The Lorentz transformation and other results of special relativity
seem to have the built in assumption that the two systems that are
being compared have compatible time and space units, otherwise the
results are nonsense. But I can't recall seeing a discussion of how
to compare the units of two frames in relative motion with each other
so that equations come out right.

On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
> The Lorentz transformation and other results of special relativity
> seem to have the built in assumption that the two systems that are
> being compared have compatible time and space units, otherwise the
> results are nonsense. But I can't recall seeing a discussion of how
> to compare the units of two frames in relative motion with each other
> so that equations come out right.

Ricardo! I am honored to make your online acquaintance. Bodkin steered me to your post regarding this subject a few months back. Your efforts proceed mine by a month or so.

I succeeded in proving much to the consternation of this forum that the v in the Lorentz Transforms is a thoroughly inconsistent quantity when used from different inertial frames. In fact I deductively derived that v is not even a variable as used in the LTs. But rather, it is always the case that v = 0 or v = c.

On Thursday, April 14, 2022 at 5:49:57 PM UTC-7, patdolan wrote:
> On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
> > The Lorentz transformation and other results of special relativity
> > seem to have the built in assumption that the two systems that are
> > being compared have compatible time and space units, otherwise the
> > results are nonsense. But I can't recall seeing a discussion of how
> > to compare the units of two frames in relative motion with each other
> > so that equations come out right.
> Ricardo! I am honored to make your online acquaintance. Bodkin steered me to your post regarding this subject a few months back. Your efforts proceed mine by a month or so.
>
> I succeeded in proving much to the consternation of this forum that the v in the Lorentz Transforms is a thoroughly inconsistent quantity when used from different inertial frames. In fact I deductively derived that v is not even a variable as used in the LTs. But rather, it is always the case that v = 0 or v = c.
Here is the heart of my argument:

We start, as always with our premises. We premise the Lorentz Transforms by reference. We further premise the quantity v' which is equal to, and interchangeable with -v of the Lorentz transforms. Finally, we premise that v and v' are both constrained have values lying in the range 0 < ||v|| < c and 0 < ||v'|| < c. The last premise is Einsteins second postulate.

Our strategy is to express v' in terms of v, t, and x then solve for v in hopes of arriving at the semi-identity -v = v'. Deltas (∆) are omitted for brevity; their locations are easily recognized by the competent student.

We can express v' in terms of x, t, and v, as v' = -x'/t'.

We now apply the LTs to the numerator and denominator of the fraction -x'/t' to get

x'/t' = g(x-vt)/g(t-vx/c^2) voi la!

x'/t' = (x-vt)/(t-vx/c^2)

We note here that our null hypothesis states v = -v' = -x'/t'. So we substitute -x'/t' for v and we are off to the races!

x'/t' = (x+[x't/t'])/(t+xx'/t'c^2)

x't/t' + x x'^2/c^2t'^2 = x + x't/t'

xx'^2/c^2t'^2 = x

x'^2 = c^2t'^2

x'/t' = c

v' = c

This contradicts the null hypothesis we wish to prove. We have conclusively and deductively disproved that v can ever be equal to -v'. And we have proved the the LT's are algebraically inconsistent by reductio ad absurdum. This wrings down the final curtain on Special Relativity.

patdolan wrote:
> [lot of bs]
>
> We can express v' in terms of x, t, and v, as v' = -x'/t'.

So you think that for all events in the "moving" frame the
ratio between x and t coordinates is constant and equal to v'?

Seriously? You miss that (x',t') are coordinates of ANY event
as measured in S'. (x',t') can be (0,0), (pi, -10) or (7, 10^100),
or whatever... *sigh*

You don't have the faintest idea of what all of this is about,
do you?

On Thursday, April 14, 2022 at 6:08:09 PM UTC-7, Python wrote:
> patdolan wrote:
> > [lot of bs]
> >
> > We can express v' in terms of x, t, and v, as v' = -x'/t'.
> So you think that for all events in the "moving" frame the
> ratio between x and t coordinates is constant and equal to v'?
>
> Seriously? You miss that (x',t') are coordinates of ANY event
> as measured in S'. (x',t') can be (0,0), (pi, -10) or (7, 10^100),
> or whatever... *sigh*
>
> You don't have the faintest idea of what all of this is about,
> do you?
Take a breath, Python. Yes, the derivation shows absurdity because the LTs are absurd. They are absurd for the reason...well...let's let Ricardo have a turn.

patdolan wrote:
> On Thursday, April 14, 2022 at 6:08:09 PM UTC-7, Python wrote:
>> patdolan wrote:
>>> [lot of bs]
>>>
>>> We can express v' in terms of x, t, and v, as v' = -x'/t'.
>> So you think that for all events in the "moving" frame the
>> ratio between x and t coordinates is constant and equal to v'?
>>
>> Seriously? You miss that (x',t') are coordinates of ANY event
>> as measured in S'. (x',t') can be (0,0), (pi, -10) or (7, 10^100),
>> or whatever... *sigh*
>>
>> You don't have the faintest idea of what all of this is about,
>> do you?
> Take a breath, Python. Yes, the derivation shows absurdity because the LTs are absurd. They are absurd for the reason...well...let's let Ricardo have a turn.

Come on, Pat. You are assuming a constraint on (x',t') which is
asinine. Transformation equations apply to ALL values for
coordinates.

This is blatantly asinine to the last degree...

patdolan <patdolan@comcast.net> wrote:
> On Thursday, April 14, 2022 at 5:49:57 PM UTC-7, patdolan wrote:
>> On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
>>> The Lorentz transformation and other results of special relativity
>>> seem to have the built in assumption that the two systems that are
>>> being compared have compatible time and space units, otherwise the
>>> results are nonsense. But I can't recall seeing a discussion of how
>>> to compare the units of two frames in relative motion with each other
>>> so that equations come out right.
>> Ricardo! I am honored to make your online acquaintance. Bodkin steered
>> me to your post regarding this subject a few months back. Your efforts
>> proceed mine by a month or so.
>>
>> I succeeded in proving much to the consternation of this forum that the
>> v in the Lorentz Transforms is a thoroughly inconsistent quantity when
>> used from different inertial frames. In fact I deductively derived that
>> v is not even a variable as used in the LTs. But rather, it is always
>> the case that v = 0 or v = c.
> Here is the heart of my argument:
>
> We start, as always with our premises. We premise the Lorentz Transforms
> by reference. We further premise the quantity v' which is equal to, and
> interchangeable with -v of the Lorentz transforms. Finally, we premise
> that v and v' are both constrained have values lying in the range 0 <
> ||v|| < c and 0 < ||v'|| < c. The last premise is Einsteins second postulate.
>
> Our strategy is to express v' in terms of v, t, and x then solve for v in
> hopes of arriving at the semi-identity -v = v'. Deltas (∆) are omitted
> for brevity; their locations are easily recognized by the competent student.
>
> We can express v' in terms of x, t, and v, as v' = -x'/t'.

Yup. This is where you lost your head last time too.
But you feel the need to have 18 people tell you the same thing?

>
> We now apply the LTs to the numerator and denominator of the fraction -x'/t' to get
>
> x'/t' = g(x-vt)/g(t-vx/c^2) voi la!
>
> x'/t' = (x-vt)/(t-vx/c^2)
>
> We note here that our null hypothesis states v = -v' = -x'/t'. So we
> substitute -x'/t' for v and we are off to the races!
>
> x'/t' = (x+[x't/t'])/(t+xx'/t'c^2)
>
> x't/t' + x x'^2/c^2t'^2 = x + x't/t'
>
> xx'^2/c^2t'^2 = x
>
> x'^2 = c^2t'^2
>
> x'/t' = c
>
> v' = c
>
> This contradicts the null hypothesis we wish to prove. We have
> conclusively and deductively disproved that v can ever be equal to -v'.
> And we have proved the the LT's are algebraically inconsistent by
> reductio ad absurdum. This wrings down the final curtain on Special Relativity.
>

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

On Thursday, April 14, 2022 at 6:27:32 PM UTC-7, bodk...@gmail.com wrote:
> patdolan <patd...@comcast.net> wrote:
> > On Thursday, April 14, 2022 at 5:49:57 PM UTC-7, patdolan wrote:
> >> On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
> >>> The Lorentz transformation and other results of special relativity
> >>> seem to have the built in assumption that the two systems that are
> >>> being compared have compatible time and space units, otherwise the
> >>> results are nonsense. But I can't recall seeing a discussion of how
> >>> to compare the units of two frames in relative motion with each other
> >>> so that equations come out right.
> >> Ricardo! I am honored to make your online acquaintance. Bodkin steered
> >> me to your post regarding this subject a few months back. Your efforts
> >> proceed mine by a month or so.
> >>
> >> I succeeded in proving much to the consternation of this forum that the
> >> v in the Lorentz Transforms is a thoroughly inconsistent quantity when
> >> used from different inertial frames. In fact I deductively derived that
> >> v is not even a variable as used in the LTs. But rather, it is always
> >> the case that v = 0 or v = c.
> > Here is the heart of my argument:
> >
> > We start, as always with our premises. We premise the Lorentz Transforms
> > by reference. We further premise the quantity v' which is equal to, and
> > interchangeable with -v of the Lorentz transforms. Finally, we premise
> > that v and v' are both constrained have values lying in the range 0 <
> > ||v|| < c and 0 < ||v'|| < c. The last premise is Einsteins second postulate.
> >
> > Our strategy is to express v' in terms of v, t, and x then solve for v in
> > hopes of arriving at the semi-identity -v = v'. Deltas (∆) are omitted
> > for brevity; their locations are easily recognized by the competent student.
> >
> > We can express v' in terms of x, t, and v, as v' = -x'/t'.
> Yup. This is where you lost your head last time too.
> But you feel the need to have 18 people tell you the same thing?
> >
> > We now apply the LTs to the numerator and denominator of the fraction -x'/t' to get
> >
> > x'/t' = g(x-vt)/g(t-vx/c^2) voi la!
> >
> > x'/t' = (x-vt)/(t-vx/c^2)
> >
> > We note here that our null hypothesis states v = -v' = -x'/t'. So we
> > substitute -x'/t' for v and we are off to the races!
> >
> > x'/t' = (x+[x't/t'])/(t+xx'/t'c^2)
> >
> > x't/t' + x x'^2/c^2t'^2 = x + x't/t'
> >
> > xx'^2/c^2t'^2 = x
> >
> > x'^2 = c^2t'^2
> >
> > x'/t' = c
> >
> > v' = c
> >
> > This contradicts the null hypothesis we wish to prove. We have
> > conclusively and deductively disproved that v can ever be equal to -v'.
> > And we have proved the the LT's are algebraically inconsistent by
> > reductio ad absurdum. This wrings down the final curtain on Special Relativity.
> >
> --
> Odd Bodkin -- maker of fine toys, tools, tables
As you can see Ricardo, this subject still raises a firestorm of protest amongst relativists hellbent on saving appearences of the LTs. Please also see my posts on inconsistency of the sacred spacetime interval; it's an empty concept. And don't get me started on the Frisch-Smith muons...

My current project is debunking Olson Events.

Utterly demented crank Pat Dolan wrote:
> On Thursday, April 14, 2022 at 6:27:32 PM UTC-7, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Thursday, April 14, 2022 at 5:49:57 PM UTC-7, patdolan wrote:
>>>> On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
>>>>> The Lorentz transformation and other results of special relativity
>>>>> seem to have the built in assumption that the two systems that are
>>>>> being compared have compatible time and space units, otherwise the
>>>>> results are nonsense. But I can't recall seeing a discussion of how
>>>>> to compare the units of two frames in relative motion with each other
>>>>> so that equations come out right.
>>>> Ricardo! I am honored to make your online acquaintance. Bodkin steered
>>>> me to your post regarding this subject a few months back. Your efforts
>>>> proceed mine by a month or so.
>>>>
>>>> I succeeded in proving much to the consternation of this forum that the
>>>> v in the Lorentz Transforms is a thoroughly inconsistent quantity when
>>>> used from different inertial frames. In fact I deductively derived that
>>>> v is not even a variable as used in the LTs. But rather, it is always
>>>> the case that v = 0 or v = c.
>>> Here is the heart of my argument:
>>>
>>> We start, as always with our premises. We premise the Lorentz Transforms
>>> by reference. We further premise the quantity v' which is equal to, and
>>> interchangeable with -v of the Lorentz transforms. Finally, we premise
>>> that v and v' are both constrained have values lying in the range 0 <
>>> ||v|| < c and 0 < ||v'|| < c. The last premise is Einsteins second postulate.
>>>
>>> Our strategy is to express v' in terms of v, t, and x then solve for v in
>>> hopes of arriving at the semi-identity -v = v'. Deltas (∆) are omitted
>>> for brevity; their locations are easily recognized by the competent student.
>>>
>>> We can express v' in terms of x, t, and v, as v' = -x'/t'.
>> Yup. This is where you lost your head last time too.
>> But you feel the need to have 18 people tell you the same thing?
>>>
>>> We now apply the LTs to the numerator and denominator of the fraction -x'/t' to get
>>>
>>> x'/t' = g(x-vt)/g(t-vx/c^2) voi la!
>>>
>>> x'/t' = (x-vt)/(t-vx/c^2)
>>>
>>> We note here that our null hypothesis states v = -v' = -x'/t'. So we
>>> substitute -x'/t' for v and we are off to the races!
>>>
>>> x'/t' = (x+[x't/t'])/(t+xx'/t'c^2)
>>>
>>> x't/t' + x x'^2/c^2t'^2 = x + x't/t'
>>>
>>> xx'^2/c^2t'^2 = x
>>>
>>> x'^2 = c^2t'^2
>>>
>>> x'/t' = c
>>>
>>> v' = c
>>>
>>> This contradicts the null hypothesis we wish to prove. We have
>>> conclusively and deductively disproved that v can ever be equal to -v'.
>>> And we have proved the the LT's are algebraically inconsistent by
>>> reductio ad absurdum. This wrings down the final curtain on Special Relativity.
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
> As you can see Ricardo, this subject still raises a firestorm of protest amongst relativists hellbent on saving appearences of the LTs. Please also see my posts on inconsistency of the sacred spacetime interval; it's an empty concept. And don't get me started on the Frisch-Smith muons...

This is hopelesss, but anyway... Have you tried to apply this "logic"
to Galilean Transformation Equations?

On Thursday, April 14, 2022 at 6:02:37 PM UTC-7, cretin pat dolan repeated the same imbecilities as earlier:

> Here is the heart of my argument:
<same cretinisms as before>

On Thursday, April 14, 2022 at 9:13:09 PM UTC-3, Ricardo Jimenez wrote:

> The Lorentz transformation and other results of special relativity
> seem to have the built in assumption that the two systems that are
> being compared have compatible time and space units, otherwise the
> results are nonsense. But I can't recall seeing a discussion of how
> to compare the units of two frames in relative motion with each other
> so that equations come out right.

Ricardo, como has visto, nadie le ha prestado atención a tu duda.

Creo que la respuesta es que la garantía de usar las mismas unidades de medida
vienen dadas por la base de partida: las transformadas galileanas:

t' = t satisface la condición de iguales unidades temporales.

x - x' = v.t satisface la condición de iguales unidades de longitud/distancia.

Aquí puede derivarse que v = (x - x')/t satisface la condición de igual unidad de velocidad para ambas tramas.

On Friday, 15 April 2022 at 03:08:09 UTC+2, Python wrote:
> patdolan wrote:
> > [lot of bs]
> >
> > We can express v' in terms of x, t, and v, as v' = -x'/t'.
> So you think that for all events in the "moving" frame the

Oh, stinker Python is opening its muzzle again,
and trying to pretend he knows something.
Tell me, poor stinker, what is your definition of
a "theory" in the terms of Peano arithmetic?
See: if a theorem is going to be a part of a theory,
it has to be formulable in the language of the
theory. Do you get it? Or are you too stupid even for
that, poor stinker?

Maciej Wozniak schwrote:
> On Friday, 15 April 2022 at 03:08:09 UTC+2, Python wrote:
>> patdolan wrote:
>>> [lot of bs]
>>>
>>> We can express v' in terms of x, t, and v, as v' = -x'/t'.
>> So you think that for all events in the "moving" frame the
>
>
> Oh, stinker [...]

OMG! Now you're choking on mentioning x, t or x'! Giving
names to numbers! It is Marxist indoctrination, right Maciej?

On Friday, 15 April 2022 at 07:10:42 UTC+2, Python wrote:
> Maciej Wozniak schwrote:
> > On Friday, 15 April 2022 at 03:08:09 UTC+2, Python wrote:
> >> patdolan wrote:
> >>> [lot of bs]
> >>>
> >>> We can express v' in terms of x, t, and v, as v' = -x'/t'.
> >> So you think that for all events in the "moving" frame the
> >
> >
> > Oh, stinker [...]
>
> OMG! Now you're choking on mentioning x, t or x'! Giving

Oh, stinker Python is opening its muzzle again,
and trying to pretend he knows something.
Tell me, poor stinker, what is your definition of
a "theory" in the terms of Peano arithmetic?
See: if a theorem is going to be a part of a theory,
it has to be formulable in the language of the
theory. Do you get it? Or are you too stupid even for
that, poor stinker?

Richard Hertz wrote:
>
> On Thursday, April 14, 2022 at 9:13:09 PM UTC-3, Ricardo Jimenez wrote:
>
> > The Lorentz transformation and other results of special relativity
> > seem to have the built in assumption that the two systems that are
> > being compared have compatible time and space units, otherwise the
> > results are nonsense. But I can't recall seeing a discussion of how
> > to compare the units of two frames in relative motion with each other
> > so that equations come out right.
>
> Ricardo, como has visto, nadie le ha prestado atención a tu duda.
>
> Creo que la respuesta es que la garantía de usar las mismas unidades de medida
> vienen dadas por la base de partida: las transformadas galileanas:
>
> t' = t satisface la condición de iguales unidades temporales.
>
> x - x' = v.t satisface la condición de iguales unidades de longitud/distancia.
>
> Aquí puede derivarse que v = (x - x')/t satisface la condición de igual unidad de velocidad para ambas tramas.

you cut and paste the spanish language?

--
The Starmaker -- To question the unquestionable, ask the unaskable,
to think the unthinkable, mention the unmentionable, say the unsayable, and challenge
the unchallengeable.

Ricardo Jimenez wrote:

> The Lorentz transformation and other results of special relativity
> seem to have the built in assumption that the two systems that are
> being compared have compatible time and space units, otherwise the
> results are nonsense. But I can't recall seeing a discussion of how
> to compare the units of two frames in relative motion with each other
> so that equations come out right.

Which “units of two frames”?

PointedEars
--
I heard that entropy isn't what it used to be.

(from: WolframAlpha)

patdolan <patdolan@comcast.net> wrote:
> On Thursday, April 14, 2022 at 6:27:32 PM UTC-7, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Thursday, April 14, 2022 at 5:49:57 PM UTC-7, patdolan wrote:
>>>> On Thursday, April 14, 2022 at 5:13:09 PM UTC-7, Ricardo Jimenez wrote:
>>>>> The Lorentz transformation and other results of special relativity
>>>>> seem to have the built in assumption that the two systems that are
>>>>> being compared have compatible time and space units, otherwise the
>>>>> results are nonsense. But I can't recall seeing a discussion of how
>>>>> to compare the units of two frames in relative motion with each other
>>>>> so that equations come out right.
>>>> Ricardo! I am honored to make your online acquaintance. Bodkin steered
>>>> me to your post regarding this subject a few months back. Your efforts
>>>> proceed mine by a month or so.
>>>>
>>>> I succeeded in proving much to the consternation of this forum that the
>>>> v in the Lorentz Transforms is a thoroughly inconsistent quantity when
>>>> used from different inertial frames. In fact I deductively derived that
>>>> v is not even a variable as used in the LTs. But rather, it is always
>>>> the case that v = 0 or v = c.
>>> Here is the heart of my argument:
>>>
>>> We start, as always with our premises. We premise the Lorentz Transforms
>>> by reference. We further premise the quantity v' which is equal to, and
>>> interchangeable with -v of the Lorentz transforms. Finally, we premise
>>> that v and v' are both constrained have values lying in the range 0 <
>>> ||v|| < c and 0 < ||v'|| < c. The last premise is Einsteins second postulate.
>>>
>>> Our strategy is to express v' in terms of v, t, and x then solve for v in
>>> hopes of arriving at the semi-identity -v = v'. Deltas (∆) are omitted
>>> for brevity; their locations are easily recognized by the competent student.
>>>
>>> We can express v' in terms of x, t, and v, as v' = -x'/t'.
>> Yup. This is where you lost your head last time too.
>> But you feel the need to have 18 people tell you the same thing?
>>>
>>> We now apply the LTs to the numerator and denominator of the fraction -x'/t' to get
>>>
>>> x'/t' = g(x-vt)/g(t-vx/c^2) voi la!
>>>
>>> x'/t' = (x-vt)/(t-vx/c^2)
>>>
>>> We note here that our null hypothesis states v = -v' = -x'/t'. So we
>>> substitute -x'/t' for v and we are off to the races!
>>>
>>> x'/t' = (x+[x't/t'])/(t+xx'/t'c^2)
>>>
>>> x't/t' + x x'^2/c^2t'^2 = x + x't/t'
>>>
>>> xx'^2/c^2t'^2 = x
>>>
>>> x'^2 = c^2t'^2
>>>
>>> x'/t' = c
>>>
>>> v' = c
>>>
>>> This contradicts the null hypothesis we wish to prove. We have
>>> conclusively and deductively disproved that v can ever be equal to -v'.
>>> And we have proved the the LT's are algebraically inconsistent by
>>> reductio ad absurdum. This wrings down the final curtain on Special Relativity.
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
> As you can see Ricardo, this subject still raises a firestorm of protest
> amongst relativists hellbent on saving appearences of the LTs. Please
> also see my posts on inconsistency of the sacred spacetime interval; it's
> an empty concept. And don't get me started on the Frisch-Smith muons...
>
> My current project is debunking Olson Events.
>

“As you can see, Ricardo, all I have to do is to say something stupid to
generate a firestorm of them noting that I’ve repeated something stupid.
You see how easy it is to produce attention, endless streams of attention?
Just say something stupid, repeatedly even better.”

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

On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
<PointedEars@web.de> wrote:

>Ricardo Jimenez wrote:
>
>> The Lorentz transformation and other results of special relativity
>> seem to have the built in assumption that the two systems that are
>> being compared have compatible time and space units, otherwise the
>> results are nonsense. But I can't recall seeing a discussion of how
>> to compare the units of two frames in relative motion with each other
>> so that equations come out right.
>
>Which “units of two frames”?
>
>
>PointedEars

The units of time and distance. Many accounts assume you choose them
so that the speed of light turns out to be 1 but that just determines
them up to a constant factor so they might be different in the two
frames which would mean that the factor would have to occur in the
Lorentz transformation also. My guess is that everything works out
correctly if all observers chose the same number of vibrations of the
cesium atom as the basic unit of time. However, I don't know how to
prove that works or if it is necessary to add it to the ever growing
list of unstated assumptions that underlie relativity theory.

Ricardo Jimenez wrote:

> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
> <PointedEars@web.de> wrote:
>> Ricardo Jimenez wrote:
>>> The Lorentz transformation and other results of special relativity
>>> seem to have the built in assumption that the two systems that are
>>> being compared have compatible time and space units, otherwise the
>>> results are nonsense. But I can't recall seeing a discussion of how
>>> to compare the units of two frames in relative motion with each other
>>> so that equations come out right.
>> Which “units of two frames”?
>
> The units of time and distance. Many accounts assume you choose them
> so that the speed of light turns out to be 1

That is just done to simplify the calculation and does not carry physical
significance.

> but that just determines them up to a constant factor so they might be
> different in the two frames which would mean that the factor would have to
> occur in the Lorentz transformation also.

Utter nonsense.

> My guess is that everything works out correctly if all observers chose the
> same number of vibrations of the cesium atom as the basic unit of time.
> However, I don't know how to prove that works or if it is necessary to add
> it to the ever growing list of unstated assumptions that underlie
> relativity theory.

Argument from ignorance.

It follows from the Galilean principle of relativity that a ruler will
measure the same length of, and a clock will measure time the same way for,
co-moving objects, no matter whether we choose to define them to be in
inertial motion or at rest. So we can tacitly assume that both observers
will use rulers and clocks of the same making, therefore the same units are
implied. But actually the units are irrelevant: they will always be units
of length and units of time, and you can (and should) convert them before
doing the calculation, respectively.

PointedEars
--
Q: What did the nuclear physicist order for lunch?
A: Fission chips.

(from: WolframAlpha)

Ricardo Jimenez <rickyjim@earthlink.net> wrote:
> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
> <PointedEars@web.de> wrote:
>
>> Ricardo Jimenez wrote:
>>
>>> The Lorentz transformation and other results of special relativity
>>> seem to have the built in assumption that the two systems that are
>>> being compared have compatible time and space units, otherwise the
>>> results are nonsense. But I can't recall seeing a discussion of how
>>> to compare the units of two frames in relative motion with each other
>>> so that equations come out right.
>>
>> Which “units of two frames”?
>>
>>
>> PointedEars
>
> The units of time and distance. Many accounts assume you choose them
> so that the speed of light turns out to be 1 but that just determines
> them up to a constant factor so they might be different in the two
> frames which would mean that the factor would have to occur in the
> Lorentz transformation also. My guess is that everything works out
> correctly if all observers chose the same number of vibrations of the
> cesium atom as the basic unit of time. However, I don't know how to
> prove that works or if it is necessary to add it to the ever growing
> list of unstated assumptions that underlie relativity theory.
>

I think you have the right guess. The statement specifically is that the
laws of physics are the same in any local inertial reference frame. This
means, for example, that any spring-mass oscillator will have the same
period that’s predicted by Hooke’s law and the solution of simple harmonic
motion. (And of course, related laws that govern cesium atoms, and so on.)

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

On Friday, April 15, 2022 at 7:35:29 AM UTC-7, Ricardo Jimenez wrote:
> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
> <Point...@web.de> wrote:
>
> >Ricardo Jimenez wrote:
> >
> >> The Lorentz transformation and other results of special relativity
> >> seem to have the built in assumption that the two systems that are
> >> being compared have compatible time and space units, otherwise the
> >> results are nonsense. But I can't recall seeing a discussion of how
> >> to compare the units of two frames in relative motion with each other
> >> so that equations come out right.
> >
> >Which “units of two frames”?
> >
> >
> >PointedEars
> The units of time and distance. Many accounts assume you choose them
> so that the speed of light turns out to be 1 but that just determines
> them up to a constant factor so they might be different in the two
> frames which would mean that the factor would have to occur in the
> Lorentz transformation also. My guess is that everything works out
> correctly if all observers chose the same number of vibrations of the
> cesium atom as the basic unit of time. However, I don't know how to
> prove that works or if it is necessary to add it to the ever growing
> list of unstated assumptions that underlie relativity theory.

Let me reveal an essential unstated assumption of the Lorentz Transforms.

0th postulate of Special Relativity: ∆x/∆t = ∆x'/∆t' must be true between inertial frames for all relative velocities from 0 to c.

But as demonstrated earlier, this unstated postulate is only true when

∆x/∆t = ∆x'/∆t' = c [ the second postulate ] or when ∆x/∆t = ∆x'/∆t' = 0

The 0th postulate is demonstrably false for all other values of ∆x/∆t and ∆x'/∆t' between 0 and c.

patdolan <patdolan@comcast.net> wrote:
> On Friday, April 15, 2022 at 7:35:29 AM UTC-7, Ricardo Jimenez wrote:
>> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
>> <Point...@web.de> wrote:
>>
>>> Ricardo Jimenez wrote:
>>>
>>>> The Lorentz transformation and other results of special relativity
>>>> seem to have the built in assumption that the two systems that are
>>>> being compared have compatible time and space units, otherwise the
>>>> results are nonsense. But I can't recall seeing a discussion of how
>>>> to compare the units of two frames in relative motion with each other
>>>> so that equations come out right.
>>>
>>> Which “units of two frames”?
>>>
>>>
>>> PointedEars
>> The units of time and distance. Many accounts assume you choose them
>> so that the speed of light turns out to be 1 but that just determines
>> them up to a constant factor so they might be different in the two
>> frames which would mean that the factor would have to occur in the
>> Lorentz transformation also. My guess is that everything works out
>> correctly if all observers chose the same number of vibrations of the
>> cesium atom as the basic unit of time. However, I don't know how to
>> prove that works or if it is necessary to add it to the ever growing
>> list of unstated assumptions that underlie relativity theory.
>
> Let me reveal an essential unstated assumption of the Lorentz Transforms.
>
> 0th postulate of Special Relativity: ∆x/∆t = ∆x'/∆t' must be true
> between inertial frames for all relative velocities from 0 to c.
>
> But as demonstrated earlier, this unstated postulate is only true when
>
> ∆x/∆t = ∆x'/∆t' = c [ the second postulate ] or when ∆x/∆t = ∆x'/∆t' = 0

Just in case it isn’t obvious to you, these quantities ∆x, ∆t, ∆x', ∆t' are
coordinate differences between ANY TWO EVENTS as viewed in the two
reference frames. ANY TWO EVENTS. That means many more pairs of events than
those that are related by a traverse of speed v.

You, though, apparently don’t know what those variables signify and have
taken ∆x/∆t to be v just because you’ve seen that somewhere.

This is the kind of “logic” invoked by people like you who unfortunately
don’t know enough about the subject to understand what the symbols refer
to. This is the kind of “demonstration” that rightly gets called “not even
wrong”.

Pat, could you *try* to be a little less lazy in the thinking you put
together before you post, please?

>
> The 0th postulate is demonstrably false for all other values of ∆x/∆t and
> ∆x'/∆t' between 0 and c.
>

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

On Friday, April 15, 2022 at 8:32:55 AM UTC-7, utter cretin pat dolan repeated his older imbecilities:

> 0th postulate of Special Relativity: ∆x/∆t = ∆x'/∆t' must be true between inertial frames for all relative velocities from 0 to c.
>

Repeating the same cretinisms over and over doesn't make them true

On Friday, April 15, 2022 at 8:41:52 AM UTC-7, bodk...@gmail.com wrote:
> patdolan <patd...@comcast.net> wrote:
> > On Friday, April 15, 2022 at 7:35:29 AM UTC-7, Ricardo Jimenez wrote:
> >> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
> >> <Point...@web.de> wrote:
> >>
> >>> Ricardo Jimenez wrote:
> >>>
> >>>> The Lorentz transformation and other results of special relativity
> >>>> seem to have the built in assumption that the two systems that are
> >>>> being compared have compatible time and space units, otherwise the
> >>>> results are nonsense. But I can't recall seeing a discussion of how
> >>>> to compare the units of two frames in relative motion with each other
> >>>> so that equations come out right.
> >>>
> >>> Which “units of two frames”?
> >>>
> >>>
> >>> PointedEars
> >> The units of time and distance. Many accounts assume you choose them
> >> so that the speed of light turns out to be 1 but that just determines
> >> them up to a constant factor so they might be different in the two
> >> frames which would mean that the factor would have to occur in the
> >> Lorentz transformation also. My guess is that everything works out
> >> correctly if all observers chose the same number of vibrations of the
> >> cesium atom as the basic unit of time. However, I don't know how to
> >> prove that works or if it is necessary to add it to the ever growing
> >> list of unstated assumptions that underlie relativity theory.
> >
> > Let me reveal an essential unstated assumption of the Lorentz Transforms.
> >
> > 0th postulate of Special Relativity: ∆x/∆t = ∆x'/∆t' must be true
> > between inertial frames for all relative velocities from 0 to c.
> >
> > But as demonstrated earlier, this unstated postulate is only true when
> >
> > ∆x/∆t = ∆x'/∆t' = c [ the second postulate ] or when ∆x/∆t = ∆x'/∆t' = 0
> Just in case it isn’t obvious to you, these quantities ∆x, ∆t, ∆x', ∆t' are
> coordinate differences between ANY TWO EVENTS as viewed in the two
> reference frames. ANY TWO EVENTS. That means many more pairs of events than
> those that are related by a traverse of speed v.
>
> You, though, apparently don’t know what those variables signify and have
> taken ∆x/∆t to be v just because you’ve seen that somewhere.
>
> This is the kind of “logic” invoked by people like you who unfortunately
> don’t know enough about the subject to understand what the symbols refer
> to. This is the kind of “demonstration” that rightly gets called “not even
> wrong”.
>
> Pat, could you *try* to be a little less lazy in the thinking you put
> together before you post, please?

You objection does not, nor ever has, made sense Bodkin. Deductive reasoning is always performed in a manner and mode that derives the most general results possible. I have strictly adhered to this logical principle in all my work. The work and its results are consummately valid in all respects.

That you don't like those results is a matter of psychology. Not logic, algebra or physics.
> >
> > The 0th postulate is demonstrably false for all other values of ∆x/∆t and
> > ∆x'/∆t' between 0 and c.
> >
> --
> Odd Bodkin -- maker of fine toys, tools, tables

patdolan <patdolan@comcast.net> wrote:
> On Friday, April 15, 2022 at 8:41:52 AM UTC-7, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> On Friday, April 15, 2022 at 7:35:29 AM UTC-7, Ricardo Jimenez wrote:
>>>> On Fri, 15 Apr 2022 13:15:45 +0200, Thomas 'PointedEars' Lahn
>>>> <Point...@web.de> wrote:
>>>>
>>>>> Ricardo Jimenez wrote:
>>>>>
>>>>>> The Lorentz transformation and other results of special relativity
>>>>>> seem to have the built in assumption that the two systems that are
>>>>>> being compared have compatible time and space units, otherwise the
>>>>>> results are nonsense. But I can't recall seeing a discussion of how
>>>>>> to compare the units of two frames in relative motion with each other
>>>>>> so that equations come out right.
>>>>>
>>>>> Which “units of two frames”?
>>>>>
>>>>>
>>>>> PointedEars
>>>> The units of time and distance. Many accounts assume you choose them
>>>> so that the speed of light turns out to be 1 but that just determines
>>>> them up to a constant factor so they might be different in the two
>>>> frames which would mean that the factor would have to occur in the
>>>> Lorentz transformation also. My guess is that everything works out
>>>> correctly if all observers chose the same number of vibrations of the
>>>> cesium atom as the basic unit of time. However, I don't know how to
>>>> prove that works or if it is necessary to add it to the ever growing
>>>> list of unstated assumptions that underlie relativity theory.
>>>
>>> Let me reveal an essential unstated assumption of the Lorentz Transforms.
>>>
>>> 0th postulate of Special Relativity: ∆x/∆t = ∆x'/∆t' must be true
>>> between inertial frames for all relative velocities from 0 to c.
>>>
>>> But as demonstrated earlier, this unstated postulate is only true when
>>>
>>> ∆x/∆t = ∆x'/∆t' = c [ the second postulate ] or when ∆x/∆t = ∆x'/∆t' = 0
>> Just in case it isn’t obvious to you, these quantities ∆x, ∆t, ∆x', ∆t' are
>> coordinate differences between ANY TWO EVENTS as viewed in the two
>> reference frames. ANY TWO EVENTS. That means many more pairs of events than
>> those that are related by a traverse of speed v.
>>
>> You, though, apparently don’t know what those variables signify and have
>> taken ∆x/∆t to be v just because you’ve seen that somewhere.
>>
>> This is the kind of “logic” invoked by people like you who unfortunately
>> don’t know enough about the subject to understand what the symbols refer
>> to. This is the kind of “demonstration” that rightly gets called “not even
>> wrong”.
>>
>> Pat, could you *try* to be a little less lazy in the thinking you put
>> together before you post, please?
>
> You objection does not, nor ever has, made sense Bodkin. Deductive
> reasoning is always performed in a manner and mode that derives the most
> general results possible. I have strictly adhered to this logical
> principle in all my work.

No, you have not. You have repeatedly claimed that in the Lorentz
transforms ∆x/∆t=v, which is a statement that is not only factually false
but conveys a complete lack of understanding of what the symbols mean.

Being able to manipulate symbols algebraically will still not allow you to
prove anything while you do not know what those symbols mean.

Since you say that this observation makes no sense to you, this further
underscores the fact that you do not understand what you do not understand,
even on simple fronts.

> The work and its results are consummately valid in all respects.
>
> That you don't like those results is a matter of psychology. Not logic,
> algebra or physics.
>>>
>>> The 0th postulate is demonstrably false for all other values of ∆x/∆t and
>>> ∆x'/∆t' between 0 and c.
>>>
>> --
>> Odd Bodkin -- maker of fine toys, tools, tables
>

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