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From: rbwinn on 23 Jul 2010 00:47
The most famous experiment regarding relativity of time
conducted in my lifetime was in 1958 when scientists put a cesium
clock in the nosecone of a Vanguard missile and then retrieved it
after the flight of the missile to compare it with an identical clock
kept on the ground. They reported that the clock in the missile was
slower than the clock on the ground by exactly the amount predicted by
Einstein's theory of relativity. Since that time we have a multitude
of similar experiments using satellites, etc., all with the same
reported results.
The problem I see with this is that scientists used a set of
equations to represent relativity that require a length contraction.
Scientists who lived before 1887 such as Galileo and Newton would
probably have been able to solve the mathematics of this event
correctly if they had seen the experiment because they were using the
correct equations, the Galilean transformation equations, but with the
wrong interpretation of time. Had they seen an experiment proving
that velocity affected the times on clocks, they would doubtlessly
have tried to incorporate this information into the equations they
were using instead of abandoning the Galilean transformation equations
altogether the way more modern scientists did when absolute time did
not describe the results of the Michelson-Morley experiment.
There seems little doubt that the clock in the nosecone of the
Vanguard missile was slower than the clock on the ground by the amount
the equations showed, but we are still brought to a conflict of
interpretation by the fact that the equations used to make the
calculation of time for the slower clock incorporate a length
contraction which does not manifest itself in the parameters of the
experiment. For instance, suppose that the Vanguard missile had been
put in orbit around the earth instead of falling back to earth and
recovered. How do we then calculate the rate of the clock in the
nosecone?
According to Galileo's principle of equivalence, if the
missile were put in orbit around the earth at the altitude of the
moon, then it would have the same speed in its orbit that the moon has
in its orbit. If the orbits were opposite in direction, then
scientists can calculate for themselves what their theory of
relativity would predict for times on the clock in the nosecone and a
clock on the moon. The Galilean transformation equations and Newton's
equations show that a clock on the moon and a clock in the nosecone
would read the same. Both clocks would be slightly slower than a
clock on earth. So now let us consider a third satellite at the same
altitude that has an astronaut.
"Calculate your speed," the astronaut is instructed. The
astronaut knows his exact altitude. From this he knows the exact
length of his orbit. He times one orbit with the clock in his
satellite and divides that time into the length of his orbit. Does he
get a length contraction or does he get a faster speed for his
satellite than an observer on the ground making the same calculation?
You cannot make this calculation with Einstein's theory of
relativity. It requires a length contraction and the same speed
calculated from the satellite as observed from the ground. So,
although Einstein's equations give an answer that agrees with
experimental data for time, the equations do not agree with reality
with regard to distance.
Scientists threw away the equations that solve this problem
correctly in 1887. If you mention the Galilean transformation
equations to a scientist, the only response you will get is, "Absolute
time, absolute time, absolute time."
The Galilean transformation equations are not showing absolute
time. The reason Newton and Galileo used that interpretation was
because they had never seen an experiment showing that time was
relative. So let us apply the Galilean transformation equations to
the clock in the Vanguard missile the way it should have been
solved.

x'=x-vt
y'=y
z'=z
t'=t

Notice the last equation. t is the time shown by the clock on
the ground, the one that runs faster than the clock in the nosecone.
Study the equation very carefully. What does the equation say about
t'?
The equation says that t'=t, the time on the clock on the
ground. The time on the clock in the nosecone is not t', the time
used by scientists in their calculations using Einstein's theory.
Whatever scientists may or may not have done, if we are going to use
the Galilean transformation equations, we have to use a different
variable than t' for time on the clock in the nosecone. So we say
that time on the clock in the nosecone is n'. According to n', the
speed of the missile is faster than an observer on ground would
observe using his clock, which shows t.
The results of the Michelson-Morley experiment indicate that
x=ct, x'=cn', meaning that

n'=t(1-v/c)

To show why t'=t does not mean absolute time, let us consider
two different planets and their moons. If we take the planet Jupiter
and one of its moons, then t is time on a clock on Jupiter, and n' is
time on a clock on the moon of Jupiter. The orbiting clock has a
slower time.
If we take Mars and one of its moons, then t is time on a clock on
Mars, and n' is time on a clock on the moon of Mars. n' is a slower
time than t.
But Mars has a faster speed in its orbit around the sun than
Jupiter does, meaning that t in the equations for Mars is a slower
time than t in the equations for Jupiter. The Galilean transformation
equations show relativity of time if they are used correctly.
I know this will be upsetting to scientists and cause them to
exhibit all sorts of bad behavior. That will not change the
mathematics. The Galilean transformation equations can describe
relativity of time. Perhaps in a few centuries there will be
scientists capable of using the correct equations. Perhaps not. They
have certainly managed to impress themselves using the wrong ones.
From: Inertial on 23 Jul 2010 01:12
>"rbwinn" wrote in message
>news:d9d01d61-d162-4090-b2c8-a1528ce45568(a)t5g2000prd.googlegroups.com...

Looks like you've given up arguing your case in your other thread, so you're
starting all over again

[snip history]
> According to Galileo's principle of equivalence, if the
>missile were put in orbit around the earth at the altitude of the
>moon, then it would have the same speed in its orbit that the moon has
>in its orbit.

And according to SR

> If the orbits were opposite in direction, then
>scientists can calculate for themselves what their theory of
>relativity would predict for times on the clock in the nosecone and a
>clock on the moon.

Because you can't

> The Galilean transformation equations and Newton's
>equations show that a clock on the moon and a clock in the nosecone
>would read the same.

And a clock on the earth. All clocks are the same in Newton's equations and
Galilean transforms. The absence of any other term in t' = t (like x, or y
,or z , or v) show that time is the same everywhere and at all speeds.

> Both clocks would be slightly slower than a
> clock on earth.

No.. that is NOT what Galilean transformation equations and Newtons
equations show at all. Please .. don't like so blatantly.

> So now let us consider a third satellite at the same
> altitude that has an astronaut.

You can't even get what Galilean transforms and Newton say about two
satellites around earth right yet.

> "Calculate your speed," the astronaut is instructed.

Which astronaut?

> The
>astronaut knows his exact altitude. From this he knows the exact
>length of his orbit. He times one orbit with the clock in his
>satellite and divides that time into the length of his orbit. Does he
>get a length contraction or does he get a faster speed for his
>satellite than an observer on the ground making the same calculation?

Do you know? Looks at GPS satellites which take these calculations into
account .. both the effect on time of gravitational potential (altitude to
you) and of relative speed

> You cannot make this calculation with Einstein's theory of
>relativity.

Wrong.

> It requires a length contraction and the same speed
>calculated from the satellite as observed from the ground. So,
>although Einstein's equations give an answer that agrees with
>experimental data for time, the equations do not agree with reality
>with regard to distance.

Wrong.

> Scientists threw away the equations that solve this problem
>correctly in 1887.

Wrong

> If you mention the Galilean transformation
>equations to a scientist, the only response you will get is, "Absolute
>time, absolute time, absolute time."

That's correct. If you claim otherwise, you do not understand Galilean
transforms and their implications

> The Galilean transformation equations are not showing absolute
> time.

Wrong. That is EXACTLY what they show. t' = t for *any* pair of inertial
frames means that time is the same in *any* pair of inertial frames, and so
the same in *all* frames.

> The reason Newton and Galileo used that interpretation was
> because they had never seen an experiment showing that time was
> relative.

Its not a matter of interpretation .. its what the transform means. You
cannot possibly 'interpret' it differently

> So let us apply the Galilean transformation equations to
>the clock in the Vanguard missile the way it should have been
>solved.
>
> x'=x-vt
> y'=y
> z'=z
> t'=t

t' = t means the same time in all frames. .No slow down. Galilean
transforms rejected.

> Notice the last equation.

The one that refutes your lies

> t is the time shown by the clock on
> the ground, the one that runs faster than the clock in the nosecone.

Galilean transforms say the clock in the nosecone does NOT run faster,
assuming it is a correct clock.

>Study the equation very carefully. What does the equation say about
> t'?

It is the same as t'

> The equation says that t'

the time in the nosecone

> =t, the time on the clock on the
> ground.

So the time in the ground is the same as that in the nosecoe

.. The time on the clock in the nosecone is not t',

Wrong

> the time
> used by scientists in their calculations using Einstein's theory.

We are talking galilean transforms .. so that is irrelevant

> Whatever scientists may or may not have done, if we are going to use
> the Galilean transformation equations, we have to use a different
> variable than t' for time on the clock in the nosecone.

It doesn't matter what variable we use .. the transforms say the time in one
frame is the same as the time in another.
Whatever-symbol-you-use-for-time-in-one-frame =
Whatever-symbol-you-use-for-time-in-the-other-frame

>So we say
> that time on the clock in the nosecone is n'.

So .. you change the name of t' to n' .. now you have

n' = t

> According to n', the
>speed of the missile is faster than an observer on ground would
>observe using his clock, which shows t.

No . because n' = t if you use Galilean transforms.

If n' M< t, then you are NO LONGER using galilean transforms

> The results of the Michelson-Morley experiment indicate that
>x=ct, x'=cn', meaning that

Nope. MMX says nothing of the sort.

> n'=t(1-v/c)

That does not follow.

> To show why t'=t does not mean absolute time,

it does. So you can't

[snip proof that really shows that Galilean transforms do not apply .. it t'
<> t]

[snip nonsense[

You've proved Galilean transforms are not valid. Thanks.

From: rbwinn on 23 Jul 2010 01:24
On Jul 22, 10:12 pm, "Inertial" <relativ...(a)rest.com> wrote:
> >"rbwinn"  wrote in message
> >news:d9d01d61-d162-4090-b2c8-a1528ce45568(a)t5g2000prd.googlegroups.com...
>
> Looks like you've given up arguing your case in your other thread, so you're
> starting all over again
>
> [snip history]
>
> >        According to Galileo's principle of equivalence, if the
> >missile were put in orbit around the earth at the altitude of the
> >moon, then it would have the same speed in its orbit that the moon has
> >in its orbit.
>
> And according to SR
>
> >  If the orbits were opposite in direction, then
> >scientists can calculate for themselves what their theory of
> >relativity would predict for times on the clock in the nosecone and a
> >clock on the moon.
>
> Because you can't
>
> >  The Galilean transformation equations and Newton's
> >equations show that a clock on the moon and a clock in the nosecone
> >would read the same.
>
> And a clock on the earth.  All clocks are the same in Newton's equations and
> Galilean transforms.  The absence of any other term in t' = t (like x, or y
> ,or z , or v) show that time is the same everywhere and at all speeds.
>
> >  Both clocks would be slightly slower than a
> > clock on earth.
>
> No.. that is NOT what Galilean transformation equations and Newtons
> equations show at all.  Please .. don't like so blatantly.
>
> >  So now let us consider a third satellite at the same
> > altitude that has an astronaut.
>
> You can't even get what Galilean transforms and Newton say about two
> satellites around earth right  yet.
>
> >         "Calculate your speed," the astronaut is instructed.
>
> Which astronaut?
>
> >  The
> >astronaut knows his exact altitude.   From this he knows the exact
> >length of his orbit.  He times one orbit with the clock in his
> >satellite and divides that time into the length of his orbit.  Does he
> >get a length contraction or does he get a faster speed for his
> >satellite than an observer on the ground making the same calculation?
>
> Do you know?  Looks at GPS satellites which take these calculations into
> account .. both the effect on time of gravitational potential (altitude to
> you) and of relative speed
>
> >       You cannot make this calculation with Einstein's theory of
> >relativity.
>
> Wrong.
>
> >  It requires a length contraction and the same speed
> >calculated from the satellite as observed from the ground.  So,
> >although Einstein's equations give an answer that agrees with
> >experimental data for time, the equations do not agree with reality
> >with regard to distance.
>
> Wrong.
>
> >       Scientists threw away the equations that solve this problem
> >correctly in 1887.
>
> Wrong
>
> >  If you mention the Galilean transformation
> >equations to a scientist, the only response you will get is, "Absolute
> >time, absolute time, absolute time."
>
> That's correct.  If you claim otherwise, you do not understand Galilean
> transforms and their implications
>
> >       The Galilean transformation equations are not showing absolute
> > time.
>
> Wrong.  That is EXACTLY what they show.  t' = t for *any* pair of inertial
> frames means that time is the same in *any* pair of inertial frames, and so
> the same in *all* frames.
>
> >  The reason Newton and Galileo used that interpretation was
> > because they had never seen an experiment showing that time was
> > relative.
>
> Its not a matter of interpretation .. its what the transform means.  You
> cannot possibly 'interpret' it differently
>
> >  So let us apply the Galilean transformation equations to
> >the clock in the Vanguard missile the way it should have been
> >solved.
>
> >                                            x'=x-vt
> >                                            y'=y
> >                                            z'=z
> >                                             t'=t
>
> t' = t means the same time in all frames. .No slow down.  Galilean
> transforms rejected.
>
> >       Notice the last equation.
>
> The one that refutes your lies
>
> >   t is the time shown by the clock on
> > the ground, the one that runs faster than the clock in the nosecone.
>
> Galilean transforms say the clock in the nosecone does NOT run faster,
> assuming it is a correct clock.
>
> >Study the equation very carefully.    What does the equation say about
> > t'?
>
> It is the same as t'
>
> >       The equation says that t'
>
> the time in the nosecone
>
> > =t, the time on the clock on the
> > ground.
>
> So the time in the ground is the same as that in the nosecoe
>
> . The time on the clock in the nosecone is not t',
>
> Wrong
>
> > the time
> > used by scientists in their calculations using Einstein's theory.
>
> We are talking galilean transforms .. so that is irrelevant
>
> > Whatever scientists may or may not have done, if we are going to use
> > the Galilean transformation equations, we have to use a different
> > variable than t' for time on the clock in the nosecone.
>
> It doesn't matter what variable we use .. the transforms say the time in one
> frame is the same as the time in another.
> Whatever-symbol-you-use-for-time-in-one-frame =
> Whatever-symbol-you-use-for-time-in-the-other-frame
>
> >So we say
> > that time on the clock in the nosecone is n'.
>
> So .. you change the name of t' to n' .. now you have
>
> n' = t
>
> >  According to n', the
> >speed of the missile is faster than an observer on ground would
> >observe using his clock, which shows t.
>
> No . because n' = t if you use Galilean transforms.
>
> If n' M< t, then you are NO LONGER using galilean transforms
>
> >     The results of the Michelson-Morley experiment indicate that
> >x=ct, x'=cn', meaning that
>
> Nope.  MMX says nothing of the sort.
>
> >                                                n'=t(1-v/c)
>
> That does not follow.
>
> >         To show why t'=t does not mean absolute time,
>
> it does.  So you can't
>
> [snip proof that really shows that Galilean transforms do not apply .. it t'
> <> t]
>
> [snip nonsense[
>
> You've proved Galilean transforms are not valid.  Thanks.

You did not prove anything except that you are capable of saying,
Absolute time, absolute time, absolute time, just as I predicted. I
already proved what I set out to prove. If there is nothing here
except people like you then it can wait for a few centuries.
From: eric gisse on 23 Jul 2010 01:28
Inertial wrote:
[snip all]

rbwinn has nothing to say except the same thing he has been saying for ~15
years now. Just like the seto, the rbwinn deserves short bursts of contempt
or straight up killfiling.
From: Inertial on 23 Jul 2010 01:31
"rbwinn" wrote in message
news:3f93b27a-8cb5-48fa-a85a-bd880bd73984(a)x20g2000pro.googlegroups.com...
> You did not prove anything

You proved Galilean transforms do not work .. in that they are contradicted
by what we observe experimentally. I just pointed out that you did it.

> except that you are capable of saying,
> Absolute time, absolute time, absolute time, just as I predicted.

Funny .. you predict that the thing the shows Galilean transforms are
incorrect would be used to prove Galilean transforms are incorrect. Not a
terribly clever prediction.

Nor does your prediction reduce in anyway the validity of the arguments that
refute Galilean transforms, because those transforms mean that time is the
same in all frames (and we know, and you admit, that time is NOT the same in
all frames).

> I already proved what I set out to prove.

No .. You proved Galilean transforms do not work. That's not what you set
out to prove. But you did it anyway.


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