My son, who’s fifteen, said he liked my post on lasers. That’s good, because I effectively wrote it thinking of him as part of the audience. He also said it stimulated him to considering taking on studies in engineering later. That’s great. I hope he does, so he doesn’t have to go through what I am going through right now. Indeed, when everything is said and done, you do want your kids to take on as much math and science they can handle when they’re young because, afterwards, it’s tough to catch up.

Now, I struggled quite a bit with bringing relativity into the picture while pondering the ‘essence’ of a photon in my previous post. Hence, I’d thought it would be good to return to the topic of (special) relativity and write another post to (1) refresh my knowledge on the topic and (2) try to stimulate him even more. Indeed, regardless of whether one does or doesn’t understand any of what I write below here, relativity theory sounds fascinating, doesn’t it? 🙂 So, this post intends to present, in a nutshell, what (special) relativity theory is all about.

**What relativity does**

The thing that’s best known about Einstein’s (special) theory of relativity is the following: the mass of an object, as measured by the (inertial) observer, increases with its speed. The formula for this is m = γm_{0}, and the γ factor here is the so-called *Lorentz* factor: γ = (1–u^{2}/*c*^{2})^{–1/2}. Let me give you that diagram of the Lorentz factor once again, which shows that *very *considerable speeds are required before relativity effects kick in. However, when they do, they kick in with a vengeance, it seems, which makes *c* the limit !

Now, you may or may not be familiar with two other things that come out of relativity theory as well:

- The first is
*length contraction*: objects are measured to be shortened in the direction of motion with respect to the (inertial) observer. The formula to be used incorporates the*reciprocal*of the Lorentz factor: L = (1/γ)L_{0}. For example, a stick of one meter in a space ship moving at a velocity*v*= 0.6*c*will appear to be only 80 cm to the external/inertial observer seeing it whizz past… That is if he can see anything at all of course: he’d have to take like a photo-finish picture as it zooms past ! 🙂 - The second is
*time dilation*, which is also rather well known – just like the mass increase effect – because of the so-called*twin paradox*: time will appear to be*slower*in that space ship and, hence, if you send one of two twins away on a space journey, traveling at relativistic speeds (i.e. a velocity sufficiently close to*c*to make the relativistic effect significant), he will come back younger than his brother. The formula here is equally simple: t = γt_{0}. Hence, one second in the space ship will be measured as 1.25 seconds by the external observer. Hence, the moving clock will appear to run slower – again: to the external (inertial) observer that is.

These simple rules, which comes out of Einsteins’ special relativity theory, give rise to all kinds of paradoxes. You know what a paradox is: a paradox (in physics) is something that, at first sight, does not make sense but that, when the issue is examined more in detail, does get resolved and actually helps us to better understand what’s going on.

You know the twin paradox already: only of the two twins can be the younger (or the older) when they meet again. However, because one can also say it’s the guy staying on Earth that’s moving (and, hence, is ‘traveling’ at relativistic speed) – so then the reference frame of the guy in the spaceship is the so-called inertial frame, one can say the guy who stayed behind (on Earth) should be the youngest when they meet after the journey. I am not ashamed to say that this actually *is *a paradox that is difficult to understand. So let me first start with another.

**The ladder paradox**

While the twin paradox examines the time dilation effect, the ladder paradox examines the length contraction effect. The situation is similar as the one for the twin paradox. However, because we don’t have accelerating and decelerating rockets and all that (cf. the twin paradox), I find this paradox not only more straightforward but also more amusing. Look at the left-hand side first. We have a garage which has both a front and back door. A ladder passes through it, and it seems to fit in the garage as you can see. Now, that may or may not be because of the length contraction effect, of course. Whatever. In any case, it seems we can (very) quickly close both doors of the garage to prove that it fits. Now look at the right-hand side. Here we are moving the garage over the ladder (I know, not very convenient, but just go along with the story). So now the ladder frame is the inertial reference frame and the garage is the moving frame. So, according to that length contraction ‘law’, it’s the garage that gets shorter and it turns out the ladder doesn’t fit any more. Hence, the paradox: does the ladder fit or not? The answer must be unambiguous, no? Yes or no. So what is it?

The paradox pushes us to consider all kinds of important questions which are usually just glossed over. How does we decide if the ladder fits? Well… By closing both the front and back door of course, you’ll say. But then you mean closing them *simultaneously*, and absolute simultaneity does not exist: two events that appear to happen at the same time in one reference frame may not happen at the same time in another. Only the *space-time* interval between two events is absolute, in the sense that it’s the same in whatever reference frame we’re measuring it, *not *the individual space and individual time intervals. Hence, if you’re in the garage shutting those doors at the same time, then that’s your time, but if I am moving with the ladder, I will *not* see those two doors shutting as something that’s simultaneous. More formally, and using the definition of space-time intervals (and assuming only one space dimension x), we have:

*c*Δt^{2 }– Δx^{2 }= *c*Δt’^{2 }– Δx’^{2}.

In this equation, we’ll take the x and t coordinates to be those of the inertial frame (so that’s the garage on the left-hand side), while the the primed coordinates (x’ and t’) are the coordinates as measured in the other reference frame, i.e. the reference frame that moves *from the perspective of the inertial frame*. Indeed, note that we cannot say that one reference frame moves while the other stands still as we we’re talking *relative* speeds here: one reference frame moves in respect to the other, and vice versa. In any case, the equation with the space-time intervals above implies that:

*c*(Δt^{2 }*– *Δt’^{2}) – (Δx^{2 }– Δx’^{2}) = 0

However, that does *not* imply that the two terms on the left-hand side of the above equation are zero *individually*. In fact, they aren’t. Hence, while it must be true that *c*(Δt^{2 }*– *Δt’^{2}) = Δx^{2 }– Δx’^{2}, we have:

Δt^{2 }*– *Δt’^{2 }≠ 0 and Δx^{2 }– Δx’^{2} ≠ 0 or Δt^{2 }*≠ *Δt’^{2 }and Δx^{2 }≠ Δx’^{2}

To put it simply, if you’re in the garage, and I am moving with the ladder (we’re talking the left-hand side situation) now, you’ll claim that you were able to shut both doors momentarily, so that Δt^{2 }= 0. I’ll say: *bollocks! *Which is rude. I should say: my Δt’^{2 }is *not *equal to zero. Hence, from my point of view, I always saw one of the two doors open and, hence, I don’t think the ladder fits. Hence, what I am seeing, effectively, is the situation on the right-hand side: your garage looks too short for my ladder.

You’ll say: what is this? The ladder fits or it doesn’t, does it? The answer is: no. It *is *ambiguous. It *does *depend on your reference frame. It fits in your reference frame but it does not fit in mine. In order to get a non-ambiguous answer you have to stop moving, or I have to stop moving– whatever: the point is that we need to merge our reference frames.

Hence, paradox solved. In fact, now that I think of it, it’s kinda funny that we don’t have such paradoxes for the relativistic mass formula. No one seems to wonder about the apparent contradiction that, if you’re moving away from me, you look heavier than me but that, vice versa, I also look heavier to you. So we *both *look heavier as seen from our *own *respective reference frames. So who’s heavier then? Perhaps no one developed a paradox because it is kinda impolite to compare personal weights? 🙂

Of course, I am joking, but think of it: it has to do with our preconceived notions of time and space. Things like inertia (mass is a measure for inertia) don’t grab our attention as much. In any case, now it’s time to discuss time dilation.

Oh ! And do think about that photo-finish picture ! It’s related to the problem of defining what constitutes a length really. 🙂

**The twin paradox**

I find the twin paradox much more difficult to analyze, and I guess many people do because it’s the one that usually receives all of the attention. [Frankly, I hadn’t heard of this ladder paradox before I started studying physics.] Feynman hardly takes the time to look at it. He basically notes that the situation is not unlike an unstable particle traveling at relativistic speeds: when it does, it lasts (much) longer that its lifetime (measured in the inertial reference frame) suggests. Let me actually just quote Feynman’s account of it:

“Peter and Paul are supposed to be twins, born at the same time. When they are old enough to drive a space ship, Paul flies away at very high speed. Because Peter, who is left on the ground, sees Paul going so fast, all of Paul’s clocks appear to go slower, his heart beats go slower, his thoughts go slower, everything goes slower, from Peter’s point of view. Of course, Paul notices nothing unusual, but if he travels around and about for a while and then comes back, he will be younger than Peter, the man on the ground! That is actually right; it is one of the consequences of the theory of relativity which has been clearly demonstrated. Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a “paradox” only by the people who believe that the principle of relativity means that *all motion* is relative; they say, “Heh, heh, heh, from the point of view of Paul, can’t we say that *Peter* was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet.” But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship—the rockets went off, things jammed up against one wall, and so on—while Peter felt nothing.

So the way to state the rule is to say that *the man who has felt the accelerations*, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an “absolute” sense, and it is certainly correct. When we discussed the fact that moving mu-mesons live longer, we used as an example their straight-line motion in the atmosphere. But we can also make mu-mesons in a laboratory and cause them to go in a curve with a magnet, and even under this accelerated motion, they last exactly as much longer as they do when they are moving in a straight line. Although no one has arranged an experiment explicitly so that we can get rid of the paradox, one could compare a mu-meson which is left standing with one that had gone around a complete circle, and it would surely be found that the one that went around the circle lasted longer. Although we have not actually carried out an experiment using a complete circle, it is really not necessary, of course, because everything fits together all right. This may not satisfy those who insist that every single fact be demonstrated directly, but we confidently predict the result of the experiment in which Paul goes in a complete circle.”

[…] Well… I am not sure I am “among those who insist that every single fact be demonstrated directly”, but you’ll admit that Feynman is quite terse here (or more terse than usual, I should say). That being said, I understand why: the calculations involved in demonstrating that the paradox is what it is, i.e. an *apparent *contradiction only, are *not* straightforward. I’ve *googled *a bit but it’s all quite confusing. Good explanations usually involve the so-called *Minkowski *diagram, also known as the spacetime diagram. You’ve surely seen it before–when the light cone was being discussed and what it implies for the concepts of past, present and future. It’s a way to represent those spacetime intervals. The Minkowski diagram–from the perspective of the twin brother on Earth (hence, we only have *unprimed *coordinates x and (*c*)t)– is shown below. Don’t worry about those simultaneity planes as for now. Just try to understand the diagram. The twin brother that stays just moves along the vertical axis: x = 0. His space-traveling brother travels out to some point and then turns back, so he first travels northeast on this diagram and then takes a turn northwest, to meet up again with his brother on Earth.

The point to note is that the twin brother is not traveling along *one* straight line, but along two. Hence, the argument that we can just as well say *his *frame of reference is inertial and that of his brother is the moving one is not correct. As Wikipedia notes (from which I got this diagram): “The trajectory of the ship is equally divided between *two* different inertial frames, while the Earth-based twin stays in the same inertial frame.”

Still, the situation is essentially symmetric and so we could draw a similar-looking spacetime diagram for the primed coordinates, i.e. x’ and ct’, and wonder what’s the difference. That’s where these planes of simultaneity come in. Look at the wonderful animation below: A, B, C are simultaneous events when I am standing still (*v* = 0). However, when I move at considerable speed (*v* = 0.3*c*), that’s no longer the case: it takes more time for news to reach me from ‘point’ A and, hence, assuming news travels at the speed of light, event A appears to happen later. Conversely, event C (in spacetime) appears to have happened *before *event B. Now that explains these blue so-called simultaneity planes on the diagram above: they’re the white lines traveling from the past to the future on the animation below, but for the trip out only (*v *> 0). For the trip back, we have the red lines, which correspond to the *v* = –0.5*c* situation below. So that’s the return trip (*v *< 0).

What you see is that, “during the U-turn, the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the Earth-based twin.” Hence, “when one transfers from the outgoing frame to the incoming frame there is a jump discontinuity *in the age of the Earth-based twin*.” [I took the quotes taken from Wikipedia here, where you can find the original references.] Now, you will say, *that* is also symmetric if we switch the reference frames. Yes… *Except for the sign*. So, yes, it is the traveling brother who effectively skips some time. Paradox solved.

**Now… For some real fun…**

Now, for some real fun, I’d like to ask you how the world would look like when you were traveling through it riding a photon. So… Think about it. Think hard. I didn’t *google *at first and I must admit the question really started wracking my brain. There are some many effects to take into account. One basic property, of course, must be that time stands still around you. You see the world as it was when you reached ** v** = c. Well… Yes and no. The fact of the matter is that, because of all the relativistic effects (e.g. aberration, Doppler shift, intensity shifts,…), you actually don’t see a whole lot. One visualization of it (visual effects of relativistic speeds) seems to indicate that (most) science fiction movies actually present the correct picture (if the animation shows the correct visualization, that is): we’re staring into one bright flash of light ahead of us as we’re getting close to

*v*= c. Interesting…

Finally, you should also try to find out what actually happens to the clocks during the deceleration and acceleration as the space ship of that twin brother turns. You’re going to find it fascinating. At the same time, the math behind is, quite simply, *daunting *and, hence, I won’t even try go into the math of this thing. 🙂

**Conclusion**

So… Well… That’s it really. I now realize why I never quite got this as a kid. These paradoxes do require some deep thinking and imagination and, most of all, some tools that one just couldn’t find as easily as today.

The Web definitely *does *make it easier to study without the guidance of professors and the material environment of a university, although I don’t think it can be a substitute for discipline. When everything is said and done, it’s still hard work. *Very *hard work. But I hope you get there, Vincent ! 🙂 And please do look at that Youtube video by clicking the link above. 🙂

**Post scriptum**: Because the resolution of the video above is quite low, I looked for others, for example one that describes the journey from the Sun to the Earth, which–as expected–takes about 8 minutes. While it has higher resolution, it is *far *less informative. I’ll let you *google *some more. Please tell me if you found something nice. 🙂