patdolan wrote:

> On Saturday, November 20, 2021 at 7:56:31 AM UTC-8, bodk...@gmail.com
> wrote:
>> If you’re invoking the square root on complex numbers then you have to
>> follow the rules of that function on the foliated complex plane, not the
>> rules you would apply on the real number line.
>>
>> See a book on complex variables and the square root foliation.
>>
> I did follow the foliation rules to the letter. I made two branch cuts on
> the complex plane and solved. You didn't recognize this. […]

Blind leading the blind :-D

PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)

On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Paul Alsing wrote:
>
> > On Friday, November 19, 2021 at 7:04:54 PM UTC-8, patdolan wrote:
> >> On Friday, November 19, 2021 at 3:00:32 PM UTC-8, Thomas 'PointedEars'
> >> Lahn wrote:
> >> > >
> >> > > 1^4 = i^4
> >> >
> >> > That much is true.
> >> >
> >> > > sqrt[ 1^4 ] = sqrt[ i^4 ]
> >
> >> > Your logic is flawed.
> >
> >> How, long ears. How?
> >
> > Because i is not a number, it is a definition.
> No; that statement is
>
> 1. utterly wrong;
> 2. not the reason.
> > The term *i* is "defined" to be 'the square root of -1',
> No, it is NOT; that is a common misconception (unfortunately, it can also be
> found in otherwise great textbooks such as Griffiths’ “Introduction to
> Quantum Mechanics”, at least in the second edition).
>
> Instead, 𝕚 is defined to be the solution (*historically*: “root”) of the
> equation
>
> x² + 1 = 0.
>
> [This equation does not have a solution in the real numbers, because the
> square of every real number is non-negative. Thus, the set of complex
> numbers were invented as a solution to the problem, as it had been done
> with ℝ for ℚ, ℚ for ℤ, and ℤ for ℕ, before.]
>
> This means that 𝕚 is defined as a number whose square is −1: 𝕚² ≔ −1. This
> is not the same thing because there are *two* complex numbers which satisfy
> this condition: 𝕚 and −𝕚.
>
> By definition, 𝕚² = −1. Then (−𝕚)² = (−1 · 𝕚)² = (−1)² i² = 1 · (−1) = −1.
> ∎
>
> Concisely, we can write this as √(−1) = ±𝕚. However, care must be taken how
> the “±” notation is used; and *that* is why Pat Dolan’s logic is flawed.

Okay Long Ears, do you your hero Feynman say. Don't just say it. Show it! Show how you rescue +/-1=-/+1 by the using the notation "with care". Go.
> > It is not a real number,
> True.
> > it is an imaginary number...
> Yes, but: “imaginary” does NOT mean “not real” in the sense of “does not
> exist” in mathematics; it means that it is not an element of the set of real
> numbers, ℝ: 𝕚 ∉ ℝ. It is an element of a superset of the real numbers, the
> set of complex numbers ℂ ⊃ ℝ, where ℂ ≔ {a + b 𝕚 | a, b ∈ ℝ; 𝕚² = −1}:
> 𝕚 ∈ ℂ∖ℝ.
>
>
> PointedEars
> --
> Q: How many theoretical physicists specializing in general relativity
> does it take to change a light bulb?
> A: Two: one to hold the bulb and one to rotate the universe.
> (from: WolframAlpha)

On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Paul Alsing wrote:
>
> > On Friday, November 19, 2021 at 7:04:54 PM UTC-8, patdolan wrote:
> >> On Friday, November 19, 2021 at 3:00:32 PM UTC-8, Thomas 'PointedEars'
> >> Lahn wrote:
> >> > >
> >> > > 1^4 = i^4
> >> >
> >> > That much is true.
> >> >
> >> > > sqrt[ 1^4 ] = sqrt[ i^4 ]
> >
> >> > Your logic is flawed.
> >
> >> How, long ears. How?
> >
> > Because i is not a number, it is a definition.

> No; that statement is
>
> 1. utterly wrong;
> 2. not the reason.

> > The term *i* is "defined" to be 'the square root of -1'

> No, it is NOT; that is a common misconception (unfortunately, it can also be
> found in otherwise great textbooks such as Griffiths’ “Introduction to
> Quantum Mechanics”, at least in the second edition).

Well, you have a LOT of web pages that you need to contact because I can show you a whole lot of them that disagree with you...

https://en.wikipedia.org/wiki/Imaginary_number

"i, which is defined by its property i2 = −1"

https://en.wikipedia.org/wiki/Imaginary_unit

"The imaginary unit or unit imaginary number (i) is a solution to the quadratic equation x2 + 1 = 0"

https://byjus.com/maths/value-of-i/

"The value of i is √-1"

https://mathworld.wolfram.com/i.html

"The imaginary number i (also called the imaginary unit) is defined as the square root of -1"

https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/

"the “imaginary” number i has the property that the square of i is -1"

https://www.mathopenref.com/i.html

"It (i) stands for the square root of negative one"

.... and there are thousands more, but at this point in time, I'm bored...

Paul Alsing wrote:

> On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> > The term *i* is "defined" to be 'the square root of -1'
>> No, it is NOT; that is a common misconception (unfortunately, it can also
>> be found in otherwise great textbooks such as Griffiths’ “Introduction to
>> Quantum Mechanics”, at least in the second edition).
>
> Well, you have a LOT of web pages that you need to contact because I can
> show you a whole lot of them that disagree with you...

As you can see above, either you misunderstand your own sources or they are
wrong.
> https://en.wikipedia.org/wiki/Imaginary_number
>
> "i, which is defined by its property i2 = −1"

This is the same as I said.
> https://en.wikipedia.org/wiki/Imaginary_unit
>
> "The imaginary unit or unit imaginary number (i) is a solution to the
> quadratic equation x2 + 1 = 0"

This is the same as I said.
> https://byjus.com/maths/value-of-i/
>
> "The value of i is √-1"

This is wrong.
> https://mathworld.wolfram.com/i.html
>
> "The imaginary number i (also called the imaginary unit) is defined as the
> square root of -1"

This is wrong, unfortunately. It should be noted, though, that Wolfram
MathWorld is but compilation of statements from textbooks, and some
textbooks oversimplify there, as I just said.
> https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
>
> "the “imaginary” number i has the property that the square of i is -1"

This is the same as I said.
> https://www.mathopenref.com/i.html
>
> "It (i) stands for the square root of negative one"

This is wrong.
> ... and there are thousands more, but at this point in time, I'm bored...

No, you are incompetent and can’t read.

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

Paul Alsing wrote:

> On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> > The term *i* is "defined" to be 'the square root of -1'
>> No, it is NOT; that is a common misconception (unfortunately, it can also
>> be found in otherwise great textbooks such as Griffiths’ “Introduction to
>> Quantum Mechanics”, at least in the second edition).
>
> Well, you have a LOT of web pages that you need to contact because I can
> show you a whole lot of them that disagree with you...

As you can see above, either you misunderstand your own sources or they are
wrong.
> https://en.wikipedia.org/wiki/Imaginary_number
>
> "i, which is defined by its property i2 = −1"

Copy & pray aside, this is the same as I said.
> https://en.wikipedia.org/wiki/Imaginary_unit
>
> "The imaginary unit or unit imaginary number (i) is a solution to the
> quadratic equation x2 + 1 = 0"

Copy & pray aside, this is the same as I said.
> https://byjus.com/maths/value-of-i/
>
> "The value of i is √-1"

This is wrong.
> https://mathworld.wolfram.com/i.html
>
> "The imaginary number i (also called the imaginary unit) is defined as the
> square root of -1"

This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.

It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
> https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
>
> "the “imaginary” number i has the property that the square of i is -1"

This is the same as I said.
> https://www.mathopenref.com/i.html
>
> "It (i) stands for the square root of negative one"

This is wrong.
> ... and there are thousands more, but at this point in time, I'm bored...

No, you are incompetent and can’t read.

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

Paul Alsing wrote:

> On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> > The term *i* is "defined" to be 'the square root of -1'
>> No, it is NOT; that is a common misconception (unfortunately, it can also
>> be found in otherwise great textbooks such as Griffiths’ “Introduction to
>> Quantum Mechanics”, at least in the second edition).
>
> Well, you have a LOT of web pages that you need to contact because I can
> show you a whole lot of them that disagree with you...

As you can see below, either you misunderstand your own sources or they are
wrong.
> https://en.wikipedia.org/wiki/Imaginary_number
>
> "i, which is defined by its property i2 = −1"

Copy & pray aside, this is the same as I said.
> https://en.wikipedia.org/wiki/Imaginary_unit
>
> "The imaginary unit or unit imaginary number (i) is a solution to the
> quadratic equation x2 + 1 = 0"

Copy & pray aside, this is the same as I said.
> https://byjus.com/maths/value-of-i/
>
> "The value of i is √-1"

This is wrong.
> https://mathworld.wolfram.com/i.html
>
> "The imaginary number i (also called the imaginary unit) is defined as the
> square root of -1"

This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.

It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
> https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
>
> "the “imaginary” number i has the property that the square of i is -1"

This is the same as I said.
> https://www.mathopenref.com/i.html
>
> "It (i) stands for the square root of negative one"

This is wrong.
> ... and there are thousands more, but at this point in time, I'm bored...

No, you are incompetent and can’t read.

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Paul Alsing wrote:
>
> > On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars'
> > Lahn wrote:
> >> > The term *i* is "defined" to be 'the square root of -1'
> >> No, it is NOT; that is a common misconception (unfortunately, it can also
> >> be found in otherwise great textbooks such as Griffiths’ “Introduction to
> >> Quantum Mechanics”, at least in the second edition).
> >
> > Well, you have a LOT of web pages that you need to contact because I can
> > show you a whole lot of them that disagree with you...
> As you can see below, either you misunderstand your own sources or they are
> wrong.
> > https://en.wikipedia.org/wiki/Imaginary_number
> >
> > "i, which is defined by its property i2 = −1"
> Copy & pray aside, this is the same as I said.
> > https://en.wikipedia.org/wiki/Imaginary_unit
> >
> > "The imaginary unit or unit imaginary number (i) is a solution to the
> > quadratic equation x2 + 1 = 0"
> Copy & pray aside, this is the same as I said.
> > https://byjus.com/maths/value-of-i/
> >
> > "The value of i is √-1"
> This is wrong.
> > https://mathworld.wolfram.com/i.html
> >
> > "The imaginary number i (also called the imaginary unit) is defined as the
> > square root of -1"
> This is wrong, unfortunately. The argument made there smells of theory
> finding and is questionable.
>
> It should be noted, though, that Wolfram MathWorld is mainly a compilation
> of statements from textbooks, and some textbooks oversimplify there, as I
> just said.
> > https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
> >
> > "the “imaginary” number i has the property that the square of i is -1"
> This is the same as I said.
> > https://www.mathopenref.com/i.html
> >
> > "It (i) stands for the square root of negative one"

> This is wrong.
Ya know what is REALLY WRONG LongEars is what you typed earlier:

"Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict. "

Man! were you wrong about mathematics being unambiguous and unforgivingly strict. This just might make you the chump #2 of the day, behind chump #1 Bodkin.

> > ... and there are thousands more, but at this point in time, I'm bored....
> No, you are incompetent and can’t read.
>
>
> PointedEars
> --
> Q: What happens when electrons lose their energy?
> A: They get Bohr'ed.
>
> (from: WolframAlpha)

Le 20/11/2021 à 23:48, Richard Hachel a écrit :
> Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois même
> plus un post par jour.

justement, la proportion de pénibles n'en serait que plus grande, et de
ce fait d'autant plus repoussante.

--
Fabrice

On Sunday, November 21, 2021 at 1:15:03 AM UTC-8, robby wrote:
> Le 20/11/2021 à 23:48, Richard Hachel a écrit :
> > Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois même
> > plus un post par jour.
> justement, la proportion de pénibles n'en serait que plus grande, et de
> ce fait d'autant plus repoussante.

Si vous avez des idées différentes des miennes, n'hésitez pas à les exprimer, robby.
>
>
> --
> Fabrice

patdolan <patdolan@comcast.net> wrote:
> On Saturday, November 20, 2021 at 7:43:48 AM UTC-8, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Friday, November 19, 2021 at 2:37:48 PM UTC-8, bodk...@gmail.com wrote:
>>>> patdolan <patd...@comcast.net> wrote:
>>>>> Critical Theory comes in several flavors: Critical Law Theory, Critical
>>>>> Literature Theory, Critical History Theory, Critical Race Theory. To
>>>>> these we now add Critical Relativity Theory!
>>>>>
>>>>> In the spirit of Derrida we shall deconstruct the clumsy reasoning of
>>>>> special relativity and separate said reasoning from the algebraic symbols
>>>>> and equations that express it, keeping in mind that mathematics is just
>>>>> another form of rhetorical expression wherein falsity can be expressed
>>>>> every bit as plausibly as the truth.
>>>>>
>>>>> “There is only the text.”— J. Derrida
>>>>>
>>>>> According to special relativity two observers in motion with respect to
>>>>> each other will disagree on each other’s length. They will also disagree
>>>>> on the proper flow of time. But they will always agree on the velocity
>>>>> they have with respect to one another. This is exceedingly strange. How
>>>>> can it be that two relative quantities, space and time, combine to
>>>>> produce an absolute quantity called relative velocity? It is true that SR
>>>>> does have a formula for calculating coordinate velocity; just like it has
>>>>> formulas for calculating coordinate space and coordinate time. But the
>>>>> Einstein velocity addition formula ONLY applies to a third object in
>>>>> motion wrt a pair of FoRs. If that third object happens to be at rest
>>>>> wrt one of the FoRs then Einsteinian velocity reduces to Galilean
>>>>> velocity, albeit subject to the speed limit c. Relativists simply
>>>>> assumed without further justification that if FoR-1 measures a velocity v
>>>>> between itself and FoR-2 then FoR-2 must measure the same numerical value
>>>>> v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
>>>>> trivial assumptions can be monumental when constructing a theory of
>>>>> motion. But a 26 year old would probably not yet have the requisite
>>>>> philosophical sophistication needed to recognize this.
>>>>>
>>>>> Einstein’s choice to make velocity strictly Galilean when calculating the
>>>>> velocity between pairs of FoRs ( yes, it is a choice because it does not
>>>>> follow from either the first or the second postulates ) can be
>>>>> expressed mathematically as
>>>>>
>>>>> ∆x’/∆t’ = ∆x/∆t = v (1)
>>>>>
>>>>> I now raise equation (1) to the level of a postulate and declare it to be
>>>>> the third, and heretofore hidden, postulate of special relativity. In
>>>>> recognizing its own structural Galileanism through this new postulate,
>>>>> special relativity can finally claim to be woke.
>>>>>
>>>>> The problem with the third postulate is that even though it is already
>>>>> assumed in every equation of special relativity, it turns out to be true
>>>>> only when v = c ( the second postulate ) or when v = 0. The third
>>>>> postulate can be demonstrated invalid for all values of v in between.
>>>>> The invalidity of the third postulate causes special relativity fall on
>>>>> it’s algebraic face. Big Bang move over…Not recognizing and
>>>>> acknowledging the third postulate was Einstein’s biggest blunder.
>>>>>
>>>>> Time for some examples.
>>>>>
>>>>>
>>>>> DIRK & DONO
>>>>>
>>>>> Consider two FoRs whose x-axes are parallel and lie very close to one
>>>>> another. The relative velocity between these two FoRs is .866c (γ = 2).
>>>>> Dirk assumes the Lotus position at the origin of one FoR whilst Dono
>>>>> assumes the fetal position at the origin of the other. Both Dirk and
>>>>> Dono and their clocks are glued to the origins of their respective FoRs.
>>>>>
>>>>> Dirk opens one eye and takes note of the meter marks on Dono’s contracted
>>>>> x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
>>>>> contracted to only half as long as his own meter marks. Dirk opens the
>>>>> other eye and observes that Dono’s clock is ticking at only half the rate
>>>>> of his own clock. Dirk begins to count Dono’s meter marks as they race
>>>>> past Dirk’s position. After one year (by Dirk’s clock) of counting
>>>>> Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
>>>>> marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
>>>>> elapsed on Dono’s clock. Dirk now calculates what his coordinate
>>>>> velocity should be according to Dono
>>>>>
>>>>> ( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
>>>>>
>>>>> [ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
>>>>>
>>>>> “Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
>>>>> stipulated to be an absolute .866c with respect to one another, when
>>>>> measured in either FoR.”
>>>>>
>>>>> That is true, according to (1). But remember that (1) is an arbitrary
>>>>> choice made by Einstein when he built his theory. It is no more
>>>>> legitimate a choice than the Dolan FoR coordinate velocity transform (3)
>>>>> for determining one’s FoR coordinate velocity. It is also no less
>>>>> inconsistent. For we immediately see by inspection that the Dolan FoR
>>>>> coordinate velocity is always greater than the relative velocity by a
>>>>> factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
>>>>> velocities clangs with just as much antinomy as Dolan’s transform, as we
>>>>> shall see. Special relativity’s dirty little secret is that it’s hidden
>>>>> third postulate (1) destroys the theory from within.
>>>>>
>>>>> I can see by Dono’s tightening fetal position that he still doesn’t
>>>>> believe me. Very well. We shall prove SR’s mathematical inconsistency
>>>>> in the next example.
>>>>>
>>>>>
>>>>> SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
>>>>>
>>>>> Consider a pair of FoRs whose relative velocity v is some value other
>>>>> than c and other than zero. This is expressed mathematically as
>>>>>
>>>>> v = ∆x’/∆t’ = ∆x/∆t != c (4)
>>>>>
>>>>> and
>>>>>
>>>>> v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
>>>>>
>>>>> The Lorentz transforms allow us to construct the FoR coordinate
>>>>> velocities for pairs of FoRs
>>>>>
>>>>> ∆x’ = γ( ∆x - v∆t )
>>>>>
>>>>> ∆t’ = γ( ∆t - ∆xv/c^2 )
>>>>>
>>>>> ∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
>>>>>
>>>>> We now endeavor to solve (6) for v in hopes of demonstrating the internal
>>>>> consistency of special relativity, i.e, that v = v for all pair of FoRs.
>>>>> We saw how this was not the case in DIRK & DONO.
>>>>>
>>>>> The reader will find that trying to solve for v is hopeless unless we
>>>>> make the substitution
>>>>>
>>>>> v = ∆x/∆t or v = ∆x’/∆t’
>>>>>
>>>>> It does not matter which we use—either substitution is permitted by the
>>>>> third postulate (1).
>>>>>
>>>>> With the substitution made, we eventually arrive at the preposterous results
>>>>>
>>>>> ∆x’/∆t’ = c (7)
>>>>>
>>>>> or
>>>>>
>>>>> ∆x’/∆t’ = 0 (8)
>>>>>
>>>>> The derivation is left as an exercise for the reader. For those needing
>>>>> help with the derivation ( I'm looking at you, Dono ) I will be happy to
>>>>> provide it in another post.
>>>>>
>>>>> The laughable results (7) and (8) directly contradict our assumptions (4)
>>>>> and (5). Furthermore, the results impose the requirement that v is not
>>>>> even a variable—v turns out to be a constant always equal to c or zero.
>>>>> Absolutely absurd.
>>>>>
>>>>> QED.
>>>>>
>>>>> Algebraic relativity is thus reduced to ridiculous rubble by means
>>>>> mathematical reductio ad absurdum. The root cause of special relativity’s
>>>>> spectacular algebraic failure lies in the propositional calculus. I am
>>>>> happy to expatiate on that subject in another post if there is interest.
>>>>>
>>>>> Let’s do one more example—this one ripped from the headlines of experimental physics.
>>>>>
>>>>>
>>>>> MUONS, SCHMUONS!
>>>>>
>>>>> Imagine that you are a hadron in deep space minding your own business
>>>>> when all of a sudden you turn to see a Lorentz-flattened earth coming at
>>>>> you at a velocity of .866c. In the impending collision the first thing
>>>>> to strike you is an air molecule high in the earth’s atmosphere. That
>>>>> molecules knocks the stuffing out of you. What is left of you is now a
>>>>> muon which means that you only have 2.2 microseconds more to live.
>>>>>
>>>>> It turns out that the surface of the flattened earth is exactly 571.56
>>>>> meters away from you, according to your own muon FoR. By a remarkable
>>>>> coincidence there is a flattened scintillator and a flattened clock in a
>>>>> flattened laboratory directly below you on the surface of the flattened
>>>>> earth. You note the time on the flattened lab clock. Because you are a
>>>>> muon, you are your own 2.2 microsecond alarm clock.
>>>>>
>>>>> The surface of the flattened earth continues to speed towards you at
>>>>> .866c. 2.2 microseconds later the flattened earth, lab and scintillator
>>>>> smash into you. Just as you expire in the scintillator you note that
>>>>> only 1.1 microseconds has elapsed on the lab clock.
>>>>
>>>> You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
>>>
>>> No synchronization required. The muon's clock doesn't even enter into
>>> the argument at all. Who said that the lab clock was ever at t=0 or any
>>> other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
>> Elapsed from what moment? How does the lab clock know when to start
>> ticking. It isn’t anywhere near the collision of the hadron and the
>> atmospheric molecule. Is it supposed to get an instantaneous signal from
>> that event?
>>
>
>
>> This is what I mean when I say you haven’t bothered to read any book about
>> relativity. Your “argument” is based on two and only two clocks passing
>> each other, both deeming the other slowed, because that’s the comic book
>> version of relativity. But that isn’t what relativity says. Ask yourself
>> why relativity talks about a *lattice* of synchronized clocks in each
>> frame. Why would you need the lattice?
>>
>> It’s so that there is a clock NEAR the event that you’re tracking the time
>> of.
>>> Quantities of elapsed time are easily determined in special relativity.
>>> Synchronization of clocks doesn't even enter into the argument It's all
>>> about one clock, the lab clock. One clock, two observers (the muon and
>>> the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
>>>
>>> Tell us how synchronization has any bearing on the argument. The muon
>>> sees the lab clock tick off 1.1 microseconds on it's trip down.
>> No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
>> the lab clock. The best the muon can do is either a) monitor its own clock,
>> because its own clock is at ONE location in its own frame, or b) have a
>> LATTICE of synchronized clocks traveling along with the muon, where one of
>> them is far away from the muon, near the lab clock at the time the muon is
>> created.
>>
>> If you have two events that occur at DIFFERENT PLACES in some frame of
>> reference, you need two DIFFERENT clocks near those events. If those clocks
>> are synchronized in this frame, THEN you can measure the elapsed time
>> between the events.
>>
>> You can’t measure the elapsed time between two events that occur in
>> different places with ONE clock.
>>
>> You, Pat Dolan, don’t understand the first thing about relativity. And it
>> shows.
>>
>> And yet you bluster simultaneously that a) you understand relativity very
>> well, and b) that it is incomprehensible.
>
> Bodkin, please declare to this forum that your argument also applies to
> the scientists in the lab--especially to the scientist in the lab of Dr's
> Frisch and Smith.
>
> https://www.youtube.com/watch?v=rbzt8gDSYIM
>
> Frisch and Smith didn't have clocks up there alongside all the muons they
> measured either. So there is NO WAY according to Bodkin that Frisch and
> Smith could have possibly determined whether or not muons experienced
> relativistic time dilation.

On Sunday, November 21, 2021 at 5:44:24 AM UTC-8, bodk...@gmail.com wrote:
> patdolan <patd...@comcast.net> wrote:
> > On Saturday, November 20, 2021 at 7:43:48 AM UTC-8, bodk...@gmail.com wrote:
> >> patdolan <patd...@comcast.net> wrote:
> >>> On Friday, November 19, 2021 at 2:37:48 PM UTC-8, bodk...@gmail.com wrote:
> >>>> patdolan <patd...@comcast.net> wrote:
> >>>>> Critical Theory comes in several flavors: Critical Law Theory, Critical
> >>>>> Literature Theory, Critical History Theory, Critical Race Theory. To
> >>>>> these we now add Critical Relativity Theory!
> >>>>>
> >>>>> In the spirit of Derrida we shall deconstruct the clumsy reasoning of
> >>>>> special relativity and separate said reasoning from the algebraic symbols
> >>>>> and equations that express it, keeping in mind that mathematics is just
> >>>>> another form of rhetorical expression wherein falsity can be expressed
> >>>>> every bit as plausibly as the truth.
> >>>>>
> >>>>> “There is only the text.”— J. Derrida
> >>>>>
> >>>>> According to special relativity two observers in motion with respect to
> >>>>> each other will disagree on each other’s length. They will also disagree
> >>>>> on the proper flow of time. But they will always agree on the velocity
> >>>>> they have with respect to one another. This is exceedingly strange. How
> >>>>> can it be that two relative quantities, space and time, combine to
> >>>>> produce an absolute quantity called relative velocity? It is true that SR
> >>>>> does have a formula for calculating coordinate velocity; just like it has
> >>>>> formulas for calculating coordinate space and coordinate time. But the
> >>>>> Einstein velocity addition formula ONLY applies to a third object in
> >>>>> motion wrt a pair of FoRs. If that third object happens to be at rest
> >>>>> wrt one of the FoRs then Einsteinian velocity reduces to Galilean
> >>>>> velocity, albeit subject to the speed limit c. Relativists simply
> >>>>> assumed without further justification that if FoR-1 measures a velocity v
> >>>>> between itself and FoR-2 then FoR-2 must measure the same numerical value
> >>>>> v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
> >>>>> trivial assumptions can be monumental when constructing a theory of
> >>>>> motion. But a 26 year old would probably not yet have the requisite
> >>>>> philosophical sophistication needed to recognize this.
> >>>>>
> >>>>> Einstein’s choice to make velocity strictly Galilean when calculating the
> >>>>> velocity between pairs of FoRs ( yes, it is a choice because it does not
> >>>>> follow from either the first or the second postulates ) can be
> >>>>> expressed mathematically as
> >>>>>
> >>>>> ∆x’/∆t’ = ∆x/∆t = v (1)
> >>>>>
> >>>>> I now raise equation (1) to the level of a postulate and declare it to be
> >>>>> the third, and heretofore hidden, postulate of special relativity. In
> >>>>> recognizing its own structural Galileanism through this new postulate,
> >>>>> special relativity can finally claim to be woke.
> >>>>>
> >>>>> The problem with the third postulate is that even though it is already
> >>>>> assumed in every equation of special relativity, it turns out to be true
> >>>>> only when v = c ( the second postulate ) or when v = 0. The third
> >>>>> postulate can be demonstrated invalid for all values of v in between.
> >>>>> The invalidity of the third postulate causes special relativity fall on
> >>>>> it’s algebraic face. Big Bang move over…Not recognizing and
> >>>>> acknowledging the third postulate was Einstein’s biggest blunder.
> >>>>>
> >>>>> Time for some examples.
> >>>>>
> >>>>>
> >>>>> DIRK & DONO
> >>>>>
> >>>>> Consider two FoRs whose x-axes are parallel and lie very close to one
> >>>>> another. The relative velocity between these two FoRs is .866c (γ = 2).
> >>>>> Dirk assumes the Lotus position at the origin of one FoR whilst Dono
> >>>>> assumes the fetal position at the origin of the other. Both Dirk and
> >>>>> Dono and their clocks are glued to the origins of their respective FoRs.
> >>>>>
> >>>>> Dirk opens one eye and takes note of the meter marks on Dono’s contracted
> >>>>> x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
> >>>>> contracted to only half as long as his own meter marks. Dirk opens the
> >>>>> other eye and observes that Dono’s clock is ticking at only half the rate
> >>>>> of his own clock. Dirk begins to count Dono’s meter marks as they race
> >>>>> past Dirk’s position. After one year (by Dirk’s clock) of counting
> >>>>> Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
> >>>>> marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
> >>>>> elapsed on Dono’s clock. Dirk now calculates what his coordinate
> >>>>> velocity should be according to Dono
> >>>>>
> >>>>> ( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
> >>>>>
> >>>>> [ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
> >>>>>
> >>>>> “Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
> >>>>> stipulated to be an absolute .866c with respect to one another, when
> >>>>> measured in either FoR.”
> >>>>>
> >>>>> That is true, according to (1). But remember that (1) is an arbitrary
> >>>>> choice made by Einstein when he built his theory. It is no more
> >>>>> legitimate a choice than the Dolan FoR coordinate velocity transform (3)
> >>>>> for determining one’s FoR coordinate velocity. It is also no less
> >>>>> inconsistent. For we immediately see by inspection that the Dolan FoR
> >>>>> coordinate velocity is always greater than the relative velocity by a
> >>>>> factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
> >>>>> velocities clangs with just as much antinomy as Dolan’s transform, as we
> >>>>> shall see. Special relativity’s dirty little secret is that it’s hidden
> >>>>> third postulate (1) destroys the theory from within.
> >>>>>
> >>>>> I can see by Dono’s tightening fetal position that he still doesn’t
> >>>>> believe me. Very well. We shall prove SR’s mathematical inconsistency
> >>>>> in the next example.
> >>>>>
> >>>>>
> >>>>> SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
> >>>>>
> >>>>> Consider a pair of FoRs whose relative velocity v is some value other
> >>>>> than c and other than zero. This is expressed mathematically as
> >>>>>
> >>>>> v = ∆x’/∆t’ = ∆x/∆t != c (4)
> >>>>>
> >>>>> and
> >>>>>
> >>>>> v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
> >>>>>
> >>>>> The Lorentz transforms allow us to construct the FoR coordinate
> >>>>> velocities for pairs of FoRs
> >>>>>
> >>>>> ∆x’ = γ( ∆x - v∆t )
> >>>>>
> >>>>> ∆t’ = γ( ∆t - ∆xv/c^2 )
> >>>>>
> >>>>> ∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
> >>>>>
> >>>>> We now endeavor to solve (6) for v in hopes of demonstrating the internal
> >>>>> consistency of special relativity, i.e, that v = v for all pair of FoRs.
> >>>>> We saw how this was not the case in DIRK & DONO.
> >>>>>
> >>>>> The reader will find that trying to solve for v is hopeless unless we
> >>>>> make the substitution
> >>>>>
> >>>>> v = ∆x/∆t or v = ∆x’/∆t’
> >>>>>
> >>>>> It does not matter which we use—either substitution is permitted by the
> >>>>> third postulate (1).
> >>>>>
> >>>>> With the substitution made, we eventually arrive at the preposterous results
> >>>>>
> >>>>> ∆x’/∆t’ = c (7)
> >>>>>
> >>>>> or
> >>>>>
> >>>>> ∆x’/∆t’ = 0 (8)
> >>>>>
> >>>>> The derivation is left as an exercise for the reader. For those needing
> >>>>> help with the derivation ( I'm looking at you, Dono ) I will be happy to
> >>>>> provide it in another post.
> >>>>>
> >>>>> The laughable results (7) and (8) directly contradict our assumptions (4)
> >>>>> and (5). Furthermore, the results impose the requirement that v is not
> >>>>> even a variable—v turns out to be a constant always equal to c or zero.
> >>>>> Absolutely absurd.
> >>>>>
> >>>>> QED.
> >>>>>
> >>>>> Algebraic relativity is thus reduced to ridiculous rubble by means
> >>>>> mathematical reductio ad absurdum. The root cause of special relativity’s
> >>>>> spectacular algebraic failure lies in the propositional calculus. I am
> >>>>> happy to expatiate on that subject in another post if there is interest.
> >>>>>
> >>>>> Let’s do one more example—this one ripped from the headlines of experimental physics.
> >>>>>
> >>>>>
> >>>>> MUONS, SCHMUONS!
> >>>>>
> >>>>> Imagine that you are a hadron in deep space minding your own business
> >>>>> when all of a sudden you turn to see a Lorentz-flattened earth coming at
> >>>>> you at a velocity of .866c. In the impending collision the first thing
> >>>>> to strike you is an air molecule high in the earth’s atmosphere. That
> >>>>> molecules knocks the stuffing out of you. What is left of you is now a
> >>>>> muon which means that you only have 2.2 microseconds more to live.
> >>>>>
> >>>>> It turns out that the surface of the flattened earth is exactly 571..56
> >>>>> meters away from you, according to your own muon FoR. By a remarkable
> >>>>> coincidence there is a flattened scintillator and a flattened clock in a
> >>>>> flattened laboratory directly below you on the surface of the flattened
> >>>>> earth. You note the time on the flattened lab clock. Because you are a
> >>>>> muon, you are your own 2.2 microsecond alarm clock.
> >>>>>
> >>>>> The surface of the flattened earth continues to speed towards you at
> >>>>> .866c. 2.2 microseconds later the flattened earth, lab and scintillator
> >>>>> smash into you. Just as you expire in the scintillator you note that
> >>>>> only 1.1 microseconds has elapsed on the lab clock.
> >>>>
> >>>> You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
> >>>
> >>> No synchronization required. The muon's clock doesn't even enter into
> >>> the argument at all. Who said that the lab clock was ever at t=0 or any
> >>> other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
> >> Elapsed from what moment? How does the lab clock know when to start
> >> ticking. It isn’t anywhere near the collision of the hadron and the
> >> atmospheric molecule. Is it supposed to get an instantaneous signal from
> >> that event?
> >>
> >
> >
> >> This is what I mean when I say you haven’t bothered to read any book about
> >> relativity. Your “argument” is based on two and only two clocks passing
> >> each other, both deeming the other slowed, because that’s the comic book
> >> version of relativity. But that isn’t what relativity says. Ask yourself
> >> why relativity talks about a *lattice* of synchronized clocks in each
> >> frame. Why would you need the lattice?
> >>
> >> It’s so that there is a clock NEAR the event that you’re tracking the time
> >> of.
> >>> Quantities of elapsed time are easily determined in special relativity.
> >>> Synchronization of clocks doesn't even enter into the argument It's all
> >>> about one clock, the lab clock. One clock, two observers (the muon and
> >>> the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
> >>>
> >>> Tell us how synchronization has any bearing on the argument. The muon
> >>> sees the lab clock tick off 1.1 microseconds on it's trip down.
> >> No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
> >> the lab clock. The best the muon can do is either a) monitor its own clock,
> >> because its own clock is at ONE location in its own frame, or b) have a
> >> LATTICE of synchronized clocks traveling along with the muon, where one of
> >> them is far away from the muon, near the lab clock at the time the muon is
> >> created.
> >>
> >> If you have two events that occur at DIFFERENT PLACES in some frame of
> >> reference, you need two DIFFERENT clocks near those events. If those clocks
> >> are synchronized in this frame, THEN you can measure the elapsed time
> >> between the events.
> >>
> >> You can’t measure the elapsed time between two events that occur in
> >> different places with ONE clock.
> >>
> >> You, Pat Dolan, don’t understand the first thing about relativity. And it
> >> shows.
> >>
> >> And yet you bluster simultaneously that a) you understand relativity very
> >> well, and b) that it is incomprehensible.
> >
> > Bodkin, please declare to this forum that your argument also applies to
> > the scientists in the lab--especially to the scientist in the lab of Dr's
> > Frisch and Smith.
> >
> > https://www.youtube.com/watch?v=rbzt8gDSYIM
> >

On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Paul Alsing wrote:
>
> > On Saturday, November 20, 2021 at 2:54:59 PM UTC-8, Thomas 'PointedEars'
> > Lahn wrote:
> >> > The term *i* is "defined" to be 'the square root of -1'
> >> No, it is NOT; that is a common misconception (unfortunately, it can also
> >> be found in otherwise great textbooks such as Griffiths’ “Introduction to
> >> Quantum Mechanics”, at least in the second edition).
> >
> > Well, you have a LOT of web pages that you need to contact because I can
> > show you a whole lot of them that disagree with you...
> As you can see below, either you misunderstand your own sources or they are
> wrong.
> > https://en.wikipedia.org/wiki/Imaginary_number
> >
> > "i, which is defined by its property i2 = −1"
> Copy & pray aside, this is the same as I said.
> > https://en.wikipedia.org/wiki/Imaginary_unit
> >
> > "The imaginary unit or unit imaginary number (i) is a solution to the
> > quadratic equation x2 + 1 = 0"
> Copy & pray aside, this is the same as I said.
> > https://byjus.com/maths/value-of-i/
> >
> > "The value of i is √-1"
> This is wrong.
> > https://mathworld.wolfram.com/i.html
> >
> > "The imaginary number i (also called the imaginary unit) is defined as the
> > square root of -1"
> This is wrong, unfortunately. The argument made there smells of theory
> finding and is questionable.
>
> It should be noted, though, that Wolfram MathWorld is mainly a compilation
> of statements from textbooks, and some textbooks oversimplify there, as I
> just said.
> > https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
> >
> > "the “imaginary” number i has the property that the square of i is -1"
> This is the same as I said.
> > https://www.mathopenref.com/i.html
> >
> > "It (i) stands for the square root of negative one"
> This is wrong.
> > ... and there are thousands more, but at this point in time, I'm bored....
> No, you are incompetent and can’t read.
>
>
> PointedEars
> --
> Q: What happens when electrons lose their energy?
> A: They get Bohr'ed.
>
> (from: WolframAlpha)

Like I said... you have a LOT of web pages and textbooks to set straight... so you better get busy!

On Sunday, November 21, 2021 at 8:04:18 AM UTC-8, patdolan wrote:
> If F&S know the time of flight and the velocity of the muons in their lab then F&S can
> also know the time of flight and velocity of the earth in the muon's FoR.
> Will you [please teach me the] calculation of the latter?

That is one of the oldest freshman befuddlements in the book. Let S be inertial coordinates in which a particle P on the earth’s surface is at rest, and let S’ be a system in which the muon is at rest. The speed of each system in terms of the other is the same. The muon, directly approaching P at high speed, is created at event E1, which is simultaneous with event E2 of P in terms of S, and with event E3 of P in terms of S’. Say the absolute interval between E1 and E2 is 14000 meters, and the absolute interval between E1 and E3 is 600 meters. Thus E3 is 46.62 microseconds later than E2. The times of flight are just distance divided by speed. There's nothing paradoxical about this. The two absolute space-like intervals (14000 and 600) are between two different pairs of events.

Of course, for each particle, when it is a given proper time from collision, in terms of it's own rest frame coordinates the spatial distance to the other particle is the same.

patdolan <patdolan@comcast.net> wrote:
> On Sunday, November 21, 2021 at 5:44:24 AM UTC-8, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Saturday, November 20, 2021 at 7:43:48 AM UTC-8, bodk...@gmail.com wrote:
>>>> patdolan <patd...@comcast.net> wrote:
>>>>> On Friday, November 19, 2021 at 2:37:48 PM UTC-8, bodk...@gmail.com wrote:
>>>>>> patdolan <patd...@comcast.net> wrote:
>>>>>>> Critical Theory comes in several flavors: Critical Law Theory, Critical
>>>>>>> Literature Theory, Critical History Theory, Critical Race Theory. To
>>>>>>> these we now add Critical Relativity Theory!
>>>>>>>
>>>>>>> In the spirit of Derrida we shall deconstruct the clumsy reasoning of
>>>>>>> special relativity and separate said reasoning from the algebraic symbols
>>>>>>> and equations that express it, keeping in mind that mathematics is just
>>>>>>> another form of rhetorical expression wherein falsity can be expressed
>>>>>>> every bit as plausibly as the truth.
>>>>>>>
>>>>>>> “There is only the text.”— J. Derrida
>>>>>>>
>>>>>>> According to special relativity two observers in motion with respect to
>>>>>>> each other will disagree on each other’s length. They will also disagree
>>>>>>> on the proper flow of time. But they will always agree on the velocity
>>>>>>> they have with respect to one another. This is exceedingly strange. How
>>>>>>> can it be that two relative quantities, space and time, combine to
>>>>>>> produce an absolute quantity called relative velocity? It is true that SR
>>>>>>> does have a formula for calculating coordinate velocity; just like it has
>>>>>>> formulas for calculating coordinate space and coordinate time. But the
>>>>>>> Einstein velocity addition formula ONLY applies to a third object in
>>>>>>> motion wrt a pair of FoRs. If that third object happens to be at rest
>>>>>>> wrt one of the FoRs then Einsteinian velocity reduces to Galilean
>>>>>>> velocity, albeit subject to the speed limit c. Relativists simply
>>>>>>> assumed without further justification that if FoR-1 measures a velocity v
>>>>>>> between itself and FoR-2 then FoR-2 must measure the same numerical value
>>>>>>> v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
>>>>>>> trivial assumptions can be monumental when constructing a theory of
>>>>>>> motion. But a 26 year old would probably not yet have the requisite
>>>>>>> philosophical sophistication needed to recognize this.
>>>>>>>
>>>>>>> Einstein’s choice to make velocity strictly Galilean when calculating the
>>>>>>> velocity between pairs of FoRs ( yes, it is a choice because it does not
>>>>>>> follow from either the first or the second postulates ) can be
>>>>>>> expressed mathematically as
>>>>>>>
>>>>>>> ∆x’/∆t’ = ∆x/∆t = v (1)
>>>>>>>
>>>>>>> I now raise equation (1) to the level of a postulate and declare it to be
>>>>>>> the third, and heretofore hidden, postulate of special relativity. In
>>>>>>> recognizing its own structural Galileanism through this new postulate,
>>>>>>> special relativity can finally claim to be woke.
>>>>>>>
>>>>>>> The problem with the third postulate is that even though it is already
>>>>>>> assumed in every equation of special relativity, it turns out to be true
>>>>>>> only when v = c ( the second postulate ) or when v = 0. The third
>>>>>>> postulate can be demonstrated invalid for all values of v in between.
>>>>>>> The invalidity of the third postulate causes special relativity fall on
>>>>>>> it’s algebraic face. Big Bang move over…Not recognizing and
>>>>>>> acknowledging the third postulate was Einstein’s biggest blunder.
>>>>>>>
>>>>>>> Time for some examples.
>>>>>>>
>>>>>>>
>>>>>>> DIRK & DONO
>>>>>>>
>>>>>>> Consider two FoRs whose x-axes are parallel and lie very close to one
>>>>>>> another. The relative velocity between these two FoRs is .866c (γ = 2).
>>>>>>> Dirk assumes the Lotus position at the origin of one FoR whilst Dono
>>>>>>> assumes the fetal position at the origin of the other. Both Dirk and
>>>>>>> Dono and their clocks are glued to the origins of their respective FoRs.
>>>>>>>
>>>>>>> Dirk opens one eye and takes note of the meter marks on Dono’s contracted
>>>>>>> x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
>>>>>>> contracted to only half as long as his own meter marks. Dirk opens the
>>>>>>> other eye and observes that Dono’s clock is ticking at only half the rate
>>>>>>> of his own clock. Dirk begins to count Dono’s meter marks as they race
>>>>>>> past Dirk’s position. After one year (by Dirk’s clock) of counting
>>>>>>> Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
>>>>>>> marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
>>>>>>> elapsed on Dono’s clock. Dirk now calculates what his coordinate
>>>>>>> velocity should be according to Dono
>>>>>>>
>>>>>>> ( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
>>>>>>>
>>>>>>> [ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
>>>>>>>
>>>>>>> “Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
>>>>>>> stipulated to be an absolute .866c with respect to one another, when
>>>>>>> measured in either FoR.”
>>>>>>>
>>>>>>> That is true, according to (1). But remember that (1) is an arbitrary
>>>>>>> choice made by Einstein when he built his theory. It is no more
>>>>>>> legitimate a choice than the Dolan FoR coordinate velocity transform (3)
>>>>>>> for determining one’s FoR coordinate velocity. It is also no less
>>>>>>> inconsistent. For we immediately see by inspection that the Dolan FoR
>>>>>>> coordinate velocity is always greater than the relative velocity by a
>>>>>>> factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
>>>>>>> velocities clangs with just as much antinomy as Dolan’s transform, as we
>>>>>>> shall see. Special relativity’s dirty little secret is that it’s hidden
>>>>>>> third postulate (1) destroys the theory from within.
>>>>>>>
>>>>>>> I can see by Dono’s tightening fetal position that he still doesn’t
>>>>>>> believe me. Very well. We shall prove SR’s mathematical inconsistency
>>>>>>> in the next example.
>>>>>>>
>>>>>>>
>>>>>>> SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
>>>>>>>
>>>>>>> Consider a pair of FoRs whose relative velocity v is some value other
>>>>>>> than c and other than zero. This is expressed mathematically as
>>>>>>>
>>>>>>> v = ∆x’/∆t’ = ∆x/∆t != c (4)
>>>>>>>
>>>>>>> and
>>>>>>>
>>>>>>> v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
>>>>>>>
>>>>>>> The Lorentz transforms allow us to construct the FoR coordinate
>>>>>>> velocities for pairs of FoRs
>>>>>>>
>>>>>>> ∆x’ = γ( ∆x - v∆t )
>>>>>>>
>>>>>>> ∆t’ = γ( ∆t - ∆xv/c^2 )
>>>>>>>
>>>>>>> ∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
>>>>>>>
>>>>>>> We now endeavor to solve (6) for v in hopes of demonstrating the internal
>>>>>>> consistency of special relativity, i.e, that v = v for all pair of FoRs.
>>>>>>> We saw how this was not the case in DIRK & DONO.
>>>>>>>
>>>>>>> The reader will find that trying to solve for v is hopeless unless we
>>>>>>> make the substitution
>>>>>>>
>>>>>>> v = ∆x/∆t or v = ∆x’/∆t’
>>>>>>>
>>>>>>> It does not matter which we use—either substitution is permitted by the
>>>>>>> third postulate (1).
>>>>>>>
>>>>>>> With the substitution made, we eventually arrive at the preposterous results
>>>>>>>
>>>>>>> ∆x’/∆t’ = c (7)
>>>>>>>
>>>>>>> or
>>>>>>>
>>>>>>> ∆x’/∆t’ = 0 (8)
>>>>>>>
>>>>>>> The derivation is left as an exercise for the reader. For those needing
>>>>>>> help with the derivation ( I'm looking at you, Dono ) I will be happy to
>>>>>>> provide it in another post.
>>>>>>>
>>>>>>> The laughable results (7) and (8) directly contradict our assumptions (4)
>>>>>>> and (5). Furthermore, the results impose the requirement that v is not
>>>>>>> even a variable—v turns out to be a constant always equal to c or zero.
>>>>>>> Absolutely absurd.
>>>>>>>
>>>>>>> QED.
>>>>>>>
>>>>>>> Algebraic relativity is thus reduced to ridiculous rubble by means
>>>>>>> mathematical reductio ad absurdum. The root cause of special relativity’s
>>>>>>> spectacular algebraic failure lies in the propositional calculus. I am
>>>>>>> happy to expatiate on that subject in another post if there is interest.
>>>>>>>
>>>>>>> Let’s do one more example—this one ripped from the headlines of
>>>>>>> experimental physics.
>>>>>>>
>>>>>>>
>>>>>>> MUONS, SCHMUONS!
>>>>>>>
>>>>>>> Imagine that you are a hadron in deep space minding your own business
>>>>>>> when all of a sudden you turn to see a Lorentz-flattened earth coming at
>>>>>>> you at a velocity of .866c. In the impending collision the first thing
>>>>>>> to strike you is an air molecule high in the earth’s atmosphere. That
>>>>>>> molecules knocks the stuffing out of you. What is left of you is now a
>>>>>>> muon which means that you only have 2.2 microseconds more to live.
>>>>>>>
>>>>>>> It turns out that the surface of the flattened earth is exactly 571.56
>>>>>>> meters away from you, according to your own muon FoR. By a remarkable
>>>>>>> coincidence there is a flattened scintillator and a flattened clock in a
>>>>>>> flattened laboratory directly below you on the surface of the flattened
>>>>>>> earth. You note the time on the flattened lab clock. Because you are a
>>>>>>> muon, you are your own 2.2 microsecond alarm clock.
>>>>>>>
>>>>>>> The surface of the flattened earth continues to speed towards you at
>>>>>>> .866c. 2.2 microseconds later the flattened earth, lab and scintillator
>>>>>>> smash into you. Just as you expire in the scintillator you note that
>>>>>>> only 1.1 microseconds has elapsed on the lab clock.
>>>>>>
>>>>>> You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
>>>>>
>>>>> No synchronization required. The muon's clock doesn't even enter into
>>>>> the argument at all. Who said that the lab clock was ever at t=0 or any
>>>>> other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
>>>> Elapsed from what moment? How does the lab clock know when to start
>>>> ticking. It isn’t anywhere near the collision of the hadron and the
>>>> atmospheric molecule. Is it supposed to get an instantaneous signal from
>>>> that event?
>>>>
>>>
>>>
>>>> This is what I mean when I say you haven’t bothered to read any book about
>>>> relativity. Your “argument” is based on two and only two clocks passing
>>>> each other, both deeming the other slowed, because that’s the comic book
>>>> version of relativity. But that isn’t what relativity says. Ask yourself
>>>> why relativity talks about a *lattice* of synchronized clocks in each
>>>> frame. Why would you need the lattice?
>>>>
>>>> It’s so that there is a clock NEAR the event that you’re tracking the time
>>>> of.
>>>>> Quantities of elapsed time are easily determined in special relativity.
>>>>> Synchronization of clocks doesn't even enter into the argument It's all
>>>>> about one clock, the lab clock. One clock, two observers (the muon and
>>>>> the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
>>>>>
>>>>> Tell us how synchronization has any bearing on the argument. The muon
>>>>> sees the lab clock tick off 1.1 microseconds on it's trip down.
>>>> No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
>>>> the lab clock. The best the muon can do is either a) monitor its own clock,
>>>> because its own clock is at ONE location in its own frame, or b) have a
>>>> LATTICE of synchronized clocks traveling along with the muon, where one of
>>>> them is far away from the muon, near the lab clock at the time the muon is
>>>> created.
>>>>
>>>> If you have two events that occur at DIFFERENT PLACES in some frame of
>>>> reference, you need two DIFFERENT clocks near those events. If those clocks
>>>> are synchronized in this frame, THEN you can measure the elapsed time
>>>> between the events.
>>>>
>>>> You can’t measure the elapsed time between two events that occur in
>>>> different places with ONE clock.
>>>>
>>>> You, Pat Dolan, don’t understand the first thing about relativity. And it
>>>> shows.
>>>>
>>>> And yet you bluster simultaneously that a) you understand relativity very
>>>> well, and b) that it is incomprehensible.
>>>
>>> Bodkin, please declare to this forum that your argument also applies to
>>> the scientists in the lab--especially to the scientist in the lab of Dr's
>>> Frisch and Smith.
>>>
>>> https://www.youtube.com/watch?v=rbzt8gDSYIM
>>>
>
>>> Frisch and Smith didn't have clocks up there alongside all the muons they
>>> measured either. So there is NO WAY according to Bodkin that Frisch and
>>> Smith could have possibly determined whether or not muons experienced
>>> relativistic time dilation.
>> Yes they can, but not by the method you floundered around with. First of
>> all, these muon experiments typically measure the speed directly by having
>> a stack of vertically separated scintillators, so that time of flight is
>> readily available. Secondly, the lifetime is measured the same way it is
>> for radioisotopes, activity rate at different samplings. It’s actually
>> straightforward.
>
> If F&S know the time of flight and the velocity of the muons in their lab
> then F&S can also know the time of flight and velocity of the earth in the muon's FoR.

Paul Alsing wrote:

> On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> Paul Alsing wrote:
>> > ... and there are thousands more, but at this point in time, I'm
>> > bored...
>> No, you are incompetent and can’t read.
>
> Like I said... you have a LOT of web pages and textbooks to set
> straight... so you better get busy!

Fallacy: Shifting the burden of proof.

I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.

And apparently it has escaped your attention that most of the sources that
you quoted so far say *exactly* what I said, and are explicitly or
implicitly contradicting what you claimed, which is contradictory *in
itself* [that on the one hand 𝕚 would NOT be a number, but merely
“imaginary”; and that on the other hand 𝕚 *would* be a number defined as
√(−1).]

Apparently you are that incompetent that not only you do not realize when
you are contradicting yourself, but also that you do not realize that yourr
own sources are contradicting yourself

In that case this discussion would be over as you would have have
demonstrated that you had disconnected from reality, and there is no
argument by which you can be convinced.

PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)

Paul Alsing wrote:

> On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> Paul Alsing wrote:
>> > ... and there are thousands more, but at this point in time, I'm
>> > bored...
>> No, you are incompetent and can’t read.
>
> Like I said... you have a LOT of web pages and textbooks to set
> straight... so you better get busy!

Fallacy: Shifting the burden of proof.

I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.

And apparently it has escaped your attention that most of the sources that
you quoted so far say *exactly* what I said, and are explicitly or
implicitly *contradicting* what you claimed, which is contradictory *in
itself*: that on the one hand 𝕚 would NOT be a number, but merely
“imaginary”; and that on the other hand 𝕚 *would* be a number defined as
√(−1).

Apparently you are that incompetent that not only you do not realize when
you are contradicting yourself, but also that you do not realize that your
own sources are contradicting you.

In that case this discussion would be over as you would have demonstrated
that you had disconnected from reality, and there is no argument by which
you can be convinced of anything that is contradictory to your claims.

<https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect>

PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)

On Sunday, November 21, 2021 at 8:57:27 AM UTC-8, Townes Olson wrote:
> On Sunday, November 21, 2021 at 8:04:18 AM UTC-8, patdolan wrote:
> > If F&S know the time of flight and the velocity of the muons in their lab then F&S can
> > also know the time of flight and velocity of the earth in the muon's FoR.
> > Will you [please teach me the] calculation of the latter?
>
> That is one of the oldest freshman befuddlements in the book. Let S be inertial coordinates in which a particle P on the earth’s surface is at rest, and let S’ be a system in which the muon is at rest.

The speed of each system in terms of the other is the same.
Prove it! The deconstruction of this very statement is the whole point of the original post.

The muon, directly approaching P at high speed, is created at event E1, which is simultaneous with event E2 of P in terms of S, and with event E3 of P in terms of S’. Say the absolute interval between E1 and E2 is 14000 meters, and the absolute interval between E1 and E3 is 600 meters. Thus E3 is 46.62 microseconds later than E2. The times of flight are just distance divided by speed. There's nothing paradoxical about this. The two absolute space-like intervals (14000 and 600) are between two different pairs of events.

What does all this even mean??? Are you sure you know, Townes? Use some equations and more words. 14000 m??? 600 m??? 46.62 microseconds??? From where does all this come???

And try to remember that Pat Dolan is holding court in this forum at the moment. That means you are in the Big Leagues for the time being, Townes. You need to bring your A game.
>
> Of course, for each particle, when it is a given proper time from collision, in terms of it's own rest frame coordinates the spatial distance to the other particle is the same.

On Sunday, November 21, 2021 at 4:07:13 PM UTC-8, patdolan wrote:
> > > If F&S know the time of flight and the velocity of the muons in their lab then F&S can
> > > also know the time of flight and velocity of the earth in the muon's FoR.
> > > Will you [please teach me the] calculation of the latter?
> >
> > Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> > and let S’ be a system in which the muon is at rest. The muon, directly approaching
> > P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> > terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
> > The two absolute space-like intervals are between two different pairs of events.
>
> What does all this even mean??? Use some equations and more words.

Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).

On Sunday, November 21, 2021 at 5:29:37 PM UTC-8, Townes Olson wrote:
> On Sunday, November 21, 2021 at 4:07:13 PM UTC-8, patdolan wrote:
> > > > If F&S know the time of flight and the velocity of the muons in their lab then F&S can
> > > > also know the time of flight and velocity of the earth in the muon's FoR.
> > > > Will you [please teach me the] calculation of the latter?
> > >
> > > Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> > > and let S’ be a system in which the muon is at rest. The muon, directly approaching
> > > P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> > > terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
> > > The two absolute space-like intervals are between two different pairs of events.
> >
> > What does all this even mean??? Use some equations and more words.
>
> Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).

I ask for clarification on E1, E2 and E3. Instead I get E4!

So Bodkin...is it Mitch & Townes, Townes & Mitch?

On Sunday, November 21, 2021 at 5:36:32 PM UTC-8, patdolan wrote:
> > > > > If F&S know the time of flight and the velocity of the muons in their lab then F&S can
> > > > > also know the time of flight and velocity of the earth in the muon's FoR.
> > > > > Will you [please teach me the] calculation of the latter?
> > > >
> > > > Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> > > > and let S’ be a system in which the muon is at rest. The muon, directly approaching
> > > > P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> > > > terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
> > > > The two absolute space-like intervals are between two different pairs of events.
> > >
> > > What does all this even mean??? Use some equations and more words.
> >
> > Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
>
> I ask for clarification on E1, E2 and E3. Instead I get E4!

What is it that you don't understand about those events? They are explicitly defined above. Again, the muon is created at E1 and it collides with the earth particle P at E4. Event E2 of P is simultaneous with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms of S'. Do you understand this?

Townes Olson <townesolson7@gmail.com> wrote:
> On Sunday, November 21, 2021 at 5:36:32 PM UTC-8, patdolan wrote:
>>>>>> If F&S know the time of flight and the velocity of the muons in
>>>>>> their lab then F&S can
>>>>>> also know the time of flight and
>>>>>> velocity of the earth in the muon's FoR.
>>>>>> Will you [please teach me the] calculation of the latter?
>>>>>
>>>>> Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
>>>>> and let S’ be a system in which the muon is at rest. The muon, directly approaching
>>>>> P at high speed, is created at event E1, which is simultaneous with event E2 of P in
>>>>> terms of S, and with event E3 of P in terms of S’... There's nothing
>>>>> paradoxical about this.
>>>>> The two absolute space-like intervals are between two different pairs of events.
>>>>
>>>> What does all this even mean??? Use some equations and more words.
>>>
>>> Sure. Let E4 be the event where the muon collides with particle P on
>>> earth's surface, and let v denote the mutual speed between the muon and
>>> P, and in terms of S let D denote the distance traveled by the muon
>>> from E1 to E4, so its time of flight is D/v. In terms of S' (in which
>>> the muon is at rest) the distance traveled by the earth particle P from
>>> E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
>>
>> I ask for clarification on E1, E2 and E3. Instead I get E4!
>
> What is it that you don't understand about those events? They are
> explicitly defined above. Again, the muon is created at E1 and it
> collides with the earth particle P at E4. Event E2 of P is simultaneous
> with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
> of S'. Do you understand this?
>

He’s walking you through something standard, Pat.

--
Odd Bodkin — Maker of fine toys, tools, tables

On Sunday, November 21, 2021 at 6:07:14 PM UTC-8, bodk...@gmail.com wrote:
> Townes Olson <townes...@gmail.com> wrote:
> > On Sunday, November 21, 2021 at 5:36:32 PM UTC-8, patdolan wrote:
> >>>>>> If F&S know the time of flight and the velocity of the muons in
> >>>>>> their lab then F&S can
> >>>>>> also know the time of flight and
> >>>>>> velocity of the earth in the muon's FoR.
> >>>>>> Will you [please teach me the] calculation of the latter?
> >>>>>
> >>>>> Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> >>>>> and let S’ be a system in which the muon is at rest. The muon, directly approaching
> >>>>> P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> >>>>> terms of S, and with event E3 of P in terms of S’... There's nothing
> >>>>> paradoxical about this.
> >>>>> The two absolute space-like intervals are between two different pairs of events.
> >>>>
> >>>> What does all this even mean??? Use some equations and more words.
> >>>
> >>> Sure. Let E4 be the event where the muon collides with particle P on
> >>> earth's surface, and let v denote the mutual speed between the muon and
> >>> P, and in terms of S let D denote the distance traveled by the muon
> >>> from E1 to E4, so its time of flight is D/v. In terms of S' (in which
> >>> the muon is at rest) the distance traveled by the earth particle P from
> >>> E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
> >>
> >> I ask for clarification on E1, E2 and E3. Instead I get E4!
> >
> > What is it that you don't understand about those events? They are
> > explicitly defined above. Again, the muon is created at E1 and it
> > collides with the earth particle P at E4. Event E2 of P is simultaneous
> > with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
> > of S'. Do you understand this?
> >
> He’s walking you through something standard, Pat.

I know. But it's like he doesn't understand that the very assumption he relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating. Use your influence to get him off my thread until he gets a clue. Naturally I would like to use young Townes for more cannon fodder; but only after he understands the topography of this battlefield.
>
> --
> Odd Bodkin — Maker of fine toys, tools, tables

On Sunday, November 21, 2021 at 2:44:23 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> Paul Alsing wrote:
>
> > On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas 'PointedEars'
> > Lahn wrote:
> >> Paul Alsing wrote:
> >> > ... and there are thousands more, but at this point in time, I'm
> >> > bored...

> >> No, you are incompetent and can’t read.
> >
> > Like I said... you have a LOT of web pages and textbooks to set
> > straight... so you better get busy!

> Fallacy: Shifting the burden of proof.
>
> I do not have to find sources to support my claim; *you* made the claim.
> I showed you the problems with your claims, and their solution.

And I have shown you hundreds of sources that agree with me... and *you* have provided exactly *zero* sources that agree with you!

I stand by my claim that a vast majority of textbooks in the world state that the definition of "i" is "the square root of -1"... if you think that I am incorrect I expect you to provide evidence otherwise... which you have failed to do. Your babble is not evidence.

As always, either put up or shut up.

Without evidence, you are done here. But then again, you have been done here more often than not.

On Sunday, November 21, 2021 at 6:29:54 PM UTC-8, patdolan wrote:
> > >>>>>> If F&S know the time of flight and the velocity of the muons in
> > >>>>>> their lab then F&S can
> > >>>>>> also know the time of flight and
> > >>>>>> velocity of the earth in the muon's FoR.
> > >>>>>> Will you [please teach me the] calculation of the latter?
> > >>>>>
> > >>>>> Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> > >>>>> and let S’ be a system in which the muon is at rest. The muon, directly approaching
> > >>>>> P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> > >>>>> terms of S, and with event E3 of P in terms of S’... There's nothing
> > >>>>> paradoxical about this.
> > >>>>> The two absolute space-like intervals are between two different pairs of events.
> > >>>>
> > >>>> What does all this even mean??? Use some equations and more words.
> > >>>
> > >>> Sure. Let E4 be the event where the muon collides with particle P on
> > >>> earth's surface, and let v denote the mutual speed between the muon and
> > >>> P, and in terms of S let D denote the distance traveled by the muon
> > >>> from E1 to E4, so its time of flight is D/v. In terms of S' (in which
> > >>> the muon is at rest) the distance traveled by the earth particle P from
> > >>> E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
> > >>
> > >> I ask for clarification on E1, E2 and E3. Instead I get E4!
> > >
> > > What is it that you don't understand about those events? They are
> > > explicitly defined above. Again, the muon is created at E1 and it
> > > collides with the earth particle P at E4. Event E2 of P is simultaneous
> > > with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
> > > of S'. Do you understand this?
>
> The very assumption you rely on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.

That's not the assumption, that's the conclusion of the calculation. In tiny baby steps: In terms of S the events E1, E2, E3, and E4 of the scenario you stipulated have the (x,t) coordinates (0,0), (D,0), (D,vD), and (D,D/v) respectively. These are stipulated in your question, given the speed and time of flight of the muon in terms of S. Given this, you asked for the speed and time of flight of P in terms of S', which is easily computed as above. What part of this do you consider to be debatable?

> ... get him off my thread...

I don't think that's going to work, because the answer is here for all to see, so even if you had the power to ban me, you can't un-ring the bell. You would have to somehow get my post deleted from the servers, but that would be nearly impossible. I';m afraid the jig is up.

> Thank you in advance.

You're welcome.

On Sunday, November 21, 2021 at 6:50:19 PM UTC-8, Townes Olson wrote:
> On Sunday, November 21, 2021 at 6:29:54 PM UTC-8, patdolan wrote:
> > > >>>>>> If F&S know the time of flight and the velocity of the muons in
> > > >>>>>> their lab then F&S can
> > > >>>>>> also know the time of flight and
> > > >>>>>> velocity of the earth in the muon's FoR.
> > > >>>>>> Will you [please teach me the] calculation of the latter?
> > > >>>>>
> > > >>>>> Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
> > > >>>>> and let S’ be a system in which the muon is at rest. The muon, directly approaching
> > > >>>>> P at high speed, is created at event E1, which is simultaneous with event E2 of P in
> > > >>>>> terms of S, and with event E3 of P in terms of S’... There's nothing
> > > >>>>> paradoxical about this.
> > > >>>>> The two absolute space-like intervals are between two different pairs of events.
> > > >>>>
> > > >>>> What does all this even mean??? Use some equations and more words.
> > > >>>
> > > >>> Sure. Let E4 be the event where the muon collides with particle P on
> > > >>> earth's surface, and let v denote the mutual speed between the muon and
> > > >>> P, and in terms of S let D denote the distance traveled by the muon
> > > >>> from E1 to E4, so its time of flight is D/v. In terms of S' (in which
> > > >>> the muon is at rest) the distance traveled by the earth particle P from
> > > >>> E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
> > > >>
> > > >> I ask for clarification on E1, E2 and E3. Instead I get E4!
> > > >
> > > > What is it that you don't understand about those events? They are
> > > > explicitly defined above. Again, the muon is created at E1 and it
> > > > collides with the earth particle P at E4. Event E2 of P is simultaneous
> > > > with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
> > > > of S'. Do you understand this?
> >
> > The very assumption you rely on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
>
> That's not the assumption, that's the conclusion of the calculation. In tiny baby steps: In terms of S the events E1, E2, E3, and E4 of the scenario you stipulated have the (x,t) coordinates (0,0), (D,0), (D,vD), and (D,D/v) respectively. These are stipulated in your question, given the speed and time of flight of the muon in terms of S. Given this, you asked for the speed and time of flight of P in terms of S', which is easily computed as above. What part of this do you consider to be debatable?

Lemesee...Dirk foundered on ALGEBRAIC RELATIVITY'S ORIGINAL SIN...and Bodkin cried uncle on MUONS, SCHMUONS....so Townes, my boy, why don't you have a go at DIRK & DONO and see what you can make of that as yet unaddressed proof of my troika. In the spirit Derrida, try to address the text that you find--resist casting my examples in your own E1, E2, etc. gibberish. Now run along my boy and come back when you think you have something to show us.
>
> > ... get him off my thread...
>
> I don't think that's going to work, because the answer is here for all to see, so even if you had the power to ban me, you can't un-ring the bell. You would have to somehow get my post deleted from the servers, but that would be nearly impossible. I';m afraid the jig is up.
>
> > Thank you in advance.
>
> You're welcome.