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'strajectory 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 eventcorresponding 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?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-
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?
Let's explore the history of this spacetime event.
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'svelocity 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
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?
On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote: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
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
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 evente2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.
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.
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 correspondingA 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
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.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
Your e4 is simultaneous is in both your S and your S'. Correct?
On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote: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
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
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.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
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 creationis 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/(
On Wednesday, May 3, 2023 at 8:14:04 AM UTC-7, patdolan wrote: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
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
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 evente2 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
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 must1) 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.
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'svelocity 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
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'.
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).
On Wednesday, May 3, 2023 at 7:47:32 PM UTC-7, patdolan wrote: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
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
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 e2simultaneous 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
event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.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
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 correspondingA 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
simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.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
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) atIn 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/
at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.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
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.
Then I will...
On Thursday, May 4, 2023 at 6:06:40 PM UTC-7, patdolan wrote: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
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
e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.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
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) atIn 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
v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.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/
I'm watching the Kraken game. I'll write something tomorrow. Channel 99 CBUT from Vancouver, Ross.I have provided a specific experiment to test the predictions made by SR. FirstDone. See above, i.e., the lab clock reads -4.4 usec at e2 and -1.1 usec at e3. We covered this before.
apply the theory of SR to the numbers I have provided to obtain a result.
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 6:06:40 PM UTC-7, patdolan wrote: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
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
e2 simultaneous with e1 in terms of S, the lab event e3 simultaneous with e1 in terms of S', and the collision event e4.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
D/g.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
v at event e2, and -D/(vg^2) at event e3. This was all covered (twice) up above.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/
I have provided a specific experiment to test the predictions made by SR. FirstDone. See above, i.e., the lab clock reads -4.4 usec at e2 and -1.1 usec at e3. We covered this before.
apply the theory of SR to the numbers I have provided to obtain a result.
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.
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 correspondingA 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
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 fromLetting 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
Which is equal to 2.2v
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 ]
and it reads -D/(vg^2) = -1.1 usec at event e3.
Why is there an extra gamma in the denominator at e3?
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