Critical Relativity Theory

[patdolan]

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. Nothing strange here; the labs clock is traveling at γ=2 with respect to you so it only logs half as much elapsed time as you.

One of the lab scientists now cries out “Just a minute! That’s not how this story goes. First of all, it is the hadron that is Lorentz-flattened, not us. And second of all, that hadron used it’s flattened FoR to determine that is was only 571.56 meters away from our scintillator at the point it became a muon. Our unflattened earth FoR clearly indicated that the hadron was actually 1143.12 meters away when it became a muon. That’s why it took 4.4 microseconds, traveling at .866 c, to reach our scintillator. And that’s what our lab clock shows. Not 1.1 microseconds, like that dumb muon is claiming.”

Special relativity leaves us in a quandary. How much time actually did elapse on the lab clock from the moment of the muon’s inception to the moment of its scintillating demise? Was it 1.1 microseconds? Or was it 4.4 microseconds? It has to be one or the other, it can’t be both. However, different elapsed times on the same lab clock can be calculated by the muon and by the earthbound scientists with equal justification. Special relativity is incapable of providing a unique elapsed time on the lab clock. I challenge anyone to show that special relativity can provide a unique elapsed time in the preceding case.

Special relativity’s greatest experimental confirmation turns out to be it’s second greatest falsification. Gravity has been special relativity’s greatest falsification since 1907 ( check out the author’s brilliant post on special relativity vs. Kepler’s 3rd law ).

[Richard Hertz]

Dono is a fucking slug, which moves at 0.000000000000000000000000000000000000000000001c, and this when is in a hurry.

You have to select another imbecile here. I propose Moroney, the FAKE EE.

[Odd Bodkin]

This was discussed recently here about a month or so ago, which you can
find by searching for postulate in thread titles. You have missed all the
clear explanation why this is a non-issue.

Rather than assuming you’re the first to arrive at this thought, why don’t
you take the time to research if anyone else has already thought about it?

Oh I know why. Because then you wouldn’t get an attention fix.

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

[Dono.]

On Monday, November 15, 2021 at 7:43:32 PM UTC-8, cretin patyycakes dolan wrote:
> snip imbecilities<

[Dirk Van de moortel]

Op 16-nov.-2021 om 04:43 schreef patdolan:

You can only write
∆x’/∆t’ = v
for events taking place on an object at rest in FoR-1, i.o.w.
for events taking place a the same location in For-1, i.o.w.
for events that satisfy
∆x = 0.

You can only write
∆x/∆t = v
for events taking place on an object at rest in FoR-2, i.o.w.
for events taking place a the same location in For-2, i.o.w.
for events that satisfy
∆x' = 0.

Unless v = 0 (and thus FoR-1 = FoR-2), there are no objects
that are at rest in both FoR-1 and FoR-2, i.o.w. there are
no distinct events that happen at the same place in both
FoR-1 and FoR-2.

So, yes
0/∆t’ = 0/∆t = 0
for all values of ∆t' and ∆t.
A profound discovery, congratulations!

So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about.

But by the way, perhaps you just want to say that the origins
of FoR-1 and FoR-2 have mutual velocity v w.r.t. each other.
So you want to say that
∆x’/∆t’ = v
for events taking place on an object at rest in FoR-1 (∆x=0)
and that
∆x/∆t = v
for events taking place on an object at rest in FoR-2 (∆x'=0)
*without* combining the equations?

In that case you have the x and x' axes pointing in different
directions, and the standard Lorentz transformation then takes
the form
∆x’ = - g ( ∆x - v ∆t )
∆t’ = g ( ∆t - v/c^2 ∆x )
and the inverse
∆x = - g ( ∆x' - v ∆t' )
∆t = g ( ∆t' - v/c^2 ∆x' )
where
g = 1 / sqrt( 1 - v^2/c^2 )

See if this helps.

Dirk Vdm

[Wade Earl]

Dirk Van de moortel wrote:

> n that case you have the x and x' axes pointing in different directions,
> and the standard Lorentz transformation then takes the form
> ∆x’ = - g ( ∆x - v ∆t ) ∆t’ = g ( ∆t - v/c^2 ∆x )
> and the inverse
> ∆x = - g ( ∆x' - v ∆t' ) ∆t = g ( ∆t' - v/c^2 ∆x' )
> where
> g = 1 / sqrt( 1 - v^2/c^2 )
>
> See if this helps.

not sure, you are working in 3D, meanwhile the g is 1D.

[patdolan]

On Tuesday, November 16, 2021 at 6:54:15 AM UTC-8, bodk...@gmail.com wrote:

Bodkin, I searched on the word "postulate" as you suggested. From Aug 1, 2021 until today. I am the only poster that the search returned.

[patdolan]

> So when you write
> ∆x’/∆t’ = ∆x/∆t = v ,
> you clearly have no idea what you are talking about.
>

c is a velocity too.

∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what I am talking about?

Prove to this forum that a pair of observers glued to the origins of their respective FoRs measure the same relative velocity between their FoRs. I have disproved this in three different ways.

You need to think about this a little more, Dirk. Why don't you try assuming the Lotus position. See if that helps.

[patdolan]

Oh, and nice pick up on my removing the "-" from my derivation. It is the one and only liberty I took for the sake of clarity. So, yes, you did at least prove that the x-axes do point in opposite directions. At least you can feel good about that.

[Dono.]

On Tuesday, November 16, 2021 at 8:27:43 AM UTC-8, pattycakes dolan wrote:
> I have disproved this in three different ways.

Three different way, one common imbecility

[Odd Bodkin]

Search for “unstated special relativity axiom”.

Would the next spoon feeding require a rubber-coated spoon?

--
Odd Bodkin -- maker of fine toys, tools, tables

[patdolan]

Bodkin,

I give Ricardo Jimenez priority as being first to publish. But I reserver for myself the claim of being first to rigorously prove what the brilliant and far-sighted Ricardo only suspected.

PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular reasoning on this issue. And rotchm! rotchm had an absolute panic attack and started delivering a course on vector algebra in order to avoid the subject entirely.

I should like to meet the Jimenez fellow online sometime.

PPS--how's the re-think going Dirk? Let me know if you get stuck again.

[Odd Bodkin]

patdolan <patd...@comcast.net> wrote:
> On Tuesday, November 16, 2021 at 8:43:21 AM UTC-8, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Tuesday, November 16, 2021 at 6:54:15 AM UTC-8, bodk...@gmail.com wrote:
>>>
>>> Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
>>> 2021 until today. I am the only poster that the search returned.
>>>
>> Search for “unstated special relativity axiom”.
>>
>> Would the next spoon feeding require a rubber-coated spoon?
>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>
> Bodkin,
>
> I give Ricardo Jimenez priority as being first to publish.

And you seem not to have read anything in the thread showing Ricardo where
he was astray.

> But I reserver for myself the claim of being first to rigorously prove
> what the brilliant and far-sighted Ricardo only suspected.
>
> PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular
> reasoning on this issue. And rotchm! rotchm had an absolute panic
> attack and started delivering a course on vector algebra in order to
> avoid the subject entirely.
>
> I should like to meet the Jimenez fellow online sometime.
>
> PPS--how's the re-think going Dirk? Let me know if you get stuck again.
>

[patdolan]

On Tuesday, November 16, 2021 at 9:10:40 AM UTC-8, bodk...@gmail.com wrote:
> patdolan <patd...@comcast.net> wrote:
> > On Tuesday, November 16, 2021 at 8:43:21 AM UTC-8, bodk...@gmail.com wrote:
> >> patdolan <patd...@comcast.net> wrote:
> >>> On Tuesday, November 16, 2021 at 6:54:15 AM UTC-8, bodk...@gmail.com wrote:
> >>>
> >>> Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
> >>> 2021 until today. I am the only poster that the search returned.
> >>>
> >> Search for “unstated special relativity axiom”.
> >>
> >> Would the next spoon feeding require a rubber-coated spoon?
> >>
> >> --
> >> Odd Bodkin -- maker of fine toys, tools, tables
> >
> > Bodkin,
> >
> > I give Ricardo Jimenez priority as being first to publish.
> And you seem not to have read anything in the thread showing Ricardo where
> he was astray.

That's is correct Bodkin. Because RJ appears to have answered every objection and parried every passe'. Do you disagree?

Bodkin: Yes.
Dolan: Where?
Bodkin: I won't spoon feed you.

[Odd Bodkin]

patdolan <patd...@comcast.net> wrote:
> On Tuesday, November 16, 2021 at 9:10:40 AM UTC-8, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Tuesday, November 16, 2021 at 8:43:21 AM UTC-8, bodk...@gmail.com wrote:
>>>> patdolan <patd...@comcast.net> wrote:
>>>>> On Tuesday, November 16, 2021 at 6:54:15 AM UTC-8, bodk...@gmail.com wrote:
>>>>>
>>>>> Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
>>>>> 2021 until today. I am the only poster that the search returned.
>>>>>
>>>> Search for “unstated special relativity axiom”.
>>>>
>>>> Would the next spoon feeding require a rubber-coated spoon?
>>>>
>>>> --
>>>> Odd Bodkin -- maker of fine toys, tools, tables
>>>
>>> Bodkin,
>>>
>>> I give Ricardo Jimenez priority as being first to publish.
>> And you seem not to have read anything in the thread showing Ricardo where
>> he was astray.
>
> That's is correct Bodkin. Because RJ appears to have answered every
> objection and parried every passe'. Do you disagree?

Yes. Read the conversation.

>
> Bodkin: Yes.
> Dolan: Where?
> Bodkin: I won't spoon feed you.

Well, I’m not going to read for you.

>
>>> But I reserver for myself the claim of being first to rigorously prove
>>> what the brilliant and far-sighted Ricardo only suspected.
>>>
>>> PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular
>>> reasoning on this issue. And rotchm! rotchm had an absolute panic
>>> attack and started delivering a course on vector algebra in order to
>>> avoid the subject entirely.
>>>
>>> I should like to meet the Jimenez fellow online sometime.
>>>
>>> PPS--how's the re-think going Dirk? Let me know if you get stuck again.
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>

[Dirk Van de moortel]

Op 16-nov.-2021 om 17:27 schreef patdolan:

>
>> So when you write ∆x’/∆t’ = ∆x/∆t = v , you clearly have no idea
>> what you are talking about.
>>
> c is a velocity too.
>
> ∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what I
> am talking about?
>
> Prove to this forum that a pair of observers glued to the origins of
> their respective FoRs measure the same relative velocity between
> their FoRs. I have disproved this in three different ways.

After all these years, you still haven't got the faintest clue.
In three different ways you've shown again what a MEGA-duncehead
you are.

Dirk Vdm

[patdolan]

That's all ya got Dirk? Okaaaaaaaay.....my paper has just passed it's first (sub-)peer review. On to bigger venues!

[Dirk Van de moortel]

Op 16-nov.-2021 om 18:50 schreef patdolan:

> On Tuesday, November 16, 2021 at 9:33:01 AM UTC-8, Dirk Van de
> moortel wrote:
>> Op 16-nov.-2021 om 17:27 schreef patdolan:
>>>
>>>> So when you write ∆x’/∆t’ = ∆x/∆t = v , you clearly have no
>>>> idea what you are talking about.
>>>>
>>> c is a velocity too.
>>>
>>> ∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what
>>> I am talking about?
>>>
>>> Prove to this forum that a pair of observers glued to the origins
>>> of their respective FoRs measure the same relative velocity
>>> between their FoRs. I have disproved this in three different
>>> ways.
>> After all these years, you still haven't got the faintest clue. In
>> three different ways you've shown again what a MEGA-duncehead you
>> are.

Make that four.

Dirk Vdm

[Wade Earl]

Dirk Van de moortel wrote:

>>>> Prove to this forum that a pair of observers glued to the origins of
>>>> their respective FoRs measure the same relative velocity between
>>>> their FoRs. I have disproved this in three different ways.
>>> After all these years, you still haven't got the faintest clue. In
>>> three different ways you've shown again what a MEGA-duncehead you are.
>
> Make that four.

you don't understand, look

So the conspiracy theorists were right again
https://www.bitchute.com/video/zw87PJZ3pea9/

[Thomas 'PointedEars' Lahn]

patdolan 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

LOL.

> and separate said reasoning from the algebraic symbols and equations
> that express it,

Since it is fundamentally based in geometry, namely frames of reference
which can understood as (moving) coordinate systems, that approach is
hopeless.

> keeping in mind that mathematics is just another form of rhetorical
> expression wherein falsity can be expressed every bit as plausibly
> as the truth.

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.

In the words of Richard Feynman:

‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols? Mathematics is just a language, and I want to be able
to translate the language." And, in fact, I can — and with patience,
I think I partly did. I could go a little further and explain more
detail — that this means if it’s twice as far away the force is
one-fourth as much, and so on — and can convert all these into words.
I would be, in other words, kind to the layman, as they all sit, hopeful
that you will explain something. Various different people get different
reputations for their skill at explaining to the layman in layman’s
language these difficult and abstruse subjects.

The layman then searches for book after book with the hope that he will
avoid the complexity which ultimately sets in, even by the best expositor
of this type. He reads the things, hoping — he finds, as he reads, a
generally increased confusion, one complicated statement after the other,
one difficult-to-understand thing after the other, all apparently
disconnected from one another — and it becomes a little obscure, and
he hopes that maybe in some other book there’s some explanation which
avoids — I mean, the man almost made it, you see — maybe another fellow
makes it right.

I don’t think it’s possible, because there’s another feature: mathematics
is not just a language; mathematics is a language plus reasoning; it’s
like a language plus logic. Mathematics is a tool for reasoning. It’s in
fact a big collection of the results of some person’s careful thought and
reasoning. […]’

<https://www.feynmanlectures.caltech.edu/fml.html#2>

> “There is only the text.”— J. Derrida

LOL.

> 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.

No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
provided that they are both looking in the same direction and choose their
coordinate systems such that the positive part in on the same (usually
right-hand) side.

> This is exceedingly strange.

There is nothing strange about it.

> How can it be that two relative quantities, space and time, combine to
> produce an absolute quantity called relative velocity?

It is not absolute at all. If it were absolute, then every observer would
assign the same velocity to the same object, regardless of their relative
state of motion. The very point of the theories of relativity is that this
is not so.

However, if you consider the relevant equations of the (inverse) Lorentz
transformation for relative (inertial) motion *only* along the x-axis,

t = γ (t' + v/c² x')
x = γ (x' + v t'),

where for simplicity we say that v would be the velocity (usually just a
speed, but this particular motion is one-dimensional) that an observer that
we define as stationary would assign to a clock that, at rest in its own
frame shows time t, then you can see that for the temporal interval

Δt = γ (Δt' + v/c² Δx') = γ Δt'

and for the spatial interval

Δx = γ (Δx' + v Δt') = γ v Δt',

then the velocity of that moving clock for the stationary observer is still

Δx/Δt = v.

Similarly for an observer at rest in the "moving" frame, when they are
talking about a clock that is at rest in the stationary frame, they
transform the coordinates by the (original) Lorentz transformation

t' = γ (t − v/c² x)
x' = γ (x − v t),

– you can obtain that by solving the equations for the other quantity, not
just by arbitrary substitution – which for (their) time intervals becomes

Δt' = γ (Δt − v/c² Δx) = γ Δt
Δx' = γ (Δx − v Δt) = −γ v Δt.

Then the velocity that they assign to the stationary clock is

Δx'/Δt' = −v,

as we would expect.

> 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.

But the second FoR is implied when we are talking about a velocity. There
has to be some frame of reference relative to which we are defining the
value.

> If that third object happens to be at rest wrt one of the FoRs then
> Einsteinian velocity reduces to Galilean velocity,

Yes, but that is a rather pointless statement because “at rest” means that
the relative velocity is zero.

It is more useful to realize that as one of the relative speeds *approaches*
zero the composed velocity *approaches* the Galilean term:

u(v, u') = (v + u')/(1 + v u'/c²)

lim_{v u' → 0} u(v, u') = v + u',

as v u'/c² → 0 then due to the large value of c.

[This equation did not simply fell out of the sky. It is, again, a
(mathematical) *consequence* of the Lorentz transformation, and Einstein
derives it in “On the Electrodynamics of Moving Bodies” (1905). I have
also derived it here many times.]

> albeit subject to the speed limit c.

That is a contradiction in terms.

> 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.

No.

> 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)

Wrong, see above.

> 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.

But you do not understand special relativity correctly (yet).
Your/an argument/conclusions from your ignorance is a useless fallacy.

> [tl;dr]


PointedEars
--
“Nature uses only the longest threads to weave her patterns
so that each small piece of her fabric reveals the organization
of the entire tapestry.”
—Richard Feynman, theoretical physicist, “Messenger Lecture” 1 (1964)

[Thomas 'PointedEars' Lahn]

Thomas 'PointedEars' Lahn wrote:

> patdolan wrote:
>> 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.
>
> No, the velocity that observer A ascribes to observer B is necessarily the
> arithmetic inverse of the velocity that observer ascribes to observer B,

I meant to write

„No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer B ascribes to observer A,“

> provided that they are both looking in the same direction and choose their
> coordinate systems such that the positive part in on the same (usually
> right-hand) side.

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)

[patdolan]

On Wednesday, November 17, 2021 at 3:09:39 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> patdolan 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
> LOL.
> > and separate said reasoning from the algebraic symbols and equations
> > that express it,
> Since it is fundamentally based in geometry, namely frames of reference
> which can understood as (moving) coordinate systems, that approach is
> hopeless.
> > keeping in mind that mathematics is just another form of rhetorical
> > expression wherein falsity can be expressed every bit as plausibly
> > as the truth.


> 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.

1^4 = i^4

sqrt[ 1^4 ] = sqrt[ i^4 ]

+/-( 1^2 ) = +/-( i^2 )

+/-( 1 ) = +/-( -1 )

+/- 1 = -/+ 1

>
> In the words of Richard Feynman:
>
> ‘You might say, “All right, then, there’s no explanation of the law;
> at least tell me what the law is — why not tell me in words instead of
> in the symbols?

I did this. See MUONS, SCHMUONS! at the bottom of my post. Read through it slow, Long Ears. Then take up my challenge if you dare: Prove that relativity provides only one and unique answer for the elapsed time on the lab clock. I can assure the world and posterity that you can't and won't do prove one, unique time.

Thomas, I have no interest in debating the algebra with you. You are not appreciating the meta-algebra involved. Nor will you anytime soon. Just address the MUONS, SCHMUONS! challenge. Or go silent. You have always been a second stringer in this forum.

[Maciej Wozniak]

On Thursday, 18 November 2021 at 00:09:39 UTC+1, Thomas 'PointedEars' Lahn wrote:
> patdolan 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
> LOL.
> > and separate said reasoning from the algebraic symbols and equations
> > that express it,
> Since it is fundamentally based in geometry, namely frames of reference
> which can understood as (moving) coordinate systems, that approach is
> hopeless.
> > keeping in mind that mathematics is just another form of rhetorical
> > expression wherein falsity can be expressed every bit as plausibly
> > as the truth.
> 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.

Wishes.

> I don’t think it’s possible, because there’s another feature: mathematics
> is not just a language; mathematics is a language plus reasoning

So is natural language. It's just that it can use complicated terms
and complicated rules.

> No, the velocity that observer A ascribes to observer B is necessarily the
> arithmetic inverse of the velocity that observer ascribes to observer B,

In the meantime in the reral world, both observers observe
GPS clocks measuring t'=t, just like all serious clocks
always did.

> It is not absolute at all. If it were absolute, then every observer would
> assign the same velocity to the same object, regardless of their relative
> state of motion. The very point of the theories of relativity is that this
> is not so.

Your tales of observers assigning this and that didn't match the real
observers even in Galileo's time. Your little theories are - simply - too
primitive; or, as you wish, the real observers are too complcated.

> “Nature uses only the longest threads to weave her patterns
> so that each small piece of her fabric reveals the organization
> of the entire tapestry.”
> —Richard Feynman, theoretical physicist, “Messenger Lecture” 1 (1964)

Mystical bullshit.

[Maciej Wozniak]

On Thursday, 18 November 2021 at 00:35:02 UTC+1, Thomas 'PointedEars' Lahn wrote:
> Thomas 'PointedEars' Lahn wrote:
> > patdolan wrote:
> >> 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.
> >
> > No, the velocity that observer A ascribes to observer B is necessarily the
> > arithmetic inverse of the velocity that observer ascribes to observer B,
> I meant to write
>
> „No, the velocity that observer A ascribes to observer B is necessarily the
> arithmetic inverse of the velocity that observer B ascribes to observer A,“

In the meantime in the real world, however, forbidden by
your moronic religion GPS clocks keep measuring

[Richard Hachel]

Le 18/11/2021 à 00:09, Thomas 'PointedEars' Lahn a écrit :

> u(v, u') = (v + u')/(1 + v u'/c²)

Yes, linear addition (because µ=0° and cosµ=1).

and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]

-----

otherwise : speeds general addition :

u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] /
(1+cosµ.u'.v/c²)

R.H.

[Dono.]

No cretinoid, this not not the way velocity composition works. You need to stop making up shit (and eating it)

[Richard Hachel]

LOL.

R.H.

[Python]

Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...

ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?

[Maciej Wozniak]

sqrt, of course, is a multivalue function, returning a set of
numbers; right, poor halfbrain?

[Python]

A multivalued function is NOT a function returning a set of numbers.
BTW, this is off-topic.

> poor halfbrain?

Nice sig, Woz.

[Maciej Wozniak]

https://en.wikipedia.org/wiki/Multivalued_function
quoting
"also called [..], set-valued function"
No, I can't agree it's off topic, since sqrt, a well-known multivalued
function is invoked.

[Richard Hachel]

Ben non, c'est la même que je donne depuis des décennies déjà.

J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.

Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.

Le temps n'érode pas les choses justes.

Il en va de même pour la contraction des distances et la dilatation des
durées.

J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).

Les bonnes formules sont :
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)

Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.

Ca n'a aucune espèce d'intérêt.

L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.

R.H.

[patdolan]

This just goes to prove that whether it's French, English, German, Swahili, or any other language, relativity just doesn't make sense. C'est pas?

[Odd Bodkin]

Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?

[Richard Hachel]

Le 18/11/2021 à 23:52, "Dono." a écrit :

> On Thursday, November 18, 2021 at 2:46:13 PM UTC-8, Richard Hachel wrote:
>> Le 18/11/2021 à 00:09, Thomas 'PointedEars' Lahn a écrit :
>>
>> > u(v, u') = (v + u')/(1 + v u'/c²)
>> Yes, linear addition (because µ=0° and cosµ=1).
>>
>> and perpendicular addition µ=90° cosµ=0 --->
>> u=sqrt|v²+u'²-v²u'²/c²]
>>
>> -----
>>
>> otherwise : speeds general addition :
>>

>> u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²))²] / (1+cosµ.u'.v/c²)

>>
>>
>> R.H.
> No cretinoid, this not not the way velocity composition works. You need to stop
> making up shit (and eating it)

Formule d'addition générale des vitesses relativistes.

<http://news2.nemoweb.net/jntp?KOAY_Ps7TstyxoZ8n1LolayUV30@jntp/Data.Media:1>

R.H.

[patdolan]

So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge. Not worth your time? Or perhaps the challenge is just too much for you; that's what I and everyone else suspect. To remind you of the challenge: Prove that special relativity provides only one, unique solution for how much time elapses on the laboratory clock. Choose which one.

[Dono.]

On Friday, November 19, 2021 at 7:43:42 AM UTC-8, Richard Hachel wrote:
> Le 18/11/2021 à 23:52, "Dono." a écrit :
> > On Thursday, November 18, 2021 at 2:46:13 PM UTC-8, Richard Hachel wrote:
> >> Le 18/11/2021 à 00:09, Thomas 'PointedEars' Lahn a écrit :
> >>
> >> > u(v, u') = (v + u')/(1 + v u'/c²)
> >> Yes, linear addition (because µ=0° and cosµ=1).
> >>
> >> and perpendicular addition µ=90° cosµ=0 --->
> >> u=sqrt|v²+u'²-v²u'²/c²]
> >>
> >> -----
> >>
> >> otherwise : speeds general addition :
> >>
> >> u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²))²] / (1+cosµ.u'.v/c²)
> >>
> >>
> >> R.H.
> > No cretinoid, this not not the way velocity composition works. You need to stop
> > making up shit (and eating it)
> Formule d'addition générale des vitesses relativistes.
>

Rubbish

[Odd Bodkin]

There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.

Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.

> Not worth your time? Or perhaps the challenge is just too much for you;
> that's what I and everyone else suspect. To remind you of the challenge:
> Prove that special relativity provides only one, unique solution for how
> much time elapses on the laboratory clock. Choose which one.
>

[Maciej Wozniak]

[patdolan]

I understand relativity far better than you Bodkin. MUONS, SCHMUONS demonstrates exactly how much better I understand it than either you or Einstein.

I request a voice vote of this forum. Who knows relativity better? Me or Einstein & Bodkin?

[Odd Bodkin]

No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.

So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)

>
> I request a voice vote of this forum. Who knows relativity better? Me
> or Einstein & Bodkin?
>

[Maciej Wozniak]

They're not, indeed; that's why forbidden by your moronic

[patdolan]

You are quite a dance expert yourself, Bodkin. You specialty is the fan dance but you never get quite naked when it comes to showing what you can and cannot do with relativity.

[Odd Bodkin]

And this is where your game is silly. This isn’t a game of poker, with
hidden cards and a lot of bluffing. Relativity is fully documented and
thoroughly described in a century’s worth of textbooks. So when you say
something and declare “This is relativity”, it’s OBVIOUS when you say
something stupid. It’s OBVIOUS when you haven’t read textbooks.

So here you are, pretending to play poker, and you’ve laid out your cards,
a 2, a 5, a 7, a Jack, and a Queen, and you announce “Straight flush”. And
you do this in a room full of people who instantly know you are an idiot
and don’t know what you’re talking about. And then you double down and say,
“So what cards have YOU got, chump?”

>>>
>>> I request a voice vote of this forum. Who knows relativity better? Me
>>> or Einstein & Bodkin?
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>

[patdolan]

Bodkin, this is one of your most penetrating insights. It accurately portrays this, and every other difference of opinion in Science. In fact, it is Science itself! You are right, I do have an excellent hand. And I've laid it down on the table. I call.

But you have phrased it better. In your own words "So what cards do you have, chump?"

[Maciej Wozniak]

And in the meantime in the real world, forbidden by your insane religion

[Luigi Cotta]

patdolan wrote:

> Bodkin, this is one of your most penetrating insights. It accurately
> portrays this, and every other difference of opinion in Science. In
> fact,
> it is Science itself! You are right, I do have an excellent hand. And
> I've laid it down on the table. I call.
>
> But you have phrased it better. In your own words "So what cards do you
> have, chump?"

tensors, 4D.

[Odd Bodkin]

You have a 2, 5, 7, J, Q hand which you are calling an “excellent hand”

>
> But you have phrased it better. In your own words "So what cards do you have, chump?"

It’s pretty simple, Pat. You asked the question, what is the “real”
distance the muon travels through, and what is the “real” time interval of
time that it traveled through decay? This question in and of itself proves
to all at the table that you have no idea what you’re talking about.

One of the foundational observations in relativity is that, given two
events (in your case, the creation of the muon for one event, and the
collision of the muon with the earth for the other event), the distance
between those events is frame dependent and the time between those events
is frame dependent. There IS NO “real” distance between the events and
there IS NO “real” time interval between the events. The distances and time
intervals have values that are accidents of the choice of reference frame.

If you had read a book on special relativity — ANY book on special
relativity — this would have been pounded on as a central theme.

For you to pose a “challenge” about what the “real” values are just shows
(once again) that you opened your mouth when your foot was too close to it.


If you’d like a simple and accessible book to understanding special
relativity, I might suggest General Relativity from A to B. Start there and
maybe you will say fewer stupid things.

>
>>>>> I request a voice vote of this forum. Who knows relativity better? Me
>>>>> or Einstein & Bodkin?
>>>>>
>>>> --
>>>> Odd Bodkin -- maker of fine toys, tools, tables
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>

[Odd Bodkin]

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. “You”
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.

“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”

“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”

No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.

You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.

> Nothing strange here; the labs clock is traveling at γ=2 with respect to
> you so it only logs half as much elapsed time as you.
>
> One of the lab scientists now cries out “Just a minute! That’s not how
> this story goes. First of all, it is the hadron that is
> Lorentz-flattened, not us. And second of all, that hadron used it’s
> flattened FoR to determine that is was only 571.56 meters away from our
> scintillator at the point it became a muon. Our unflattened earth FoR
> clearly indicated that the hadron was actually 1143.12 meters away when
> it became a muon. That’s why it took 4.4 microseconds, traveling at .866
> c, to reach our scintillator. And that’s what our lab clock shows. Not
> 1.1 microseconds, like that dumb muon is claiming.”
>
> Special relativity leaves us in a quandary. How much time actually did
> elapse on the lab clock from the moment of the muon’s inception to the
> moment of its scintillating demise? Was it 1.1 microseconds? Or was it
> 4.4 microseconds? It has to be one or the other, it can’t be both.
> However, different elapsed times on the same lab clock can be calculated
> by the muon and by the earthbound scientists with equal justification.
> Special relativity is incapable of providing a unique elapsed time on the
> lab clock. I challenge anyone to show that special relativity can
> provide a unique elapsed time in the preceding case.
>
> Special relativity’s greatest experimental confirmation turns out to be
> it’s second greatest falsification. Gravity has been special
> relativity’s greatest falsification since 1907 ( check out the author’s
> brilliant post on special relativity vs. Kepler’s 3rd law ).

[Thomas 'PointedEars' Lahn]

patdolan wrote:

> On Wednesday, November 17, 2021 at 3:09:39 PM UTC-8, Thomas 'PointedEars'
> Lahn wrote:
>> patdolan wrote:
>> > […] mathematics is just another form of rhetorical expression wherein

>> > falsity can be expressed every bit as plausibly as the truth.
>>

>> 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.
>
> 1^4 = i^4

That much is true.

> sqrt[ 1^4 ] = sqrt[ i^4 ]
>
> +/-( 1^2 ) = +/-( i^2 )
>
> +/-( 1 ) = +/-( -1 )
>
> +/- 1 = -/+ 1

Your logic is flawed.

>> In the words of Richard Feynman:
>>
>> ‘You might say, “All right, then, there’s no explanation of the law;
>> at least tell me what the law is — why not tell me in words instead of
>> in the symbols?
>
> I did this.

I know that you did. Feynman explains there why your request cannot be
fulfilled.

> See MUONS, SCHMUONS! at the bottom of my post. Read through it slow,
> Long Ears.

I have better things to do than discussing with you at this low a level.

> [tl;dr]


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven

[patdolan]

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. 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. The lab scientists, using relativity and the standard muon life-span, calculate the same lab clock ticks off 4.4 seconds for the same trip.

Now take a deep breath and form an argument that addresses the facts of the case, not imaginary t=0 stuff.

[patdolan]

On Friday, November 19, 2021 at 3:00:32 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
> patdolan wrote:
>
> > On Wednesday, November 17, 2021 at 3:09:39 PM UTC-8, Thomas 'PointedEars'
> > Lahn wrote:
> >> patdolan wrote:
> >> > […] mathematics is just another form of rhetorical expression wherein
> >> > falsity can be expressed every bit as plausibly as the truth.
> >>
> >> 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.
> >
> > 1^4 = i^4
>
> That much is true.
>
> > sqrt[ 1^4 ] = sqrt[ i^4 ]
> >
> > +/-( 1^2 ) = +/-( i^2 )
> >
> > +/-( 1 ) = +/-( -1 )
> >
> > +/- 1 = -/+ 1
>
> Your logic is flawed.

How, long ears. How?

[Paul Alsing]

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. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...

[Maciej Wozniak]

On Saturday, 20 November 2021 at 07:24:32 UTC+1, 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. The term *i* is "defined" to be 'the square root of -1',

:) And how is "the square root" defined, Al,
poor halfbrain?

[patdolan]

[ This is going to be fun, tee-hee-hee...] So Paul A. #2, when you multiply a number by a definition is the product a number or another definition?

[Python]

There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).

[Maciej Wozniak]

It's not true it was recently, but anyway poor halfbrain Al
may always start asking a wiser one about some explainations.

[patdolan]

I have accounted for both branches of the double branch sqrt function in my derivation and conclusion and made the correct branch cuts:

+/-1 = -/+1

Just as I would have accounted for all three branches of the cube root function, four branches for the...

I am an algebraic King Cobra. You are just a python. Never mess with the king.

[Odd 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.

[Odd Bodkin]

patdolan <patd...@comcast.net> wrote:
> On Friday, November 19, 2021 at 3:00:32 PM UTC-8, Thomas 'PointedEars' Lahn wrote:
>> patdolan wrote:
>>
>>> On Wednesday, November 17, 2021 at 3:09:39 PM UTC-8, Thomas 'PointedEars'
>>> Lahn wrote:
>>>> patdolan wrote:
>>>>> […] mathematics is just another form of rhetorical expression wherein
>>>>> falsity can be expressed every bit as plausibly as the truth.
>>>>
>>>> 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.
>>>
>>> 1^4 = i^4
>>
>> That much is true.
>>
>>> sqrt[ 1^4 ] = sqrt[ i^4 ]
>>>
>>> +/-( 1^2 ) = +/-( i^2 )
>>>
>>> +/-( 1 ) = +/-( -1 )
>>>
>>> +/- 1 = -/+ 1
>>
>> Your logic is flawed.
>
> How, long ears. How?

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.

Idiot.

Half-ignorant idiot.

Blissfully ignorant of being ignorant, half-ignorant idiot.

>>
>>>> In the words of Richard Feynman:
>>>>
>>>> ‘You might say, “All right, then, there’s no explanation of the law;
>>>> at least tell me what the law is — why not tell me in words instead of
>>>> in the symbols?
>>>
>>> I did this.
>>
>> I know that you did. Feynman explains there why your request cannot be
>> fulfilled.
>>
>>> See MUONS, SCHMUONS! at the bottom of my post. Read through it slow,
>>> Long Ears.
>>
>> I have better things to do than discussing with you at this low a level.
>>
>>> [tl;dr]
>>
>>
>> PointedEars
>> --
>> «Nec fasces, nec opes, sola artis sceptra perennant.»
>> (“Neither high office nor power, only the scepters of science survive.”)
>>
>> —Tycho Brahe, astronomer (1546-1601): inscription at Hven
>

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

[patdolan]

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. I will be more explicit in my conclusion:

+/-1+/-i0 = -/+1-/+i0

[patdolan]

The muon is bathed in the light cone from the lab clock a the moment of it's inception. The muon notes what the light cone is indicating at that moment while it still 572m/1143m away from the clock. The muon notes the light from the lab clock again when it is in the scintillator. This is T1. From T1 the muon subtracts 2.2 microseconds to get the elapsed time on the lab clock for the muon's trip. This elapsed time will be 1.1 microseconds according to relativity and the muon. No lattice of clocks, no synchronization no nothing. Easy-peasy Bodkin.

[patdolan]

*the difference of the times is T1: light cone time at inception - light cone in scintillator = T1

[patdolan]

So on more time just to make it absolutely clear for LongEars:

The muon is bathed in the light cone from the lab clock a the moment of it's inception. The muon notes what the light cone is indicating at that moment while it still 572m/1143m away from the clock. The muon notes the light from the lab clock again when it is in the scintillator. This is T-inception. The muon notes the light from the lab clock again when it is in the scintillator. This is T-demise.

The muon subtracts another 2.2 microseconds to give the total elapsed time on the lab clock:

Total elapsed time on lab clock for relativistic muon trip from muon's pov = (T-demise - T-inception) - 2.2 microseconds = 1.1 microseconds

[Paul Alsing]

Well, the *king* is dead... dead wrong!

[patdolan]

Very cheeky, Paul A#2. I recall that Paul A#1 used to be quite cheeky also until he was put in his place. Paul A#1 lives in a country that still respects Kings and Queens.

[Paul Alsing]

Ya know, when you have both feet in your mouth at the same time, you don't have a leg to stand on...

[patdolan]

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.

Declare this, you checkmated chump.

[Odd Bodkin]

And how long did it take that light to get there?

> The muon notes the light from the lab clock again when it is in the
> scintillator. This is T1. From T1 the muon subtracts 2.2 microseconds
> to get the elapsed time on the lab clock for the muon's trip. This
> elapsed time will be 1.1 microseconds according to relativity and the muon.

No, not according to relativity. You are an idiot.

> No lattice of clocks, no synchronization no nothing. Easy-peasy Bodkin.
>

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

[pehache]

Le 19/11/2021 à 00:21, Richard Hachel a écrit :

>
> LOL.
> R.H.
>

Tu devrais inviter Pencho Valev à boire le thé chez toi. Ca ferait des
vacances à usenet.

--
"...sois ouvert aux idées des autres pour peu qu'elles aillent dans le
même sens que les tiennes.", ST sur fr.bio.medecine

[patdolan]

On Saturday, November 20, 2021 at 12:47:54 PM UTC-8, pehache wrote:
> Le 19/11/2021 à 00:21, Richard Hachel a écrit :
>
> >
> > LOL.
> > R.H.
> >
>
> Tu devrais inviter Pencho Valev à boire le thé chez toi. Ca ferait des
> vacances à usenet.
>

Pencho et moi buvons et mangeons de la relativité pour le petit-déjeuner, le déjeuner et le dîner. Toi et Bodkin pouvez prendre le thé avec vos assiettes de corbeau.

[Richard Hachel]

[Richard Hachel]

Le 20/11/2021 à 21:47, pehache a écrit :

> Le 19/11/2021 à 00:21, Richard Hachel a écrit :
>
>>
>> LOL.
>> R.H.
>>
>
> Tu devrais inviter Pencho Valev à boire le thé chez toi. Ca ferait des
> vacances à usenet.

Nan écoute pehache, soit un peu beau joueur.

Et même si je racontais des conneries comme Pencho Valev, comme tu dis,
pffff... Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois
même plus un post par jour.

Je dis ça, je dis rien.

R.H.

[Thomas 'PointedEars' Lahn]

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.

> 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)

[Thomas 'PointedEars' Lahn]

Thomas 'PointedEars' Lahn wrote:

> 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. ∎

Therefore, I should have been more precise and should have written
“_a_ solution” instead.


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

[Thomas 'PointedEars' Lahn]

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)

[patdolan]

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.

[Paul Alsing]

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...

[Thomas 'PointedEars' Lahn]

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)

[patdolan]

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.

[robby]

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

[patdolan]

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

[Odd Bodkin]

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.

But what is clear through out this is that you never READ exactly how
Frisch and Smith did this experiment, and you’ve never READ how relativity
describes this phenomenon, both of which are common expositions available
in lots of books.

If you’d only do that, you’d bluster less with loaded squirt guns you’re
holding backwards.

[patdolan]

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 now deign to delight this forum with your calculation of the latter? Thank you, in advance. Also, please use an online translator to render a version of your answer in French for those Francophones looking in on our debate. Thank you again in advance.

[Paul Alsing]

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

[Townes Olson]

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.

[Odd Bodkin]

That depends on what you mean “know”. If you mean, what does relativity
CALCULATE these to be, then yes, that’s certainly doable. You’ve in fact
already listed those results. If you want to be sure, of course, it’s a
mystery to me why you don’t read any of the copious presentations of this
very thing in many textbooks. Why you need to have explanations REPRINTED
here, is frankly beyond me, other than you just trying to get people to do
things you’re too lazy to do yourself.

[Thomas 'PointedEars' Lahn]

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)

[patdolan]

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.

[Townes Olson]

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).

[patdolan]

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

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

[Townes Olson]

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?

[Odd Bodkin]

He’s walking you through something standard, Pat.

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

[patdolan]

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.

[Paul Alsing]

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.

[Townes Olson]

> 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.

[patdolan]

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.