On Friday, September 1, 2023 at 10:16:50 PM UTC-7, Bill wrote:
> On Friday, September 1, 2023 at 7:54:08 PM UTC-7, patdolan wrote:
> > > > > > > > Frisch and Smith state that the muons are traveling one thousand feet per usec.
> > > > > > > > at between .990c and .995c.
> > > > > > >
> > > > > > > Right, the ones that decay in the scintillator are the ones that were moving that fast when they entered the calibrated stack of iron, and got slowed down to zero speed when reaching the scintillator. You were suggesting the "extra" mesons at sea level were just produced by cosmic rays between the mountain top and sea level, or else just above the mountain top with a higher speed that made them undetected, but both of those suggestions are utterly braindead, because (1) they explicitly point out that they accounted for the (very small) number of muons created betqween mountain top and sea level, which turns out to be negligibly small, and (2) the whole point is to filter out mesons outside that speed range.
> > > > > >
> > > > > > You are forgetting all the muons with velocities greater than .995c that pass
> > > > > > through the scintillator and don't get caught there.
> > > > >
> > > > > No, those are not "forgotten". Sheesh. The depth of iron bars was re-calibrated at sea level to account for the effect of the atmosphere down to sea level, so that the mesons being slowed to zero at sea level are the same slice of the population that were at 0.995c and slowed to zero at the mountain top. In other words, the mesons that were filtered out at the mountain top were also filtered out at sea level. They were not "forgotten". I say again: Sheesh.
> > > >
> > > > The iron depth recalibration for sea level is obviously wrong.
> > >
> > > No, that's based on knowledge of how passage through iron and air affects the energy of the particles. If you're interested in that aspect of the phenomena, go ahead and study it.
> > >
> > > > honest experimentalists would add iron on top of Mt. Washington to
> > > > where there are no more "got-aways".
> > >
> > > Huh? That makes no sense at all. The objective is to examine time dilation, which depends on velocity, so we need to select particles with a particular velocity (energy), and then examine this same population at sea level, which requires compensating for the slowing of the air mass, i.e., it requires us not to be ignorant of this aspect of particle behavior.
> > >
> > > > [The experimenters] probably tinkered with the iron pile for months trying to get
> > > > their numbers right...
> > >
> > > That's doesn't make any sense. They state specifically how much they change the iron layers to compensate for the 6000 ft of air, based on knowledge of how passage through iron and air affect the energies of those particles. This presented in plain daylight for any knowledgeable person to check. My goodness, you objection comes down to "the bastards probably falsified the results".
> > > >
> > > > And as is the case with Eddington an his bent starlight, so too with the FSX:
> > > > no one has ever been able to duplicate their results.
> > >
> > > Both of those assertions are blatantly false. In particular, duplicating the Frisch ex is a undergraduate lab exercise. Basing your belief system on lies is not a good idea.
> >
> > On the right side of Figure 6a of the Frisch-Smith paper we find that 350 out
> > of 568 muons disintegrated in less than the published 2.2 usecs. muon half-life.
> Huh? What's shown on the right side of that figure is the expected number of surviving muons after a noted amount of proper time, given that they have a proper half-life of 2.2 usecs. The elapsed coordinate time is 6.4 usecs, so if that was also the elapsed proper time the expected number of survivors would be about 27, whereas if the proper time of the muons at that speed is 0.7 usecs (which is the predicted relativistic proper time at that speed) the expected number would be about 412. The measured number is 412.
> > the average height above the summit of Mt. WA at which these muons
> > were born was no more than 2,200 ft.
> No, almost all were created much higher, but the height at which they were created is fairly irrelevant. The density of muons in the narrow speed/energy range around 0.995 (before entering the iron layers) at the mountain top is what's measured at the mountain top, with that specific depth of iron..
> > Most first time readers of the FSX are overwhelmed by the huge number
> > of moving parts of this experiment.
> Huh? This is an extremely simple demonstration, performed with very simple equipment.
> > 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.
Legion, below is your best relativity work product ever...for anyone who wants an airtight defeater for SR.

> Let 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.

Don't worry. We will get to your ingenious 4-event spacetime scenario in due course. But over on the Pion thread. That's the thread where SR "supposedly" still has a fighting chance. Have you watched Double-Talk Don Lincoln's Fermilab video as instructed? We will fight this battle on his pion hole in the ground. Forget about the Frisch and Smith's Muons of Mt. Washington. They are as lost to your cause as my Lorentz contraction velocity article is to Wikipedia. Let dispense with them one last time. It will only take one Proustian sentence:

Thanks to your cosmic ray flux vs. altitude figures, we conclude that the iconic Frisch-Smith muon data is easily explained as the simple ratio of the muon production within an atmospheric layer beginning at sea level and extending 2,200 feet (or 2.2 usecs. ) above Cambridge Massachusetts, to the muon production within a 2,200 foot high atmospheric layer beginning at the summit of Mt. Washington.

> > 1) What did the muon calculate its proper altitude above the earth to be at the spacetime event corresponding to its creation?
> Again, 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 micro-seconds.
> > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> By the way, based on your confusion over the Frisch experiment, I suspect one of your underlying problems is that you don't understand the proper half-life of an existing muon that was created, say, 5 usec ago. This is probably why you are confused about when the muons "must have been created" in the Frisch experiment.
> > how far will the average muon travel before disintegrating?
> Ah, here you confirm my diagnosis of the source of your confusion. The question for you is, for an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?

On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > Let 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?
> > Again, 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 micro-seconds.
> > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv.. Do you understand these things?
>
> We will get to your ingenious 4-event spacetime scenario in due course.

LOL. Running away as always.

> The Frisch-Smith muon data is easily explained...

No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?

On Saturday, September 2, 2023 at 12:57:08 AM UTC-7, Bill wrote:
> On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > > Let 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?
> > > Again, 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 micro-seconds.
> > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> >
> > We will get to your ingenious 4-event spacetime scenario in due course.
> LOL. Running away as always.
>
> > The Frisch-Smith muon data is easily explained...
>
> No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?
Legion, you logical fool. It is YOU who does not understand that concealed in that question is the assumption that SR is the correct interpretation of the experimental results.

Here is my answer. Just like a very old German Shepherd of 20 years, a muon that has reached the venerable old age of 5 usec has done so by stochastic chance.

The majority of the FSX muon disintegration blips (350 out of 568) occurs between a fraction of a usec and 2 usecs. This is consistent with those muons being created within atmospheric layer that lies between the summit of Mt. Washington and 2,200 feet above the summit. The other 200 disintegration blips are long-lived stochastic outliers also created in said layer. When you protest "That's a lot of outliers" I retort that just as many short-lived outliers were created in said layer. But they did not live long enough to make it down to the summit. Now do you understand?

On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> On Saturday, September 2, 2023 at 12:57:08 AM UTC-7, Bill wrote:
> > On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > > > Let 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?
> > > > Again, 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 micro-seconds.
> > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > >
> > > We will get to your ingenious 4-event spacetime scenario in due course.
> > LOL. Running away as always.
> >
> > > The Frisch-Smith muon data is easily explained...
> >
> > No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?
> Legion, you logical fool. It is YOU who does not understand that concealed in that question is the assumption that SR is the correct interpretation of the experimental results.
>
> Here is my answer. Just like a very old German Shepherd of 20 years, a muon that has reached the venerable old age of 5 usec has done so by stochastic chance.
>
> The majority of the FSX muon disintegration blips (350 out of 568) occurs between a fraction of a usec and 2 usecs. This is consistent with those muons being created within atmospheric layer that lies between the summit of Mt. Washington and 2,200 feet above the summit. The other 200 disintegration blips are long-lived stochastic outliers also created in said layer. When you protest "That's a lot of outliers" I retort that just as many short-lived outliers were created in said layer. But they did not live long enough to make it down to the summit. Now do you understand?
Or they were blocked by the iron pile. The Gaussian for muons is quite broad.

On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> On Saturday, September 2, 2023 at 12:57:08 AM UTC-7, Bill wrote:
> > On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > > > Let 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?
> > > > Again, 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 micro-seconds.
> > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > >
> > > We will get to your ingenious 4-event spacetime scenario in due course.
> > LOL. Running away as always.
> >
> > > The Frisch-Smith muon data is easily explained...
> >
> > No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?
> Legion, you logical fool. It is YOU who does not understand that concealed in that question is the assumption that SR is the correct interpretation of the experimental results.
>
> Here is my answer. Just like a very old German Shepherd of 20 years, a muon that has reached the venerable old age of 5 usec has done so by stochastic chance.
>
> The majority of the FSX muon disintegration blips (350 out of 568) occurs between a fraction of a usec and 2 usecs. This is consistent with those muons being created within atmospheric layer that lies between the summit of Mt. Washington and 2,200 feet above the summit. The other 200 disintegration blips are long-lived stochastic outliers also created in said layer. When you protest "That's a lot of outliers" I retort that just as many short-lived outliers were created in said layer. But they did not live long enough to make it down to the summit. Now do you understand?
Or the short-lived outliers were blocked by the iron pile before making it to the iron pile. The Gaussian for muons is quite broad.

On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> On Saturday, September 2, 2023 at 12:57:08 AM UTC-7, Bill wrote:
> > On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > > > Let 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?
> > > > Again, 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 micro-seconds.
> > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > >
> > > We will get to your ingenious 4-event spacetime scenario in due course.
> > LOL. Running away as always.
> >
> > > The Frisch-Smith muon data is easily explained...
> >
> > No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?
> Legion, you logical fool. It is YOU who does not understand that concealed in that question is the assumption that SR is the correct interpretation of the experimental results.
>
> Here is my answer. Just like a very old German Shepherd of 20 years, a muon that has reached the venerable old age of 5 usec has done so by stochastic chance.
>
> The majority of the FSX muon disintegration blips (350 out of 568) occurs between a fraction of a usec and 2 usecs. This is consistent with those muons being created within atmospheric layer that lies between the summit of Mt. Washington and 2,200 feet above the summit. The other 200 disintegration blips are long-lived stochastic outliers also created in said layer. When you protest "That's a lot of outliers" I retort that just as many short-lived outliers were created in said layer. But they did not live long enough to make it down to the summit. Now do you understand?

Or the short-lived outliers were blocked by the iron pile before making it to the scintillator. The Gaussian for muons is quite broad.

On Saturday, September 2, 2023 at 7:52:42 AM UTC-7, patdolan wrote:
> On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> > On Saturday, September 2, 2023 at 12:57:08 AM UTC-7, Bill wrote:
> > > On Friday, September 1, 2023 at 10:47:24 PM UTC-7, patdolan wrote:
> > > > > Let 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?
> > > > > Again, 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 micro-seconds.
> > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > >
> > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > LOL. Running away as always.
> > >
> > > > The Frisch-Smith muon data is easily explained...
> > >
> > > No, you misunderstand, and the string of words you typed is wrong for the reasons patiently explained to you several times. Again, you won't be able to understand until you can answer this very simple question: For an existing muon that was created 5 proper usec ago, what is its expected proper time (from now) to decay?
> > Legion, you logical fool. It is YOU who does not understand that concealed in that question is the assumption that SR is the correct interpretation of the experimental results.
> >
> > Here is my answer. Just like a very old German Shepherd of 20 years, a muon that has reached the venerable old age of 5 usec has done so by stochastic chance.
> >
> > The majority of the FSX muon disintegration blips (350 out of 568) occurs between a fraction of a usec and 2 usecs. This is consistent with those muons being created within atmospheric layer that lies between the summit of Mt. Washington and 2,200 feet above the summit. The other 200 disintegration blips are long-lived stochastic outliers also created in said layer. When you protest "That's a lot of outliers" I retort that just as many short-lived outliers were created in said layer. But they did not live long enough to make it down to the summit. Now do you understand?
> Or the short-lived outliers were blocked by the iron pile before making it to the scintillator. The Gaussian for muons is quite broad.
I move that this forum vote yet again on the whose muon narrative, Legion's or mine, is more believe...wait!...I change the question on which to vote....

I rise to request that this forum vote on the following question: whose muon narrative, Frisch-Smith's or mine, requires the fewest assumptions about the FSX data?

On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> > > > Let 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?
> > > > Again, 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 micro-seconds.
> > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > >
> > > We will get to your ingenious 4-event spacetime scenario in due course.
> > LOL. Running away as always.

Still running.

On May 13 , patdolan wrote:
> I'll respond later. I am dead tired after a day of razor clamming on the coast.

That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.

> > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > ago, what is its expected time (from now) to decay?
>
> Just like a very old German Shepherd of 20 years, a muon that has reached the
> venerable old age of 5 usec has done so by stochastic chance.

So, the answer to the question would be.... ? (Day 3 for this one.)

On Saturday, September 2, 2023 at 8:38:47 AM UTC-7, Bill wrote:
> On Saturday, September 2, 2023 at 7:43:15 AM UTC-7, patdolan wrote:
> > > > > Let 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?
> > > > > Again, 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 micro-seconds.
> > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > >
> > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > LOL. Running away as always.
> Still running.
>
> On May 13 , patdolan wrote:
> > I'll respond later. I am dead tired after a day of razor clamming on the coast.
>
> That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.
>
> > > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > > ago, what is its expected time (from now) to decay?
> >
> > Just like a very old German Shepherd of 20 years, a muon that has reached the
> > venerable old age of 5 usec has done so by stochastic chance.
> So, the answer to the question would be.... ? (Day 3 for this one.)
Ah! then you refuse to even consider, much less respond to my brilliant answer?

Legion, you have found a way to run away in place.

Clamming season is over until fall. But you knew that, being a local.

On Saturday, September 2, 2023 at 8:54:20 AM UTC-7, patdolan wrote:
> > > > > > Let 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?
> > > > > > Again, 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 micro-seconds.
> > > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > > >
> > > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > > LOL. Running away as always.
> > Still running.
> >
> > On May 13 , patdolan wrote:
> > > I'll respond later. I am dead tired after a day of razor clamming on the coast.
> >
> > That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.

Still running.

> > > > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > > > ago, what is its expected time (from now) to decay?
> > >
> > > Just like a very old German Shepherd of 20 years, a muon that has reached the
> > > venerable old age of 5 usec has done so by stochastic chance.
> >
> > So, the answer to the question would be.... ? (Day 3 for this one.)
>
> Ah! then you refuse to even consider, much less respond to my brilliant answer?

You provided no answer. Again, for an existing muon (say, at rest in a given frame) that was created 5 usec ago, what is its expected time (from now) to decay?

On Saturday, September 2, 2023 at 9:28:13 AM UTC-7, Bill wrote:
> On Saturday, September 2, 2023 at 8:54:20 AM UTC-7, patdolan wrote:
> > > > > > > Let 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?
> > > > > > > Again, 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 micro-seconds.
> > > > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > > > >
> > > > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > > > LOL. Running away as always.
> > > Still running.
> > >
> > > On May 13 , patdolan wrote:
> > > > I'll respond later. I am dead tired after a day of razor clamming on the coast.
> > >
> > > That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.
> Still running.
> > > > > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > > > > ago, what is its expected time (from now) to decay?
> > > >
> > > > Just like a very old German Shepherd of 20 years, a muon that has reached the
> > > > venerable old age of 5 usec has done so by stochastic chance.
> > >
> > > So, the answer to the question would be.... ? (Day 3 for this one.)
> >
> > Ah! then you refuse to even consider, much less respond to my brilliant answer?
> You provided no answer. Again, for an existing muon (say, at rest in a given frame) that was created 5 usec ago, what is its expected time (from now) to decay?
Okay, I'll play.

-2.8 usecs. Or 2.8 usecs ago.

And may I request and answer from you? How much lifetime is left for an average man born in the USA 93 years ago?

On Saturday, September 2, 2023 at 9:35:40 AM UTC-7, patdolan wrote:
> > For an existing muon (say, at rest in a given frame) that was created
> 5 usec ago, what is its expected time (from now) to decay?
>
> -2.8 usecs. Or 2.8 usecs ago.

Huh? The question is, if the muon sitting next to you was created 5 usec ago, and you start your timer right now, what is the expected reading on your timer for the muon to decay? The answer is a positive number (unless you think your timer will be going backwards).

> How much lifetime is left for an average man born in the USA 93 years ago?

Life expectancy is 2.96 years.

On Saturday, September 2, 2023 at 9:35:40 AM UTC-7, patdolan wrote:
> On Saturday, September 2, 2023 at 9:28:13 AM UTC-7, Bill wrote:
> > On Saturday, September 2, 2023 at 8:54:20 AM UTC-7, patdolan wrote:
> > > > > > > > Let 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?
> > > > > > > > Again, 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 micro-seconds.
> > > > > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > > > > >
> > > > > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > > > > LOL. Running away as always.
> > > > Still running.
> > > >
> > > > On May 13 , patdolan wrote:
> > > > > I'll respond later. I am dead tired after a day of razor clamming on the coast.
> > > >
> > > > That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.
> > Still running.
> > > > > > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > > > > > ago, what is its expected time (from now) to decay?
> > > > >
> > > > > Just like a very old German Shepherd of 20 years, a muon that has reached the
> > > > > venerable old age of 5 usec has done so by stochastic chance.
> > > >
> > > > So, the answer to the question would be.... ? (Day 3 for this one.)
> > >
> > > Ah! then you refuse to even consider, much less respond to my brilliant answer?
> > You provided no answer. Again, for an existing muon (say, at rest in a given frame) that was created 5 usec ago, what is its expected time (from now) to decay?
> Okay, I'll play.
>
> -2.8 usecs. Or 2.8 usecs ago.
>
> And may I request and answer from you? How much lifetime is left for an average man born in the USA 93 years ago?
At least 20 more years if he does planks and/or other exercises targeting core strength. I've been doing planks for four months now and they work wonders.

On Saturday, September 2, 2023 at 11:16:54 AM UTC-7, Laurence Clark Crossen wrote:
> On Saturday, September 2, 2023 at 9:35:40 AM UTC-7, patdolan wrote:
> > On Saturday, September 2, 2023 at 9:28:13 AM UTC-7, Bill wrote:
> > > On Saturday, September 2, 2023 at 8:54:20 AM UTC-7, patdolan wrote:
> > > > > > > > > Let 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?
> > > > > > > > > Again, 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 micro-seconds.
> > > > > > > > > > 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, and it reads -D/(vg^2) at event e3. The elapsed proper time in the lab between e2 and e3 is Dv. Do you understand these things?
> > > > > > > >
> > > > > > > > We will get to your ingenious 4-event spacetime scenario in due course.
> > > > > > > LOL. Running away as always.
> > > > > Still running.
> > > > >
> > > > > On May 13 , patdolan wrote:
> > > > > > I'll respond later. I am dead tired after a day of razor clamming on the coast.
> > > > >
> > > > > That was May 13, which was 110 days ago. Here we are on Day 111 of your recuperation from the clamming, and still no answer.
> > > Still running.
> > > > > > > For an existing muon (say, at rest in a given frame) that was created 5 usec
> > > > > > > ago, what is its expected time (from now) to decay?
> > > > > >
> > > > > > Just like a very old German Shepherd of 20 years, a muon that has reached the
> > > > > > venerable old age of 5 usec has done so by stochastic chance.
> > > > >
> > > > > So, the answer to the question would be.... ? (Day 3 for this one..)
> > > >
> > > > Ah! then you refuse to even consider, much less respond to my brilliant answer?
> > > You provided no answer. Again, for an existing muon (say, at rest in a given frame) that was created 5 usec ago, what is its expected time (from now) to decay?
> > Okay, I'll play.
> >
> > -2.8 usecs. Or 2.8 usecs ago.
> >
> > And may I request and answer from you? How much lifetime is left for an average man born in the USA 93 years ago?
> At least 20 more years if he does planks and/or other exercises targeting core strength. I've been doing planks for four months now and they work wonders.

A high kinetic energy is the same particle.