Is the Duel on the Moving Train Fair in Relativistic Terms?

[Saw]

This is a recapitulation of a subject that has incidentally appeared in another thread (Light shone in a train bouncing off mirrors…) that started over another issue (why does light take the direction of the source but not its speed?), although there may be a connection between the two things.

I have already received some helpful comments, but I still have doubts. If you find this introduction too long, please jump to the next post, which is where I place the question that worries me now.

The subject is to judge if the duel described by Brian Greene in the Fabric of Cosmos (when explaining the relativity of simultaneity) is “fair”, in its original version and with a variation.

There is a duel on a train, between duellers that we will call Back and Front, situated at the tail and the tip of the car, respectively. For their duel, they employ laser guns, of identical construction.

There are two referees: Althonhare, on the train, and myself, Saw, on the ground. The signal for the commencement of the duel is given at the precise instant when Althon and myself are lined up. Of course, it is impossible that the two referees occupy the same position in space and so the perception of this alignment would require light traveling some distance from one to the other. But we stipulate that the distance is so small that its consequences are negligible. We all agree that the alignment (which is a single event) is simultaneous in both frames.

In order to give the duellers the signal for shooting, a pile of gunpowder, set midway between them (i.e., where Althon, on the car, and myself, on the ground, are standing at that instant), explodes and thus sends flashes of light in both directions.

Was the duel fair?

Each referee has a different opinion:

* Althon is certain that light from the flare reaches the duelers simultaneously, so he raises the green flag and declares it a fair draw.

[More technically: In the train frame, the light pulses travel equal paths in their respective directions and do it at the same speed. So the two distant events (= arrival of the signals to shoot) are SIMULTANEOUS.)

* According to the author, I wildly squeal foul play, claiming that Back got the light signal from the explosion before Front did. I explain that, because the train was moving forward, Back was heading toward the light while Front was moving away from it. This means that the light did not have to travel quite as far to reach Back, since he moved closer to it; moreover, the light had to travel farther to reach Front, since he moved away from it. Since the speed of light, moving left or right from anyone’s perspective, is constant, I am supposed to claim that it took the light longer to reach Front, since it had to travel farther, rendering the duel unfair.

[Idem: In the ground frame, light towards Back travels a shorter path, since its target is heading towards it, while light towards Front travels a longer path, since its target is escaping away. As both pulses travel at the same speed, the one that hits Back arrives earlier than the one that hits Front. The two distant events are NOT SIMULTANEOUS.]

Who is right? The matter is not trivial, because the judge who has made a mistake will be sanctioned by the competent supervision body and he may not be able to exercise his profession any more.

Fortunately (?), Einstein comes to the rescue of both with a salomonic opinion:

“Einstein’s unexpected answer is that they both are (…) they simply have different perspectives on the same sequence of events. The shocking thing that Einstein revealed is that their different perspectives yield different but equally valid claims of what events happen at the same time. Of course, at everyday speeds like that of the train, the disparity is small – Saw claims that Front got the light less than a trillionth of a second after Back- but were the train moving faster, near light speed, the time difference would be substantial” (literal quotation, I just changed the names).

But I am a prudent referee. I try to do my job properly. I know that the words of the law (you shall raise a green flag if the duel is “fair”) have a practical purpose and I must interpret them in the light of this practical purpose. The duel is “fair” if the two duellers are given equal opportunities of hitting each other, in practical terms. So my opinion that “Back got the signal earlier” is only relevant for the matter to be judged if it gives Back a real advantage (or disadvantage!).

To this effect, I consider the following scenarios:

(a) Back sees the signal earlier and fires earlier as well. Can his laser pulse hit Front before the latter receives the signal?

(b) If not, they have both received the signals and shot their laser pulses. The latter will also take some time to reach their targets. During this time interval, theoretically, any dueler could try the usual trick: fire and stand aside, before being wounded. There is very little time for this deed, especially if we talk about laser guns, but we can imagine the distance arbitrary long for this purpose. Is this time interval longer for Back than for Front or vice versa?

(c) If not, we imagine Back badly wounded, kneeling on the floor of the car, but brave enough to fire a second shot. Can it reach Front before the latter has the opportunity to fire his own second shot…?

You can think of other scenarios if you wish. The more, the better. That is the point of the exercise.

Then I call a group of experts on SR and ask their answers to the practical questions. If the answers, as I expect, are negative (there is no breach of the principle of equal opportunities), then I do not “wildly squeal foul play” as Brian Greene suggests. Instead, I gently and gallantly raise a green flag, in agreement with Althonhare’s opinion.

JesseM confirmed that the answers to questions (a) and (c) is:

No. If two events are simultaneous in any frame, that must mean there is a spacelike separation between the two events, meaning neither event lies in the other event's future light cone.

And to question (b):

Same time for both, since they're both at rest in the train frame, and they both fire at the same moment in this frame.

We still discussed for a while whether the question “is the duel fair?” is purely legal or also physical, but I think in the end we agreed that:

- The legal problem is the same as the physical problem. The role of the law is to solve problems, just as the role of physics is to solve problems.

- The legal problem is not to establish some arbitrary conventional rules for the game. The rules, on top of being agreed upon, must ensure that both duellers have “equal opportunities” to shoot and avoid being shot. For example, after receiving the signal for shooting and before being shot, the duellers may stand aside, duck down or try any other tricks. Do they dispose of the same time for this purpose?

- This legal need can be translated into a well-defined physical expression:

"the duel is fair if each dueller experiences the same proper time between firing their own gun and the laser from the other guy's gun reaching their position"

.

(I initially contested the definition, but I was wrong. In the end I agree that it was perfectly appropriate.)

In other words, the law requires that the duellers can do equal number of “tricks” to win. Physics answers that that means that their clocks register the same number of “ticks”, but in particular SR specifies that the relevant ticks are “proper” ticks, those of the duellers or of assistants situated where they receive the signals and the shots.

These are the calculations assuming that the relative velocity of the train wrt the ground is 0.5 c and that the train is 2 ls long in its rest frame (1.732 ls in the ground frame):

- In the train frame: Back and Front get the light signals simultaneously (after 1s) and their shots take 2 s to reach their opponents. So the proper time interval is 2s for both duellers. The duel is fair.

- In the ground frame, the duel is also fair because the proper time interval, although another one, is also the same for both duellers:

* Back got the light signal earlier (0.577 s) because he was heading towards the signal (thus making its path shorter), which looks like an advantage, but then it also happens that he is shot earlier (2.886 s), for the same reason, because he was heading towards the laser (thus making its path shorter), which is a disadvantage. The interval is 2.886 - 0.577 = 2.309 s.

* Front got the light signal later (1.732 s) because she was racing away from the light (thus making its path longer), which looks like a disadvantage, but then it also happens that he receives the shot earlier (3.732 s), for the same reason, because she was escaping from the bullet (thus making its path longer). The interval is 3.732 – 1.732 = 2.309 s.

Then we engaged in a challenging discussion about what it means that, as observed in the ground frame, the duellers get the light signals non-simultaneously.

My final understanding, as referee in the ground, is very prudent: the physical fact about it is that, if the clocks of my assistants (situated by Back and Front when they received the light signals) had been synchronized following the Einstein convention, they would show different readings; but this is just a piece of the puzzle; if I want to rule, I have to compare the clock readings at reception of the light signal with the clock readings of other assistants witnessing the reception of the shots; the latter are also non-simultaneous; the difference is the same for both duellers = they disposed of the same proper ticks to do their proper tricks.

Thus I rule, in accordance with the advice of SR experts, that the duel was fair.

Now we face a variation of the same duel: the signals for shooting are still light signals but the duellers shoot mechanical bullets.

What I have studied about SR tells me that the solution must be identical, since mechanical objects obey the same laws of physics:

- In the train frame: just like the light signals travel equal paths at the same speed (c) and arrive simultaneously at the duellers, their bullets also travel the same paths at the same speed and hit simultaneously the duellers.

- In the ground frame, you can reach the same conclusion through two routes:

* If I apply the relativistic formula for the addition of velocities to the light signals, I will discover that they still travel wrt me at c. And if I apply it to the bullets, I will find a differential element. Each bullet travels at a different speed wrt me, but this fact does not change the judgment: the bullet from Back travels wrt me faster than the bullet from Front, but less fast than it would if we had applied the Galilean addition formula (which compensates for the fact that it departs earlier) and the bullet from Front travels less slowly than it would with the Galilean formula (which compensates for the fact that it departs later).

* If, with the aid of my assistants, I measure directly the speed of the light signals and the bullets, I will find that they travel at exactly the speeds that stem from the above formulas.

However, I have doubts: due to some reasoning about the difference between what is “real” and “conceptual” (which I will not repeat here, it’s just too long), I fear that, in this new case, the proper time intervals (during which the duellers must do their tricks) should be different. I don’t know who would have the advantage and, in any case, the difference would be tiny. But this would make the duel, by definition, unfair, in the opinion of both referees.

Nevertheless, I conclude: if the experts tell me that the solution to the thought experiment I describe in the next post is “yes” (and maybe you give some sort of explanation for that, it doesn’t have to be very profound, I want to finish my job!), I will rule as proposed by SR.

Please give me a hand!

 

[Saw]

Imagine that the referee sends, at the same time, light and mechanical signals to the duellers and that the latter shoot with both bullets and laser beams. It just happens that the light travels both ways through a tube of a certain “imaginary” material. Light slows down when crossing certain media, as a function of their index of refraction. The bullets are as fast as can be achievable and our imaginary material has an imaginary refraction index such that, when we observe the light traveling to Back, such light is slowed down to the point of being always aligned with the corresponding bullet. If we observe the light in the other direction or in its way back, will it also be aligned with the bullet?

What is SR’s answer to this thought-experiment?

Will, in these circumstances, light and matter travel in exact harmony in both directions?

If yes, is there any standard explanation of the cause for that, maybe related to the way that matter is accelerated?

 

[ZikZak]

It sounds like you're understanding the situation pretty well. It's not clear exactly what you're asking in the second duel, but:

When light travels through a medium, it travels at less than c. Now, in vacuum, it's perfectly legitimate to say "light travels at c" without referring to any reference frame because c is an invariant. But if the light ray is traveling in a medium at less than c, you now must specify the reference frame in which it travels at speed c'. In this case, when you use the material with refractive index n, the light ray travels at speed c'=c/n, in the rest frame of the material. In a different frame, the relativistic velocity addition formula must then be used. Indeed, this is exactly what the 1851 Fizeau Water Experiment demonstrates.

So the light ray in medium and the bullet will always be aligned. And incidentally, since they are always aligned in the rest frame, they must always be aligned in every frame by invariance of coincidence.

 

[Dale]

The rules, on top of being agreed upon, must ensure that both duellers have “equal opportunities” to shoot and avoid being shot.

I don't think so. I think if they agree on the rules then the rules are fair. I mean, maybe one dueller would prefer the flash to be a little further away in order to avoid the possibility that the flash affects his vision. Maybe one dueller is so much better than another that he is willing to accept some handicap. Maybe one dueller is so bent on exacting revenge that he considers accepting a handicap to be a reasonable price for the chance. I say if the duellers agree on the rules then they are fair, regardless of other concerns especially relativistic ones.

 

[Saw]

When light travels through a medium, it travels at less than c. Now, in vacuum, it's perfectly legitimate to say "light travels at c" without referring to any reference frame because c is an invariant. But if the light ray is traveling in a medium at less than c, you now must specify the reference frame in which it travels at speed c'. In this case, when you use the material with refractive index n, the light ray travels at speed c'=c/n, in the rest frame of the material. In a different frame, the relativistic velocity addition formula must then be used. Indeed, this is exactly what the 1851 Fizeau Water Experiment demonstrates.

I’ve revised what Einstein says in his 1916 book about the Fizeau’s experiment.

First, it sems that Einstein accepts that there is some quantitative difference between the outcome of the relativistic and Fizeau’s formula. He claims his relativistic formula is superior because it is based on a better conceptual model, so the error would be for Fizeau. Is this understanding right? If so, can you clarify to me the nature of the difference?

Fizeau’s experiment is, yes, the same as the one posed here:

- Fizeau’s tube is for Enstein the railway embankment, which we call in our duel the “ground”.
- The water flowing inside the tube in Fizeau’s experiment plays the part of the imaginary material that I’ve invented, which is at rest with “the train”.
- The light signal is traveling for Fizeau through the water and for us through the imaginary material and along the train.
- The bullet is traveling in full harmony with the slowed down light signal, regardless its direction.

Then you say:

- The train is traveling at v (0.5 c in our example) wrt the ground.
- The light is travelling:
* In the train frame = the frame of the imaginary material, “at speed c'=c/n”, n being the refractive index of that material. Let us say that c’ = 0.9 c.
* In the ground frame, the speed of light is determined by the relativistic addition formula. Let us focus on the light signal to Front. Under the Galilean addition formula, this light signal would travel wrt the ground at v + c’ = 0.5 + 0.9 = 1.4c. Under the relativistic formula, it would travel wrt the ground at (v + c’)/(1+vc’) = (0.9 + 0.5)/(1+0.5*0.9) = 0.966 c. Also the speed of the bullets (which travel in harmony with the light signal) would be 0.966 c wrt the ground.

Well, all this would mean that the speed of light is:

- In vacuum, invariant for all observers as if it didn’t acquire the motion of the source; on the contrray, it is as if it acquired the motion of each observer.
- In materials, more and more varying for different observers as we increase the density of medium, until ideally, when it reached an infinitely small speed in a highly dense medium, it would behave virtually as if it had “acquired the motion of the source”.

But please note that it also means that the speed of the bullet:

- As it becomes higher wrt to the train, it also becomes less and less variant for different observers, until ideally, very close to the speed of light, it would become virtually invariant for all observers.
- As it becomes lower wrt the train, it also becomes more and more variant for all observers, recovering its normal motion pattern of acquiring the speed of the source…

For the first fact, there is a connection with a physical fact: light travels through a less dense or denser medium, although it is not evident why traveling through a medium should make it less invariant: one can understand that it is slower but not that it is measured as more or less variant by different observers.

But for the second fact, there is no link at all with any physical explanation: the only thing we find in Einstein’s text is a very vague reference to the “electromagnetic nature of matter” and a referral to Lorentz’s investigations.

How do you feel about all this?

 

[Mentz114]

How do you feel about all this?

It's very boring.

Also distasteful. Duels are violent and in no way related to justice of except the mythical lawless wild west kind of 'justice'.

 

[Saw]

It's very boring.

Also distasteful. Duels are violent and in no way related to justice of except the mythical lawless wild west kind of 'justice'.

Mentz114, I have explained (maybe not well enough) why there is here a physical problem:

"will the clock readings of the duellers show the same proper time intervals between firing and being shot?"

and why the explanation of SR (which may be right after all) appears to suffer from a logical gap, which has been also illustrated with a thought experiment, which in turn may have served many people to visualize how an often forgotten aspect of SR (the relativistic formula for addition of velocities) works.

That is all quite related to physics. Your comment is not. If I and many people appreciate this forum, it is because it is serious and posters are not teenagers who clatter it with nonsensical out-of-the point remarks. I would appreciate if you could delete your post, so that I can delete this and thus not waste people's time forcing them to read such "improper" things.

 

[Dale]

- In the ground frame, the duel is also fair because the proper time interval, although another one, is also the same for both duellers:

* Back got the light signal earlier (0.577 s) because he was heading towards the signal (thus making its path shorter), which looks like an advantage, but then it also happens that he is shot earlier (2.886 s), for the same reason, because he was heading towards the laser (thus making its path shorter), which is a disadvantage. The interval is 2.886 - 0.577 = 2.309 s.

* Front got the light signal later (1.732 s) because she was racing away from the light (thus making its path longer), which looks like a disadvantage, but then it also happens that he receives the shot earlier (3.732 s), for the same reason, because she was escaping from the bullet (thus making its path longer). The interval is 3.732 – 1.732 = 2.309 s.

This calculation is incorrect. The proper time interval is invariant, so it is also 2 s in the ground frame (that is what invariant means). You are calculating the coordinate time in the ground frame, not the proper time.

You don't need to do any calculations. Once you have specified the rules in terms of frame-invariant quantities then you know that, by definition, all frames will agree.

 

[JesseM]

"will the clock readings of the duellers show the same proper time intervals between firing and being shot?"

Isn't it pretty obvious that they will, just based on considering things in the rest frame of both duellers, and assuming their guns are physically identical and they both fire simultaneously in this frame? Once you've figured out the proper times in this frame, you know they must be the same in every other frame, since different frames never disagree about frame-invariant quantities like the proper time on a worldline between two events on that worldline.

 

[Saw]

This calculation is incorrect. The proper time interval is invariant, so it is also 2 s in the ground frame (that is what invariant means). You are calculating the coordinate time in the ground frame, not the proper time. You don't need to do any calculations. Once you have specified the rules in terms of frame-invariant quantities then you know that, by definition, all frames will agree

Thanks a lot. I had made a mental note about a problem here, but forgot to pursue it. Yes, in the ground frame, those values are of the clocks of different assistants of mine: one witnesses the arrival of the signal to Back, but only another one, situated more to the right (the train moves rightwards wrt he ground), witnesses the event when Back is wounded. The same happens with Front, logically. That means different clocks, with different worldlines, so we are talking about coordinate time.

But then I try to find a “proper time interval” in the ground frame and don’t find it. How do you get the proper 2s in the ground frame? Or do you mean that the ground agrees that the proper time interval measured in the train is what it is measured in the train?

On the other hand, the truth is that the coordinate time intervals measured by my assistants on the ground are also equal for both duellers. However, JessesM’s definition of fairness had focused on proper times. Why? In the end, the duel is fair if both duellers have equal chances to do their “tricks”. So if I, from the ground, could by chance only measure coordinate time intervals (see the question above), for me, it would also be a valid duel, because what matters for my purpose is that both duellers have equal chances to kill each other, not that there are more or fewer chances that they kill each other… But what if that was the question? What if a judge asked: is this situation likely to end up with a death? Would proper or coordinate time prevail? What is more “real”, the “invariant” quantity (the proper time interval), which is the only one that is identical cross-frame? But I don’t want to ask too many questions at the same time…

 

[Saw]

Isn't it pretty obvious that they will, just based on considering things in the rest frame of both duellers, and assuming their guns are physically identical and they both fire simultaneously in this frame?

Ok. I will explain why I have my doubts. Unfortunately, I can't summarize it. This is the reason:

Before the SR experts, I had interviewed a panel of classical scientists, who still adhere to the Galilean/Newtonian ideas.

They had argued that a homogeneous duel (either mechanical signals and mechanical shots or light signals plus laser shots) is fair, but an inhomogeneous one (light signals and mechanical shots is not). They justified it as follows:

a) Homogeneous duel:

a.1) Mechanical signals (e.g., balls) and shots (e.g.: bullets):

Before being projected, all elements share the same “state of motion” of the train. Any motion that is imparted to them is added or “superimposed” to that original state of motion. Therefore, relative to the train, everything happens as if the duellers were on the ground. Nothing changes and the duel is as fair as it would be on the ground. They call this the principle of relativity and it looks quite logical to me.

a.2) Light signals and laser shots:

Light has its own state of motion, which is not affected by the motion of the train. They justify it because light, just like sound travels through the air, would travel through a medium called the aether. Only wrt to that medium would light’s speed be c. Because of this, they explain that, if, for example, the aether is at rest with the ground and the train has been accelerated towards Front, the light signal will reach Back earlier (as it travels at c+v with regard to him) and only later Front (wrt to whom it travels at c-v).

“So you claim that Back has an advantage, don’t you?”, I ask.

“Only momentarily, since you have to take into account that Back receives a fast signal but also a fast shot, just like Front receives a slow signal but also a slow shot. One thing compensates the other and in the end both duellers dispose of the same time to do their tricks.”

I check with some real numbers and, actually, with a train being 2 ls long, both duellers would dispose of a time interval of 2.66 s to do their tricks. That is fair.

b) Inhomogeneous duel (light signals and mechanical shots)

With respect to this configuration, the classical scientists argue that the duel is not fair:

- The light signal would arrive at Back earlier and later at Front.
- And this advantage for Back (firing earlier) would not be canceled out by the correlative disadvantage of receiving Front’s shot earlier, since the bullets, unlike the light signals, do travel wrt the train at the same speed in both directions.

“Hmm”, I object, “are you sure of that? I think the next team, the SR team, will claim that the referee of the train should measure that the two signals are simultaneous for the duellers, because of the relativity of simultaneity or something like that…”

“Time and simultaneity are absolute”, they object.

“Please, don’t tell me fairy tales. Time is just a concept. In the end, if you really want to be sure about how the light signals and the mechanical shots behave, you have to compare their qualities with those of other real objects, with which you make them participate in a sort of competition. How do you measure the motion of objects?”

They explain the following:

Speed is the rate of change of distance over time. We divide this concept into two components: “distance” and “rate of change”. Different frames measure different distances, though with homogeneous sticks. But different frames directly measure the same rate of change. How? Imagine a ball bouncing between the top and the bottom of a spaceship, far from any gravity source (thus we can imagine that it moves inertially from top to bottom). It can act as a clock, in order to measure that “rate of change”: different observers disagree on the distance that the ball traverses on their respective X axes, though they share the same Y axis; however, the difference as to the X axis would only be relevant if the ball went out of the clock and hit the observers on their eyes; but, for our current purpose, which is to measure “change”, such difference is irrelevant, since what we are interested in now is the “number of ticks” and the “ticks counter” (the floor of the spacechip) has, by definition, for all observers, the same state of motion as the ticker, which forcefully means that all observers will measure the same number of ticks…

That looks ingenious, but I still make some objections:

- What happens when the ticker oscillates? The answer is that, with every collision, it loses kinetic energy and slows down, maybe erratically and in any case it is doomed to stop, more sooner than later…
- Is the ball fast? No, it is very slow, which doesn’t enable us to obtain small units for the sake of precision in measurements…
- How do you measure long distances when you cannot lay one stick after the other…?

Since all the answers to these practical questions are unsatisfactory, I dismiss the classical lot and receive the SR team. Although they start by telling me that “time is relative”, I directly ask them the practical question: how do you measure time and distances/length?

The answer is they use light for all purposes: they measure time rate with light clocks (atomic clocks) and synchronize clocks and measure length and distances with light signals, following the Einstein convention.

The choice of light seems definitely preferable to me:

For measuring the rate of change, it is a perfect ticker (fast and massless, so immune to collisions).
For the measurement of length, it may reach anywhere in the cosmos.
For synch purposes, it might even serve to synchronize clocks between different frames, since its motion is not affected by the motion of the source or the reflecting surface…

Yes, the problem is that it seems to move “on its own”, it doesn’t share the state of motion of the source and hence it will render different measurements for each observer, but there will be no problem with these intermediary disagreements if we find a conversion rule that leads to final agreement on events, on what happens.

I am shown the equations for transforming intervals and coordinates. The mathematical derivation looks logical. You take as start-point the Galilean Transformations, you keep c and v as invariants, but you allow x and t to change… The resulting formulas seem to work: each observer attributes to the other TD, LC and lack of synchronization, but the disagreement in each piece of the puzzle does not impede final agreement on certain events.

But what events? All the thought experiments take as ticker or protagonist a light signal. In fact, if the protagonist had been a ball, the thought experiments would have ended soon with a classical solution: someone would have said that the matter ticker takes the motion of the source and the fun would be over… Therefore, it seems to me that the mathematical reasoning “only” allows for this prediction: it guesses the x and t values of light signals in your coordinate system on the basis of the x’ and t’ values of any other frame’s light signals in his coordinate system… = it precisely guesses the events of other measurement instruments, but does it exactly predict the events of a bullet?

I had dismissed the classical advisors because their measurement instruments are inaccurate, but their prediction about what “happens” with the bullets, at least on a one-way trip, based on the idea that matter takes the motion of the source, seemed spotless: if we hire bouncing bullets of an elastic material as a ticker between the walls of the train car, we’ll make a mistake; but if we see them as “real life” objects, it still seems that they suffer the same push in both directions, which means there is no advantage in this respect for any dueller; instead, in their one-way to the duellers, the light signals “behave” differently for each observer, so one of them has an objective advantage. This might lead to the conclusion that the clock of one of the duellers (maybe Back’s, maybe Front’s) registers a higher difference for the “Interval” that we have taken as reference (the interval between firing and being shot).

Conclusion: I would rule that the duel is by definition unfair, unless I am told that SR says something more, namely it has discovered that, with regard to the issue of taking (A) or not taking (B) the motion of the source, both matter and light behave “alike”… but following which model, A or B?

If it were A (also light takes the motion of the source), we wouldn’t need SR and this has been ruled out, at least for light, by experiments.
If it were B, it could not be simply B, because the relativistic formula for the addition of velocities [(v - w)/(1-vw), assuming c=1 and velocities are measured as a fraction of c] only departs from the classical formula (v + w) and assimilates matter to light to the extent that the product “vw” is a substantial fraction of 1. So (although this is still confusing to me) I draw this funny conclusion: the bullet takes the motion of the source when it is at rest in the train, but when it is shot it starts not taking the motion of the source and moving “on its own”, this effect being smaller the lower that v (speed of the train) or w (speed of the bullet wrt the train) are, and the bigger the higher that v or w are…

That is why I posted the question that you can find above about the thought experiment of light traveling through an imaginary very dense material, in harmony with the bullets.

 

[JesseM]

For the first fact, there is a connection with a physical fact: light travels through a less dense or denser medium, although it is not evident why traveling through a medium should make it less invariant: one can understand that it is slower but not that it is measured as more or less variant by different observers.

But for the second fact, there is no link at all with any physical explanation: the only thing we find in Einstein’s text is a very vague reference to the “electromagnetic nature of matter” and a referral to Lorentz’s investigations.

Maybe the problem here is that you're imagining the velocity addition formula somehow depends on physical properties of the bullet? It doesn't, it can be derived directly from the Lorentz transform based only on knowing the coordinates of a moving object in one frame and then figuring out what the equivalent coordinates would be in another frame; so, if two observers in a classical Newtonian universe chose to use coordinate systems that were related to one another by the Lorentz transformation, they would find coordinate velocities in one system related to coordinate velocities in another exactly the same way, as long as you're using this coordinate transformation you don't have to worry about any physical properties of the moving object whatsoever. Of course, in a classical Newtonian universe the distances and times in these two coordinate systems would not match up with distances and times as measured by normal rulers and clocks at rest with respect to each observer, but you're always free to use a "weird" coordinate system if you choose. And in relativity of course coordinate systems based on normal rulers and clocks (and the Einstein clock synchronization convention) will naturally be related to one another by the Lorentz transform, and the laws of physics will take the same form in different inertial coordinate systems constructed this way, so in this context it is the most "natural" coordinate transformation.

 

[atyy]

a.2) Light signals and laser shots:

Light has its own state of motion, which is not affected by the motion of the train. They justify it because light, just like sound travels through the air, would travel through a medium called the aether. Only wrt to that medium would light’s speed be c. Because of this, they explain that, if, for example, the aether is at rest with the ground and the train has been accelerated towards Front, the light signal will reach Back earlier (as it travels at c+v with regard to him) and only later Front (wrt to whom it travels at c-v).

For a train at constant velocity, the Newtonian panel will find that the train is an aether frame.

 

[Dale]

But then I try to find a “proper time interval” in the ground frame and don’t find it. How do you get the proper 2s in the ground frame?

http://en.wikipedia.org/wiki/Proper_time" is defined as dτ² = dt² - dx²/c² which is proportional to the spacetime interval ds² = -c²dt² + dx² and is real for timelike intervals. All frames agree on these quantities.

 

[Saw]

http://en.wikipedia.org/wiki/Proper_time" is defined as dτ² = dt² - dx²/c² .

Ok, and if I measure distances in light seconds, I understand that c becomes = 1 and thus the formula for proper time simplifies to dτ² = dt² - dx².

So in the duel the “proper time interval” between the two relevant events (when a dueller fires and when he is shot) is:

* In the train frame: since here the two events occur at the same place (dx = 0), the expression reduces to dτ² = dt² and this in turn to dt = the difference between the readings of the clock of the dueller at the time of each event (when he fires and when he is hit by the other’s shot).

In particular, both duellers shoot at t1 = 1 and are shot at t2 = 3, so the interval is for both of them 2s.

* In the ground frame:

- As to Back:
. An assistant of mine observes that Back fires at t1 = 0.557 s, Back moves on and another assistant of mine observes that Back is shot at t2 = 2.886 s. The difference = coordinate time = dt = 2.309 s.
. The distance between the two events = dx = 1.1545 s.

- As to Front:
. An assistant of mine observes that Front fires at t1 = 1.732 s, Front moves on and another assistant of mine observes that Front is shot at t2 = 4.041 s (not 3.732 as I had said before). The difference = coordinate time = dt = 2.309 s.
. The distance between the two events = dx = 1.1545 s.

Both for Back and Front, dτ² = dt² - dx² = (2.309)² – (1.1545)² = 5.3 – 1.3 = 4, so τ = 2 s.

Thanks for the correction again.

 

[Dale]

Both for Back and Front, dτ² = dt² - dx² = (2.309)² – (1.1545)² = 5.3 – 1.3 = 4, so τ = 2 s.

You got it exactly.

 

[Saw]

you're always free to use a "weird" coordinate system if you choose. And in relativity of course coordinate systems based on normal rulers and clocks (and the Einstein clock synchronization convention) will naturally be related to one another by the Lorentz transform, and the laws of physics will take the same form in different inertial coordinate systems constructed this way, so in this context it is the most "natural" coordinate transformation.

Well, the “ball clock” I talked about in the introduction is “weird”, because it has major physical shortcomings: the ticker is slow and its collisions with the walls of the instrument make it lose kinetic energy. Therefore, if we used this instrument and drew our coordinate systems with the values measured with it and related the x readings of different frames with the Galilean Transformations, we would make imperfect predictions.

But would you agree that if those physical shortcomings did not exist, the “ball clock” would render homogeneous time measurements for all observers and hence the classical coordinate system and the GTs would be preferable, because they are simpler? If those physical cons did not exist, wouldn’t the pros of the instrument make it preferable, as well as the geometry and mathematics it is associated with?

I say all this after assuming that the light instrument is preferable because it does its job more efficiently. In this sense, yes, it is more “natural”. I just want to point out that the choice of the instrument (which no doubt has certain physical characteristics) conditions the choice of the diagram and of the corresponding mathematics... But all the rest is unclear. I'll think of an example.

 

[JesseM]

Well, the “ball clock” I talked about in the introduction is “weird”, because it has major physical shortcomings: the ticker is slow and its collisions with the walls of the instrument make it lose kinetic energy. Therefore, if we used this instrument and drew our coordinate systems with the values measured with it and related the x readings of different frames with the Galilean Transformations, we would make imperfect predictions.

But would you agree that if those physical shortcomings did not exist, the “ball clock” would render homogeneous time measurements for all observers and hence the classical coordinate system and the GTs would be preferable, because they are simpler? If those physical cons did not exist, wouldn’t the pros of the instrument make it preferable, as well as the geometry and mathematics it is associated with?

You didn't really specify the details of how the ball clock would work in different frames. Does each observer have his own distinct ball bouncing up and down in his own rest frame? Is each observer's ball bouncing the same distance from floor to ceiling, and at the same speed, relative to the coordinates of SR inertial frames? If all this is true, then ball clocks will show time dilation just like light clocks. Since all the laws of physics are Lorentz-symmetric, all physical clocks will exhibit time dilation, the only way this wouldn't be the case would be if some of the laws of physics were not Lorentz-symmetric (like if we lived in a Newtonian universe where the fundamental laws were Galilei-symmetric).

I just want to point out that the choice of the instrument (which no doubt has certain physical characteristics) conditions the choice of the diagram and of the corresponding mathematics...

No, that is definitely a major misunderstanding, if there were any physical instrument that didn't function identically in the different inertial frames related by the Lorentz transformation, this would be a violation of the first postulate and would therefore prove relativity wrong. There is no reason to believe that such a thing is possible in reality, and if anyone could show it was they would make headlines and probably win the Nobel prize.

 

[Saw]

You didn't really specify the details of how the ball clock would work in different frames. Does each observer have his own distinct ball bouncing up and down in his own rest frame?

Yes. There are two observers armed with ball clocks, moving relative to each other as inertial frames.

Is each observer's ball bouncing the same distance from floor to ceiling,

Yes, and since this distance is in the axis perpendicular to the direction of relative motion between the two observers, there is no issue of length contraction here. Even if the observers are SR observers, they’ll agree that this y distance is the same for both of them.

and at the same speed, relative to the coordinates of SR inertial frames?

Well, if we introduce speed, we complicate things, because speed is distance divided by time and then we must wonder… “time”, but measured how? It will have to be measured by another observer, armed with another instrument. But which instrument? One like this very instrument? Then the question is what kind of time this other instrument measures, homogeneous or inhomogeneous for all observers...? We go into a vicious circle.

I prefer to talk, instead of time, about “change”. Motion can be regarded as the fact that distance is changing (growing or shrinking) between two objects. So the mind decomposes this phenomenon into its two elements and measures them independently. After a while, you look at the object in question and observe how much distance it has traversed and at the same time you observe how much “change” has happened. You measure the change with some other object that is ticking or oscillating inside a box (every tick or toc is a unit of “change”). Then you put the two things together and say that, for example, two kilometres of distance fit within one minute of change. That is why speed is a division. You distribute the space units between the units of change.

And how do you count units of change? For a change to exist, you need something moving (the ticker) and a counter, which is the walls of the clock.

The ticker is a moving thing. It is true that the two observers moving wrt each other have different perspectives and use different coordinate systems. How can it be that they agree on what happens to the tickers of both of them?

Well, to start with, they do not disagree on everything. The motion of the ticker can be decomposed into two vectors: its motion through the Y axis (perpendicular to the relative direction of motion of the two observers) and through the X axis. For the observer at rest with the clock, there is only motion through the Y axis, for an external observer there is also motion through the X axis. But at least we have thus reduced the scope of the disagreement: for example, for me your ticker has lateral motion while mine hasn’t any; for you my ticker has lateral motion while yours hasn’t any; but we both agree that the two tickers have the same Y motion.

Is this discrepancy about the X motion relevant? It depends on what purpose we consider.

If we regard the ticker as a “real life” object, then the discrepancy is important. If the ticker accidentally escaped out of the clock through a hole (breaking into “real life”), it would hurt the local observer more or less hard, but it would hurt an external observer, approaching in another frame, much harder, because its speed and hence its energy content is different for him.

However, here we do not have that purpose. We are not analyzing the damaging capacity of the ticker for our eyes. We are considering only the “instrumental” role of the ticker, how many “changes” it measures, as counted by the owner of the clock, the observer at rest with it.

What is important for this purpose? Not the motion of the ticker wrt a foreign observer but wrt to its own clock:

- What was the initial state of motion of the ticker wrt its clock? In both cases, in the case of both tickers, nil. Before being set in motion, the ticker was fully at rest with the frame where it ticks.

- What is the state of motion it acquires? Both tickers are pushed by an agent located at their respective frames, so they receive the same push relative to their own frames.

- Who counts the ticks? Their own clock, which is supposed to share their own state of motion, except for the transversal motion, which is the same for both observers in spite of their different perspectives.

Therefore, both observers should agree that their tickers tick homogeneously. Their reasoning will be: “Ok, your ticker is pushed wrt to me harder or more weakly than wrt you, but I admit that here the relevant thing is how fast it will travel wrt you, since you are the counter and not me. Ok, your ticker has to traverse in my CS an X path that it doesn’t traverse in yours, but it is true that the floor and ceiling of your clock also traverse the same path, in harmony with the ticker, and after all you are the counter. So I agree that your ticker ticks wrt your counter the same number of ticks that mine ticks wrt to my own counter”.

Nevertheless, here we have eliminated the distorting effect of collisions when the ticker bounces off and this may disqualify the instrument for practical purposes.

And, yes, there may be another reason why these ideal instruments would not render homogeneus results for all observers: it may be that two apparently equal pushes (for example, the balls are shot from identical little cannons) do not generate the same result, the same motion wrt to the local frame, this being due to the electromagnetic nature of the interaction... But again that would be a physical reason at the heart of the option for Lorentzian diagrams and equations...

 

[JesseM]

Well, if we introduce speed, we complicate things, because speed is distance divided by time and then we must wonder… “time”, but measured how? It will have to be measured by another observer, armed with another instrument. But which instrument? One like this very instrument? Then the question is what kind of time this other instrument measures, homogeneous or inhomogeneous for all observers...? We go into a vicious circle.

All that matters is that the devices you want to serve as clocks are constructed identically in each frame. For example, instead of worrying about speed, each observer could use physically identical springs which have been compressed an equal amount (as measured by their own rulers) to set the balls in motion from an initial state of rest, and in this case the Lorentz-invariance of the laws of physics ensures that these ball clocks will all have the same ratio of "ticks" to the ticks of a light clock at rest in the same frame. Likewise, there are various ways to construct physical clocks that don't involve looking at the time for some object (or light particle) to get from location to another, like clocks based on the decay rate of radioactive isotopes. If you aren't concerned too much about accuracy, you could even use biological rhythms like heart rate; these would experience time dilation too. There is no way that different observers could use identical physical procedures to construct clocks in their own frame and yet not have clocks in different frames show time dilation by the amount predicted by the Lorentz transformation.

And, yes, there may be another reason why these ideal instruments would not render homogeneus results for all observers: it may be that two apparently equal pushes (for example, the balls are shot from identical little cannons) do not generate the same result, the same motion wrt to the local frame, this being due to the electromagnetic nature of the interaction

As long as the frames' coordinates are the standard SR inertial frames related by the Lorentz transform, then identical cannons would have to fire the balls at the same coordinate speed according to relativity, anything else would violate the first postulate.

 

[Saw]

All that matters is that the devices you want to serve as clocks are constructed identically in each frame. For example, instead of worrying about speed, each observer could use physically identical springs which have been compressed an equal amount (as measured by their own rulers) to set the balls in motion from an initial state of rest, and in this case the Lorentz-invariance of the laws of physics ensures that these ball clocks will all have the same ratio of "ticks" to the ticks of a light clock at rest in the same frame. Likewise, there are various ways to construct physical clocks that don't involve looking at the time for some object (or light particle) to get from location to another, like clocks based on the decay rate of radioactive isotopes. If you aren't concerned too much about accuracy, you could even use biological rhythms like heart rate; these would experience time dilation too. There is no way that different observers could use identical physical procedures to construct clocks in their own frame and yet not have clocks in different frames show time dilation by the amount predicted by the Lorentz transformation.

Well, your comments have made me think and revise my position, as usual. If I still have some concerns, it’s a question of small nuances… Let me just share what has struck me in the popular introductions to SR and hear your reaction.

Let us analyze the typical example of the light clock, but only after analyzing the ball clock.

The “ball clock”

We construct “ball clocks” with identical springs and other identical materials when at rest wrt each other in the same frame. We set them in motion.

I accelerate away with my own spaceship clock. What happens immediately? My spaceship clock stops functioning. While the ball is traveling towards the ceiling of the train, the ceiling has moved away. The ball feels a fictitious force, as if it were being pushed against the tail of the train: it was moving by inertia and it has not yet been accelerated to acquire the new state of motion of the train. In normal circumstances, the ball would in the end acquire the new state of motion by crashing against the walls of the train and my own hand if I catch it when it is slow enough (it’d lose its excess of kinetic energy in part by transferring it to the train and myself, in part by getting hotter and making the train hotter). Since I had stipulated that this weird ball does not lose kinetic energy in collisions, I have a problem here. Let us eliminate this artificial postulate, if you wish.

I have gone around the universe and I’m coming back to where you are. During this trip I have suffered (since I have rotated) multiple accelerations and the clock has misfunctioned during each of them. But now, finally, the effect of all accelerations has vanished and I am approaching you by sheer inertia in a straight line.

The ball is quiet in my hand (it is there and it doesn’t try to escape away = it now shares my state of motion). Since the spring is pointing in the direction perpendicular to our relative direction of motion, it doesn’t matter if its length is measured with my rulers in my new frame or with your rulers in your frame: we all agree it has the same length.

If I activated the clock before meeting you again, I would have a big problem: how do I synchronize your clock with mine when away from you? Ok, that is another of the practical drawbacks of the “mechanical” system. But let us stipulate that we activate our respective clocks right at the time of our new encounter.

Now, we freeze the film and we wonder, “have our respective balls reached the same height wrt our respective ceilings?” = “are they relatively changing (your ball wrt your ceiling, my ball wrt to my ceiling) at the same rate?”

Well, I thought: you cannot answer the question with a pure geometrical or mathematical approach, unless you feed the drawing or the equation with some physical assumption:

* YOUR ball:

- In your CS = K: It has followed a vertical path.
- In my CS = K’: It has followed a diagonal path bent to the left.

So the diagonal path that I paint in my K’ coordinate system is incontestably LONGER than the vertical path that you paint in your K coordinate system.

Mathematically, the relation is given by the Pythagorean Theorem:

(My Diagonal Path for your ball)² = (Your vertical path for your ball) ² + (The distance that we have separated from each other)²

But this reasoning is all referred to YOUR ball. Does it tell us anything about where MY ball is at this time? No, it doesn’t!

* MY ball:

- In your CS = K: It has followed a diagonal path bent to the right.
- In my CS = K’: It has followed a vertical path.

So the diagonal path that you paint in your K coordinate system is incontestably LONGER than the vertical path that I paint in my K’ coordinate system.

Mathematically:

(Your Diagonal Path for my ball)² = (My vertical path for my ball) ² + (The distance that we have separated from each other)²

But this reasoning is all referred to MY ball. Does it tell us anything about where YOUR ball is at this time? No, it doesn’t!

So we are left with a gap of knowledge. However, we do need to establish a comparison between the performances of the two balls. In particular, as noted above, what is relevant for our purpose is knowing if your vertical path for your ball (= the vertical component of my diagonal path for your ball) is equal to my vertical path for my ball (= the vertical component of your diagonal path for my ball).

Here is where the physical assumption pops in. Classical relativity fills the gap by assuming that the two balls have received, in their respective environments, equal pushes wrt to those environments. So they have, in their respective frames, the same, let us say, kinetic energy content. Thus the theory concludes that the two balls must have reached their respective targets simultaneously = they are ticking (measuring change) at the same rate.

Things are obscured if, as Newton did, we postulate that “time is absolute”, as if it were an entity. But if we stuck to the physical assumption that I have just explained, the conclusion seemed reasonable SO FAR: the two balls tick homogeneously because of physical reasons. A different thing is that, anyhow, a “ball clock” is not practical, because it is slow and erratic, and so we shift to the light clock, which is preferable.

The light clock

Popular accounts of the thought experiment tell you “first dispense with your logic” and “second believe this: because I measure that the light of your clock follows a diagonal path in my CS, I must think that your clock slows down wrt mine and because you measure that my clock follows a diagonal path in your CS, you must think that it is my clock the one that slows down”.

That is a fallacy. Of course, your photon has traveled in my CS a longer path than in your own CS, just as my photon has traveled a longer path in your CS than in mine, but that doesn’t tell us anything about which photon is closer to its own target at a given moment.

Only if you introduce in the measurement system the synchronisation method of the Einstein convention, as well as the measurement of length with that very same convention, does the system, as a whole, make sense. In fact, it is my impression that we could do without the concept of length in SR: the x coordinate is just a reminder of the “delay time” that you have attributed to the distant clock when you synchronized it with yours…

The system (I learn from your comments) is quite immune to physical circumstances. If in the former case (the ball clock) we needed, to draw our conclusion, to make a physical assumption (the balls take the motion of their respective sources), now we work precisely on the basis of the opposite assumption (light does not take the motion of the source) and it is completely indifferent whether one frame moves more or less with regard to any hypothetical standard like the old aether: even if there existed an aether, the system would work anyhow.

But if I have learned anything from our previous discussion is that SR itself “relativizes” the value of the frame-dependent quantities, like simultaneity and time dilation, which is also a consequence of the synchronization method: you do not claim that the duel is unfair just because different observers have different opinions about whether the signals arrived to the duellers simultaneously; instead, you look at the proper time available for each dueller, which is invariant for all observers.

I visualize it as follows: you and I are joined by a road on whose length we agree; then a lake appears between the two of us and we cannot use the road to measure our separation any more; so each of us circumvents the lake by building two new roads that form right angled triangles, of which the height would be the length of the original road; you use for that purpose the left side of the lake, I use the right side; we may use triangles with different bases and hypotenuses, but we make our calculations and keep agreeing on the length of the original road.

That is why I thought: SR has “down-graded” simultaneity as a variant. A distant clock is synchronized by means of a method based on the two way trip of light. Thus no observer can claim that he has the absolute truth about where a light beam is during a one-way trip. This looks like a gap in our knowledge. In my example, this is like having recovered the knowledge about which distance separates us, but having definitely lost the information about where exactly was that little restaurant by the original road, which we used to visit in the past and is now covered by water.

Can this be problematic? Well, I thought it might be if we still maintain the physical assumption I made before: two bullets shot from the train receive identical pushes wrt their respective frames. In that case, the duel with light signals and bullets would not be fair: even if we don’t know it, the light signal would have arrived “really” earlier to one dueller (which is an advantage) and this would not be compensated by a correlative disadvantage (the bullet aimed at that dueller would not arrive “really” earlier than the other bullet).

The problem would not exist, however, if we did revise the concept of the “ball (or bullet) clock”: the ball or the bullet would not receive so identical pushes in both directions, because each of them is accelerated through an electromagnetic interaction and this means that at the root of its motion there would exist a “light clock”…

Conclusion:

- I see you may be very right in holding that the principle of relativity may explain everything, without the need of physical assumptions.
- But I still feel more comfortable if I find some physical explanation…

As long as the frames' coordinates are the standard SR inertial frames related by the Lorentz transform, then identical cannons would have to fire the balls at the same coordinate speed according to relativity, anything else would violate the first postulate.

Well, with this explanation of the "light clock" being at the heart of the electromagnetic interaction, I would agree that identical guns fire the bullets at the same speed in both directions wrt the train and at the speeds given by the relativistic addition formula wrt the ground. The "difference" between the two pushes would lie somewhere hidden in the realm of a non-visible "reality" (just like the "real" earlier of the light signals), without practcal consequences.

I assume all this sounds unnecessary to you, but did I say anything that you as scientist would consider, let us say, "unacceptable"?

 

[Saw]

Hmm… As I was too long without supervision, I have supervised myself a little. I have checked what I had been told, that is to say, that the relativistic formula for the addition of velocities…

can be derived directly from the Lorentz transform based only on knowing the coordinates of a moving object in one frame and then figuring out what the equivalent coordinates would be in another frame

Yes, I understand the derivation of the formula goes as follows:

- Observers in the train note the x’ and t’ coordinates at two events: when the bullet is fired by Back and when it hits Front.
- dx’ / dt´= the speed of the bullet in the train frame = u’.
- To transform into the ground frame, you apply the LT to both numerator and denominator and thus put everything in terms of dx and dt.
- Gamma disappears from the division.
- You look for ways to put dx/dt in the equation and replace this expression by u, meaning the speed of the bullet in the ground frame.
- You solve for u (speed of the bullet in the ground frame) and you get it in terms of v and u’ (speed of the bullet in the train frame).

In this sense, the formula is irrefutable.

But all that this means is that, if you have plotted the track of the bullet in the train frame with the train light clocks and rulers, those readings made in the train are a sufficient clue to figure out the readings that should have been obtained by clocks and rulers of the ground.

However, does this predict the coordinates that should have been obtained in the train for the bullet projected in the opposite direction? Must we assume that this second set of coordinates that will be measured by the light clocks and rulers of the train will be identical to the first set?

Both Galilean relativity and SR predict so, but they do it on the basis of physical (albeit different) assumptions… It seems clear to me that you cannot predict so, unless you make some physical assumption about how the interaction that accelerates the bullet develops:

- Galilean relativity assumed that the bullet receives identical pushes in both directions wrt the train. Ergo, the same should happen with a car train at rest on the ground: the bullets would receive identical pushes wrt the ground. Ergo, since a bullet can be clock, if it bounces to and fro, the old theory also predicted that clocks in different frames should measure the same time (see above for the logical link between the two ideas).
- SR, instead, postulates (and corroborates with experiment) that light clocks of different frames measure different times, because light does not take the motion of the source. But if it wants to postulate that the light clocks “mirror” the motion of the bullet as well, if a “bullets clock” should suffer the same effect, then SR must assume that the classical postulate (identical pushes) was not correct: for example, if, contrary to the classical prediction, a ball signaling the commencement of the duel to Back had received a weaker push, the corresponding bullet aimed by Front at Back should also have received a weaker push; only thus should Back have in the end the same opportunities (tic-tocs of his clock) as Front, only thus could we say that the light clocks and rulers render the same “proper time” for Back and Front between firing and being fired.

There may be a reasonable explanation for this: the pushes are produced by electromagnetic interactions. The pushes are produced by a light clock: if the light signaling the commencement of the duel for Back faced a heading target, then also, at the microscopic level, the electromagnetic force in the atoms of the powder pushing the atoms of the bullet towards Back would face a heading target. I don’t see anything wrong with testing this line of thought, which was by the way hinted at by Einstein himself in his 1916 book. Of course, you can dispense with it by taking a purely “descriptive” approach (things “are” as we observe them because nature conforms to the principle of relativity), but nothing bad can come from speculating over the ultimate causes for that (why things “should be” like that)…

Anyhow, I will not insist on this.

Just a technical question: are these two diagrams I’ve drawn for the duel with lasers and with bullets, respectively, right? Thanks for any correction.

 

[Saw]

I forgot to say that I've chosen as speed of the bullet in the train frame 0.5 c, which transforms into 0 c towards Back and 0.8 c towards Front in the ground frame.

 

[Saw]

I would like to reinforce what I said before with some comments on the meaning on the so called transformation for “coordinates”, the Lorentz Transformations, and the other equations of SR, the so called transformations for “intervals” of time and length.

Let us take the specific example of these two events:

- Event 1: the two referees are together and, right at that instant, the train referee emits the signals towards the duellers.
- Event 2: Train Front receives his signal and instantaneously fires, right when he is by Ground Front (I call Ground Front the observer in the ground frame, who could be standing at the right edge of a bigger car, painted in the drawing with blue colour, by the platform).

The readings of the observers of each frame who watch those two events will be:

- at Event 1, the witnesses are the two referees; by stipulation, their clocks read 0s and they are situated at the origins of their coordinate systems, so Ground x1 = Train x1 = 0 and Ground t1 = Train t1 = 0;
- at Event 2, the witnesses are Ground Front, whose Ground t2 = 1.732 s and whose Ground x2 = 1.732 ls, and Train Front, whose Train t2 = 1 s and whose Train x2 = 1 ls.

These numbers are related by the LTs, which [if we measure distances in light-seconds and, consequently, have c = 1 and if v=0.5 and so gamma = 1/sqrt(1-v²) = 1.154] look as follows:

Ground t2 = gamma * (Train t2 + v * Train x2) = 1.154 (1+ 0.5*1) = 1.732 s
Ground x2 = gamma * (Train x2 + v * Train t2) = 1.154 (1+ 0.5*1) = 1.732 s

Train t2 = gamma * (Ground t2 - v * Ground x2) = 1.154 (1.732 - 0.5*1.732) = 1 s
Train x2 = gamma * (Ground x2 - v * Ground t2) = 1.154 (1.732 - 0.5*1.732) = 1 ls

Thus both frames measure the same speed for light = 1c.

The same result can be obtained with the concepts of Relativity of Simultaneity (which I will call “Bad Synchronization” = BS), Time Dilation (TD) and Length Contraction (LC) + considering the motion of the other frame:

From the perspective of the train frame:

- First, the problem is that Ground Front thinks that, at Event 1, he was at a distance from the referees of twice the length of the right side of my brown car. However, in fact the distance was 1.5 times that length. He suffers LC, because he measures length with sticks that are too short and and thus obtains too many length units. I must shrink his measurement by multiplying it by sqrt(1-v²) = 1.732 * 0.866 = 1.5 ls.

- For the same reason, at the time of Event 1, he was not by the place of my frame that he thinks he was. He was closer to the origin, right by another colleague of mine, whose clock read zero, while Ground Front’s read 0.866 s. Obviously, the poor guy, Ground Front, has also done a BS. In particular, his clock is ahead by vx, x being the distance that he measured with his short sticks. His clock read 0.866 s when it should read 0.866 – 1.732*0.5 = 0.866 – 0.866 = 0 s, as my assistant’s.

- Thus the real time interval that Ground Front should have measured must be calculated eliminating that “artificial advantage” that he introduced = 1.732 – 0.866 = 0.866 s. But there is another problem: this interval (0.866 s) is too short, since it is affected by TD. Hence I must enlarge it by multiplying it by 1/sqrt(1-v²) = 0.866 s / 0.866 s = 1 s.

- Finally, Ground Front’s last mistake is that he thinks that he is stationary, but is in fact moving to the left, heading towards the light signal. In particular, during the trip time of the light signal, he displaced towards it by 0.5 c * 1 s = 0.5 ls. So the real displacement of the light signal in my frame was L (1.5) – vt (0.5*1) = 1.5 – 0.5 = 1 ls.

Thus, after considering all those elements, Train Front concludes what the LT had told him, that is to say, that at Event 2 (arrival of the light signal) his clock is located at 1 ls from the origin and reads 1 s, which also corroborates that the speed of light for him is c.

From the perspective of the ground frame:

For Ground Front, the reasoning is the reverse one:

- Train Front suffers LC: at Event 1, he was closer to the origin than what he thinks. He thinks that he was 1 ls away, but the problem is that his sticks have shrunk and that is why he measures too many length units. Actually the distance was shorter = 1 ls * sqrt(1-v²) = 0.866 ls.

- Train Front has done a BS: at Event 1, Train Front, being closer to the origin than what he thought, was by a colleague of mine whose clock read 0s, while Train Front’s was behind by vx = 0.5 * 1 = – 0.5 s.

- Train Front suffers TD: if we eliminate that “artificial handicap” of Trains Front’s clock, we infer it has really measured an interval of 1.5 s. But this time is too short, because it is time dilated. Actually it should be 1.5 / sqrt(1-v²) = 1.5 * 1.154 = 1.732 ls.

- Train Front is not stationary, but moves to the right, escaping from the light signal: in particular, during the trip time of the signal, Train Front has moved away in my frame L (0.866) + vt (0.5 *1.732) = 0.866 + 0.866 = 1.732 ls.

Thus, after considering all those elements, Ground Front concludes what the LT had told him, that is to say, that at Event 2 (arrival of the light signal) his clock is located at 1.732 ls from the origin and reads 1.732 ls, which also corroborates that the speed of light for him is c.

Conclusions:

1) If someone asks me again, “do you believe that there is no objective truth about which observer is right in terms of the simultaneity of two events?”, I will copy and paste this text:

Your question is not well phrased. If you rephrase it in one of these manners, here are my answers:

a) Do you believe that the x and t values of two clocks that meet at a given moment will be related by the LT? Yes, I do.

b) Do you believe that the isolated judgment that two events are not or are simultaneous (just like the isolated judgment that another frame suffers LC or TD) reflect any objective truth about what has happened and may serve to solve any physical problem by itself? No, I don´t.

c) Do you believe that, if you combine the FOUR elements in your judgment (BS, LC, TD + motion of the other frame) you get the same result as through the LT? Yes, I do.

2) Incidentally, I have confirmed my impression that the distinction between transformations from “coordinates” and transformations for “intervals” is quite artificial. One of them (the LT) gives the "final" outcome of the maximum information we can obtain about physical facts, the others are just “provisional” judgments, which become only valid if put together and when put together give the same result as the LT…

3) If you apply this reasoning to the extraterrestrial muons experiment, for instance, what do you think it proves? Wouldn’t you conclude that simply by applying SR formulas there is no reason to infer that the extraterrestrial muons live longer than the Earthlings? Wouldn’t you admit that, if you wish to affirm so or you hold that other experiments prove so, that must be due to a physical reason that is not contained in the mere formulas?

 

[Saw]

If you apply this reasoning to the extraterrestrial muons experiment, for instance, what do you think it proves? Wouldn’t you conclude that simply by applying SR formulas there is no reason to infer that the extraterrestrial muons live longer than the Earthlings? Wouldn’t you admit that, if you wish to affirm so or you hold that other experiments prove so, that must be due to a physical reason that is not contained in the mere formulas?

Having had no comment, I’ve been thinking about the muon experiment.

The standard explanation of this experiment is as follows (I quote JesseM from another thread for convenience, but this is what you can find in all books):

In different frames there are different explanations--in the Earth's frame you must use time dilation to explain how a muon created by a cosmic ray hitting the upper atmosphere can make it all the way to the surface before decaying, while in the muon's frame you must use length contraction. There is no preferred frame, so neither explanation is "more correct" than the other.

Another example of the standard explanation:

http://hyperphysics.phy-astr.gsu.edu/HBASE/Relativ/muon.html

Earth-Frame Observer:

“The muon’s clock is time dilated or running slow by the factor T = gamma * T₀”

Muon-frame Observer:

“The muon sees distance as length-contracted so that L = L_O / gamma”

When you scrutinize this explanation you find that L₀ or “proper length” is defined as the distance between “two simultaneous events”. But which events and for whom are they simultaneous? Obviously, the events are the acts of measuring this distance by different observers at rest with the Earth-frame at its two edges, the surface of the Earth and the upper atmosphere, and the judgment of simultaneity is made from the perspective of the Earth. One Earth-observer, while measuring this length, would observe the birth of the muon and the other, simultaneously, while putting the ruler on the surface of the Earth, might witness the birth of another terrestrial muon, created artificially in a laboratory. This lab muon would decay before the arrival of the extraterrestrial muon.

But why should we imagine that the Muon-frame cannot measure a proper length? If the muon had a long enough ruler at rest with it, it would also measure a proper length. It would only happen that the muon-frame would have a different view on simultaneity. The muon-observer at the surface of the Earth would do its job at a time that would also be simultaneous with the muon’s birth, in the muon-frame, but would be a later time in the Earth-frame. Thus, if a second terrestrial muon were created in the lab at that time, the extraterrestrial muon would appreciate that the latter is the one that suffers TD. Likewise, it would judge that the Earth sees the proper distance measured by the extraterrestrial muon as length contracted (i.e., too long).

This would not change the facts, but it would change the explanation. The extraterrestrial muon, from its own perspective, would confirm that the first lab muon has decayed before its own arrival to the surface but it would provide a different explanation: the first lab muon has decayed because it is not really its twin brother but by far its elder; instead, the second terrestrial lab muon has not decayed because it suffers TD.

In conclusion, it seems to me that:

- The standard explanation does not really take the two perspectives. Actually, it is taking the simultaneity perspective of the Earth.
- Only from this perspective is it true that the muon appears to suffer TD and that the muon appears to see the distance to the Earth as shorter.
- In fact, if you take the simultaneity perspective of the muon, it is the other way round: it appears to the muon that the Earth suffers TD and the Earth sees the distance measured by the muon as too long, as if the muon measured with short sticks (LC).

Right?

 

[JesseM]

Another example of the standard explanation:

http://hyperphysics.phy-astr.gsu.edu/HBASE/Relativ/muon.html

Earth-Frame Observer:

“The muon’s clock is time dilated or running slow by the factor T = gamma * T₀”

Muon-frame Observer:

“The muon sees distance as length-contracted so that L = L_O / gamma”

When you scrutinize this explanation you find that L₀ or “proper length” is defined as the distance between “two simultaneous events”. But which events and for whom are they simultaneous? Obviously, the events are the acts of measuring this distance by different observers at rest with the Earth-frame at its two edges, the surface of the Earth and the upper atmosphere, and the judgment of simultaneity is made from the perspective of the Earth.

In the length contraction equation each observer picks events on the worldline of either end of the object that are simultaneous in their own frame. So although the observer at rest in the Earth frame would pick events in the upper atmosphere and at the surface which were simultaneous in the Earth frame in order to define L₀, the observer at rest in the muon frame would not say the distance between those same two events is L; rather, L refers to the distance between two different events in the upper atmosphere and on the Earth's surface which are simultaneous in the muon's rest frame.

But why should we imagine that the Muon-frame cannot measure a proper length? If the muon had a long enough ruler at rest with it, it would also measure a proper length.

As I said, the muon frame can certainly measure the proper length between events which are simultaneous in the muon's frame, this is exactly what the meaning of L is in the length contraction equation.

The muon-observer at the surface of the Earth would do its job at a time that would also be simultaneous with the muon’s birth, in the muon-frame, but would be a later time in the Earth-frame.

Not clear what you mean by "the muon-observer at the surface of the Earth would do its job"--what job is that? Do you just mean picking an event on the surface which is simultaneous with the birth of the muon in the upper atmosphere in that muon's own rest frame? If so, you've got the order backwards, in the Earth's frame the event on the Earth's surface would happen before the event of the birth of the muon in the upper atmosphere. Remember, if two clocks are synchronized in their own rest frame, then in a frame where they're moving along the axis joining them, it's the clock in the rear whose time is ahead of the clock in the lead. So if two clocks are at rest and synchronized and a distance L apart in the muon rest frame, and one clock reads 12:00 at the moment it's next to the birth of the muon, while the other clock reads 12:00 when it's at the surface of the Earth, in the Earth frame the clock which is next to the muon's birth is trailing the clock which reaches the surface first, so when the clock in the upper atmosphere reads 12:00 the clock in the lead reads an earlier time (and thus has not yet reached the surface, so the event of it reaching the surface happens later in the Earth frame).

Thus, if a second terrestrial muon were created in the lab at that time

Created at what time? The time that is simultaneous with the event of the extraterrestrial muon being created, in the extraterrestrial muon's rest frame?

the extraterrestrial muon would appreciate that the latter is the one that suffers TD.

By "the latter", do you mean the extraterrestrial muon? If so, no, in the rest frame of the extraterrestrial muon it is the terrestrial muon that takes longer to decay than the extraterrestrial one. If this wasn't true it would be a clear violation of the first postulate, since the first postulate says all laws of physics have to look the same in all frames, so if it's true in one frame (like the Earth's frame) that a muon in motion takes longer to decay than a muon at rest, this must be true in all other frames (including the rest frame of the extraterrestrial muon).

Likewise, it would judge that the Earth sees the proper distance measured by the extraterrestrial muon as length contracted (i.e., too long).

This is a very convoluted sentence, I don't really understand what it means for the extraterrestrial muon to "judge" that the Earth "sees" the distance measured by in the extraterrestrial muon's frame to be too long. All talk of "observers" in SR is just supposed to be a shorthand way of talking about measurements made in a specific frame, so it's completely confusing to talk about an observer making judgments about a different frame, much less an observer judging something about a second observer making a judgment about the first observer's frame, I can't make heads or tails of what you're trying to convey here. If you want to compare distances or times in different frames, just talk about what happens when we (the omniscient problem-analyzer) compare some value obtained by observer #1 with some value obtained by observer #2 (each value being specific to their own rest frame). And if we do try to phrase things this way, it's not clear to me if you want to compare the distance from the surface to the upper atmosphere in both frames, or if you want to compare the distance in each frame between two specific events on the surface and in the upper atmosphere which are simultaneous in the extraterrestrial muon's frame but not the Earth's frame, or something else entirely.

This would not change the facts, but it would change the explanation. The extraterrestrial muon, from its own perspective, would confirm that the first lab muon has decayed before its own arrival to the surface but it would provide a different explanation: the first lab muon has decayed because it is not really its twin brother but by far its elder; instead, the second terrestrial lab muon has not decayed because it suffers TD.

You're supposing two terrestrial muons are created at rest relative to Earth? If so, in the extraterrestrial muon's rest frame both terrestrial muons have the same velocity and thus experience the same time dilation factor in this frame. If you mean that the first terrestrial muon was created simultaneously with the extraterrestrial muon's birth in the Earth's rest frame, while the second terrestrial muon was created simultaneously with the extraterrestrial muon's birth in the extraterrestrial muon's rest frame--and it would really be better to spell this sort of thing out if you want your posts to be understandable!--then I agree the first terrestrial muon will have already decayed by the time the extraterrestrial muon reaches the surface, but in the extraterrestrial muon's rest frame that's because this first terrestrial muon was created significantly earlier than the event of the extraterrestrial muon's birth.

- The standard explanation does not really take the two perspectives. Actually, it is taking the simultaneity perspective of the Earth.
- Only from this perspective is it true that the muon appears to suffer TD and that the muon appears to see the distance to the Earth as shorter.

Again, it makes no sense to me to talk about what the muon "sees" from the perspective of the Earth's frame, the terminology of talking about what different observers or objects "see" is understood to mean what happens in their own rest frame.

- In fact, if you take the simultaneity perspective of the muon, it is the other way round: it appears to the muon that the Earth suffers TD and the Earth sees the distance measured by the muon as too long, as if the muon measured with short sticks (LC).

It's true that in the muon's frame the Earth suffers TD, but this is irrelevant to the problem as usually stated since we're just interested in how the muon manages to reach the surface without decaying first. In the muon's rest frame, this is just because the distance between the upper atmosphere and the surface is shrunk (i.e. the distance in this frame is smaller than the distance from the upper atmosphere to the surface in the Earth's frame), so the coordinate time between the muon passing the upper atmosphere and reaching the surface is smaller in this frame than it is in the Earth's frame, which means that even though the muon's clock is not dilated relative to coordinate time in its own rest frame, it can still reach the surface before decaying. In contrast, in the Earth's rest frame the coordinate time for the muon to travel between the upper atmosphere and the surface is greater than the time it would take a muon to decay if it were at rest in that frame, but the extraterrestrial muon has a high velocity in this frame so its decay time is extended long enough for it to reach the surface.

 

[Saw]

JesseM, thanks.

Let us see if my understanding of the standard explanation is better now.

The story is painted in the attached diagram.

Description of the diagram

The numbers are not at all realistic, since muons created by cosmic rays travel at almost the speed of light, the half-life of a muon is about 2.2 microseconds…, but I personally find these numbers easier and, with them, the reasoning is the same.

Some clarifications:

- A Red and a Blue Frame are moving wrt each other at v = 0.5 c and so sqrt(1-v^2) = 0.866 and gamma = 1/sqrt(1-v^2) = 1.1545.
- The Red Frame plays the part of the Earth or Lab, which is placed at the left edge, and there is a red atmosphere that is 1 ls long in the Red Frame and 0.866 ls long in the Blue Frame.
- The Blue Frame has a blue stick that is 2.309 ls long in its rest frame and 2 ls long in the Red Frame.
- We arbitrarily fix the “half-life” of a moun in its own rest frame as 1.8 s. For simplicity, we leave out statistics and assume that half-life means “life”, as if all muons decayed after that (proper) time interval.
- A red observer hovering at the edge of the red atmosphere and a blue observer hovering some distance from the edge of the blue atmosphere set their clocks to zero when they meet and synchronize clocks with the networks of assistants of their respective frames.

There are four events:

Event 0: A 1st Red Muon is born at the Red Lab at red t = 1 s, blue t = 0.577 s.
Event 0 bis: A 2nd Red Muon is born at the Red Lab at red t = 1.5 s, blue t = 1.154 s.
Event 1: The Blue Muon is born at red t = 1 s (same red time as the birth of the 1st Lab Muon) and blue t = 1.154 s (same blue time as the birth of the 2nd Red Muon).
Event 2: The Blue Muon lands on the Red Lab at red t = 3 s, blue t = 2.886 s.

The measurements

If we consider specific measurements of each frame, we get, with the numbers of the example, these results:

In the Red Frame:

The Red Lab is stationary and it is the Blue Muon the one that moves to the left.

When the Blue Muon is born (Event 1),

* the Blue Muon occupies the origin of the Red CS, so red x1 = 0 ls, which is the upper end of the red atmosphere and
* the clock of the red observer located there reads t1 = 1 s.

In the Red Frame, all clocks are synchronous, so the Red Lab's clock also reads at this time red t1 = 1 s; by then, the left edge of the blue stick has reached precisely the Red Lab and the 1st Red Muon is born (Event 0). (Events 0 and 1 are simultaneous in the Red Frame.)

When the Blue Muon collides with the Red Lab (Event 2), after moving to the left:

* red x2 of the BM = -1 ls and
* red t2 = 3 s

Thus the red intervals are:

* red dx = -1 – 0 = -1 ls and
* red dt = 3 – 1 = 2 s.

The velocity of the Blue Muon in the Red Frame = - 1 ls / 2 s = -0.5 c.

In the Blue Frame:

In this Frame, the Blue Muon does not move, it is stationary and what approaches is the Red Lab.

When the Blue Muon is born (Event 1):

blue x1 of the Red Lab = - 0.289 ls and
blue t1 = 1.154 s (which is later than red t1 = 1s).

In the Blue Frame, all clocks are synchronous, so the clock of the blue observer who is by then by the Red Lab also reads blue t1 = 1.154; by that time, the Red Lab, as it is later, is not by the left edge of the blue stick any more, it has moved a little further to the right and is closer to the Blue Muon and by this time the 2nd Red Muon is born (Event 0 bis). Events 0 bis and 1 are simultaneous in the Blue Frame.

When the Red Lab collides with the Blue Muon:

* blue x2 for the Red Lab = 0.577 ls and
* blue t2 = 2.866 s

Thus the blue intervals are:

* blue dt = 2.866 – 1.154 = 1.732 s
* blue dx = 0.577 – (-0.289) = 0.866 ls

The velocity of the Red Lab in the Blue Frame = 0.866 ls / 1.732 s = 0.5 c

Similarities between the two stories

As noted, the stories of each frame are different, but they have elements in common:

- In both cases, the start time is the birth of the Blue Muon and the end time is the collision between the Red Lab and the Blue Muon.

- In both cases, at the start time, the separation between the two actors, the Red Lab and the Blue Muon, is the length of the red atmosphere.

The concepts and the formulas used in the standard explanation

As to time:

The dt is “proper time”, because it is measured by the same blue clock, which is present at the two events which have been taken as reference in the Blue Frame, birth of the Blue Muon and its collision with the Red Lab. It is the T₀ of the formula.

The dt is “coordinate time”, because it is measured by two red observers at different places. It is the T of the formula.

The formula relating the two is:

T = gamma * T₀

or

Blue Muon’s proper time = T₀ = dt / gamma = 2 s * sqrt(1-v^2) = 2 s * 0.866 = 1.732 s

As to length:

The formula does not focus on distances traversed but on the length of the red atmosphere. It is the following:

L = L₀ / gamma = 1 * sqrt(1-v^2) = 1 * 0.866 = 0.866 ls

L₀ is the length of the red atmosphere as measured in the Red Frame. It has been measured on its two edges simultaneously, for example at red t1 = 1s. Thus the act of measuring the right edge coincides with the birth of the Blue Muon (Event 1) and the act of measurement the left edge coincides with the birth of the 1st Red Muon (Event 0).

L is the length of the red atmosphere as measured in the Blue Frame. It has also been measured on its two edges simultaneously, for example at blue t = 1.154 by the Blue Muon at the time of its birth (Event 1) and by another blue observer who witnesses the birth of the 2nd Red Muon (Event 0 bis).

So both length magnitudes are measured in the corresponding frame simultaneously in either end, although at one edge each each frame measures at a different event, precisely because simultaneity is relative. I suppose that L₀ is called “proper length” because the measurer is the frame that the object in question (the red atmosphere) is at rest with.

Right?

If these data / concepts are correct, I would still like to comment the language of the standard explanation…

 

[Saw]

Note: I've edited the previous posts to correct some mistakes with the colors at the end...

 

[JesseM]

JesseM, thanks.

Let us see if my understanding of the standard explanation is better now.

The story is painted in the attached diagram.

Description of the diagram

The numbers are not at all realistic, since muons created by cosmic rays travel at almost the speed of light, the half-life of a muon is about 2.2 microseconds…, but I personally find these numbers easier and, with them, the reasoning is the same.

Some clarifications:

- A Red and a Blue Frame are moving wrt each other at v = 0.5 c and so sqrt(1-v^2) = 0.866 and gamma = 1/sqrt(1-v^2) = 1.1545.
- The Red Frame plays the part of the Earth or Lab, which is placed at the left edge, and there is a red atmosphere that is 1 ls long in the Red Frame and 0.866 ls long in the Blue Frame.
- The Blue Frame has a blue stick that is 2.309 ls long in its rest frame and 2 ls long in the Red Frame.
- We arbitrarily fix the “half-life” of a moun in its own rest frame as 1.8 s. For simplicity, we leave out statistics and assume that half-life means “life”, as if all muons decayed after that (proper) time interval.
- A red observer hovering at the edge of the red atmosphere and a blue observer hovering some distance from the edge of the blue atmosphere set their clocks to zero when they meet and synchronize clocks with the networks of assistants of their respective frames.

Just a minor question about how to visualize this scenario which doesn't have anything to do with the math--you say "hovering at some distance from the edge of the blue atmosphere", but do you really mean to imply two different atmospheres in motion relative to each other? Earlier you just referred to a "blue stick" moving relative to the Earth, which is a little easier to visualize. And you don't specify how far this blue observer is from the end of the blue stick, but judging from the diagram, in the blue observer's rest frame he's 0.577 light-seconds from the "lower" end (defining 'lower' and 'upper' relative to the axis of the lab and the top of the red atmosphere) and 1.732 light-seconds from the "upper" end, yes?

There are four events:

Event 0: A 1st Red Muon is born at the Red Lab at red t = 1 s, blue t = 0.577 s.
Event 0 bis: A 2nd Red Muon is born at the Red Lab at red t = 1.5 s, blue t = 1.154 s.
Event 1: The Blue Muon is born at red t = 1 s (same red time as the birth of the 1st Lab Muon) and blue t = 1.154 s (same blue time as the birth of the 2nd Red Muon).
Event 2: The Blue Muon lands on the Red Lab at red t = 3 s, blue t = 2.886 s.

Just for completeness let's add the x-coordinates of these events in each frame:
Event 0: x=-1 in red frame, x=-0.577 in blue frame.
Event 0 bis: x=-1 in red frame, x=-0.289 in blue frame
Event 1: x=0 in red frame, x=0.577 in blue frame
Event 2: x=-1 in red frame, x=0.577 in blue frame

The measurements

If we consider specific measurements of each frame, we get, with the numbers of the example, these results:

In the Red Frame:

The Red Lab is stationary and it is the Blue Muon the one that moves to the left.

When the Blue Muon is born (Event 1),

* the Blue Muon occupies the origin of the Red CS, so red x1 = 0 ls, which is the upper end of the red atmosphere and
* the clock of the red observer located there reads t1 = 1 s.

In the Red Frame, all clocks are synchronous, so the Red Lab's clock also reads at this time red t1 = 1 s; by then, the left edge of the blue stick has reached precisely the Red Lab and the 1st Red Muon is born (Event 0). (Events 0 and 1 are simultaneous in the Red Frame.)

When the Blue Muon collides with the Red Lab (Event 2), after moving to the left:

* red x2 of the BM = -1 ls and
* red t2 = 3 s

Thus the red intervals are:

* red dx = -1 – 0 = -1 ls and
* red dt = 3 – 1 = 2 s.

The velocity of the Blue Muon in the Red Frame = - 1 ls / 2 s = -0.5 c.

In the Blue Frame:

In this Frame, the Blue Muon does not move, it is stationary and what approaches is the Red Lab.

When the Blue Muon is born (Event 1):

blue x1 of the Red Lab = - 0.289 ls and
blue t1 = 1.154 s (which is later than red t1 = 1s).

In the Blue Frame, all clocks are synchronous, so the clock of the blue observer who is by then by the Red Lab also reads blue t1 = 1.154; by that time, the Red Lab, as it is later, is not by the left edge of the blue stick any more, it has moved a little further to the right and is closer to the Blue Muon and by this time the 2nd Red Muon is born (Event 0 bis). Events 0 bis and 1 are simultaneous in the Blue Frame.

When the Red Lab collides with the Blue Muon:

* blue x2 for the Red Lab = 0.577 ls and
* blue t2 = 2.866 s

Thus the blue intervals are:

* blue dt = 2.866 – 1.154 = 1.732 s
* blue dx = 0.577 – (-0.289) = 0.866 ls

The velocity of the Red Lab in the Blue Frame = 0.866 ls / 1.732 s = 0.5 c

It would take me a while to actually go through with the Lorentz transform to check all these numbers, so I'll just trust you on this--the diagrams look right from what I did check (the lengths of the atmosphere and rod, the speed they're moving, and the times the muons are created).

Similarities between the two stories

As noted, the stories of each frame are different, but they have elements in common:

- In both cases, the start time is the birth of the Blue Muon and the end time is the collision between the Red Lab and the Blue Muon.

- In both cases, at the start time, the separation between the two actors, the Red Lab and the Blue Muon, is the length of the red atmosphere.

The concepts and the formulas used in the standard explanation

As to time:

The dt is “proper time”, because it is measured by the same blue clock, which is present at the two events which have been taken as reference in the Blue Frame, birth of the Blue Muon and its collision with the Red Lab. It is the T₀ of the formula.

The dt is “coordinate time”, because it is measured by two red observers at different places. It is the T of the formula.

The formula relating the two is:

T = gamma * T₀

or

Blue Muon’s proper time = T₀ = dt / gamma = 2 s * sqrt(1-v^2) = 2 s * 0.866 = 1.732 s

As to length:

The formula does not focus on distances traversed but on the length of the red atmosphere. It is the following:

L = L₀ / gamma = 1 * sqrt(1-v^2) = 1 * 0.866 = 0.866 ls

L₀ is the length of the red atmosphere as measured in the Red Frame. It has been measured on its two edges simultaneously, for example at red t1 = 1s. Thus the act of measuring the right edge coincides with the birth of the Blue Muon (Event 1) and the act of measurement the left edge coincides with the birth of the 1st Red Muon (Event 0).

L is the length of the red atmosphere as measured in the Blue Frame. It has also been measured on its two edges simultaneously, for example at blue t = 1.154 by the Blue Muon at the time of its birth (Event 1) and by another blue observer who witnesses the birth of the 2nd Red Muon (Event 0 bis).

So both length magnitudes are measured in the corresponding frame simultaneously in either end, although at one edge each each frame measures at a different event, precisely because simultaneity is relative. I suppose that L₀ is called “proper length” because the measurer is the frame that the object in question (the red atmosphere) is at rest with.

Right?

Yes, that all looks right to me. Although as a matter of terminology, usually the phrase "proper length" refers not to the length of a physical object, but to the distance between two events in the frame where they are simultaneous. If you happen to pick events at either end of an object such that the events are simultaneous in the object's rest frame, in that case the proper length between the events will be the same as the object's rest length.

 

[Saw]

Just a minor question about how to visualize this scenario which doesn't have anything to do with the math--you say "hovering at some distance from the edge of the blue atmosphere", but do you really mean to imply two different atmospheres in motion relative to each other? Earlier you just referred to a "blue stick" moving relative to the Earth, which is a little easier to visualize.

Yes, I had started to construct the example as two planets with their respective atmospheres. Then I tried to simplify and put the blue stick instead, but forgot to change that reference to a blue atmosphere.

And you don't specify how far this blue observer is from the end of the blue stick, but judging from the diagram, in the blue observer's rest frame he's 0.577 light-seconds from the "lower" end (defining 'lower' and 'upper' relative to the axis of the lab and the top of the red atmosphere) and 1.732 light-seconds from the "upper" end, yes?

Yes!

Just for completeness let's add the x-coordinates of these events in each frame:

Event 0: x=-1 in red frame, x=-0.577 in blue frame.
Event 0 bis: x=-1 in red frame, x=-0.289 in blue frame
Event 1: x=0 in red frame, x=0.577 in blue frame
Event 2: x=-1 in red frame, x=0.577 in blue frame

Thanks.

usually the phrase "proper length" refers not to the length of a physical object, but to the distance between two events in the frame where they are simultaneous. If you happen to pick events at either end of an object such that the events are simultaneous in the object's rest frame, in that case the proper length between the events will be the same as the object's rest length.

Ok, it is slippery concept… I understand that:

(i) L and L₀ are both distances between simultaneous events, although, at one end of the distance, each frame considers as simultaneous a different event;

(ii) the red atmosphere is the object that happens to extend, in both cases, for both frames, between the shared event and the corresponding different events and

(iii) L₀ is special or at least it occupies the position that it occupies in the equation (which makes it longer) because it is measured in the frame of that particular object (the red atmosphere) that lies between the two ends of both distances, L and L₀.

Well, we could even dispense with the red atmosphere, it could be there or not, but what seems important is that the problem is defined as "will the Blue Muon survive between the interval defined by its birth (collision with the upper atmosphere or simply passing by red x1) and collision with the Red Lab (or simply passing by red x2)?" (That is something I also wanted to comment. Instead of asking "how the muon manages to make it to the surface", I think it is helpful to define the problem in the above mentioned "neutral", non-frame dependent manner and then analyze it from the two perspectives.)

In any case, it seemed to me that L₀ deserved a special name, because the problem is defined as the ability of the Blue Muon to survive between two Events (1 and 2) that are marked by the Blue Muon's meetings with, if not necessarily two parts of an object, at least two points of the Red Frame. How do you differentiate L₀ ? Just with the symbol? No special word to refer to it?

Many texts call “proper length” what is “rest length”. I’ve googled now and found that terminology in “Special relativity and how it works”, by Moses Fayngold, who distinguishes between “proper distance” (defined as the distance between two simultaneous events) and “proper length” (defined as the length of an object in its rest frame).

Wikipedia also warns in a Note that “Proper length has also been used in a (…) restricted sense to help with discussions of length contraction by textbooks, where it is defined as the length of an object when measured by someone at rest relative to that object”.

If you reject this terminology, I would agree with you, because in the example L₀ is something more than rest length, as commented: it is the distance between the two red points that pass by the Blue Muon in the interval he disposes of, as measured in the rest frame of those points, yes, but not any other distance (and not any other length, if there happens to be a physical object linking the two points).

Anyhow, Wikipedia prefers this definition: “proper length is an invariant quantity which is the rod distance between spacelike-separated events in a frame of reference in which the events are simultaneous”. At this stage I'm too tired to make sense of it. Would this apply to our example?

And, in our example, would there be any invariant length magnitude, whatever it is called? What would be here the length equivalent of proper time?

 

[JesseM]

Ok, it is slippery concept… I understand that:

(i) L and L₀ are both distances between simultaneous events, although, at one end of the distance, each frame considers as simultaneous a different event;

Not really disagreeing, but I would phrase it a little differently--I would say that in the length contraction equation, L and L₀ just represent the length of the same object in two different frames. This is of course equal to the distance between simultaneous events on either end of the object, but I'd say "length" is conceptually different in that it doesn't require you to think of any specific events, regardless of whether you pick simultaneous events at t1 or t2 or t3 the distance is the same, as long as you're dealing with a rigid object moving inertially.

(ii) the red atmosphere is the object that happens to extend, in both cases, for both frames, between the shared event and the corresponding different events and

By "shared event", you mean the idea of taking one event on the worldline of one end of the object, then looking at two separate events on the worldline of the other end, one of which is simultaneous with the first event in the object's rest frame, the other of which is simultaneous with the first event in the frame where the object is moving?

(iii) L₀ is special or at least it occupies the position that it occupies in the equation (which makes it longer) because it is measured in the frame of that particular object (the red atmosphere) that lies between the two ends of both distances, L and L₀.

In the object's rest frame, yes.

Well, we could even dispense with the red atmosphere, it could be there or not, but what seems important is that the problem is defined as "will the Blue Muon survive between the interval defined by its birth (collision with the upper atmosphere or simply passing by red x1) and collision with the Red Lab (or simply passing by red x2)?" (That is something I also wanted to comment. Instead of asking "how the muon manages to make it to the surface", I think it is helpful to define the problem in the above mentioned "neutral", non-frame dependent manner and then analyze it from the two perspectives.)

Isn't the question "does the muon make it from its birth to the surface" stated in a non-frame-dependent manner too?

In any case, it seemed to me that L₀ deserved a special name, because the problem is defined as the ability of the Blue Muon to survive between two Events (1 and 2) that are marked by the Blue Muon's meetings with, if not necessarily two parts of an object, at least two points of the Red Frame. How do you differentiate L₀ ? Just with the symbol? No special word to refer to it?

Usually it's clear from the context, but if you want a term, I'd call it the "rest length".

Many texts call “proper length” what is “rest length”. I’ve googled now and found that terminology in “Special relativity and how it works”, by Moses Fayngold, who distinguishes between “proper distance” (defined as the distance between two simultaneous events) and “proper length” (defined as the length of an object in its rest frame).

You're right, I wasn't thinking about the idea that "proper distance" might be different from "proper length". Still, if "proper distance" refers to the distance between two events in the frame where they're simultaneous, and "rest length" refers to the distance between ends of an object in the frame where the object is at rest, it might be better to avoid the term "proper length" as it could be potentially confusing.

If you reject this terminology, I would agree with you, because in the example L₀ is something more than rest length, as commented: it is the distance between the two red points that pass by the Blue Muon in the interval he disposes of, as measured in the rest frame of those points, yes, but not any other distance (and not any other length, if there happens to be a physical object linking the two points).

Usually "points" refers to "points in spacetime", i.e. events, which don't have a rest frame; I assume you mean here to refer to something more like markers which persist over time (defining extended worldlines rather than single points in spacetime) and which are at rest in the red frame? I would say that "rest length" need not imply we are talking about two points that are ends of a continuous physical object, for example we can talk about the "rest length" of between two inertial ships which are at rest relative to one another but have a large empty gap between them. All it really means is that if you have two worldlines that represent inertial objects at rest relative to one another, the "rest length" is the distance between these objects in their mutual rest frame.

Anyhow, Wikipedia prefers this definition: “proper length is an invariant quantity which is the rod distance between spacelike-separated events in a frame of reference in which the events are simultaneous”. At this stage I'm too tired to make sense of it. Would this apply to our example?

In this case they're using "proper length" to mean the same thing as "proper distance", referring to two specific events. As I said, conceptually "rest length" does not refer to any specific pair of events, but if you happen to pick events on the worldlines of two objects which are simultaneous in the objects' mutual rest frame, then the proper distance between these two events will be identical to the rest length between the two objects.

And, in our example, would there be any invariant length magnitude, whatever it is called? What would be here the length equivalent of proper time?

A spacelike path has an invariant length just like a timelike path has an invariant proper time. Assuming tachyons don't exist a spacelike path can't represent a worldline, but a "straight" spacelike path can represent a set of events which occur on an extended physical object at a single moment in time in some inertial frame (like all events whose spatial position x lies between the left end x0 and right end x1 of a rod, and which all share the same time coordinate t).

 

[Saw]

As to the definition of the problem

Isn't the question "does the muon make it from its birth to the surface" stated in a non-frame-dependent manner too?

Strictly speaking, in the muon-frame the muon does not “make it to the surface”. In the muon-frame, the muon does not move, it is stationary and the “surface makes it to the muon”. I would suggest leaving aside both expressions, therefore, and favour a neutral one, whose language is not made from any perspective, either the Earth’s or the muon’s, either the Red Frame’s or the Blue Frame’s, like:

“Will the Blue Muon survive between its birth at collision with the upper atmosphere (Event 1) and collision of its birth-place with the surface (Event 2)?”

The utility of this approach was shown to me by a comment you made. I had objected to the standard explanation that, from the perspective of the Blue Muon, there is also time-dilation in the Earth or Red Frame and the proof is, for example, that the Blue Muon will judge that in its Blue Frame the 1st Red Muon has decayed before Event 2 only because in that Blue Frame the birth of the 1st Red Muon (Event 0, blue time = 0.577 s) took place significantly earlier than the birth of the Blue Muon (Event 1, blue time = 1.154 s), but the Blue Frame also judges that the 1st Red Muon suffers TD. You said, more or less: that is true, but it’s not relevant to the resolution of the problem as usually stated, which concerns exclusively the survival or non-survival of the Blue Muon. I agree with that.

Completing the idea, I would add: If we now think of a solution in terms of length, the same comment can be made. Someone could come and say: but also in the Earth-frame or Red Frame, the blue objects suffer LC… And the answer should be: that is true, but irrelevant to the problem, which concerns the survival of the Blue Muon between two Events, 1 and 2, which are identified by the collision of such Blue Muon with two edges of an object at rest in the Red Frame (or two any other markers at rest in that frame).

Thus we have a parallelism between the solution to the problem and its definition: the solution is one and “invariant” (if in the duel example, it was “the duel is fair”, here it is “the muon survives”) and the definition is also one and “invariant”.

Would you agree to that?

 

[Saw]

As to the right term for L₀:

Usually it's clear from the context, but if you want a term, I'd call it the "rest length".

Let us divide the concept into two parts: (i) is L₀ a distance or a length? and (ii) is it the distance or length of what?

If we start with the second issue, let us call the concept X, just to avoid the terms length or distance.

L₀ is the “rest X”, but the rest X of what? I think it’s important to note that this rest X is qualified by something else.

In the definition of the problem, we have identified two “physical elements”, which are present for both frames: the Blue Muon and two Red Markers at rest in the Red Frame.

[The two Red Markers can be the two edges of the red atmosphere or, as you suggest, two elements at rest in the Red Frame with some gap in between. I had also suggested two “spatial points” (as you put it, “markers which persist over time”), but it is true that then, without some physical reference, it would be impossible to identify whether the Red Markers meet or not with the Blue Muon.]

Simplifying we could say that the Blue Frame contributes to the problem with a single element and the Red Frame contributes with two separated sub-elements. We could also say that the Blue Muon is the “time element” (whose characteristic is that it “ticks” as it decays) and the Red Markers are the “spatial element” (their role is that there’s space between them).

Thus, in the resolution of the problem, we construct the formulas focusing on the characteristics of those two elements:

- In the T formula, T₀ is the proper time interval as measured in the frame of the “time element” of the problem.
- In the X formula, L₀ is the X between the two separated sub-elements of the “spatial element” of the problem, as measured in their rest frame.

but I'd say "length" is conceptually different in that it doesn't require you to think of any specific events, regardless of whether you pick simultaneous events at t1 or t2 or t3 the distance is the same, as long as you're dealing with a rigid object moving inertially

I think that your comment contains a big truth. A length is something objective, in the sense that the two edges of the physical element that is measured (a rigid rod moving inertially or two objects hovering inertially always at the same distance wrt each other) “are” there, somewhere, no matter whether you measure them or not.

However, in SR this objective fact does not impede disagreement on the magnitude of lengths. In the Red Frame, the Red Markers are separated 1 ls from each other. In the Blue Frame, by 0.866 ls. Both frames agree that the respective lengths they measure do not change, but they measure different values.

And I think it is important to note that this discrepancy on the length values is linked to the relativity of simultaneity. The two frames disagree on the length of the red atmosphere because they measure it as follows:

- In the Red Frame, following the same convention that is used for synch purposes: a red observer sends a light signal to the end of the red atmosphere and, if it returns after 2 s, she concludes that the red atmosphere is 1 ls long. Of course, the red observer could also put one 1-metre sticks after another from the surface to the upper edge, but the length of those 1-metre sticks has also been determined following the Einstein convention.

- In the Blue Frame, the Blue Muon could do three things:

* The Blue Muon sees that the upper atmosphere collides with it at Event 1, at blue t = 1.154 s. It asks which other blue observer is observing at the same blue time the surface of the atmosphere and it is told that it is one who is separated 0.289 ls from the left edge of the blue stick, at Event 0 bis, who is 0.866 ls away from the Blue Muon. But the choice of that blue observer is determined by the fact that all blue clocks have been synchronized following the Einstein convention.

* The Blue Muon measures with its clock that the two edges have passed by in 1.732 s and it infers that the length of the red atmosphere is L = v*T₀ = 0.5 * 1.732 = 0.866 ls. But this formula includes v and v is dx / dt. Curiously enough, if these dx and dt have been measured in the Red Frame and communicated to the Muon, they are both tainted by the judgment on simultaneity of the Red Frame, but combined in the formula with the blue proper time, they give as output a length that is tainted by the Blue Muon’s judgment on simultaneity. If, instead, v has been measured by the Blue Muon, then both the blue dx and the blue dt are tainted by the blue judgment on simultaneity.

* The Blue Muon hears that the length of the red atmosphere in the Red Frame is 1 ls. It multiplies this red length by sqrt(1 – v^2) and gets 0.866 ls. But this formula also includes v and so what I said before also applies.

So it seems to me that:

- while it is true that, in order to phrase the problem, we need some physical object, that is to say, a length,
- the values that appear in the “LC formula” are “proper distances between events”, in so far as they are made by each frame considering two events that are simultaneous in the relevant frame.

Thus one could say that L and L₀ refer to the length of the “spatial element” of a problem, but they are both proper distances (measured by picking two simultaneous events in the relevant frame), L₀ being qualified by the fact that it is the one measured in the rest frame of that “spatial element”.

Well, in the end, this is a long way to say that “rest length” for L₀ is OK for me, only noting that both L and L₀ are tainted by the RS...

Is this right for you?

 

[JesseM]

Strictly speaking, in the muon-frame the muon does not “make it to the surface”. In the muon-frame, the muon does not move, it is stationary and the “surface makes it to the muon”.

I didn't mean "makes it" to specifically refer to movement, I just meant that the muon lasts long enough to come into contact with the surface before it decays.

I would suggest leaving aside both expressions, therefore, and favour a neutral one, whose language is not made from any perspective, either the Earth’s or the muon’s, either the Red Frame’s or the Blue Frame’s, like:

“Will the Blue Muon survive between its birth at collision with the upper atmosphere (Event 1) and collision of its birth-place with the surface (Event 2)?”

That's not really neutral, because in the Earth's frame the "birth-place" of the muon (i.e., the position-coordinate in that frame where the muon was born) remains next to the upper atmosphere. You could replace it with "collision of the muon with the surface" to make it neutral though.

The utility of this approach was shown to me by a comment you made. I had objected to the standard explanation that, from the perspective of the Blue Muon, there is also time-dilation in the Earth or Red Frame and the proof is, for example, that the Blue Muon will judge that in its Blue Frame the 1st Red Muon has decayed before Event 2 only because in that Blue Frame the birth of the 1st Red Muon (Event 0, blue time = 0.577 s) took place significantly earlier than the birth of the Blue Muon (Event 1, blue time = 1.154 s), but the Blue Frame also judges that the 1st Red Muon suffers TD. You said, more or less: that is true, but it’s not relevant to the resolution of the problem as usually stated, which concerns exclusively the survival or non-survival of the Blue Muon. I agree with that.

Completing the idea, I would add: If we now think of a solution in terms of length, the same comment can be made. Someone could come and say: but also in the Earth-frame or Red Frame, the blue objects suffer LC… And the answer should be: that is true, but irrelevant to the problem, which concerns the survival of the Blue Muon between two Events, 1 and 2, which are identified by the collision of such Blue Muon with two edges of an object at rest in the Red Frame (or two any other markers at rest in that frame).

Thus we have a parallelism between the solution to the problem and its definition: the solution is one and “invariant” (if in the duel example, it was “the duel is fair”, here it is “the muon survives”) and the definition is also one and “invariant”.

Would you agree to that?

What specifically are you asking if I agree with? I agree that the question of the muon surviving to meet up with the Earth can be stated in a frame-invariant way, and that the question of the length of a rod at rest in the muon's frame or the lifetime of another muon at rest on the Earth's surface isn't really relevant to this question.

 

[JesseM]

Thus, in the resolution of the problem, we construct the formulas focusing on the characteristics of those two elements:

- In the T formula, T₀ is the proper time interval as measured in the frame of the “time element” of the problem.
- In the X formula, L₀ is the X between the two separated sub-elements of the “spatial element” of the problem, as measured in their rest frame.

Yes, but in the T formula you're talking about a time between a single pair of events, and in the X formula you're talking about the distance between two extended worldlines of markers at rest relative to one another. If you instead picked two events and defined L₀ as the distance between them in the frame where they're simultaneous (the 'proper distance'), and L as the distance between them in a frame moving at speed v relative to the first, then the formula would be L = L₀ * gamma, not L = L₀ / gamma as in the normal length contraction formula. For some discussion of the conceptual difference between length contraction and time dilation, as well as the "spatial analogue of time dilation" which is what I called the first equation above, see this thread, especially the diagram neopolitan posted (which I had drawn for him earlier) in post #5.

I think that your comment contains a big truth. A length is something objective, in the sense that the two edges of the physical element that is measured (a rigid rod moving inertially or two objects hovering inertially always at the same distance wrt each other) “are” there, somewhere, no matter whether you measure them or not.

As opposed to what? The two events in the time dilation equation (or the spatial analogue of time dilation) are also objective, they happened whether you measured them or not, the difference is just that events are instantaneously brief while worldlines extend through time.

However, in SR this objective fact does not impede disagreement on the magnitude of lengths. In the Red Frame, the Red Markers are separated 1 ls from each other. In the Blue Frame, by 0.866 ls. Both frames agree that the respective lengths they measure do not change, but they measure different values.

And I think it is important to note that this discrepancy on the length values is linked to the relativity of simultaneity. The two frames disagree on the length of the red atmosphere because they measure it as follows:

- In the Red Frame, following the same convention that is used for synch purposes: a red observer sends a light signal to the end of the red atmosphere and, if it returns after 2 s, she concludes that the red atmosphere is 1 ls long. Of course, the red observer could also put one 1-metre sticks after another from the surface to the upper edge, but the length of those 1-metre sticks has also been determined following the Einstein convention.

The modern convention is to define a meter in terms of light speed, but this wasn't always true; there would be other rigorous ways to define it, like some multiple of the radius of the first orbital of a hydrogen atom that's at rest in whatever frame you're using.

- In the Blue Frame, the Blue Muon could do three things:

* The Blue Muon sees that the upper atmosphere collides with it at Event 1, at blue t = 1.154 s. It asks which other blue observer is observing at the same blue time the surface of the atmosphere and it is told that it is one who is separated 0.289 ls from the left edge of the blue stick, at Event 0 bis, who is 0.866 ls away from the Blue Muon. But the choice of that blue observer is determined by the fact that all blue clocks have been synchronized following the Einstein convention.

* The Blue Muon measures with its clock that the two edges have passed by in 1.732 s and it infers that the length of the red atmosphere is L = v*T₀ = 0.5 * 1.732 = 0.866 ls. But this formula includes v and v is dx / dt. Curiously enough, if these dx and dt have been measured in the Red Frame and communicated to the Muon, they are both tainted by the judgment on simultaneity of the Red Frame, but combined in the formula with the blue proper time, they give as output a length that is tainted by the Blue Muon’s judgment on simultaneity. If, instead, v has been measured by the Blue Muon, then both the blue dx and the blue dt are tainted by the blue judgment on simultaneity.

Yes, the blue frame would need two synchronized clocks at different positions to measure the speed of the red atmosphere.

So it seems to me that:

- while it is true that, in order to phrase the problem, we need some physical object, that is to say, a length,
- the values that appear in the “LC formula” are “proper distances between events”, in so far as they are made by each frame considering two events that are simultaneous in the relevant frame.

But not between any specific set of events; like I said, the length is just the distance between any arbitrary events on either end of the object which are simultaneous in that frame.

Thus one could say that L and L₀ refer to the length of the “spatial element” of a problem, but they are both proper distances (measured by picking two simultaneous events in the relevant frame), L₀ being qualified by the fact that it is the one measured in the rest frame of that “spatial element”.

Like I said, you can always pick a pair of event such that the proper distance between those events is the same as the length of the object, but I still would consider them conceptually different because proper distance depends on picking a specific pair of events while length does not.