http://hem.bredband.net/b153434/Works/Einstein.htm
Divine Albert: "In a space which is free of gravitational fields we
introduce a Galilean system of reference K (x, y, z, t), and also a
system of co-ordinates K' (x', y', z', t') in uniform rotation
relatively to K. Let the origins of both systems, as well as their
axes of Z, permanently coincide. We shall show that for a space-time
measurement in the system K' the above definition of the physical
meaning of lengths and times cannot be maintained. For reasons of
symmetry it is clear that a circle around the origin in the X, Y plane
of K may at the same time be regarded as a circle in the X', Y' plane
of K'. We suppose that the circumference and diameter of this circle
have been measured with a unit measure infinitely small compared with
the radius, and that we have the quotient of the two results. If this
experiment were performed with a measuring-rod at rest relatively to
the Galilean system K, the quotient would be [pi]. With a measuring-
rod at rest relatively to K', the quotient would be greater than [pi].
This is readily understood if we envisage the whole process of
measuring from the "stationary" system K, and take into consideration
that the measuring-rod applied to the periphery undergoes a Lorentzian
contraction, while the one applied along the radius does not. Hence
Euclidean geometry does not apply to K'."

Why does Divine Albert believe that, "with a measuring-rod at rest
relatively to K', the quotient would be greater than [pi]"? Obviously
because "the measuring-rod applied to the periphery undergoes a
Lorentzian contraction". Then why does the segment of the periphery,
to which the measuring-rod is applied and which moves together with
the measuring-rod, fail to undergo a Lorentzian contraction?

In 1902, in "La Science et l'hypothese", Henri Poincare, in order to
justify non-Euclidean geometries, presented a parabole. Bidimensional
creatures live on a disc. The disc is heated under its center so that
the temperature is high at the center and decreases towards the
periphery. The creatures use rigid measuring rods in order to
determine the geometry of their world. They know nothing about the
heater and accordingly discover that the ratio of the circumference
and the diameter is greater than [pi]. The creatures conclude that
Euclidean geometry cannot be true on their disc.

Pentcho Valev
pvalev***yahoo.com

On Jan 30, 2:30 am, Pentcho Valev <pva...***yahoo.com> wrote:
> http://hem.bredband.net/b153434/Works/Einstein.htm
> Divine Albert: "In a space which is free of gravitational fields we
> introduce a Galilean system of reference K (x, y, z, t), and also a
> system of co-ordinates K' (x', y', z', t') in uniform rotation
> relatively to K. Let the origins of both systems, as well as their
> axes of Z, permanently coincide. We shall show that for a space-time
> measurement in the system K' the above definition of the physical
> meaning of lengths and times cannot be maintained. For reasons of
> symmetry it is clear that a circle around the origin in the X, Y plane
> of K may at the same time be regarded as a circle in the X', Y' plane
> of K'. We suppose that the circumference and diameter of this circle
> have been measured with a unit measure infinitely small compared with
> the radius, and that we have the quotient of the two results. If this
> experiment were performed with a measuring-rod at rest relatively to
> the Galilean system K, the quotient would be [pi]. With a measuring-
> rod at rest relatively to K', the quotient would be greater than [pi].
> This is readily understood if we envisage the whole process of
> measuring from the "stationary" system K, and take into consideration
> that the measuring-rod applied to the periphery undergoes a Lorentzian
> contraction, while the one applied along the radius does not. Hence
> Euclidean geometry does not apply to K'."
>
> Why does Divine Albert believe that, "with a measuring-rod at rest
> relatively to K', the quotient would be greater than [pi]"? Obviously
> because "the measuring-rod applied to the periphery undergoes a
> Lorentzian contraction". Then why does the segment of the periphery,
> to which the measuring-rod is applied and which moves together with
> the measuring-rod, fail to undergo a Lorentzian contraction?

The bulk of the disc prevents it. The circumference wants to contract
but it can't - it develops internal strain instead. If the centrifugal
forces could be ignored, this strain could cause the disc to crack
along a vaguely radial direction if the angular velocity was large
enough. To an observer sitting on the disc's circumference OTOH his
measuring stick does not shorten (obviously but the circumference
would elongate, hence the disc would according to him also develop
internal strain (as it must - the presence or absence of such strain
cannot depend on the observer).

This is really a circular version of Bell's "spaceship paradox" in
which each circumference segment plays the role of the rope connecting
Bell's spaceship pair. Check Bell's "Speakable and Unspeakable in
Quantum Mechanics".

--
Jan

On Jan 30, 5:30***am, Pentcho Valev <pva...***yahoo.com> wrote:
> http://hem.bredband.net/b153434/Works/Einstein.htm
> Divine Albert: "In a space which is free of gravitational fields we
> introduce a Galilean system of reference K (x, y, z, t), and also a
> system of co-ordinates K' (x', y', z', t') in uniform rotation
> relatively to K. Let the origins of both systems, as well as their
> axes of Z, permanently coincide. We shall show that for a space-time
> measurement in the system K' the above definition of the physical
> meaning of lengths and times cannot be maintained. For reasons of
> symmetry it is clear that a circle around the origin in the X, Y plane
> of K may at the same time be regarded as a circle in the X', Y' plane
> of K'. We suppose that the circumference and diameter of this circle
> have been measured with a unit measure infinitely small compared with
> the radius, and that we have the quotient of the two results. If this
> experiment were performed with a measuring-rod at rest relatively to
> the Galilean system K, the quotient would be [pi]. With a measuring-
> rod at rest relatively to K', the quotient would be greater than [pi].
> This is readily understood if we envisage the whole process of
> measuring from the "stationary" system K, and take into consideration
> that the measuring-rod applied to the periphery undergoes a Lorentzian
> contraction, while the one applied along the radius does not. Hence
> Euclidean geometry does not apply to K'."
>
> Why does Divine Albert believe that, "with a measuring-rod at rest
> relatively to K', the quotient would be greater than [pi]"? Obviously
> because "the measuring-rod applied to the periphery undergoes a
> Lorentzian contraction". Then why does the segment of the periphery,
> to which the measuring-rod is applied and which moves together with
> the measuring-rod, fail to undergo a Lorentzian contraction?
>
> In 1902, in "La Science et l'hypothese", Henri Poincare, in order to
> justify non-Euclidean geometries, presented a parabole. Bidimensional
> creatures live on a disc. The disc is heated under its center so that
> the temperature is high at the center and decreases towards the
> periphery. The creatures use rigid measuring rods in order to
> determine the geometry of their world. They know nothing about the
> heater and accordingly discover that the ratio of the circumference
> and the diameter is greater than [pi]. The creatures conclude that
> Euclidean geometry cannot be true on their disc.
>
> Pentcho Valev
> pva...***yahoo.com

xxein: We are K' to any other observer (gravity or not). For your
question (sans gravity) who would not be K'? How do WE get pi?

Perhaps the answer is that we measure the circumference in segments
with the ruler. Likewise, we measure a radius in the same way. As
you might realise, any rotation wrt the direction of motion will cause
both measurements to vary in exactly the same way. But even without a
rotation, there is a necessary rotation of the ruler and you can still
see the mapping of the ruler and and the circumference/radius as being
a co-temporary measurement regardless of the actual shape (< circular)
to K (zero v in a simplified and flat ether system).

Bell is both right and wrong. K' is not a circle/sphere, but pi still
remains the differential measure for those of K'.

We ARE K'. We cannot rationalize any differently. A logic prevails.
There is no paradox.

"xxein" <xxein***comcast.net> wrote in message
news:50c4d3b3-f2c5-4cdd-8095-1760a4851e85***i72g2000hsd.googlegroups.com...
On Jan 30, 5:30 am, Pentcho Valev <pva...***yahoo.com> wrote:
> http://hem.bredband.net/b153434/Works/Einstein.htm
> Divine Albert: "In a space which is free of gravitational fields we
> introduce a Galilean system of reference K (x, y, z, t), and also a
> system of co-ordinates K' (x', y', z', t') in uniform rotation
> relatively to K. Let the origins of both systems, as well as their
> axes of Z, permanently coincide. We shall show that for a space-time
> measurement in the system K' the above definition of the physical
> meaning of lengths and times cannot be maintained. For reasons of
> symmetry it is clear that a circle around the origin in the X, Y plane
> of K may at the same time be regarded as a circle in the X', Y' plane
> of K'. We suppose that the circumference and diameter of this circle
> have been measured with a unit measure infinitely small compared with
> the radius, and that we have the quotient of the two results. If this
> experiment were performed with a measuring-rod at rest relatively to
> the Galilean system K, the quotient would be [pi]. With a measuring-
> rod at rest relatively to K', the quotient would be greater than [pi].
> This is readily understood if we envisage the whole process of
> measuring from the "stationary" system K, and take into consideration
> that the measuring-rod applied to the periphery undergoes a Lorentzian
> contraction, while the one applied along the radius does not. Hence
> Euclidean geometry does not apply to K'."
>
> Why does Divine Albert believe that, "with a measuring-rod at rest
> relatively to K', the quotient would be greater than [pi]"? Obviously
> because "the measuring-rod applied to the periphery undergoes a
> Lorentzian contraction". Then why does the segment of the periphery,
> to which the measuring-rod is applied and which moves together with
> the measuring-rod, fail to undergo a Lorentzian contraction?
>
> In 1902, in "La Science et l'hypothese", Henri Poincare, in order to
> justify non-Euclidean geometries, presented a parabole. Bidimensional
> creatures live on a disc. The disc is heated under its center so that
> the temperature is high at the center and decreases towards the
> periphery. The creatures use rigid measuring rods in order to
> determine the geometry of their world. They know nothing about the
> heater and accordingly discover that the ratio of the circumference
> and the diameter is greater than [pi]. The creatures conclude that
> Euclidean geometry cannot be true on their disc.
>
> Pentcho Valev
> pva...***yahoo.com

xxein: We are K' to any other observer (gravity or not). For your
question (sans gravity) who would not be K'? How do WE get pi?

http://en.wikipedia.org/wiki/Computing_%CF%80

> On Jan 30, 5:30 am, Pentcho Valev <pva...***yahoo.com> wrote:
>> http://hem.bredband.net/b153434/Works/Einstein.htm
>> Divine Albert: "In a space which is free of gravitational fields we
>> introduce a Galilean system of reference K (x, y, z, t), and also a
>> system of co-ordinates K' (x', y', z', t') in uniform rotation
>> relatively to K. Let the origins of both systems, as well as their
>> axes of Z, permanently coincide. We shall show that for a space-time
>> measurement in the system K' the above definition of the physical
>> meaning of lengths and times cannot be maintained. For reasons of
>> symmetry it is clear that a circle around the origin in the X, Y plane
>> of K may at the same time be regarded as a circle in the X', Y' plane
>> of K'. We suppose that the circumference and diameter of this circle
>> have been measured with a unit measure infinitely small compared with
>> the radius, and that we have the quotient of the two results. If this
>> experiment were performed with a measuring-rod at rest relatively to
>> the Galilean system K, the quotient would be [pi]. With a measuring-
>> rod at rest relatively to K', the quotient would be greater than [pi].
>> This is readily understood if we envisage the whole process of
>> measuring from the "stationary" system K, and take into consideration
>> that the measuring-rod applied to the periphery undergoes a Lorentzian
>> contraction, while the one applied along the radius does not. Hence
>> Euclidean geometry does not apply to K'."
>>
>> Why does Divine Albert believe that, "with a measuring-rod at rest
>> relatively to K', the quotient would be greater than [pi]"? Obviously
>> because "the measuring-rod applied to the periphery undergoes a
>> Lorentzian contraction". Then why does the segment of the periphery,
>> to which the measuring-rod is applied and which moves together with
>> the measuring-rod, fail to undergo a Lorentzian contraction?

The rod when measuring the radius has its long edge aligned in the direction
of a radial line, and so the long edge is not in the diretio of motion (it
is a right angles to it) so you get no contraction in length of ruler (but
you do get one in the width).

When the rod is turned around to measure along the periphery, the long edge
of the rod aligned in the direction of motion and so it is contracted, but
this time the width is not.

Est-il possible que les abérations que ces deux timbrés viennent d'écricre
puissent être traduites en un français concis ? ? ? SVP

"Jeckyl" <noone***nowhere.com> a écrit dans le message news:
13q2ch3l52qce11***corp.supernews.com...
> > On Jan 30, 5:30 am, Pentcho Valev <pva...***yahoo.com> wrote:
> >> http://hem.bredband.net/b153434/Works/Einstein.htm
> >> Divine Albert: "In a space which is free of gravitational fields we
> >> introduce a Galilean system of reference K (x, y, z, t), and also a
> >> system of co-ordinates K' (x', y', z', t') in uniform rotation
> >> relatively to K. Let the origins of both systems, as well as their
> >> axes of Z, permanently coincide. We shall show that for a space-time
> >> measurement in the system K' the above definition of the physical
> >> meaning of lengths and times cannot be maintained. For reasons of
> >> symmetry it is clear that a circle around the origin in the X, Y plane
> >> of K may at the same time be regarded as a circle in the X', Y' plane
> >> of K'. We suppose that the circumference and diameter of this circle
> >> have been measured with a unit measure infinitely small compared with
> >> the radius, and that we have the quotient of the two results. If this
> >> experiment were performed with a measuring-rod at rest relatively to
> >> the Galilean system K, the quotient would be [pi]. With a measuring-
> >> rod at rest relatively to K', the quotient would be greater than [pi].
> >> This is readily understood if we envisage the whole process of
> >> measuring from the "stationary" system K, and take into consideration
> >> that the measuring-rod applied to the periphery undergoes a Lorentzian
> >> contraction, while the one applied along the radius does not. Hence
> >> Euclidean geometry does not apply to K'."
> >>
> >> Why does Divine Albert believe that, "with a measuring-rod at rest
> >> relatively to K', the quotient would be greater than [pi]"? Obviously
> >> because "the measuring-rod applied to the periphery undergoes a
> >> Lorentzian contraction". Then why does the segment of the periphery,
> >> to which the measuring-rod is applied and which moves together with
> >> the measuring-rod, fail to undergo a Lorentzian contraction?
>
> The rod when measuring the radius has its long edge aligned in the
direction
> of a radial line, and so the long edge is not in the diretio of motion (it
> is a right angles to it) so you get no contraction in length of ruler (but
> you do get one in the width).
>
> When the rod is turned around to measure along the periphery, the long
edge
> of the rod aligned in the direction of motion and so it is contracted, but
> this time the width is not.
>
>

"Maxime" <m.lenord***nospam.fr> wrote in message
news:47a139dd$0$907$ba4acef3***news.orange.fr...
> Est-il possible que les abérations que ces deux timbrés viennent d'écricre
> puissent être traduites en un français concis ? ? ? SVP

I would do it for you. But my french is not good enough to translate for
you. The best I could do is to use Google langauge tools, but you could do
that yourself

"Jeckyl" <noone***nowhere.com> a écrit dans le message news:
13q2eq0etl39m6e***corp.supernews.com...
> "Maxime" <m.lenord***nospam.fr> wrote in message
> news:47a139dd$0$907$ba4acef3***news.orange.fr...
> > Est-il possible que les abérations que ces deux timbrés viennent
d'écricre
> > puissent être traduites en un français concis ? ? ? SVP
>
> I would do it for you. But my french is not good enough to translate for
> you. The best I could do is to use Google langauge tools, but you could
do
> that yourself

:-))

Un vraie humaniste ce type là !

Je ne savait pas qu'Einstein avait écrit ses textes en anglais.

Max

"Maxime" <m.lenord***nospam.fr> wrote in message
news:47a13cbb$0$876$ba4acef3***news.orange.fr...
>
> "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
> 13q2eq0etl39m6e***corp.supernews.com...
>> "Maxime" <m.lenord***nospam.fr> wrote in message
>> news:47a139dd$0$907$ba4acef3***news.orange.fr...
>> > Est-il possible que les abérations que ces deux timbrés viennent
> d'écricre
>> > puissent être traduites en un français concis ? ? ? SVP
>>
>> I would do it for you. But my french is not good enough to translate for
>> you. The best I could do is to use Google langauge tools, but you could
> do
>> that yourself
>
> :-))
>
> Un vraie humaniste ce type là !
>
> Je ne savait pas qu'Einstein avait écrit ses textes en anglais.

A lot of Einstein's work has been translated into English. I think there
would be translations into French somewhere. I do not know where to find
them. Good luck.

"Jeckyl" <noone***nowhere.com> a écrit dans le message news:
13q2feu6lnec84***corp.supernews.com...
> "Maxime" <m.lenord***nospam.fr> wrote in message
> news:47a13cbb$0$876$ba4acef3***news.orange.fr...
> >
> > "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
> > 13q2eq0etl39m6e***corp.supernews.com...
> >> "Maxime" <m.lenord***nospam.fr> wrote in message
> >> news:47a139dd$0$907$ba4acef3***news.orange.fr...
> >> > Est-il possible que les abérations que ces deux timbrés viennent
> > d'écricre
> >> > puissent être traduites en un français concis ? ? ? SVP
> >>
> >> I would do it for you. But my french is not good enough to translate
for
> >> you. The best I could do is to use Google langauge tools, but you
could
> > do
> >> that yourself
> >
> > :-))
> >
> > Un vraie humaniste ce type là !
> >
> > Je ne savait pas qu'Einstein avait écrit ses textes en anglais.
>
> A lot of Einstein's work has been translated into English. I think there
> would be translations into French somewhere. I do not know where to find
> them. Good luck.

Je ne sais pas si tu à la moindre conscience du crospost que tu viens de
faire sur fr.sci.philo et d'autres hiérarchies fr.

Max

"Maxime" <m.lenord***nospam.fr> wrote in message
news:47a13f5c$0$860$ba4acef3***news.orange.fr...
>
> "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
> 13q2feu6lnec84***corp.supernews.com...
>> "Maxime" <m.lenord***nospam.fr> wrote in message
>> news:47a13cbb$0$876$ba4acef3***news.orange.fr...
>> >
>> > "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
>> > 13q2eq0etl39m6e***corp.supernews.com...
>> >> "Maxime" <m.lenord***nospam.fr> wrote in message
>> >> news:47a139dd$0$907$ba4acef3***news.orange.fr...
>> >> > Est-il possible que les abérations que ces deux timbrés viennent
>> > d'écricre
>> >> > puissent être traduites en un français concis ? ? ? SVP
>> >>
>> >> I would do it for you. But my french is not good enough to translate
> for
>> >> you. The best I could do is to use Google langauge tools, but you
> could
>> > do
>> >> that yourself
>> >
>> > :-))
>> >
>> > Un vraie humaniste ce type là !
>> >
>> > Je ne savait pas qu'Einstein avait écrit ses textes en anglais.
>>
>> A lot of Einstein's work has been translated into English. I think there
>> would be translations into French somewhere. I do not know where to find
>> them. Good luck.
>
> Je ne sais pas si tu à la moindre conscience du crospost que tu viens de
> faire sur fr.sci.philo et d'autres hiérarchies fr.

Blame the original poster who posted to the French groups. I only replied
to a message that was already cross-posted.

"Jeckyl" <noone***nowhere.com> a écrit dans le message news:
13q2gpvagbcae3c***corp.supernews.com...
> "Maxime" <m.lenord***nospam.fr> wrote in message
> news:47a13f5c$0$860$ba4acef3***news.orange.fr...
> >
> > "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
> > 13q2feu6lnec84***corp.supernews.com...
> >> "Maxime" <m.lenord***nospam.fr> wrote in message
> >> news:47a13cbb$0$876$ba4acef3***news.orange.fr...
> >> >
> >> > "Jeckyl" <noone***nowhere.com> a écrit dans le message news:
> >> > 13q2eq0etl39m6e***corp.supernews.com...
> >> >> "Maxime" <m.lenord***nospam.fr> wrote in message
> >> >> news:47a139dd$0$907$ba4acef3***news.orange.fr...
> >> >> > Est-il possible que les abérations que ces deux timbrés viennent
> >> > d'écricre
> >> >> > puissent être traduites en un français concis ? ? ? SVP
> >> >>
> >> >> I would do it for you. But my french is not good enough to
translate
> > for
> >> >> you. The best I could do is to use Google langauge tools, but you
> > could
> >> > do
> >> >> that yourself
> >> >
> >> > :-))
> >> >
> >> > Un vraie humaniste ce type là !
> >> >
> >> > Je ne savait pas qu'Einstein avait écrit ses textes en anglais.
> >>
> >> A lot of Einstein's work has been translated into English. I think
there
> >> would be translations into French somewhere. I do not know where to
find
> >> them. Good luck.
> >
> > Je ne sais pas si tu à la moindre conscience du crospost que tu viens de
> > faire sur fr.sci.philo et d'autres hiérarchies fr.
>
> Blame the original poster who posted to the French groups. I only replied
> to a message that was already cross-posted.

Bravo le sens de responsabilité !

Max