Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.

I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.

To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity.. We begin…

A) Spacetime interval from muon inception until decay from the point of view of the lab technician.

A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:

∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6

B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.

A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:

∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5

Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.

On 28-Jan-22 2:52 pm, patdolan wrote:
> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>
> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>
> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
>
> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
>
> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
>
> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
>
> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
>
> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
>
> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
>
>
> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.

In the muon's frame, the two events occur in the same place (i.e. where
the muon is), and there is no distance between them.

Sylvia.

On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>
> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>
> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity.. We begin…
>
> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
>
> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
>
> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
>
> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
>
> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
>
> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
>
>
> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.

Hmm....

Root-mean-squared - about geometric average - leads to revisiting the derivation,
what with respect to output, has that two points each have local conformal metrics,
while, the distance between them is a symmetric reflection, that at the same time
they're also each symmetric reflections. I.e., re-derivation after symmetry of the
notation in a potential field of coordinates and classical field of coordinates, that
the ds^2 as a line element is both ways - makes for why it's fundamental for
the nonadiabatic, that also happens to be conservative in symmetry flex.

On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >
> > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >
> > To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity.. We begin…
> >
> > A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> >
> > A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> >
> > ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> >
> > B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> >
> > A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> >
> > ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> >
> >
> > Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> Hmm....
>
> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> what with respect to output, has that two points each have local conformal metrics,
> while, the distance between them is a symmetric reflection, that at the same time
> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> notation in a potential field of coordinates and classical field of coordinates, that
> the ds^2 as a line element is both ways - makes for why it's fundamental for
> the nonadiabatic, that also happens to be conservative in symmetry flex.
No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?

I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?

(I will redeem my shame with this question)

On Thursday, January 27, 2022 at 8:22:38 PM UTC-8, patdolan wrote:
> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> > On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> > >
> > > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> > >
> > > To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> > >
> > > A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> > >
> > > A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> > >
> > > ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> > >
> > > B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> > >
> > > A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> > >
> > > ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> > >
> > >
> > > Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> > Hmm....
> >
> > Root-mean-squared - about geometric average - leads to revisiting the derivation,
> > what with respect to output, has that two points each have local conformal metrics,
> > while, the distance between them is a symmetric reflection, that at the same time
> > they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> > notation in a potential field of coordinates and classical field of coordinates, that
> > the ds^2 as a line element is both ways - makes for why it's fundamental for
> > the nonadiabatic, that also happens to be conservative in symmetry flex..
> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
>
> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
>
> (I will redeem my shame with this question)

Muons oscillate.

On Thursday, January 27, 2022 at 8:26:52 PM UTC-8, Ross A. Finlayson wrote:
> On Thursday, January 27, 2022 at 8:22:38 PM UTC-8, patdolan wrote:
> > On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> > > On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > > > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> > > >
> > > > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> > > >
> > > > To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> > > >
> > > > A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> > > >
> > > > A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> > > >
> > > > ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> > > >
> > > > B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> > > >
> > > > A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> > > >
> > > > ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> > > >
> > > >
> > > > Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> > > Hmm....
> > >
> > > Root-mean-squared - about geometric average - leads to revisiting the derivation,
> > > what with respect to output, has that two points each have local conformal metrics,
> > > while, the distance between them is a symmetric reflection, that at the same time
> > > they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> > > notation in a potential field of coordinates and classical field of coordinates, that
> > > the ds^2 as a line element is both ways - makes for why it's fundamental for
> > > the nonadiabatic, that also happens to be conservative in symmetry flex.
> > No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> >
> > I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> >
> > (I will redeem my shame with this question)
> Muons oscillate.
Ross, old chap, you seem to be turning into a multisyllabic Mitch. Take better care of yourself.

On 28-Jan-22 3:22 pm, patdolan wrote:
> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>>>
>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>>>
>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
>>>
>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
>>>
>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
>>>
>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
>>>
>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
>>>
>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
>>>
>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
>>>
>>>
>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
>> Hmm....
>>
>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
>> what with respect to output, has that two points each have local conformal metrics,
>> while, the distance between them is a symmetric reflection, that at the same time
>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
>> notation in a potential field of coordinates and classical field of coordinates, that
>> the ds^2 as a line element is both ways - makes for why it's fundamental for
>> the nonadiabatic, that also happens to be conservative in symmetry flex.
> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
>
> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
>
> (I will redeem my shame with this question)

It's a rather ambiguous question. If the muon pulls out a telescope at
the moment of its creation, and watches the clock, then it will see the
hands move by some amount between the events.

If the question is about what the hands point to in the frame of the
muon, at the two events, then there's a different answer, albeit one
almost devoid of physical meaning.

Neither represents a period of time in the frame of the muon, and thus
neither have any relevance to the space-time interval calculation for
the muon, which depends only on the time between the two events in the
muon's frame.

Sylvia.

On Thursday, January 27, 2022 at 8:31:26 PM UTC-8, patdolan wrote:
> On Thursday, January 27, 2022 at 8:26:52 PM UTC-8, Ross A. Finlayson wrote:
> > On Thursday, January 27, 2022 at 8:22:38 PM UTC-8, patdolan wrote:
> > > On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> > > > On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > > > > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> > > > >
> > > > > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> > > > >
> > > > > To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> > > > >
> > > > > A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> > > > >
> > > > > A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> > > > >
> > > > > ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> > > > >
> > > > > B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> > > > >
> > > > > A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> > > > >
> > > > > ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> > > > >
> > > > >
> > > > > Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> > > > Hmm....
> > > >
> > > > Root-mean-squared - about geometric average - leads to revisiting the derivation,
> > > > what with respect to output, has that two points each have local conformal metrics,
> > > > while, the distance between them is a symmetric reflection, that at the same time
> > > > they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> > > > notation in a potential field of coordinates and classical field of coordinates, that
> > > > the ds^2 as a line element is both ways - makes for why it's fundamental for
> > > > the nonadiabatic, that also happens to be conservative in symmetry flex.
> > > No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> > >
> > > I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> > >
> > > (I will redeem my shame with this question)
> > Muons oscillate.
> Ross, old chap, you seem to be turning into a multisyllabic Mitch. Take better care of yourself.

No, Mitch is a bit more of a balker, with underdefined, overloaded, and ambiguous terms,
I just point to modern neutrino physics, and encourage you to read about F-mean.

https://en.wikipedia.org/wiki/Quasi-arithmetic_mean
https://en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities

Though, I understand why it might feel that way.

If you think about ds^2 as about how the triangle inequality builds the line integral,
i.e. I assume you understand the path is defined that way, or derived, then you
should understand that it's not the same as "-ds"^2. This is about the outer product
besides the inner product.

The quasi-invariant for measure theory is a very modern thing, and foundations
of mathematics is to support building it up. If these are unknowns to you I'd
encourage learning more about them and particularly what are the key derivations
as what so result in things.

Or, you know, feel free to model muons as you have it but you should know that
muon physics has come a long way since the 1960's.

Muon physics fits under "exotic" particles and are defined first-class to placehold
their properties. A true theory of everything has them all derived.

I.e. there's no need of "muons" to frame your problem, and, muons oscillate,
which you've left out.

It's like the old question: "is the Kerr metric: defined or derived?". Which you
can read among my framework of opinion in philosophy of science.

On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
> On 28-Jan-22 3:22 pm, patdolan wrote:
> > On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> >> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> >>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >>>
> >>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >>>
> >>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> >>>
> >>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> >>>
> >>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> >>>
> >>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> >>>
> >>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> >>>
> >>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> >>>
> >>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> >>>
> >>>
> >>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> >> Hmm....
> >>
> >> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> >> what with respect to output, has that two points each have local conformal metrics,
> >> while, the distance between them is a symmetric reflection, that at the same time
> >> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> >> notation in a potential field of coordinates and classical field of coordinates, that
> >> the ds^2 as a line element is both ways - makes for why it's fundamental for
> >> the nonadiabatic, that also happens to be conservative in symmetry flex.
> > No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> >
> > I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> >
> > (I will redeem my shame with this question)
> It's a rather ambiguous question. If the muon pulls out a telescope at
> the moment of its creation, and watches the clock, then it will see the
> hands move by some amount between the events.
>
> If the question is about what the hands point to in the frame of the
> muon, at the two events, then there's a different answer, albeit one
> almost devoid of physical meaning.
>
> Neither represents a period of time in the frame of the muon, and thus
> neither have any relevance to the space-time interval calculation for
> the muon, which depends only on the time between the two events in the
> muon's frame.
>
> Sylvia.
I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.

On Thursday, January 27, 2022 at 8:54:05 PM UTC-8, patdolan wrote:
> On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
> > On 28-Jan-22 3:22 pm, patdolan wrote:
> > > On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> > >> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > >>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> > >>>
> > >>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> > >>>
> > >>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> > >>>
> > >>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> > >>>
> > >>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> > >>>
> > >>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> > >>>
> > >>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> > >>>
> > >>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> > >>>
> > >>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> > >>>
> > >>>
> > >>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> > >> Hmm....
> > >>
> > >> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> > >> what with respect to output, has that two points each have local conformal metrics,
> > >> while, the distance between them is a symmetric reflection, that at the same time
> > >> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> > >> notation in a potential field of coordinates and classical field of coordinates, that
> > >> the ds^2 as a line element is both ways - makes for why it's fundamental for
> > >> the nonadiabatic, that also happens to be conservative in symmetry flex.
> > > No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> > >
> > > I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> > >
> > > (I will redeem my shame with this question)
> > It's a rather ambiguous question. If the muon pulls out a telescope at
> > the moment of its creation, and watches the clock, then it will see the
> > hands move by some amount between the events.
> >
> > If the question is about what the hands point to in the frame of the
> > muon, at the two events, then there's a different answer, albeit one
> > almost devoid of physical meaning.
> >
> > Neither represents a period of time in the frame of the muon, and thus
> > neither have any relevance to the space-time interval calculation for
> > the muon, which depends only on the time between the two events in the
> > muon's frame.
> >
> > Sylvia.
> I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.

Sylvia, let's break the situation down into small components in hopes of providing some physical meaning. Let's start with what the muon will observe the Lab clock's rate to be, compared to it's own rate. With a relative velocity of .867c, wouldn't it be plausible that the muon would conclude that the Lab clock was ticking off time at a rate only one half of its own?

On 28-Jan-22 3:54 pm, patdolan wrote:
> On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
>> On 28-Jan-22 3:22 pm, patdolan wrote:
>>> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
>>>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
>>>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>>>>>
>>>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>>>>>
>>>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
>>>>>
>>>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
>>>>>
>>>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
>>>>>
>>>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
>>>>>
>>>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
>>>>>
>>>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
>>>>>
>>>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
>>>>>
>>>>>
>>>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
>>>> Hmm....
>>>>
>>>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
>>>> what with respect to output, has that two points each have local conformal metrics,
>>>> while, the distance between them is a symmetric reflection, that at the same time
>>>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
>>>> notation in a potential field of coordinates and classical field of coordinates, that
>>>> the ds^2 as a line element is both ways - makes for why it's fundamental for
>>>> the nonadiabatic, that also happens to be conservative in symmetry flex.
>>> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
>>>
>>> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
>>>
>>> (I will redeem my shame with this question)
>> It's a rather ambiguous question. If the muon pulls out a telescope at
>> the moment of its creation, and watches the clock, then it will see the
>> hands move by some amount between the events.
>>
>> If the question is about what the hands point to in the frame of the
>> muon, at the two events, then there's a different answer, albeit one
>> almost devoid of physical meaning.
>>
>> Neither represents a period of time in the frame of the muon, and thus
>> neither have any relevance to the space-time interval calculation for
>> the muon, which depends only on the time between the two events in the
>> muon's frame.
>>
>> Sylvia.
> I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.

In the frame of the laboratory, it takes about 3.8 microseconds for the
light from the clock to reach the point where the muon is created. So
the image of the clock where the muon is created is 3.8 microseconds
behind what the clock is showing. It takes 4.4 microseconds for the muon
to get to the laboratory during which time the clock advances by that
amount.

So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
travel. Since the clock is moving in the muon's frame, this will be
related to the time taken in the muon's frame by the relativistic
Doppler affect.

There is no problem with physical meaning of this.

But once one moves from seeing light, to calculating what is happening
"now" in some frame, one has moved away from things that can be
measured, and are instead looking at only part of the calculation of
what would be measured. One can arrive at the same conclusion of what
would be measured by different routes, with different intermediate
values, whose physical meaning is a matter for philosophy[*].

This all comes down to algebraic juggling of the Lorentz transformation,
and really conveys no new information.

Sylvia.

[*] This is a polite way of saying it's worthless.

On Thursday, January 27, 2022 at 9:25:59 PM UTC-8, Sylvia Else wrote:
> On 28-Jan-22 3:54 pm, patdolan wrote:
> > On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
> >> On 28-Jan-22 3:22 pm, patdolan wrote:
> >>> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> >>>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> >>>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >>>>>
> >>>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >>>>>
> >>>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> >>>>>
> >>>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> >>>>>
> >>>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> >>>>>
> >>>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> >>>>>
> >>>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> >>>>>
> >>>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> >>>>>
> >>>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> >>>>>
> >>>>>
> >>>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> >>>> Hmm....
> >>>>
> >>>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> >>>> what with respect to output, has that two points each have local conformal metrics,
> >>>> while, the distance between them is a symmetric reflection, that at the same time
> >>>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> >>>> notation in a potential field of coordinates and classical field of coordinates, that
> >>>> the ds^2 as a line element is both ways - makes for why it's fundamental for
> >>>> the nonadiabatic, that also happens to be conservative in symmetry flex.
> >>> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> >>>
> >>> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> >>>
> >>> (I will redeem my shame with this question)
> >> It's a rather ambiguous question. If the muon pulls out a telescope at
> >> the moment of its creation, and watches the clock, then it will see the
> >> hands move by some amount between the events.
> >>
> >> If the question is about what the hands point to in the frame of the
> >> muon, at the two events, then there's a different answer, albeit one
> >> almost devoid of physical meaning.
> >>
> >> Neither represents a period of time in the frame of the muon, and thus
> >> neither have any relevance to the space-time interval calculation for
> >> the muon, which depends only on the time between the two events in the
> >> muon's frame.
> >>
> >> Sylvia.
> > I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.
> In the frame of the laboratory, it takes about 3.8 microseconds for the
> light from the clock to reach the point where the muon is created. So
> the image of the clock where the muon is created is 3.8 microseconds
> behind what the clock is showing. It takes 4.4 microseconds for the muon
> to get to the laboratory during which time the clock advances by that
> amount.
>
> So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
> travel. Since the clock is moving in the muon's frame, this will be
> related to the time taken in the muon's frame by the relativistic
> Doppler affect.
>
> There is no problem with physical meaning of this.
>
> But once one moves from seeing light, to calculating what is happening
> "now" in some frame, one has moved away from things that can be
> measured, and are instead looking at only part of the calculation of
> what would be measured. One can arrive at the same conclusion of what
> would be measured by different routes, with different intermediate
> values, whose physical meaning is a matter for philosophy[*].

Does this criticism of your also apply to Frisch and Smith's original work on muon lifespan dilation? If not, why not? Perhaps it is the case that, as you point out, special relativity's greatest confirmation is little more that one particular measurement by one particular route and is therefor more of a result of philosophy[*] than solid metrology. Do you disagree?
>
> This all comes down to algebraic juggling of the Lorentz transformation,
> and really conveys no new information.
>
> Sylvia.
>
> [*] This is a polite way of saying it's worthless.

On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, crank patdolan wrote:
> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>
> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>

This is nonsense, Pattycakes, spacetime is frame invariant.

On Thursday, January 27, 2022 at 9:39:15 PM UTC-8, Dono. wrote:
> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, crank patdolan wrote:
> > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >
> > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >
> This is nonsense, Pattycakes, spacetime is frame invariant.
I have already conceded to Sylvia that the algebra of spacetime intervals is consistent. I am now pressing Sylvia on the point that the LTs and the spacetime intervals they imply cannot actually represent a consistent universe. The muon and the lab technician will always disagree about how much time has elapsed on the Lab clock at the moment of the muon's decay in the lab's scintillator. I am slowly bringing Sylvia along with baby steps to this realization. The usually precise and terse Sylvia is having difficulty at the moment--her word salads are the tell.

On Thursday, January 27, 2022 at 9:25:59 PM UTC-8, Sylvia Else wrote:
> On 28-Jan-22 3:54 pm, patdolan wrote:
> > On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
> >> On 28-Jan-22 3:22 pm, patdolan wrote:
> >>> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> >>>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> >>>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >>>>>
> >>>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >>>>>
> >>>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> >>>>>
> >>>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> >>>>>
> >>>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> >>>>>
> >>>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> >>>>>
> >>>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> >>>>>
> >>>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> >>>>>
> >>>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> >>>>>
> >>>>>
> >>>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> >>>> Hmm....
> >>>>
> >>>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> >>>> what with respect to output, has that two points each have local conformal metrics,
> >>>> while, the distance between them is a symmetric reflection, that at the same time
> >>>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> >>>> notation in a potential field of coordinates and classical field of coordinates, that
> >>>> the ds^2 as a line element is both ways - makes for why it's fundamental for
> >>>> the nonadiabatic, that also happens to be conservative in symmetry flex.
> >>> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> >>>
> >>> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> >>>
> >>> (I will redeem my shame with this question)
> >> It's a rather ambiguous question. If the muon pulls out a telescope at
> >> the moment of its creation, and watches the clock, then it will see the
> >> hands move by some amount between the events.
> >>
> >> If the question is about what the hands point to in the frame of the
> >> muon, at the two events, then there's a different answer, albeit one
> >> almost devoid of physical meaning.
> >>
> >> Neither represents a period of time in the frame of the muon, and thus
> >> neither have any relevance to the space-time interval calculation for
> >> the muon, which depends only on the time between the two events in the
> >> muon's frame.
> >>
> >> Sylvia.
> > I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.
> In the frame of the laboratory, it takes about 3.8 microseconds for the
> light from the clock to reach the point where the muon is created. So
> the image of the clock where the muon is created is 3.8 microseconds
> behind what the clock is showing. It takes 4.4 microseconds for the muon
> to get to the laboratory during which time the clock advances by that
> amount.
>
> So the muon sees the clock advance by 3.8 + 4.4 microseconds during its

Sylvia, you appear to have made as big a blunder as I did earlier this evening when you type "So the muon sees the clock advance by 3.8 + 4.4 microseconds during its travel". Surely the lab clock has been slowed by a factor of gamma as far a the muon is concerned. N'es pas? Would you like to submit another total time?
> travel. Since the clock is moving in the muon's frame, this will be
> related to the time taken in the muon's frame by the relativistic
> Doppler affect.
>
> There is no problem with physical meaning of this.
>
> But once one moves from seeing light, to calculating what is happening
> "now" in some frame, one has moved away from things that can be
> measured, and are instead looking at only part of the calculation of
> what would be measured. One can arrive at the same conclusion of what
> would be measured by different routes, with different intermediate
> values, whose physical meaning is a matter for philosophy[*].
>
> This all comes down to algebraic juggling of the Lorentz transformation,
> and really conveys no new information.
>
> Sylvia.
>
> [*] This is a polite way of saying it's worthless.

On Thursday, January 27, 2022 at 9:54:51 PM UTC-8, hardened patdolan wrote:
> I am now pressing Sylvia on the point that the LTs and the spacetime intervals they imply cannot actually represent a consistent universe.

Rubbish.

On Friday, 28 January 2022 at 05:06:23 UTC+1, Sylvia Else wrote:

> In the muon's frame, the two events occur in the same place (i.e. where
> the muon is), and there is no distance between them.

In the meantime in the real world, however, forbidden
by Your insane religion TAI keep measuring t'=t, just like
all serious clocks always did.

On Friday, 28 January 2022 at 06:25:59 UTC+1, Sylvia Else wrote:

> So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
> travel.

And as muon sees - it's an OBSERVATIONAL FACT!!! Thus,
observational facts confirm that it has to be so.

On Thursday, January 27, 2022 at 11:30:32 PM UTC-8, maluw...@gmail.com wrote:
> On Friday, 28 January 2022 at 06:25:59 UTC+1, Sylvia Else wrote:
>
> > So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
> > travel.
> And as muon sees - it's an OBSERVATIONAL FACT!!! Thus,
> observational facts confirm that it has to be so.
Peace maluw, peace. Sylvia has made a monstrous mistake when she proclaims the sum 3.8 + 4.4. One thing we know from Lorentz time dilation is that the muon must experience the lab clock as running half the speed of its own internal clock because of a gamma equal to 2. We also know that the muon's internal clock is wired to trigger a time bomb which brings the muon certain death after 2.2 microseconds of proper time. From this we can logically conclude that the maximum the Sylvia sum can be is 3.8 + 1.1

Sylvia now finds her skirts tangled in Lorentz's briar patch. There is no escape. And I quite suspect that rather than wait around for we hyenas to pounce, she will instead choose to run away naked, never to be heard from again on this thread. I know her ways.

Do you hear a rooting and snorting sound? It might be Bodkin. Let us see what the morrow brings...

On Thursday, January 27, 2022 at 8:53:12 PM UTC-8, Ross A. Finlayson wrote:
> On Thursday, January 27, 2022 at 8:31:26 PM UTC-8, patdolan wrote:
> > On Thursday, January 27, 2022 at 8:26:52 PM UTC-8, Ross A. Finlayson wrote:
> > > On Thursday, January 27, 2022 at 8:22:38 PM UTC-8, patdolan wrote:
> > > > On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> > > > > On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> > > > > > Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> > > > > >
> > > > > > I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> > > > > >
> > > > > > To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> > > > > >
> > > > > > A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> > > > > >
> > > > > > A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> > > > > >
> > > > > > ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> > > > > >
> > > > > > B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> > > > > >
> > > > > > A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> > > > > >
> > > > > > ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> > > > > >
> > > > > >
> > > > > > Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> > > > > Hmm....
> > > > >
> > > > > Root-mean-squared - about geometric average - leads to revisiting the derivation,
> > > > > what with respect to output, has that two points each have local conformal metrics,
> > > > > while, the distance between them is a symmetric reflection, that at the same time
> > > > > they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> > > > > notation in a potential field of coordinates and classical field of coordinates, that
> > > > > the ds^2 as a line element is both ways - makes for why it's fundamental for
> > > > > the nonadiabatic, that also happens to be conservative in symmetry flex.
> > > > No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> > > >
> > > > I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> > > >
> > > > (I will redeem my shame with this question)
> > > Muons oscillate.
> > Ross, old chap, you seem to be turning into a multisyllabic Mitch. Take better care of yourself.
> No, Mitch is a bit more of a balker, with underdefined, overloaded, and ambiguous terms,
> I just point to modern neutrino physics, and encourage you to read about F-mean.
>
> https://en.wikipedia.org/wiki/Quasi-arithmetic_mean
> https://en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities
>
> Though, I understand why it might feel that way.
>
> If you think about ds^2 as about how the triangle inequality builds the line integral,
> i.e. I assume you understand the path is defined that way, or derived, then you
> should understand that it's not the same as "-ds"^2. This is about the outer product
> besides the inner product.
>
>
>
> The quasi-invariant for measure theory is a very modern thing, and foundations
> of mathematics is to support building it up. If these are unknowns to you I'd
> encourage learning more about them and particularly what are the key derivations
> as what so result in things.
>
> Or, you know, feel free to model muons as you have it but you should know that
> muon physics has come a long way since the 1960's.
>
> Muon physics fits under "exotic" particles and are defined first-class to placehold
> their properties. A true theory of everything has them all derived.
>
> I.e. there's no need of "muons" to frame your problem, and, muons oscillate,
> which you've left out.
>
>
>
> It's like the old question: "is the Kerr metric: defined or derived?". Which you
> can read among my framework of opinion in philosophy of science.

https://news.medill.northwestern.edu/chicago/fermilab-following-clues-to-unravel-the-latest-mysteries-of-the-universe/

"New research regarding muons — small, electron-like fundamental particles
that only survive for 2 millionths of a second before they decay — recently
made headlines around the world. Their behavior suggests an 18th particle
because the muon shows sensitivity to forces that were not accounted for by
current theories."

Srivastava: "Of course muons can oscillate."

https://3quarksdaily.com/3quarksdaily/2021/04/the-slightly-wrong-physics-of-spinning-muons.html

Quoting from this blog:
"An explainer comic from the American Physical Society stressed
how “new discoveries are on the horizon,” and Résonaances, a
blog widely read by physicists, noted the deluge of new g-2 papers,
and found hints that might “open a new experimental era that is
almost too beautiful to imagine.”"

"Essentially, this oscillation shows how the muon spin precesses
relative to the magnetic field, which is quantified via the muon’s
magnetic dipole moment."

"As every young physics student learns, certain absurdities lurk
in quantum analogies that are reified too ambitiously. The muon
is not a loop of wire, even if it is sometimes helpful to imagine it
as one. Many revolutions in physics (including, famously, Newton’s
law of universal gravitation and Maxwell’s equations of electrodynamics)
required their creators to leave old analogies behind."

https://physics.aps.org/articles/v14/47

On 28-Jan-22 5:12 pm, patdolan wrote:
> On Thursday, January 27, 2022 at 9:25:59 PM UTC-8, Sylvia Else wrote:
>> On 28-Jan-22 3:54 pm, patdolan wrote:
>>> On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
>>>> On 28-Jan-22 3:22 pm, patdolan wrote:
>>>>> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
>>>>>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
>>>>>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
>>>>>>>
>>>>>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
>>>>>>>
>>>>>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
>>>>>>>
>>>>>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
>>>>>>>
>>>>>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
>>>>>>>
>>>>>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
>>>>>>>
>>>>>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
>>>>>>>
>>>>>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
>>>>>>>
>>>>>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
>>>>>>>
>>>>>>>
>>>>>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
>>>>>> Hmm....
>>>>>>
>>>>>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
>>>>>> what with respect to output, has that two points each have local conformal metrics,
>>>>>> while, the distance between them is a symmetric reflection, that at the same time
>>>>>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
>>>>>> notation in a potential field of coordinates and classical field of coordinates, that
>>>>>> the ds^2 as a line element is both ways - makes for why it's fundamental for
>>>>>> the nonadiabatic, that also happens to be conservative in symmetry flex.
>>>>> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
>>>>>
>>>>> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
>>>>>
>>>>> (I will redeem my shame with this question)
>>>> It's a rather ambiguous question. If the muon pulls out a telescope at
>>>> the moment of its creation, and watches the clock, then it will see the
>>>> hands move by some amount between the events.
>>>>
>>>> If the question is about what the hands point to in the frame of the
>>>> muon, at the two events, then there's a different answer, albeit one
>>>> almost devoid of physical meaning.
>>>>
>>>> Neither represents a period of time in the frame of the muon, and thus
>>>> neither have any relevance to the space-time interval calculation for
>>>> the muon, which depends only on the time between the two events in the
>>>> muon's frame.
>>>>
>>>> Sylvia.
>>> I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.
>> In the frame of the laboratory, it takes about 3.8 microseconds for the
>> light from the clock to reach the point where the muon is created. So
>> the image of the clock where the muon is created is 3.8 microseconds
>> behind what the clock is showing. It takes 4.4 microseconds for the muon
>> to get to the laboratory during which time the clock advances by that
>> amount.
>>
>> So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
>
> Sylvia, you appear to have made as big a blunder as I did earlier this evening when you type "So the muon sees the clock advance by 3.8 + 4.4 microseconds during its travel". Surely the lab clock has been slowed by a factor of gamma as far a the muon is concerned. N'es pas? Would you like to submit another total time?

One event is when the light departs from clock in time to arrive at the
muon's creation.

A second event is when the muon arrives in the vicinity of the clock.

The clock must advance by the time between those two events in the
clock's frame, or it's a useless clock. The muon must see the clock
advance by that amount since it is present when the light arrives at the
muon's creation, and when the muon arrives at the clock.

The amount the clock is seen to advance by the muon is not the same as
the time elapsed in the frame of the muon, and gamma has nothing to do
with it.

Sylvia

On 28-Jan-22 4:54 pm, patdolan wrote:
> On Thursday, January 27, 2022 at 9:39:15 PM UTC-8, Dono. wrote:
>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, crank patdolan
>> wrote:
>>> Inspired by Richard's many challenge problems, I now submit a
>>> little conundrum of my own for this forum's consideration.
>>>
>>> I have made the startling discovery that the concept of the
>>> spacetime interval, conceived by Minchumpski, does not work in
>>> the general case. That is to say observers in different frames of
>>> reference will disagree on the spacetime interval between two
>>> time-like events. Minchumpski's formula only works for space-like
>>> events. I made this discovery whilst toying with the Bodkin
>>> Equalities. I may expatiate on this in another place.
>>>
>> This is nonsense, Pattycakes, spacetime is frame invariant.
> I have already conceded to Sylvia that the algebra of spacetime
> intervals is consistent. I am now pressing Sylvia on the point that
> the LTs and the spacetime intervals they imply cannot actually
> represent a consistent universe. The muon and the lab technician
> will always disagree about how much time has elapsed on the Lab clock
> at the moment of the muon's decay in the lab's scintillator. I am
> slowly bringing Sylvia along with baby steps to this realization.

I think not. All you're doing is continuing to illustrate your muddled
thinking.

Sylvia.

On Friday, 28 January 2022 at 09:45:26 UTC+1, Sylvia Else wrote:
> On 28-Jan-22 5:12 pm, patdolan wrote:
> > On Thursday, January 27, 2022 at 9:25:59 PM UTC-8, Sylvia Else wrote:
> >> On 28-Jan-22 3:54 pm, patdolan wrote:
> >>> On Thursday, January 27, 2022 at 8:43:56 PM UTC-8, Sylvia Else wrote:
> >>>> On 28-Jan-22 3:22 pm, patdolan wrote:
> >>>>> On Thursday, January 27, 2022 at 8:11:58 PM UTC-8, Ross A. Finlayson wrote:
> >>>>>> On Thursday, January 27, 2022 at 7:53:01 PM UTC-8, patdolan wrote:
> >>>>>>> Inspired by Richard's many challenge problems, I now submit a little conundrum of my own for this forum's consideration.
> >>>>>>>
> >>>>>>> I have made the startling discovery that the concept of the spacetime interval, conceived by Minchumpski, does not work in the general case. That is to say observers in different frames of reference will disagree on the spacetime interval between two time-like events. Minchumpski's formula only works for space-like events. I made this discovery whilst toying with the Bodkin Equalities. I may expatiate on this in another place.
> >>>>>>>
> >>>>>>> To demonstrate the nonsense of the uniqueness of spacetime interval ∆s for a give pair of events, we shall take as an example the most iconic of all proofs of special relativity: the time dilation of the muon’s mean lifetime. We shall take as our two spacetime events 1) the inception of a muon above the surface of the earth and 2) its decay in a laboratory scintillator at the surface of the earth. We shall calculate the spacetime interval between these two events from the points of view (frames of reference) of the scintillator lab technician and the muon. We will then show that the two intervals are not equal, in violation of special relativity. We begin…
> >>>>>>>
> >>>>>>> A) Spacetime interval from muon inception until decay from the point of view of the lab technician.
> >>>>>>>
> >>>>>>> A hadron from deep space collides with an air molecule in the earth’s atmosphere producing a muon at an altitude of 1143.12 meters above the laboratory. This newly formed muon retains enough velocity from the collision so that it is now traveling at .867c relative to the earth and directly towards the laboratory scintillator. At this velocity and distance the muon will take 4.4 microseconds to come to a rest inside the scintillator. But this muon is extremely average and so it will decay in 2.2 microseconds proper time. However, the muon makes it twice as far as it ought, all the way to the scintillator because gamma = 2. So in the lab technician’s frame of reference the muon lives for 4.4 microseconds. Let’s calculate the spacetime interval between these two events from the point of view of the lab technician:
> >>>>>>>
> >>>>>>> ∆s^2 = [ 3x10^8 m/s X 4.4x10^-6 s ]^2 - [ 2.6x10^8 m/s x 4.4x10^-6 s ]^2 = 4.2x10^6
> >>>>>>>
> >>>>>>> B) Spacetime interval calculation from muon inception until decay from the point of view of the muon.
> >>>>>>>
> >>>>>>> A hadron in deep space finds itself directly in the path of a speeding earth which is traveling at .867c relative to the hadron. As the earth collides with the hadron, the first point of contact is an air molecule in the earth’s atmosphere which according to the hadron is only 571.56 meters above the surface of the Lorentz-contracted earth and its Lorentz-contracted atmosphere. This collision turns the hadron into a muon. The earth continue to barrel towards the newly born muon. 2.2 microseconds later when the muon finally impacts the surface of the earth, it is at a point where there is a scintillator inside a laboratory. The muon comes to rest inside the scintillator just as it decays. Let’s calculate the spacetime interval between these two events from the point of view of the muon:
> >>>>>>>
> >>>>>>> ∆s^2 = [ 3x10^8 m/s X 2.2x10^-6 s ]^2 - [ 2.6x10^8 m/s x 2.2x10^-6 s ]^2 = 1.09x10^5
> >>>>>>>
> >>>>>>>
> >>>>>>> Same two events. Two entirely different spacetime intervals. The spacetime interval is officially in shambles. You are welcome.
> >>>>>> Hmm....
> >>>>>>
> >>>>>> Root-mean-squared - about geometric average - leads to revisiting the derivation,
> >>>>>> what with respect to output, has that two points each have local conformal metrics,
> >>>>>> while, the distance between them is a symmetric reflection, that at the same time
> >>>>>> they're also each symmetric reflections. I.e., re-derivation after symmetry of the
> >>>>>> notation in a potential field of coordinates and classical field of coordinates, that
> >>>>>> the ds^2 as a line element is both ways - makes for why it's fundamental for
> >>>>>> the nonadiabatic, that also happens to be conservative in symmetry flex.
> >>>>> No Ross. It's as simple as Sylvia has made it out to be. She has beaten me a second time. Or has she?
> >>>>>
> >>>>> I believe Sylvia will concur that that the Lab technician will have observed that 4.4 microseconds have elapsed on his laboratory clock between the two events. I put this question to Sylvia: How many seconds will the muon observe have elapsed on the laboratory clock between the two events?
> >>>>>
> >>>>> (I will redeem my shame with this question)
> >>>> It's a rather ambiguous question. If the muon pulls out a telescope at
> >>>> the moment of its creation, and watches the clock, then it will see the
> >>>> hands move by some amount between the events.
> >>>>
> >>>> If the question is about what the hands point to in the frame of the
> >>>> muon, at the two events, then there's a different answer, albeit one
> >>>> almost devoid of physical meaning.
> >>>>
> >>>> Neither represents a period of time in the frame of the muon, and thus
> >>>> neither have any relevance to the space-time interval calculation for
> >>>> the muon, which depends only on the time between the two events in the
> >>>> muon's frame.
> >>>>
> >>>> Sylvia.
> >>> I am intrigued by the movement of the hands that you say are devoid of physical meaning. Can we put the space-time interval aside for a moment and you give us your speculations about the movement of hands observed by the muon. I'm open to any speculation you might make.
> >> In the frame of the laboratory, it takes about 3.8 microseconds for the
> >> light from the clock to reach the point where the muon is created. So
> >> the image of the clock where the muon is created is 3.8 microseconds
> >> behind what the clock is showing. It takes 4.4 microseconds for the muon
> >> to get to the laboratory during which time the clock advances by that
> >> amount.
> >>
> >> So the muon sees the clock advance by 3.8 + 4.4 microseconds during its
> >
> > Sylvia, you appear to have made as big a blunder as I did earlier this evening when you type "So the muon sees the clock advance by 3.8 + 4.4 microseconds during its travel". Surely the lab clock has been slowed by a factor of gamma as far a the muon is concerned. N'es pas? Would you like to submit another total time?
> One event is when the light departs from clock in time to arrive at the
> muon's creation.
>
> A second event is when the muon arrives in the vicinity of the clock.
>
> The clock must advance by the time between those two events in the
> clock's frame, or it's a useless clock. The muon must see the clock
> advance by that amount since it is present when the light arrives at the
> muon's creation, and when the muon arrives at the clock.
> The amount the clock is seen to advance by the muon is not the same as
> the time elapsed in the frame of the muon

Muon sees it! So, it's an observational fact! Don't dare
denying what is clearly seen! By a muon!

Well, if a human was seeing it, it could still be a delusion
of some kind. But how could a muon experience delusions?
If a muon sees - it must be true.

Le 28/01/2022 à 04:52, patdolan a écrit :

> I now submit a little conundrum of my own for this forum's consideration.

You are welcome.

R.H.

Le 28/01/2022 à 05:06, Sylvia Else a écrit :

> In the muon's frame, the two events occur in the same place (i.e. where
> the muon is), and there is no distance between them.

Yes.

Thanks Sylvia.
This seems trivial, but it is very important to clarify.
We have the same thing in the accelerated mediums (see the traveler of Tau
Ceti).
We can consider that the rocket is fixed in its frame of reference (even
accelerated) and that it is space that moves around (space is a reference
mollusc).
But there is here a very important thing to notice and on which I press.
The proper time of an object (or of an observer for itself) is what is
called real time (Tr).
Why?
Because he does not move. He is still in the same place. It does not
undergo any temporal aberration due to chronotropy and anisochrony.
It is not like an observer placed in a laboratory and who observes a
proton moving at speed v for a certain proper time of the proton (Tr)
which is the real one.
The physicist does not measure me Tr, nor Vr. (real time, real speed).
His vision is distorted, and he measures To and Vo.
Two things distort its measurements and concepts. The anisochrony which
falsifies them as (1+cosµ.v/c), and the chronotropy which falsifies them
as sqrt(1-v²/c²). Change of position and change of speed distort the
observations. Only proper times and proper velocities are true.
Note that the proper speed (real speed of a particle) is the ratio between
the distance traveled in the laboratory, and the time measured in the
reference frame of the particle.
It's a weirdness of relativistic physics, and it looks like you're
dividing carrots by turnips, but that's how the universe is made.
And that's exactly how it should be done.

>
> Sylvia.

R.H.

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