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

[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.

 

[Saw]

As to the “spatial analogue of TD”

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.

Maybe all I say below is incorrect as shown in the above comment, but I don’t understand the comment. I found the thread very interesting, but it’s a complicated discussion and find it very hard to follow it. If you had time, it’d be interesting to see a sort of summary of the conclusion.

As to the definition of the problem

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.

Sorry if I had misunderstood you.

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.

Thanks for the correction. That was the original wording, but then I thought it was not appropriate to assume in the question that the muon is alive at the time of collision with the surface, since that is precisely the answer we seek. Shall we use the following sentence?

“Will the Blue Muon, after its birth at Event 1, last long enough so as to collide with the surface at Event 2?”

As to the objectivity or frame-invariance of the “proper time” value

When I talked about the objectivity of a physical length, you asked:

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.

Yes, yes, of course. For me the “proper time” value of the Blue Muon is perfectly objective. It is even “more” objective than length, in the following sense:

“Proper time” is a simultaneity-free concept. It has been obtained by a single clock, so there is no issue about whether a good or bad synchronization existed. In my opinion, this gives “proper time” a privileged position in the resolution of the problem: it is a self-sufficient value. If you happen to know it, that is it, you have the solution.

In fact, what I see is a disparity of rank between “proper time” and “rest length”. Regarding length, I appreciate that there is an underlying unique reality, but I do not find any measured value where that objective reality shines up in the same manner as it brightly shines up in the case of proper time, since none of the length values for the Red Markers is simultaneity-free.

But is “rest length” really a simultaneity-dependent judgment?

You seem to accept that the length measurement about the Red Markers is really simultaneity-dependent in the Blue Frame.

But you seem to deny it for the “rest length” measured in the Red Frame:

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.

Well, but:

- Currently, we measure length in terms of a two-way trip of light.
- If we adopted any other convention, like the one you suggest and SR is true, the coordinate system and the equations constructed on the basis of such convention should be identical and Lorentz-invariant, shouldn’t they? Just like the choice between light clocks, mechanical clocks or muon clocks does not change anything (other than precision), the choice between light rods or hydrogen-orbit clocks should not change anything (other than precision). So if length measurement with the go-and-return trip of light leads both frames to measuring values that are tainted by the relativity of simultaneity, so should any other method.

As to the difference between “proper distance” and “rest length”

If I follow you well, you point out that:

- When we talk about “proper distance”, we refer to the distance between two specific events that are simultaneous in the frame where the measurement is made.

- When we talk about the “length”, for example, of the separation between the two Red Markers, since that separation persists over time, the events taken as reference by each frame may be arbitrarily chosen.

- Because of this, both things are “conceptually” different.

Well, the difference certainly exists. Both frames can measure the length of the separation between the Red Markers 100 times in a day and on each occasion they will choose different events as references and, in spite of that, each frame will always measure the same length value for the Red Markers: 1 ls in the Red Frame, 0.866 ls in the Blue Frame.

But conceptual distinctions are valid to the extent that, in a given context, they have practical consequences. When analyzing two elements under comparison, you must identify what is relevant and what is irrelevant for the practical purpose under consideration. If the elements that are relevant are identical in both terms of the comparison, then the two concepts are also functionally identical. If elements that are irrelevant are different in each term of the comparison, the two concepts are still functionally identical.

In this case, it is my impression that:

- Whenever the Red Frame or the Blue Frame measure the length between the Red Markers, they must look at two specific events. Although that could happen on any another occasion, if we take the example of how they do it “at Event 1”, we see that they pick: on the one hand, like I said, a shared Event, Event 1, the birth of the Blue Muon, at the upper atmosphere; on the other hand, on the other end, each frame picks a different event: the Red Frame picks Event 0, the Blue Frame picks Event 0 bis. This discrepancy, which is due to the RS, is certainly relevant for the resolution of the problem: because of this, none of the two discrepant values about the reality of the length in question can lead by itself to an invariant solution, for the simple reason that the values are divergent and the solution is unique.

- Instead, the fact that five minutes earlier or two days later the two frames picked or will pick other events as reference and reach the same conclusions about the length of the Red Markers seems to a good extent irrelevant. Not totally irrelevant, perhaps: if they did in the past, they do not need to repeat it now. That is true. But still the fact that the two frames can measure the length between the Red Markers every five minutes, on the basis of arbitrarily chosen events, does not make any of the values thus obtained less simultaneity-dependent and hence less inapt to provide the solution to the problem by itself.

Conclusion: L and L₀ are both "distances between events" (hence equally tainted by the RS), which is relevant for the resolution of the problem (it makes each of them inapt to solve it isolatedly) and the fact that they may have been measured at the very time when the story starts or at any other past or future time is relevant for convenience purposes, but irrelevant for the resolution of the problem.

 

[Saw]

I think that all I said before may be clearer if I spell out what the “mechanics of the resolution of the muon problem in SR” are, in my opinion, as an alternative to the language of the Standard Explanation.

The problem was:

Will the Blue Muon, after its birth at the upper red atmosphere at Event 1, last long enough so as to collide with the surface at Event 2?

The solution is:

We must look at the Blue Muon’s proper time interval between Event 1 and Event 2. If it is less than the life-expectancy of any muon in any frame (in this example, arbitrarily, 1.8 s) then the Blue Muon will last long enough.

To be noted: since “proper time” is an invariant concept, it is agreed upon by both frames. The same explanation is given from both perspectives. Of course, both perspectives are correct, in so far as they provide… not two solutions, as the Standard Explanation claims, but the same unique solution.

That is the gist of the thread. In order to do what the aim of physics is (solving practical problems), the physicist must look for the invariant, frame-independent quantities, since it cannot be (unless one believes in parallel universes for some metaphysical reason) that practical problems (is the duel fair?; am I alive?; does the muon survive?; should I get a golden frequent-flyer card?) have different solutions depending on the observer. (By “solutions” I mean the ultimate answer to the question, not the paths for reaching the solution, which may of course be different for different observers.)

But through which path do we reach the solution?

As you nicely put it, “we” are you and I, the “omniscient problem-analyzers”. We dispose, as you also point out, of all the “measurements made in specific frames”. But in practice specific observers will not dispose of all the data. That is precisely the challenge: often must they work on the basis of partial information and find a trick in order to, in spite of all, hit on the answer.

Some examples:

(a) The observer disposes of the two red values, red dt (coordinate time = 2s) and red dx (rest length of the Red Markers = 1 ls)

She combines them into the formula for proper time:

dτ or T₀ = sqrt (red dt^2 – red dx^2) = sqrt(2^2 – 1^2) = sqrt(4-1) =sqrt(3) = 1.732 s

That is the proper time of the Blue Muon, that is the solution.

(b) The observer disposes of the red dt (2 s) and v (0.5 c).

She combines them into the formula:

Red dt = (blue dt ot T₀) * gamma --> (blue dt ot T₀) = red dt / gamma = 2 * sqrt(1-v^2) = 2*0.866 = 1.732 s

That is the proper time of the Blue Muon, that is the solution.

To be noted: This second formula apparently only uses as input the red dt, but that is an appearance. It also uses v and the observer can only know v if he or she also knows:

- Red dx, in which case we are again in hypothesis (a). In fact, if you replace in the 2nd formula v by red dx / red dt (which is what one knows here), you get the 1st formula or
- Blue dt and blue dx, in which case what the formula is still doing is extracting the proper time of the Blue Muon out of the combination of the other values.

(c) The observer knows red dx (1 ls) and v.

She combines them into the formula:

(L or blue dx) = (L₀ or red dx) * sqrt (1 – v^2) = 1 * 0.866 = 0.866 ls

But that is not yet the solution. The observer must multiply L by v and thus get the blue dt = 1.732 s. That is the solution.

To be noted: the same comment about the contents of v applies.

(d) The observer knows blue dx and v.

She multiplies blue dx by v and gets the proper time of the Blue Muon = 1.732 s. That is the solution.

To be noted: the same comment about the contents of v applies.

(e) The observer knows blue dt and… any other value.

In this case, she does not need any formula. She simply picks the blue dt, the proper time of the Blue Muon, and that is the solution.

How can this be summarised?

Now it is time to voice around how we solved the problem. I think this is a simple and honest description = the Blue Muon lasts long enough after Event 1 so as to be present at Event 2 because its proper time between these two Events is 1.732 s (less than its life expectancy), which value can be obtained:

(a) Directly, on the basis of the blue time measurement of a blue clock accompanying the Blue Muon.

(b) Indirectly, through different combinations of the coordinate time and (either rest or non-rest) length values into the appropriate formulas.

What is all this pointing at?

There are many pointers here pointing at the same truth.

The pointers are countless:

- The phrasing of the problem doesn’t make any comparison: it doesn’t ask whether the Blue Muon’s clock runs more or less slowly than a Red Muon’s; it doesn’t ask whether the red atmosphere is longer or shorter in any frame. In fact, if you try to make such comparisons during the reasoning process, you must be told, as we agreed, that such comparison is inappropriate, it is out of question. The phrasing of the problem is simultaneity-free.

- The solution to the problem is always, ultimately, proper time, which is simultaneity-free.

- The two frames disagree on whether Event 1 was simultaneous with Event 0 or Event 0 bis. But curiously enough those two controversial Events are absent both from the phrasing of the problem and of the solution.

- The coordinate time (red dt), unlike blue dt (proper time of the Blue Muon), does not solve the problem by itself; to be of any use, the red dt must be combined with either of the length values. That is because, admittedly, red dt is a simultaneity-dependent concept.

- None of the length values (red dx and blue dx) solve the problem by themselves, unless combined with some other values. That must be because both are simultaneity-dependent values.

- In fact, all the SR formulas really look like an intelligent way to get a simultaneity-free solution to a simultaneity-free problem by combining (and offsetting to the appropriate extent) different simultaneity-depending values and thus “decontaminating” them of relative simultaneity.

- In the duel example it is shown that the fact that the two frames have different judgments about whether the two duellers received the signal for shooting at the same time is not the solution to the problem. The solution is (ultimately) given by a simultaneity-free concept (proper time). Likewise, in this muon example, it is shown that the concepts of time dilation and length contraction, since they are the off-spring of the relativity of simultaneity, are not the solution to the problem, either, contrary to what the Standard Explanation declares. All relative concepts (RS, TD and LC) are intellectual tools that turn out to be helpful only if (i) proper time is unknown and (ii) if so, they are mixed together in a recipe so as to produce the right chemical reaction...

The truth may be that proper time “mirrors” a physical reality, while the other concepts (RS, TD and LC) don’t, unless in mutual combination. But why so? We can discuss it later, if you wish. I’m going now for a trip. Regards.

 

[JesseM]

Maybe all I say below is incorrect as shown in the above comment, but I don’t understand the comment. I found the thread very interesting, but it’s a complicated discussion and find it very hard to follow it. If you had time, it’d be interesting to see a sort of summary of the conclusion.

Well, if you write the time dilation equation as \Delta t' = \Delta t * \gamma, then \Delta t represents the time between a pair of events in the frame where they occur at the same coordinate position, and \Delta t' represents the time between the same two events in a frame moving at speed v relative to the first. So, the "spatial analogue" of this would be to take a single pair of events in a frame where they occur at the same coordinate time and call the distance between them \Delta x, then let \Delta x' be the distance between the same two events in a frame moving at speed v relative to the first; in this case the equation will look just like the time dilation equation with t replaced with x, or \Delta x' = \Delta x * \gamma. This is obviously different from the length contraction equation \Delta x' = \Delta x / \gamma; in the length contraction equation we are not looking at the spatial distance between a single pair of events in two different frames, instead we are considering two different worldlines which represent objects at rest in the unprimed frame (so they are lines of constant x in the unprimed frame), and letting \Delta x be the distance between the worldlines at any given instant in the unprimed frame, and \Delta x' be the distance between the worldlines at any given instant in the primed frame. You could also imagine a "temporal analogue for length contraction" by considering two different lines of constant t in the unprimed frame (surfaces of simultaneity in this frame), and considering the time between them in the unprimed frame vs. the primed frame; this would give you the equation \Delta t' = \Delta t / \gamma, which is different from the usual time dilation equation. All this is illustrated in the diagram in post #5 of that previous thread, even if you don't want to read the whole thread you may find it clarifies things to take a look at that diagram.

Thanks for the correction. That was the original wording, but then I thought it was not appropriate to assume in the question that the muon is alive at the time of collision with the surface, since that is precisely the answer we seek. Shall we use the following sentence?

“Will the Blue Muon, after its birth at Event 1, last long enough so as to collide with the surface at Event 2?”

Sounds good.

As to the objectivity or frame-invariance of the “proper time” value

When I talked about the objectivity of a physical length, you asked:

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.

Yes, yes, of course. For me the “proper time” value of the Blue Muon is perfectly objective. It is even “more” objective than length, in the following sense:

“Proper time” is a simultaneity-free concept. It has been obtained by a single clock, so there is no issue about whether a good or bad synchronization existed. In my opinion, this gives “proper time” a privileged position in the resolution of the problem: it is a self-sufficient value. If you happen to know it, that is it, you have the solution.

In fact, what I see is a disparity of rank between “proper time” and “rest length”. Regarding length, I appreciate that there is an underlying unique reality, but I do not find any measured value where that objective reality shines up in the same manner as it brightly shines up in the case of proper time, since none of the length values for the Red Markers is simultaneity-free.

Well, the perspective from differential geometry is that the objective geometry of a surface is defined in terms of some measure of "distance" along arbitrary paths. If you have a coordinate system on the path, then you can come up with a line element at every point, and then integrate the line element along an extended path to find the length of that path. In general relativity the line element at every point is given by the metric, and may very from point to point; but if you are talking about an inertial coordinate system in the flat spacetime of SR, the line element at every point is given by ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2. This means that if you've parametrized a curve through spacetime in terms of these coordinates, you can integrate the line element to find the "length" of the curve (the integral of ds/dp where p is your parameter where each value of p gives a point on the curve with some functions x(p), y(p), z(p), and t(p) that constitute the parametrization of that curve), as discussed in quasar987's example in post #3 here where he uses the line element for spherical coordinates in 3D Euclidean space (whereas the line element above is for inertial coordinates in 4D Minowski spacetime). If you integrate over curve that's a timelike worldline, then the square of the answer is just equal to -c^2 times the square of the proper time along that curve. This answer is intrinsic to the geometry of the spacetime--from a GR perspective, you could equally well use a non-inertial coordinate system, but the metric would give you the correct line element for this coordinate system which would ensure that when you integrate over the same curve expressed in the new coordinate system, you get the same answer for the value of the integral, which again is proportional to the proper time along that curve. And if you integrate along a spacelike curve, this is no longer the proper time (the answer will end up being in units of distance), but it's still every bit as geometric and coordinate-independent. Just as you can conceptualize the proper time along a path in terms of how much time would elapse on a clock that has that path as its worldline, so you could imagine that if we could build "FTL odometers" that tick units of distance as they travel along spacelike paths, then the distance along a spacelike path would be what would be measured by an FTL odometer that had that path as its worldline.

Also note that if you pick a "straight" path through spacetime as seen in any inertial coordinate system, then you don't actually have to parametrize the path and integrate ds/dp, instead you can just look at the coordinates x_0, y_0, z_0, t_0 of one end, the coordinates x_1, y_1, z_1, t_1 of the other end, and with \Delta x = x_1 - x_0 and so forth, the integral of ds/dp will just be equal to \sqrt{-c^2 \Delta t + \Delta x + \Delta y + \Delta z}. As before, for a timelike path, the square of this answer is just -c^2 times the square of the proper time. And as before, the answer is coordinate-invariant: you can look at the same path in different inertial coordinate systems, and although different coordinate systems assign different coordinates to the two ends of the path and thus have different values for \Delta x, \Delta y, \Delta z, and \Delta t, they will all calculate the same answer for \sqrt{-c^2 \Delta t + \Delta x + \Delta y + \Delta z}. And in the case where we are talking about a straight path between two events with a spacelike separation, then the answer is just equal to the "proper distance", i.e. the distance between the events in the frame where they are simultaneous. As I mentioned before, if your two events are events on either end of a rigid object that are simultaneous in the object's rest frame, that the proper distance between these events is the same as the proper length of the rigid object.

But is “rest length” really a simultaneity-dependent judgment?

You seem to accept that the length measurement about the Red Markers is really simultaneity-dependent in the Blue Frame.

But you seem to deny it for the “rest length” measured in the Red Frame

It depends what you mean by "simultaneity-dependent". Different observers define the length of the same object differently, but on the other hand different observers all agree on which frame is the object's own rest frame, so they all agree on its rest length. Also note that when you're talking about an object's own rest frame, there's no need to exclusively talk about the distance between simultaneous events when talking about rest length; since both ends of the object have constant coordinate position, you can pick an event on one end that happens at some coordinate time t0 and an event on the other end that happens at a different coordinate time t1, the coordinate distance between these events in the rest frame is still equal to the rest length (although for events which are not simultaneous in this frame, the coordinate distance between the events is not equal to the proper distance between the events).

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.

Well, but:

- Currently, we measure length in terms of a two-way trip of light.
- If we adopted any other convention, like the one you suggest and SR is true, the coordinate system and the equations constructed on the basis of such convention should be identical and Lorentz-invariant, shouldn’t they? Just like the choice between light clocks, mechanical clocks or muon clocks does not change anything (other than precision), the choice between light rods or hydrogen-orbit clocks should not change anything (other than precision). So if length measurement with the go-and-return trip of light leads both frames to measuring values that are tainted by the relativity of simultaneity, so should any other method.

Yes, I agree with both of these.

As to the difference between “proper distance” and “rest length”

If I follow you well, you point out that:

- When we talk about “proper distance”, we refer to the distance between two specific events that are simultaneous in the frame where the measurement is made.

You can talk about "proper distance" between events even if the events were not simultaneous in the frame where the measurement was made, it just means that you have to do a Lorentz transformation to figure out how far apart the events would be in a frame where they were simultaneous. I'm not sure if this is different from what you say above.

- When we talk about the “length”, for example, of the separation between the two Red Markers, since that separation persists over time, the events taken as reference by each frame may be arbitrarily chosen.

More importantly, the two frames are not talking about the same pair of events when we talk about the two different distances in the length contraction equation; if they both wanted to measure the distance between the same pair of events, then you'd have to use the "spatial analogue of time dilation" equation which I gave earlier, \Delta x' = \Delta x * \gamma. Instead, each frame can pick their own arbitrary pair of events on either end of the object which are simultaneous in their own frame, and define the "length" as the distance between these events.

Well, the difference certainly exists. Both frames can measure the length of the separation between the Red Markers 100 times in a day and on each occasion they will choose different events as references and, in spite of that, each frame will always measure the same length value for the Red Markers: 1 ls in the Red Frame, 0.866 ls in the Blue Frame.

But conceptual distinctions are valid to the extent that, in a given context, they have practical consequences. When analyzing two elements under comparison, you must identify what is relevant and what is irrelevant for the practical purpose under consideration. If the elements that are relevant are identical in both terms of the comparison, then the two concepts are also functionally identical. If elements that are irrelevant are different in each term of the comparison, the two concepts are still functionally identical.

In this case, it is my impression that:

- Whenever the Red Frame or the Blue Frame measure the length between the Red Markers, they must look at two specific events.

But they cannot both look at the same specific pair of events, as they would in the case of the time dilation equation.

Although that could happen on any another occasion, if we take the example of how they do it “at Event 1”, we see that they pick: on the one hand, like I said, a shared Event, Event 1, the birth of the Blue Muon, at the upper atmosphere; on the other hand, on the other end, each frame picks a different event: the Red Frame picks Event 0, the Blue Frame picks Event 0 bis.

Exactly.

This discrepancy, which is due to the RS, is certainly relevant for the resolution of the problem: because of this, none of the two discrepant values about the reality of the length in question can lead by itself to an invariant solution, for the simple reason that the values are divergent and the solution is unique.

I don't understand what you mean by "lead to an invariant solution" here. In the case of the time dilation equation, where we are dealing with a single pair of events viewed in two different frames, do you think the two values for the time between these events "lead to an invariant solution"? If so, what is it?

Instead, the fact that five minutes earlier or two days later the two frames picked or will pick other events as reference and reach the same conclusions about the length of the Red Markers seems to a good extent irrelevant.

Irrelevant to what? My point wasn't exactly about repeating length measurements at different times, but mainly just that "proper distance" involves talking about a specific pair of events that all observers agree on, while "length" does not.

Conclusion: L and L₀ are both "distances between events" (hence equally tainted by the RS)

But not the same pair of events viewed in different frames as with the time dilation equation.

which is relevant for the resolution of the problem (it makes each of them inapt to solve it isolatedly)

What specific problem are you talking about, and what do you mean by "solve it isolatedly"?

 

[JesseM]

I think that all I said before may be clearer if I spell out what the “mechanics of the resolution of the muon problem in SR” are, in my opinion, as an alternative to the language of the Standard Explanation.

The problem was:

Will the Blue Muon, after its birth at the upper red atmosphere at Event 1, last long enough so as to collide with the surface at Event 2?

The solution is:

We must look at the Blue Muon’s proper time interval between Event 1 and Event 2. If it is less than the life-expectancy of any muon in any frame (in this example, arbitrarily, 1.8 s)

The life-expectancy of a muon in a given frame depends on whether it's moving in that frame. Perhaps you meant something more like "if it is less than the expected proper time along a muon's lifetime from birth to death", which is a frame-invariant number.

Anyway stating it in frame-invariant terms as above is one way to look at the problem, but not the only way. You can equally well state the answer by saying "we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will reach the surface." The point is that although different frames disagree on the coordinate time between event 1 and event 2, and the life-expectancy of that particular muon, they will all end up making the same prediction about whether the muon survives long enough to collide with the surface. That's the magic of using different frames in SR, you always find that they agree about local events!

That is the gist of the thread. In order to do what the aim of physics is (solving practical problems), the physicist must look for the invariant, frame-independent quantities, since it cannot be (unless one believes in parallel universes for some metaphysical reason) that practical problems (is the duel fair?; am I alive?; does the muon survive?; should I get a golden frequent-flyer card?) have different solutions depending on the observer. (By “solutions” I mean the ultimate answer to the question, not the paths for reaching the solution, which may of course be different for different observers.)

Again, as long as your "solution" is an answer to a question about local events, like whether the muon in fact comes into contact with the surface, then it doesn't matter if your method of solving the problem is stated in a frame-dependent way, since all frames do in fact agree about local events. Perhaps this is what you meant by "not the paths for reaching the solution", but then why did you take care to state your answer above in a way that contained no frame-dependent quantities? Do you disagree that my alternate answer, ""we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will survive to collide with the surface", is a perfectly valid one that will reach the same conclusion about the frame-invariant issue of whether the blue muon collides with the surface?

But through which path do we reach the solution?

As you nicely put it, “we” are you and I, the “omniscient problem-analyzers”. We dispose, as you also point out, of all the “measurements made in specific frames”. But in practice specific observers will not dispose of all the data. That is precisely the challenge: often must they work on the basis of partial information and find a trick in order to, in spite of all, hit on the answer.

Some examples:

(a) The observer disposes of the two red values, red dt (coordinate time = 2s) and red dx (rest length of the Red Markers = 1 ls)

She combines them into the formula for proper time:

dτ or T₀ = sqrt (red dt^2 – red dx^2) = sqrt(2^2 – 1^2) = sqrt(4-1) =sqrt(3) = 1.732 s

That is the proper time of the Blue Muon, that is the solution.

(b) The observer disposes of the red dt (2 s) and v (0.5 c).

She combines them into the formula:

Red dt = (blue dt ot T₀) * gamma --> (blue dt ot T₀) = red dt / gamma = 2 * sqrt(1-v^2) = 2*0.866 = 1.732 s

That is the proper time of the Blue Muon, that is the solution.

Those are valid ways of figuring out the answer, but again, it is not necessary to figure out the proper time for the muon between being created and hitting the surface. She could equally well take the lifetime of the muon in its rest frame, 1.8s, and the velocity of the muon in her frame, 0.5c, and plug into the time dilation equation to find that the muon should survive for 1.8/0.866 = 2.08 seconds in her frame, which is greater than the time dt for the muon to reach the surface in her frame.

What is all this pointing at?

There are many pointers here pointing at the same truth.

The pointers are countless:

- The phrasing of the problem doesn’t make any comparison: it doesn’t ask whether the Blue Muon’s clock runs more or less slowly than a Red Muon’s; it doesn’t ask whether the red atmosphere is longer or shorter in any frame. In fact, if you try to make such comparisons during the reasoning process, you must be told, as we agreed, that such comparison is inappropriate, it is out of question. The phrasing of the problem is simultaneity-free.

- The solution to the problem is always, ultimately, proper time, which is simultaneity-free.

I thought the "solution" was the same as the "ultimate answer to the question", which was just the fact that the muon does, in fact, survive to collide with the surface. Calculating the proper time is just part of your method of determining that this local fact is true, as opposed to the local fact "the muon annihilates at some point in the atmosphere above the surface". And as I said, since all frames agree on the outcome of local events, there is no reason to privilege your particular method over other methods of arriving at the same ultimate answer to the question of whether the muon survives to collide with the surface or not.

- In fact, all the SR formulas really look like an intelligent way to get a simultaneity-free solution to a simultaneity-free problem by combining (and offsetting to the appropriate extent) different simultaneity-depending values and thus “decontaminating” them of relative simultaneity.

I disagree that this is necessary, again I think you are confusing the method with the ultimate answer. The method I suggested, ""we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will reach the surface", is a perfectly valid one, and does not involve taking simultaneity out of the problem, although the final answer--"the muon does in fact collide with the surface"--is a local fact that's frame-independent and thus simultaneity-independent.

 

[Saw]

Back again, hello JesseM and anybody else.

The spatial analogue of time dilation (SATD) and the temporal analogue of length contraction (TALC)

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.

Well, if you write the time dilation equation as \Delta t' = \Delta t * \gamma, then \Delta t represents the time between a pair of events in the frame where they occur at the same coordinate position, and \Delta t' represents the time between the same two events in a frame moving at speed v relative to the first. So, the "spatial analogue" of this would be to take a single pair of events in a frame where they occur at the same coordinate time and call the distance between them \Delta x, then let \Delta x' be the distance between the same two events in a frame moving at speed v relative to the first; in this case the equation will look just like the time dilation equation with t replaced with x, or \Delta x' = \Delta x * \gamma. This is obviously different from the length contraction equation \Delta x' = \Delta x / \gamma; in the length contraction equation we are not looking at the spatial distance between a single pair of events in two different frames, instead we are considering two different worldlines which represent objects at rest in the unprimed frame (so they are lines of constant x in the unprimed frame), and letting \Delta x be the distance between the worldlines at any given instant in the unprimed frame, and \Delta x' be the distance between the worldlines at any given instant in the primed frame. You could also imagine a "temporal analogue for length contraction" by considering two different lines of constant t in the unprimed frame (surfaces of simultaneity in this frame), and considering the time between them in the unprimed frame vs. the primed frame; this would give you the equation \Delta t' = \Delta t / \gamma, which is different from the usual time dilation equation. All this is illustrated in the diagram in post #5 of that previous thread, even if you don't want to read the whole thread you may find it clarifies things to take a look at that diagram.

I have tried to draw an adaptation of your diagram and concepts to the problem we are analyzing here (attached).

I had argued that a problem like this involves two objective frame-invariant elements:

- A temporal element, which is contributed by the Blue Frame: the fact that the Blue Muon decays or ticks at a certain rate.

- A spatial element, which is contributed by the Red Frame (the Earth): the Red Markers (the two edges of the red atmosphere).

I had also argued that the so called TD and LC formulas must be used in the following manner:

- Time formula:

T₀ = T * sqrt (1 – v^2) = 2 * 0.866 = 1.732 s

or

dt = dt * sqrt (1 – v^2) = 2 * 0.866 = 1.732 s

That is to say: the “time element” of the problem, which is the “proper time” of the Blue Muon, occupies the left side of the formula, which makes it shorter.

- Length or X formula:

L = L₀ * sqrt (1 – v^2) = 1 * 0.866 = = 0.866 ls

or

dx = dx * sqrt (1 – v^2) = 1 * 0.866 = 0.866 ls

That is to say: the “spatial element” of the problem, which is the “rest length” of the Red Markers, occupies the right side of the formula, which makes it longer, while the coordinate length measured in the Blue Frame is shorter.

If my adaptation of your concepts is correct:

- My “length or X formula” is not the SATD, because the latter plays with the value dx = 1.154 s and my formula does not play with that value. My “length formula” would be the so called standard LC equation.

If you want my opinion, the value 1.154 ls that is associated to the SATD would have its place in the very LC equation but in another problem, where it was analyzed the survival of the 1st Red Muon between the events of its collision with the two edges of the left half of the blue stick (the Blue Markers). In that case, the very same reasoning would lead to place the rest length of the Blue Markers in the right side of the formula, as follows:

L = L₀ * sqrt (1 – v^2) = 1.154 * 0.866 = 1 ls

- My “time formula” ends up being the TALC, if I got it right at all. But again I think it cannot be otherwise, because I have to play with the value dt = 1.732 s. If instead I played with the value dt = 2.309 s, we would be again in a different problem, analyzing the survival of the 1st Red Muon between the events of its collisions with the Blue Markers. In that case, the very same reasoning would lead to place the proper time of the 1st Red Muon in the left side of the formula, as follows:

T₀ = T * sqrt (1 – v^2) = 2.309 * 0.866 = 2 s

Conclusions:

Time element of a problem = proper time = left side = shorter
Spatial element = rest length = right side = longer
The two values of the frame owning the time element (in our example 1.732 s and 0.866 ls) are shorter as if both the time and length units were dilatated vis-à-vis the units of the other frame.
The two values of the frame owning the spatial element (here 2 s and 1 ls) are longer as if both the time and length units were contracted vis-à-vis the units of the other frame.

For sure, however, there are other ways to put it. But would you agree that this one I proposed is a correct problem-solving technique or rule of thumb for guidance on how to employ the formulas to solve a problem?

Definition of the question or problem

You mention several times that stating the question in a frame-invariant way is not necessary and that what is important is that the question is about a “local event” (if the Blue Muon will survive at Event 2).

Yes, I agree. That is clearer. Maybe, since events “belong” to all frames (all frames agree that they happen), could we say for even more precision that the question is about a “single event” = whether Event 2 will count with the presence of the Blue Muon?

We could call this Question 1. As opposed to what?

Another possible formulation (Question 2) would be:

“does a Blue Muon decay faster than a Red Muon or more slowly = is the internal clock of a Blue Muon faster or slower than the internal clock of a Red Muon?”

Obviously, this Question 2 is a comparison between “pairs of events that are distant from each other”. To answer it, we would need to be constantly comparing the Blue Muon ticks or decay steps with those of a Red Muon. This, in turn, involves a judgment on simultaneity, on which the two frames disagree. Thus, if we made that judgment, each frame would be entitled to claim that it is the other’s clock rate or decay process the one that is slower.

But I think it is important to note that here the question is less ambitious, it is just Question 1. Question 2 would only be asked in a twin paradox scenario, where one of the muons were accelerated back to re-encounter another twin brother created at the same Event 1. That would be more complicated to analyze and I propose to leave it aside for the time being.

What is the answer?

I think you are confusing the method with the ultimate answer. The method I suggested, ""we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will reach the surface", is a perfectly valid one, and does not involve taking simultaneity out of the problem, although the final answer--"the muon does in fact collide with the surface"--is a local fact that's frame-independent and thus simultaneity-independent.

Well, yes, if we want to be precise in the use of words… the answer would simply be:

“yes, the muon is alive at Event 2”

Then we can discuss the method for reaching that simple answer = how do we learn that the answer is yes? In particular, I’d like to answer your comments regarding the roles played in the “method” by length, proper time, coordinate time… But I’ll do it in another post, otherwise this one would be too long.

 

[Saw]

Proper distance between the non-simultaneous (in any frame) Events 1 and 2

Also note that if you pick a "straight" path through spacetime as seen in any inertial coordinate system, then you don't actually have to parametrize the path and integrate ds/dp, instead you can just look at the coordinates x_0, y_0, z_0, t_0 of one end, the coordinates x_1, y_1, z_1, t_1 of the other end, and with \Delta x = x_1 - x_0 and so forth, the integral of ds/dp will just be equal to \sqrt{-c^2 \Delta t + \Delta x + \Delta y + \Delta z}. As before, for a timelike path, the square of this answer is just -c^2 times the square of the proper time. And as before, the answer is coordinate-invariant: you can look at the same path in different inertial coordinate systems, and although different coordinate systems assign different coordinates to the two ends of the path and thus have different values for \Delta x, \Delta y, \Delta z, and \Delta t, they will all calculate the same answer for \sqrt{-c^2 \Delta t + \Delta x + \Delta y + \Delta z}. And in the case where we are talking about a straight path between two events with a spacelike separation, then the answer is just equal to the "proper distance", i.e. the distance between the events in the frame where they are simultaneous. As I mentioned before, if your two events are events on either end of a rigid object that are simultaneous in the object's rest frame, that the proper distance between these events is the same as the proper length of the rigid object.

Applying this information to the problem at hand (Question 1 referred to in my most recent post), I understand that:

- Question 1 is about the spacetime interval between Events 1 and 2.

- If we measure distances with light seconds, then c=1 and the formula for the spacetime interval simplifies to:

ds² = -dt² + dx²

- The distance separating these two Events is timelike since dt > dx.

- Also with c=1 the formula for the proper time interval between Events 1 and 2 is:

dτ² = dt² - dx²

- In our example, calculations for proper time would be:

With red values: dτ² = dt² - dx² = 2^2 – 1^2 = 4 – 1 = 3  dτ = sqrt(3) = 1.732 s
With blue values: dτ² = dt² - dx² = 1.732^2 – 0 = 3  dτ = sqrt(3) = 1.732 s

- It seems that instead the spacetime interval would be… sqrt(-3)? Is that right? If so, in an example like this one, where the distance between Events 1 and 2 is timelike, what is the meaning and the added value of the concept of spacetime interval vis-à-vis proper time?

Proper distance between simultaneous events versus length

It depends what you mean by "simultaneity-dependent". Different observers define the length of the same object differently, but on the other hand different observers all agree on which frame is the object's own rest frame, so they all agree on its rest length. Also note that when you're talking about an object's own rest frame, there's no need to exclusively talk about the distance between simultaneous events when talking about rest length; since both ends of the object have constant coordinate position, you can pick an event on one end that happens at some coordinate time t0 and an event on the other end that happens at a different coordinate time t1, the coordinate distance between these events in the rest frame is still equal to the rest length (although for events which are not simultaneous in this frame, the coordinate distance between the events is not equal to the proper distance between the events).

There are several issues here:

- “All observers agree on which one the rest frame of an object is and on the rest length of an object”.

Yes, for example, all observers agree that the Red Markers’ rest frame is the Red Frame and that the rest length of the Red Markers is 1 ls. But all observers also agree that the non-rest length of the Red Markers in the Blue Frame is 0.866 ls.

In fact, all observers agree on all values. All values are correlated by the same formulas, so they can all be figured out from any perspective.

However, a given value will be “different” if it has a problem-solving capacity “by itself”, without the need of combining it with other values. This is not the case of “rest length”. If you know that the rest length of the Red Markers is 1 ls, have you solved the problem? No, you would need to know something more, wouldn’t you?

- “There's no need to exclusively talk about the distance between simultaneous events when talking about rest length (…) you can pick an event on one end that happens at some coordinate time t0 and an event on the other end that happens at a different coordinate time t1, the coordinate distance between these events in the rest frame is still equal to the rest length”

Yes, this is what is usually exemplified by saying that you can put the measuring tape on one end of the object at t0 and then quietly walk to the other end while stretching out the tape, until you measure the other end, quite later, at t1. Thus the two acts of length measurement are not simultaneous. That is true. But you had also agreed that (i) the units of the measurement tape have been established with a measurement method which is based on the time it takes for light to traverse a distance in a two-way trip and (ii) that any other measurement convention should render essentially analogous results, if SR is true. Thus the length of the measurement instruments is tainted by the relativity of simultaneity and so must be any measurement made with those instruments.

- Furthermore, it is also true that a length can be measured at ay time, not necessarily at the time when the problem we are analyzing starts (birth of the Blue Muon, Event 1). If we talk about the length of the Red Markers, the Blue Frame will use two synchronous clocks for that purpose, while the Red Frame may use the method described above. But on all those occasions, the measurement will be tainted by the relativity of simultaneity: in the case of the Blue Frame, it is obvious; in the case of the Red Frame, we need the reasoning of the preceding paragraph to reach such conclusion, but we do reach it as well. Certainly, the fact that the measurement can be made at any time makes life easier for both frames. The Red Frame may have done it a century ago. The Blue Frame may also have done it in another historical meeting with the Red Markers. But those historical measurements will also be tainted by the relativity of simultaneity.

Conclusion: rest length is not a value which is self-sufficient in problem-solving terms and it is tainted by the relativity of simultaneity.

Method based on coordinate time

The life-expectancy of a muon in a given frame depends on whether it's moving in that frame. Perhaps you meant something more like "if it is less than the expected proper time along a muon's lifetime from birth to death", which is a frame-invariant number. Anyway stating it in frame-invariant terms as above is one way to look at the problem, but not the only way. You can equally well state the answer by saying "we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will reach the surface." The point is that although different frames disagree on the coordinate time between event 1 and event 2, and the life-expectancy of that particular muon, they will all end up making the same prediction about whether the muon survives long enough to collide with the surface. That's the magic of using different frames in SR, you always find that they agree about local events!

Yes, I agree this is also a valid method. You will see that it is in the list of sub-methods I mentioned in post #37.

You may disagree with the language I used in the following respect: in my language your method is a sort of intermediate step leading to proper time, while you may argue that it hits at the solution without the need to go through proper time.

Ok, but you’ve not given importance to my other point, which –I admit- is nothing revolutionary. In fact, I’ve borrowed it from you. In the example of the duel, you rapidly identified “proper time” as the key for solving the problem. I myself had doubts and wondered in a post of this thread if coordinate time was not an equally valid reference as well, since in the ground frame the coordinate time that the two duellers dispose of to do their tricks, between firing and being fired, is also identical for both duellers (2 s). But then I realized that “proper time” looks like a more straight path (if not better) for obtaining the answer in so far as it is self-sufficient. This is not so apparent in the duel example (although it is also valid there), but it’s clearer here in the muon example. For example:

- If you are a red scientist, you know that Red Muons decay in your frame after 1.8 s and you are told that the proper time of the Blue Muon between Events 1 and 2 is 1.732 s, would you say: “No, sorry, that’s not enough. I can’t solve the problem. You must provide me as well with v and my red coordinate time between Events 1 and 2. If that coordinate time is less than 1.8 s * sqrt(1-v^2) = 2.08 s, then the Blue Muon survives at Event 2”? I think you would rather say directly: “Yes, the Blue Muon survives.”

- If you are given the coordinate time of the trip of the Blue Muon between Events 1 and 2 = 2s, would you dare to answer the question? No. You would also ask for either the rest length of the Red Markers, which would enable you to do your calculation, or either v as measured by the Blue Muon as the ratio between its proper time and its coordinate length for the Red Markers.

So maybe we could agree on the following: proper time, if it is not the only method for solving the problem, not even the method to which other methods ultimately lead, it is at least a sort of “more direct” method, in the sense that, unlike other values, it is "self-sufficient" in providing the answer to the problem, at the same time that it is simultaneity-free, whatever the significance of those two facts?

The concept of invariant quantities

It transpires in these comments that the label “invariant quantities” employed SR may not be the best one, since it does not reveal the true nature of “proper time”, for example.

In SR all observers agree on the occurrence of events. Since measurement instruments are physical and what happens inside them are also physical events, all observers agree that all instruments measure the values that they measure. For example, just as the Red Frame agrees that the Blue Muon’s proper time is 1.732 s, the Blue Frame agrees that the Red Lab’s coordinate time is 2 s. Just as the Blue Frame agrees that the rest length of the Red Markers is 1 ls, the Red Frame agrees that such length has a coordinate value of 0.866 s in the Blue Frame.

So “universal agreement” or “frame-invariance” is a characteristic that does not express well enough in what sense proper time, for instance, is qualified.

To find that out, we would need to start talking about what we have not talked yet, that is, the physical or logical explanation of the solution: why the method of looking at certain values, either a single value (if it is self-sufficient) or several combined values (in the other cases) provides the answer “yes, the muon survives”.

 

[Saw]

Not having elicited any comment, I’ve been trying to look for faults in my own language or reasoning.

I may be giving the impression that I might object that time dilation is a fact. If so, it was not my intention. I believe time dilation is real, as proved by this very muon experiment.

Certainly, I’ve argued that all the values measured by the two frames involved in this experiment, except for proper time, are “simultaneity-dependent”. Thus it may look as if I was arguing that, should RS be “taken out of the problem”, all disagreement would be removed and the two frames would even agree that the tick rates of their clocks (and muons) are identical, in some absolute sense. But that is not really my intention.

The problem is that the very concept of RS is intermixed with that of TD and LC. To realize so, one just has to look at the way that measurements are made in practice:

- When the Red Frame measures that the red dt between Events 1 and Event 2 is the difference between red t2 = 3 s and red t1 =1 s (2 s), it says so because it has previously established that the red clock at the Red Lab should be set at 1 s (= red t1) when it received a synchronizing light signal from the origin of the red coordinate system. Thus it is true that the red time judgment of the time elapsed between those two Events is tainted by the RS. But it is also true that the Einstein convention for synchronizing clocks is also dependent on the tick rate of the red clocks: the clock at the Red Lab has been set at t1 = 1 s because the same light signal has bounced back, returned to the origin of the red coordinate system and found that the red clock at the origin reads 2 s (which, divided by 2, gives 1 s). Therefore, the red dt for the lapse between Events 1 and 2 is also conditioned by the tick rate of all red clocks.

- When the Red Frame measures that the rest length of the red atmosphere (or the proper distance between Events 1 and 0, which are simultaneous in the Red Frame) is 2 ls, the same reasoning applies: this is tainted by the RS, but for this very reason it is also tainted by the tick rate of the red clocks.

- When the Blue Frame measures that the length of the red atmosphere in the Blue Frame (or the proper distance between Events 1 and 0 bis, which are simultaneous in the Blue Frame) is 0.866 ls, this is tainted by the RS and the tick rate of blue clocks.

Thus the statement I had made ...

all the SR formulas really look like an intelligent way to get a simultaneity-free solution to a simultaneity-free problem by combining (and offsetting to the appropriate extent) different simultaneity-depending values and thus “decontaminating” them of relative simultaneity.

may not be very accurate, to the extent that it focuses on the simultaneity issue, as if it were independent from tick rate, when actually it is not.

But before trying an alternate wording, I think it is necessary to deal with “physical explanation” that is behind the solution of the problem.

We are saying here that the solution is “yes, the muon survives” because the measurement instruments measure so and so in each frame.

Obviously, this is based on the assumption that what (physically) happens in the measurement instruments “mirrors” what (physically) happens inside the muon. We may not perfectly know all the intricacies of the physical mechanisms responsible for the decay of the muon (the weak interactions), just as we don’t perfectly know all the details of the interactions that take place inside a clock, but we assume that the former and the latter are essentially analogous: one “mirrors” the other.

If we consider one single frame, this idea of the measurement as a mirror is not a big deal: the choice of the clock is purely arbitrary. Any clock, any oscillating process and any unit is a valid reference, a good mirror. For example: the red scientists check that the Red Muon decays after 1.8 s.

If the muon is now a Blue Muon that moves wrt the Red Frame, then the question is whether this fact (the muon is at rest in a different frame, it has a different state of motion) will change anything, that is to say, whether the Blue Muon will also decay after 1.8 s as measured by a blue clock of identical construction as the red clock.

The assumption of both Galilean relativity and SR is that “nothing changes”: the correlation between the decay rate of the Blue Muon and the tick rate of the blue clock will still be the same and the Blue Muon will only decay after 1.8 s as measured by the blue clock, so that the clock’s tick rate still “mirrors” the decay rate of the muon.

Then we find the surprise that the red scientists measure for the trip of the Blue Muon between the Red Markers a time lapse of 2 s, which is the difference between the reading of the clock at Event 1 (t1 = 1 s) and the reading of another clock, located somewhere else, at the Red Lab, the other edge of the red atmosphere, at Event 2 (t2 = 3 s). And now the question is: does what happens in the red clocks “mirror” what happens with the Blue Muon?

The obvious answer is that it doesn’t. With this information you do not solve the problem. You need to combine it with more information: the fact that the clock at the Red Lab has been set at 1 s at another Event (Event 0), which is deemed to be simultaneous with Event 1 in the Red Frame. This information is written precisely in the red landmark that shows that the distance between the red clocks is precisely 1 ls = red dx. Once that you know this, you can use any of the formulas [either dτ = sqrt (red dt^2 – red dx^2) or, what amounts to the same, since it also implicitly uses red dx within the v value, blue T₀ = red T * sqrt(1-v^2)] and only then, yes, can you conclude that the joint consideration of these two red measurements (red dt and red dx) “mirrors” what happens with the muon.

Why so? I would say that this is so because the red set of clocks and rods mirrors what happens in the blue set of clocks and rods. Since the latter, in turn, mirrors what happens with the muon (let us not forget that blue dx = 0), the former also reproduces that image.

Conclusion:

- We have a problem defined by reference to certain objective realities: how many interactions take place within the Blue Muon between its collision (at birth) with the upper red atmosphere and its collision with the red surface.
- One method for finding the solution is to look at the Blue Muon’s proper time because that objective reality directly mirrors, if the principle of relativity is true, what happens with the Blue Muon.
- Another method is looking at an appropriate combination of the red coordinate time and the red rest length, because, even if none of those values individually mirrors what happens with the Blue Muon, their combination does mirror an equal combination of the blue time and length measurements and thus, indirectly, what happens inside the Blue Muon.

In view of this, I would put the above statement as follows: all the SR formulas really look like an intelligent way to get a prediction about an objective reality (will the Blue Muon survive?) by either

(i) looking at a different objective reality that mirrors what happens with the Blue Muon (the blue proper time) or

(ii) looking at other realities that do not (and what is more important are not meant to and do not intend to and nobody should say that they do) individually mirror what happens with the muon, but contain information that, if appropriately combined through the right formulas, does in the end reveal or mirror what happens with the muon.

In the light of this, linking with the problem posed above about the accuracy of the term "invariant", I would propose to call the proper time not simply an invariant quantity on which "all frames agree", but a quantity that, unlike others, individually mirrors the objective reality that is the object of the problem, whereas the other quantities are only portions of another mirror, but are not meant to individually reflect what happens with the object in question.

 

[JesseM]

Not having elicited any comment, I’ve been trying to look for faults in my own language or reasoning.

Don't worry, it's on my mental to-do list to respond again to this thread...it's just a little intimidating because the posts have gotten so lengthy.

 

[Saw]

Don't worry, it's on my mental to-do list to respond again to this thread...

Thanks a lot for that.

it's just a little intimidating because the posts have gotten so lengthy.

And sorry for that. I am just too wordy, but I'll try to be more synthetic...

 

[JesseM]

And sorry for that. I am just too wordy, but I'll try to be more synthetic...

No problem, my own responses tend to be wordy too so I totally understand...as someone once said, "apologies for the length of this letter, but I didn't have time to write a shorter one."

 

[Saw]

I was going to post this comment in the thread about Time Dilation II with cos, but I was disappointed to discover it was closed. So I post it here because these quotes of that thread are very useful to illustrate what we discuss here.

In fact, this may serve as a summary of previous posts of this thread and thus anyone who wants to comment here can do it on the basis of this single post:

Say two cars start at the same position A on flat 2D surface, and their odometers show the same reading. Then they drive along different paths along the surface to a different position B, where they meet and compare odometer readings. One car took a straight-line path between A and B, the other took a non-straight path, so since we know a straight line is the shortest distance between any two points, naturally the odometer reading of the car that took the straight path won't have increased as much as the odometer reading of the car that took a non-straight path. Where did the missing odometer increments go, or what caused the difference in readings?

The ends of two tape measures are together. For the first 6 inches, the tapes are side-by-side and measure the same distance. Then one continues in a taut straight line, from its 6-inch mark to its 18-inch mark. The other is slack and takes a different route. Further along, both tape measures come together and run side-by-side one more. But where the taut tape measures 18 to 24 inches along this section, the slack tape measures 20 to 26 inches. Where did the missing two inches go, or what caused the difference in readings

These are very good analogies. But, as it happens with any analogy, they can be misleading if you don’t spell out all the details behind. In particular, I think we should spell out the following:

- SR only solves problems about whether certain events will happen or not in a specific point of spacetime (I call that Question 1 problems). It doesn’t solve hypothetical problems where the issue at stake is whether two events are happening simultaneously in different locations of space (Question 2 problems), precisely because (i) it is impossible to obtain absolute judgments about that issue (i.e., non-frame-dependent measurements of simultaneity); (ii) the only thing you can obtain is relative or frame-dependent measurements about such question and (iii) the ultimate solution to a practical problem cannot be different for different observers.

- Currently, all practical problems I can think of are problems about single events, not about whether two spatially separated events happen simultaneously in an absolute sense. So that sort of “limitation” in the problem-solving capacity of SR is not dramatic. But it is not out of question to recognize it exists; on the contrary it avoids much confusion.

With this background, the analogies you mention are much clearer to me.

For example: a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

The methods for reaching this solution can be two:

- A direct or “straight path” or “taut chord” expedient is looking at the ship’s clock and comparing it with the Earth’s clock located at that place.

- An indirect or “bent path” or “slack chord” expedient is looking at the x and t measurements of the Earth frame, namely dt = 2s and dx = 1 ls, combine them in the formula based on the Pythagorean Theorem and thus guess that the proper time of the ship’s clock = sqrt(2^2 – 1^2) = sqrt(3) = 1.732 s, less than the Earth's coordinate time = 2 s.

But why is the Earth’s frame forced to take the indirect or “scenic” route while the ship’s frame has the privilege of using the direct route? It is so because the problem we have posed ourselves is related to what happens to the ship’s clock between the events of its meetings with two spatially separated Earth’s clocks. Thus the ship’s measurement does not have to rely on the concept of simultaneity, while the Earth’s does. Thus the ship can take the direct route for solving the problem, while the Earth is forced to make a complex judgment with two legs, both legs being based on simultaneity-dependent concepts (dt and dx).

In fact, if we had posed ourselves a different problem, where the question is if an Earth’s clock reads more or less time than the clock of a second ship coming behind in perfect formation with the first ship, then the roles of the ships and the Earth would be exchanged and the ships’ measurements would be the indirect route while the Earth’s measurement would be the straight route.

 

[JesseM]

Back again, hello JesseM and anybody else.

The spatial analogue of time dilation (SATD) and the temporal analogue of length contraction (TALC)

I have tried to draw an adaptation of your diagram and concepts to the problem we are analyzing here (attached).

I had argued that a problem like this involves two objective frame-invariant elements:

- A temporal element, which is contributed by the Blue Frame: the fact that the Blue Muon decays or ticks at a certain rate.

- A spatial element, which is contributed by the Red Frame (the Earth): the Red Markers (the two edges of the red atmosphere).

I had also argued that the so called TD and LC formulas must be used in the following manner:

- Time formula:

T₀ = T * sqrt (1 – v^2) = 2 * 0.866 = 1.732 s

or

dt = dt * sqrt (1 – v^2) = 2 * 0.866 = 1.732 s

Yes, although usually these formulas are written as if you know the proper time interval between the events--the lifetime of the blue muon as measured by a clock traveling alongside it, i.e. its lifetime in the blue frame--and want to find the time interval between the same events in the frame where the clock is moving. So we can just take these equations and solve for the red times (by dividing both sides by 0.866), giving the usual form of the equations:

T = T₀ / 0.866

or

dt = dt / 0.866

Obviously this is just a sort of "cosmetic" change but it may help avoid confusion if you see the time dilation written in this more standard form where you're dividing by sqrt(1 - v^2) rather than multiplying by it.

- Length or X formula:

L = L₀ * sqrt (1 – v^2) = 1 * 0.866 = = 0.866 ls

or

dx = dx * sqrt (1 – v^2) = 1 * 0.866 = 0.866 ls

That is to say: the “spatial element” of the problem, which is the “rest length” of the Red Markers, occupies the right side of the formula, which makes it longer, while the coordinate length measured in the Blue Frame is shorter.

If my adaptation of your concepts is correct:

- My “length or X formula” is not the SATD, because the latter plays with the value dx = 1.154 s and my formula does not play with that value. My “length formula” would be the so called standard LC equation.

Yes, dx = 1.154[/color] ls would be the distance in the blue frame between two events in the red frame which had a distance of 1 ls[/color] and were simultaneous in the red frame.

If you want my opinion, the value 1.154 ls that is associated to the SATD would have its place in the very LC equation but in another problem, where it was analyzed the survival of the 1st Red Muon between the events of its collision with the two edges of the left half of the blue stick (the Blue Markers). In that case, the very same reasoning would lead to place the rest length of the Blue Markers in the right side of the formula, as follows:

L = L₀ * sqrt (1 – v^2) = 1.154 * 0.866 = 1 ls

It's not the number 1.154 l.s. which is itself relevant to deciding whether you're using the SAFTD or the LC equation though, it's more of a conceptual matter of what the two dx values represent--if you're talking about the instantaneous distance between two ends of a pair of inertial markers in two different frames (one of which is the markers' rest frame), then you're using the LC equation, while if you're talking about the coordinate distance between two events in two different frames (one of which is the frame where the events happen simultaneously), then you're using the SAFTD equation. In both cases, if you let the unprimed dx represent the distance in the frame that is "special" for the markers/events--the frame where the markers are at rest, or the frame where the events are simultaneous--and you let the primed dx' represent the distance in the "non-special" frame, then LC should look like this:

dx' = dx * sqrt(1 - v^2)

while the SAFTD should look like this:

dx' = dx / sqrt(1 - v^2)

(of course in both cases you're free to solve for the unprimed dx by dividing/multiplying both sides by gamma, but the equations still look different if you keep track of where the unprimed dx is vs. the primed dx')

My “time formula” ends up being the TALC, if I got it right at all.

No, it's just the usual time formula with but rearranged so you're solving for time in the "special" frame where the events are simultaneous, which to be consistent I would label as the unprimed frame. In other words, these are both rearranged versions of the time dilation equation:

dx' = dx / sqrt(1 - v^2)
dx = dx' * sqrt(1 - v^2)

Whereas these are two rearranged versions of the TAFLC:

dx' = dx * sqrt(1 - v^2)
dx = dx' / sqrt(1 - v^2)

...where again, I'm assuming in all cases that the unprimed dx is measured in the "special" frame (in the case of the TAFLC, the 'special frame' would be the one where the two straight spacelike paths, which are analogous to the two straight timelike worldlines of the markers in the case of length contraction, actually represent surfaces of constant t in this frame, whereas dx' in the 'non-special' frame represents the time interval between these two spacelike paths in the frame where they're slanted and don't have a constant t' coordinate. If this is confusing I really wouldn't worry too much about it, TAFLC is a concept of my own invention that was useful in a discussion with neopolitan but isn't something that would be needed in any practical SR problem I can think of).

The two values of the frame owning the time element (in our example 1.732 s and 0.866 ls) are shorter as if both the time and length units were dilatated vis-à-vis the units of the other frame.

That's true because of the particular things you're choosing to measure--the time between two events which are colocated in the blue frame (the creation and decay of a muon at rest in the blue frame), and the length of an object which is at rest in the red frame (the red atmosphere). If you were instead having the blue frame be "special" with regard to both the times and lengths you were measuring (i.e. if you were still talking about the time between events on the worldline of a muon at rest in the blue frame, but you were also talking about the length of an object at rest in the blue frame), then the time would be dilated in the red frame while the length would be contracted in the red frame.

Definition of the question or problem

You mention several times that stating the question in a frame-invariant way is not necessary and that what is important is that the question is about a “local event” (if the Blue Muon will survive at Event 2).

Yes, I agree. That is clearer. Maybe, since events “belong” to all frames (all frames agree that they happen), could we say for even more precision that the question is about a “single event” = whether Event 2 will count with the presence of the Blue Muon?

We could call this Question 1.

Yes, that's a good physical question about frame-invariant events. And as I said, you may use frame-dependent time and distance measurements in the course of figuring out the answer.

Another possible formulation (Question 2) would be:

“does a Blue Muon decay faster than a Red Muon or more slowly = is the internal clock of a Blue Muon faster or slower than the internal clock of a Red Muon?”

Obviously, this Question 2 is a comparison between “pairs of events that are distant from each other”. To answer it, we would need to be constantly comparing the Blue Muon ticks or decay steps with those of a Red Muon. This, in turn, involves a judgment on simultaneity, on which the two frames disagree. Thus, if we made that judgment, each frame would be entitled to claim that it is the other’s clock rate or decay process the one that is slower.

Right.

But I think it is important to note that here the question is less ambitious, it is just Question 1. Question 2 would only be asked in a twin paradox scenario, where one of the muons were accelerated back to re-encounter another twin brother created at the same Event 1. That would be more complicated to analyze and I propose to leave it aside for the time being.

Even in a twin paradox you aren't quite addressing question 2, because different frames can disagree about which clock was running faster during a particular phase of the trip, they only agree on the total elapsed time when the two clocks reunite at a single location (which therefore becomes another question about purely local events).

 

[JesseM]

Proper distance between the non-simultaneous (in any frame) Events 1 and 2

Applying this information to the problem at hand (Question 1 referred to in my most recent post), I understand that:

- Question 1 is about the spacetime interval between Events 1 and 2.

- If we measure distances with light seconds, then c=1 and the formula for the spacetime interval simplifies to:

ds² = -dt² + dx²

- The distance separating these two Events is timelike since dt > dx.

- Also with c=1 the formula for the proper time interval between Events 1 and 2 is:

dτ² = dt² - dx²

- In our example, calculations for proper time would be:

With red values: dτ² = dt² - dx² = 2^2 – 1^2 = 4 – 1 = 3  dτ = sqrt(3) = 1.732 s
With blue values: dτ² = dt² - dx² = 1.732^2 – 0 = 3  dτ = sqrt(3) = 1.732 s

- It seems that instead the spacetime interval would be… sqrt(-3)? Is that right? If so, in an example like this one, where the distance between Events 1 and 2 is timelike, what is the meaning and the added value of the concept of spacetime interval vis-à-vis proper time?

Well, there's a value of having a notion of "distance" in spacetime which can be measured along both timelike and spacelike paths, because being able to define a notion of "distance along arbitary continuous paths in a manifold" is the key to how differential geometry defines the intrinsic curvature of a surface without needing to assign points on the surface coordinates in a higher-dimensional "embedding space" (like the 3D x-y-z space we might use to define the shape of a 2D sphere using an equation like x^2 + y^2 + z^2 = 1), and this is how general relativity deals with the notion of "curved spacetime". The geometry of spacetime is such that if distance along spacelike paths is real than distance along timelike paths must be imaginary, and vice versa. I think you're asking why ds is defined so distance along timelike paths is imaginary rather than the other way around, and I'm pretty sure the answer is that it's purely a matter of convention, I think I've actually seen some authors define ds the other way around so it has units of time, i.e. ds^2 = dt^2 - (1/c^2)*(dx^2 + dy^2 + dz^2). It seems more common to do it so ds has units of distance, but in any case it's simple to just take the ds along a timelike path and divide by i*c to get the proper time.

Proper distance between simultaneous events versus length

There are several issues here:

- “All observers agree on which one the rest frame of an object is and on the rest length of an object”.

Yes, for example, all observers agree that the Red Markers’ rest frame is the Red Frame and that the rest length of the Red Markers is 1 ls. But all observers also agree that the non-rest length of the Red Markers in the Blue Frame is 0.866 ls.

True, it's just that no one has bothered to come up with a catchy name for the concept of "length in a frame where the object is moving at 0.5c" like they have with length in an object's rest frame.

However, a given value will be “different” if it has a problem-solving capacity “by itself”, without the need of combining it with other values. This is not the case of “rest length”. If you know that the rest length of the Red Markers is 1 ls, have you solved the problem? No, you would need to know something more, wouldn’t you?

I suppose, but if you aren't given something like the proper time between the muon's creation and it reaching the surface at the start of the problem, then you still have to use multiple frame-dependent quantities (like dx and dt between these events) to derive it, so nothing is lost by skipping it and just using these frame-dependent quantities to derive the answer to the original "question 1" (does the muon survive to hit the surface) directly.

“There's no need to exclusively talk about the distance between simultaneous events when talking about rest length (…) you can pick an event on one end that happens at some coordinate time t0 and an event on the other end that happens at a different coordinate time t1, the coordinate distance between these events in the rest frame is still equal to the rest length”

Yes, this is what is usually exemplified by saying that you can put the measuring tape on one end of the object at t0 and then quietly walk to the other end while stretching out the tape, until you measure the other end, quite later, at t1. Thus the two acts of length measurement are not simultaneous. That is true. But you had also agreed that (i) the units of the measurement tape have been established with a measurement method which is based on the time it takes for light to traverse a distance in a two-way trip and (ii) that any other measurement convention should render essentially analogous results, if SR is true. Thus the length of the measurement instruments is tainted by the relativity of simultaneity and so must be any measurement made with those instruments.

I'm not clear on exactly what is meant by the phrase "tainted by the relativity of simultaneity". If you want to define the meter in terms of the distance light travels in a certain clock time, then it seems to me you can avoid simultaneity issues by measuring the two-way time for light to depart from one end, get reflected at the other end, and return to the first end, as measured by a single clock at that end...then the length would be defined as (1/2)*(time interval)*c. As long as coordinate distance has this property (and it should have this property even if you don't define distance in terms of c, like if you use multiples of atomic orbitals for atoms at rest in your frame), then you can synchronize clocks at rest in your frame any way you like, and although the one-way speed of light may not be c, the average two-way speed of light when it departs from a single location and returns to it will still be c.

Method based on coordinate time

The life-expectancy of a muon in a given frame depends on whether it's moving in that frame. Perhaps you meant something more like "if it is less than the expected proper time along a muon's lifetime from birth to death", which is a frame-invariant number. Anyway stating it in frame-invariant terms as above is one way to look at the problem, but not the only way. You can equally well state the answer by saying "we must look at the coordinate time between event 1 and event 2 in whatever frame we're using, and if it less than the life-expectancy of the blue muon in this same frame, then the blue muon will reach the surface." The point is that although different frames disagree on the coordinate time between event 1 and event 2, and the life-expectancy of that particular muon, they will all end up making the same prediction about whether the muon survives long enough to collide with the surface. That's the magic of using different frames in SR, you always find that they agree about local events!

Yes, I agree this is also a valid method. You will see that it is in the list of sub-methods I mentioned in post #37.

You may disagree with the language I used in the following respect: in my language your method is a sort of intermediate step leading to proper time, while you may argue that it hits at the solution without the need to go through proper time.

Ok, but you’ve not given importance to my other point, which –I admit- is nothing revolutionary. In fact, I’ve borrowed it from you. In the example of the duel, you rapidly identified “proper time” as the key for solving the problem. I myself had doubts and wondered in a post of this thread if coordinate time was not an equally valid reference as well, since in the ground frame the coordinate time that the two duellers dispose of to do their tricks, between firing and being fired, is also identical for both duellers (2 s). But then I realized that “proper time” looks like a more straight path (if not better) for obtaining the answer in so far as it is self-sufficient. This is not so apparent in the duel example (although it is also valid there), but it’s clearer here in the muon example. For example:

- If you are a red scientist, you know that Red Muons decay in your frame after 1.8 s and you are told that the proper time of the Blue Muon between Events 1 and 2 is 1.732 s, would you say: “No, sorry, that’s not enough. I can’t solve the problem. You must provide me as well with v and my red coordinate time between Events 1 and 2. If that coordinate time is less than 1.8 s * sqrt(1-v^2) = 2.08 s, then the Blue Muon survives at Event 2”? I think you would rather say directly: “Yes, the Blue Muon survives.”

True, if you are given the proper time at the start of the problem it's simple to answer whether it survives. But if you aren't, then you don't necessarily need to do the intermediate step of figuring out the proper time between events 1 and 2 in order to get the answer to whether the muon survives.

- If you are given the coordinate time of the trip of the Blue Muon between Events 1 and 2 = 2s, would you dare to answer the question? No. You would also ask for either the rest length of the Red Markers, which would enable you to do your calculation

Yes, then you could figure out distance/time in the red frame and therefore the speed of the blue muon, which would give you its time dilation factor so you could see if it would decay in more or less than 2 seconds in the red frame.

or either v as measured by the Blue Muon as the ratio between its proper time and its coordinate length for the Red Markers.

Right, because the speed of the red markers in the blue muon rest frame will be the same as the speed of the blue muon in the red rest frame, though the directions will be different.

So maybe we could agree on the following: proper time, if it is not the only method for solving the problem, not even the method to which other methods ultimately lead, it is at least a sort of “more direct” method, in the sense that, unlike other values, it is "self-sufficient" in providing the answer to the problem, at the same time that it is simultaneity-free, whatever the significance of those two facts?

Yeah, I agree with that.

The concept of invariant quantities

It transpires in these comments that the label “invariant quantities” employed SR may not be the best one, since it does not reveal the true nature of “proper time”, for example.

In SR all observers agree on the occurrence of events. Since measurement instruments are physical and what happens inside them are also physical events, all observers agree that all instruments measure the values that they measure. For example, just as the Red Frame agrees that the Blue Muon’s proper time is 1.732 s, the Blue Frame agrees that the Red Lab’s coordinate time is 2 s. Just as the Blue Frame agrees that the rest length of the Red Markers is 1 ls, the Red Frame agrees that such length has a coordinate value of 0.866 s in the Blue Frame.

I would say it's true that all observers agree on what the length and distances are in any particular choice of frame...it's kind of weird to talk about frames agreeing on what's true in other frames, that would be a bit like asking if the metric system "agrees" that the freezing point of water is 32 degrees in the Fahrenheit scale.

So “universal agreement” or “frame-invariance” is a characteristic that does not express well enough in what sense proper time, for instance, is qualified.

I think you're making things too complicated here, "frame-invariance" just means that if you do calculations of a quantity using only the coordinates of one frame, and then repeat using only the coordinates of another frame, you're guaranteed to get the same answer both times.

 

[DrGreg]

- A direct or “straight path” or “taut chord” expedient is looking at the ship’s clock and comparing it with the Earth’s clock located at that place.

- An indirect or “bent path” or “slack chord” expedient is looking at the x and t measurements of the Earth frame, namely dt = 2s and dx = 1 ls, combine them in the formula based on the Pythagorean Theorem and thus guess that the proper time of the ship’s clock = sqrt(2^2 – 1^2) = sqrt(3) = 1.732 s, less than the Earth's coordinate time = 2 s.

But why is the Earth’s frame forced to take the indirect or “scenic” route while the ship’s frame has the privilege of using the direct route?

Actually, it's the other way round. The inertial Earth takes the direct "taut" route; the ship, which is not inertial for at least part of its journey, takes the indirect "slack" route. But in the topsy-turvy geometry of spacetime, straight lines are longer than bent or curved lines, unlike the normal Euclidean geometry we are familiar with.

So my analogy isn't perfect. It demonstrates how two measurements can differ, but it gets it wrong which is the longest.

The reason why in spacetime straight worldlines have longer length (i.e. more proper time τ) than bent lines is because there is minus sign in

dτ² = dt² − dx² / c²​

whereas, in Euclidean geometry, length is given by the Pythagoras formula with a plus sign

dr² = dx² + dy²​

 

[DrGreg]

Actually, it's the other way round. The inertial Earth takes the direct "taut" route; the ship, which is not inertial for at least part of its journey, takes the indirect "slack" route. But in the topsy-turvy geometry of spacetime, straight lines are longer than bent or curved lines, unlike the normal Euclidean geometry we are familiar with.

So my analogy isn't perfect. It demonstrates how two measurements can differ, but it gets it wrong which is the longest.

The reason why in spacetime straight worldlines have longer length (i.e. more proper time τ) than bent lines is because there is minus sign in

dτ² = dt² − dx² / c²​

whereas, in Euclidean geometry, length is given by the Pythagoras formula with a plus sign

dr² = dx² + dy²​

Oops!

My response above to post #46 was too hasty -- I hadn't read the entire post, and I assumed we were talking about a ship traveling from Earth and then back again. But I hadn't read this:

For example: a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

In this case, my analogy doesn't help much. Both routes are "taut", but as they don't meet again, you can't directly compare one against the other.

An analagous geometrical question is this:

There are two rulers at an angle to each other, whose zero marks coincide. Is the 10cm marker on one ruler at a greater vertical height than the other?

I hope you will agree that there is insufficient information to answer this question, specifically I haven't said which way is "up". Similarly in your question you haven't specified in which frame you want your answer, the earth's, the ship's or someone else's.

 

[atyy]

So “universal agreement” or “frame-invariance” is a characteristic that does not express well enough in what sense proper time, for instance, is qualified.

We assume that standard ideal clocks exist. These are clocks that when located at the same point in spacetime tick at the same rate. After they have been separated and read differently, we can always bring them back together to another point in spacetime and check if they still tick at the same rate. In this way we will know whether a particular real clock is standard and ideal. (This is just to start. There are some other requirements too)

The accumulated proper time along any path (no matter how short or long) in spacetime is the total number of ticks a standard ideal clock makes when it takes that path. Each tick is an event. So the proper time is related to events.

 

[Saw]

Actually, it's the other way round. The inertial Earth takes the direct "taut" route; the ship, which is not inertial for at least part of its journey, takes the indirect "slack" route. But in the topsy-turvy geometry of spacetime, straight lines are longer than bent or curved lines, unlike the normal Euclidean geometry we are familiar with.

So my analogy isn't perfect. It demonstrates how two measurements can differ, but it gets it wrong which is the longest.

The reason why in spacetime straight worldlines have longer length (i.e. more proper time τ) than bent lines is because there is minus sign in

dτ² = dt² − dx² / c²​

whereas, in Euclidean geometry, length is given by the Pythagoras formula with a plus sign

dr² = dx² + dy²​

Thanks. Well, I realize now we were talking about different things.

You were talking about a twin paradox scenario, where the Earth is inertial all the time, whereas the ship accelerates away and accelerates back to return to the Earth. Yes, in that case the Earth takes the direct "taut" route while the ship takes the indirect "slack" route through the two legs of its trip.

Instead my example was a "pure-blooded" SR case with no acceleration (and hence, I understand, analyzable with pure Euclidean geometry) :

a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

But the example is more interesting if you include a moun in it, as we were doing in the latest posts. Imagine that there is a lab in the ship and that a muon is artificially created in the ship's lab when the latter meets the Earth observer. Imagine as well that a muon's lifetime in its rest frame is, arbitrarily, for convenience, 1.8 s. Will the muon be still "alive" when it arrives at position x2 = 1 ls at t2 = 2s in the Earth frame?

Well, the detailed arguments given above, which JesseM is patiently commenting, are that:

- Given the reality under consideration (a muon is decaying), the ticks of the ship's clock accompanying such muon must directly be, if the principle of relativity is true, a "mirror" of what happens with the muon.

- Instead, the reading of the clock of the Earth frame that waits for the muon at x2 depends on two things: the Einstein convention for synchronization (which in turn also depends on the tick rate of the Earth's clock) and its own tick rate. The Einstein's convention has determined that the clock at x2 will be set at, let us say, 0 s when it received the synch light signal and later on 1 s is added to it when it is learned that such synch signal has taken 2 s to return to the Earth. This 1 s that the light signal is deemed to have taken for the trip to the clock at x2 is written as a reminder in the landmark for that clock in the Earth frame, with the label "x2 = 1 ls". When the Earth wants to solve the problem, in my opinion, the Earth combines the two values, coordinate time = dt (2 s) and rest length = dx (1 ls), in the formula sqrt (2^2 - 1^2) in order to get (this "in order to" is objected by JesseM) 1.732 s, which is the proper time of the muon. That means the muon survives. So the Earth has reached the same solution but through an indirect path that combines two simultaneity-dependent values, dt and dx. Thus the Earth values dt and dx also "mirror" what happens with the muon but only if put together in the right equation.

Geometrically speaking, I thought: if you look at the spacetime diagram of the Earth frame, there is a right angle triangle where (i) the base is the rest length, dx = 1 ls; (ii) the height is the coordinate time, dt = 2s and the hypothenuse is the wordline of the muon. If you now change the place of the dt value and you attribute this value to the hypothenuse (the muon's wordline), you can solve for the value of the height (now vacant) and you get sqrt (2^2 - 1^2) = 1.732 s. Intuitively, this looks like if the Earth were admitting that it has reached the solution by moving farther away through the hypothenuse and coming back through the base, or vice versa... But I accept in advance that this looks like an ad hoc construction and is speculation...

 

[JesseM]

For example: a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

Well, "less time" according to the Earth frame's definition of simultaneity. Suppose there is some physical marker at rest at position x=1 in the Earth's frame. In the ship's frame, the event of the marker whizzing by the ship coincides with the event of the ship's clock reading 1.732 seconds. But in the ship's frame these events are not simultaneous with the event of the Earth's clock reading 2 s; instead, they are simultaneous with the event of the Earth's clock reading 1.5 s, so in the ship's frame the Earth's clock reads less time at the moment the ship passes this marker. Of course, if the marker has its own clock which is synchronized with the Earth's clock in the Earth's rest frame, then it's an objective fact that when the ship passes the marker, the ship's clock reads 1.732 s and the marker's clock reads 2 s (but in the ship's frame, the marker's clock already read 0.5 s at the moment the ship was passing next to Earth).

The methods for reaching this solution can be two:

- A direct or “straight path” or “taut chord” expedient is looking at the ship’s clock and comparing it with the Earth’s clock located at that place.

- An indirect or “bent path” or “slack chord” expedient is looking at the x and t measurements of the Earth frame, namely dt = 2s and dx = 1 ls, combine them in the formula based on the Pythagorean Theorem and thus guess that the proper time of the ship’s clock = sqrt(2^2 – 1^2) = sqrt(3) = 1.732 s, less than the Earth's coordinate time = 2 s.

The formula for proper time isn't really based on the Pythagorean Theorem, although it's the spacetime analogy for the Pythagorean Theorem.

But why is the Earth’s frame forced to take the indirect or “scenic” route while the ship’s frame has the privilege of using the direct route?

What do you mean by that? In the ship's frame, how would you calculate the time on the marker's clock "directly"? It seems to me you'd have to start with the fact that the marker's clock showed a reading of 0.5 s at t'=0 in the ship's frame, then calculate the proper time elapsed on the marker's worldline between t'=0 and the event of it reaching the ship, and add that to the initial 0.5 s.

It is so because the problem we have posed ourselves is related to what happens to the ship’s clock between the events of its meetings with two spatially separated Earth’s clocks. Thus the ship’s measurement does not have to rely on the concept of simultaneity, while the Earth’s does.

The ship needs to use the concept of simultaneity if it wants to predict the time elapsed on the Earth clock or the marker clock though.

In fact, if we had posed ourselves a different problem, where the question is if an Earth’s clock reads more or less time than the clock of a second ship coming behind in perfect formation with the first ship, then the roles of the ships and the Earth would be exchanged and the ships’ measurements would be the indirect route while the Earth’s measurement would be the straight route.

In this case, when calculating things in the Earth frame you'd need to know the initial time on the second ship's clock at t=0, and the elapsed time on the second ship's clock between that time and the moment the second ship reached the Earth.

 

[Saw]

JesseM, I agree with the numbers in your latest post and of course with the idea that each frame finds the other suffers TD and LC and each frame has a different view on simultaneity. But we had reached an agreement on how those concepts play in the definition and resolution of a particular practical, real-life problem, which I would not like to lose.

The problem was first described in post #27 and refined in successive posts. Unfortunately then in post #46 I placed here an answer that was intended for another thread and used for that purpose a simplified version of the problem, which (I think) has made us lose perspective. My fault.

I’ll try to summarise the conclusions reached so far, so as to introduce the reply to other posters:

- The question or problem was: will the Blue Muon, created at Event 1 (collision with the red upper atmosphere) “survive” to be “present” at Event 2 (collision with the Red Lab)?

To be noted: the definition of the question is frame-invariant, not only in the sense that its phrasing does not choose a perspective about who moves and who is stationary, but also in the sense (which is more important) that it relies on single events, on whose “happening” all frames agree, even if they assign different coordinates to them; the question is NOT whether two spatially separated events are simultaneous or not, although that plays a part in the resolution of the problem, as explained below.

- The answer or solution is: yes, the muon “survives”.

To be noted: all frames agree on this answer.

- The methods for finding the answer or solution:

* If we know that the life expectancy of a Red Muon at rest in the Red Lab is “dt” and we know that the the proper time of the Blue Muon as measured by a blue clock is less than “dt”, then we know the answer is yes.

The reason is the principle of relativity: relative motion does not change the results of experiments = if it happened to affect the muon, it would affect the clock at rest with it in the same manner, and vice versa.

To be noted: “proper time” is self-sufficient in the resolution of the problem and proper time is a simultaneity-free concept.

* If we know the red dt (coordinate time difference between Event 1 at the birth place of the Blue Muon, which is simultaneous with Event 0 at the Red Lab, and Event 2) and the red dx (the rest length of the red atmosphere), then we also know the answer. We can explain why in two ways: either that (i) by combining these two values in the appropriate formula, we get the proper time of the Blue Muon (my view) or that (ii) we conclude that the Blue Muon’s coordinate time is less than the life expectancy of a muon moving at the given velocity in the Red Frame (an alternative view that for you stands on equal footing).

To be noted: The red dt (coordinate time), which is simultaneity-dependent, is not self-sufficient in the resolution of the problem. Neither is the red dx (rest length). You have to combine the two values to find the answer.

Open points:

Is rest length also a simultaneity-dependent concept?

Is one way of explaining the solution more meaningful in some sense than the other? For example, how do you link your explanation (ii) with the principle of relativity?

How to label “proper time” in view of its capacity for self-sufficiently solving problems like this?

* There are other routes but the same reasoning applies to them.

Well, given this basic agreement (if I summarised correctly), I draw conclusions as to the significance of SR, which in my opinion would avoid many misunderstandings and objections that one often finds. (At least this has served myself to overcome my initial reluctance to accept it.) I have already advanced them, but will try to put them in a simpler manner.

The accumulated proper time along any path (no matter how short or long) in spacetime is the total number of ticks a standard ideal clock makes when it takes that path. Each tick is an event. So the proper time is related to events.

Yes, I agree with that. That is why, since the problem is here about “local events” in JesseM’s terminology, I suggested that the proper time reading of a clock is different (it self-sufficiently solves the problem) because it directly “mirrors” those events, in accordance with the principle of relativity.

An analagous geometrical question is this: There are two rulers at an angle to each other, whose zero marks coincide. Is the 10cm marker on one ruler at a greater vertical height than the other? I hope you will agree that there is insufficient information to answer this question, specifically I haven't said which way is "up".

I agree with that, of course.

Similarly in your question you haven't specified in which frame you want your answer, the earth's, the ship's or someone else's.

That is the question. After some mutual efforts, in this thread we have defined “answer” as the ultimate practical solution to a practical problem. I understand that this “answer” cannot be frame-dependent: it must be the same in the earth’s, in the ship’s or in anybody’s frame. During the calculation process of the answer, certainly, different frames may play with different measurements about what is “relatively up” or what is “relatively simultaneous” or whose clock “relatively dilates”, but this does not lead to disagreement on the final answer to the problem.

 

[DrGreg]

That is the question. After some mutual efforts, in this thread we have defined “answer” as the ultimate practical solution to a practical problem. I understand that this “answer” cannot be frame-dependent: it must be the same in the earth’s, in the ship’s or in anybody’s frame. During the calculation process of the answer, certainly, different frames may play with different measurements about what is “relatively up” or what is “relatively simultaneous” or whose clock “relatively dilates”, but this does not lead to disagreement on the final answer to the problem.

Sorry, I must have been half asleep yesterday. The question I answered was not the question you asked -- I misread it. I withdraw my last post, it is not applicable to what you really said! Sorry.

 

[JesseM]

Sorry to take a few days to reply again, but in this case I see my response can be fairly short since there doesn't seem to be any real disagreement on the main issues:

I’ll try to summarise the conclusions reached so far, so as to introduce the reply to other posters:

- The question or problem was: will the Blue Muon, created at Event 1 (collision with the red upper atmosphere) “survive” to be “present” at Event 2 (collision with the Red Lab)?

To be noted: the definition of the question is frame-invariant, not only in the sense that its phrasing does not choose a perspective about who moves and who is stationary, but also in the sense (which is more important) that it relies on single events, on whose “happening” all frames agree, even if they assign different coordinates to them; the question is NOT whether two spatially separated events are simultaneous or not, although that plays a part in the resolution of the problem, as explained below.

- The answer or solution is: yes, the muon “survives”.

To be noted: all frames agree on this answer.

- The methods for finding the answer or solution:

* If we know that the life expectancy of a Red Muon at rest in the Red Lab is “dt” and we know that the the proper time of the Blue Muon as measured by a blue clock is less than “dt”, then we know the answer is yes.

The reason is the principle of relativity: relative motion does not change the results of experiments = if it happened to affect the muon, it would affect the clock at rest with it in the same manner, and vice versa.

To be noted: “proper time” is self-sufficient in the resolution of the problem and proper time is a simultaneity-free concept.

* If we know the red dt (coordinate time difference between Event 1 at the birth place of the Blue Muon, which is simultaneous with Event 0 at the Red Lab, and Event 2) and the red dx (the rest length of the red atmosphere), then we also know the answer. We can explain why in two ways: either that (i) by combining these two values in the appropriate formula, we get the proper time of the Blue Muon (my view) or that (ii) we conclude that the Blue Muon’s coordinate time is less than the life expectancy of a muon moving at the given velocity in the Red Frame (an alternative view that for you stands on equal footing).

To be noted: The red dt (coordinate time), which is simultaneity-dependent, is not self-sufficient in the resolution of the problem. Neither is the red dx (rest length). You have to combine the two values to find the answer.

Yes, agree with all of this.

Open points:

Is rest length also a simultaneity-dependent concept?

Well, what do you think about my proposal of defining "rest length" in terms of the two-way time for light to get from one end to the other and back, as measured by a clock at the end the light departs from and returns to?

Is one way of explaining the solution more meaningful in some sense than the other? For example, how do you link your explanation (ii) with the principle of relativity?

I guess I would just say that the principle of relativity tells us that the laws of physics work the same way when expressed in the coordinates of any inertial frame, and that one of those laws is that the time on a clock moving at v in a given frame is dilated by the gamma factor, so we can dilate the blue muon's lifetime by the gamma factor and see whether it's longer or shorter than the coordinate time needed for it to reach the surface.

 

[Saw]

As to rest length

Well, what do you think about my proposal of defining "rest length" in terms of the two-way time for light to get from one end to the other and back, as measured by a clock at the end the light departs from and returns to?

This method is the Einstein convention, that is to say, precisely the method used for synchronizing two distant clocks. If you use the same signal to both (i) measure rest length and (ii) synchronize the clock located at the other end of the rod and the signal returns after 2 s as measured by the origin clock, you will conclude that the distant clock is separated from you by 1 ls and that it should read 1 s at the time when it received the signal…

That is why I said that the x value, the separation in length units of a distant clock from the origin of a coordinate system, is like a reminder of the time it took for the synch signal to get there, which is in turn a component of the t value that such distant clock will show later on.

Does all that justify saying that rest length is a “simultaneity-dependent” value? Well, I am not very happy with the expression, because, as noted in another post, the concept of simultaneity is also intertwined with clock rate … In the end, too, I do not see a clear utility for this distinction. It might have one if, as I proposed in other threads, it were true that the length of a given rod, after acceleration (the pure change of frame, leaving aside the temporary physical effect of the acceleration process) had to be “recalibrated” = a new length should be attributed to the rod, just as you would also have to re-synchronize clocks in that frame. The latter (clock re-synchronization) was accepted in one thread, but the former (length re-calibration) seems to be rejected (another thread). If you confirm that is wrong, I’ll leave the idea, then.

As to how to link the principle of relativity with the idea that from the Red Frame you need to combine two values to get the answer to the problem:

I guess I would just say that the principle of relativity tells us that the laws of physics work the same way when expressed in the coordinates of any inertial frame, and that one of those laws is that the time on a clock moving at v in a given frame is dilated by the gamma factor, so we can dilate the blue muon's lifetime by the gamma factor and see whether it's longer or shorter than the coordinate time needed for it to reach the surface.

Yes, that must be the way to put it in technical terms. I was just trying to get a more didactic expression for a non-specialized audience. Do you think the following would be correct?

- A measurement is something that happens in an “instrument”, which mirrors what happens in a “real-life” object we’re interested in (though in standard units).
- The ticks of a clock accompanying the Blue Muon mirror the decay ticks of the same. This would be so both for classical and special relativity.
- But SR notes that relative motion brings about a number of discrepancies between the measurements of different frames (I don't need to repeat them). Thus, for example, the difference between the readings of the two Red clocks that the Blue Muon passes by do not directly mirror what happens to the Blue Muon.
- However, the combined red readings of dt and dx do mirror the combined blue readings of dt and dx, through the right formula. For example, sqrt(red dt^2 – red dx^2) = sqrt(blue dt^2 – blue dx^2) = 1.732 s in this example.
- This is because SR uses a sort of MM-experiment instrument, where two light pulses (or two tennis balls, although that would be less precise and less workable), leaving simultaneously from the origin and traveling perpendicular to each other, always reunite at the origin. That is to say, even if two instruments of this sort give out different values for length/distance (in the X axis, where there is relative motion) and time, they have something in common: same Y length and the light pulse returns to the origin simultaneously in all cases…

I admit all this is very rough, but I would welcome criticism... if it is not so rough that it can't even be criticized!