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PHYS 200: Fundamentals of Physics I
Lecture 13
 Lorentz Transformation
Overview
This lecture offers detailed analysis of the Lorentz transformations which relate the coordinates of an event in two frames in relative motion. It is shown how length, time and simultaneity are relative.
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htmlFundamentals of Physics IPHYS 200  Lecture 13  Lorentz TransformationChapter 1. Describing an Event with Two Observers [00:00:00]Professor Ramamurti Shankar: So, what did we do the last time? Here is what we were talking about. We were trying to describe an event as seen by two observers. One will be called S and here is the x axis for S; the other is called S′. S′ is sliding to the right at velocity u. Therefore, at a certain time–Let me say S is me and S′ is you. When you pass me, I’m sitting at the origin of my coordinates, x = 0. You’re at the origin of your moving coordinates, x ′ = 0, and when we crossed each other, that time we set as 0 in our own clocks. We synchronized them so that the event, x = 0, t = 0, also had coordinate x′ = 0, t′ = 0. When you and I crossed, we pushed our stopwatches, we synchronized our clocks, our origins coincided, and that’s when it began. So, after some time, you are here. This is your frame and this is my frame. If something happens here, something could be anything. I hope you understand what an event is. That’s not a relativistic notion. That’s the very old notion. Something happens at some place; firecracker goes off. I ask you, “When did it happen?” and “Where did it happen?” and you are given the coordinates. So, everything that happens is an event. You saw Elvis, I say, “Where did you see Elvis?” At the supermarket. Okay. When did you see Elvis? I would also ask you what you were smoking because this third question, that’s not an extra coordinate but in this case I would like, well, I should tell you beyond all the laughter that’s a serious issue. Why is that not a coordinate? Why not ask more and more questions and call them all coordinates? The reason the other things are not coordinates is very important. Why is time now suddenly a coordinate? Well, as you could have always asked 100 years ago when did something happen, and the reason is that you will see in the new relativistic physics, x and t, according to one person, get mixed up into x′ and t ′ of the other person. The fact that space and time coordinates can be combined to give the new space and time coordinates is why it’s called a coordinate. I’ll say more about it later. Now, by the way, I’ve posted some notes on this topic for those of you who want to have a second pass at it. I also don’t know how much the textbook covers this topic because I spend a lot of time on this topic, not in proportion to what’s in the book. So, you can look at those notes which are posted today. Before all the Einstein stuff, if something occurred here, you would say it’s at a distance x′ from your origin. I will say it’s at a distance x from my origin. This difference between our origins would be ut. Therefore, before Einstein, I would say x′ is x  ut and you would say x should be what you think is a coordinate + ut. This is just going back and forth from you to me and you can already notice that to go back and forth from you to me, we just change the sign of the velocity, because I’m going to the right and you’re going to the left [Note: Left and right mixed up here.]. That’s why the formulas have sign of u reversed. What is the big change after Einstein? First is, I admit, the possibility that maybe you will think the time elapsed since we synchronized our clocks is not necessarily the same. I leave that option open that t and t′ are not the same. Second thing I do is–I have already explained to you that the velocity of light coming out the same for different people is very counterintuitive, because if you are going to the right, I expect you to get a smaller speed and yet to get the same answer. So, we know that it’s because we don’t agree anymore on clocks and meter sticks being the same. So in particular, you will say, “I don’t buy your prediction that I expect to get the answer to be x  ut. I’m going to put a fudge factor, gamma, which depends on the velocity between you and me, because your lengths are not my lengths.” Likewise, I will tell you, “I don’t agree with your expectation that I should get x′ + ut′. I don’t believe your lengths are really my lengths, but I’m going to put the same fudge factor, gamma.” This is a very interesting result. If I think your meter sticks are short so that I want to blow up the answer with some amount; you are allowed to say the same about me. It’s one of the big paradoxes in relativity that we can accuse each other of using meter sticks, which are short, and we’ll go into a little bit more how that’s even possible. But the postulates tell us that if my fudge factor is gamma, yours should also be the same gamma. So, gamma is what we don’t know. I showed you a trick to find gamma. I said, let’s imagine that this event here, which could’ve been a firecracker, is triggered by a light pulse that was emitted when you and I met here. The light pulse goes racing and sets off an explosion here. That’s the event we are talking about. That’s not a generic event. It’s a particular event in which x has to be c times t because that’s the equation for a light pulse, according to me. Light goes at the velocity of c, and the event has got the same velocity, according to you. This event here should be connected by a light pulse that for x/t is c, and x′/t′ is the same c. Then, I said, take these three equations, multiply the lefthand side by the righthand side, put in these numbers and extract gamma, and I will not repeat that part, and gamma was this. Once you’ve got this, you can take this gamma and put it back here. Let’s see what we get. So I’m going to put what we get here into this board. You guys got all of that? Okay. So, I’m going to put it back here, because this is just background from last lecture. What we have then is x′ = x  ut divided by this famous square root, and if you go and solve for t′ in the other equation, it’s a simple algebraic manipulation, so I don’t want to waste my time doing that. You will find t′ = t  ux/c^{2} divided by the same square root. These are priceless. This is really all of relativity, all the funny stuff you hear about E = Mc^{2}, funny clocks, the Twin Paradox. Everything comes from this equation. And you people should be very pleased that within, what, six weeks of starting Physics 200, you have all the things you need to understand these equations, to understand where they came from. That’s also the remarkable thing about relativity. A lot of the modern things in physics require a lot of mathematics. In fact, if you want to take a beam and you want to load it with weight and you want to see the stresses and strains, the math in all that is 1,000 times more difficult than this one, and yet the mathematics of relativity is very, very simple, and accessible to all of you, and there is nothing that I know of that you don’t know as far as how to get these equations. This is what you do. Everything follows from taking these equations and analyzing them, extracting the consequences and that’s really more an issue of courage than just intelligence because once you’ve got these equations, you will have to take them, you’ll have to follow them where they take you. That’s what we’re going to do. The rest of the whole remaining lecture, it’s all about getting information from these equations. Now, some of you are not used to writing equations. I’ve noticed, when you see something like this, it’s not clear what is being stated. So, let me remind you one more time. u is a fixed number. That’s your speed relative to mine. I see something happening, I give it a pair of numbers, xt, you give the same event, another pair of numbers, x ′, t′, and this is how they are related. So, I gave you an analogy, but let me repeat the analogy. If you’re now talking about the xy plane rather than the xt plane, that guard is not an event. That guard is a coordinate of a point, maybe where something is sitting. I give to it a pair of numbers, x and y. Now, you have a different coordinate system, rotated relative to mine, by some angle θ. To that same point, same location, you measure x′ up to here and t′ from there. I’m sorry, x′ and y′ are measured along your axis, and the formula there is x′ = x cos θ + y sin θ and y′ is x sin θ + y cos θ. θ is the analog of the velocity now. If θ is 0, you and I agree completely. If θ is not 0, your axis is rotated relative to mine, and the same point has two different numbers, xy for me, x′/y′ for you, and the relation between them is this. So, you plug in my x and y, you can get your x′ and y′. Let me give a concrete example. Let’s take θ to π/4 or 45°. Then, as you know, at 45, sine and cosine, they’re all the same. One over root 2. So, in that case, x′ is x over root 2 + y over root 2, and y′ is x over root 2 + y over root 2. That’s a special case when θ is 45°. So, for every angle, cosine and sine will reduce to usual simple numbers that happen to be the [one over] square root of 2 here. Then, it tells you my x and y are related to your x′ and y′ in this manner, and you can test it. For example, take a point with coordinates (1,1). (x,y) is 1,1. If (x,y) is (1,1), and my axis is related to yours by 45°, (1,1) lies right there. So, what I expect is your y′, there’s no y′, and the coordinate should be entirely x′. If I put in (1,1) and indeed you find if I put x = y = 1, y′ becomes 0. And how about x′? It becomes 1+1 over root 2; that means 2 over root 2 is the square root of 2. The square root of 2, of course, would be the lengthy measure this way because the length of the vector is square root of 2, and the vector is entirely along the x direction. So, the length you will get for this coordinate will be root 2. So, come back to these equations. They’re the same thing. If you want, I can write x′ as x divided by this number  u divided by this number times t. So, you can think of 1 over this number here as the analog of cos θ and u over this number as the analog of sine. So, it’s just that instead of cosine and sine all being upstairs, some of the numbers are down, some of the numbers are up. But for a given u, these are also numbers constant depending on u, and x′ is the linear combination of x and t, and t′ is some other linear combination of x and t. But I should warn you that it is not an ordinary rotation. In other words, you cannot treat this as a cosine and that as a sine. They’re not cosine or sine of anything, because if they were, this squared plus that squared should add up to 1, and they won’t. So, don’t even try that. This is not an ordinary rotation but it is still what you can call a linear transformation. Linear transformation means the new numbers are related to the first powers of the old numbers. They are linear. They don’t involve t^{2} and x^{2}. Okay, so that is the content of the Lorentz transformation. Everybody should understand what they do by way of going back and forth. But now, you can go backwards. You can say, “Well, how do I write x in terms of x′ and t′?” There are two options open for you. One is, these are simultaneous equations. You’ve got to find a way to solve for x and t in terms of x′ and t′ and all these funny functions involving u. You treat them all as constants and juggle them around, multiply by this, divide by that, but you shouldn’t do that because you know what the answer should be. The answer should be the same as what I got with the velocity reversed. So, x should be x′ + ut′ or the square root. If I write the square root, it means I don’t feel like putting what’s in it. It’s the same old thing, so you got tired of writing that. t′ will be t + ux over c^{2} divided by the square root. So, one should be able to go back and forth, just like in this formula. If you like, I gave it to you as an example long back as a homework problem. You can write a new formula that says x = x′ cos θ  y sin θ and y′ is equal to something. I mean, y is equal to something something, and the way to get that is to either solve the equations or realize that if I go from me to you by a θ, I go from you to me by θ, and all you have to do is change sin θ to sin θ and leave the cos θ alone. Similarly here, you change the sign of the velocity u to get the reverse transformations. Okay, so that is the Lorentz transformation. Now, we are going to start milking the transformation. Everything is going to be applying it to understand various things. The first step in getting the mileage out of the Lorentz transformation is, take a pair of events. I urge you, whenever you get a problem with relativity, to think in terms of events and quite often think in terms of pair of events. Two events. So, event one is, let me give concrete events, and most of the relativistic examples involve some degree of violence and so this one involves a gun. So, I take this gun. I’m not going to point it at any of you guys. I fire the gun. That’s event 1. The bullet hits the wall; that’s event 2. You can take two events connected by a bullet leaving me and hitting the wall, or you can take two events not connected to anything, okay. You can take two unrelated events. It doesn’t matter. We’re going to call them event 1 and event 2. So, event 1 will have coordinate x_{1}, t_{1}, right? And according to you, x_{1}′, t_{1}′. So, write the Lorentz transformation that first relates x_{1}′ is x_{1}  ut_{1} over the square root. Then right here, t_{1}′ is t_{1}ux_{1} or c^{2} divided by a square root. Similarly, take the second event and write the law for that. Well, it’s the same thing with the new numbers in it. So, t_{2}′ is equal to t_{2}  ux_{2} over c^{2} divided by a square root. Now, take the difference of 2  1. You can take 1  2 but it’s very common to define the difference to be 2  1, and call it Δx_{2}′. Δx′ is x_{2} ′  x_{1}′. All the Δs will be defined to be 2nd  the 1st. I’m telling you to take x_{1}′, subtract it from x_{2}′ on the lefthand side and call it Δ of x′. Δ of x′ is the difference in the spatial coordinates of the two events according to you. If you come to the righthand side, now this is something you guys should be able to do in your head. I want to subtract that from that. They share the same denominator, so let me put the denominator there. In the numerator you will get x_{2}  x_{1}, which in my convention I will call it Δx and the other one will be u Δt. What this tells you is that differences in coordinates also obey the same Lorentz transformation. Instead of saying the coordinate of an event was x, t and you get x′, t′ from the Lorentz transformation, if you take a pair of events and they are separated in space by Δx and in time by Δt according to me, the separation according to you, your Δx′s and Δt′s are given by similar formulas as the Lorentz transformation; put a Δ everywhere. This just came from taking the difference of two equations applied to the two separate events. All right. You can also do this backwards, if you like. If you want the differences that I get in terms of yours and yours in terms of mine, you have to reverse the sign of u. So, I won’t do that again. Yep? Student: Shouldn’t that be t? Professor Ramamurti Shankar: Here? Student: Yes. Professor Ramamurti Shankar: This was the formula for x in terms of x′ and t′, and t ′ in terms of t and x. Student: [inaudible] Professor Ramamurti Shankar: Here? Professor Ramamurti Shankar: Okay, I’ve got to give it some numbers. This is a, this is b, this is c. Where is the problem? One of you [inaudible] Professor Ramamurti Shankar: Oh, yes yes yes. Of course. Yes. Thank you. Okay, now a public apology is forthcoming. You wanted me to do this. Student: Yeah. Professor Ramamurti Shankar: Yes? Thank you very much. Okay, do not hesitate to do this, okay? In fact, when I said I don’t know anymore than you do about this, it looks like I know less than you do about this. So, the fact that I know this doesn’t mean I’m going to get it right. So, all of you who screwed up in the Midterm, remember there is hope for you. Maybe you won’t do well as students but you can become a professor here. Apparently, it’s all right for us to get things wrong. That’s correct. Very good. Okay, so if you’re following me that well, I’m very happy now to know you must be following what I’m saying here. So now, let us–Everything now I keep telling you I’m building it up in a big way but everything really is going to come from this version of the Lorentz transformation for differences. Look, suppose you were smart enough like Einstein and you did use these equations for gamma. What do you do next? You write them down. You can publish them and say according to me, this is a rule for transformation, but you cannot stop now. You have to say what are the implications of my equations because these equations are dramatic variations of Newtonian laws. For example, let us do one thing. Two events are separated by 10 meters according to me. Δx′, if Δx is 10 meters, then Δx′ is not 10 meters. It is 10 minus something divided by something. So, that distance between two events is changing. That’s not supposed to happen just because you get into a train. Right? If I hold my hands up and say I caught a fish that big, that distance should be the same for me and the train or if you look at me riding in the train from outside the train you should find the same distance. We’re saying it’s not true. Likewise, if I say two events took place when I came to Yale, I’m certain there I got a degree four years later, that’s supposed to be true for anybody. But that’s also not true. If Δt is four years, Δx is whatever you like. Maybe you never left New Haven, so Δx is 0. That’s four years, that’s not four years. These are all drastic consequences and we’ve got to explore the consequences. The first thing I’m going to do is to make sure that the velocity transformations from one frame to other will be the requirements we set on it. So, here’s what I’m going to do. Event 1: I fired the gun. Okay. Let me just call it Event 1, I fire the gun. Event 2: a bullet hits the wall. So, the separation between these two events is Δx for me and the time it took the bullet is Δt. You can also see the bullet from your train and you think the distance between me and the wall is something and the time between the firing and hitting the wall is something else. What’s the velocity of the bullet according to you and me? v velocity of bullet, for me, would be Δx/ Δt. Okay, now I should be a little more careful with my rotation. Usually Δ is used for infinitesimal numbers, especially when you’re going to define the velocity, velocity is the limit of these guys going to 0. But it is the property of Lorentz transformation that at this level, when I say differences in time and differences in space, they are not necessarily small. Nowhere was I assuming that the difference in x was a small or the difference in [inaudible] was small [inaudible]. They could be events separated by five light years, you could still use them. But at this stage, don’t think Δx has to be infinitesimal. It is simply a shorthand for difference. If I had all the time in the world, I would write everything as x_{2}  x_{1}, but I’m using Δ as a shorthand. But when I’m going to find the velocity of a bullet, which could even be the instantaneous velocity, then, here I do want to take the limit in which they go to 0 so it becomes dx/dt . So, from now on, for this purpose of velocity calculation, you should take them to be infinitesimal and approaching 0, but not in general. These equations are valid but arbitrarily big intervals in time and space. Now, w better be the velocity for you. That is, I am S and you are S′. w is Δx′ over Δt′ with suitable limits. Well, from these transformational laws, you can get them because here is Δx′ and here is Δt′. Let me divide this by this on the lefthand side to get w. When I write Δx′ over Δt′, you guys take the limits of everything going to 0. I don’t feel like writing that. In fact, it’s true even without the limit, but let’s apply it in the end to an instantaneous velocity. What happens on the righthand side? Here is what Δx′ is equal to. Can you do this in your head? If this Δx  u Δt, this denominator cancels between dividing this by this, so it looks like–I just divided this guy by this guy, because that’s when you divide the lefthand side by the lefthand side, you have to divide the righthand side by the righthand side. Now, you should have an idea of what I’m planning to do. What do we do next to this expression here? Yes? Student: [inaudible] Professor Ramamurti Shankar: A limit of what? If we just take a limit of Δt going to 0, you’re just going to get 0. That’s not the limit you want to take. You’ve got to bring velocity into the picture. Yes? Student: u over t? Professor Ramamurti Shankar: Pardon me? Student: u over the t? Professor Ramamurti Shankar: No. For any value of u– Student: [inaudible] Professor Ramamurti Shankar: Yes. What you want to do now is to divide the top and bottom of this by Δt because you got the velocity of the bullet according to you. I want to get the velocity of the bullet according to me into the picture. That is Δx over Δt, but this is not a typical derivative in calculus, okay? You’ve got to divide everything by Δt, then you will get Δx over Δt there  u divided by 1  u over c^{2} times Δx over Δt. This is true even for finite differences. Now, take the limit. Then, this becomes the velocity of the bullet  u divided by 1  uv over c^{2}. So, you’ve got to draw a box around this guy. This is another great formula. So, I hope you know why I divided by Δt. I want to find the velocity of the bullet according to you and according to me. For you, I took the distance over time. Well, I’ve got all these distances and times in the top and bottom. I divide by the time so everything turns into velocity. So, this is the velocity of the bullet according to me. If you’re going to the right, at a speed u, in the good old days, what would I expect? You’ve got to understand that. My expectation in the old days would be v  u, right? Whether it’s going as speed v, they’re going as speed u to the right. So, you will see a diminished speed by an amount equal to your speed. But now is the twist. There’s something in the bottom, and the stuff in the bottom is a number less than 1, because of 1  something. We’re going to jack up velocity. Therefore, the velocity you will actually measure is somewhat bigger than what I expected in the old days. If you ever want to get back to the good old days in any relativistic calculation, you should let the velocity of light go to infinity. Of course, it’s not infinity. What you really mean is u/c and v/c are negligible. That means my velocity, the bullet velocity, they’re all small compared to the velocity of light; we get back the answer from the old days. This is the answer from the new days. Now, let’s find the beauty of this result. Let’s get the reverse result. Let us solve for the velocity I get in terms of the velocity you get. Again, I think you realize you can do the algebra, but you guys should know that to go from me back to you, I should simply reverse the sign of the relative velocity. This is the backwards result. In other words, you’re in a train and you fired a bullet towards the front of the train. What speed do I attribute to the bullet from the ground? Well, I add the bullet speed to the train speed. That’s what the numerator is. The numerator says the answer is somewhat less than that by this number. Yes? Student: [inaudible] Professor Ramamurti Shankar: Thank you very much. Yeah. That’s absolutely correct. What I really need to do–quite correct–it’s not simply to change u to u, but to change the velocities that you were seeing and put them in the place of velocities that I was seeing. That’s correct. This is what you want to get. Now, let’s look at the strength of this result. If someone tells you according to relatively, nothing can go faster than the speed of light, you can try to beat the system as follows. You can come to me and say, “Can there be a gun whose bullets go at threefourths the velocity of light?” and I would say, “yes,” and you’d say, “How about a train that goes at threefourths the velocity of light?” and I would say that seems to be allowed. Then you can say, “Well, let me get into this train at threefourths the velocity of light and fire a bullet at threefourths the velocity, then from the ground it should appear to be going at 1.5 times the velocity of light.” Well, that’s the naïve expectation. But if you do it now, let’s put w = ¾c and u = ¾c, the old answer is disaster; it says 1.5c, but the correct answer is that plus (¾)^{2}. So, what is this guy? This is 1.5c divided by 1 + 9/16. You multiply everything by 16. On the top you’ll get 24, the bottom will get 25 [times] c. You see, you can jack up the velocity as seen by the ground but it’ll never be 1.5. It’ll still be less than the velocity of light. And the last thing you want to check of this velocity addition formula is why we started the whole thing. In other words, suppose I see a pulse of light. Let’s go back now to this formula. w is v  u over 1  uv over c^{2}. Rather than applying it to a bullet or whatnot, let’s apply it to the light pulse itself. So, I saw a light pulse so I say the object I saw, which is what v stands for, had a value c. And this is the preEinstein expectation. w should be c  u because the pulse appears to travel slower to a person moving in the same direction as the pulse. But our new formula says it is really that uc over c^{2}. So, that canceled one part of c, multiply top and bottom by c, and you will find it is c. So, the velocity of light will always come out to be the same. That is built into the formula but it is a good thing to test. So, what you find is, velocities don’t add in the simple way. If they did, you are in trouble. You cannot get an upper limit if they added in the simple way because you can put a rocket inside another rocket inside another rocket and add up all the speeds and even though each one is less than 0, the total could be whatever you like. But they don’t add that way. They add this way so that the answer is either bigger than what you think if v and u are opposite, or smaller than what you think if the velocities in the top are of the same sign. No matter what you do, the answer will always be less than c. Chapter 2. The Relativity of Simultaneity [00:33:58]Okay, so that is the first conclusion from this. Second conclusion, which is again very staggering, is that simultaneity is the relative concept. In other words, if two events occur at the same time for me, they don’t occur at the same time for you. That is very surprising. For example, if you have twins born, one in New– sorry, twins cannot be born in New York and Los Angeles. So, I’ve got to pick two kids of different motherhood, just happened to be born at the same time. When I say “time,” we’re not talking about trivial threehour time difference between Los Angeles and New York. That’s an artificial thing. In the Einstein world, you imagine, we all have clocks that read the same time. So, by that measure, the two kids are born at the same time but if you watch that from a moving train or a moving rocket, you will disagree on that. You will, in fact, say they are not simultaneous. How does that come? Well, it just comes from going to this formula. Δt′ is Δt  u Δa x over c^{2} divided by this number and all I’m telling you is, even though Δt is 0, Δt′ is not 0. Another shock. Simultaneity is not absolute. We used to think it’s absolute. In other words, two events are occurring in Los Angeles and New York. They can be arranged to be simultaneous for me, living on the planet, let’s say, but if you go in a rocket, I expect you to agree they were simultaneous. I mean, how can they be different? Two things are happening right now in different places, it’s got to be right now for you. But it’s not. It just comes from that formula. Once again, if the velocity of light is made much bigger than everything in the problem, simply to set it to infinity, you will find Δt′ is Δt. That goes back to Galilean times or the old preEinstein times. That’s why in relativity, space and time form a new spacetime and time is called a fourth component because if c goes to infinity, Δ′ is always Δt and t′ is always t. If the coordinate never mixes with anything else, it doesn’t deserve to be called a coordinate, whereas, it’s the x and y coordinates that mix with each other when you do rotations. After Einstein, the space and time coordinates mix with each other to give you new space and time coordinates under Lorentz transformation, which means, when seen in a moving train. That’s why time is now elevated to another dimension because it transforms very much the way x and y did. The details of the transformation are different. We have sines and cosines replaced by u and u over the square root but it’s still mathematically the way components of vectors will transform. Okay, so simultaneity is relative. So, you have to ask yourself, “How did that happen?” So, here’s the famous example that is given in all the books. Maybe it was given by Einstein, I’m not sure. So, here’s a train. Let’s say the train is at rest and you are standing in the middle and you want to arrange for two events to be simultaneous. Oh by the way, there is one catch. If the two events occurred at the same time and the same place, Δx = 0, Δt = 0; then, the transformation will tell you Δx′ is 0 and Δt′ is 0. Because two events occurring at the same time, at the same place means, for example, my two hands came and clapped, the two hands were at the same time at the same place. If someone said, well, they were not at the same place at the same time, it means I didn’t clap. Two cars collide. So, if two things occur at the same time at the same place, something happened that they met in spacetime. You cannot find another observer who says they did not meet in spacetime, or they’ll say the two cars did not collide. Even after relativity, it is true that two events at the same time and same place occur the same time and same place for all people. Same time alone is not enough and same place alone is not enough. You see that? If two cars came and they collided, they were at the same time at the same place. If the cars were at the same time, here and there, that’s no accident. If they were at the same place because if this car went and two minutes later that car went at the same place, nothing happens. That’s why you say this poor guy was in the wrong place at the wrong time. You don’t want to say this guy was in the wrong place. What does that mean? You can go to the Battle of Gettysburg site and stand now and nothing happens to you because it’s the same place but it’s the wrong time. Okay? We all realize, same place, same time, is a congruence in spacetime and even after Einstein, you cannot say something did not happen. Okay. We leave that to politicians please. It did happen. It still happened and the equations have the property but I’ve got several events separated in space but at the same time. You’re in the train. Your job is to make two things happen at the same time and here’s what you do. You send a beam of light that splits into a, you know, here’s a blown up picture. Beam of light comes and splits like that and goes to the back and front of the train. You’re in the middle of the train, and you know that events will be simultaneous for you. The pulse travels on either side and hits the two things at the two ends and sets off two explosions and you have done the best you can to have simultaneous events. Okay now, I see you from the ground. So, all this happening in the moving train, you have every right to say you’re not moving but to me–relative to me you are moving. That’s a fact. And I look at how well you did with this. So, you sent off these two light pulses. Then what happened? This wall of the train is moving away from the light pulse. This wall of the train, the rear end, is rushing to meet the light pulse. Now, the velocity of light is the same for everybody so I can think in terms of what happens. If one wall is rushing to meet the light pulse and one wall is running away from the light pulse, I know very clearly the firecracker in the back of the train will go off first and in the front of the train will go off later. That means that Δt will not be 0. In fact, you can see Δt will be negative because the event with the greater x coordinate occurs later. That’s also an agreement of the formula. If Δt = 0, then Δx′ will be some negative number. Now, why do we bring in the light pulse? We bring in the light pulse because we can tell that they were simply not simultaneous because about the light we know this. Its velocity is the same for everybody. That’s why all arguments in relativity involve doing things with light pulses or communicating with a light pulse because we know what light does. It travels at one and the same velocity for all people. That’s a postulate. Therefore, we know you couldn’t have done any better in making them simultaneous and I simply disagree with you. I’d say they just did not happen at the same time, and there’s no question of who is right. Operationally, for me, those two events were separated in time by some amount and for you they were not. That’s a great new idea that things did not happen the same time for all people, and relativity tells you that they didn’t and tells you by how much they were different. Chapter 3. Time Dilation [00:41:57]Okay. Then, I take the next surprise here. The next surprise has to do with clocks. The claim is that clocks will not run at the same rate. Okay, when we bought the clocks, we compared them, we got them from the same shop, they were completely in synch and you put one in your train and you got into your train and the claim is that I will find that your clocks are running slow. Let me show that to you. Again, if you want to do anything with relativity, at least learn this one thing. Go back to Lorentz transformation and think in terms of events that are just going to come out of the wash. If you try something original on your own, you people do all kinds of stuff. I’ve seen you going in a cyclical circular or arguments from which you cannot even come out until somebody rescues you. So don’t do that. So, I’m looking at a clock. You’ve got to ask yourself, how do I turn this issue of time into a pair of events? I have my clock, so it goes ticktock, ticktock. I pick two events. Event 1: a clock says tick, and event 2: clock says tock. I tell you why I have tick and tock because I want to have two distant kind of events so that we can talk about them. So, this event is a clock that I’m carrying with me. The clock is with me. So, let me put a clock at the center of my coordinates system or the origin. It doesn’t matter where it is. Let me put it at the origin. The spacetime coordinate x = 0, t = 0 is the first tick of the clock. How about the next tick? The next tick, this is x and t coordinates, the tock of the clock, let me call τ_{0} is the time of the clock. It’s a time period; how many seconds elapsed between the tick and the tock. The main question is, “What is the location of the second tick of the clock?” Where does that happen? Yes? Student: [inaudible] Professor Ramamurti Shankar: It doesn’t move with respect to me and I’m talking about a clock that I’m holding in my hand so if the first event took place at x = 0, second one also takes place at x = 0. You understand? I’m holding the clock. I’ve traveled forward in time. It’s τ_{0} seconds later but the clock has not gone anywhere so the two events, the two tickings of the clock, are separated in space by 0 and in time by τ_{0}. That means Δx = 0, Δt = τ_{0}. What do you get? According to you, the time difference between the two ticks is Δt, which is τ_{0}  u etc., times 0 divided by this factor. So, that means the time difference between the two ticks would be bigger. If τ_{0} is 1 second, if you divide 1 by [root of] 1  u^{2} over c^{2}, you’re going to find out it is less. There’s more. For example, if this factor in the denominator is .5, then Δt would be 2 times τ_{0}. In other words, when your clock has taken 1 second to go from 1 tick to the next, I will say according to me the real time elapsed is 2 seconds. It can be 10 seconds, it can 100 seconds. You can make Δt as big as you like by letting you approach c. Yes? Student: [inaudible] Professor Ramamurti Shankar: Thank you. I’m having this problem today with the prime. t′ is τ_{0} over this. If you put τ_{0}, I put 1 second. Question there, guys? Student: [inaudible] Professor Ramamurti Shankar: No, u did not go to 0. u is not 0. u Δx over c^{2,} it is the Δx that went to 0. Go the Lorentz transformation. Δt ′ is Δt  u Δx over c^{2}. Δx = 0. That’s where I’m telling the equation the clock is at rest with respect to me. That means that– Student: [inaudible] Professor Ramamurti Shankar: No, no. In this example, let the unprimed observer be the one holding the clock. Then Δx is, you know Δt is whatever the time of the clock is. 1 second. Δt′ is seen by anybody else going at a speed u relative to me. You can also do it backwards by taking the clock in the hands of the S′ but you’ve got to do a little more work. It’s a lot easier here because I get to pump into the equation two facts at one shot, that the time between the two ticks is 1 second and it’s at rest with respect to me, which is why the Δx vanished between the two ticks. If I want to use your equations, Δx′ won’t be 0 because the clock has moved and you can do it. I’ve, I think, shown in my notes how to do that but it’s unnecessary. This is the main result. Now, here is the paradox. You can go back to the backwards equations. Let’s do this informally, okay. Let’s not write it again. It will look like this. Right. You know that. Let’s take a clock you are carrying. It ticks off 1 second, so between the tick and the next tick, the time difference is 1 second, space difference is 0, I’ll find Δt is 1 over a square root. So, I will say your clock is slow, you’ll say my clock is slow. How is that possible? How can it be that we accuse each other of having–yes? Student: [inaudible] Professor Ramamurti Shankar: Oh, you’re absolutely right. The predictions are in agreement with that principle, but I ask you to think about the conflict you have, yes? Student: [inaudible] Professor Ramamurti Shankar: Oh, that’s very good. Student: [inaudible] Professor Ramamurti Shankar: No, that is correct. Let me repeat what he said. He said, if there are two people, they were right next to each other, they are at the same height and they move apart and each person looks at the other person and finds the person diminished in size because of the distance, and each person can tell the other person you look small from where I am. In fact, even more paradoxical is how in this world two people can simultaneously look down on each other. You figure that out. There are two people who simultaneously have higher opinion of themselves to the other person and are the same thing, they look down on each other. I have not found how many space dimensions I should embed but these people to make it that possible, but this is certainly correct. A spatial resolution does give the impression. So, here’s the answer that’s usually given to explain this to you. If I take a real clock like this watch here and I ask you why does it look slow to you when I’m moving relative to you, it’s difficult because it’s got electronics and stuff and I still don’t know how to set the clock on my VCR. I’m not going to figure that out. So, for all of us guys who are challenged, there is a clock that’s particularly simple, and the clock works like this. It has got two mirrors and a light pulse just goes up and down between the mirrors and every time it completes a round trip, it sets off some detector and it goes “click.” Now, this is very important to mention that this is my x coordinate and that’s the y coordinate and I did not show you here there are logical arguments and why, if you and I are moving along x, why our y coordinates, if they agree in the beginning, continue to agree. That’s because if you say something is two meters tall and they pass each other, you cannot have a disagreement on the fact that heights are the same. But as in a linear dimension there is room for discussion, in the transverse dimension there isn’t. So, anyway, I have this clock that goes up and down. Light pulse goes up and down. It goes a certain distance, L, in a vertical direction, and 2L/c is the time period of my clock. That’s the time for a round trip. You look at my clock and what do you think is happening? Remember, you are moving to the right, relative to me, so according to you my clock is moving to the left, and it looks like this. Light beam is going on a zigzag path, according to you. You agree that’s very clear. For me, the pulse is going up and down, for you it’s going on the hypotenuse. You can imagine how with the hypotenuse u^{2} and c^{2} are going to come from drawing the sides of the triangle. That’s why the time would come out longer. Now, why do they like this clock? Because we know everything about the operation of the clock. We know that this path is longer than the straight up and down path because the transfers’ coordinate is known to us. Furthermore, we know the velocity of light is the same in all frames of reference; so, light going on a longer path is simply going to take longer. So, I know your clock would slow down. Now, here is the beauty. The beauty is that, if you have a clock that’s going up and down, straight up and down, when I see it, it’s going to look like this. So, it’s as simple as saying that if you had these two clocks, what I think is going up and down and you think it’s a zigzag this way, and I look at your clock, it’s a zigzag the other way, it’s perfectly okay for me to say to you, your light pulse is going on a hypotenuse and mine is going up and down, and why you can say that to me. So, at least this light clock explains to you why clocks will appear to be slow. Now, the question you can ask is, what if I have some other clock with gears and wheels and teeth and whatnot? How does that slow down? The answer is, we don’t know exactly how to explain that clock but I know that if you carry a light clock and another clock made of wheels and gears, they should run at the same rate. They should run at the same rate because if they ran at different rates, one slowed down and one didn’t, in comparing the two clocks you can determine your velocity, and that we know is impossible by the postulate that you cannot detect uniform velocity. Therefore, if the light clock does something, all clocks must do the same thing regardless of their mechanism, and that includes biological clocks. So, if you have a clock which is just yourself, I look at you, you know, over 15 or 20 years I notice some changes. You become taller, then your hair turns white, your teeth fall off. That’s a clock and that clock should also slow down. I don’t care how your life systems work but you are a clock and you’ve got to slow down. That’s why we can make predictions about what happened to living systems even though that’s not our main business. So, in particular, if you take a clock, which is made up of mechanical parts and a human being, and they travel at high speeds, the aging of the human being should slow down just like the clock’s ticking will slow it down, without knowing the reasons for it. Chapter 4. The Twin Paradox [00:53:41]So, this leads to a very famous paradox called the Twin Paradox. The Twin Paradox says that you’ve got two guys, twins. Now, this is actually a valid example. They were born at the same time, same place, and then the one goes on a trip at some speed u, and goes around and comes back. So, I think the person’s been gone for 20 years, my twin. So, let’s say he was 20 when he left, so I expect him to be 40 when he gets back. But he’ll come back younger because as a clock, he has slowed down. So, what I think is 20 years, this could be, my time is 20 years, but his time could be 10 because his factor downstairs is .5. So, he can come back being younger than me. Yes? Student: [inaudible] Professor Ramamurti Shankar: Oh, but we’ve got to be careful. So, his point had something to do with acceleration, but the real question is this. See, if I’m on the ground and I send my twin on the trip and we meet, now the clocks are being compared. As long as the zigzag is going this way and that zigzag is going that way, you can say what you like. But what if the clocks are brought headtohead and compared? Can they both be slower than the other? We know that’s impossible. So, instead of clocks, we use human beings and dramatize the paradox and say, “Who will be younger?” Me or the twin who went on the rocket? Well, there can only be one answer to that question, and yet no matter what the twin says, I have no reason to believe that I moved. You are the one who went on the trip the opposite way so you’ve got to be younger than me. Yes? Student: [inaudible] Professor Ramamurti Shankar: Very good. Okay. I couldn’t have said it better. Let me repeat it. The point is, as long as you’ve got oppositely moving clocks, they can believe what they want of each other, but if you want the clocks to be subject to a comparison, then you can never compare objects that are constantly moving at opposite velocities. So, somebody’s got to turn around and come back. So, in the case of the twin, he and I were abreast, the twin got into the rocket to set up the velocity difference between him and me. So, in the early stages, he is the one undergoing acceleration and not me. Likewise, the twin went somewhere and stopped and turned around and came back. The twin is suffering the deceleration. So, there are two periods at least, during the voyage, when the twin has no right to claim that he is not moving, that I am. That’s why we don’t have the same status. Whereas for me, the whole time Newton’s laws and the laws of inertia were operating for me, whereas for him and the rocket during takeoff, things started flying off and during landing things started flying off. He cannot possibly claim he has the same status as me. That’s why in that problem, the relationship is not symmetrical, and one person can say I did not move because I was always inertial, the other will have, I have to concede that he moved. Yes? Student: [inaudible] Professor Ramamurti Shankar: Yeah. Then if you do it, if you give them symmetrical acceleration, both leave like this and both come back. Then, both will agree on their age but they will disagree with the person on the ground. Suppose there’s some triplets. You have triplets and two kids are sent out that way, third one stays back, the third one that stays back will be older than the others. So, this is not science fiction at all. If you want to be alive for the year 3000, you can do it. You just have to get into a rocket, at sufficient velocity close to the speed of light, so that this is 3000 years, okay, and you figure out how long you’ve got to live. Maybe another 50 years. Do the math, you find the speed, get into the rocket and go and come back. Now, this experiment is done all the time with microscopic particles, you know. They are accelerated in Fermilab, for example. They go around in a ring and just by virtue of their motion, they live a very long time. So, particles are supposed to have a short lifetime which you calculate in their own rest frame, live much longer because they are moving, and one way to keep them moving is to put them on an accelerated ring and they live for a very long time. Yes? Student: [inaudible] Professor Ramamurti Shankar: No. No, the rocket will think I’m 50 years older. If you went on the rocket, it’s 50 years for you but for people on the ground it’s 3,000 years. Student: [inaudible] Professor Ramamurti Shankar: Yeah, but if you don’t want to age at all, you can arrange it. You want, what, five years? Okay. You just fixed the number. You want to age only .5 years, that’s another number. You’ll never run out of numbers. You can let u approach c as much as you like and you can live as long as you want and you can come back whenever you want. Other issues like going backwards in time–I’ll come back to that later. There are problems with that. We’ll come to that, but you can stop time as far as we know, and we see it all the time. The muons produced in the upper atmosphere, their lifetime is something, 10 to  6 seconds or whatever. Even if they travel at the speed of light, we see their time is not long enough to cross the atmosphere but they make it here. How did they make it? They make it because their lifetime is computed in their frame of reference. They still think they lived only their lifetime, but I think they lived a much longer time. That’s how they made it to here. But how about from their point of view? They only lived a short time and how did they go 100 kilometers? That’s the next thing I’m going to tell you. According to them, the atmosphere is not 100 kilometers but maybe 8 kilometers long. Now, that’s the last point which has to do with the length contraction. So length contraction says, by the way, so thing to remember what time dilatation is: Every clock runs the fastest in its own rest frame. In any other frame of reference in which a clock has any velocity, it will appear to be slower than the advertised time period of that clock. Chapter 5. Length Contraction [01:00:02]The last thing I want to do is length contraction. Length contraction says, if you have a meter stick and you and I bought them from the same store, once you’re in this plane or this rocket, I will claim your meter stick is actually shorter than a meter. In fact, a shortening factor is this. This is the shortening factor. In fact, way back, in the derivation of the laws of transformation, the fudge factor I calculated, taking your lengths into my lengths, is precisely connected with this factor. If you say the length to something, I will tell you that the length is actually less than that because your meter sticks are short. But let me prove that to you. So, let’s take a rod that’s moving. It’s moving at a speed u; you’re carrying the rod and I want to find its length. What should I do? That’s the question, okay? So, the rod is moving at speed u. I want to find its length. So, Δx′ is Δx  u Δt over the square root. I’m going to make a pair of events to find the length of the rod. Event 1, remember this rod is zooming past me, Event 1 is when this front end of the rod hits a certain marking on my graduated x axis and event 2 is when the back end of the train hits my origin, and the distance between those two is the length of the rod, provided one condition. How do you find the length of a moving rod? You find this end and that end at the same time. Otherwise you will screw up, right? If you find this end now, go on a lunch break and come back and the road is over there. And I said that’s the rear end of the rod, and you get back to be the length; that’s not how you find the length of a moving object. You find the length by measuring the coordinates of the two ends at the same time. Consequently, at the same time means Δt must be 0; Δx is the length according to you. According to the person going with the rod, the two events, since they take place at the two ends of the rod, are separated by the length of the rod because the rod, according to you, is not going anywhere. Suppose you have the rod. It’s not moving relative to you. Something happens at one end, something happens with the other end. What’s the spatial distance between them? It’s the length of the rod. That’s the meaning of the length of the rod. Therefore, if you crossmultiply, you get this result, L is L_{0} times–;So, a rod will appear longest in its rest frame and to anybody else it will appear shorter. Clocks appear fastest in the rest frame; rods appear longest in the rest frame. For anybody to whom a rod is moving, the rod appears short. So, here’s another paradox then. I make a hole in my x axis. I make the hole to be half a meter long. This is half a meter. You and I bought a meter stick but you are moving at a velocity where this number has become half. u over c is 1 over root 2. u^{2} over c^{2} is whatever. Look, it’s whatever it takes to make this factor half. So, I expect your meter stick to have shrunk to half its length. So, I make a hole on the table, half a meter long, the rod goes by, I think it would fall into the hole. That’s my expectation. You come and say, my meter stick is a meter stick. The hole that you dug that you think is half a meter long actually is a quarter meter long. So, there’s no way my meter stick is going through that hole in the table because it’s four times as long as the hole in the table. Do you understand that? Normally, you make a meter stick, you put it on top of a hole of the same length, it will fall down. But if it’s a moving meter stick, which has contracted according to me from one meter to half a meter, half a meter hole is enough for it to fall. But from your vantage point, my half a meter hole looks like a quarter of a meter hole. Your meter stick still looks like a meter. The question is, when this experiment occurs, will it fall or not? Hey, these are all paradoxes. If you invented the theory, you have got to defend these paradoxes. You cannot get two answers to one question. That’s a logical issue. So, the answer to these problems is, if you know one point of view in which it is correct, that’s the right answer. Then, the other person has to figure out what happened. So, in my point of view, if I made a half a meter hole and your rod has shrunk to half the length, it’s going to fall. It will fall. The question is, how does the other person reconcile himself to the fact that here is, according to that person, the hole is a quarter meter long and the rod is one meter long? Well, I should–for it to fall, you’ve got to give it a little tilt, okay, so that it can actually fall in. I give it a little here. How can this object fall into that? The way other person explains it is, he said first, this end went in here. Other end was sticking way out. Sometime later, the second end went in here. That’s how we got to the hole. Namely, the two ends did not go at the same time. That way, it’s certainly possible for a onemeter stick to go through a quartercentimeter hole in the table. I mean, give the rod a little tilt so it can fall into the hole. The tip enters first; the tail is way outside but after a while the tail goes in and in that manner; the rod goes in. So, they can reconcile this paradox by saying the two ends did not go at the same time. Also, if you ask the other person who thinks something is two meters long and the other person thinks it’s half a meter long, how did they reconcile it? I say, I measure the two ends at the same time but you will tell me, “You did not measure the ends of my meter stick at the same time. You measured one end and you goofed off and you came back and measured the second end. By the time the rod had slipped to the right, you measured that end and you goofed off. You waited until it came there and then you measured the other end. That’s where you got half instead of one.” And nobody is right or wrong. I did measure the two ends simultaneously according to me but you don’t have to agree they were simultaneous. So, it’s by going soft on what simultaneity means that you are able to reconcile this. Because in Newtonian days, simultaneity is absolute; length is absolute. In relativity, simultaneity is relative and length is also relative because the operational way to find the length of a body will not satisfy all observers. If I find the length of your moving meter stick, the two ends at the same time, you will say I didn’t measure them at the same time and the formulas relative will allow that. That’s how we live with the fact that we accuse each other of having relatively short meter sticks. I blame you on your measurement, but that blame is not a genuine blame because you can dig a hole equal to the reduced length and things will be falling. So, you have every right to say things are shorter, and I will say I didn’t shrink; I fell in because my nose went in first and then a little later the tail went. There are various paradoxes. I’ll just leave you with one. You have a garage. Your parents built you a garage, 12 meters long. You bought a car 14 meters long. Can you park it? Well, if you go at sufficiently high velocities, so the 14 gets shrunk to 12, there will be a brief instant in which both ends of your car are inside the garage. Of course, you will have to smash through the back end of the garage but you will maintain, “Yes, I smashed into that but my car was in the garage for some time” and the parent will say, no, the front end went in and smashed the rear. The back hadn’t even come in. A little later the back came. So, everyone will agree you broke the garage, you broke the car, and the lengths will contract relatively to each other and the phenomena is that according to you, there was a time the whole car was in the garage, that according to the parents it was not. Okay. [end of transcript] Back to Top 
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