from everyday experiene (where v ) that the Galilean transformations work just fine.. Often times, you will see the Lorentz transformation written in the more symmetri form Δx = (Δx' + β ( Δt')) Δt = (( Δt') + β Δx') where β = v. The Lorentz transformation an also be written as a symmetri matrix Δx Δt = β β Δx' Δt' (). One important point is that we must hek is that two suessive Lorentz transformations (from S to S 2 and then from S 2 to S ) again yield a Lorentz transformation (from S to S ). This must be true beause we showed that any two frames must be related by the Lorentz transformation equations. If we omposed two Lorentz transformations and found that the transformation from S to S was not a Lorentz transformation for some new v, then the whole theory would be inonsistent, and we would have to drop one of our postulates. We an easily prove that the omposition of two Lorentz transformations is a two Lorentz transformation using Mathematia and the above matrix form. L = L2 = Lfin = - β 2, - β2 2, - β 2, β - β 2, β2 - β2 2, β - β 2, β - β 2, β2 - β2 2, β - β 2, - β 2 ; - β2 2 ; - β 2 ; Solve[Lfin == Simplify[L.L2, Assumptions 0 < β < && 0 < β2 < ], β] β β + β2 + β β2 (Reall that β = v.) This shows that after using one Lorentz transformation with veloity v and then another Lorentz transformation with veloity v 2, then the initial and final frames are related by a Lorentz transformation with veloity v = v +v 2 v v2. We will derive this result again as a "Veloity Addition Formula" setion below One of the inredible disovery that we an easily hek with the Lorentz transformation is that () (Δx) 2-2 (Δt) 2 = (Δx') 2-2 (Δt') 2 (5) In other words, subtrating off the square of the time differene (times ) from the square of the distane yields an invariant quantity! As you an imagine, this symmetry is immensely useful in problems. Δx = (ΔxP + v ΔtP); Δt = ΔtP + v 2 ΔxP ; FullSimplify - 2 Δt 2 + Δx 2 /. - 2 ΔtP 2 + ΔxP 2 - v2 2

7 Leture nb 7 Using the Lorentz Transformation Example A train with proper length L moves with speed 5 with respet to the ground. A ball is thrown from the bak of the train to the front. The speed of the ball with respet to the train is. As viewed by someone on the ground, how muh time does the ball spend in the air, and how far does it travel? Solution The fator assoiated with the train s speed 5 is =. The two events we are onerned with are ball leaving 2 bak of train and ball arriving at front of train. The spae and time separation between these events is easy to alulate in the train s frame, where Δx T = L and Δt T = L. The Lorentz transformation giving the oordinates on the ground is Therefore, = L / Δx G = (Δx T + v Δt T ) Δt G = Δt T + v Δx T 2 Δx G = 2 L + 5 Δt G = 2 L + 5 L = 7 L L = L 2 Of ourse, we ould have gotten this answer using our tatis from the previous letures; the Lorentz transformation ontains no new information. For example, we ould ompute that the time interval seen on the ground would be the time dilation of the time it takes on the train - taking into aount the Head Start effet between the two loks, Δt G = Δt T + L v = L + L 5 = L. After that, the distane traveled by the ball as seen by the ground would equal the length of the ontrated train ( L ) plus the extra distane traveled by the train while the ball is in the air (v Δt G ) to yield the total length Δx G = L + v Δt G = 2 L + 5 L = 7 L. There are other ways to ompute these values, but all of them are muh more painful then solving the problem in the train frame and then plugging and hugging into the Lorentz transformation equations. (In most problems, there will be one frame where the values are extremely easy to ompute, so this strategy works brilliantly!) Veloity Addition Formula Complementary Setion: Longitudinal Veloity Addition Consider the following setup. An objet moves at speed v with respet to frame S'. And frame S' moves at speed v 2 with respet to frame S, in the same diretion as the motion of the objet. What is the speed, u, of the objet with respet to frame S? (6) (7) The Lorentz transformation may be used to easily answer this question. The relative speed of the frames is v 2.

8 8 Leture nb may easily question. speed Consider two events along the objet's path (for example, say it makes some beeps). We are given that Δx' = v Δt'. Our goal is to find u Δx. The Lorentz transformation from S' to S is Δt Δx = 2 (Δx' + v 2 Δt') (8) Δt = 2 Δt' + v 2 Δx' 2 (9) where 2 = - v22 2 /2. Therefore, u Δx = Δt Δx' +v Δt' 2 Δx' Δt' + v2 2 = v +v 2 v v2 + This is the veloity-addition formula (for adding veloities in the same diretion). Let s look at some of its properties. It is symmetri with respet to v and v 2, as it should be, beause we ould swith the roles of the objet and frame S. For v v 2 2, it redues to u v + v 2, whih we know holds perfetly well for everyday speeds. If v = or v 2 =, then u =, as should be the ase, beause anything that moves with speed in one frame moves with speed in another. The maximum (or minimum) of u in the region - v, v 2 equals (or -), whih an be seen by noting that the partial derivatives u = - v 2 2 and u = - v 2 2 are never zero in the interior of the region. v + v v v 2 + v v If you take any two veloities that are less than, and add them aording to the above equation, then you will obtain a veloity that is again less than. This shows that no matter how muh you keep aelerating an objet (that is, no matter how many times you give the objet a speed v with respet to the frame moving at speed v 2 that it was just in), you an t bring the speed up to the speed of light. We ll give another argument for this result when we disuss energy. Let s onsider when the veloity-addition formula is needed. Consider the following two senarios. 2 (20) The veloity-addition formula is required in both the upper and lower senario when our goal is to find the speed of A with respet to C. More generally, the veloity-addition formula applies when we ask, "If A moves at v with respet to B, and B moves at v 2 with respet to C (whih means, of ourse, that C moves at speed v 2 with respet to B), then how fast does A move with respet to C?" The veloity-addition formula does not apply if we ask the more mundane question, "What is the relative speed of A and C, as viewed by B?" The answer to this is simply v + v 2 in the bottom senario, but for the top senario we would have to transform into B's frame, where C would have veloity v 2 pointing to the left, but to determine A' s veloity whih would require using the veloity-addition formula.

9 Leture nb 9 Complementary Setion: The Sign of the Veloity Addition Formula Equal Speeds Example A and B travel at and 5 5, respetively. How fast should C travel between them, so that she sees A and B approahing her at the same speed? What is this speed? Solution Method : Suppose C is traveling with v, then the A s veloity in C s frame would equal 5 -v - 5 v 2. (In other words, this would be what the veloity of A would look like for an observer on a train moving to the right at veloity v.) We have defined positive veloity to the right, so that we are relativistially boosting the veloities to the left by -v (as a hek, note that C s final veloity in this boosted frame will be 0). Similarly, the veloity of B in C's frame point would be 5 -v - 5 v 2 pointing to the right, or Beause we want for A and B to both approah C with the same veloity, we set whih we solve using Mathematia Solve 5 - v 5 - v 2 v v 5 - v 2, v, v v - = v 5 2 v- 5-5 v v- 5-5 v 2 pointing to the left. 2 (2) We find the two solutions v = 5, 7 ; sine v = 7 implies traveling faster than the speed of light, the only physial solution is v = 5. The veloities of A or B in C's frame are found by substituting v = 5 above, yielding = so that A and B travel at speed towards C. 5 5 Method 2: Using v A = 5 and v B = 5, the relative speed of A in B s frame equals = 5. From C s point of view (where A and B are approahing eah other at speed v ), this 5 is the result of relativistially adding v with another v, whih implies 5 = 2 v + v 2 2. Solving this yields v =, 5. Only the former is physial, so 5 that A and B approah eah other in C s frame at speed 5. We ould then use the equation C s value v = 5, as found above v - 5 v 2 = 5 to solve for

10 0 Leture nb Extra Problem: Fizeau Experiment Example The seond postulate of relativity says that the speed of light in vauum is always (in an inertial frame). However, the speed of light in a medium (suh as water) is given by, where n is the index of refration of the n medium. For water, n is about. Imagine aiming a beam of light rightward into a pipe of water moving rightward with speed v. Naively, the speed of the light with respet to the ground should be + v. Find the orret speed using the veloity addition formula. n Then, in the ase where v (whih is ertainly a valid approximation in the ase of moving water), show that to leading order in v, the speed takes the form of + A v. What is the value of A? n Solution Sine the light moves at speed with respet to the water, rand the water moves at speed v with respet to the n ground, the veloity addition formula gives the speed of the light with respet to the ground as V = n +v + n (v) 2 = n +v + v n Taking the first order Taylor approximation in v, we obtain the expression so that A = - n 2. Series n + v + v n, {v, 0, } n + - n 2 v + O[v] 2 (22) V n + - n 2 v (2) We an hek this result in a few speial ases. If n =, whih means that we have a vauum instead of water, we obtain a speed of V =. This is orret beause we know that light always moves with speed in vauum. If n is very large (implying a very dense medium), we obtain a speed of V + v. This is orret beause it is the naive n addition of the speed,s whih we know works perfetly fine when both speeds are muh less than. In 85, well before Einstein s veloity addition formulas was known, Fizeau performed an experiment to measure the speed (with respet to the ground) of light in moving water. His setup involved an interferometer similar to the one Mihelson and Morley used in their experiment. He obtained a result onsistent with the approximate formula Equation (2), so he onjetured that the formula held (exatly) in general. Many people then made unsuessful attempts (involving frame dragging of the "ether," for example (this was before it was known that there is no ether)) to explain why the parameter A took on the value - n 2 instead of the naive value of. In retrospet, of ourse, failure was the likely result of their (ommendable) efforts to generate an exat theory from an approximate result. It was more than half a entury until Einstein produed the theory of speial relativity in 905, from whih the orret explanation of A s value followed via the veloity addition formula (along with the approximation we made in Equation (2)). Conversely, the result of Fizeau s experiment was highly influential in Einstein s formulation of speial relativity. Many Veloity Additions

11 Leture nb Example An objet moves at speed β = v with respet to S, whih moves at speed β 2 with respet to S 2, whih moves at speed β with respet to S, and so on, until finally S N- moves at speed β N- with respet to S N. Show that the speed, β (N), of the objet with respet to S N an be written as where β (N) = P N + -P N - P N + +P N - (2) P + N = Π j= ( + β j ) (25) P - N = Π j= ( - β j ) (26) Solution We proeed by indution. When N =, β (N) = β as expeted. Now assume that the formula holds for N, and let us show that it holds for N + as well. Using the veloity addition formula, the veloity in the N + frame equals β N+ +β (N) +β N+ β (N) = β N++ as desired. This proves that the result holds for all N. PN + -PN - PN + +PN - +β N+ PN + -PN - PN + +PN - = β N+(P N + +P N - )+(P N + -P N - ) (P N + +P N - )+β N+ (P N + -P N - ) = P N + (+β N+ )-P N - (-β N+ ) P N + (+β N+ )+P N - (-β N+ ) = P + N+-P- N+ P+ N+ +P- N+ = β (N+) Notie that β (N) is symmetri in the β j s, as we expet from the symmetry of the veloity addition formula (in other words, it doesn t matter the order in whih the β j are applied). Additionally, if at least one of the β j equal, then P N - = 0 and β (N) =, as it should. Similarly, if at least one of the β j equal -, then P N + = 0 so that β (N) = -, as it should. (If one of the β j equals while another equals -, then we get the indeterminate 0. This is aused by the 0 fat that depending on how you approah this limit, you ould get different answers; this same indeterminate answer is seen in the normal veloity addition formula when v = and v 2 = -.) To get a sense for this effet, suppose that all of the β j = for all j, then the first few resulting β s would be 2 Table[(Produt[ + β, {nn}] - Produt[ - β, {nn}]) / (Produt[ + β, {nn}] + Produt[ - β, {nn}]), {nn,, 6}] /. β 0.5 {0.5, 0.8, , , , } The following diagram lets you visualize this effet for the ase where all β j s have the same value, but you an manipulate what that value is (using the top slide in Mathematia). You an also hoose vary the number of (27)

12 2 Leture nb manipulate (using top ). vary platforms using the bottom slides. Green shows the length ontration of eah platform (in the bottom platform s frame), red shows the time dilation effet (in the bottom platform s frame when the top platform measures seond), and blue indiates the perent of the speed of light at whih eah person is traveling. fration of the speed of light number of platforms in motion 50 seond %.57 seond % seond % 2.69 seond % seond 0.% There are several interesting points to notie. First, no matter how fast the 0 β < value is, none of the platforms an ever reah the speed of light. Seond, when β 0 we see the standard addition of veloities, with the veloities of the platforms from the bottom up taking the values 0, β, 2 β, β...; as β is inreased, this linear trend starts to break down as subsequently adding another β results in a hange that is smaller than β. As your speed approahes the speed of light, adding β leads to a negligible hange. Visualizing Speial Relativity Find speial relativity onfusing? Want to really "see" the effets with your eyes? Veloity Raptor has got your bak (onsider it as optional researh). In this game, you must use the powers given to you by Speial Relativity to fight the fores of evil. While I enourage you to play this game in its entirety (it takes about 0 minutes), I urge you to at least try to ahieve level 26. For in playing level 25 and level 26, you will how you would physially observe with your eyes

13 Leture nb playing you you the world at veloities near the speed of light. And one you see this, never use that reasoning in this lass. Beause we don t are about when photons atually hit your eyes; we desribe "seeing" in Speial Relativity as desribing what atually happens in a frame. (In other words, we use "seeing" to desribe a room full of synhronized loks in level 8, whereas what you would physially observe with your eyes (whih we don t are about!) is shown in level 26).