Galilean transformation equations and Robert B. Winn

[Python]

Robert B. Winn being in jail, or in a psychiatric ward these days,
or dead, here are again the equations he stubbornly refuses to use:

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

[Dirk Van de moortel]

Op 21-apr-2017 om 22:23 schreef Python:

> Robert B. Winn being in jail, or in a psychiatric ward these days,
> or dead,

or any linear combination thereof,

Dirk Vdm

[Odd Bodkin]

Was it something I said?

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

[Robert Winn]

The reports of my demise are greatly exaggerated.

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

Now if you have a slower clock somewhere, then you have to use different variables for the time of that clock.

x'=x-v'n
y'=y
z'=z
n'=n

See, a second on a fast clock is not the same as a second on a slower clock. I know how difficult this might be for people who believe in a length contraction to visualize.

[Gary Harnagel]

On Sunday, June 25, 2017 at 9:30:13 AM UTC-6, Robert Winn wrote:
>
> On Friday, April 21, 2017 at 1:23:54 PM UTC-7, Python wrote:
> >
> > Robert B. Winn being in jail, or in a psychiatric ward these days,
> > or dead, here are again the equations he stubbornly refuses to use:
> >
> > x'=x-vt
> > y'=y
> > z'=z
> > t'=t
>
>
> The reports of my demise are greatly exaggerated.
>
> x'=x-vt
> y'=y
> z'=z
> t'=t
>
> Now if you have a slower clock somewhere, then you have to use different
> variables for the time of that clock.
>
> x'=x-v'n
> y'=y
> z'=z
> n'=n

Wrong! Not only wrong, it's asinine.

> See, a second on a fast clock is not the same as a second on a slower clock.

Only Windy Winn fawns over his slower clock. Normal people use accurate
clocks.

I know how difficult this might be for Windy Winn who denies reality to
visualize.

[Robert Winn]

Einstein had a slower clock, but you say there are no slower clocks, only accurate clocks and inaccurate clocks. So how do you think Einstein came to have a slower clock?

[Gary Harnagel]

No, he didn't have a slower clock. He was a THEORETICAL physicist.

> but you say there are no slower clocks,

You are fatally mistaken. I said only that fools would believe what they say.

> only accurate clocks and inaccurate clocks. So how do you think Einstein
> came to have a slower clock?

He probably bought it at Walgreens, like YOU did.

[Robert Winn]

No, I think you are wrong about that. Einstein definitely said he had a slower clock. I have to agree with him. The moving clock was slower.

[Gary Harnagel]

You are the one in error, mon ami. And Einstein was wrong about moving
clocks running slow. Didn't you get the memos? They've been floating around
here for quite some time.

> Einstein definitely said he had a slower clock.

No, he didn't say that. He said moving clocks run slow, and he was wrong
about that. What he SHOULD have said is that moving clocks APPEAR to run
slow.

> I have to agree with him.

Then you are wronger than he was. He had an excuse, YOU don't.

> The moving clock was slower.

No, it wasn't. You are BADLY mistaken. Scientists today KNOW better. YOU
are decades out of date. The world of science has moved on, yet there you
sit in your jockies pretending that you're smarter than everyone else.
You're not.

“Being ignorant is not so much a shame, as being unwilling to learn.”
-- Benjamin Franklin

“Talk no more so exceeding proudly; let not arrogancy come out of your
mouth” – 1 Samuel 2:3

“A fool is someone whose arrogance is only surpassed by his ignorance.”
― Orrin Woodward

[Robert Winn]

Orrin Woodward? Now you are taking a name in vain. I sell books for Orrin Woodward. Orrin Woodward has a degree in engineering. Do you think he is going to agree with Einstein?
But anyway, how would you like to get out of debt? make more money? pay less taxes? I can sell you the Green Box for $120 and get you started on the road to financial freedom.

[mlwo...@wp.pl]

W dniu niedziela, 25 czerwca 2017 17:56:20 UTC+2 użytkownik Gary Harnagel napisał:

> Only Windy Winn fawns over his slower clock. Normal people use accurate
> clocks.

Accurate enough to indicate t'=t, ignoring mad
prophecies of your idiot guru.

[Gary Harnagel]

No, I'm not. He described YOU quite accurately.

> I sell books for Orrin Woodward.

He might not let you if he found out how arrogant, ignorant and dishonest
you are.

> Orrin Woodward has a degree in engineering.
> Do you think he is going to agree with Einstein?

I have a degree in engineering and I don't agree with YOU :-)

> But anyway, how would you like to get out of debt?

I'm not in debt.

> make more money?

I'm doing okay.

> pay less taxes?

That's "Pay less TAX." The plural implies the amount is individual,
countable monetary units, so you should say, "Pay FEWER taxes."

> I can sell you the Green Box for $120 and get you started on the road to
> financial freedom.

Is that why you're living in a trailer on a ranch in Arizona?

[Nicolaas Vroom]

On Monday, 26 June 2017 05:19:57 UTC+2, Gary Harnagel wrote:
> On Sunday, June 25, 2017 at 8:45:59 PM UTC-6, Robert Winn wrote:
> >
> > On Sunday, June 25, 2017 at 6:00:20 PM UTC-7, Gary Harnagel wrote:
> > >
> > > > only accurate clocks and inaccurate clocks. So how do you
> > > > think Einstein came to have a slower clock?
> > >
> > > He probably bought it at Walgreens, like YOU did.
> >
> > No, I think you are wrong about that.
>
> You are the one in error, mon ami. And Einstein was wrong about moving
> clocks running slow. Didn't you get the memos? They've been
> floating around here for quite some time.

Somewhere along the line I'am missing something.
Can some one give me a link to the memos mentioned?

> > Einstein definitely said he had a slower clock.
>
> No, he didn't say that. He said moving clocks run slow, and he was wrong
> about that. What he SHOULD have said is that moving clocks APPEAR to run
> slow.
>

Consider two identical clocks A and B in front of me, on a table.
They are identical in the sense that after one year the two clocks still
show the same time/reading.
Using a "tool" clock B is moved away and when clock B is brought back,
from now on: when clock A shows 12 o'clock clock B shows 11.55.
That means there is a constant time difference between the two clocks.
Clock B runs behind.
That means during the time that clock B travelled (away and back)
clock B did run slower than clock A.
To describe this phemomena (experiment) IMO the wording APPEAR is wrong.

Nicolaas Vroom

[Gary Harnagel]

On Tuesday, June 27, 2017 at 6:52:06 AM UTC-6, Nicolaas Vroom wrote:
>
> On Monday, 26 June 2017 05:19:57 UTC+2, Gary Harnagel wrote:
> >
> > On Sunday, June 25, 2017 at 8:45:59 PM UTC-6, Robert Winn wrote:
> > >
> > > On Sunday, June 25, 2017 at 6:00:20 PM UTC-7, Gary Harnagel wrote:
> > > >
> > > > > only accurate clocks and inaccurate clocks. So how do you
> > > > > think Einstein came to have a slower clock?
> > > >
> > > > He probably bought it at Walgreens, like YOU did.
> > >
> > > No, I think you are wrong about that.
> >
> > You are the one in error, mon ami. And Einstein was wrong about moving
> > clocks running slow. Didn't you get the memos? They've been
> > floating around here for quite some time.
>
> Somewhere along the line I'am missing something.
> Can some one give me a link to the memos mentioned?

Too numerous to bother with. Just ask your questions.

> > > Einstein definitely said he had a slower clock.
> >
> > No, he didn't say that. He said moving clocks run slow, and he was wrong
> > about that. What he SHOULD have said is that moving clocks APPEAR to run
> > slow.
> >
>
> Consider two identical clocks A and B in front of me, on a table.
> They are identical in the sense that after one year the two clocks still
> show the same time/reading.
> Using a "tool" clock B is moved away and when clock B is brought back,
> from now on: when clock A shows 12 o'clock clock B shows 11.55.
> That means there is a constant time difference between the two clocks.
> Clock B runs behind.

Yes, it is running at the same rate as A, but it is behind.

> That means during the time that clock B travelled (away and back)
> clock B did run slower than clock A.

But you have no way of directly determining that. You ASSUME that it was
running slower in transit, but there is another explanation, one that
doesn't run into a paradox (i.e., while B was in transit, he saw A's
clock running slower).

> To describe this phemomena (experiment) IMO the wording APPEAR is wrong.
>
> Nicolaas Vroom

Nope. What is wrong is claiming that clocks in motion actually run slower.
The first problem is, What do you mean by "in motion"? B can justifiably
claim that HE is stationary and A is in motion (which is why A's clock
appears to run slower to B).

[mlwo...@wp.pl]

W dniu wtorek, 27 czerwca 2017 14:52:06 UTC+2 użytkownik Nicolaas Vroom napisał:

> Consider two identical clocks A and B in front of me, on a table.
> They are identical in the sense that after one year the two clocks still
> show the same time/reading.
> Using a "tool" clock B is moved away and when clock B is brought back,
> from now on: when clock A shows 12 o'clock clock B shows 11.55.

So, they have non-identical readings, and that denies
your "identical" condition.

[Nicolaas Vroom]

On Tuesday, 27 June 2017 15:50:13 UTC+2, Gary Harnagel wrote:
> On Tuesday, June 27, 2017 at 6:52:06 AM UTC-6, Nicolaas Vroom wrote:
> >
> > On Monday, 26 June 2017 05:19:57 UTC+2, Gary Harnagel wrote:
> >
> > > No, he didn't say that. He said moving clocks run slow, and he
> > > was wrong about that. What he SHOULD have said is that moving
> > > clocks APPEAR to run slow.
> > >
> >
> > Consider two identical clocks A and B in front of me, on a table.
> > They are identical in the sense that after one year the two clocks
> > still show the same time/reading.
> > Using a "tool" clock B is moved away and when clock B is brought back,
> > from now on: when clock A shows 12 o'clock clock B shows 11.55.
> > That means there is a constant time difference between the two clocks.
> > Clock B runs behind.
>
> Yes, it is running at the same rate as A, but it is behind.
>
> > That means during the time that clock B travelled (away and back)
> > clock B did run slower than clock A.
>
> But you have no way of directly determining that.

That is correct if you mean during the time the two clocks were not
at my desk.
After clock B returned and before looking there are three possible answers:
1) They both show the same time.
2) Clock B shows an earlier time and runs behind.
3) Clock A shows an earlier time and runs behind.
In case 2) clock B is considered the moving clock
In case 3) clock A is considered the moving clock

> You ASSUME that it was
> running slower in transit, but there is another explanation, one that
> doesn't run into a paradox (i.e., while B was in transit, he saw A's
> clock running slower).

I do not see any paradox. This is a physical experiment and it requires
a physical explanation.

While B is in transit it is "difficult" to see A's clock
The same for A to see B's clock.
Ofcourse each can transmit at regular intervals a timing system.

> > To describe this phemomena (experiment) IMO the wording APPEAR is wrong.
> >
>

> Nope. What is wrong is claiming that clocks in motion actually run slower.
> The first problem is, What do you mean by "in motion"? B can justifiably
> claim that HE is stationary and A is in motion

That is "correct" before they meet again.

> (which is why A's clock APPEARS to run slower to B).

And vice versa.
The wording APPEARS is tricky when they are in transit.
IMO after they meet the word APPEAR can not be used any more.

Nicolaas Vroom.

[Paparios]

Consider, for example, the GPS satellite atomic clocks. Before launching, the frequency of those clocks is set to 10.2299999954326 MHz. Then, in your GPS receiver on the ground, the signals received from the GPS satellite are measured to be exactly of 10.23 Mhz. Nothing has changed in the orbiting atomic clock, but on the ground its frequency is measured at 10.23 Mhz, which is the geometrical projection in spacetime from the frequency of the orbiting satellite.

[Gary Harnagel]

On Tuesday, June 27, 2017 at 8:33:11 AM UTC-6, Nicolaas Vroom wrote:
>
> On Tuesday, 27 June 2017 15:50:13 UTC+2, Gary Harnagel wrote:
> >
> > On Tuesday, June 27, 2017 at 6:52:06 AM UTC-6, Nicolaas Vroom wrote:
> > >
> > > Consider two identical clocks A and B in front of me, on a table.
> > > They are identical in the sense that after one year the two clocks
> > > still show the same time/reading.
> > > Using a "tool" clock B is moved away and when clock B is brought back,
> > > from now on: when clock A shows 12 o'clock clock B shows 11.55.
> > > That means there is a constant time difference between the two clocks.
> > > Clock B runs behind.
> >
> > Yes, it is running at the same rate as A, but it is behind.
> >
> > > That means during the time that clock B travelled (away and back)
> > > clock B did run slower than clock A.
> >
> > But you have no way of directly determining that.
>
> That is correct if you mean during the time the two clocks were not
> at my desk.
> After clock B returned and before looking there are three possible answers:
> 1) They both show the same time.
> 2) Clock B shows an earlier time and runs behind.
> 3) Clock A shows an earlier time and runs behind.
> In case 2) clock B is considered the moving clock
> In case 3) clock A is considered the moving clock

A and B cannot get back together again unless one of them changes frames.
The one that changes frames will lag the other.

> > You ASSUME that it was
> > running slower in transit, but there is another explanation, one that
> > doesn't run into a paradox (i.e., while B was in transit, he saw A's
> > clock running slower).
>
> I do not see any paradox. This is a physical experiment and it requires
> a physical explanation.

And there is one: one that does not require either clock to actually be
running slower.

> While B is in transit it is "difficult" to see A's clock
> The same for A to see B's clock.
> Ofcourse each can transmit at regular intervals a timing system.

Sure. Not really so difficult after all. Of course, Doppler effects and
time delays would have to be taken into account.

> > > To describe this phemomena (experiment) IMO the wording APPEAR is wrong.
> > >
> >
> > Nope. What is wrong is claiming that clocks in motion actually run slower.
> > The first problem is, What do you mean by "in motion"? B can justifiably
> > claim that HE is stationary and A is in motion
>
> That is "correct" before they meet again.

Which can't happen unless one of them changes frames.

> > (which is why A's clock APPEARS to run slower to B).
>
> And vice versa.
> The wording APPEARS is tricky when they are in transit.
> IMO after they meet the word APPEAR can not be used any more.
>
> Nicolaas Vroom.

They will be reading different times, but you still can't assume either
of them was "running slow." As Paparios says, it's a geometric projection
effect.

[Jim Petroff]

Paparios wrote:

> 10.2299999954326 MHz. Then, in
> your GPS receiver on the ground, the signals received from the GPS
> satellite are measured to be exactly of 10.23 Mhz.

Talking MHz, there is essentially no any difference between a 10.23 MHz
and a 10.229(99) MHz. Which reveals you must be a guitar player in
Electronics and Embedded Systems.

[Tom Roberts]

On 6/27/17 6/27/17 7:52 AM, Nicolaas Vroom wrote:
> On Monday, 26 June 2017 05:19:57 UTC+2, Gary Harnagel wrote:
>> No, he didn't say that. He said moving clocks run slow, and he was wrong
>> about that. What he SHOULD have said is that moving clocks APPEAR to run
>> slow.

Yes.

> Consider two identical clocks A and B in front of me, on a table.
> They are identical in the sense that after one year the two clocks still
> show the same time/reading.
> Using a "tool" clock B is moved away and when clock B is brought back,
> from now on: when clock A shows 12 o'clock clock B shows 11.55.
> That means there is a constant time difference between the two clocks.
> Clock B runs behind.

OK. That is some incredibly powerful "tool", far beyond what is possible today.
But gedankens can invoke such "tools". I assume both clocks are ideal and undamaged.

> That means during the time that clock B travelled (away and back)
> clock B did run slower than clock A.

No. The problem is as much in your wording as in your physics, but BOTH are wrong.

Wording: When you say "that clock runs slow", you mention ONLY the clock, and
thus cannot impose some OTHER reference, all you can use is the clock ITSELF.
And the clock just ticks along at its usual rate. Indeed put a frequency
standard right next to it, co-moving with it, and it will show that the clock
DOES tick at its usual rate. So moving clocks DO NOT RUN SLOW.

This was not fully understood until after 1916 when GR was
published. Even today, elementary or popular writings often
take that shortcut to avoid the lengthy and complicated
explanation I give here.

One can say "Measured in the inertial frame of clock A, clock B did tick more
slowly than clock A". But you didn't say that. Remember there is no "absolute"
frame, and if you make a measurement in some frame, you MUST mention the frame
used. It seems you are implicitly assuming a measurement in the frame of clock
A, but you did NOT mention that.

Physics: Your scenario does NOT display "time dilation", it demonstrates a
different geometrical phenomenon: different timelike paths through spacetime can
have different elapsed proper times. Every clock ALWAYS ticks at its usual rate,
regardless of where it might be located or how it might be moved [#]. If you
want to compare a moving or distant clock to a clock nearby, you must use
SIGNALS from the other clock to this one, and you must consider how those
SIGNALS are affected (by the motion or distance) -- do that and you find the
ENTIRE effect is due to the way those SIGNALS are measured, leaving "no room"
for the clocks to tick at different rates.

[#] That is \integral d\tau, when integrated along the path of
the clock between successive ticks, ALWAYS remains the same,
regardless of the path or the geometry of the manifold along
the path. Here \tau is the clock's proper time.

The fact that clocks A and B experience different elapsed proper times along
their different paths through spacetime is no more remarkable than the fact that
for triangle UVW the paths UV and UWV have different lengths. Indeed if your
"tool" used instantaneous accelerations and inertial motion between them, your
scenario is just a triangle in spacetime, with clock a traveling UV and clock B
traveling UWV (U is when they separate, V is when they rejoin, and W is a
turn-aound in clock B's path that enables it to come back to clock A).

> To describe this phemomena (experiment) IMO the wording APPEAR is wrong.

Yes. Say "is measured (in the appropriate inertial frame)" instead. This is not
mere appearance, and this effect can and does have real physical consequences
(e.g. pion beam lines that are a kilometer long -- in the lab it takes the pions
3.3 microseconds to traverse the beamline, but the pions experience elapsed
proper times considerably less than their 26 nanosecond lifetime). There are
other, more direct experimental implementations of the "twin paradox"....

Tom Roberts

[mlwo...@wp.pl]

And, unfortunately, "orbiting atomic clock" measures the same
frequency. Good bye, the Shit.

[Nicolaas Vroom]

On Tuesday, 27 June 2017 17:32:03 UTC+2, Gary Harnagel wrote:
> On Tuesday, June 27, 2017 at 8:33:11 AM UTC-6, Nicolaas Vroom wrote:
> >
> >
> > After clock B returned and before looking there are three possible answers:
> > 1) They both show the same time.
> > 2) Clock B shows an earlier time and runs behind.
> > 3) Clock A shows an earlier time and runs behind.
> > In case 2) clock B is considered the moving clock
> > In case 3) clock A is considered the moving clock
>
> A and B cannot get back together again unless one of them changes frames.
> The one that changes frames will lag the other.

Cannot they both change frames?
Does not any change in speed implies a change in frame?
Anyway what has changing frames to do with this experiment?
IMO the most important issue is (in case 2) that clock B must change its speed
(direction of speed)

> > > You ASSUME that it was
> > > running slower in transit, but there is another explanation, one that
> > > doesn't run into a paradox (i.e., while B was in transit, he saw A's
> > > clock running slower).
> >
> > I do not see any paradox. This is a physical experiment and it requires
> > a physical explanation.
>
> And there is one: one that does not require either clock to actually be
> running slower.

I do not understand how this explanation explains the outcome of the
experiment i.e. that when the two clocks meet their readings are different.

> > While B is in transit it is "difficult" to see A's clock
> > The same for A to see B's clock.
> > Ofcourse each can transmit at regular intervals a timing system.
>
> Sure. Not really so difficult after all. Of course, Doppler effects and
> time delays would have to be taken into account.
>
> > > > To describe this phemomena (experiment) IMO the wording APPEAR is wrong.
> > > >
> > >
> > > Nope. What is wrong is claiming that clocks in motion actually run slower.
> > > The first problem is, What do you mean by "in motion"? B can justifiably
> > > claim that HE is stationary and A is in motion
> >
> > That is "correct" before they meet again.
>
> Which can't happen unless one of them changes frames.
>

If you mean unless at least one changes its speed.
In this experiment clock B has at least change its speed twice.

> > > (which is why A's clock APPEARS to run slower to B).
> >
> > And vice versa.
> > The wording APPEARS is tricky when they are in transit.
> > IMO after they meet the word APPEAR can not be used any more.
>

> They will be reading different times, but you still can't assume either
> of them was "running slow." As Paparios says, it's a geometric projection
> effect.

Why can not I say that one clock is running slower as the other (while moving)?
Such an description is in accordance with the final observation i.e. that
the readings are different.

IMO the only lesson learned from this experiment that when you want to understand
the laws of nature you should not use clocks which move relatif to each other
and if you do then you should adjust the readings of each clock.

Nicolaas Vroom

[Gary Harnagel]

On Wednesday, June 28, 2017 at 2:57:53 AM UTC-6, Nicolaas Vroom wrote:
>
> On Tuesday, 27 June 2017 17:32:03 UTC+2, Gary Harnagel wrote:
> >
> > On Tuesday, June 27, 2017 at 8:33:11 AM UTC-6, Nicolaas Vroom wrote:
> > >
> > >
> > > After clock B returned and before looking there are three possible
> > > answers:
> > > 1) They both show the same time.
> > > 2) Clock B shows an earlier time and runs behind.
> > > 3) Clock A shows an earlier time and runs behind.
> > > In case 2) clock B is considered the moving clock
> > > In case 3) clock A is considered the moving clock
> >
> > A and B cannot get back together again unless one of them changes frames.
> > The one that changes frames will lag the other.
>
> Cannot they both change frames?

Sure, but then one cannot say anything definite without describing exactly
how each did so. You would also have to define another frame that remained
inertial. Why would you try to insert such complications into a simple
statement?

> Does not any change in speed implies a change in frame?

Yes (we're discussing SR here).

> Anyway what has changing frames to do with this experiment?

It should be obvious that A and B cannot get back together without doing
so, and THAT is the crux of the matter (look at a Minkowski diagram of the
situation).

> IMO the most important issue is (in case 2) that clock B must change its
> speed (direction of speed)

THAT changes frames.

> > > > You ASSUME that it was
> > > > running slower in transit, but there is another explanation, one that
> > > > doesn't run into a paradox (i.e., while B was in transit, he saw A's
> > > > clock running slower).
> > >
> > > I do not see any paradox. This is a physical experiment and it requires
> > > a physical explanation.
> >
> > And there is one: one that does not require either clock to actually be
> > running slower.
>
> I do not understand how this explanation explains the outcome of the
> experiment i.e. that when the two clocks meet their readings are different.

Look at a Minkowski diagram: the paths have different lengths.

> > > While B is in transit it is "difficult" to see A's clock
> > > The same for A to see B's clock.
> > > Ofcourse each can transmit at regular intervals a timing system.
> >
> > Sure. Not really so difficult after all. Of course, Doppler effects and
> > time delays would have to be taken into account.
> >
> > > > > To describe this phemomena (experiment) IMO the wording APPEAR is
> > > > > wrong.
> > > >
> > > > Nope. What is wrong is claiming that clocks in motion actually run
> > > > slower. The first problem is, What do you mean by "in motion"? B
> > > > can justifiably claim that HE is stationary and A is in motion
> > >
> > > That is "correct" before they meet again.
> >
> > Which can't happen unless one of them changes frames.
>
> If you mean unless at least one changes its speed.
> In this experiment clock B has at least change its speed twice.

Exactly!

> > > > (which is why A's clock APPEARS to run slower to B).
> > >
> > > And vice versa.
> > > The wording APPEARS is tricky when they are in transit.
> > > IMO after they meet the word APPEAR can not be used any more.
> >
> > They will be reading different times, but you still can't assume either
> > of them was "running slow." As Paparios says, it's a geometric projection
> > effect.
>
> Why can not I say that one clock is running slower as the other (while
> moving)?

Because you cannot ascertain WHICH clock is "moving." Speed is purely
relative.

> Such an description is in accordance with the final observation i.e. that
> the readings are different.

But it's not in accordance with what actually happens. Remember? Which
clock ends up behind the other depends upon who changed frames.

> IMO the only lesson learned from this experiment that when you want to

> understand the laws of nature you should not use clocks which move relative

> to each other and if you do then you should adjust the readings of each
> clock.
>
> Nicolaas Vroom

Sounds like you are reinventing the first postulate (PoR) here.

[Nicolaas Vroom]

On Wednesday, 28 June 2017 05:27:28 UTC+2, tjrob137 wrote:
> On 6/27/17 6/27/17 7:52 AM, Nicolaas Vroom wrote:
> > Consider two identical clocks A and B in front of me, on a table.
> > They are identical in the sense that after one year the two clocks still
> > show the same time/reading.
> > Using a "tool" clock B is moved away and when clock B is brought back,
> > from now on: when clock A shows 12 o'clock clock B shows 11.55.
> > That means there is a constant time difference between the two clocks.
> > Clock B runs behind.
>
> OK. That is some incredibly powerful "tool", far beyond what is possible
> today. But gedankens can invoke such "tools". I assume both clocks are
> ideal and undamaged.

Anyway I assume that you agree with the possible outcome of such an
experiment. A better word for tool is maybe actor.

> > That means during the time that clock B travelled (away and back)
> > clock B did run slower than clock A.
>
> No. The problem is as much in your wording as in your physics, but BOTH
> are wrong.
>
> Wording: When you say "that clock runs slow", you mention ONLY the clock,
> and thus cannot impose some OTHER reference, all you can use is the
> clock ITSELF.

I write on purpose that clock B runs slower than clock A.
That means I compare the condition of two clocks based on its final outcome.

> And the clock just ticks along at its usual rate.

No. Based on the final observations at least one does not.

> Indeed put a frequency standard right next to it, co-moving with it,
> and it will show that the clock DOES tick at its usual rate.
> So moving clocks DO NOT RUN SLOW.

When you put an other frequency standard A' near clock A and one B' near B
(co moving with B) the pairs A',A and B',B each pair will show the same result
but disagree which each other: the pair B',B will run behind.

> This was not fully understood until after 1916 when GR was
> published. Even today, elementary or popular writings often
> take that shortcut to avoid the lengthy and complicated
> explanation I give here.

I do not understand what GR has to do with this.

> One can say "Measured in the inertial frame of clock A, clock B did tick more
> slowly than clock A". But you didn't say that. Remember there is no "absolute"
> frame, and if you make a measurement in some frame, you MUST mention the frame
> used. It seems you are implicitly assuming a measurement in the frame of clock
> A, but you did NOT mention that.

The frame I use is a table which initialy and finaly contains two clocks:
A and B.

> Physics: Your scenario does NOT display "time dilation", it demonstrates
> a different geometrical phenomenon: different timelike paths through
> spacetime can have different elapsed proper times. Every clock ALWAYS
> ticks at its usual rate,
> regardless of where it might be located or how it might be moved [#]. If you
> want to compare a moving or distant clock to a clock nearby, you must use
> SIGNALS from the other clock to this one, and you must consider how those
> SIGNALS are affected (by the motion or distance) -- do that and you find the
> ENTIRE effect is due to the way those SIGNALS are measured, leaving "no room"
> for the clocks to tick at different rates.

IMO the physical explanation that at the end of the experiment the two
clocks show a different reading is only because the two clocks experience different accelerations. These different accelerations influence the
internal operation of each clock differently.
IMO when you move a clock which operation is based on light signals which
the speed of light (a little less) it does (almost) not tick at all.
(comparing equal distances)

> There are
> other, more direct experimental implementations of the "twin paradox"....

I do not understand why you call this a paradox.
If my clock A shows 10 ticks and the moving clock B shows 6 ticks when
the moving clock returns you know there is a time keeping problem.
When the moving clock B shows 3 ticks at furthest distance and this distance
is known (in straight line and B's speed is constant) then IMO it only makes
sense to use my 10 ticks in order to calculate B's average speed.
(because the clock B is running slow)

Nicolaas Vroom.

[ldz...@gmail.com]

Well, I have not become a millionaire yet. I just bought the green box a month ago. You would only be my second customer.

[ldz...@gmail.com]

The equations that describe what you describe are the equations that scientists threw away in 1887, the Galilean transformation equations.

[ldz...@gmail.com]

So go ahead and explain what you mean by this changing of frames. As far as I can tell, a clock in S' never leaves S'.

>
> > > (which is why A's clock APPEARS to run slower to B).
> >
> > And vice versa.
> > The wording APPEARS is tricky when they are in transit.
> > IMO after they meet the word APPEAR can not be used any more.
> >
> > Nicolaas Vroom.
>
> They will be reading different times, but you still can't assume either
> of them was "running slow." As Paparios says, it's a geometric projection
> effect.

Geometric projection effect. That sounds kind of complicated. Could we say that one frame of reference just has fewer transitions of a cesium atom?

[ldz...@gmail.com]

Well, that all sounds very technical, but if you are going to define a second as a certain number of transitions of a cesium atom, then you still come up with the fact that one cesium atom has slower transitions than another cesium atom. I think Galileo had a better definition of a second than scientists of today have.

[ldz...@gmail.com]

The definition of a frame of reference appears to me to need to be clarified. Einstein appeared to understand it better. He said there was a frame of reference of a stationary clock called S and a frame of reference of a moving clock called S'. If the moving clock slows down and stops moving relative to S, it still has its own frame of reference S'. The only thing we can say with regard to S and S' is that neither is moving relative to the other. There is no change of frames of reference.

[Tom Roberts]

On 6/28/17 6/28/17 - 7:42 AM, ldz...@gmail.com wrote:
> The equations that describe what you describe are the equations that scientists threw away in 1887, the Galilean transformation equations.
>
> x'=x-vt
> y'=y
> z'=z
> t'=t

Scientists did not "throw them away", it's just that we now recognize they are
only an approximation to a MUCH better theory that uses Lorentz transforms
between (locally) inertial frames.

And that recognition did not happen in 1887, more like the decade or two after
1905 -- it was a process, not a single event; it started with Einstein's 1905
paper. Note the key point of that paper is not his postulates, it is his
approach: recognizing the importance of symmetries in physics.

Tom Roberts

[mlwo...@wp.pl]

W dniu środa, 28 czerwca 2017 18:03:00 UTC+2 użytkownik tjrob137 napisał:
> On 6/28/17 6/28/17 - 7:42 AM, ldz...@gmail.com wrote:
> > The equations that describe what you describe are the equations that scientists threw away in 1887, the Galilean transformation equations.
> >
> > x'=x-vt
> > y'=y
> > z'=z
> > t'=t
>
> Scientists did not "throw them away",

Yes, they did.

it's just that we now recognize they are
> only an approximation to a MUCH better theory

Unfortunately, the practice fuck even the MOST
best theories sometimes, like it did with your
Shit.

[ldz...@gmail.com]

Lorentz came up with his equations before 1905. His interpretation of his equations was different from Einstein's.

[Tom Roberts]

On 6/28/17 6/28/17 7:33 AM, Nicolaas Vroom wrote:
> On Wednesday, 28 June 2017 05:27:28 UTC+2, tjrob137 wrote:
>> Wording: When you say "that clock runs slow", you mention ONLY the clock,
>> and thus cannot impose some OTHER reference, all you can use is the clock
>> ITSELF.
>
> I write on purpose that clock B runs slower than clock A.

But that is NOT what happens. Clock B just travels a different path between the
endpoints than clock A, and the paths have different elapsed proper times.

Two cars drive from Chicago to New York, one directly and
one via New Orleans. Their odometers display different
distances. Do you seriously think the calibrations of their
odometers are different? -- Of course not, they merely
traveled different distances between the endpoints.
The twin scenario, and yours, are EXACTLY the same, except
in a space-time plane. It's also true that for the cars we
have a clear an obvious "frame" to use, that of the earth;
for the clock scenarios there is no such ready-made and
obvious frame of reference.

Comparing the two clocks' values at the endpoints is NOT sufficient to conclude
their tick rates are different. You MUST also account for their different paths,
and when you do, you find it accounts for 100% of the difference, leaving "no
room" for them to tick at different rates.

After all, this has been done, several times in several ways.

> That means I compare the condition of two clocks based on its final outcome.

OK. But the ONLY conclusions you can make from that are those supported by the
measurements you made. You did NOT measure their tick rates, and thus cannot
make any statements about their tick rates.

>> And the clock just ticks along at its usual rate.
> No. Based on the final observations at least one does not.

Not true. See above.

You did NOT compare rates, and thus cannot make any statement about their rates.

I, on the other hand, can extend your scenario to include
frequency standards co-located and co-moving with each clock.
They MEASURE each clock to always tick at its usual rate.

>> This was not fully understood until after 1916 when GR was published. Even
>> today, elementary or popular writings often take that shortcut to avoid the
>> lengthy and complicated explanation I give here.
>
> I do not understand what GR has to do with this.

It taught physicists that the integral of proper time is the relevant quantity.

>> Physics: Your scenario does NOT display "time dilation", it demonstrates a
>> different geometrical phenomenon: different timelike paths through
>> spacetime can have different elapsed proper times. Every clock ALWAYS ticks
>> at its usual rate, regardless of where it might be located or how it might
>> be moved [#]. If you want to compare a moving or distant clock to a clock
>> nearby, you must use SIGNALS from the other clock to this one, and you must
>> consider how those SIGNALS are affected (by the motion or distance) -- do
>> that and you find the ENTIRE effect is due to the way those SIGNALS are
>> measured, leaving "no room" for the clocks to tick at different rates.
>
> IMO the physical explanation that at the end of the experiment the two clocks
> show a different reading is only because the two clocks experience different
> accelerations. These different accelerations influence the internal operation
> of each clock differently.

Not in SR or GR. In both of them, it is the different PATH that matters, not the
different accelerations. Of course in SR one needs accelerations to make the
paths different. But in GR one does not. For all cases, it is the PATH LENGTH
(integral of proper time) that matters.

>> There are other, more direct experimental implementations of the "twin
>> paradox"....
>
> I do not understand why you call this a paradox.

Because that is its name.

"Paradox" has two meanings, and the relevant one is "a SEEMING contradiction
that upon analysis proves to be fully consistent".

> If my clock A shows 10 ticks and the moving clock B shows 6 ticks when the
> moving clock returns you know there is a time keeping problem.

Not "problem", but "difference". After all, this is completely unavoidable:
different paths can have different lengths, and for a timelike path that length
is elapsed proper time.

Tom Roberts

[ldz...@gmail.com]

So, in other words, you are saying that if you step on the accelerator of your car, your car gets shorter. Here is the problem you have. If you use the correct equations for relativity, they show that if one clock ticks ten times and the other only ticks six, then according to the time of the slower clock, the car is not getting shorter, it is going faster. The equations show the exact relationship.
\

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

Suppose t is the time of a clock by the side of the road that ticks ten times. Then we have a clock in the car that only ticks six times. If t is the time of the clock at the side of the road, the above equations say nothing about the clock in the car because t'=t, meaning that the clock by the side of the road ticks ten times while the car moves from A to B. If you divide the distance from A to B by ten ticks, you get the speed of the car according to the time of the clock by the side of the road. This says nothing about the speed of the car according to the clock in the car that only ticked six times. In order to calculate speed according to the time of that clock, you have to use a different set of Galilean transformation equations.

x = x' - v'n'
y = y'
z = z'
n = n'

This time the events are calculated from the frame of reference of the car. The car is taken to be standing still, and the highway is moving beneath the wheels of the car. x can be determined by measuring the circumference of a wheel of the car, and multiplying that by the number of times the wheel turns during six ticks of the clock in the car. The number of times the clock on the ground ticks is going to be ten, the same as it was in the first set of equations. The distance from A to B is going to be the same measured from either set of equations. x' is going to be 0 in either set of equations. From either set of equations, the speed of one frame of reference relative to the other is going to be the distance between A and B divided by the time of the clock being used to time the events. From either set of equations, the clock by the side of the road has the car going the distance from A to B divided by ten ticks; from either frame of reference, the clock in the car has the car going the distance from A to B divided by six ticks.
As you can see, the Galilean transformation equations describe reality. There is no paradox. Any person can understand what takes place. There is no reason to make a nuclear weapon and blow the entire thing up. Reality may be too simple for scientists to ever understand, but it exists nonetheless.

[Buck Millard]

W dniu piątek, 30 czerwca 2017 10:52:05 UTC-4 użytkownik ldzwinn napisał:

>> Tom Roberts
>
> So, in other words, you are saying that if you step on the accelerator
> of your car, your car gets shorter. Here is the problem you have. If
> you use the correct equations for relativity, they show that if one
> clock ticks ten times and the other only ticks six, then according to
> the time of the slower clock, the car is not getting shorter, it is
> going faster. The equations show the exact relationship.

Totally nonsense. It reveals you have no idea what is going on in Physics.

[ldz...@gmail.com]

[ldz...@gmail.com]

On Sunday, July 2, 2017 at 3:45:18 AM UTC-7, Buck Millard wrote:

I know what you say is going on. But since you say that the Galilean transformation equations are total nonsense, go ahead and show the proof of what you say.

[danco...@gmail.com]

On Sunday, July 2, 2017 at 6:32:13 AM UTC-7, ldz...@gmail.com wrote:
> Since you say that the Galilean transformation equations

> are total nonsense, go ahead and show the proof of what
> you say.

The problem with the Galilean transformations is not that they are total nonsense. They are a perfectly legitimate set of coordinate transformations. However, they do not exactly describe the relationship between relatively moving systems of inertial coordinates (meaning coordinate systems in which Newton's equations of mechanics hold good to the first approximation).

Remember, Maxwell's equations (1865) already showed that objects (whose size and shape is governed by electromagnetic forces) contract by a factor of sqrt(1 - (v/c)^2) when moving with speed v, and this has been confirmed experimentally. Any viable theory of relativity must account for this.

[Thomas 'PointedEars' Lahn]

Tom Roberts wrote:

> On 6/27/17 6/27/17 7:52 AM, Nicolaas Vroom wrote:
>> On Monday, 26 June 2017 05:19:57 UTC+2, Gary Harnagel wrote:
>>> No, he didn't say that. He said moving clocks run slow, and he was
>>> wrong
>>> about that. What he SHOULD have said is that moving clocks APPEAR to
>>> run slow.
>
> Yes.
>
>> Consider two identical clocks A and B in front of me, on a table.
>> They are identical in the sense that after one year the two clocks still
>> show the same time/reading.
>> Using a "tool" clock B is moved away and when clock B is brought back,
>> from now on: when clock A shows 12 o'clock clock B shows 11.55.
>> That means there is a constant time difference between the two clocks.
>> Clock B runs behind.
>
> OK. That is some incredibly powerful "tool", far beyond what is possible
> today.

Why? The Hafele–Keating experiment of 1971 already showed a difference
in *nanoseconds* as a combination of special-relativistic and general-
relativistic time dilation.

> But gedankens can invoke such "tools".

The proper term is “gedanken(-)experiment” or “thought experiment”.

„Gedanken“ means just “thought” (n.) in German. Trust me, I am German.

--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

[Tom Roberts]

On 7/3/17 7/3/17 11:30 AM, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>> On 6/27/17 6/27/17 7:52 AM, Nicolaas Vroom wrote:
>>> Using a "tool" clock B is moved away and when clock B is brought back,
>>> from now on: when clock A shows 12 o'clock clock B shows 11.55.
>>

>> OK. That is some incredibly powerful "tool", far beyond what is possible
>> today.
>
> Why? The Hafele–Keating experiment of 1971 already showed a difference
> in *nanoseconds* as a combination of special-relativistic and general-
> relativistic time dilation.

To get the result mentioned requires a spaceship BILLIONS of times more powerful
than we have today -- that is indeed "incredibly powerful".

I ignore having the airplanes circle the globe for billions of years.

>> But gedankens can invoke such "tools".
>
> The proper term is “gedanken(-)experiment” or “thought experiment”.
> „Gedanken“ means just “thought” (n.) in German. Trust me, I am German.

This is ENGLISH, not German. We often take verbal shortcuts, and this is a
common one.

Tom Roberts

[ldz...@gmail.com]

This is wonderful. At last we have encountered someone who is knowledgeable about the experiment that shows that objects contract by a factor of sqrt (1-v^2/c^2) when moving with speed v. Go ahead and describe the experiment. We have been waiting a long time for someone willing to do this.

[Thomas 'PointedEars' Lahn]

Tom Roberts wrote:

> On 7/3/17 7/3/17 11:30 AM, Thomas 'PointedEars' Lahn wrote:
>> Tom Roberts wrote:
>>> On 6/27/17 6/27/17 7:52 AM, Nicolaas Vroom wrote:
>>>> Using a "tool" clock B is moved away and when clock B is brought back,
>>>> from now on: when clock A shows 12 o'clock clock B shows 11.55.
>>> OK. That is some incredibly powerful "tool", far beyond what is possible
>>> today.
>> Why? The Hafele–Keating experiment of 1971 already showed a difference
>> in *nanoseconds* as a combination of special-relativistic and general-
>> relativistic time dilation.
>
> To get the result mentioned requires a spaceship BILLIONS of times more
> powerful than we have today -- that is indeed "incredibly powerful".

You are forgetting interplanetary and interstellar probes.

The figures above give a Lorentz factor of

γ = 12∕11.55 ≈ 1.039.

For v in c:

γ = 1∕√(1 – v²)
v = √(1 – 1∕γ²)

So you can have that γ at v ≈ 0.271 c.

The Juno spacecraft approached Jupiter in 2016 at v_max = 209'000 km∕h
≈ 0.00019365249 c₀.

The Helios 2 spacecraft approached Sol in 1976 at v_max = 252'792 km∕h
≈ 0.00023422871 c₀.

Breakthrough Starshot aims at v_max = 0.2 c₀.

So “a spaceship BILLIONS of times more powerful than we have today” is,
at best, an exaggeration.

> I ignore having the airplanes circle the globe for billions of years.

Tough luck. Consider satellites then.

>>> But gedankens can invoke such "tools".
>> The proper term is “gedanken(-)experiment” or “thought experiment”.
>> „Gedanken“ means just “thought” (n.) in German. Trust me, I am German.
>
> This is ENGLISH, not German.

Still, it is a German word that you are using. So I would ask for more
respect when using it.

> We often take verbal shortcuts,

Possible.

> and this is a common one.

Cite evidence.

[danco...@gmail.com]

It hasn't been very long since experimental evidence for length contraction was described here. For example, there was a description on May 26, just over a month ago. You didn't voice any objections or raise any follow-up questions at that time. Have you thought of some objections or follow-up questions since then? If so, I'd be happy to respond to them.

[ldz...@gmail.com]

Well, here is my objection. If you take two sets of Galilean transformation equations,

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

x = x' - v'n'
y = y'
z = z'
n = n'

Then, as you can see, x in the first set of equations equals x in the second set of equation, x' in the second set of equations equals x' in the second set of equations, etc. We can even say that t in the first set of equations represents the same amount of time that is represented by n', but that time is counted by clocks with different rates. Then it stands to reason that v' is really the same velocity as v, but they are counted by clocks with different rates. So how does a velocity counted by a clock showing n' relate to a velocity counted by a clock showing t?
Scientists tell us that by the time of either clock, the velocity of light is c. x=ct, x'=cn'.
But for light going in the opposite direction, x = -ct, x' = -cn'. The equation that includes both directions is

n' = (t-vx/c^2)

So now we figure velocities.

u=dx/dt
u'=dx'/dn'

x'=x-vt
dx' = dx - vdt
dx' = (dx/dt - v)dt
dx' = (u-v)dt

n' = (t-vx/c^2)
dn' = (dt - v dx / c^2)dt
dn' = (1 - v dx/dt / c^2)dt
dn' = (1-vu/c^2)
u' = dx'/dn' = (u-v)dt/(1-vu/c^2)
u' = (u-v)/(1-uv/c^2)

Einstein got this same answer from the Lorentz equations, but, as you can see, there is no length contraction in two sets of Galilean transformation equations, so the burden of proof is upon you to prove that there is a length contraction. If you think you can do it, do it right here, not by saying that there is an irrelevant discussion of the Lorentz equations somewhere on May 26 or somewhere on the internet. Show the proof here.

[danco...@gmail.com]

On Tuesday, July 4, 2017 at 11:51:57 AM UTC-7, ldz...@gmail.com wrote:
> Take [the] transformation equations
> x' = x-vt
> t' = (t-vx/c^2)
> [I'm relabeling your n' as t', since all the rest is superfluous.]
> So now we figure velocities...

> u' = (u-v)/(1-uv/c^2)
> Einstein got this same answer from the Lorentz equations,

> but, as you can see, there is no length contraction [according
> to the Winnian3 transformation].

Right. We've been over this before. The Winnian3 transformation gives the correct speed of light in both directions, and it gives the correct speed composition formula. However, as explained previously, it isn't quite correct, because the inverse of Winnnian3 is

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

where g = 1/sqrt(1 - (v/c)^2). So, the transformation is not symmetrical in form, and it implies a distinguished absolute rest frame, i.e., we would see a difference in how things behave depending on our absolute velocity, but no such differences have ever been detected. But without affecting either the speeds of light or the velocity composition formula, we can put a g factor in the Winnian3 transformation, to give the Winnian4 transformation

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

which has the inverse

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

So, Winnian4 is perfect. And it has the added advantage that Maxwell's equations, which have been experimentally confirmed in countless ways, are invariant under these transformations (not under Galilean transformations), and the laws of mechanics demonstrated for high speed objects are also invariant under Winnian4 transformations. In fact, every know force and physical process is invariant under Winnian4 transformations (as empirically confirmed), and they imply E = mc^2 (as empirically confirmed) and so on. It also correctly accounts perfectly for Maxwell's length contraction. So, far from constituting an objection to length contraction, this merely adds to the ways in which length contraction has been experimentally confirmed.

[ldz...@gmail.com]

Well, you have misinterpreted the equations. The inverse of the Galilean transformation equations is

x = x' - v't'
t = t'

So v'= -v, and you end up with the original Galilean transformation equation

x'=x-vt


But if you are saying that

x = x' - v'n'

then v' in this equation is not -v, but is

v' = -v(t/n')

This factors in the two different rates of the clocks.
No length contraction is shown, just clocks with different rates.

[danco...@gmail.com]

On Tuesday, July 4, 2017 at 5:42:44 PM UTC-7, ldz...@gmail.com wrote:
> The inverse of the Galilean transformation equations is...

...not relevant, because you aren't using the Galilean transformation, you are using the Winnian3 transformation

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

Notice that Winnian3 actually predicts more length contraction (albeit in an asymmetrical way) than the Winnian4 transformation, which is the transformation that agrees with experiment.

[Robert Winn]

There is no Winnian transform. The only transformation is the Galilean transformation. That is this one.

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

Notice the last equation, t'=t. See the two little horizontal lines between t' and t? They are called an equal sign. Whatever time t is, t' is the same time. So now you say you have a clock with a different rate. You cannot use the time of that clock in the above transformation. In order to use the time of this other clock, you have to convert its time to t. Then you can use that time in the above transformation.
But you say, you want to use the time of the slower or faster clock in a transformation without converting it to t. Well, then you have to use an entirely different set of Galilean transformation equations with different variables for time and velocity.

x = x' - v'n'

y = y'
z = z'
n = n'

n' is the time of the slower or faster clock in S', which is not shown in any way in the first set of Galilean transformation equations. The first set of Galilean transformation equations just says that t, the time of a clock in S, is being used in both frames of reference. The second set of Galilean transformation equations says that n', the time of a clock in S', is being used in both frames of reference.
There is no Winnian3 or Winnian4 transformation. The above two transformations include all rates of time. There can be a slower clock in S'. There can be a faster clock in S'. Scientists say that the clock in S' can be slower or faster, depending on velocity and gravitation. What you are calling the Winnian3 transformation does not exist.

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

If x'=x-vt, then t'=t. Sorry, but that is just the way it is. You have to use two sets of Galilean transformation equations, and the correct equations would be

x'=x-vt n'= t-vx/c^2


It should be noted that the equation for n' applies only to light, not to any x and t as it would in a transformation equation. n'= t-vx/c^2 is only true if x=ct or if x = -ct.

[danco...@gmail.com]

On Wednesday, July 5, 2017 at 6:19:31 AM UTC-7, Robert Winn wrote:
> The only transformation is the Galilean transformation... x'=x-vt, t'=t

But if x,t is an inertial coordinate system then the x',t' coordinates given by the Galilean transformation do not constitute another inertial coordinate system. It they did, there would be no time dilation, the speed of light would be noticeably different in different directions, energy would not have inertia, so E would not equal mc^2, Maxwell's equations would be wrong, and so on. All of this is abundantly falsified by experiment.

> ...the correct equations would be

> x'=x-vt n'= t-vx/c^2
> It should be noted that the equation for n' applies
> only to light, not to any x and t as it would in a
> transformation equation. n'= t-vx/c^2 is only true
> if x=ct or if x = -ct.

That's a fatal flaw, and makes your claims logically incoherent. The whole point is that we know Maxwell's equations and the isotropic equations of mechanics hold good in one system x,t, and we have found that all those same equations also hold good in terms of another system x',t' moving with any given speed v in terms of x,t. This implies, for example, that an ideal clock at rest in x,t reads t, and an identically constructed clock at rest in x',t' reads t'. We also note that the speed of light is c (all directions) in terms of both coordinate systems. The question is, how are x,t and x',t' related to each other? It is self-contradictory to say we have one transformation for mechanics and a different one for optics.

[Robert Winn]

It is all one transformation. If you will not admit that different clocks can have different rates, then there is no way to explain transformation equations to you. So far, there is no scientist who will admit that clocks can have different rates. We consider the Hafele-Keating experiment. A clock flown on an airplane has more or less time than a clock kept on the ground. How do we explain it?
Both clocks have the same rate, say scientists. That is why they register different amounts of time. So there is no way to communicate with scientists concerning this particular thing. As far as I can tell, it is useless to even try.

[Tom Roberts]

On 7/2/17 7/2/17 - 5:42 AM, ldz...@gmail.com wrote:
> On Saturday, July 1, 2017 at 9:55:39 AM UTC-7, tjrob137 wrote:

>> [...]

> So, in other words, you are saying that if you step on the accelerator of
> your car, your car gets shorter.

Not at all! I am saying what I said. You are not competent to summarize my
statements.

In fact, when you step on the accelerator of your car, if it is front-wheel
drive it will get longer, and if it is rear-wheel drive it will get shorter --
mechanical strains ENORMOUSLY exceed any effects of relativity (which don't make
things "get shorter" anyway).

As I keep saying, "length contraction" and "time dilation" do
NOT affect the objects involved, they are merely geometrical
projections that observers in other frames inherently make
when they measure a moving object.

Tom Roberts

[Tom Roberts]

On 6/28/17 6/28/17 - 8:31 AM, ldz...@gmail.com wrote:
> if you are going to define a second as a certain number of transitions of a
> cesium atom, then you still come up with the fact that one cesium atom has
> slower transitions than another cesium atom.

Nope. BY DEFINITION any free Cs133 atom has a hyperfine transition frequency of
9,192,631,770 Hz -- that is what these words MEAN.

> I think Galileo had a better definition of a second than scientists of today
> have.

You are wrong. Flat-out wrong.

Tom Roberts

[Tom Roberts]

Yes. But Lorentz's interpretation involves a completely unobservable aether that
is an "unmoved mover" [#] -- such fictions are quite difficult for people to
swallow, especially when there is an alternative (Einstein's) that does not
involve such nonsense. Moreover, Lorentz's approach does not include Lorentz
symmetry, which (after Einstein) has become a founding principle of all modern
physics. So Lorentz's approach is just a historical footnote, ignored by
mainstream physics (it is basically known only to kooks and people interested in
the details of the history).

[#] C.f. Aristotle.

Tom Roberts

[Robert Winn]

Well, I don't think so. Galileo believed the earth was rotating on its axis, and a second was 1/86400 of the time it took for the earth to rotate once on its axis. Now I know how wrong this must seem to scientists of today, just as it seemed wrong to scientists of his time. He was put under house arrest for expressing this opinion.
But I see nothing wrong with Galileo's idea. If you accept his idea of what a second is, then a slower or faster rate of transitions of a cesium atom results in something other than a Galilean second. These faster or slower seconds would then have to be converted into Galilean seconds in order to be used in the Galilean transformation equations if we are using the rotation of the earth as our standard for time, which is what Galileo did.
But you claim it is wrong to do that because modern scientists do not do it. I do other things that modern scientists do not do. But Galileo was a scientist, and a good one. I have yet to see a modern scientist try to refute Galileo's equations.

[Odd Bodkin]

For most everyday uses, calling a second 1/86400th of the rotation of
the earth is completely acceptable and you are free to use that.

However, one should be aware that the speed of the earth's rotation
changes from day to day and from year to year. As a result, the second
-- if defined as a fraction of that -- will also vary a little from day
to day and from year to year. In fact, because the earth is transferring
its angular momentum to the Moon, the time of rotation of the earth is
slowing by about 20 microseconds every year. For everyday uses like a
welder might use, this gradual deterioration of the second probably
doesn't matter. The precision a welder needs is nothing like a ten
thousandth of a millionth of a second.

But there are some applications -- which you probably don't care about
-- where that variation of the second from day to day will cause
problems. And so this is one of the requirements of setting a standard
for a unit -- the reproducible precision of a standard should be greater
than any application that would require it, not just everyday applications.

Heck, if all we needed for standards was to satisfy everyday
applications, then a watch with a second hand and a tape measure and
conventional iron disk would be all we would need. And that may suit you
just fine, personally. Just don't expect your needs to match the needs
of everyone else.

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

[anne]

watch with a second hand, ok, tape measure, fine, but whats the iron disk for?

[Robert Winn]

I don't make those geometric projections, never have. I have always used these equations:

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

You say it is not allowed, but I have always done it anyway. At one point, I tried to use these equations:

x'=(x-vt)/sqrt(1-v^2/c^2)
y'=y
z'=z
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)

I could tell there was something wrong. It took me a while, but I finally figured out what the error was. If t' was less than t, as Einstein said it was, then velocity as calculated from S' would be different from velocity calculated from S. The length contraction immediately disappeared from my thinking, never to return. So what were the correct equations?
Well, as it so happens, they are the equations that scientists threw away in 1887, the Galilean transformation equations. As proof of what I say, look at the inverse Lorentz equations.

x = (x'+ vt')/sqrt(1-v^2/c^2)

y = y'
z = z'

t = (t' + vx'/c^2)/sqrt(1-v^2/c^2)

The exact same velocity is used, meaning that these equations do not agree with experiment, however useful they may have been to scientists up until now.
So now we conduct the experiment. An astronaut is put in orbit around the earth. All scientists seem to agree that a clock in the satellite will not agree with a clock on earth. They all seem to agree that the clock in the satellite will be faster. The Lorentz equations predict a slower clock, but faster or slower, the principle is the same. If you keep the velocity the same as timed by either clock, the distance traveled by the satellite has to change. So you have a mathematical error. Einstein tried to resolve this by saying that maybe the value of Pi changes. No, the value of Pi does not change. It is a mathematical error. The length of the orbit does not change. It is fixed by the radius of the orbit, which is determined by the altitude of the satellite. That altitude will be the same measured from the satellite as measured from earth. So suppose we run the same experiment with a car. If a clock in the car is slower, as the Lorentz equations predict it will be, then if the car is going 60 miles per hour according to the time of a clock on the ground, it will be going faster than 60 miles per hour according to the time of a clock in the car. We can prove this. We time the speed of the car between two mile markers with a clock in the car and a clock on the ground. If the clock in the car is slower as the Lorentz equations say it will be, then we get a faster velocity for the car using the time of that clock.
Now I will show you why scientists love the Lorentz equations. If we say. Well maybe Robert B. Winn is right about this, so then we use the Lorentz equations to check his hypothesis. We say that the velocity as seen from S' is v' and calculate what it is.

x = (x'-v't')/sqrt(1-v'^2/c^2)

y = y'
z = z'

t = (t' - v'x'/c^2)/sqrt(1 - v'^2/c^2)

What do you know? v'= -v, the same as Isaac Newton got in his interpretation of the Galilean transformation equations. So it must be right. There really is a length contraction.
The problem Robert B. Winn sees is the same as before. If two clocks have different rates, then they do not have times that result in the same velocity. It is a mechanical impossibility.
Now it is true that there is only one velocity, but you cannot reach it with two different times without converting one to the other in reality. So I have to go to what I can see.

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

Then you have another set of Galilean transformation equations representing distances and times shown by a clock with a different rate.

x = x' - v'n'
y = y'
z = z'
n = n'

t is the time of a clock in S. n' is the time of a clock in S'.
If not, prove this wrong. Adding other forces and tensions does not change the basic problem you have.
If you do not want to discuss this, I will certainly understand. As far as I can tell the position of science will never change with regard to this, so what are you, a rebel who will consider something other than orthodoxy? If science is a religion, then you had better obey the commandments of science and scientists.

[Robert Winn]

On Wednesday, July 5, 2017 at 9:44:18 AM UTC-7, tjrob137 wrote:

No, I am not flat out wrong. I am using the Galilean transformation equations. Galileo was the scientist who first used the rotation of the earth to define time. So I am doing the same thing Galileo did, I am using the rotation of the earth to measure time. I know how wrong that must seem to you, but I am doing it. Why? As strange as it may seem, all scientists used to do what I am doing. There was a time when scientists did not even know about hyperfine transitions of free Cs133 atoms. But they did know the earth was rotating on its axis. Galileo was one of those scientists. So we are going to call Galileo's definition of a second a Galilean second. Now we compare your definition of a second to a Galilean second.
Well, we have some experiments that show the difference. Hafele and Keating transported cesium clocks, (there are your hyperfine transitions of Cs133 atoms), in airplanes. If they flew them around the earth one way, the hyperfine transitions were slower than the hyperfine transitions of Cs133 atoms in an identical clock on the ground. If they flew them around the earth the other way, the hyperfine transitions were faster than the hyperfine transitions of Cs133 atoms in an identical clock on the ground. So we will say that the hyperfine transitions of the clock on the ground agree with a Galilean second. If you want to use the seconds counted by hyperfine transitions of Cs133 atoms in the airplane, you need to convert them to Galilean seconds, otherwise you are going to end up with a length contraction of some kind. That is all I am going to say about it.

[Robert Winn]

I am not saying that you do not have an extremely accurate approximation of times and distances for some of the things scientists do, but I am saying you have an approximation, not exact equations.

[danco...@gmail.com]

On Wednesday, July 5, 2017 at 8:18:43 AM UTC-7, Robert Winn wrote:
> It is all one transformation.

Not true. You've specified two different transformations, one for light (optics, electromagnetism) and one for everything else (mechanics, etc). But both of your proposed transformations fail to give the correct relationhip between inertial coordinate systems.

> There is no scientist who will admit that clocks can
> have different rates.

Not true. I've told you several times precisely how dtau/dt varies with velocity and position in a gravitational field, where tau is the elapsed time reading on a clock and t is a suitable coordinate time. You can also find this in any book on relativity.

> We consider the Hafele-Keating experiment. A clock flown
> on an airplane has more or less time than a clock kept on
> the ground. How do we explain it?

We can easily predict precise what the clocks will show by computing the values of dtau/dt given by general relativity for the respective clocks. It's quite simple.

> Both clocks have the same rate, say scientists.

That is simply stating the obvious, i.e., we always have dtau/dtau = 1. In other words, if we measure the rate of a clock by the clock itself, it will always run at the same rate. Now, it's worth noting that an ideal clock will work perfectly to boil an egg if both are co-moving, so it isn't entirely tautological, but none of this changes the fact that dtau/dt varies with speed and gravitational potential, where t is a suitable coordinate time.

> That is why they register different amounts of time.

Of course. The integrated dtau over the paths of the clocks is different, and the values of dtau/dt are different along their paths. This is all obvious.

> So there is no way to communicate with scientists
> concerning this particular thing.

You shouldn't have trouble communicating about any factual statements with any physicist. You may encounter individuals who enjoy telling you that dtau/dtau = 1, and you may enjoy arguing with them about this, but it's not particularly helpful. What you should focus on is dtau/dt for a suitable coordinate time t (such as the Schwarzschild coordinate time for the region around the Earth, which enables you the analyze things like Hafele-Keating).

[Robert Winn]

Well, I am aware of all that, but we can take a billion or so years of time, or even 20 trillion like the national debt, and get an average for what a Galilean second would be. Then we could compare that to hyperfine transitions. But that would be blasphemy against modern science. I understand how you feel about this. If science is going to be a religion, there has to be orthodoxy.

[Odd Bodkin]

Well, you could certainly do that average. But then again, what have you
gained? The average second determined by this suggestion is now a little
different than the second that is 1/86400th of this particular day. And
so you're not really using this particular day to set today's standard
at all. All you're doing is using some average, and you have no PHYSICAL
artifact to check against, which is an essential feature of a standard.
I mean, think about it. Suppose you wanted to calibrate a clock today so
that it would read seconds correctly. How would you do it? Would you let
it count for a day and see if it got to 86,400 at the end of the day,
where that means the sun is at the same place in the sky? No, that won't
work, because today might not be the same duration as this mystical
average. In fact, you've got no artifact that will consistently mark
that average duration of a day, right? All you've got is this day, or
the next day, and neither of them is going to be the same as that average.

What you need is something that is going to consistently produce that
average value of the day, every single day that you use it. You'd have
to do some careful criteria, like this artifact has to be local to the
clock you're calibrating and at rest relative to it. But within these
criteria, it would consistently produce a time equal to that average day
duration, averaged over 10,000 years.

Hmmm.... what can we use that will consistently tick off the same time
with extremely high precision, day after day, year after year, always
available regardless of the whims and wobbles of the Earth's rotation?

Hmmmm.... what to use, what to use.... Any ideas, Robert?

> Then we could compare that to hyperfine transitions. But that would be blasphemy against modern science. I understand how you feel about this. If science is going to be a religion, there has to be orthodoxy.
>

[Robert Winn]

Well, but I am using the Galilean transformation equations, the correct equations for relativity, and Galileo's definition of a second, notwithstanding the hyperfine transitions of Cs133 atoms. In other words, there are variations even in hyperfine transitions of atoms. So what I can do instead of what you suggest is take a sufficiently long time in which the earth rotates and average out a value for a Galilean second that would satisfy scientists and say, a Galilean second is equal to a particular number of hyperfine transitions of a Cs133 atom under certain conditions of gravitation, etc., and scientists should be able to agree, Well, that would be close enough for modern science of today, although maybe not for some future time.
Then we could talk about Galilean seconds as they are used in the Galilean transformation equations. This is just a suggestion. I know how touchy scientists are about this, but I thought I would suggest it anyway.

[Odd Bodkin]

On 7/5/17 2:48 PM, Robert Winn wrote:
> Well, but I am using the Galilean transformation equations, the correct equations for relativity, and Galileo's definition of a second, notwithstanding the hyperfine transitions of Cs133 atoms. In other words, there are variations even in hyperfine transitions of atoms. So what I can do instead of what you suggest is take a sufficiently long time in which the earth rotates and average out a value for a Galilean second that would satisfy scientists and say, a Galilean second is equal to a particular number of hyperfine transitions of a Cs133 atom under certain conditions of gravitation, etc., and scientists should be able to agree, Well, that would be close enough for modern science of today, although maybe not for some future time.
> Then we could talk about Galilean seconds as they are used in the Galilean transformation equations. This is just a suggestion. I know how touchy scientists are about this, but I thought I would suggest it anyway.

Well, let's investigate this a little bit.

Let's say you wanted to design a little system with a spring and a
massive ball in an evacuated tube, and you want it to oscillate at a
certain rate. Fortunately, to do this design, you can actually use some
knowledge of physics. There's things called laws of physics, and there's
one here that says that the period of the oscillation of this system is
going to be
T = 2*pi*sqrt(m/k), where m is the mass of that ball, and k is the
stiffness of the spring, which you get by pulling on the spring with
some known force and measuring how far it stretches. So if we wanted to
actually design this system to oscillate at a rate of once every 0.50
seconds, we could do it using this physical law and then choosing the
spring and the ball appropriately. Physical laws are such handy things
for designing objects and avoiding a lot of trial and error -- which is
what DESIGNING means, after all.

Now, here's the thing. Let's say we put one of these spring-mass systems
up on a satellite and let it oscillate. Will it be oscillating at once
per half second as designed? It should be, if the law of physics is any
good. Let's say there is some other system on board that is counting on
it ticking at 0.50 seconds.

But let's take your scheme of making sure that all clocks, regardless of
where they are in a gravitational field or how fast they're moving, are
synchronized with some ground master clock that is ticking away so that
86,400 of its ticks corresponds to an "average" earth day? Will the
spring oscillate at 0.50 seconds according to this earth clock? No, it
will not.

Oddly enough, it WOULD oscillate at 0.50 seconds if compared to a local
atomic clock. But as you observed, this local atomic clock on the
satellite doesn't show the same elapsed time as the ground atomic clock.

Now, I know what you're going to say, because you've said it before.
You're going to say that the rate of the mass-spring system's
oscillations is ALSO affected by the difference in speed and
gravitation, the same way the local atomic clock is. So what that means
is you're saying that the law T = 2*pi*sqrt(m/k) just doesn't work for
a mass-spring system up in a satellite and you have to find some other
law to design by. Or maybe because you don't know what the law should be
up there, you just have to do it by trial and error.

So there's an important trade-off here. What you've gained is that all
the clocks everywhere, even in satellites, are all ticking together in
synchronization with the ground clock. But you've lost your ability to
design anything using any laws of physics, because those laws don't seem
to work up there in the satellite. Your spring mass system just doesn't
oscillate at 0.50 seconds by the ground clock, even though the law of
physics says it should.

So what have you really gained, in the long run? What's so good about
having all those clocks running in synchronization always, when you've
lost the ability to design things using laws of physics?

What MIGHT be better -- and just consider this as a possibility -- is to
say that the laws of physics are the same and still work, but only
against a local clock. And if you want to see how it works against a
ground clock, well then, you know how to translate the rate as measured
by the local clock compared to the rate of the ground clock.

[Robert Winn]

Well, you are the one who makes tables, tools, and toys, no doubt about that. All I was saying was that no matter what the reasons are for something having transitions that slow down or speed up, you cannot equate slowed down or speeded up transitions to the ones you are saying are your standard unless you are willing to have a length contraction of some kind. You say you are not only willing to have a length contraction, you are going to base your entire concept of physics on it. So we will have to leave the matter there. I just have no desire to be involved in a length contraction just to please people who want to have one. I would rather just use the Galilean transformation equations.

[danco...@gmail.com]

On Wednesday, July 5, 2017 at 12:48:18 PM UTC-7, Robert Winn wrote:
> I am using the Galilean transformation equations...
> ...what I can do instead of what you suggest is take
> a sufficiently long time in which the earth rotates ...

We've already established that the Galilean transformations don't give the correct relationship between relatively moving systems of inertial coordinates. I don't see any connection between what we've been discussing and what you have said in this message. I think you must be confusing different conversations.

> Then we could talk about Galilean seconds as they are used
> in the Galilean transformation equations.

A spinning Earth, or a spinning object of any kind, is no different than the spinning hands of a clock. For example, if two initially adjacent space ships each contain identical spinning spheres on frictionless mounts, spinning at the same speed, and then one accelerates away and back, its sphere will have turned fewer times than the one that stayed home. Galileo never suspected this. So, using a spinning sphere to measure time doesn't affect our conclusions in any way. Again, we've established that the Galilean transformations do not give the correct relationship between relatively moving systems of inertial coordinates.

[Robert Winn]

Well, one set of Galilean transformation equations does not describe the sphere turning fewer times. But it will describe the number of times the one turns that you call having stayed home. Then if you want to describe the spinning of the one you say accelerated away and back, you use another set of Galilean transformation equations with different variables for time and velocity.

It takes two sets of Galilean transformation equations to describe what you say you want to describe.

[danco...@gmail.com]

No, the behavior of the spinning spheres (and everything else) in their respective rest frames does not conform to the coordinate systems given by two Galilean transformations. We've been over this before. All physical processes inside the respective ships are accurately described by the normal equations of mechanics and electrodynamics, but only if described in terms of inertial coordinate systems. We've already established that Galilean transformations doesn't yield inertial coordinate systems. So your descriptions would need to include fictitious forces, and the only way to know what fictitious forces to apply would be to first analyze the behavior in terms of inertial coordinates, which are related by Lorentz transformations. That's what we mean when we say the Galilean transformations aren't correct.

[Odd Bodkin]

Agreed. And so it's a happy thing that length contraction has actually
been observed, something that you are a) unaware of, b) unwilling to
take any effort to become aware of, c) likely to discount even if you
were made aware of it.

Your stance on it is not particularly relevant. All that matters to me
is that *I* am aware of it, and so your arguments that say length
contraction would be a bad thing kind of fall on uncaring ears.

> You say you are not only willing to have a length contraction, you are going to base your entire concept of physics on it.

Well, let's just say that it's one important observational fact. I also
believe in the existence of nucleons but length contraction isn't the
basis for that belief. Likewise, I believe in quantum mechanical
tunneling but length contraction isn't the basis of that either. And so
on for perhaps a hundred other things that are true but don't have
length contraction as their basis. So I think you are overestimating the
importance of the simple observational fact of length contraction.

> So we will have to leave the matter there. I just have no desire to be involved in a length contraction just to please people who want to have one. I would rather just use the Galilean transformation equations.

And so you shall use the Galilean transformation equations. Meanwhile,
other people who are aware of length contraction will use the Lorentz
transforms as the need arises, and your opinion of those people will
probably not change a thing.

[Robert Winn]

Well, I know what you mean. You mean that Isaac Newton's interpretation of the Galilean transformation equations was incorrect because he believed that all correctly working clocks in the universe would agree with each other. Since scientists have proven that all correctly working clocks do not agree with each other, that has to be described. The only way to describe it with the Galilean transformation equations is to use more than one set of Galilean transformation equations.

[Robert Winn]

So we have come to the end of the matter once again.

[danco...@gmail.com]

On Thursday, July 6, 2017 at 9:32:10 AM UTC-7, Robert Winn wrote:
> You mean that Isaac Newton's interpretation of the Galilean
> transformation equations was incorrect because he believed
> that all correctly working clocks in the universe would agree
> with each other.

Yes, although it goes deeper than that. It isn't just clock rates. The fundamental fact that was missing from Newtonian physics was that all forms of energy have inertia, from which it follows that inertial coordinate systems are related by Lorentz transformations, not Galilean transformations.

> Since scientists have proven that all correctly working
> clocks do not agree with each other, that has to be described.

"Describing", per se, is not the issue. We can describe any given set of events in terms of any system of coordinates we like. There is no law that says we must use only inertial coordinate systems. However, the laws of physics (electrodynamics, mechanics, etc) take a very simple form, homogeneous and isotropic, without the need for fictitious forces, only in terms of inertial coordinate systems. So, to apply the laws of physics (which is what physicists do), without introducing fictitious forces, we must use inertial coordinates.

> The only way to describe it with the Galilean transformation
> equations is to use more than one set of Galilean transformation
> equations.

Not true. All events can be described in terms of any coordinate system we like. It isn't even necessary to apply any transformation. Or we can apply a single Galilean transformation, or we can re-scale the time coordinate, or we can apply a Lorentz transformation, etc. It doesn't matter. We can describe all events in any coordinate system. But... not every coordinate system is an inertial coordinate system, so the laws of physics, without fictitious forces, do not apply in any system other than an inertial coordinate system. And inertial coordinate systems are related by Lorentz transformations, not Galilean transformations.

[Odd Bodkin]

Well, I suppose so. But on the other hand, it isn't really a matter of
opinion. Let's supposed you were unaware of the existence of Sri Lanka.
And you then posted a lot of usenet posts about the world being a better
place without a mistaken belief in this mythical Sri Lanka. Then other
people might come up to you and assure you that Sri Lanka was in fact a
real place, not a myth at all. You might then ask them to describe on
usenet the features of this place called Sri Lanka. Those people might
decline, encouraging you to google it yourself. You very likely would
respond that you don't have enough interest to google Sri Lanka, and
even if you did you were sure that nothing but irrelevant information
would appear. And so there would be some sense on your part of a
standoff, where you didn't believe in Sri Lanka and other people were
quite positive of its reality.

But the bottom line would remain that Sri Lanka really does exist,
regardless whether anyone has proven it to you on usenet or not.

[Robert Winn]

I believe what you are doing is called improper equivalence, or some such thing. I posted equations. If you want to prove me wrong, prove the equations wrong.

[Robert Winn]

So you are saying Isaac Newton could not talk about inertia because he was using the Galilean transformation equations, and the Lorentz equations had not been invented yet. What do you plan to do, go back to all of Newton's writings and delete the word inertia everywhere Newton used it?

[Tom Roberts]

On 7/5/17 7/5/17 12:57 PM, Robert Winn wrote:
> On Wednesday, July 5, 2017 at 9:44:18 AM UTC-7, tjrob137 wrote:
>> You are wrong. Flat-out wrong.
>

> Well, I don't think so.

That's just because you are so personally ignorant that you do not even know
what you don't know. How sad.

> Galileo believed [...second derived from earth's rotation]

> But you claim it is wrong to do that because modern scientists do not do it.

NOT AT ALL! You clearly have no idea how science is performed, or how scientists
obtain definitions for things like the second.

I claim that is wrong because it has been SHOWN to be wrong, and that the
earth's rotation VARIES. We now have MUCH better clocks than the earth's rotation.

Your ignorance changes nothing. Perhaps you should go STUDY to reduce it, rather
than wasting your time posting nonsense to the 'net.

Tom Roberts

[mlwo...@wp.pl]

W dniu piątek, 7 lipca 2017 05:29:26 UTC+2 użytkownik tjrob137 napisał:
> On 7/5/17 7/5/17 12:57 PM, Robert Winn wrote:
> > On Wednesday, July 5, 2017 at 9:44:18 AM UTC-7, tjrob137 wrote:
> >> You are wrong. Flat-out wrong.
> >
> > Well, I don't think so.
>
> That's just because you are so personally ignorant that you do not even know
> what you don't know. How sad.
>
> > Galileo believed [...second derived from earth's rotation]
> > But you claim it is wrong to do that because modern scientists do not do it.
>
> NOT AT ALL! You clearly have no idea how science is performed, or how scientists
> obtain definitions for things like the second.
>
> I claim that is wrong because it has been SHOWN to be wrong, and that the
> earth's rotation VARIES. We now have MUCH better clocks than the earth's rotation.

"Better" in the meaning better matching your moronic
mumble.

[Robert Winn]

No, I am not a scientist. I can tell that the earth rotates. I can divide the time of a rotation into fractions. You claim that is not "accurate". What is not accurate is using two different rates of time in the same set of transformation equations and then being surprised that there is a "length contraction". You are right about one thing. Trying to communicate with a scientist is a waste of time.

[Upech Bezradne]

Robert Winn wrote:

> No, I am not a scientist. I can tell that the earth rotates.

No you can't. You may safely assume that the Universe with all in it
rotates, and the earth being fixed. Same equations.

> I can
> divide the time of a rotation into fractions. You claim that is not
> "accurate". What is not accurate is using two different rates of time
> in the same set of transformation equations and then being surprised
> that there is a "length contraction".

You just said that you divide the same time, hence your time cannot
possibly pass at different rates. Same rate, but divided, is same rate.

> You are right about one thing.
> Trying to communicate with a scientist is a waste of time.

However with you, impossible.

[Robert Winn]

As I said, I am not a scientist. I do not have to pretend that the earth is standing still and that galaxies are whizzing around at faster than the speed of light. As Galileo noted long ago, the earth is rotating on its axis. If scientists do not wish to concede that fact, let scientists do whatever they choose to do. I do not really care how irrelevant scientists wish to become.

[mlwo...@wp.pl]

W dniu piątek, 7 lipca 2017 12:34:24 UTC+2 użytkownik Upech Bezradne napisał:
> Robert Winn wrote:
>
> > No, I am not a scientist. I can tell that the earth rotates.
>
> No you can't.

Yes, I can.

> You may safely assume that the Universe with all in it
> rotates, and the earth being fixed. Same equations.

I may safely assume it, but I won't assume it.

[Gary Harnagel]

On Friday, July 7, 2017 at 5:06:57 AM UTC-6, Robert Winn wrote:
>
> As I said, I am not a scientist.

Then why are you so full of "answers"?

“A wise person is full of questions. A dull person is full of answers.”
– Paulo Coelho

> I do not have to pretend that the earth is standing still

GOOD for you. You are now up to date with Galileo.

> and that galaxies are whizzing around at faster than the speed of light.

No one has seen a galaxy moving faster than the speed of light. Are you
nuts?

> As Galileo noted long ago, the earth is rotating on its axis.

I thought he just said that it moves.

> If scientists do not wish to concede that fact, let scientists do whatever
> they choose to do.

They choose to ignore fools like you.

> I do not really care how irrelevant scientists wish to become.

And they don't care how stupid you wish to be.

[danco...@gmail.com]

On Thursday, July 6, 2017 at 7:04:54 PM UTC-7, Robert Winn wrote:
> On Thursday, July 6, 2017 at 11:05:30 AM UTC-7, danco...@gmail.com wrote:
> > The fundamental fact that was missing from Newtonian physics
> > was that all forms of energy have inertia, from which it follows
> > that inertial coordinate systems are related by Lorentz
> > transformations, not Galilean transformations.
>

> So you are saying Isaac Newton could not talk about inertia

> because he was using the Galilean transformation equations...

No, Newton could (and did) talk about inertia, but only about the inertia of the rest mass associated with matter. He did not know about the inertia of energy. For example, Newton thought that a material object subjected to a constant force would undergo constant acceleration, but we now know this is not true, because of the inertia of energy. Newton didn't notice this, because the effect is imperceptibly small at low speeds.

> Since scientists have proven that all correctly working
> clocks do not agree with each other, that has to be described.

"Describing", per se, is not the issue. We can describe any given set of events in terms of any system of coordinates we like. There is no law that says we must use only inertial coordinate systems. However, the laws of physics (electrodynamics, mechanics, etc) take a very simple form, homogeneous and isotropic, without the need for fictitious forces, only in terms of inertial coordinate systems. So, to apply the laws of physics (which is what physicists do), without introducing fictitious forces, we must use inertial coordinates.

> The only way to describe it with the Galilean transformation
> equations is to use more than one set of Galilean transformation
> equations.

Not true. All events can be described in terms of any coordinate system we like. It isn't even necessary to apply any transformation at all. But not every coordinate system is an inertial coordinate system. The laws of physics, without fictitious forces, do not apply in any system other than an inertial coordinate system. And inertial coordinate systems are related by Lorentz transformations, not Galilean transformations.

[Robert Winn]

On Friday, July 7, 2017 at 6:53:45 AM UTC-7, Gary Harnagel wrote:
> On Friday, July 7, 2017 at 5:06:57 AM UTC-6, Robert Winn wrote:
> >
> > As I said, I am not a scientist.
>
> Then why are you so full of "answers"?
>
> “A wise person is full of questions. A dull person is full of answers.”
> – Paulo Coelho
>
> > I do not have to pretend that the earth is standing still
>
> GOOD for you. You are now up to date with Galileo.
>
> > and that galaxies are whizzing around at faster than the speed of light.
>
> No one has seen a galaxy moving faster than the speed of light. Are you
> nuts?
>
> > As Galileo noted long ago, the earth is rotating on its axis.
> I
>

> > If scientists do not wish to concede that fact, let scientists do whatever
> > they choose to do.
>
> They choose to ignore fools like you.
>
> > I do not really care how irrelevant scientists wish to become.
>
> And they don't care how stupid you wish to be.

You just proved yourself wrong by answering my post. If you are ignoring someone, you do not answer posts they make on the internet.

[Robert Winn]

Well,as I said, you are probably going to want to go through Isaac Newton's writings and delete the word inertia every time he said it. Newton said that moving objects have inertia. An object in motion tends to stay in motion. Newton had equations for all of these things. Of course, he did not have the permission of modern scientists to do what he did.

[Gary Harnagel]

You would possibly be correct IF I were a scientist. OTOH, Churchill had
YOU figured out:

“When eagles are silent, the parrots begin to jabber.”
-- Winston Churchill

[Robert Winn]

Oh, you are an eagle, not a scientist. I did not know. Well, I just asked a question a long time ago. You are allowed to answer the question even if you are not a scientist. Here was my original question, asked about twenty years ago.
If a clock in a satellite has a different rate from a clock on earth, how fast is the satellite going?
I was told that the satellite had the same speed as calculated from either clock. That answer did not satisfy my mind, so I tried to find the correct equations to answer the question I had asked. Eventually, I did find the correct equations.
The correct equations were these:

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

The equations have to be used twice, once for the time of each clock. If the clock in the satellite is slower than a clock on earth, then the velocity obtained using the time of that clock is faster than the velocity obtained using the time of a clock on earth. If the clock in the satellite is faster than a clock on earth, then the velocity obtained using the time of that clock is slower than the velocity obtained using the time of a clock on earth.
So that is where the matter has stood since that time. Scientists and their admirers do not want to discuss this. Scientists and their admirers do not have to discuss this. Scientists and their admirers can use whatever equations they want to use. What scientists and their admirers cannot do is blow smoke in my face. Go blow your smoke in someone else's face. Thank you very much.

[Gary Harnagel]

On Saturday, July 8, 2017 at 7:02:50 AM UTC-6, Robert Winn wrote:
>
> On Friday, July 7, 2017 at 8:25:10 PM UTC-7, Gary Harnagel wrote:
> >
> > On Friday, July 7, 2017 at 9:04:13 PM UTC-6, Robert Winn wrote:
> > >
> > > You just proved yourself wrong by answering my post. If you are ignoring
> > > someone, you do not answer posts they make on the internet.
> >
> > You would possibly be correct IF I were a scientist. OTOH, Churchill had
> > YOU figured out:
> >
> > “When eagles are silent, the parrots begin to jabber.”
> > -- Winston Churchill
>
> Oh, you are an eagle,

And you are a parrot.

> not a scientist.

You apparently don't understand the meanings of "possibly" and "if."

> I did not know.

Well, that's a given.

> Well, I just asked a question a long time ago. You are allowed to answer
> the question even if you are not a scientist. Here was my original question,
> asked about twenty years ago.
> If a clock in a satellite has a different rate from a clock on earth, how
> fast is the satellite going?

Your question is too crude to have a definite answer and your thinking
is too dull understand why.

> I was told that the satellite had the same speed as calculated from either
clock. That answer did not satisfy my mind,

Why does "satisfy" and "correct" have anything to do with each other"

> [Regurgitated piddlin' pony plop deleted for sanitary reasons]

[danco...@gmail.com]

On Friday, July 7, 2017 at 8:10:27 PM Robert Winn wrote:
> You are probably going to want to go through

> Isaac Newton's writings and delete the word
> inertia every time he said it.

Newton's understanding of inertia was incomplete, so when we use the Newtonian equations today we understand that they are only approximately correct, and only for low speeds.

> Newton said that moving objects have inertia.

Newton said (rightly) that all objects have inertia, whether moving or not. Inertia is the resistance of an object to changes in its state of motion or rest. Remember, an object at rest in terms of one inertial coordinate system is moving in terms of another inertial coordinate system.

> An object in motion tends to stay in motion.

Right, and an object at rest tends to stay at rest.

> Newton had equations for all of these things.

Indeed he did, and they were believed to be exactly correct until people began to observe the dynamics of objects with greater precision and at extremely high speeds. It was then discovered that all of Newton's equations are incorrect. For example, Newton said F = d(mv)/dt, and since he believed the inertial mass m of an object was constant, this implied F = ma, but this is not exactly correct for all velocities. Based on observations of the relationship between force and acceleration (in the same direction) for an object of rest mass m moving (in the same direction) at speed v is not F = ma but rather

F = ma/[1 - (v/c)^2]^(3/2) (1)

Likewise all of the other equations of Newtonian physics had to be revised in order to be consistent not only with low speed behavior but with high speed behavior as well. All we need to do in order to correct all of Newton's equations of mechanics is to simply replace m with m0/sqrt[1 - (v/c)^2] where m0 is the rest mass of the object. For example, inserting this into Newton's equation F = d(mv)/dt gives equation (1).

> Of course, he [Newton] did not have the permission of

> modern scientists to do what he did.

Obviously.

> Since scientists have proven that all correctly working
> clocks do not agree with each other, that has to be described.

"Describing" is not the issue. We can describe any given set of events in terms of any system of coordinates we like, regardless of whether those coordinates are related by Galilean, Lorentzian, or any other linear transformation. However, all the laws of physics (electrodynamics, mechanics, etc) take a particularly simple form, homogeneous and isotropic, without the need for fictitious forces, only in terms of inertial coordinate systems. So, to apply the laws of physics without the need for fictitious forces, we must use inertial coordinates, which are related by Lorentz transformations, not Galilean transformations, for the reason explained above. Do you understand this?

[Robert Winn]

Well, so you do not want to discuss mathematics or physics. As I told you before, go blow smoke in someone else's face. I already have the correct equations to answer the question I had.

[Robert Winn]

I understand that scientists of today worship the Lorentz equations. To prove that all I have to do is show that the Galilean transformation equations, not the Lorentz equations, describe reality as we observe it. Slower clocks really do indicate faster velocities, faster clocks really do indicate slower velocities. But not to scientists. They have an equation that says that the times of slower and faster clocks result in the same velocities.
So we all understand that scientists are off on a tangent from which they never will return. If that is what scientists want to do, they should be allowed to do it, but I am not going to encourage them.

[danco...@gmail.com]

On Saturday, July 8, 2017 at 11:59:36 AM Robert Winn wrote:
> The Galilean transformation equations, not the Lorentz

> equations, describe reality as we observe it.

A coordinate transformation is nothing but the relationship between two systems of coordinates. A system of coordinates is nothing but a set of labels (space and time coordinates) that we may assign to events. It makes no sense to say one class of coordinate transformations "describes reality" and another does not. In fact, it doesn't even make sense to say that one coordinate system describes reality and another does not. We can describe reality (or our observations) in terms of any coordinate system we like.

I've explained the special properties of one particular class of coordinate systems, called inertial coordinates, that makes them very convenient, but there is nothing to stop us from using less convenient coordinate systems, or systems that are very close approximations to inertial coordinates.

> Slower clocks really do indicate faster velocities, faster
> clocks really do indicate slower velocities. But not to
> scientists.

I doubt that anyone (even you) really believes that the speed of a car is different, depending on whether our clock is running fast or slow. The speed of the car is whatever it is. Sure, it may have a different numerical value if we express it in terms of miles per hour, meters per second, or furlongs per fortnight, but that doesn't change the speed.

Your claim that time dilation is due to differently constructed clocks is simply infantile (excuse me for saying), and it is totally unrelated to the actual observed phenomenon of relativistic time dilation, which has been carefully explained to you several times. Also, your rejection of Maxwell's equations and the experimentally confirmed length contraction is without any rational basis.

[Gary Harnagel]

I don't want to discuss WRONG mathematics and WRONG physics, which is all
you got.

> As I told you before, go blow smoke in someone else's face.

Why would I want to do what YOU do?

> I already have the correct equations to answer the question I had.

You don't and they don't. So wipe the self-satisfied, dimwitted smile off
your face.

> I understand that scientists of today worship the Lorentz equations.

No, they don't, which proves that YOU understand NOTHING. YOU are the
fool that is off ... and a LOT worse than a tangent.

[Robert Winn]

OK, Harnagel, here are the equations I use for relativity.

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

Go ahead and show the mistake.

[Ursler Zilberschlag]

Robert Winn wrote:

> OK, Harnagel, here are the equations I use for relativity.
> x'=x-vt y'=y z'=z t'=t Go ahead and show the mistake.

t'=t

[Robert Winn]

Sorry, Harnagel, that is not a mistake. It is the only way the equations will work. If you do not have the same rate of time in both frames of reference, you are going to end up with a length contraction.

[Gary Harnagel]

Actually, you DON'T use them, which is the Galilean transform equations.
I'll explain to you why the GT is wrong, though. If the GT were correct,
light from a moving source would travel at c with respect to (wrt) the
source and at c + v wrt a stationary observer. The Lunar Laser Ranging
Experiments measure the travel time for a laser pulse to travel to the
moon and back. The distance between the observatory and the moon varies
on a daily basis because the surface of the earth is moving at several
hundred miles/hour towards or away from the moon, or by the Principle
of Relativity, one sees the moon moving toward or away from the earth at
said several hundred miles/hour:

http://tmurphy.physics.ucsd.edu/apollo/basics.html

If the GT were correct, the computed path of the moon would not be a
reasonable orbit. Rather, it would look like this:

https://drive.google.com/file/d/0B_DPK0NJYoyXZ200UTRJMEpPbnc/view

That's only ONE refutation of the GT. Another is that NASA communicates
with its spacecraft all over the solar system and computes distances
using short pules of electromagnetic energy. A pulse is sent from
stations on earth, received by the spacecraft and a return pulse is sent
back to earth with onboard transmitters. These spacecraft are moving
at up to 50 km/sec wrt earth, many times faster that the earth-moon
system. NASA can check on spacecraft positions with ephemerides. If
the GT were correct, calculated spacecraft positions from communication
signals would be up to 100,000 km off from ephemerides positions, but
would be right on if invariant c were assumed (which NASA does because
they aren't fools).

Other experimental evidence exists, some of it quite old:

http://www.datasync.com/~rsf1/desit-1e.htm

"An Astronomical Proof for the Constancy of the Speed of Light" By W. de
Sitter

http://adsabs.harvard.edu/abs/1977PhRvL..39.1051B

"Is the speed of light independent of the velocity of the source" by
K. Brecher

"Recent observations of regularly pulsating X-ray sources in binary star
systems are analyzed in the framework of the 'emission' theory of light.
Assuming that light emitted by a source moving at velocity v with respect
to an observer has a speed equal to c + kv in the observer's rest frame,
the arrival time of pulses from the binary X-ray sources is found to
imply that k is less than 2 billionths. This appears to be the most direct
and sensitive demonstration that the speed of light is independent of the
velocity of the source."

https://en.wikisource.org/wiki/Effect_of_Reflection_from_a_Moving_Mirror_on_the_Velocity_of_Light

"EFFECT OF REFLECTION FROM A MOVING MIRROR ON THE VELOCITY OF LIGHT"
BY A. A. MICHELSON

So, Windy, each of these experiments refutes the GT, but you don't really
care about that, do you? All you want is to hear the wind whistling
through your tonsils.

[mlwo...@wp.pl]

W dniu niedziela, 9 lipca 2017 14:54:10 UTC+2 użytkownik Gary Harnagel napisał:

https://en.wikisource.org/wiki/Effect_of_Reflection_from_a_Moving_Mirror_on_the_Velocity_of_Light
>
> "EFFECT OF REFLECTION FROM A MOVING MIRROR ON THE VELOCITY OF LIGHT"
> BY A. A. MICHELSON
>
> So, Windy, each of these experiments refutes the GT

In the imagination of fanatic idiots, like you.

[Ursler Zilberschlag]

mlwozniak wrote:

> W dniu niedziela, 9 lipca 2017 14:54:10 UTC+2 użytkownik Gary Harnagel
> napisał:
> https://en.wikisource.org/wiki/Effect_of_Reflection_from_a_Moving_Mirror_on_the_Velocity_of_Light
>> "EFFECT OF REFLECTION FROM A MOVING MIRROR ON THE VELOCITY OF LIGHT"

>> BY A. A. MICHELSONSo, Windy, each of these experiments refutes the GT

>
> In the imagination of fanatic idiots, like you.

Stop fucking posting about Physics. Post something you are good at.
I might be interested.

[mlwo...@wp.pl]

No.