Clocks don't measure time!What clocks actually measure is the space-time interval between events on the clock's world-line. That turns out to reduce to what we call "time" if you and the clock are not moving relative to each other. But if you and the clock are moving relative to each other, then things get weird because, well, clocks don't measure time. In fact, time is not even a well-defined concept when things are moving relative to each other!
Two surprising consequences of this: first, the speed of light is not constant. In fact, "the speed of light" is not even a well-defined concept because time is not a well-defined concept, and speed is defined in terms of time. (In fact, "the speed of X" is not a well-defined concept for any X, because, well, you know.)
(What is true is that the measuring the "speed" of light will give you the same result no matter what reference frame you're in. But that turns out to be a consequence of the fact that clocks measure space-time intervals rather than time. It is not, as is often taught, the foundational principle of relativity. Einstein himself got this wrong.)
The second surprising consequence is that the most common resolution of the "twin paradox" is mistaken. Maudlin quotes Feynman as the prototypical example:
This is called a “paradox” only by people who believe that the principle of relativity means that 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 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 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.Maudlin then minces no words:
Everything in this "explanation" is wrong.That sort of clarity is rare.
(It's actually easy to see that any explanation in terms of acceleration must be wrong because it is easy to set up a "twin paradox" that involves no accelerations: use three clocks all moving in inertial trajectories along the same line. Clock A starts to the left of clock B and is moving to the right (relative to B). Clock C starts to the right of clock B and is moving to the left (relative to B). The initial positions are such that A will be co-located with B before it is co-located with C. When A meets B, A is set to the time shown on B. Then, when A meets C, C is set to the time shown on A. When C meets B, the reading on C will be less than the reading on B despite the fact that none of the clocks have undergone any acceleration.)
Anyway, I thought this was cool and so I thought I'd take a moment to share it.
In your 3 clocks example, I'm trying to think through the C perspective. You've described it from the perspective of B, which is clear. And of course only one thing can happen, so the math surely must work out.
ReplyDeleteBut from C's perspective: it is stationary. It notices A & B meet, and synchronize their clocks. Then it sees A moving very fast towards it, and B moving more slowly towards it. So A's clock is ticking slower. So then A meets C, and C sets its own clock. But it knows, at this point, that B's clock is already ahead and has experienced more time, than the time it now sees on A's clock.
OK, so C's clock is now "behind" B's clock. But B still has to move and catch up with C, and so from now on, B's clock will be ticking slower than C's clock.
The math must work out, that there is never enough time (or distance) for C's clock -- which starts behind but is ticking faster -- to catch up to B's clock, which starts ahead but is ticking slower. Regardless of the relative speeds of A and B, as they head towards C. They both could be going near light speed. Or A could be closing near light speed, while B is just taking an ordinary train ride. Either way, there must not be enough time for C's clock to overtake B's clock.
But I'll admit that I haven't worked out the math yet.
"the speed of light is not constant ... measuring the "speed" of light will give you the same result no matter what reference frame you're in ... It is not, as is often taught, the foundational principle of relativity. Einstein himself got this wrong."
ReplyDeleteI'm not sure that I fully appreciate this philosophical point. To be fair to Einstein, the universe "could have been" Newtonian. And it was the experimental evidence of the constant ("measurement of the") speed of light, that required the development of relativity. So at least historically, it was a "foundational principle".
But if you mean that we can now identify an equivalent framework, which in hindsight is clearer and simpler but with the same predictions ... perhaps that is true.
In some ways it reminds me of the reasonable claim that pi is wrong, and we really should base all our math on tau instead. Of course, pi isn't really "wrong". But tau is "cleaner", and knowing what we know now, would have been a better choice. If that's the kind of sense in which you mean "the speed of light is not constant", then ... maybe I need to read the book.
> it was the experimental evidence of the constant ("measurement of the") speed of light, that required the development of relativity
ReplyDeleteWell, kinda sorta. The Michaelson-Morley experiment didn't really prove that light has the same speed in all reference frames, it only showed that earth doesn't move with respect to the medium through which light travels (the hypothetical "luminiferous aether"). The idea that the speed of light is the same in all frames is one explanation of this (the correct one as it ultimately turns out) but it's not the only possibility.
But the whole light thing is kind of a red herring. What relativity is really about is the geometric structure of space-time. That structure has an invariant velocity as a *mathematical* consequence. It was not necessary (as far as anyone can tell) that there exist a *physical* phenomenon that actually propagates at this velocity. Light cones would exist even in the absence of light. It is just a happy coincidence that light exists and conveniently maps out the structure of space-time for us in the way that it does.
Or it could be that this *is* necessary, and there's a Nobel prize waiting for the person who figures out why.
"earth doesn't move with respect to the medium ... the speed of light is the same in all frames ... [is] not the only possibility."
ReplyDeleteYes, yes! Absolutely. Fair enough. But that's why I would say that making that assumption, and then exploring the consequences, does make the "constant speed of light" the "foundational principle" of (special) relativity. At least historically.
"What relativity is really about is the geometric structure of space-time."
Yeah, ok, but now you've (I think) jumped to general relativity, which took another decade. And the math is way harder.
I can totally accept that, now that we know what the final answer is, we can go back and refactor or reformulate the theory according to "cleaner" foundational principles. That can be an interesting exercise. (Just like with tau.)
BTW: I haven't read Maudlin's book (yet). But I once came across Lewis Carroll Epstein's Relativity Visualized (1985). From what you describe of Maudlin's book, they seem to have similar goals: explain relativity without math. Epstein even directly tackles the philosophy of science: what does it mean to have an "explanation"? We can have the math of relativity, but if you want to "understand" it, what would that mean?
ReplyDeleteI still remember Epstein's "explanation": there is no such thing as "speed". Or, more precisely, there is only one speed in the universe. Every entity in the universe is always travelling at the exact same speed -- but through "space-time", not just through space. You can't change your speed; you can only change your direction.
If you point your single speed in the direction of space, like a photon, then you get none left over to go through time. At the other extreme, if you sit still in one place so that you don't go through space at all, then you will find time passing at the maximum rate. Or, you can choose something in the middle, where some of your (one and only) velocity will allow you to travel through space, while the remaining (now smaller) leftover amount will be used to travel through time (so, slower than if you were still).
The idea of only having a single speed, and only being able to choose a "direction" -- through four dimensional space time instead of mere three dimensional space -- is a powerful but useful "myth", as Epstein describes it. It is neither "right" nor "wrong". Instead, it is an intuitive "explanation" that matches the mathematics.
It sounds like Maudlin has explored down a similar kind of road.
Everything in this "explanation" is wrong.
ReplyDeleteI think this is a bit extreme. A better statement would be "this explanation does not generalize". Or, expanding somewhat: whenever you have two people who separate, then later come back together, and they find that their elapsed times are different, there must be some difference between them. So the general form of Feynman's logic is correct. But the *specific* difference that he cites in his scenario--acceleration--does not have to be the difference in every scenario; and furthermore, it doesn't even have to be the critical difference in his scenario. See further comments below.
(It's actually easy to see that any explanation in terms of acceleration must be wrong because it is easy to set up a "twin paradox" that involves no accelerations: use three clocks all moving in inertial trajectories along the same line. Clock A starts to the left of clock B and is moving to the right (relative to B). Clock C starts to the right of clock B and is moving to the left (relative to B). The initial positions are such that A will be co-located with B before it is co-located with C. When A meets B, A is set to the time shown on B. Then, when A meets C, C is set to the time shown on A. When C meets B, the reading on C will be less than the reading on B despite the fact that none of the clocks have undergone any acceleration.
This is not really a "twin paradox" because no pair of clocks meets twice. The whole point of the twin paradox is that the same pair of clocks meets twice. However, it is true that the worldlines realized by your three clocks form the same kind of triangle in spacetime that is formed by the two twins in the standard twin paradox, and it shows that the difference in elapsed times is just a case of the flat spacetime version of the triangle inequality, i.e., it can be explained entirely by geometry.
The role that acceleration plays in the standard twin paradox is to allow one single clock to make the "bend" between two sides of the triangle (instead of clock C having to be set to clock A's time when they meet). So the acceleration is not irrelevant--in flat spacetime, it's impossible for two clocks to meet twice if both of them always move inertially, so if two clocks do in fact meet twice, at least one has to accelerate somewhere.
The really interesting thing, IMO, is to look for examples in curved spacetime (i.e., where gravity is present, such as in the vacuum region around a planet or star) where two clocks *can* meet twice even though both of them are moving inertially, and where their elapsed times between meetings differ.
I'll admit that I haven't worked out the math yet.
ReplyDeleteSpoiler alert: it does work out. :-)
The math must work out, that there is never enough time (or distance) for C's clock -- which starts behind but is ticking faster -- to catch up to B's clock, which starts ahead but is ticking slower. Regardless of the relative speeds of A and B, as they head towards C. They both could be going near light speed. Or A could be closing near light speed, while B is just taking an ordinary train ride.
Yes, all of this is correct. That is because the things you mention are not all independent variables. As I said in my response to Ron just now, we have a triangle in spacetime, whose sides are the segments of B's, A's, and C's worldlines between the respective meetings. This triangle is invariant (i.e., its geometric properties--the side lengths and angles--are the same in every inertial frame), and its geometry determines everything in the problem. And a requirement of that geometry is the triangle inequality (which I mentioned to Ron), which in the spacetime version says that the sum of the lengths of any two sides of this triangle must be *less* than the length of the third (instead of greater, as it is in ordinary Euclidean geometry--the difference is because of the minus sign in the metric).
Changing the relative velocities changes the shape of the triangle, but it can't change the fact that the triangle inequality has to hold.
If you point your single speed in the direction of space, like a photon, then you get none left over to go through time. At the other extreme, if you sit still in one place so that you don't go through space at all, then you will find time passing at the maximum rate. Or, you can choose something in the middle, where some of your (one and only) velocity will allow you to travel through space, while the remaining (now smaller) leftover amount will be used to travel through time (so, slower than if you were still).
ReplyDeleteThe problem with this viewpoint is that "sitting still in one place" is relative. You might think you are sitting still in one place, but relative to a cosmic ray particle you are moving at only a smidgen less than the speed of light. There is no single invariant "speed" that you have that determines "how much" of your speed through spacetime is "speed through time" vs. "speed through space". In fact, the only object that does have such an invariant speed is the photon! (The underlying mathematical reason for this is that Lorentz transformations rotate timelike vectors but dilate null vectors.)
If you are very, very careful never to draw any inferences that are made invalid by the above issue, you could possibly use this viewpoint for something. But I have seen a lot of people try precisely this in discussions on Physics Forums (usually after seeing one of Brian Greene's TV specials or reading one of his articles where he describes the viewpoint), and nobody has ever succeeded: they always end up drawing some inference that is wrong, but which the viewpoint makes seem correct. And then it goes downhill from there.
Btw, while I have seen this viewpoint in plenty of pop science articles, I have never seen it used in any actual scientific paper. Which leads me to believe that the viewpoint actually isn't useful for doing physics; it just makes a good "oh, wow!" kind of thing for selling pop science books.
Instead, it is an intuitive "explanation" that matches the mathematics.
ReplyDeleteActually, it doesn't. This is a pet peeve of mine so I'll elaborate a bit. :-)
Timelike worldlines, like yours and mine, have a thing called "4-velocity" at any point, which is the unit vector pointing into the future that is tangent to the worldline at that point. Since it's a unit vector, it can be viewed as describing the object "traveling at the speed of light through spacetime" (since the speed of light is 1 in the natural units that are usually used in relativity, and in which the 4-velocity is described as a unit vector).
Null worldlines, like those of photons, also have tangent vectors, but those tangent vectors have *zero* length! (That's why they are called "null".) So you *cannot* view the photon's tangent vector as describing its "speed through spacetime", because that would require it to have unit length, not zero length.
The reason we usually say that photons have speed 1 (in natural units) is that, if you draw a triangle in spacetime with one timelike side, one spacelike side, and the third side as a segment of a photon's worldline, and you make the timelike and spacelike sides orthogonal (so that they represent "time" and "space" grid lines in some inertial frame), then the lengths of the timelike and spacelike sides will be equal (so their quotient, which is the "speed" of the photon in the chosen inertial frame, is 1). But it's obvious that this rigmarole has nothing to do with the tangent vector to the photon's worldline or with a photon's "speed through spacetime".
Thanks for the interesting comments, Peter!
ReplyDeletethe triangle inequality (which I mentioned to Ron), which in the spacetime version says that the sum of the lengths of any two sides of this triangle must be *less* than the length of the third
ReplyDeleteActually, I misstated this. The correct statement is that the sum of two sides that are placed so that the future endpoint of one is the past endpoint of the other, must be less than the third side. In the case under discussion, these are the A and C sides, which must be less than the B side.
@Peter:
ReplyDelete> if two clocks do in fact meet twice, at least one has to accelerate somewhere
That's true. But it's possible to tweak the scenario so that both clocks undergo the same accelerations (not at the same time of course) and still reunite with different ages. Figuring out how makes a good exercise.