A very typical atmospheric muon simultaneously comes to rest and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. Let's explore the history of this spacetime event. The muon's trajectory was normal to the surface of the earth.

The muon's velocity relative to the earth was measured to be 0.867c which results in gamma = 2. The muon's clock showed an elapsed proper time of 2..2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.

1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?

2) What did the scientists in the scintillator lab calculate the muon's altitude to be at the spacetime event corresponding to the muon's creation?

3) What did the lab scientist read on the scintillator lab's clock at the spacetime event corresponding to the muon's creation?

patdolan wrote:

> 1) What did the muon calculate its proper altitude above the earth to be
> at the spacetime event corresponding to its creation?

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Typical evil, illegal Obama strategy spinning out of control and
backfiring.

On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote:
> A very typical atmospheric muon simultaneously comes to rest and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. Let's explore the history of this spacetime event. The muon's trajectory was normal to the surface of the earth.
>
> The muon's velocity relative to the earth was measured to be 0.867c which results in gamma = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
>
> 1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?
>
> 2) What did the scientists in the scintillator lab calculate the muon's altitude to be at the spacetime event corresponding to the muon's creation?
>
> 3) What did the lab scientist read on the scintillator lab's clock at the spacetime event corresponding to the muon's creation?
One thing you can say for Eponymous: when he sets up the chess board, it's always a one move checkmate. I give relativists credit for recognizing this. But I also condemn them for their cowardice when it comes to takin' their whupin' for their error-filled beliefs.

On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote:
> Let's explore the history of this spacetime event.

A spacetime event doesn't have a history, it is simply a specific "point" in spacetime. Of course, the causal past of an event consists of the entirety of the past light cone and its interior.

> A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.

Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) that you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2.

> 1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?

Muon's don't calculate things, and even if they did, it would be irrelevant.. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.

> 2) What did the scientists in the scintillator lab calculate the muon's altitude to be at the spacetime event corresponding to the muon's creation?

As noted above, in terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds.

> 3) What did the lab scientist read on the scintillator lab's clock at the spacetime event corresponding to the muon's creation?

In terms of S, the lab clock read -D/v at event e2. Of course, it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.

On Wednesday, May 3, 2023 at 5:56:42 PM UTC-7, Trevor Lange wrote:
> On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote:
> > Let's explore the history of this spacetime event.
> A spacetime event doesn't have a history, it is simply a specific "point" in spacetime. Of course, the causal past of an event consists of the entirety of the past light cone and its interior.
>
> > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
>
> Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) that you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.

A point of clarification please. Your e4 is simultaneous is in both your S and your S'. Correct? I will take the liberty of anticipating your answering in the affirmative and go on to cast this fact in terms of my own narrative, which most will agree is far more informative, heuristic, natural, numerical and entertaining than your own style of composure, Legion Lange. To wit:

The lab clock reads 0.00 at the moment of the muon's disintegration in the scintillator. And the muon's egg-timer-to-disintegration has ticked down from 2.2 usec (which the anthropomorphized muon set when it was formed high in the atmosphere) to 0.00 usec at the moment of its disintegration in the scintillator. So both the lab clock and the muon's egg timer read 0.00 usec at the moment of disintegration. This is your e4 and it is the END OF THE BRIEF HISTORY OF MURRAY THE MUON.

From here Legion, we can examine Murray's life history in both your S and your S'.

In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2.
> > 1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?
> Muon's don't calculate things, and even if they did, it would be irrelevant. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > 2) What did the scientists in the scintillator lab calculate the muon's altitude to be at the spacetime event corresponding to the muon's creation?
> As noted above, in terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds.
> > 3) What did the lab scientist read on the scintillator lab's clock at the spacetime event corresponding to the muon's creation?
> In terms of S, the lab clock read -D/v at event e2. Of course, it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.

On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote:
> > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> >
> > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) that you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
>
> Your e4 is simultaneous is in both your S and your S'. Correct?

Simultaneous with what? Event e4 is the collision event. It is not simultaneous with any of the other events in terms of any standard system of inertial coordinates.

Again, in terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2. Of course, it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.

On Wednesday, May 3, 2023 at 8:05:27 PM UTC-7, Trevor Lange wrote:
> On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote:
> > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > >
> > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) that you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
> >
> > Your e4 is simultaneous is in both your S and your S'. Correct?
> Simultaneous with what? Event e4 is the collision event. It is not simultaneous with any of the other events in terms of any standard system of inertial coordinates.

Just checking Legion, just checking. You are known to be a slippery fellow who delights in introducing your own notation for purposes of obfuscation. Everybody says so.

As you can see out your window, twilight is almost over in Seattle. The Night is falling fast. Ross is probably tucked up in his bed already and is reading hizself to sleep with the CRC Standard Mathematical Tables--you'll notice he never records his YouTube videos at night. I am a bit tire too. If I wake up between now and dawn I may continue the bifurcated and inconsistent histories of Murray the muon. For sure tomorrow.
>
> Again, in terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2. Of course, it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.

On Wednesday, May 3, 2023 at 5:56:42 PM UTC-7, Trevor Lange wrote:
> On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote:
> > Let's explore the history of this spacetime event.
> A spacetime event doesn't have a history, it is simply a specific "point" in spacetime. Of course, the causal past of an event consists of the entirety of the past light cone and its interior.
>
> > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
>
> Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) that you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2.
> > 1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?
> Muon's don't calculate things, and even if they did, it would be irrelevant. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > 2) What did the scientists in the scintillator lab calculate the muon's altitude to be at the spacetime event corresponding to the muon's creation?
> As noted above, in terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds.
> > 3) What did the lab scientist read on the scintillator lab's clock at the spacetime event corresponding to the muon's creation?
> In terms of S, the lab clock read -D/v at event e2. Of course, it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.
No, no, no Legion! This simply won't do. All you have done is copied out the schema of the standard Minkowski treatment of spacetime while avoiding my numerical values like a vampire avoids a garlic smeared crucifix. Do you not realize that you must sink your hands into the numerical calculations up to the elbows in order for me to trap you and Einstein in foolhardy inconsistency. You want the spacetime event e1 (the creation of Murray the muon) to not be simultaneous in the lab frame S and Murray's frame S'. Fine, I'm down with that. In fact it is essential to my demonstration. You want the attitudes of Murray's nativity to have different values in each frame too. Good! This is also vital to my demonstration. Go ahead. Put numbers to these things. Then watch me show how we lose all hope of ever demonstrating that Murray's egg timer and the lab clock each correspond to 0.0 usec at e4 (the disintegration of Murray in the lab).

On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote:
> A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.

Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.

In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.

> You want the spacetime event e1 (the creation of the muon) to not be simultaneous
> in the lab frame S and in the muon's frame S'.

Simultaneous with what? Again, e1 is simultaneous with e2 in terms of S, and it is simultaneous with e3 in terms of S'.

> Watch me show how we lose all hope of ever demonstrating that the muon's egg timer
> and the lab clock each correspond to 0.0 usec at e4 (the disintegration of the muon in the lab).

Again, the elapsed proper time of the muon from creation to collision is 2.2 usec, so if it's co-moving clock to reads zero at the collision, it read -2.2 usec at its creation. Likewise if the lab clock reads zero at the collision, it must read -D/v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.

On Thursday, May 4, 2023 at 5:09:40 PM UTC-7, Trevor Lange wrote:
> On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote:
> > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
>
> In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.
>
> > You want the spacetime event e1 (the creation of the muon) to not be simultaneous
> > in the lab frame S and in the muon's frame S'.
>
> Simultaneous with what? Again, e1 is simultaneous with e2 in terms of S, and it is simultaneous with e3 in terms of S'.
>
> > Watch me show how we lose all hope of ever demonstrating that the muon's egg timer
> > and the lab clock each correspond to 0.0 usec at e4 (the disintegration of the muon in the lab).
>
> Again, the elapsed proper time of the muon from creation to collision is 2.2 usec, so if it's co-moving clock to reads zero at the collision, it read -2.2 usec at its creation. Likewise if the lab clock reads zero at the collision, it must read -D/v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.

Legion, do you understand what you are doing? Do you even understand the fallacious style of argumentation that you are using? Here is your *argument*

1) You make the claim to this forum that the Theory of Special Relativity is true.
2) Then you write down the Theory of Special Relativity.
3) After this you reach the conclusion that the Theory of Special Relativity is true.

In order to be scientific, Legion, a theory must be a) falsifiable, and b) must make predictions.
I have provided a specific experiment along the lines of Smith-Frisch, to test the predictions made by SR. First apply the theory of SR to the numbers I have provided to obtain a result. Then I will apply the theory of SR to the same number to arrive at an equally valid result, but completely at odds with the first result. This will demonstrate the inconsistency of SR and thereby falsify it. That is my program here.

On Thursday, May 4, 2023 at 6:06:40 PM UTC-7, patdolan wrote:
> > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
> >
> > In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2.. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.
> >
> > > Watch me show how we lose all hope of ever demonstrating that the muon's egg timer
> > > and the lab clock each correspond to 0.0 usec at e4 (the disintegration of the muon in the lab).
> >
> > Again, the elapsed proper time of the muon from creation to collision is 2.2 usec, so if it's co-moving clock to reads zero at the collision, it read -2.2 usec at its creation. Likewise if the lab clock reads zero at the collision, it must read -D/v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.
>
> I have provided a specific experiment to test the predictions made by SR. First
> apply the theory of SR to the numbers I have provided to obtain a result.

Done. See above, i.e., the lab clock reads -4.4 usec at e2 and -1.1 usec at e3. We covered this before.

> Then I will...

Rather than constantly making grand announcements about what you are about to do, it would be more interesting if you would actually do it.

On Thursday, May 4, 2023 at 7:08:11 PM UTC-7, Trevor Lange wrote:
> On Thursday, May 4, 2023 at 6:06:40 PM UTC-7, patdolan wrote:
> > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
> > >
> > > In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v at event e2, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.
> > >
> > > > Watch me show how we lose all hope of ever demonstrating that the muon's egg timer
> > > > and the lab clock each correspond to 0.0 usec at e4 (the disintegration of the muon in the lab).
> > >
> > > Again, the elapsed proper time of the muon from creation to collision is 2.2 usec, so if it's co-moving clock to reads zero at the collision, it read -2.2 usec at its creation. Likewise if the lab clock reads zero at the collision, it must read -D/v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.
> >
> > I have provided a specific experiment to test the predictions made by SR. First
> > apply the theory of SR to the numbers I have provided to obtain a result.
> Done. See above, i.e., the lab clock reads -4.4 usec at e2 and -1.1 usec at e3. We covered this before.
>
> > Then I will...
>
> Rather than constantly making grand announcements about what you are about to do, it would be more interesting if you would actually do it.
I'm watching the Kraken game. I'll write something tomorrow. Channel 99 CBUT from Vancouver, Ross.

On Thursday, May 4, 2023 at 7:08:11 PM UTC-7, Trevor Lange wrote:
> On Thursday, May 4, 2023 at 6:06:40 PM UTC-7, patdolan wrote:
> > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
> > >
> > > In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g.

Which is equal to 2.2v

This is the magnitude of the interval from e1 to e3. In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock read -D/v

Which is equal to -2.2g = -4.4 usec. [ okay ]

at event e2, and it reads -D/(vg^2) [ huh? ]

Which is equal to -2.2/g = -1.1 usec.

at event e3.

Question: why is there an extra gamma in the denominator at e3?

The elapsed proper time in the lab between e2 and e3 is Dv.
> > >
> > > > Watch me show how we lose all hope of ever demonstrating that the muon's egg timer
> > > > and the lab clock each correspond to 0.0 usec at e4 (the disintegration of the muon in the lab).
> > >
> > > Again, the elapsed proper time of the muon from creation to collision is 2.2 usec, so if it's co-moving clock to reads zero at the collision, it read -2.2 usec at its creation. Likewise if the lab clock reads zero at the collision, it must read -D/v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.
> >
> > I have provided a specific experiment to test the predictions made by SR. First
> > apply the theory of SR to the numbers I have provided to obtain a result.
> Done. See above, i.e., the lab clock reads -4.4 usec at e2 and -1.1 usec at e3. We covered this before.
>
> > Then I will...
>
> Rather than constantly making grand announcements about what you are about to do, it would be more interesting if you would actually do it.

On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> >
> > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
>
> Which is equal to 2.2v

Right! Bravo!

> > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
>
> Which is equal to -2.2g = -4.4 usec. [ okay ]

Right again! You're on fire.

> > and it reads -D/(vg^2) = -1.1 usec at event e3.
>
> Why is there an extra gamma in the denominator at e3?

It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D(vg^2) at event e3, which is simultaneous with e1 in terms of S'.

On Saturday, May 6, 2023 at 12:46:52 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > >
> > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> >
> > Which is equal to 2.2v
> Right! Bravo!
>
> > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> >
> > Which is equal to -2.2g = -4.4 usec. [ okay ]
> Right again! You're on fire.
>
> > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> >
> > Why is there an extra gamma in the denominator at e3?
>
> It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D(vg^2) at event e3, which is simultaneous with e1 in terms of S'.

Let me write out those ( space, time ) coordinates you just gave in S:
e1: 0, 0
e2: 2.2vg, 0
e3: 2.2vg, 2.2(v^2)g
e4: 2.2vg, 2.2g

Also, do you realize that D(vg^2) = 2.2( v^2 )( g^3 )

D/v - Dv = Show me

On Saturday, May 6, 2023 at 12:46:52 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > >
> > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> >
> > Which is equal to 2.2v
> Right! Bravo!
>
> > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> >
> > Which is equal to -2.2g = -4.4 usec. [ okay ]
> Right again! You're on fire.
>
> > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> >
> > Why is there an extra gamma in the denominator at e3?
>
> It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'.

Fixed the obvious typo, in agreement with what has been typed for you a hundred times before (including the very expression that you quoted).

On Saturday, May 6, 2023 at 2:09:25 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 12:46:52 PM UTC-7, Trevor Lange wrote:
> > On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > >
> > > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > >
> > > Which is equal to 2.2v
> > Right! Bravo!
> >
> > > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> > >
> > > Which is equal to -2.2g = -4.4 usec. [ okay ]
> > Right again! You're on fire.
> >
> > > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> > >
> > > Why is there an extra gamma in the denominator at e3?
> >
> > It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'.
>
> Fixed the obvious typo, in agreement with what has been typed for you a hundred times before (including the very expression that you quoted).

Ha! I caught you, Legion Lange. I don't give a damn about the minus sign typo.

On Saturday, May 6, 2023 at 2:34:28 PM UTC-7, patdolan wrote:
> On Saturday, May 6, 2023 at 2:09:25 PM UTC-7, Trevor Lange wrote:
> > On Saturday, May 6, 2023 at 12:46:52 PM UTC-7, Trevor Lange wrote:
> > > On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > > >
> > > > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2.. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > > >
> > > > Which is equal to 2.2v
> > > Right! Bravo!
> > >
> > > > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> > > >
> > > > Which is equal to -2.2g = -4.4 usec. [ okay ]
> > > Right again! You're on fire.
> > >
> > > > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> > > >
> > > > Why is there an extra gamma in the denominator at e3?
> > >
> > > It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'.
> >
> > Fixed the obvious typo, in agreement with what has been typed for you a hundred times before (including the very expression that you quoted).
> Ha! I caught you, Legion Lange. I don't give a damn about the minus sign typo.

Huh? There was no minus sign typo. The typo was the missing division ( / ) symbol. So, is anything still unclear to you?

On Saturday, May 6, 2023 at 2:43:38 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 2:34:28 PM UTC-7, patdolan wrote:
> > On Saturday, May 6, 2023 at 2:09:25 PM UTC-7, Trevor Lange wrote:
> > > On Saturday, May 6, 2023 at 12:46:52 PM UTC-7, Trevor Lange wrote:
> > > > On Saturday, May 6, 2023 at 11:58:13 AM UTC-7, patdolan wrote:
> > > > > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > > > >
> > > > > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > > > >
> > > > > Which is equal to 2.2v
> > > > Right! Bravo!
> > > >
> > > > > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> > > > >
> > > > > Which is equal to -2.2g = -4.4 usec. [ okay ]
> > > > Right again! You're on fire.
> > > >
> > > > > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> > > > >
> > > > > Why is there an extra gamma in the denominator at e3?
> > > >
> > > > It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'.
> > >
> > > Fixed the obvious typo, in agreement with what has been typed for you a hundred times before (including the very expression that you quoted).
> > Ha! I caught you, Legion Lange. I don't give a damn about the minus sign typo.
> Huh? There was no minus sign typo. The typo was the missing division ( / ) symbol. So, is anything still unclear to you?

Yes. D/v - Dv = D/(vg^2) is still very unclear to me.

D/(vg^2) = 2.2 usec/2 = 1.1 usec

4.4 usec - (4.4 usec)v^2 = 1.1 usec ??

Show me!

On Saturday, May 6, 2023 at 3:28:44 PM UTC-7, patdolan wrote:
> > > > > > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > > > > >
> > > > > > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > > > > >
> > > > > > Which is equal to 2.2v
> > > > > Right! Bravo!
> > > > >
> > > > > > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> > > > > >
> > > > > > Which is equal to -2.2g = -4.4 usec. [ okay ]
> > > > > Right again! You're on fire.
> > > > >
> > > > > > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> > > > > >
> > > > > > Why is there an extra gamma in the denominator at e3?
> > > > >
> > > > > It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'. Is anything still unclear to you?
>
> Yes. D/v - Dv = D/(vg^2) is still very unclear to me.

Well, D/v - Dv = D(1/v - v) = D((1-v^2)/v) = D/(v g^2). Remember, g=1/sqrt(1-v^2). This is just simple algebra. What part is unclear?
> D/(vg^2) = 2.2 usec/2 = 1.1 usec

Right! That's the elapsed lab time from e3 to e4, and since you stipulated the clock reads 0.0 at e4, it read -1.1 usec at e3. Likewise the lab clock reads -4.4 usec at e2. The muon clock reads -2.2 usec at e1 and 0.0 at e4. We covered all this before (several times). Is anything still unclear to you?

On Saturday, May 6, 2023 at 3:44:51 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 3:28:44 PM UTC-7, patdolan wrote:
> > > > > > > > > A muon simultaneously [sic] comes to rest [sic] and disintegrates in a laboratory scintillator on the surface of the earth just as the laboratory clock strikes 0.000 hours. The muon's trajectory was normal to the surface of the earth. The muon's velocity relative to the earth [sic] was measured to be [v=]0.867c which results in g = 2. The muon's clock showed an elapsed proper time of 2.2 microseconds between the spacetime event corresponding to its creation and the spacetime event corresponding to its disintegration in the lab scintillator.
> > > > > > > >
> > > > > > > > Letting S and S' denote standard inertial coordinate systems in which the lab and the muon in flight are at rest, respectively, four relevant events (among infinitely many) you might be interested in are the muon's creation event e1, the lab event e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4. In terms of S the height of the muon at creation is D = (2.2)(vg) where g = 1/sqrt(1-v^2). This is the magnitude of the interval from e1 to e2. In terms of S', the spatial distance between the muon and the lab at the time of the muon's creation is D/g. This is the magnitude of the interval from e1 to e3.
> > > > > > >
> > > > > > > Which is equal to 2.2v
> > > > > > Right! Bravo!
> > > > > >
> > > > > > > > In terms of S, the altitude of the muon's creation event is D, which in your example is 2.2(vg) light microseconds. In terms of S, the lab clock reads -D/v at event e2 ...
> > > > > > >
> > > > > > > Which is equal to -2.2g = -4.4 usec. [ okay ]
> > > > > > Right again! You're on fire.
> > > > > >
> > > > > > > > and it reads -D/(vg^2) = -1.1 usec at event e3.
> > > > > > >
> > > > > > > Why is there an extra gamma in the denominator at e3?
> > > > > >
> > > > > > It isn't an "extra" factor. For convenience, place the origin of S at the muon creation event, so the events e1, e2, e3, and e4 have the coordinates (0,0), (D,0), (D,Dv), and (D,D/v) respectively. So the elapsed time in the lab from e3 to e4 is D/v - Dv, which equals D/(vg^2). Thus for the clock to read 0.0 at the collision, it must read -D/(vg^2) at event e3, which is simultaneous with e1 in terms of S'. Is anything still unclear to you?
> >
> > Yes. D/v - Dv = D/(vg^2) is still very unclear to me.
> Well, D/v - Dv = D(1/v - v) = D((1-v^2)/v) = D/(v g^2). Remember, g=1/sqrt(1-v^2). This is just simple algebra. What part is unclear?
> > D/(vg^2) = 2.2 usec/2 = 1.1 usec
> Right! That's the elapsed lab time from e3 to e4, and since you stipulated the clock reads 0.0 at e4, it read -1.1 usec at e3. Likewise the lab clock reads -4.4 usec at e2. The muon clock reads -2.2 usec at e1 and 0.0 at e4.. We covered all this before (several times). Is anything still unclear to you?
At laboratory event e2, which is simultaneous with e1 in the lab's frame, the scientist start's the lab's one and only stopwatch.

At laboratory event e3, which is simultaneous with e1 in the muon's frame, the scientist starts the lab's one and only stopwatch.

What does that stopwatch read at the muon's disintegration at e4?

On Saturday, May 6, 2023 at 7:50:26 PM UTC-7, patdolan wrote:
> At laboratory event e2, which is simultaneous with e1 in the lab's frame,
> the scientist start's the lab's one and only stopwatch.

So, at e2 the lab guy initializes his stopwatch to 0.0 and lets it run.

> At laboratory event e3, which is simultaneous with e1 in the muon's frame,
> the scientist starts the lab's one and only stopwatch.

So, by the time he reaches e3, the stopwatch will have ticked off 3.3 usec, and you are saying at this point he re-initializes the stop watch to 0.0 and lets it run.
> What does that stopwatch read at the muon's disintegration at e4?

It reads 1.1 usec, consistent with the fact that the elapsed proper time from e2 to e3 is 3.3 usec, and the elapsed proper time from e3 to e4 is 1.1 usec, and the total elapsed proper time from e2 to e4 is 4.4 usec. Also, the elapsed proper time on the muon's clock from e1 to e4 is 2.2 usec. Is anything still unclear to you?

On Saturday, May 6, 2023 at 8:07:37 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 7:50:26 PM UTC-7, patdolan wrote:
> > At laboratory event e2, which is simultaneous with e1 in the lab's frame,
> > the scientist start's the lab's one and only stopwatch.
> So, at e2 the lab guy initializes his stopwatch to 0.0 and lets it run.
> > At laboratory event e3, which is simultaneous with e1 in the muon's frame,
> > the scientist starts the lab's one and only stopwatch.
> So, by the time he reaches e3, the stopwatch will have ticked off 3.3 usec, and you are saying at this point he re-initializes the stop watch to 0.0 and lets it run.
> > What does that stopwatch read at the muon's disintegration at e4?
> It reads 1.1 usec, consistent with the fact that the elapsed proper time from e2 to e3 is 3.3 usec, and the elapsed proper time from e3 to e4 is 1.1 usec, and the total elapsed proper time from e2 to e4 is 4.4 usec. Also, the elapsed proper time on the muon's clock from e1 to e4 is 2.2 usec. Is anything still unclear to you?

Yes Legion. Teach me the "many worlds" interpretation of SR which you have just propounded. This is the weakness of your "four event formula" that Roberts, Bodkin and Paul Anderson wouldn't even take the time to study. But I DID take the time. And now I present you with it's flaw. Is this unclear to you?

On Saturday, May 6, 2023 at 8:19:48 PM UTC-7, patdolan wrote:
> > > At laboratory event e2, which is simultaneous with e1 in the lab's frame,
> > > the scientist start's the lab's one and only stopwatch.
> > So, at e2 the lab guy initializes his stopwatch to 0.0 and lets it run.
> > > At laboratory event e3, which is simultaneous with e1 in the muon's frame,
> > > the scientist starts the lab's one and only stopwatch.
> > So, by the time he reaches e3, the stopwatch will have ticked off 3.3 usec, and you are saying at this point he re-initializes the stop watch to 0..0 and lets it run.
> > > What does that stopwatch read at the muon's disintegration at e4?
> > It reads 1.1 usec, consistent with the fact that the elapsed proper time from e2 to e3 is 3.3 usec, and the elapsed proper time from e3 to e4 is 1..1 usec, and the total elapsed proper time from e2 to e4 is 4.4 usec. Also, the elapsed proper time on the muon's clock from e1 to e4 is 2.2 usec. Is anything still unclear to you?
>
> Yes. Please teach me the "many worlds" interpretation of SR which you have just propounded.

There is no "many worlds" interpretation of special relativity, and I obviously have not propounded any such thing. What specifically is unclear to you about the spatio-temporal relations in the above scenario?

> And now I present you with it's flaw: [then.... crickets]

Apparently your message was truncated at this crucial point, just as you were about to point out a flaw in the simple and clear explanation of the scenario. Could you try posting again, and this time be sure to actually present the flaw?

[I know, I know... you'll get to it tomorrow... as always. yawn]

On Saturday, May 6, 2023 at 8:32:58 PM UTC-7, Trevor Lange wrote:
> On Saturday, May 6, 2023 at 8:19:48 PM UTC-7, patdolan wrote:
> > > > At laboratory event e2, which is simultaneous with e1 in the lab's frame,
> > > > the scientist start's the lab's one and only stopwatch.
> > > So, at e2 the lab guy initializes his stopwatch to 0.0 and lets it run.
> > > > At laboratory event e3, which is simultaneous with e1 in the muon's frame,
> > > > the scientist starts the lab's one and only stopwatch.
> > > So, by the time he reaches e3, the stopwatch will have ticked off 3.3 usec, and you are saying at this point he re-initializes the stop watch to 0.0 and lets it run.
> > > > What does that stopwatch read at the muon's disintegration at e4?
> > > It reads 1.1 usec, consistent with the fact that the elapsed proper time from e2 to e3 is 3.3 usec, and the elapsed proper time from e3 to e4 is 1.1 usec, and the total elapsed proper time from e2 to e4 is 4.4 usec. Also, the elapsed proper time on the muon's clock from e1 to e4 is 2.2 usec. Is anything still unclear to you?
> >
> > Yes. Please teach me the "many worlds" interpretation of SR which you have just propounded.
>
> There is no "many worlds" interpretation of special relativity, and I obviously have not propounded any such thing. What specifically is unclear to you about the spatio-temporal relations in the above scenario?
>
> > And now I present you with it's flaw: [then.... crickets]
>
> Apparently your message was truncated at this crucial point, just as you were about to point out a flaw in the simple and clear explanation of the scenario. Could you try posting again, and this time be sure to actually present the flaw?
>
> [I know, I know... you'll get to it tomorrow... as always. yawn]
Rest up Legion. You'll have a big day ahead of you Monday...maybe Sunday night. I'm headed to the Hen in Green Lake to hear music that is too old played by a band that is too young. Come up from Renton if you like. It will only take you 20 minutes this time of night. You will recognize me instantly. You too Ross. I'll buy you guys a drink. But nothing fancy. I only buy well.