Special Relativity, Lecture 6: The Twin Paradox - 3rd Year Student Lecture

— Welcome to the sixth lecture of uh special relativity. So, today we will see uh some very nice consequences of the Lorentz transformations. So, over the last few lectures in the last few lectures, we derived in two different ways these Lorentz transformations, which are basically the transformations between coordinates of a space-time between two frames, a frame O and a frame O prime, which is moving at a speed V with respect to the system O. Okay? Now, something that I would like to um to to kind of uh stress in this lecture is the following. I am assuming So, basically, people have different motivations in life. There is a lot of people uh and that's kind of geeks, nerds, physicists, you know, they overlap very much. And for us, what is important is to understand the truth and to understand how the universe works. Okay? Now, we are lazy. We are mostly lazy. However, if we need to work a bit hard to understand how the universe works, that wins. Okay? And the important thing is that these transformations are how the universe works. Okay? So, the Galilean transformations, they are fine, but they are just an approximation. Okay? They are an approximation if the speeds between the two systems are really small compared to the speed of light. But although you need an effort to understand a special relativity, a very important thing about special relativity, it really described how space-time works. Okay? Now, since 2,000 years, scientists have been wondering what is the nature of a space and time. And now, or like 100 years ago, for the first time, we are really understanding it. Okay? So, this is an historic moment. So, we are all very lucky that we can uh study this. Okay? So, I know some things are not super easy, but please, like, if your aim, right, it is to really understand how the universe works, then you should need and you do need to understand this. Okay? Um Okay. So, this, on the other hand, these transformations have crazy consequences. And today, we will describe So, we have been seeing some of the consequences. And uh let's describe uh another consequence today, which is uh super nice. So, we will see what is known as time dilation. And imagine that O prime carries a clock. Okay? That ticks with period delta. Okay? So, delta could be 1 second, could be 1 week, could be 1 month. Every second, tick tick. The question is what is this period according to O? Okay? And this is very easy to answer with the formulas we have just arrived. Because the ticks these ticks or the clicks are events at C T prime X prime equals N C delta zero with N equals zero, one, two, three, sorry. Three dot dot. Okay? So, each of these events, they represent the clock of O prime ticking at times equals delta, two delta, three delta, four delta. Is that okay? Yeah? Beautiful.

Beautiful. So, uh for O, these events are very easy to see where they are at because we just use this transformation. Okay? We just use the Lorentz transformation without thinking too much about it. And this is given by gamma. One V over C one and C delta zero. Okay? But then, we just can compute the time T that these clicks corresponds to. And we see that the time between clicks for O is gamma times delta. Okay? Because of this factor gamma here. Now, this factor gamma is bigger than one. Okay? For any speed. So, that means that this is bigger than delta. All right? For instance, if the speed is if the speed V is equal to a square root of three over two times the speed of light, then this gamma is exactly two. Okay? Just to give an example. And by increasing the speed and getting closer and closer to the speed of light, you can make this gamma bigger and bigger. Okay? Beautiful. Now, when you think of uh clocks in a special relativity, you are thinking about something that could be very sophisticated. It doesn't have to be sophisticated. Imagine that O prime is in a spaceship. Okay? And delta is 1 year. And every time that this delta comes, they celebrate their birthday. Okay? That's the click. The click is a biological thing. It could be they celebrate their birthday. However, if they are moving with respect to me, what I will see is that they celebrate their birthday every 2 years from my point of view. And they age 1 year every 2 years for me. Okay? They really do age less if they are moving with respect to me. Okay? This is real. So, the clock could be a biological clock. And this is true. Okay? And in two, three lectures, I will describe an experiment in particle physics where we see this. And we see this happening. Okay? And in the example I will give, the speed is more than 90% the speed of light. So, we will see things aging way, way less. Okay? Okay, there will be muons, not people, but it is true. Okay? And this effect, this effect leads to something quite famous, which is the twin paradox. So, the twin paradox, you have to choose two twins. One of them has to start with A. Let's say Alice. And the other one has to start with B. For instance, Bob. Okay? It could be Alfredo and Beatrice or any other names, but following the general literature, we will choose Alice and Bob. Okay? So, what happens in the paradox is that Alice They are born. Okay? So, T is equal to zero very easily. And as soon as they are born, Alice stays on Earth. Okay? And Bob goes to a far away planet. Okay? At some uh speed, which is almost the speed of light. It's very fast, right? So, Bob goes to the planet. Okay?

So, as Bob um goes to the planet, uh Alice sees that after time T in Alice frame, which is just Earth, okay? Bob reaches the planet. Okay? And then, when he arrives to the planet, okay? He comes back to Earth. Fair enough. So, from at the same speed, okay? So, he goes to the planet at the speed V, quite fast, and he comes back also at the speed V. Okay? Now, according to Alice, the time that Alice see, this is 2 T, let's say. Okay? So, Bob has reached the planet, and then he comes back. And imagine that whenever he's traveling, okay? He celebrates his birthdays. Why not? Okay? But, what Alice will see is that he celebrates a birthday every 2 years. Let's say that he moves at this uh speed. Okay? And when he comes back, he also celebrates his birthdays every 2 years. So, that when they are back together, Bob is much younger than Alice. Okay? Because of this factor. All right? So, the time for Bob, the time that has age, that Bob has age, is 2 T over this factor gamma, which is um bigger than one. Okay? Any questions so far? Yes. So, Bob experiences time the same way? That's a very good point. So, what we are seeing, right? So, now it will come the paradox. So, Bob experiences the time in exactly the same way. Okay? So, if you are moving at the speed V, no matter how fast the speed is, you will experience time in exactly the same way. Because because that was one of our postulates. Our postulates tell us that if our speed is constant, we are just an inertial frame. So, Bob is an inertial frame. He's moving at a constant speed V. No matter how close to the speed of light Bob is going, the speed is constant, is an inertial frame. There is no experiment, nothing that Bob can do that can show that Bob is actually moving, right? Because he's moving with respect to Alice. All right? So, but the point is that then Bob comes back, and when he comes back, they see each other. Okay? And Bob is younger. But, now it comes the paradox, right? Because we are seeing things from the point of view of Alice. Okay? Let's have a look at the things from the point of view of Bob. From the point of view of Bob, Bob is put in his spaceship, Alice takes Earth with her, moves at the speed V, all with the Earth. Okay? It reaches some point, and then comes back with the whole Earth. Okay? Why not? And when she does that, she should be younger. Okay? Yeah? So, the question is what is going on? Who is younger of the two? Yeah? And that is the paradox. That is the twin paradox. So, people has done uh science fiction movies with it. They have done all kind of crazy stuff. Now, beautiful. Let's solve the paradox. Let's understand the paradox, and let's see why it is not a paradox. Okay? So, the resolution of the paradox, resolution, — [snorts] — is that Bob has to come back. And in order to come back, he [snorts] needs to accelerate.

Okay? And when he needs to accelerate, he's changing frame. All right? Because his uh speed first is V, and then it's minus V. So, the frame changes. So, that the situation is not symmetric. Because Bob, when at the speed V, has to turn back, actually accelerating, and in this acceleration, he was not an inertial frame anymore. Okay? Uh so, that when they meet, Bob is actually younger. All right? Uh because the Alice is an inertial frame all the time. So, this conclusion by Alice is true, but the conclusions by Bob are not true because he was not an inertial frame. Okay? Let's do some pictures that will show that is very clear. Now, in the problem sheet, you will have to solve a beautiful problem of two twins that depart, and they send beams of light to each other every year. But, this is uh like a baby version of that. So, let's try to plot something very nice. So, this is Alice. Okay? And this is the system of Alice. Sorry, this should be a straight line. Okay? And this E is the time at which they are born, and Bob leaves. Okay? To planet P. So, this one here, let's say here, is uh this is time T for Alice. — [snorts] — And this event B is where Bob uh reaches the planet. So, this is Bob. Like this. — [snorts] — And you see that when Bob travels to the planet, he is in an inertial frame, like this, which we will call O prime. Okay? However, whenever Bob now has to go back to Earth, he has to change frame. And these angles should be the same. So, this is Bob going back to Earth. And this one here is another system, which we denote O prime prime. Okay? So, actually, in this picture, there are three inertial frames. There is the inertial frame O, which is Earth. of Bob going there to the planet, which is O prime. And there is the inertial frame of Bob returning, which is the frame O prime prime. Okay? And this event P is [snorts] Bob reaching the planet. Any questions about this? No, it's good. Yes. Yeah, I'm if you were to derive everything from Bob's perspective, — Yes. you were to do the Lorentz transformation into O prime, and then into O double prime, which you derive the exact same thing as what Alice sees. Beautiful. Yes. Now, let's try to do something like this. Fantastic. Thank you. Yes, that's fantastic. That's what we are kind of about to do, but we will just drop something. Indeed. Beautiful. Yeah. Absolutely. I mean, you could do and we will do something like that in a second. Indeed, you could have drawn this axis straight, right? And then the axis of Alice would be a little bit to the left. Indeed. But the problem is that what when Bob comes back, he needs to change reference

frame to O' prime prime. Right? And so there is this system. But let's try to see the following. So why So this event So let's draw three events. Let's call this Y which is what we call time T. Okay? And this is And this Y What does it mean? Is when Alice thinks Bob reached the planet. Okay? Uh beautiful. But let's try to answer the question that you are asking. Let's imagine now that this system Let's take the system O' prime. Okay? And in this system O' prime, let's draw the axis. Remember, it has to be So this is the axis uh T' prime equals constant. Right? So this is the ans the T' prime of the event P. Okay? So this point here if you do this point here which we call X So what this X means this X is how old Bob thinks Alice is when he reaches the planet. Okay? Answering exactly your question. So from the point of view of Bob okay? When he reaches the planet this is the line uh X' prime equals zero and this is the line T' prime equals a constant. Okay? From the point of view of Bob. So that if he wants to see how old Alice is, he has to continue this line T' prime P and see the intersection of that with X with the line of Alice. Okay? So what he thinks the age of Alex Alice is when he reaches the planet, he thinks Alice is a bit younger. Okay? And this is this factor uh is T over gamma basically. Okay? So from the point of view of Bob he thinks that Alice is younger when he reaches the planet. Okay? So from the point of view of Alice, Alice thinks Bob is younger. But from the point of view of Bob, exactly the same happens and Bob thinks Alice is younger. Okay? But now there is something very cool. So Bob is going uh to the planet, he starts feeling nostalgia for Earth. I don't know if that is an English word, but I think you understand. He misses Earth, which is funny because he was never in on Earth. He departed as soon as he was born. But let's say he wants to be on Earth, then he starts going back. Okay? Now when he starts to going to go back he changes to this system here. Okay? Now this system O' prime has some T' prime of P which is given by this line. So there is another event here and here is where it gets a bit crazy but is the resolution of the paradox. So this event set is how old Bob thinks Alice is when he starts going back. Okay? Because he has changed He has to change a reference frame in order to start going back. Now he thinks that the age of Alice went from X to Z. So from his point of view, what is happening actually, and this is true, is

that as he goes back he sees that Alice suddenly became 2 years older just because he came back. Okay? Just this turnaround from his point of view Sorry. Uh makes Alice to grow older in his view. Okay? So that when they are together, indeed, Alice is older and there is no paradox. Okay? So this turning here and Alice from the point of view of Bob jumping from X to Y is from X to Z is the resolution of the twin paradox. Which is not that paradox at all. Is that okay? Yes. So is it the case that Bob's perspective at every point other than the turnaround, Alice is aging slower than Yes. — And then the turnaround just makes it so that overall she's actually aged more. Beautiful. — over it over corrects for the 100%. That is exactly what's going on. Yes. So would it be that in the process of accelerating and decelerating, this aging actually happens like smoothly? Mhm. Rather than be uh Absolutely. Yeah. Absolutely. So here, although this is a tip you could imagine that in real life you have to start decelerating and then accelerating back. Indeed. So this would be more round if you wish. So this happens um yeah, in a smooth way. So that's what would happen in real life. Yeah. Yes. So does the length of time that she accelerates and decelerates affect That that's a beautiful question actually. So okay. So if you want to understand acceleration super precisely you would have to do to go into the realm of general relativity. However, for this problem, you can make this as a small as you want and the conclusions of the problems do not change. Um we will deal with accelerated motion actually in uh towards the in a few lectures from now and you will see. Uh so it's no gravity. Um Yeah. I mean there is a lot to say about acceleration and gravity. Uh why they are the same thing etc. But basically we will deal One can deal with acceleration without using general relativity and that's what we will do. Um yeah. But one could do a more careful analysis. But basically this conclusion doesn't depend on these details because you could do this smoothly but in such a way that is as small as possible so to say. Okay? And yeah, it's a So one paradox less in your life. Any other question? Beautiful. Very nice. Oh beautiful. Yeah. This is nice. So now let's let's solve another very nice problem. And again uh yeah, let me use this here. And here to do some problems. Uh so the problem we want to show to solve is Can you see if you are right here? Yeah? Is addition of velocities. Okay? In a special relativity. I told you that you cannot add up velocities. But what do you do? Okay? And we will find the answer to that. It's a beautiful answer. So what we are imagining now is that we have three systems O O' prime and O' prime we imagine that is traveling at velocity U with respect to O. Okay? And this one here is traveling at velocity V with respect to O' prime. Okay? And what we want to find is W which is the velocity of O' prime with respect to O. Okay? And we want to find in usual Galilean mechanics, W would be U plus V.

Here, that's not true. Okay? So that that's what we want to solve. Uh so all Yeah, just in Galilean In the Galilean problem, you would have a system O which let's say is a static. Then you would have a system O prime which is moving at a speed U. Right? And now you need to imagine a system O prime which is moving at a speed V with respect to O prime. Okay? So that in usual um in usual um Galilean mechanics, W would be the sum of these two velocities. Is that okay? Is it okay? I I'm just thinking while I'm thinking. Um if Yeah, if O prime is traveling at velocity V with respect to O, then surely O and O prime are like the same or is it because So U is traveling with respect to um with respect to O. And this guy, O prime prime, is traveling at velocity V with respect to O prime. Yeah? So basically, imagine an ant, right? And the ant is moving at 1 m per second in the direction forward to my not notebook. Okay? But it's velocity not speed that we care about the direction. Right. Right. It's always in the positive X. Yeah, I apologize. Yeah. So it's always, let's say, positive X. We are in one dimension. So we will work we will work in uh yeah, in more than one dimension later. But here it's just one dimension and they are moving in the positive direction. Yeah. Sorry, you're right. So it's velocity, let's say, U and V and both of them are positive. Okay? Yeah? So it's clear what the problem is? Beautiful. So now in a special relativity, always you need to plot diagrams. Okay? I'm plotting diagrams is super helpful. Uh it's the way to go. And let's um Yeah, let let's plot a diagram here. Yeah, let's Let me draw it here. So first, this is the system O. Okay? And O prime which is moving at a velocity U with respect to O. And O prime which is moving at a velocity W which is what I want to find, what we want to find. Or we could draw the same from the point of view of O prime. So this would be exactly what you were suggesting. And then O would be here. Right? And O prime would be here. Okay? Moving at a velocity V. Yeah? And now we look at one event and the same event from both point of view. So this is the event A. And the event E is where everyone coincides. So this is the event E. And the event A is somewhere here. Okay? Beautiful. So now we are going to be very careful. Okay? So write down a few equations. If you do not understand any of these equations, please stop me. Um Beautiful. So first we will write the coordinates of the events E and A in O prime. Okay? Well, that's very easy. So T prime of E and X prime of E have to be zero because just that's just the origin. Okay? Sorry, the origin should be like here. And T prime of A we call it T prime. And X prime of A is some X prime. But notice that because the event A is in the world line of O prime and we

are saying that O prime is moving at a speed at velocity V with respect to O prime, then this X prime has to be V T prime. Okay? Yeah? Yeah. So that's everything from the point of view of O prime. All right? Beautiful. Yeah. So now I want to know what are the same events in O. Okay? And for that we used our beloved Lorentz transformations. Okay? With the speed U which is the speed at which O prime is moving with respect to O. Okay? So we just use Lorentz. Very easy. Easy peasy. So from this point of view uh T of E is actually zero. And X of E is also zero. Very easy. So it's just the Lorentz transformations of zero zero. While C T of A is just this transform. You need to transform this, put in the matrix on the right, and then you find gamma of U. C T prime plus U over C X prime while X of A is gamma of U X prime plus U T prime. Okay? So this is just applying the Lorentz transformations Sorry. to this event A. Beautiful. But then once we have this we can compute W. Right? And here, remember um that X prime is V T prime. And W because A is in the world line of O prime we should be able to compute W by taking X A divided T A. Okay? Yeah? We just take these. Use these. And we find the following beautiful formula. U plus V divided one plus U V over C squared. This is the way you add velocities in real life. When the velocities are not a small compared to the speed of light. Okay? So velocities they don't quite add up. But there is this extra factor. Okay? And in particular uh of course, if U and V are a small compared to the speed of light you just uh add them up. But if they are not a small then you need to take this uh thing into account. And this is the way you add velocities. Something beautiful about this is that if you add up two velocities that are almost at the speed of light, the sum can never be bigger than the speed of light. So whatever you add up, you can never reach the speed of light. And that formula beautifully takes care of this. Is that okay? Questions? Yes. If you have a big negative U and a big positive V, that denominator gets Right, very large very quickly. What What's the physical phenomenon that it's kind of corresponding to? Okay. So here, in the diagrams, etc., I have assumed U and V are bigger than zero. Um because the this is um this is a bit simpler. Is a nice exercise to assume that some of them are negative, but then you would have to use like a slightly different transformations. Okay, so that is not valid. Yeah, that's

for U and V positive, let's say. Yeah. I mean, here you would have to be a bit careful here. So, and in the diagrams, one diagram would be to the left and another right. But actually, it's not super complicated. Uh basically, you can see that from that point of view uh you could say Yeah, you could start with this diagram, let's say. So, the starting point would be this diagram. Right? And then you would like to find the difference between this and this. Yeah. So, so you can find this from here, but you have to be slightly careful. Yeah. But this is a very good question. Yeah. But even so, you can never reach speeds bigger than the speed of light. Yeah. Um Yeah. I think probably this works even if U and V are negative. I would have to think a little bit, but I Probably they are is fine. Like either way. But we will they use even more than this. in three dimensions. Uh and that will be the full-fledged uh addition of velocities. Is that okay? Yeah? Beautiful. So, that all of you loves group theory. So, we will uh Do you like group theory? Or you don't like group Yeah, that's the group. Yeah. So, beautiful. So, what we will show will show and we are uh we will develop that the next lecture. But we will describe from a very nice point of view, from a kind of group theory point of view the Lorentz group in 1 + 1 dimensions. And dimensions as we will see very quickly, is something that is similar to rotations. But they are actually called hyperbolic rotations. So, the first definition in order to make that precise the first definition is the rapidity. So, the rapidity uh associated to a speed U with absolute value of U smaller than C is given by phi of U equals tangent arc tangent hyperbolic of U over C. Okay? So, if you have U between minus C and C you define another variable phi of U, which is given by this arc tangent. Now, you may have drawn before um this arc tangent in Mathematica or something. You will see that if you draw this is phi of U. This is U over C. This has singularities as U approaches C and minus C. And the arc tangent behaves something like this. Okay? So, that the rapidity and the velocity are in one-to-one correspondence in this region, in the region in which the absolute value of U is a smaller than C. Okay? So, we don't lose any information by talking about the rapidity. All right? So, instead of the velocity we will use the rapidity. Why not? Okay? What happens if we do that? Okay? So, that is that we get a few very cool formulas, very cool formulas. If you have this, for instance — [snorts] — uh you have that for instance

gamma of U defined here in terms of the rapidity is cosh of the rapidity phi. Very cool, right? And U over C gamma of U is sinh of phi. Okay? That's quite cool. Looking good. Then Then if you take these two things and put them into a into the Lorentz transformation Let's put it here. So, gamma of U one U over C one What is this? cosh phi sinh phi cosh phi. Does it ring a bell? Do you remember how a rotation looks like? Okay? A rotation looks very similar. It's cosine minus sine cos. This is like a rotation, but in a space where one dimension is different because it's the time. Okay? Yeah, we just did a change of coordinates. Rather than using U, the velocity we use this rapidity. And if we use the rapidity the Lorentz transformation is like this uh transformation. Okay? And something super cool that shows that the structure is a structure of a group is that if you uh do do the product of matrices cosh phi one sinh phi one cosh phi one Sorry. So, cosh phi two sinh phi two. Is it okay if I use this instead of writing cosh? Thank you. sinh phi two cosh phi two and you do this matrix multiplication and you use identities between coshes and sinches you get something beautiful. cosh of phi one plus phi two sinh and cosh phi one plus phi two. Okay? So, that in terms of the rapidities this matrix multiplication is just addition of phi one plus phi two. Okay? Very much like rotations. If you have a rotation by an angle pi over six and another angle pi over six, this is the same as pi over three, right? Because you can uh basically add up the angles. All right? So, this tells that these matrices form a group. And the inverse I will write one more formula. is just uh obtained by reversing phi to minus phi. So, if you just do the inverse of a Lorentz transformation, that's like inverting rapidities. And let me write for you addition of velocities in terms of rapidities because it's very cool. phi of omega So, it's the the velocity Sorry, that we the write before. This one here in terms of rapidities phi of omega is phi of U plus phi of V. That's it. Okay? So, if you use rapidities instead of velocities, the rapidities just add up. Okay? So, what happens is that the rapidities go back to Galilean transformations, right? For very small uh velocities. And that's why you get Galilean

transformations. And you can show that this is equivalent to this rule that we found before. Okay? Now, uh you may have heard about integrability. I am not sure whether you have heard that concept. So, basically, in every single problem or in a lot of serious problems in research having to do with relativity, you will be using rapidities. You will not using the velocity of particles. Okay? So, this rapidity is a very important concept. And yeah, and it's a concept that makes manifest this uh group structure of the Lorentz transformations. So, so it's quite cool. So, we are kind of back to addition, but with this rapidities. Thank you very much.