Re: 'CURVED SPACE TIME'' - IS A NONSENSE HARMFUL PHYSICS !!!

[Carmine Portacio]

reber g=emc^2 wrote:

> On Friday, January 8, 2016 at 12:00:31 PM UTC-8, Y.Porat wrote:
>> On Friday, January 8, 2016 at 7:24:54 PM UTC+2, Nimo wrote:
>> > Bcoz of light we can see the space and define its(space's) 3
>> > dimensionality property,
>> > Well, what my point is just close your eyes and your brain would
>> > shows you a finite blackness, can you define its dimentionality ( if
>> > you assume it as space)

Instead of doing that expensive experiment, just imagine that spacetime is
not curved, cannot curve, as being as it is, empty vacuo impossible to
curve. Interestingly enough, by just imagine it like that, you realize
that you already came up with many alternative to Relativity in describing
Gravity. For instance IRT E-Matrix, Ballistic Theory and of course
Divergent Matter. Equally good to substitute Relativity and its Spacetime.
Thanks, I look forward to cooperate.

[JanPB]

On Monday, January 11, 2016 at 2:40:15 PM UTC-8, Carmine Portacio wrote:
> reber g=emc^2 wrote:
>
> > On Friday, January 8, 2016 at 12:00:31 PM UTC-8, Y.Porat wrote:
> >> On Friday, January 8, 2016 at 7:24:54 PM UTC+2, Nimo wrote:
> >> > Bcoz of light we can see the space and define its(space's) 3
> >> > dimensionality property,
> >> > Well, what my point is just close your eyes and your brain would
> >> > shows you a finite blackness, can you define its dimentionality ( if
> >> > you assume it as space)
>
> Instead of doing that expensive experiment, just imagine that spacetime is
> not curved, cannot curve, as being as it is, empty vacuo impossible to
> curve.

This is irrelevant. Spacetime curvature is a mathematical concept here,
it simply means that paths of free fall converge or diverge from one another
instead of staying at constant separation distances. If you don't like the
word "curvature", use something else.

--
Jan

[mlwo...@wp.pl]

W dniu poniedziałek, 11 stycznia 2016 23:55:32 UTC+1 użytkownik JanPB napisał:
> On Monday, January 11, 2016 at 2:40:15 PM UTC-8, Carmine Portacio wrote:
> > reber g=emc^2 wrote:
> >
> > > On Friday, January 8, 2016 at 12:00:31 PM UTC-8, Y.Porat wrote:
> > >> On Friday, January 8, 2016 at 7:24:54 PM UTC+2, Nimo wrote:
> > >> > Bcoz of light we can see the space and define its(space's) 3
> > >> > dimensionality property,
> > >> > Well, what my point is just close your eyes and your brain would
> > >> > shows you a finite blackness, can you define its dimentionality ( if
> > >> > you assume it as space)
> >
> > Instead of doing that expensive experiment, just imagine that spacetime is
> > not curved, cannot curve, as being as it is, empty vacuo impossible to
> > curve.
>
> This is irrelevant. Spacetime curvature is a mathematical concept here,

Right. That means, that it's a result of
assumptions of a relativistic moron.

[Koobee Wublee]

On Monday, January 11, 2016 at 2:55:32 PM UTC-8, JanPB wrote:

> On Monday, January 11, 2016 at 2:40:15 PM, Carmine Portacio wrote:

> > Instead of doing that expensive experiment, just imagine that
> > spacetime is not curved, cannot curve, as being as it is, empty
> > vacuo impossible to curve.
>
> This is irrelevant.

Dumb ass Jan the relativistic moron, it is all very relevant. Spacetime curvature involves the spatial and the temporal curvatures. Only an Aether model can explain both of these categories of curvatures. <shrug>

Empty space has no properties as Tesla has explained so, and thus space cannot curve in vacuum. However, under Aether, space can very well be curved. The spatial curvature is all relative. That means an observer's perception of flat space can be another observer's perception of curved space.

Temporal curvature, on the other hand, must be absolute. That means all observers must agree on gravitational time dilation at every single point in space, or else, there would be no physics. Why? Before Jan the relativistic moron foams up its mouth, think with whatever left of its sick brain first. <shrug>

With that said, space curvature is relative, and temporal curvature is absolute. It becomes fvckingly ridiculous to claim a combination of time and space called spacetime. Time and space just cannot mix in real life. That is why the religion of relativity offers contradictions like the Twin paradox. <shrug>

> [rest of bullshit peddled by Jan the relativistic moron snipped]

[JanPB]

On Tuesday, January 12, 2016 at 12:19:13 AM UTC-8, Koobee Wublee wrote:
> On Monday, January 11, 2016 at 2:55:32 PM UTC-8, JanPB wrote:
> > On Monday, January 11, 2016 at 2:40:15 PM, Carmine Portacio wrote:
>
> > > Instead of doing that expensive experiment, just imagine that
> > > spacetime is not curved, cannot curve, as being as it is, empty
> > > vacuo impossible to curve.
> >
> > This is irrelevant.
>
> Dumb ass Jan the relativistic moron, it is all very relevant. Spacetime curvature involves the spatial and the temporal curvatures. Only an Aether model can explain both of these categories of curvatures. <shrug>
>
> Empty space has no properties as Tesla has explained so, and thus space cannot curve in vacuum.

As I said, this is not a valid objection. "Curvature" is just a word to describe certain
properties of the trajectories of free fall.

--
Jan

[Tom Roberts]

Or equivalently, certain properties in the set of all local distances between
points of spacetime.

Just like the "wavelike" properties of photons involve probability "waves" and
thus need no medium, spacetime curvature in GR is relationships among distances
and thus does not need "empty space" to have any "properties".


Tom Roberts

[Odd Bodkin]

On 1/12/2016 2:19 AM, Koobee Wublee wrote:
> Empty space has no properties as Tesla has explained so, and thus space cannot curve in vacuum.

Tesla was in error. As we now know.
For the record, Tesla died in 1943.
Some things have been learned in the 20th century.
Do catch up.

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

[Odd Bodkin]

On 1/12/2016 9:02 AM, Tom Roberts wrote:
> Just like the "wavelike" properties of photons involve probability
> "waves" and thus need no medium, spacetime curvature in GR is
> relationships among distances and thus does not need "empty space" to
> have any "properties".

Begs the question about what "distance" means. When you talk about a
measure between two points IN SPACE, this is making a physical statement
about that space. It's not like distance is an abstraction from the
space or, for that matter, that the points are an abstraction from
space. If you do want them to be abstractions then what is meant by
"points in space"?

[Koobee Wublee]

On Tuesday, January 12, 2016 at 7:47:21 AM UTC-8, Odd Bodkin wrote:
> On 1/12/2016 2:19 AM, Koobee Wublee wrote:

> > Empty space has no properties as Tesla has explained so, and thus
> > space cannot curve in vacuum.
>
> Tesla was in error. As we now know.

Tesla was not mystified by the religion of relativity. The Einstein dingleberries are the ones in error. <shrug>

> For the record, Tesla died in 1943. Some things have been learned
> in the 20th century. Do catch up.

Since Tesla's time, the religion of relativity has mystified the autistic retards which Odd Bodkin aka PD is one of them. <shrug>

[Koobee Wublee]

On Tuesday, January 12, 2016, Tom Roberts wrote:
> Jan Bielawski wrote:

> > Koobee Wublee wrote:

> > > Spacetime curvature involves the spatial and the temporal
> > > curvatures. Only an Aether model can explain both of these
> > > categories of curvatures. <shrug>
>
> > > Empty space has no properties as Tesla has explained so, and
> > > thus space cannot curve in vacuum.
>
> > As I said, this is not a valid objection. "Curvature" is just a
> > word to describe certain properties of the trajectories of free
> > fall.

Sounds like Jan the relativistic moron is bullshitting again. <shrug>

> Or equivalently, certain properties in the set of all local
> distances between points of spacetime.

Nonsense! GR describes properties of a point in spacetime NOT between two points. Learn the basics first, Tom. <shrug>

> Just like the "wavelike" properties of photons involve probability
> "waves" and thus need no medium, spacetime curvature in GR is
> relationships among distances and thus does not need "empty space"
> to have any "properties".

The above is a scripture of the religion of relativity where Einstein the nitwit, the plagiarist, and the liar serves as the patron god who would whisper into the ears of the relativistic morons on bullshit. <shrug>

[Odd Bodkin]

On 1/12/2016 11:20 AM, Koobee Wublee wrote:
> On Tuesday, January 12, 2016 at 7:47:21 AM UTC-8, Odd Bodkin wrote:
>> On 1/12/2016 2:19 AM, Koobee Wublee wrote:
>
>>> Empty space has no properties as Tesla has explained so, and thus
>>> space cannot curve in vacuum.
>>
>> Tesla was in error. As we now know.
>
> Tesla was not mystified by the religion of relativity.

Nor are thousands of undergraduate students. The fact that you are
mystified by it and so call it mystification is no one's problem but
your own.

> The Einstein dingleberries are the ones in error. <shrug>

Don't think so.

>
>> For the record, Tesla died in 1943. Some things have been learned
>> in the 20th century. Do catch up.
>
> Since Tesla's time, the religion of relativity has mystified the
> autistic retards which Odd Bodkin aka PD is one of them. <shrug>
>

When someone like Koobee Wublee frequently enters into the loop of
faux-logic that goes like the following, I can't help but giggle:
"This theory makes no sense whatsoever and is demonstrably internally
inconsistent. However, I am the only one on the planet to understand it
completely so that I comprehend that it is incomprehensible. All others
who claim to understand it and who therefore claim it is comprehensible
do not comprehend that it is incomprehensible, which I do comprehend."

Koobee can live in this interesting but twisted world if he likes. I
find it too weird myself.

[Tom Roberts]

On 1/12/16 1/12/16 - 9:55 AM, Odd Bodkin wrote:
> On 1/12/2016 9:02 AM, Tom Roberts wrote:
>> Just like the "wavelike" properties of photons involve probability
>> "waves" and thus need no medium, spacetime curvature in GR is
>> relationships among distances and thus does not need "empty space" to
>> have any "properties".
>
> Begs the question about what "distance" means.

OK. I should have said PHYSICAL properties.

Because space, as physicists use the term, has a well defined distance between
every pair of points within a suitably small local region. This, of course, is
known as geometry.

> When you talk about a measure
> between two points IN SPACE, this is making a physical statement about that
> space.

Hmmm. Say, rather, that it is a GEOMETRICAL statement. Geometry needs no
physical underpinning, and it seems to be the difference between physical
properties and geometrical properties that is at the heart of this confusion.

> It's not like distance is an abstraction from the space or, for that
> matter, that the points are an abstraction from space.

Hmmm. Distances and points are GEOMETRICAL properties/abstractions.

> If you do want them to be
> abstractions then what is meant by "points in space"?

They are the primitive points of the mathematical/geometrical manifold we use to
model the world we inhabit (and sometimes other potential worlds of interest).
The concept "manifold" was abstracted from observations of the world we inhabit.


Tom Roberts

[Maciej Woźniak]

Użytkownik "Tom Roberts" napisał w wiadomości grup
dyskusyjnych:CsKdnZw6eM-7ygjL...@giganews.com...

|Because space, as physicists use the term, has a well defined distance
between
|every pair of points within a suitably small local region.

:)
Or rather, it had before your insane guru started his insane
revolution.

> If you do want them to be
> abstractions then what is meant by "points in space"?

|They are the primitive points of the mathematical/geometrical manifold we
use to
|model the world we inhabit (and sometimes other potential worlds of
interest).
|The concept "manifold" was abstracted from observations of the world we
inhabit.

Not bad, as for a moron.

[Odd Bodkin]

On 1/12/2016 1:42 PM, Tom Roberts wrote:
> On 1/12/16 1/12/16 - 9:55 AM, Odd Bodkin wrote:
>> On 1/12/2016 9:02 AM, Tom Roberts wrote:
>>> Just like the "wavelike" properties of photons involve probability
>>> "waves" and thus need no medium, spacetime curvature in GR is
>>> relationships among distances and thus does not need "empty space" to
>>> have any "properties".
>>
>> Begs the question about what "distance" means.
>
> OK. I should have said PHYSICAL properties.
>
> Because space, as physicists use the term, has a well defined distance
> between every pair of points within a suitably small local region. This,
> of course, is known as geometry.

Well, now I guess I'm confused about what you mean by PHYSICAL
properties. Distance you characterize as a geometric, not a physical
property. So I guess I'm curious whether you consider velocity, which is
a composition of two geometric properties, to be a physical property.
Likewise, then, for momentum.

While we're at it, we might talk about entropy or temperature, which as
you know are not properties of physical objects but of ensembles, and
are defined in terms of a statistical statement.

As we have discussed before, if there is a distinction between what is a
PHYSICAL property and what is a MATHEMATICAL property, I rapidly lose
sight of where that boundary is. After all, if I can take a wheel and
make a measurement of its circumference while it is sitting there prone
on the floor, are you saying that I've measured no physical property of
the wheel?

>
>
>> When you talk about a measure
>> between two points IN SPACE, this is making a physical statement about
>> that
>> space.
>
> Hmmm. Say, rather, that it is a GEOMETRICAL statement. Geometry needs no

> physical properties and geometrical properties that is at the heart of
> this confusion.
>
>
>> It's not like distance is an abstraction from the space or, for that
>> matter, that the points are an abstraction from space.
>
> Hmmm. Distances and points are GEOMETRICAL properties/abstractions.
>
>
>> If you do want them to be
>> abstractions then what is meant by "points in space"?
>
> They are the primitive points of the mathematical/geometrical manifold
> we use to model the world we inhabit (and sometimes other potential
> worlds of interest). The concept "manifold" was abstracted from
> observations of the world we inhabit.

So when you say a "point in space", you're not referring to any real
location in reality at all, but only to some abstracted notion that
belongs to a model the world? Or if I take my finger and point to a
location on the surface of my desk right HERE, then that point I'm
pointing to doesn't belong to the world I inhabit? That I'm not even
REFERRING to any point in the world I inhabit?

Likewise, when you use the phrase "empty space" you were not referring
to anything that belongs to the world we inhabit? If so, then why make
reference to something that doesn't pertain to our world at all?

Surely you can reconsider....

[mlwo...@wp.pl]

W dniu wtorek, 12 stycznia 2016 21:22:46 UTC+1 użytkownik Odd Bodkin napisał:

> > Because space, as physicists use the term, has a well defined distance
> > between every pair of points within a suitably small local region. This,
> > of course, is known as geometry.
>
> Well, now I guess I'm confused about what you mean by PHYSICAL
> properties. Distance you characterize as a geometric, not a physical
> property. So I guess I'm curious whether you consider velocity, which is
> a composition of two geometric properties, to be a physical property.
> Likewise, then, for momentum.

You're confused, because you expect Tom's mumble to
be logical, precise and consistent.

[Carmine Portacio]

JanPB wrote:

>> Instead of doing that expensive experiment, just imagine that spacetime
>> is not curved, cannot curve, as being as it is, empty vacuo impossible
>> to curve.
>
> This is irrelevant. Spacetime curvature is a mathematical concept here,
> it simply means that paths of free fall converge or diverge from one
> another instead of staying at constant separation distances. If you
> don't like the word "curvature", use something else.

I can't, since whatever you mean by your "math concept", the Gravity, it
directly relates at, is terribly persistent and Real. The problem you have
is that SpaceTime relates both to Gravity, Vacuum and Time. Gravity and
Vacuum are mutually Exclusive.

[JanPB]

No, you simply need to study this some more. Spacetime is defined as
the set of all events, so in this sense it's not "empty". OTOH in GR when
people say "vacuum" they mean that momentum-energy is identically zero.
But the manifold points are still there, as mathematical entities.

Now whether all this means that there is some actual substance
representing empty space that _actually_ curves is at this point
a philosophical question.

--
Jan

[james...@gmail.com]

Gravity as a round closed 3D curve of distance is of no threat.
Man is a danger to himself. A mathematical physics of round radial gravity
or "curved distance" is vital for the acceptance principle of science future.

Distance is as flat as the flat Earth. Gravity is a dimensional force
that is revealed through a pendulum clock as what God is doing in the universe...

If you are willing to entertain it... as God's Absolute Truth
in mathematical science.

Mitchell Raemsch

[mlwo...@wp.pl]

W dniu piątek, 15 stycznia 2016 00:40:19 UTC+1 użytkownik JanPB napisał:
> No, you simply need to study this some more. Spacetime is defined as
> the set of all events, so

No, it isn't. All or almost all spaces are defined,
as a pair {[a set],[something ordering]}, and spacetime
is surely not an exception, poor idiot guru

[Koobee Wublee]

On Thursday, January 14, 2016 at 3:40:19 PM UTC-8, JanPB wrote:

> Spacetime is defined as the set of all events, so in this sense
> it's not "empty".

This is just not true. Spacetime came about when Minkowski was able to condense the following Lorentz transform into a single, precise equation. <shrug>

** c dt' = (c dt - v dx / c) / sqrt(1 - v^2 / c^2)
** dx' = (dx - v dt) / sqrt(1 - v^2 / c^2)
** dy' = dy
** dz' = dz

And

** c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2

Given several observers, it is eventual that the local rate of time flow at the event where it is mutually observed by all observers can be described as the following. <shrug>

** dTau = dt sqrt(1 - B^2)

Where

** dTau = Local rate of time flow at the observed event
** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2

Thus, there is no such thing as the proper time. dTau represents the local time and nothing more. <shrug>

> OTOH in GR when people say "vacuum" they mean that momentum-energy
> is identically zero. But the manifold points are still there, as
> mathematical entities.

This also applies to the classical Newtonian system. So, why is Jan the relativistic moron trying to make things more divine? <shrug>

> Now whether all this means that there is some actual substance
> representing empty space that _actually_ curves is at this point
> a philosophical question.

No, it is a philosophic question but a serious scientific one. GR can best be explained with the existence of the Aether. Why? The condensed spacetime equation derived from the Lorentz transform can easily be derived from Larmor's transform in the exact same form. Thus, unlike the Lorentz transform which satisfies the principle of relativity, Larmor's transform demands the existence of the absolute frame of reference. <shrug>

[Gary Harnagel]

On Friday, January 15, 2016 at 12:40:40 AM UTC-7, Koobee Wublee wrote:
>
> Given several observers, it is eventual that the local rate of time
> flow at the event where it is mutually observed by all observers can
> be described as the following. <shrug>
>
> ** dTau = dt sqrt(1 - B^2)
>
> Where
>
> ** dTau = Local rate of time flow at the observed event
> ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>
> Thus, there is no such thing as the proper time. dTau represents
> the local time and nothing more. <shrug>

"Local time" means time where the observer is located. To him, there is
no dx or dy or dz, so B = 1 and dTau = dt.

But a second observer moving at v = sqrt(dx^2 + dy^2 + dz^2) wrt the
first sees the first's time different from his by sqrt(1 - v^2/c^2),
and dTau does NOT equal dt. So your assertion is demonstrably wrong.

[mlwo...@wp.pl]

W dniu piątek, 15 stycznia 2016 14:02:53 UTC+1 użytkownik Gary Harnagel napisał:
>
> "Local time" means time where the observer is located. To him, there is
> no dx or dy or dz, so B = 1 and dTau = dt.

It's always worth reminding, that physicist's tales
of observers didn't match real observers even
in Galileo's time. Now they're complete idiocies
not worthy even laugh.

[Gary Harnagel]

It's always laughable to see what stupid, arrogant plebeians like you
say when you counterfeit intelligence.

[Koobee Wublee]

On Friday, January 15, 2016 at 5:02:53 AM UTC-8, Gary Harnagel wrote:

> On Thursday, January 14, 2016 at 11:40:40 PM, Koobee Wublee wrote:

> > Spacetime came about when Minkowski was able to condense the
> > following Lorentz transform into a single, precise equation. <shrug>
>
> > ** c dt' = (c dt - v dx / c) / sqrt(1 - v^2 / c^2)
> > ** dx' = (dx - v dt) / sqrt(1 - v^2 / c^2)
> > ** dy' = dy
> > ** dz' = dz
>
> > And
>
> > ** c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
>

> > Given several observers, it is eventual that the local rate of time
> > flow at the event where it is mutually observed by all observers can
> > be described as the following. <shrug>
>
> > ** dTau = dt sqrt(1 - B^2)
>
> > Where
>
> > ** dTau = Local rate of time flow at the observed event
> > ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>
> > Thus, there is no such thing as the proper time. dTau represents the
> > local time and nothing more. <shrug>
>
> "Local time" means time where the observer is located. To him, there is
> no dx or dy or dz, so B = 1 and dTau = dt.

This is wrong. Apparently, you have failed basic science. <shrug>

Local time in this case is the rate of time flow at where the observed event is taking place. This is what Galilean, the Lorentz, and all other transforms mean. (dt, dx, dy, dz) is the coordinate system used by the observer to observe that remote event although this event does not have to be remote in which it can be located where the observer is at. <shrug>

> But a second observer moving at v = sqrt(dx^2 + dy^2 + dz^2) wrt the
> first sees the first's time different from his by sqrt(1 - v^2/c^2),
> and dTau does NOT equal dt. So your assertion is demonstrably wrong.

This is very wrong as well. No wonder the Einstein dingleberries aka relativistic morons are so mystified. <shrug>

[JanPB]

On Thursday, January 14, 2016 at 11:40:40 PM UTC-8, Koobee Wublee wrote:
> On Thursday, January 14, 2016 at 3:40:19 PM UTC-8, JanPB wrote:
>
> > Spacetime is defined as the set of all events, so in this sense
> > it's not "empty".
>
> This is just not true.

It is. Spacetime is not the empty set. Minkowski notes this too in his
1908 paper:

"In order not to allow any yawning gap to exist, we shall suppose that at every
place and time, something perceptible exists."

This is abstracted into a point in a 4D vector space (a 4D smooth manifold
in GR). Hence this space is nonempty.

--
Jan

[Paul B. Andersen]

On 15.01.2016 08:40, Koobee Wublee wrote:
> Given several observers, it is eventual that the local rate
> of time flow at the event where it is mutually observed by all
> observers can be described as the following. <shrug>

So let this "time flow" be shown by a local clock.

>
> ** dTau = dt sqrt(1 - B^2)
>
> Where
>
> ** dTau = Local rate of time flow at the observed event
> ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2

so B^2 c^2 = v(t)^2 where v(t) is the speed of the clock showing
the "local time flow" in the frame of reference with coordinates
(t,x,y,z).

dTau/dt = sqrt(1 v(t)^2/c^2)
= rate of time flow of the local clock compared to coordinate time t.

This is equation (7) in:
https://paulba.no/pdf/TwinsByMetric.pdf

Integration of this equation yields: (equation 8 in the above)
\tau = \int_{t_0}^{t_1} sqrt(1 v(t)^2/c^2) dt

So when the local clock move with the speed v(t) from
time (t = t_0) to (t = t_1) the clock showing "local time"
will advance by \tau calculated above.

Isn't this right, Koobee?

>
> Thus, there is no such thing as the proper time. dTau represents the local time and nothing more. <shrug>

"Proper time" is what a clock show.
Is you "local time" something else?
In that case - what?

--
Paul

https://paulba.no/

[Paul B. Andersen]

On 15.01.2016 20:26, Paul B. Andersen wrote:
> On 15.01.2016 08:40, Koobee Wublee wrote:
>> Given several observers, it is eventual that the local rate
>> of time flow at the event where it is mutually observed by all
>> observers can be described as the following. <shrug>
>
> So let this "time flow" be shown by a local clock.
>
>>
>> ** dTau = dt sqrt(1 - B^2)
>>
>> Where
>>
>> ** dTau = Local rate of time flow at the observed event
>> ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>
> so B^2 c^2 = v(t)^2 where v(t) is the speed of the clock showing
> the "local time flow" in the frame of reference with coordinates
> (t,x,y,z).
>
> dTau/dt = sqrt(1 v(t)^2/c^2)

dTau/dt = sqrt(1 - v(t)^2/c^2)

> = rate of time flow of the local clock compared to coordinate time t.
>
> This is equation (7) in:
> https://paulba.no/pdf/TwinsByMetric.pdf
>
> Integration of this equation yields: (equation 8 in the above)
> \tau = \int_{t_0}^{t_1} sqrt(1 v(t)^2/c^2) dt

\tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt

[Maciej Woźniak]

Użytkownik "JanPB" napisał w wiadomości grup
dyskusyjnych:a7e52adb-e6d5-43aa...@googlegroups.com...

On Thursday, January 14, 2016 at 11:40:40 PM UTC-8, Koobee Wublee wrote:
> On Thursday, January 14, 2016 at 3:40:19 PM UTC-8, JanPB wrote:
>
> > Spacetime is defined as the set of all events, so in this sense
> > it's not "empty".
>
> This is just not true.

|It is. Spacetime is not the empty set.

No, it isn't. It isn't a set at all. You don't know basic
definitions, as expected from relativistic moron
pretending a guru.

[Gary Harnagel]

On Friday, January 15, 2016 at 11:31:05 AM UTC-7, Koobee Wublee wrote:
>
> On Friday, January 15, 2016 at 5:02:53 AM UTC-8, Gary Harnagel wrote:
> >
> > On Thursday, January 14, 2016 at 11:40:40 PM, Koobee Wublee wrote:
>
> > > Spacetime came about when Minkowski was able to condense the
> > > following Lorentz transform into a single, precise equation. <shrug>
> >
> > > ** c dt' = (c dt - v dx / c) / sqrt(1 - v^2 / c^2)
> > > ** dx' = (dx - v dt) / sqrt(1 - v^2 / c^2)
> > > ** dy' = dy
> > > ** dz' = dz
> >
> > > And
> >
> > > ** c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
> >
> > > Given several observers, it is eventual that the local rate of time
> > > flow at the event where it is mutually observed by all observers can
> > > be described as the following. <shrug>
> >
> > > ** dTau = dt sqrt(1 - B^2)
> >
> > > Where
> >
> > > ** dTau = Local rate of time flow at the observed event
> > > ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
> >
> > > Thus, there is no such thing as the proper time. dTau represents the
> > > local time and nothing more. <shrug>
> >
> > "Local time" means time where the observer is located. To him, there is
> > no dx or dy or dz, so B = 1 and dTau = dt.
>
> This is wrong.

Nope. The observer is always stationary.

> Apparently, you have failed basic science.

Apparently you have failed basic rationality <shrug>

> Local time in this case is the rate of time flow at where the observed
> event is taking place.

Baloney. If the observed event is in motion wrt the observer, local times
are different.

> This is what Galilean, the Lorentz, and all other transforms mean.
> (dt, dx, dy, dz) is the coordinate system used by the observer to
> observe that remote event although this event does not have to be
> remote in which it can be located where the observer is at. <shrug>

Which is the only situation where dTau = dt. When v <> 0, dt' =/= dt

> > But a second observer moving at v = sqrt(dx^2 + dy^2 + dz^2) wrt the
> > first sees the first's time different from his by sqrt(1 - v^2/c^2),
> > and dTau does NOT equal dt. So your assertion is demonstrably wrong.
>
> This is very wrong as well. No wonder the Einstein dingleberries aka
> relativistic morons are so mystified. <shrug>

No wonder you can't think your way out of a paper bag.

[Maciej Woźniak]

Użytkownik "Gary Harnagel" napisał w wiadomości grup
dyskusyjnych:ae6b5a96-6c20-4048...@googlegroups.com...

|Nope. The observer is always stationary.

Of course! When an observer is going through a forest,
he is stationary and trees are running around.
SO SAID THE SHIT!!!!!!!!!!!
MUST BE TRUTH!!!!!!!!!!!!!!

And when he doesn't, he's not a real observer.

[Koobee Wublee]

On Friday, January 15, 2016, Paul B. Andersen wrote:
> Koobee Wublee wrote:

> > Given several observers, it is eventual that the local rate of
> > time flow at the event where it is mutually observed by all
> > observers can be described as the following. <shrug>
>

> > ** dTau = dt sqrt(1 - B^2)
>
> > Where
>
> > ** dTau = Local rate of time flow at the observed event
> > ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>

> Integration of this equation yields:
>

> \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt

>
> So when the local clock move with the speed v(t) from
> time (t = t_0) to (t = t_1) the clock showing "local time"
> will advance by \tau calculated above.
>
> Isn't this right, Koobee?

Yes, this is very correct. <shrug>

> > Thus, there is no such thing as the proper time. dTau
> > represents the local time and nothing more. <shrug>
>
> "Proper time" is what a clock show.
> Is you "local time" something else?

Proper time is dTau which is the time at where the event is observed by an observer using the coordinate system (dt, dx, dy, dz). <shrug>

Now, let's do a recap. We now have an equation describing the intrinsic rate of time flow at the observed event (dTau) and the observer's own time (dt). The observed speed (by the observer of the event), which is v(t), is not necessarily constant. <shrug>

** dTau = dt sqrt(1 - v(t)^2 / c^2)

In the case of the twins A and B, when A is observing B, we have:

** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)

Where

** dTau = dt_BB = intrinsic rate of time flow at B
** dt = dt_AB = observed rate of time flow at B by A

On the other hand, when B is observing A (B is now the observer), we have:

** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)

Where

** dTau = dt_AA = intrinsic rate of time flow at A
** dt = dt_BA = observed rate of time flow at A by B

We end up with the following 2 equations. <shrgu>

** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)

And

** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)

Say the twins start out at rest to each other, and at (T = 0) trivially agreed by both, they move away from each other. Finally, they meet again at some future time where each twin does not have to agree with the other's time elapsed. We have:

** T_BB = Integral(0, T_AA)[1 - v(t_AB)^2 / c^2]

And

** T_AA = Integral(0, T_BB)[1 - v(t_BA)^2 / c^2]

Where

** T_BB = B's time when reuniting with A
** T_AA = A's time when reuniting with B

The fatal contradiction to the paradox should now be very apparent. The Lorentz transform and thus SR must be wrong. <shrug>

<CHECKMATE> :-)

[Gary Harnagel]

So if I'm stationary wrt the North Pole and you are standing in a forest
at the equator, who's stationary, you or me?

So if you are out in space, stationary wrt the sun, and the earth comes
hurtling toward you at 29.8 km/sec, who is stationary, you or someone
standing in a forest?

“The highest form of ignorance is when you reject something you don’t
know anything about” – Wayne Dyer

[Gary Harnagel]

On Friday, January 15, 2016 at 3:40:16 PM UTC-7, Koobee Wublee wrote:
>
> Say the twins start out at rest to each other, and at (T = 0) trivially
> agreed by both, they move away from each other. Finally, they meet
> again at some future time where each twin does not have to agree with
> the other's time elapsed. We have:
>
> ** T_BB = Integral(0, T_AA)[1 - v(t_AB)^2 / c^2]
>
> And
>
> ** T_AA = Integral(0, T_BB)[1 - v(t_BA)^2 / c^2]
>
> Where
>
> ** T_BB = B's time when reuniting with A
> ** T_AA = A's time when reuniting with B
>
> The fatal contradiction to the paradox should now be very apparent. The Lorentz transform and thus SR must be wrong. <shrug>
>
> <CHECKMATE> :-)

This has been explained to Kobbly Wobbly many times. Kobbly Wobbly is
soooo ludicrous :-))

[Maciej Woźniak]

Użytkownik "Gary Harnagel" napisał w wiadomości grup

dyskusyjnych:7c4f21f2-d3af-441c...@googlegroups.com...

On Friday, January 15, 2016 at 1:24:48 PM UTC-7, Maciej Woźniak wrote:
>
> Użytkownik "Gary Harnagel" napisał w wiadomości grup
> dyskusyjnych:ae6b5a96-6c20-4048...@googlegroups.com...
> >
> > Nope. The observer is always stationary.
>
> Of course! When an observer is going through a forest,
> he is stationary and trees are running around.
> SO SAID THE SHIT!!!!!!!!!!!
> MUST BE TRUTH!!!!!!!!!!!!!!
>
> And when he doesn't, he's not a real observer.

|So if

So if observer is going through a forest
he is stationary and he sees trees running

[Gary Harnagel]

On Saturday, January 16, 2016 at 2:35:33 AM UTC-7, Maciej Woźniak wrote:
>
> [Complete nonsensical baloney]

Like this:

> So if observer is going through a forest
> he is stationary and he sees trees running
> around.
> SO SAID THE SHIT!!!!!!!!!!!
> MUST BE TRUTH!!!!!!!!!!!!!!
> And when he doesn't, he's not a real observer.

Wozniak is a really, really stupid dishonest clown.

“The difference between genius and stupidity is that genius has its limits”
-- Albert Einstein

"Only two things are infinite, the universe and human stupidity, and I'm
not sure about the former." -- Albert Einstein

“Some people study artificial intelligence. I study natural stupidity.”
-- Carl Icahn

"Life's tough..... It's even tougher if you're stupid." -- John Wayne

“Stupid is forever, ignorance can be fixed.” -- Don Wood

“I’m allergic to stupidity. I break out in sarcasm.” – Rebels Market

[Paul B. Andersen]

On 15.01.2016 23:40, Koobee Wublee wrote:
> On Friday, January 15, 2016, Paul B. Andersen wrote:
>> Koobee Wublee wrote:
>
>>> Given several observers, it is eventual that the local rate of
>>> time flow at the event where it is mutually observed by all
>>> observers can be described as the following. <shrug>
>>
>>> ** dTau = dt sqrt(1 - B^2)
>>
>>> Where
>>
>>> ** dTau = Local rate of time flow at the observed event
>>> ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>>
>> Integration of this equation yields:
>>
>> \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt
>>
>> So when the local clock move with the speed v(t) from
>> time (t = t_0) to (t = t_1) the clock showing "local time"
>> will advance by \tau calculated above.
>>
>> Isn't this right, Koobee?
>
> Yes, this is very correct. <shrug>
>
>>> Thus, there is no such thing as the proper time. dTau
>>> represents the local time and nothing more. <shrug>
>>
>> "Proper time" is what a clock show.
>> Is you "local time" something else?
>
> Proper time is dTau which is the time at where the event
> is observed by an observer using the coordinate system (dt, dx, dy, dz). <shrug>

dTau is a differential which have no value in itself,
and so are (dt, dx, dy, dz).

\tau is the time shown by the moving clock (or twin's clock)
and the coordinates of the frame of reference is (t,x,y,z).

So lets recap:
Right above you wrote that the following was correct:

\tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt

So when the local clock move with the speed v(t) from
time (t = t_0) to (t = t_1) the clock showing "local time"
will advance by \tau calculated above.

The question you carefully avoided answering was:
Is the time \tau the proper time of the moving clock,
or is it something else?

>
> Now, let's do a recap. We now have an equation describing
> the intrinsic rate of time flow at the observed event (dTau)
> and the observer's own time (dt). The observed speed (by the observer of the event),
> which is v(t), is not necessarily constant. <shrug>
>
> ** dTau = dt sqrt(1 - v(t)^2 / c^2)

Since this equation comes from the metric in flat spacetime:
c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
it is pretty obvious that the frame of reference with coordinates
(t, x, y, z) must be an _inertial_ frame of reference.

Isn't this correct, Koobee?

>
> In the case of the twins A and B, when A is observing B, we have:
>
> ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)

This can only mean that twin A is stationary in
the _inertial_ frame frame of reference (t_AB, x_AB, y_AB, z_AB)
in which twin B is moving at the speed v(t_AB), and twin B's
clock is showing t_BB.

So this is the same as the equation above, but you
have renamed \tau = t_BB and t = t_AB.

>
> [..]

>
> On the other hand, when B is observing A (B is now the observer), we have:
>
> ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)

This can only mean that twin B is stationary in
the _inertial_ frame frame of reference (t_BA, x_BA, y_BA, z_BA)
in which twin A is moving at the speed v(t_BA), and twin A's
clock is showing t_AA.

>
> [..]

>
> We end up with the following 2 equations. <shrgu>
>
> ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>
> And
>
> ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)

Nothing wrong with this so far.
So we have:
Twin A is moving with constant speed v_BA in
twin B's inertial rest frame,
and
twin B is moving with constant speed v_AB in
twin A's inertial rest frame,
and
v_AB = - v_BA, v^2 = v_AB^2 = v_BA^2
and
twin A will see twin B's moving clock run at the rate
dt_BB/dt_AB = sqrt(1 - v^2 / c^2)
and
twin B will see twin A's moving clock run at the rate
dt_AA/dt_BA = sqrt(1 - v^2 / c^2)

We have demonstrated mutual time dilation!

>
> Say the twins start out at rest to each other,
> and at (T = 0) trivially agreed by both,
> they move away from each other. Finally,
> they meet again at some future time where each
> twin does not have to agree with the other's time elapsed.

No, this make no sense.
According to your equations above must both twins be
stationary in an inertial frame, and these two
frames of reference must be moving with the relative
constant speed v.

These twins can never meet more than once,
and the rest is nonsense.

> We have:
>
> ** T_BB = Integral(0, T_AA)[1 - v(t_AB)^2 / c^2]
>
> And
>
> ** T_AA = Integral(0, T_BB)[1 - v(t_BA)^2 / c^2]
>
> Where
>
> ** T_BB = B's time when reuniting with A
> ** T_AA = A's time when reuniting with B
>
> The fatal contradiction to the paradox should now be very apparent.
> The Lorentz transform and thus SR must be wrong. <shrug>
>
> <CHECKMATE> :-)
>

Indeed.
You can see the correct application of the equation
dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)
used on the twin 'paradox' here:

https://paulba.no/pdf/TwinsByMetric.pdf

--
Paul

https://paulba.no/

[Maciej Woźniak]

Użytkownik "Gary Harnagel" napisał w wiadomości grup

dyskusyjnych:e64fb0b2-7f32-41aa...@googlegroups.com...

|“The difference between genius and stupidity is that genius has its limits”
|-- Albert Einstein

Yes. Genius is limited by common sense, idiot thinks it's a set of
prejudices.

[Koobee Wublee]

On Saturday, January 16, 2016 at 4:59:53 AM UTC-8, Paul Andersen wrote:
> Koobee Wublee wrote:

No, Paul has agreed that dTau is the intrinsic clock tick rate at the event being observed. <shrug>

> \tau is the time shown by the moving clock (or twin's clock)
> and the coordinates of the frame of reference is (t,x,y,z).

No, no. Tau becomes what the clock would read at the event being observed by this observers using the coordinate system of (t, x, y, z). It is INDEPENDENT OF WHOEVER IS MOVING AND WHOVER IS OBSERVING THIS CLOCK. IT IS INTRINSIC. IT IS INVARIANT. IT IS WHAT IT IS. <shrug>

> Right above you wrote that the following was correct:
>
> \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt
>
> So when the local clock move with the speed v(t) from
> time (t = t_0) to (t = t_1) the clock showing "local time"
> will advance by \tau calculated above.

Yes. dTau is the local clock tick rate that is totally independent of any influence. Any observation on this dTau thing will reflect a projection of this observation. It is the observation that is subject to distortion but never dTau. <shrug>

> Is the time \tau the proper time of the moving clock, or is it
> something else?

Again, Koobee Wublee has explained many times over that Tau is what the watch/calendar shows at where the remote event being observed is, and again, Tau is mapped/projected by any observer's observation on this event. <shrug>

> > We now have an equation describing the intrinsic rate of time
> > flow at the observed event (dTau) and the observer's own time
> > (dt). The observed speed (by the observer of the event),
> > which is v(t), is not necessarily constant. <shrug>
>
> > ** dTau = dt sqrt(1 - v(t)^2 / c^2)
>
> Since this equation comes from the metric in flat spacetime:
> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
> it is pretty obvious that the frame of reference with coordinates
> (t, x, y, z) must be an _inertial_ frame of reference.

No, Paul is the one who has correctly suggested v(t) can be variable. The above question satisfies all frames of references including non-inertial ones. Does Paul have a short memory? <shrug>

> > In the case of the twins A and B, when A is observing B, we have:
>
> > ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>
> This can only mean that twin A is stationary in the _inertial_
> frame frame of reference (t_AB, x_AB, y_AB, z_AB) in which twin B
> is moving at the speed v(t_AB), and twin B's clock is showing t_BB.

No, A does not have to be stationary relative to whoever. A is the observer. A is observing B's clock. A is comparing its own clock with what A has observed on B's clock. <shrug>

> So this is the same as the equation above, but you have renamed
> \tau = t_BB and t = t_AB.

No, the equation (dTau = dt sqrt(1 - v(t)^2 / c^2)) is generic. It describes how the clock tick rate is observed to be slowed by an observer noting the speed of the observed to be v(t). So, like good engineers, when B is being observed by A, one must correctly apply the generic equation, and when A is being observed by B, the same law of physics must apply. <shrug>

> > On the other hand, when B is observing A (B is now the observer),
> > we have:
>
> > ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>
> This can only mean that twin B is stationary in the _inertial_
> frame frame of reference (t_BA, x_BA, y_BA, z_BA) in which twin A
> is moving at the speed v(t_BA), and twin A's clock is showing t_AA.

No for the same reason as Koobee Wublee has mentioned above. <shrug>

> > We end up with the following 2 equations. <shrgu>
>
> > ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>
> > And
>
> > ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>
> Nothing wrong with this so far.

Good! <shrug>

> So we have:
> Twin A is moving with constant speed v_BA in
> twin B's inertial rest frame,

No, v_BA = v(t_BA) does not have to be constant. <shrug>

> [...]

>
> We have demonstrated mutual time dilation!

This is correct. <shrug>

> > Say the twins start out at rest to each other, and at (T = 0)
> > trivially agreed by both, they move away from each other.
> > Finally, they meet again at some future time where each twin
> > does not have to agree with the other's time elapsed.
>

> According to your equations above must both twins be stationary
> in an inertial frame, and these two frames of reference must be
> moving with the relative constant speed v.

No, again v(t) = v(t_AB) = v(t_BA) does not have to be constant. SR or the Galilean relativity places no restriction on anybody to be in inertial or non-inertial frames of reference. <shrug>

> > We have:
>
> > ** T_BB = Integral(0, T_AA)[1 - v(t_AB)^2 / c^2]
>
> > And
>
> > ** T_AA = Integral(0, T_BB)[1 - v(t_BA)^2 / c^2]
>
> > Where
>
> > ** T_BB = B's time when reuniting with A
> > ** T_AA = A's time when reuniting with B
>
> These twins can never meet more than once,
> and the rest is nonsense.

Since v(t) does not have to be constant, the twins can meet again at rest to each other. Paul is once again contradicting itself. <shrug>

> > The fatal contradiction to the paradox should now be very apparent.
> > The Lorentz transform and thus SR must be wrong. <shrug>
>

> You can see the correct application of the equation
> dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)
> used on the twin 'paradox' here:

Again, Paul had v(t) not necessarily constant but has now insisted on v(t) to be constant. Paul is lost. <shrug>

> > <CHECKMATE> :-)
>
> Indeed.

Yes, indeed. Demystification should be complete to anyone with a working brain. The religion of relativity is not kosher. <shrug>

[Paul B. Andersen]

dTau is no clock rate.

>> \tau is the time shown by the moving clock (or twin's clock)
>> and the coordinates of the frame of reference is (t,x,y,z).
>
> No, no. Tau becomes what the clock would read at the eventbeing
> observed by this observers using the coordinate system of (t, x, y, z).

Yes, yes, \tau IS the time shown by the moving clock at the event
observed by the observer to have the coordinates (t, x, y, z)
in the inertial frame of reference.

> It is INDEPENDENT OF WHOEVER IS MOVING AND WHOVER IS OBSERVING THIS CLOCK.
> IT IS INTRINSIC. IT IS INVARIANT. IT IS WHAT IT IS. <shrug>

Of course the proper time \tau shown by the moving clock
at a specific event is INDEPENDENT OF WHOEVER IS MOVING
AND WHOEVER IS OBSERVING THIS CLOCK. IT IS INTRINSIC.

IT IS INVARIANT. IT IS WHAT IT IS.

>

>> Right above you wrote that the following was correct:
>>
>> \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt
>>
>> So when the local clock move with the speed v(t) from
>> time (t = t_0) to (t = t_1) the clock showing "local time"
>> will advance by \tau calculated above.
>
> Yes.

So you agree that this \tau is the proper time measured by
the moving clock between the events with the temporal coordinates
t_0 and t_1 in the inertial frame with coordinates (t, x, y, z).

> dTau is
no clock rate.

But

> the local clock tick rate that is totally independent of any influence.

Of course it is.
The rate of any proper clock is independent of anything.

> Any observation on this dTau thing will reflect a projection of this observation.
> It is the observation that is subject to distortion but never dTau. <shrug>

You cannot observe dTau.

>> Is the time \tau the proper time of the moving clock, or is it
>> something else?
>
> Again, Koobee Wublee has explained many times over that Tau is what
> the watch/calendar shows at where the remote event being observed is,
> and again, Tau is mapped/projected by any observer's observation on this event. <shrug>

Was this bablygook a yes or no to the question:
"Is the time \tau the proper time of the moving clock?" ?

>>
>>> ** dTau = dt sqrt(1 - v(t)^2 / c^2)
>>
>> Since this equation comes from the metric in flat spacetime:
>> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>> it is pretty obvious that the frame of reference with coordinates
>> (t, x, y, z) must be an _inertial_ frame of reference.
>
> No, Paul is the one who has correctly suggested v(t) can be variable.

Why "No"?
The speed of the moving clock v(t) in the inertial frame
can be variable, dv/dt may be different from zero.

> The above question satisfies all frames of references including non-inertial ones.

Does that mean that Koobee Wublee doesn't understand

that the frame of reference with coordinates (t,x,y,z)

in the metric:

c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)

is an inertial frame of reference?

Does Koobe Wublee fail to understand that the metric
of an accelerated frame is different?

>>> In the case of the twins A and B, when A is observing B, we have:
>>
>>> ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>>
>> This can only mean that twin A is stationary in the _inertial_
>> frame frame of reference (t_AB, x_AB, y_AB, z_AB) in which twin B
>> is moving at the speed v(t_AB), and twin B's clock is showing t_BB.
>
> No, A does not have to be stationary relative to whoever. A is the observer.
> A is observing B's clock. A is comparing its own clock with what A has observed on B's clock. <shrug>

Why "No"?
If t_AB is the time on A's clock when the speed of B relative A
is v(t_AB), so must A be stationary in the inertial frame of
reference with the temporal coordinate t_AB.

>> So this is the same as the equation above, but you have renamed
>> \tau = t_BB and t = t_AB.
>
> No, the equation (dTau = dt sqrt(1 - v(t)^2 / c^2)) is generic.
> It describes how the clock tick rate is observed to be slowed by
> an observer noting the speed of the observed to be v(t).

Indeed.
And

dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)

describes how the clock tick rate of B's clock
is observed to be slowed by observer A, noting
the speed of the observed B to be v(t_AB).

So it is the same equation with renamed coordinates.

Note that this equation is valid only if A is stationary
in an inertial frame.

>>> On the other hand, when B is observing A (B is now the observer),
>>> we have:
>>
>>> ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>>
>> This can only mean that twin B is stationary in the _inertial_
>> frame frame of reference (t_BA, x_BA, y_BA, z_BA) in which twin A
>> is moving at the speed v(t_BA), and twin A's clock is showing t_AA.
>
> No for the same reason as Koobee Wublee has mentioned above. <shrug>

The equation:

dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)

is valid only if B is stationary in an inertial frame.

>>> We end up with the following 2 equations. <shrgu>
>>
>>> ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>>
>>> And
>>
>>> ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>>
>> Nothing wrong with this so far.
>
> Good! <shrug>
>
>> So we have:
>> Twin A is moving with constant speed v_BA in
>> twin B's inertial rest frame,
>
> No, v_BA = v(t_BA) does not have to be constant. <shrug>

Yes.
The equations above are valid only if both A and B
are stationary in inertial frames, that is none of them
are accelerating. That means that the relative speed
between A and B is constant.

A and B can not meet more than once, and the rest is nonsense.

[Carmine Portacio]

Paul B. Andersen wrote:

>>> dTau is a differential which have no value in itself,
>>> and so are (dt, dx, dy, dz).
>>
>> No, Paul has agreed that dTau is the intrinsic clock tick rate at the
>> event being observed. <shrug>
>
> dTau is no clock rate.

Displaymanager, you just said and wrote it as a difference. Man, you use
to pretend you know what symbols and models are standing for. I am so
depressed to see, at you age, you still is not.

[Koobee Wublee]

On Sunday, January 17, 2016 at 6:07:51 AM, Paul B. Andersen wrote:
> Koobee Wublee wrote:
> > Paul Andersen wrote:
> > > Koobee Wublee wrote:

> > > > Given several observers, it is eventual that the local rate of
> > > > time flow at the event where it is mutually observed by all
> > > > observers can be described as the following. <shrug>
>
> > > > ** dTau = dt sqrt(1 - B^2)
>
> > > > Where
>
> > > > ** dTau = Local rate of time flow at the observed event
> > > > ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>
> > > Integration of this equation yields:
>
> > > \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt
>
> > > So when the local clock move with the speed v(t) from
> > > time (t = t_0) to (t = t_1) the clock showing "local time"
> > > will advance by \tau calculated above.
>
> > > Isn't this right, Koobee?
>
> > Yes, this is very correct. <shrug>

The above shows Paul has understood perfectly what the relationships between dTau and (dt, dx, dy, dz) are. <shrug>

> dTau is no clock rate.

Again, dTau is the intrinsic clock tick rate at where the observed event is observed by an observer using the coordinate system (t, x, y, z). <shrug>

The rest of Paul's insane ranting is snipped for sanitary reasons. Paul had once understood very well what dTau means as logged above but now has attempted to execute a dishonest, hypocritical retreat after a checkmate move by Koobee Wublee upon Paul's own writing above. <shrug>

Paul is another troll! <shrug>

[Paul B. Andersen]

On 18.01.2016 08:59, Koobee Wublee wrote:
>
> The rest of Paul's insane ranting is snipped for sanitary reasons.
> Paul had once understood very well what dTau means as logged above
> but now has attempted to execute a dishonest, hypocritical retreat after
> a checkmate move by Koobee Wublee upon Paul's own writing above. <shrug>

OK, Koobee Wublee, I accept you capitulation.
You were unable to find an error in what I wrote,
so you had to snip it and - as so many times before -
resort to the only form of argument you know
- argumentum ad hominem:

>
> Paul is another troll! <shrug>
>

BTW, Koobee Wublee,
you can see the correct application of the equation

dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)
used on the twin 'paradox' here:

https://paulba.no/pdf/TwinsByMetric.pdf

Every equation is numbered, and I know you can't
find an error in any of them and point out exactly
what the error is. <shrug>

--
Paul

https://paulba.no/

[Carmine Portacio]

Paul B. Andersen wrote:

> On 18.01.2016 08:59, Koobee Wublee wrote:
>>
>> The rest of Paul's insane ranting is snipped for sanitary reasons.
>> Paul had once understood very well what dTau means as logged above but
>> now has attempted to execute a dishonest, hypocritical retreat after a
>> checkmate move by Koobee Wublee upon Paul's own writing above. <shrug>
>
> OK, Koobee Wublee, I accept you capitulation.
> You were unable to find an error in what I wrote,
> so you had to snip it and - as so many times before -
> resort to the only form of argument you know - argumentum ad hominem:

Are you so stupid not seeing that a Difference is NOT a Magnitude, hence
can only be a Rate? By this move you just invalidate all your papers and
you Java Applets models. This is elementary, you can't invoke any excuse
and arguments.

[Koobee Wublee]

On Monday, January 18, 2016 at 12:48:35 PM UTC-8, Paul B. Andersen wrote:
> Koobee Wublee wrote:
> > Paul Andersen wrote:
> > > Koobee Wublee wrote:

> > > > Given several observers, it is eventual that the local rate of
> > > > time flow at the event where it is mutually observed by all
> > > > observers can be described as the following. <shrug>
>
> > > > ** dTau = dt sqrt(1 - B^2)
>
> > > > Where
>
> > > > ** dTau = Local rate of time flow at the observed event
> > > > ** B^2 c^2 = (dx^2 + dy^2 + dz^2) / dt^2
>
> > > Integration of this equation yields:
>
> > > \tau = \int_{t_0}^{t_1} sqrt(1 - v(t)^2/c^2) dt
>
> > > So when the local clock move with the speed v(t) from
> > > time (t = t_0) to (t = t_1) the clock showing "local time"
> > > will advance by \tau calculated above.
>
> > > Isn't this right, Koobee?
>

> > Yes, this is very correct. Thus, there is no such thing as the

> > proper time. dTau represents the local time and nothing more.
> > <shrug>
>

> > Now, let's do a recap. We now have an equation describing the

> > intrinsic rate of time flow at the observed event (dTau) and the
> > observer's own time (dt). The observed speed (by the observer
> > of the event), which is v(t), is not necessarily constant.
> > <shrug>
>

> > ** dTau = dt sqrt(1 - v(t)^2 / c^2)

>
> > In the case of the twins A and B, when A is observing B, we have:
>
> > ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>

> > Where
>
> > ** dTau = dt_BB = intrinsic rate of time flow at B

> > ** dt = dt_AB = observed rate of time flow at B by A

>
> > On the other hand, when B is observing A (B is now the observer),
> > we have:
>
> > ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>

> > Where
>
> > ** dTau = dt_AA = intrinsic rate of time flow at A

> > ** dt = dt_BA = observed rate of time flow at A by B

>
> > We end up with the following 2 equations. <shrgu>
>
> > ** dt_BB = dt_AB sqrt(1 - v(t_AB)^2 / c^2)
>
> > And
>
> > ** dt_AA = dt_BA sqrt(1 - v(t_BA)^2 / c^2)
>

> > Say the twins start out at rest to each other, and at (T = 0)
> > trivially agreed by both, they move away from each other.
> > Finally, they meet again at some future time where each twin does

> > not have to agree with the other's time elapsed. We have:

>
> > ** T_BB = Integral(0, T_AA)[1 - v(t_AB)^2 / c^2]
>
> > And
>
> > ** T_AA = Integral(0, T_BB)[1 - v(t_BA)^2 / c^2]
>
> > Where
>
> > ** T_BB = B's time when reuniting with A
> > ** T_AA = A's time when reuniting with B
>

> > The fatal contradiction to the paradox should now be very apparent.
> > The Lorentz transform and thus SR must be wrong. <shrug>
>

> > <CHECKMATE> :-)

>
> OK, Koobee Wublee, I accept you capitulation.

Hmmm... It looks like Paul is stuck in its own make-up world of Never-Never-Land. <shrug>

> You were unable to find an error in what I wrote, so you had to snip
> it and - as so many times before - resort to the only form of
> argument you know - argumentum ad hominem:

<shaking head>

> you can see the correct application of the equation
> dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)
> used on the twin 'paradox' here:
>
> https://paulba.no/pdf/TwinsByMetric.pdf

This is all wrong. To apply correctly to the equation that Paul has agreed on, just look at what Koobee Wublee has written in the past that is cached above. <shrug>

[Paul B. Andersen]

>> Every equation is numbered, and I know you can't
>> find an error in any of them and point out exactly
>> what the error is. <shrug>
>

> This is all wrong.

"all wrong"! :-D

Thanks for confirming that Koobee Wublee is unable
to find an error in any of the equations and point

out exactly what the error is. <shrug>

BTW, Koobee.
Since you believe that the metric:
c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)

"satisfies all frames of references including non-inertial ones"

what kind of frames of reference do you think this metric "satisfies"?
c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)

a = distance/time^2

--
Paul

https://paulba.no/

[Koobee Wublee]

> Thanks for confirming that Koobee Wublee is unable
> to find an error in any of the equations and point
> out exactly what the error is. <shrug>

<shaking head> Paul is indeed living in its own fantasy world where Paul is the king of chickens that Paul is very fond of chasing. <shrug>

> Since you believe that the metric:
> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
> "satisfies all frames of references including non-inertial ones"
>
> what kind of frames of reference do you think this metric "satisfies"?
> c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>
> a = distance/time^2

All metrics satisfy both the inertial and the non-inertial frames of reference. There is no mathematics that says otherwise. Paul still needs to remove the wrong mathematics derived from Paul's made up laws of physics. <shrug>

[Paul B. Andersen]

On 19.01.2016 19:22, Koobee Wublee wrote:
> On Tuesday, January 19, 2016 at 1:56:55 AM, Paul B. Andersen wrote:
>>
>> Thanks for confirming that Koobee Wublee is unable
>> to find an error in any of the equations and point
>> out exactly what the error is. <shrug>

Here are the equations in which Koobee is unable to find errors,
and therefore had to snip it:

>> you can see the correct application of the equation
>> dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)
>> used on the twin 'paradox' here:
>>
>> https://paulba.no/pdf/TwinsByMetric.pdf
>>
>> Every equation is numbered, and I know you can't

>> find an error in any of them and point out exactly

>> what the error is. <shrug>

Wasn't I right, or was I right? :-D

> <shaking head> Paul is indeed living in its own fantasy world where Paul is the king of chickens that Paul is very fond of chasing. <shrug>
>
>> Since you believe that the metric:
>> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>> "satisfies all frames of references including non-inertial ones"
>>
>> what kind of frames of reference do you think this metric "satisfies"?
>> c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>>
>> a = distance/time^2
>
> All metrics satisfy both the inertial and the non-inertial frames of reference.
> There is no mathematics that says otherwise. Paul still needs to remove
> the wrong mathematics derived from Paul's made up laws of physics. <shrug>

!!!! :-D <shrug>



--
Paul

https://paulba.no/

[Koobee Wublee]

> Here are the equations in which Koobee is unable to find errors,
> and therefore had to snip it:
>
> > > you can see the correct application of the equation
> > > dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)

dTau = local time at where the event is being observed by someone using coordinate system (t, x, y, z). So, Koobee Wublee has already said that above equation is correct. Paul is stuck in its toilet again. <shrug>

> > > used on the twin 'paradox' here:
>
> > > https://paulba.no/pdf/TwinsByMetric.pdf

The application of the equation above is wrong. Paul did fvck it up. <shrug>

> > All metrics satisfy both the inertial and the non-inertial frames
> > of reference. There is no mathematics that says otherwise. Paul
> > still needs to remove the wrong mathematics derived from Paul's
> > made up laws of physics. <shrug>
>
> !!!! :-D <shrug>

Paul knows it has fvcked up. <shrug>

[Paul B. Andersen]

On 20.01.2016 23:40, Koobee Wublee wrote:
> On Wednesday, January 20, 2016 at 2:01:33 AM, Paul B. Andersen wrote:
>> Koobee Wublee wrote:
>>> Paul Andersen wrote:
>>>>
>>>> you can see the correct application of the equation
>>>> dTau = dt sqrt(1 - v(t)^2 / c^2) (equation 7)

>>>> used on the twin 'paradox' here:
>>>>
>>>> https://paulba.no/pdf/TwinsByMetric.pdf
>>>>

>>>> Every equation is numbered, and I know you can't
>>>> find an error in any of them and point out exactly
>>>> what the error is. <shrug>
>

> The application of the equation above is wrong. Paul did fvck it up. <shrug>

Wasn't I right, or was I right? :-D

|>>> [unsnip what Koobee Wublee don't want us too see]

|>>>
|>>> Since you believe that the metric:
|>>> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
|>>> "satisfies all frames of references including non-inertial ones"
|>>>
|>>> what kind of frames of reference do you think this metric "satisfies"?

|>>> c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)
|>>>
|>>> a = distance/time^2

See Kobee Wublee make a giant blunder when commenting the above:

>>> All metrics satisfy both the inertial and the non-inertial frames
>>> of reference. There is no mathematics that says otherwise. Paul
>>> still needs to remove the wrong mathematics derived from Paul's
>>> made up laws of physics. <shrug>
>>
>> !!!! :-D <shrug>
>
> Paul knows it has fvcked up. <shrug>

I think we all know who "fvcked up". :-D

This is fun, Koobee, isn't it? :-D

--
Paul

https://paulba.no/

[Koobee Wublee]

On Thursday, January 21, 2016 at 9:21:16 AM, Paul B. Andersen wrote:
> Koobee Wublee wrote:
> > Paul Andersen wrote:

> > > Since you believe that the metric:
> > > c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
> > > "satisfies all frames of references including non-inertial ones"
>
> > > what kind of frames of reference do you think this metric "satisfies"?
> > > c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>

> > All metrics satisfy both the inertial and the non-inertial frames
> > of reference. There is no mathematics that says otherwise. Paul
> > still needs to remove the wrong mathematics derived from Paul's
> > made up laws of physics. <shrug>
>

> See Ko[o]bee Wublee make a giant blunder when commenting the above

What blunder, Paul? <shrug>

[Paul B. Andersen]

On 21.01.2016 22:16, Koobee Wublee wrote:
> On Thursday, January 21, 2016 at 9:21:16 AM, Paul B. Andersen wrote:
>>
>> Koobee Wublee wrote:
>>>
>>> Paul Andersen wrote:
>>>>
>>>> Since you believe that the metric:
>>>> c^2 d\tau^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>>>> "satisfies all frames of references including non-inertial ones"
>>>>
>>>> what kind of frames of reference do you think this metric "satisfies"?
>>>> c^2 d\tau^2 = (1+ax/c^2)^2 c^2 dt^2 - (dx^2 + dy^2 + dz^2)
>>>>

>>>> a = distance/time^2
>>>>
>>
>> See Koobee Wublee make a giant blunder when commenting the above:

>>
>>> All metrics satisfy both the inertial and the non-inertial frames
>>> of reference. There is no mathematics that says otherwise. Paul
>>> still needs to remove the wrong mathematics derived from Paul's
>>> made up laws of physics. <shrug>
>>
>

> What blunder, Paul? <shrug>
>

Koobee Wublee's latest blunder is that he doesn't understand
that his response above is a blunder.

--
Paul

https://paulba.no/

[Koobee Wublee]

<shrug>