On Friday, October 14, 2022 at 10:57:14 AM UTC-7, Ross A. Finlayson wrote:
> On Friday, October 14, 2022 at 10:40:35 AM UTC-7, maluw...@gmail.com wrote:
> > On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
> > > Den 14.10.2022 04:47, skrev patdolan:
> > > >> Den 12.10.2022 21:06, skrev patdolan:
> > > >>>
> > > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > So you are claiming that Fig. 1 is wrong.
> > > >>>
> > > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > > >>>
> > > >>> in Frame K, d' = d/gamma
> > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > Quote:
> > > "• Let’s have two clocks which are synchronized according
> > > to Einstein’s procedure in each of two inertial frames
> > > of reference.
> > > • Let the clocks be a proper distance d from each other
> > > in their respective frames.
> > > • Let the frames move with the relative speed v.
> > > "
> > >
> > > patdolan reads this as:
> > > "Let the clocks in one of the inertial frames be
> > > a distance d/γ from each other, and let the clocks in
> > > the other inertial frame be a distance d from each other."
> > >
> > > Fig 1 is correct because by definition:
> > > ---------------------------------------
> > > In frame K the distance between the stationary clocks A and B is d.
> > > In frame K' the distance between the stationary clocks A' and B' is d..
> > >
> > > However:
> > > In frame K the distance between the moving clocks A' and B' is d/γ.
> > > In frame K' the distance between the moving clocks A and B is d/γ.
> > > >
> > > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > > A bit confused, patdolan? :-D
> > In the meantime in the real world - forbidden by your
> > bunch of idiots GPS and TAI keep measuring t'=t
> > in forbidden by your bunch of idiots old seconds.
> > To dumb to be confused, paulanderson? :-D
> What's the clock of the satellite in the decaying orbit?
>
> Straight down is a decaying orbit, ....
>
> Here the satellite was launched from Earth so never actually
> left or joined its frame, terrestrial frame.
>
> Of course after that there's orbital frames, ..., that everything
> orbits everything.
>
> Any orbit is a decayed orbit, ....
>
> The free work that keeps orbit or gravity, objects accelerate where
> the source, ..., of the direct force vector between them, that that's
> just some non-zero well in the potential space, and finite. This is where
> "finite" means non-zero and non-infinite. This is that in deep space it's
> for example zero, the potential, the object itself not gravitating.
>
> Then, the gravitating frame is the terrestrial frame, where "the terrestrial
> frame is the center of the orbit, for example we are on it".
>
>
> There's keeping the inverse-square while the satellite "tracks the ephemeris",
> what happens in the racionization of two bodies as the same size, or potential,
> dynamics, or different sizes, statics in the gravitating and dynamics and statics
> in the traversing.
>
> There are different mechanisms of clocks and here basically they are mechanical,
> pendulum or spring, clocks, for example, or, crystal quartz or atomic, ...., clocks.

Ha! I have finally figured out Ross A. Finlayson! He is doing stream of conscious posts, a la James Joyce. Ross assiduously and faithfully types out in real time all of his thought processes as he tackles a response to a post. We who are self-aware know that this can be quite random, as ideas occur to us. But we end up editing down all of our thoughts to a short, cogent [ and artificial? ] reply. Ross does not do this. It is up to reader of Ross, as it is up to the reader of Ulysses, to comb from the stream-of-conscious-text the Wittgensteinian fact picture.

Do I win a prize Ross?

PS--Mitch is another situation, entirely.

On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
> Den 14.10.2022 04:47, skrev patdolan:
> >> Den 12.10.2022 21:06, skrev patdolan:
> >>>
> >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> So you are claiming that Fig. 1 is wrong.
> >>>
> >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> >>>
> >>> in Frame K, d' = d/gamma
> https://paulba.no/pdf/Mutual_time_dilation.pdf
> Quote:
> "• Let’s have two clocks which are synchronized according
> to Einstein’s procedure in each of two inertial frames
> of reference.
> • Let the clocks be a proper distance d from each other
> in their respective frames.
> • Let the frames move with the relative speed v.
> "
>
> patdolan reads this as:
> "Let the clocks in one of the inertial frames be
> a distance d/γ from each other, and let the clocks in
> the other inertial frame be a distance d from each other."
>
> Fig 1 is correct because by definition:
> ---------------------------------------
> In frame K the distance between the stationary clocks A and B is d.
> In frame K' the distance between the stationary clocks A' and B' is d.
>
> However:
> In frame K the distance between the moving clocks A' and B' is d/γ.
> In frame K' the distance between the moving clocks A and B is d/γ.
> >
> > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> A bit confused, patdolan? :-D
>
>
> --
> Paul
>
> https://paulba.no/

"In frame K' the distance between the moving clocks A and B is d/γ." -- Paul B. Andersen, yesterday

"Event E2: In frame K′, A will be at the position −d when B′ shows t′2 = d/v. According to the Lorentz transform, A shows t2 = d/vγ" -- Paul B. Andersen, 2008

You foolish fool, Paul B. Andersen. AGAIN you provide your own grist for the Dolo-relativiey mill.

That the distance between clocks A and B is d/γ when viewed from Frame K' means that there are dγ x-axis length units detectable between clocks A and B when viewed from Frame K'.

We now construct the coordinate relative velocity v' of clock A as viewed in Frame K' according your own calculations

v' = dγ/( d/vγ )

from which we directly arrive at the Dolatonian

v' = dγ/( d/vγ ) = vγ^2. No apostrophe d' needed.

Honestly Paul...Coach Dirk needs to bench you for a few plays. You are playing as if you are concussed.

On Friday, October 14, 2022 at 10:19:44 AM UTC-7, Dono. wrote:
> On Wednesday, October 12, 2022 at 12:06:35 PM UTC-7, patdolan wrote:
> > in Frame K, d' = d/gamma
> >
> > So by your own calculation, t2 on Clock A and E2 is
> >
> > t2 = [ d'/v - ( d'v/c^2 ) ] x gamma = ( d'/v )/gamma
> >
> > remembering that d' = d/gamma
> >
> > t2 = [ d/( v*gamma ) ] / gamma = d/[ v*gamma^2 ]
> >
> > d/t2 = v*gamma^2 at E2
> >
> Pattycakes,
>
> It is most entertaining watching your pathetic attempts at using the Lorentz transforms. You are as imbecile as Ken Shito. Maybe you two should team up?

I am a modern maestro of the LTs. I may be the discoverer of its fatal flaw, Dono. Which is also SR's fatal flaw and third--and hidden--postulate: coordinate and proper relative velocity must be equal, just like Galilean coordinate and proper relative velocity are equal. This postulate does not lead to antinomy in the case of Galilean relativity. But it crashes Einsteinian relativity.

Like any good and conscientious MD, Doctor Hachel wishes to save the patient. Dolan endeavors to euthanize it.

On 2022-10-14 17:40:33 +0000, Maciej Wozniak said:

> On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
>> Den 14.10.2022 04:47, skrev patdolan:> >> Den 12.10.2022 21:06, skrev
>> patdolan:> >>>> >>> Professor Paul, we've tarried long enough waiting
>> for you to stipulate the truth; namely, that your almost perfect
>> graphic contains one error. To wit, that in Frame K' the distance -d is
>> actually -d'. With this correction, we proceed.
>> https://paulba.no/pdf/Mutual_time_dilation.pdf> So you are claiming
>> that Fig. 1 is wrong.
>>>>>
>>>>> It is obvious that the distance | -d' | in Frame K' is equivalent to
>>>>> d/gamma in Frame K. Let's give this fact it's own line for added
>>>>> emphasis> >>>> >>> in Frame K, d' = d/gamma
>> https://paulba.no/pdf/Mutual_time_dilation.pdf> Quote:> "• Let’s have
>> two clocks which are synchronized according> to Einstein’s procedure in
>> each of two inertial frames> of reference.> • Let the clocks be a
>> proper distance d from each other> in their respective frames.> • Let
>> the frames move with the relative speed v.> ">> patdolan reads this
>> as:> "Let the clocks in one of the inertial frames be> a distance d/γ
>> from each other, and let the clocks in> the other inertial frame be a
>> distance d from each other.">> Fig 1 is correct because by definition:>
>> ---------------------------------------> In frame K the distance
>> between the stationary clocks A and B is d.> In frame K' the distance
>> between the stationary clocks A' and B' is d.>> However:> In frame K
>> the distance between the moving clocks A' and B' is d/γ.> In frame K'
>> the distance between the moving clocks A and B is d/γ.
>>>
>>> You might as well conclude that labeling the K' axis x' and the K axis
>>> x is also a stupid idea. We label the d with an apostrophe in K' to
>>> remind ourselves that d' and d will have different coordinate lengths
>>> and proper lengths in K' and K, respectively.
>> A bit confused, patdolan? :-D
> In the meantime in the real world - forbidden by your
> bunch of idiots GPS and TAI keep measuring t'=t
> in forbidden by your bunch of idiots old seconds.
> To dumb to be confused, paulanderson? :-D

514

--
Athel -- French and British, living mainly in England until 1987.

On Friday, October 14, 2022 at 12:30:53 PM UTC-7, patdolan wrote:
> On Friday, October 14, 2022 at 10:19:44 AM UTC-7, Dono. wrote:
> > On Wednesday, October 12, 2022 at 12:06:35 PM UTC-7, patdolan wrote:
> > > in Frame K, d' = d/gamma
> > >
> > > So by your own calculation, t2 on Clock A and E2 is
> > >
> > > t2 = [ d'/v - ( d'v/c^2 ) ] x gamma = ( d'/v )/gamma
> > >
> > > remembering that d' = d/gamma
> > >
> > > t2 = [ d/( v*gamma ) ] / gamma = d/[ v*gamma^2 ]
> > >
> > > d/t2 = v*gamma^2 at E2
> > >
> > Pattycakes,
> >
> > It is most entertaining watching your pathetic attempts at using the Lorentz transforms. You are as imbecile as Ken Shito. Maybe you two should team up?
> I am a modern utter cretin

Yep

On Friday, October 14, 2022 at 12:22:27 PM UTC-7, patdolan wrote:
> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
> > Den 14.10.2022 04:47, skrev patdolan:
> > >> Den 12.10.2022 21:06, skrev patdolan:
> > >>>
> > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> > So you are claiming that Fig. 1 is wrong.
> > >>>
> > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > >>>
> > >>> in Frame K, d' = d/gamma
> > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > Quote:
> > "• Let’s have two clocks which are synchronized according
> > to Einstein’s procedure in each of two inertial frames
> > of reference.
> > • Let the clocks be a proper distance d from each other
> > in their respective frames.
> > • Let the frames move with the relative speed v.
> > "
> >
> > patdolan reads this as:
> > "Let the clocks in one of the inertial frames be
> > a distance d/γ from each other, and let the clocks in
> > the other inertial frame be a distance d from each other."
> >
> > Fig 1 is correct because by definition:
> > ---------------------------------------
> > In frame K the distance between the stationary clocks A and B is d.
> > In frame K' the distance between the stationary clocks A' and B' is d.
> >
> > However:
> > In frame K the distance between the moving clocks A' and B' is d/γ..
> > In frame K' the distance between the moving clocks A and B is d/γ.
> > >
> > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > A bit confused, patdolan? :-D
> >
> >
> > --
> > Paul
> >
> > https://paulba.no/
> "In frame K' the distance between the moving clocks A and B is d/γ." -- Paul B. Andersen, yesterday
>
> "Event E2: In frame K′, A will be at the position −d when B′ shows t′2 = d/v. According to the Lorentz transform, A shows t2 = d/vγ" -- Paul B. Andersen, 2008
>
> You foolish fool, Paul B. Andersen. AGAIN you provide your own grist for the Dolo-relativiey mill.
>
> That the distance between clocks A and B is d/γ when viewed from Frame K' means that there are dγ x-axis length units detectable between clocks A and B when viewed from Frame K'.
>
> We now construct the coordinate relative velocity v' of clock A as viewed in Frame K' according your own calculations
>
> v' = dγ/( d/vγ )
>
> from which we directly arrive at the Dolatonian
>
> v' = dγ/( d/vγ ) = vγ^2. No apostrophe d' needed.
>
> Honestly Paul...Coach Dirk needs to bench you for a few plays. You are playing as if you are concussed.

Ahem...my demonstration needs a little correction\clarification.

At E1 clock A is dγ away from clock B' as viewed from Frame K'. This follows from Paul's statements

In frame K the distance between the moving clocks A' and B' is d/γ.
In frame K' the distance between the moving clocks A and B is d/γ.

So at E1 point -d (not shown) on the x-axis will be opposite point -d/γ on the x'-axis as viewed from Frame K'. It follows from this that at E1 clock B' will be opposite the point -dγ (not shown) on the x-axis when viewed from Frame K'.

We now listen again to Paul as he tells us: "At E2 in Frame K′, clock A will be at the position −d [on the x'-axis] when B′ shows t′2 = d/v. According to the Lorentz transform, A shows t2 = d/vγ"

So as viewed from Frame K', clock A has traveled a coordinate distance of -dγ on the x-axis in a coordinate time interval of t2 = d/vγ. Constructing the coordinate relative velocity v' for clock A as viewed from Frame K' we get

v' = -dγ/( d/vγ ) = -vγ^2

I'm surprised you didn't pounce on my error earlier, Paul. But now everything is right again. Dono will need to think about this all night. Paul Alsing will need to think about it all year.

On Friday, October 14, 2022 at 11:12:41 AM UTC-7, patdolan wrote:
> On Friday, October 14, 2022 at 10:57:14 AM UTC-7, Ross A. Finlayson wrote:
> > On Friday, October 14, 2022 at 10:40:35 AM UTC-7, maluw...@gmail.com wrote:
> > > On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
> > > > Den 14.10.2022 04:47, skrev patdolan:
> > > > >> Den 12.10.2022 21:06, skrev patdolan:
> > > > >>>
> > > > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > So you are claiming that Fig. 1 is wrong.
> > > > >>>
> > > > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > > > >>>
> > > > >>> in Frame K, d' = d/gamma
> > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > Quote:
> > > > "• Let’s have two clocks which are synchronized according
> > > > to Einstein’s procedure in each of two inertial frames
> > > > of reference.
> > > > • Let the clocks be a proper distance d from each other
> > > > in their respective frames.
> > > > • Let the frames move with the relative speed v.
> > > > "
> > > >
> > > > patdolan reads this as:
> > > > "Let the clocks in one of the inertial frames be
> > > > a distance d/γ from each other, and let the clocks in
> > > > the other inertial frame be a distance d from each other."
> > > >
> > > > Fig 1 is correct because by definition:
> > > > ---------------------------------------
> > > > In frame K the distance between the stationary clocks A and B is d.
> > > > In frame K' the distance between the stationary clocks A' and B' is d.
> > > >
> > > > However:
> > > > In frame K the distance between the moving clocks A' and B' is d/γ.
> > > > In frame K' the distance between the moving clocks A and B is d/γ.
> > > > >
> > > > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > > > A bit confused, patdolan? :-D
> > > In the meantime in the real world - forbidden by your
> > > bunch of idiots GPS and TAI keep measuring t'=t
> > > in forbidden by your bunch of idiots old seconds.
> > > To dumb to be confused, paulanderson? :-D
> > What's the clock of the satellite in the decaying orbit?
> >
> > Straight down is a decaying orbit, ....
> >
> > Here the satellite was launched from Earth so never actually
> > left or joined its frame, terrestrial frame.
> >
> > Of course after that there's orbital frames, ..., that everything
> > orbits everything.
> >
> > Any orbit is a decayed orbit, ....
> >
> > The free work that keeps orbit or gravity, objects accelerate where
> > the source, ..., of the direct force vector between them, that that's
> > just some non-zero well in the potential space, and finite. This is where
> > "finite" means non-zero and non-infinite. This is that in deep space it's
> > for example zero, the potential, the object itself not gravitating.
> >
> > Then, the gravitating frame is the terrestrial frame, where "the terrestrial
> > frame is the center of the orbit, for example we are on it".
> >
> >
> > There's keeping the inverse-square while the satellite "tracks the ephemeris",
> > what happens in the racionization of two bodies as the same size, or potential,
> > dynamics, or different sizes, statics in the gravitating and dynamics and statics
> > in the traversing.
> >
> > There are different mechanisms of clocks and here basically they are mechanical,
> > pendulum or spring, clocks, for example, or, crystal quartz or atomic, ...., clocks.
> Ha! I have finally figured out Ross A. Finlayson! He is doing stream of conscious posts, a la James Joyce. Ross assiduously and faithfully types out in real time all of his thought processes as he tackles a response to a post. We who are self-aware know that this can be quite random, as ideas occur to us. But we end up editing down all of our thoughts to a short, cogent [ and artificial? ] reply. Ross does not do this. It is up to reader of Ross, as it is up to the reader of Ulysses, to comb from the stream-of-conscious-text the Wittgensteinian fact picture.
>
> Do I win a prize Ross?
>
> PS--Mitch is another situation, entirely.

Indeed.

The object that reaches rest at the position in orbit
as if it had an orbit from where it rested,
immediately decays the orbit as the fall orbit.

Then, that as a general property, is that by going out of rest,
in its regular motion, the linear impulse results, that subtracts
that half part lost for decaying orbit, the other falling out of orbit.

Here for example for Patty's formula, writing vy^2 or what he has there,
is about their relative motions, and positions, and rest.

On Friday, October 14, 2022 at 9:36:14 PM UTC-7, Ross A. Finlayson wrote:
> On Friday, October 14, 2022 at 11:12:41 AM UTC-7, patdolan wrote:
> > On Friday, October 14, 2022 at 10:57:14 AM UTC-7, Ross A. Finlayson wrote:
> > > On Friday, October 14, 2022 at 10:40:35 AM UTC-7, maluw...@gmail.com wrote:
> > > > On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
> > > > > Den 14.10.2022 04:47, skrev patdolan:
> > > > > >> Den 12.10.2022 21:06, skrev patdolan:
> > > > > >>>
> > > > > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > > So you are claiming that Fig. 1 is wrong.
> > > > > >>>
> > > > > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > > > > >>>
> > > > > >>> in Frame K, d' = d/gamma
> > > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > > Quote:
> > > > > "• Let’s have two clocks which are synchronized according
> > > > > to Einstein’s procedure in each of two inertial frames
> > > > > of reference.
> > > > > • Let the clocks be a proper distance d from each other
> > > > > in their respective frames.
> > > > > • Let the frames move with the relative speed v.
> > > > > "
> > > > >
> > > > > patdolan reads this as:
> > > > > "Let the clocks in one of the inertial frames be
> > > > > a distance d/γ from each other, and let the clocks in
> > > > > the other inertial frame be a distance d from each other."
> > > > >
> > > > > Fig 1 is correct because by definition:
> > > > > ---------------------------------------
> > > > > In frame K the distance between the stationary clocks A and B is d.
> > > > > In frame K' the distance between the stationary clocks A' and B' is d.
> > > > >
> > > > > However:
> > > > > In frame K the distance between the moving clocks A' and B' is d/γ.
> > > > > In frame K' the distance between the moving clocks A and B is d/γ.
> > > > > >
> > > > > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > > > > A bit confused, patdolan? :-D
> > > > In the meantime in the real world - forbidden by your
> > > > bunch of idiots GPS and TAI keep measuring t'=t
> > > > in forbidden by your bunch of idiots old seconds.
> > > > To dumb to be confused, paulanderson? :-D
> > > What's the clock of the satellite in the decaying orbit?
> > >
> > > Straight down is a decaying orbit, ....
> > >
> > > Here the satellite was launched from Earth so never actually
> > > left or joined its frame, terrestrial frame.
> > >
> > > Of course after that there's orbital frames, ..., that everything
> > > orbits everything.
> > >
> > > Any orbit is a decayed orbit, ....
> > >
> > > The free work that keeps orbit or gravity, objects accelerate where
> > > the source, ..., of the direct force vector between them, that that's
> > > just some non-zero well in the potential space, and finite. This is where
> > > "finite" means non-zero and non-infinite. This is that in deep space it's
> > > for example zero, the potential, the object itself not gravitating.
> > >
> > > Then, the gravitating frame is the terrestrial frame, where "the terrestrial
> > > frame is the center of the orbit, for example we are on it".
> > >
> > >
> > > There's keeping the inverse-square while the satellite "tracks the ephemeris",
> > > what happens in the racionization of two bodies as the same size, or potential,
> > > dynamics, or different sizes, statics in the gravitating and dynamics and statics
> > > in the traversing.
> > >
> > > There are different mechanisms of clocks and here basically they are mechanical,
> > > pendulum or spring, clocks, for example, or, crystal quartz or atomic, ..., clocks.
> > Ha! I have finally figured out Ross A. Finlayson! He is doing stream of conscious posts, a la James Joyce. Ross assiduously and faithfully types out in real time all of his thought processes as he tackles a response to a post. We who are self-aware know that this can be quite random, as ideas occur to us. But we end up editing down all of our thoughts to a short, cogent [ and artificial? ] reply. Ross does not do this. It is up to reader of Ross, as it is up to the reader of Ulysses, to comb from the stream-of-conscious-text the Wittgensteinian fact picture.
> >
> > Do I win a prize Ross?
> >
> > PS--Mitch is another situation, entirely.
> Indeed.
>
> The object that reaches rest at the position in orbit
> as if it had an orbit from where it rested,
> immediately decays the orbit as the fall orbit.
>
>
> Then, that as a general property, is that by going out of rest,
> in its regular motion, the linear impulse results, that subtracts
> that half part lost for decaying orbit, the other falling out of orbit.
>
> Here for example for Patty's formula, writing vy^2 or what he has there,
> is about their relative motions, and positions, and rest.

Sure, I write to impress myself later.

No, I have a long, running schtick.

So here we are, all in orbit together, there is that stellar distribution in matter distribution results,
the terrestrial frame is a satellite, solar.

The moon then though or next satellite, the Earth's terrestrial frame, nearest is the Moon, in distribution.
There is Lagrange the Earth and Lagrange the Moon in Lagrange the Sun.

Though, we are all in orbit together, of the angular velocity, of a frame, and how much things weigh in
terms of gravity, which results the gradient a constant, the "constant" that is the gravitational constant,
that it is just dimensionless, angular frame.

A cooled planet like the Earth it's rotation, is mostly a gravitational constant, the flat part of the
Earth, with respect to resting on the Earth bodies in orbit, the acceleration in freefall is a constant
for everything.

So, the acceleration-in-freefall constant, or gravitational constant, here for example is the same,
in both theories classical, also, that at rest and freefall together, both fall out of orbit.

This is where, the angular to linear translation, results an equivalence in terms, in the angular, and linear,
the position and velocity is central down between acceleration and momentum, or torque,
that the solar, outside the Earth, after the Moon, is next, then for after whether escaping the radial disk,
has that for whatever reason it appear most planets around Sol are in the same elliptical plane, or that
Neptune is the only planet outside the elliptical plane, and that solar systems as their rotating stellar
frames, are most often this way, habitable, just like the Earth with one Moon.

That anything is free in motion or "yes you can spin a chair around", free floating, is that buoyancy in terms
works out about the same way, how easy it is to jump off when it's as easy as staying on.

For example there are two train tracks. Train is speeding along, other train speeds up, guy gets
off the train onto the other train. They have constant relative velocity so the "constant relative
velocity" of for example air, is where there is a distance between the cars , and whether jumping across
is free, or there is a drop or the distance is too far that underneath the trains' frames it's a static frame,
here crossing the trains, then the first train continues while the train with the guy stops.

Now the guy has got on and off the train, where the only effect of that it was observable that
the train cars passed, or met or were passing, the train was lighter when he got off but rocking.

In guy's forward motion, turning around is the rotational component, free, it's not necessary
to unwind after getting off the train. But, the wheels went all the way around, in terms of that
the input of the work to the tracks resulted the combination, of free friction and rolling from ideal
wheels and track and also wheel footprint, to get to equating wheels and feet.

Den 15.10.2022 05:47, skrev patdolan:
>> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
>>> https://paulba.no/pdf/Mutual_time_dilation.pdf
>>>
>>> Fig 1 is correct because by definition:
>>> ---------------------------------------
>>> In frame K the distance between the stationary clocks A and B is d.
>>> In frame K' the distance between the stationary clocks A' and B' is d.
>>>
>>> However:
>>> In frame K the distance between the moving clocks A' and B' is d/γ.
>>> In frame K' the distance between the moving clocks A and B is d/γ.
>>>>
>
> At E1 clock A is dγ away from clock B' as viewed from Frame K'.

Event E1 is that clock A and clock A' are adjacent.(colocated)
So your statement above is equivalent to:

"At E1 clock A' is dγ away from clock B' as viewed from Frame K'."

> This follows from Paul's statements

Which were:
In frame K the distance between the stationary clocks A and B is d.
In frame K' the distance between the stationary clocks A' and B' is d.

and
>
> In frame K the distance between the moving clocks A' and B' is d/γ.
> In frame K' the distance between the moving clocks A and B is d/γ.

'nuff said!

--
Paul

https://paulba.no/

On Saturday, October 15, 2022 at 2:45:58 AM UTC-7, Paul B. Andersen wrote:
> Den 15.10.2022 05:47, skrev patdolan:
> >> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
> >>> https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> >>>
> >>> Fig 1 is correct because by definition:
> >>> ---------------------------------------
> >>> In frame K the distance between the stationary clocks A and B is d.
> >>> In frame K' the distance between the stationary clocks A' and B' is d..
> >>>
> >>> However:
> >>> In frame K the distance between the moving clocks A' and B' is d/γ.
> >>> In frame K' the distance between the moving clocks A and B is d/γ.
> >>>>
> >
> > At E1 clock A is dγ away from clock B' as viewed from Frame K'.
> Event E1 is that clock A and clock A' are adjacent.(colocated)
> So your statement above is equivalent to:
> "At E1 clock A' is dγ away from clock B' as viewed from Frame K'."
>
> > This follows from Paul's statements
> Which were:
> In frame K the distance between the stationary clocks A and B is d.
> In frame K' the distance between the stationary clocks A' and B' is d.
> and
> >
> > In frame K the distance between the moving clocks A' and B' is d/γ..
> > In frame K' the distance between the moving clocks A and B is d/γ.
> 'nuff said!
>
> --
> Paul
>
> https://paulba.no/
But I am finding the distance between clocks A and B' at E1 as viewed from Frame K'. Your drawing ignores this distance. That distance is -dγ.

On Saturday, October 1, 2022 at 1:02:56 PM UTC-4, tjrob137 wrote:
> On 9/30/22 10:17 PM, rotchm wrote:
> >>>> Since you talked about closing speeds, can you explain what
> >>>> you mean by such an expression?
> >> ... in this post he says..."Closing speed is the rate of change of
> >> the distance between two objects as they both move.
> >
> > Yes, that's basically it. The closing speed between two entities is
> > the rate of change of the distance (separation) between them, as
> > measured by a given frame (system of coordinates).
> Yes, but only for inertial coordinates. The closing speed for a given
> pair of objects varies, depending on which inertial frame is used to
> calculate/measure it.
> > Its "v1 ± v2" depending of sign conventions.
> Only when the two objects are moving along a single axis in the inertial
> frame used. If that isn't so, the calculation is MUCH more complicated.
>
Since there is no initial observer in a gravitational field then there is no initial observer who is qualified to measure the closing speed between two objects. The only observer who is qualified is the observer who is in a state of absolute rest.....but no such observer available. Therefore closing speed between two objects is not measurable.

On Friday, October 14, 2022 at 8:47:30 PM UTC-7, crank pat dolan wrote:

> v' = -dγ/( d/vγ ) = -vγ^2
>

Pattycakes,

You have no clue what "v" means. So, everything you post is just utter imbecilities.

On Saturday, October 15, 2022 at 8:05:49 AM UTC-7, Dono. wrote:
> On Friday, October 14, 2022 at 8:47:30 PM UTC-7, crank pat dolan wrote:
>
> > v' = -dγ/( d/vγ ) = -vγ^2
> >
> Pattycakes,
>
> You have no clue what "v" means. So, everything you post is just utter imbecilities.

You must shrink the v to less than c = 1, and then there's that
c_light, and c_gravity, or c_g, are not necessarily the same.

That in relativity they're both 1.

So, "outside" relativity, the theory, some have c < c_g.

That's though including relativity, of course, then that where,
symmetry breaking, in terms of apparent v and meeting,
works down to mass energy equivalence in light speed's terms,
back in c_g's terms again.

Then the mc^2 is the linear kinetic KE or about so, of course.
That divided by its mass results about its velocity.

Then the symmetry breaking is only symmetry flex
or lets light out / leaves light behind.

This is while it is space contraction whatever is in any terms
the distance between two positions going to zero.

Thus, it is much simpler to understand, and all clear and laid out.

On Friday, October 14, 2022 at 11:12:41 AM UTC-7, patdolan wrote:
> On Friday, October 14, 2022 at 10:57:14 AM UTC-7, Ross A. Finlayson wrote:
> > On Friday, October 14, 2022 at 10:40:35 AM UTC-7, maluw...@gmail.com wrote:
> > > On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
> > > > Den 14.10.2022 04:47, skrev patdolan:
> > > > >> Den 12.10.2022 21:06, skrev patdolan:
> > > > >>>
> > > > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > So you are claiming that Fig. 1 is wrong.
> > > > >>>
> > > > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > > > >>>
> > > > >>> in Frame K, d' = d/gamma
> > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > Quote:
> > > > "• Let’s have two clocks which are synchronized according
> > > > to Einstein’s procedure in each of two inertial frames
> > > > of reference.
> > > > • Let the clocks be a proper distance d from each other
> > > > in their respective frames.
> > > > • Let the frames move with the relative speed v.
> > > > "
> > > >
> > > > patdolan reads this as:
> > > > "Let the clocks in one of the inertial frames be
> > > > a distance d/γ from each other, and let the clocks in
> > > > the other inertial frame be a distance d from each other."
> > > >
> > > > Fig 1 is correct because by definition:
> > > > ---------------------------------------
> > > > In frame K the distance between the stationary clocks A and B is d.
> > > > In frame K' the distance between the stationary clocks A' and B' is d.
> > > >
> > > > However:
> > > > In frame K the distance between the moving clocks A' and B' is d/γ.
> > > > In frame K' the distance between the moving clocks A and B is d/γ.
> > > > >
> > > > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > > > A bit confused, patdolan? :-D
> > > In the meantime in the real world - forbidden by your
> > > bunch of idiots GPS and TAI keep measuring t'=t
> > > in forbidden by your bunch of idiots old seconds.
> > > To dumb to be confused, paulanderson? :-D
> > What's the clock of the satellite in the decaying orbit?
> >
> > Straight down is a decaying orbit, ....
> >
> > Here the satellite was launched from Earth so never actually
> > left or joined its frame, terrestrial frame.
> >
> > Of course after that there's orbital frames, ..., that everything
> > orbits everything.
> >
> > Any orbit is a decayed orbit, ....
> >
> > The free work that keeps orbit or gravity, objects accelerate where
> > the source, ..., of the direct force vector between them, that that's
> > just some non-zero well in the potential space, and finite. This is where
> > "finite" means non-zero and non-infinite. This is that in deep space it's
> > for example zero, the potential, the object itself not gravitating.
> >
> > Then, the gravitating frame is the terrestrial frame, where "the terrestrial
> > frame is the center of the orbit, for example we are on it".
> >
> >
> > There's keeping the inverse-square while the satellite "tracks the ephemeris",
> > what happens in the racionization of two bodies as the same size, or potential,
> > dynamics, or different sizes, statics in the gravitating and dynamics and statics
> > in the traversing.
> >
> > There are different mechanisms of clocks and here basically they are mechanical,
> > pendulum or spring, clocks, for example, or, crystal quartz or atomic, ...., clocks.
> Ha! I have finally figured out Ross A. Finlayson! He is doing stream of conscious posts, a la James Joyce. Ross assiduously and faithfully types out in real time all of his thought processes as he tackles a response to a post. We who are self-aware know that this can be quite random, as ideas occur to us. But we end up editing down all of our thoughts to a short, cogent [ and artificial? ] reply. Ross does not do this. It is up to reader of Ross, as it is up to the reader of Ulysses, to comb from the stream-of-conscious-text the Wittgensteinian fact picture.
>
> Do I win a prize Ross?
>
> PS--Mitch is another situation, entirely.

No, thanks, I am not a Wittgensteinian, though I have read much the Brown Book,
Wittgenstein is our anti-Plato.

Now, the Tractatus, is something else, let's say that Wittgenstein 's Tractatus,
is most parts, Platonic.

I don't need Wittgenstein for free will there's Platonism including a chance there might be free will.

Clearly though Wittgenstein is free to retreat to his thinking, though it's an exercise.

Basically there's a stack of Kant and Hegel and Heidegger, though not the government
or moralism of Hegel only the Being and Nothing, for where Nothing fits in the Western,
while the Mayadhayana philosophy, explores void, as it does or always has. Then Heidegger
is given a little note for Time, what results that then there is the "Four-Dimensionality" separate
"Space-Time" what forwards to after Quantum Mechanics for example, what all is the usual
"philosophical stack" under the "logical stack" which is a scientific stack, then mathematics,
which is a physics and so on.

Thanks though Patty and about your idea here, what you do is keep refining the conditions
what applies, and find a perfect curve some solution of the component, then add that up,
and result some numerical method to compute it , then illustrate the effect and computation
at the same time as a tabletop device.

Here I've been writing how a model of orbits or dcay, is basically for an elephant,
balanced on a sphere.

It's like "how many elephants can walk on a sphere", "one".

On Saturday, October 15, 2022 at 10:50:19 AM UTC-7, Ross A. Finlayson wrote:
> On Friday, October 14, 2022 at 11:12:41 AM UTC-7, patdolan wrote:
> > On Friday, October 14, 2022 at 10:57:14 AM UTC-7, Ross A. Finlayson wrote:
> > > On Friday, October 14, 2022 at 10:40:35 AM UTC-7, maluw...@gmail.com wrote:
> > > > On Friday, 14 October 2022 at 14:40:19 UTC+2, Paul B. Andersen wrote:
> > > > > Den 14.10.2022 04:47, skrev patdolan:
> > > > > >> Den 12.10.2022 21:06, skrev patdolan:
> > > > > >>>
> > > > > >>> Professor Paul, we've tarried long enough waiting for you to stipulate the truth; namely, that your almost perfect graphic contains one error. To wit, that in Frame K' the distance -d is actually -d'. With this correction, we proceed.
> > > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > > So you are claiming that Fig. 1 is wrong.
> > > > > >>>
> > > > > >>> It is obvious that the distance | -d' | in Frame K' is equivalent to d/gamma in Frame K. Let's give this fact it's own line for added emphasis
> > > > > >>>
> > > > > >>> in Frame K, d' = d/gamma
> > > > > https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > > > Quote:
> > > > > "• Let’s have two clocks which are synchronized according
> > > > > to Einstein’s procedure in each of two inertial frames
> > > > > of reference.
> > > > > • Let the clocks be a proper distance d from each other
> > > > > in their respective frames.
> > > > > • Let the frames move with the relative speed v.
> > > > > "
> > > > >
> > > > > patdolan reads this as:
> > > > > "Let the clocks in one of the inertial frames be
> > > > > a distance d/γ from each other, and let the clocks in
> > > > > the other inertial frame be a distance d from each other."
> > > > >
> > > > > Fig 1 is correct because by definition:
> > > > > ---------------------------------------
> > > > > In frame K the distance between the stationary clocks A and B is d.
> > > > > In frame K' the distance between the stationary clocks A' and B' is d.
> > > > >
> > > > > However:
> > > > > In frame K the distance between the moving clocks A' and B' is d/γ.
> > > > > In frame K' the distance between the moving clocks A and B is d/γ.
> > > > > >
> > > > > > You might as well conclude that labeling the K' axis x' and the K axis x is also a stupid idea. We label the d with an apostrophe in K' to remind ourselves that d' and d will have different coordinate lengths and proper lengths in K' and K, respectively.
> > > > > A bit confused, patdolan? :-D
> > > > In the meantime in the real world - forbidden by your
> > > > bunch of idiots GPS and TAI keep measuring t'=t
> > > > in forbidden by your bunch of idiots old seconds.
> > > > To dumb to be confused, paulanderson? :-D
> > > What's the clock of the satellite in the decaying orbit?
> > >
> > > Straight down is a decaying orbit, ....
> > >
> > > Here the satellite was launched from Earth so never actually
> > > left or joined its frame, terrestrial frame.
> > >
> > > Of course after that there's orbital frames, ..., that everything
> > > orbits everything.
> > >
> > > Any orbit is a decayed orbit, ....
> > >
> > > The free work that keeps orbit or gravity, objects accelerate where
> > > the source, ..., of the direct force vector between them, that that's
> > > just some non-zero well in the potential space, and finite. This is where
> > > "finite" means non-zero and non-infinite. This is that in deep space it's
> > > for example zero, the potential, the object itself not gravitating.
> > >
> > > Then, the gravitating frame is the terrestrial frame, where "the terrestrial
> > > frame is the center of the orbit, for example we are on it".
> > >
> > >
> > > There's keeping the inverse-square while the satellite "tracks the ephemeris",
> > > what happens in the racionization of two bodies as the same size, or potential,
> > > dynamics, or different sizes, statics in the gravitating and dynamics and statics
> > > in the traversing.
> > >
> > > There are different mechanisms of clocks and here basically they are mechanical,
> > > pendulum or spring, clocks, for example, or, crystal quartz or atomic, ..., clocks.
> > Ha! I have finally figured out Ross A. Finlayson! He is doing stream of conscious posts, a la James Joyce. Ross assiduously and faithfully types out in real time all of his thought processes as he tackles a response to a post. We who are self-aware know that this can be quite random, as ideas occur to us. But we end up editing down all of our thoughts to a short, cogent [ and artificial? ] reply. Ross does not do this. It is up to reader of Ross, as it is up to the reader of Ulysses, to comb from the stream-of-conscious-text the Wittgensteinian fact picture.
> >
> > Do I win a prize Ross?
> >
> > PS--Mitch is another situation, entirely.
> No, thanks, I am not a Wittgensteinian, though I have read much the Brown Book,
> Wittgenstein is our anti-Plato.
>
> Now, the Tractatus, is something else, let's say that Wittgenstein 's Tractatus,
> is most parts, Platonic.
>
> I don't need Wittgenstein for free will there's Platonism including a chance there might be free will.
>
> Clearly though Wittgenstein is free to retreat to his thinking, though it's an exercise.
>
> Basically there's a stack of Kant and Hegel and Heidegger, though not the government
> or moralism of Hegel only the Being and Nothing, for where Nothing fits in the Western,
> while the Mayadhayana philosophy, explores void, as it does or always has.. Then Heidegger
> is given a little note for Time, what results that then there is the "Four-Dimensionality" separate
> "Space-Time" what forwards to after Quantum Mechanics for example, what all is the usual
> "philosophical stack" under the "logical stack" which is a scientific stack, then mathematics,
> which is a physics and so on.
>
> Thanks though Patty and about your idea here, what you do is keep refining the conditions
> what applies, and find a perfect curve some solution of the component, then add that up,
> and result some numerical method to compute it , then illustrate the effect and computation
> at the same time as a tabletop device.
>
>
> Here I've been writing how a model of orbits or dcay, is basically for an elephant,
> balanced on a sphere.
>
> It's like "how many elephants can walk on a sphere", "one".

Turtles all the way down to elephants.

One time there was a lecture and a visitor after "where's the world",
said "back of a turtle". "Oh, what's the turtle on?" "Another turtle."
"And so on?" "Of course." "Then, the turtle will never reach the ground."
"It's turtles all the way down."

Here then, only one elephant might walk balanced on a sphere, except
for example insofar as elephants might link, here it's one elephant and
any more slide off, so, then there is a stack of them and it's all the way high.

Then, this model of orbits and falling together instead of
one-way gravity, down, again the point is only to make
it free in abstraction, then what work to validate the usual
results of weight given mass in gravity, then that also under
dynamics and travel result inner centrs, that are potential stores
or reserves, if usually just unconscious , or subconscious rather,
then to what is "rest exchange momentum theory".

Newton's Zeroeth laws, ....

On Saturday, October 15, 2022 at 2:45:58 AM UTC-7, Paul B. Andersen wrote:
> Den 15.10.2022 05:47, skrev patdolan:
> >> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
> >>> https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> >>>
> >>> Fig 1 is correct because by definition:
> >>> ---------------------------------------
> >>> In frame K the distance between the stationary clocks A and B is d.
> >>> In frame K' the distance between the stationary clocks A' and B' is d..
> >>>
> >>> However:
> >>> In frame K the distance between the moving clocks A' and B' is d/γ.
> >>> In frame K' the distance between the moving clocks A and B is d/γ.
> >>>>
> >
> > At E1 clock A is dγ away from clock B' as viewed from Frame K'.
> Event E1 is that clock A and clock A' are adjacent.(colocated)
> So your statement above is equivalent to:
> "At E1 clock A' is dγ away from clock B' as viewed from Frame K'."
>
> > This follows from Paul's statements
> Which were:
> In frame K the distance between the stationary clocks A and B is d.
> In frame K' the distance between the stationary clocks A' and B' is d.
> and
> >
> > In frame K the distance between the moving clocks A' and B' is d/γ..
> > In frame K' the distance between the moving clocks A and B is d/γ.
> 'nuff said!
>
> --
> Paul
>
> https://paulba.no/

Is it the value of the clock, or, the rate of change of the value of the clock?

I.e., without meeting and resting together, in passing, what range
of time passes?

For example 2-way 60 mph traffic is passing each other at 120.

The change of the rate of change will be zero to two objects
moving in the same direction with the same velocity. That's doppler
where otherwise it's non-zero, where the escape velocity of an object
from its own frame is c_g, what defines its frame.

So, the the higher order derivatives of time, and the integral's what
parameterizes the path and time-symmetry invariance has never been
demonstrated, here is about "decelerating to rest" as acceleration as
much as acceleration from rest, it's an open system.

Now if you'll excuse me this is only conversational, but the value of the
clock is with respect to the values of all the other clocks in a field of time
and that's a usual clock hypothesis, and, what can be read from a clock
is a range of time and of a range of time, that time always flows, and time only slows.

Synchronization is a two-step process, comparing clocks later comparing clocks.
It allows though information to guide direction in travel opportunely.

Two Allais pendulums under effect meeting, they can meet and depart,
meaning in their ringing and constructive and destructive interference
canceling out, in a very contrived way, for trapeze artists linking in midair,
why objects meeting share clocks and speeds while objects passing have that
the proportional velocity reflects either clock's slowing with respect to reading
the other clock, and, reading it in a range, what establishes its value in a rate of change.

Ken Seto wrote:

> Since there is no initial observer in a gravitational field then there
> is no initial observer who is qualified to measure the closing speed
> between two objects. The only observer who is qualified is the observer
> who is in a state of absolute rest.....but no such observer available.
> Therefore closing speed between two objects is not measurable.

it's a simple addition, fucking stupid. lol.

piątek, 30 września 2022 o 04:35:14 UTC+2 rotchm napisał(a):
> dono incited me to ask such a question.
>
> In the recent thread entitled "addition of velocities", the op
> inquired about closing speeds.
>
> An aggressive debate ensued.
>
> I (and others) maintained that closing speed (of two things)
> is simply "v1 ± v2" [depending of the sign conventions].
> Or, the rate of change of the separation (distance) between the two things.
> [We were discussing trajectories along the x_axis only btw].
>
> Dono however, maintains that the above are not the meaning of closing speed as used here (relativity and kinematics in general). And he reasons that:
>
> "v1+v2 is the closing speed of two objects that STARTED SIMULTANEOSLY. "
>
> " If the two objects don't start simultaneously, one cannot define closing speed."
>
> I am wondering as to what is the consensus here concerning the concept of
> "closing speed".

The 'closing speed' is the relative speed of two bodies in normal sense - classical.

In the relativity - SR, there is so-callede relativistic speed, which is an illusion,
but accepted by this model as something real in place of the relative speed which is correct.

c+c = 2c.

On Saturday, October 15, 2022 at 2:52:31 PM UTC-4, Everet Baldini wrote:
> Ken Seto wrote:
>
> > Since there is no initial observer in a gravitational field then there
> > is no initial observer who is qualified to measure the closing speed
> > between two objects. The only observer who is qualified is the observer
> > who is in a state of absolute rest.....but no such observer available.
> > Therefore closing speed between two objects is not measurable.
> it's a simple addition, fucking stupid. lol.

Fucking idiot......in order to get v1 and v2 the observer had to be in a state of absolute rest. No such observer available. Gee you are so fucking stupid.

Ken Seto wrote:

>> > Since there is no initial observer in a gravitational field then
>> > there is no initial observer who is qualified to measure the closing
>> > speed between two objects. The only observer who is qualified is the
>> > observer who is in a state of absolute rest.....but no such observer
>> > available. Therefore closing speed between two objects is not
>> > measurable.
>> it's a simple addition, fucking stupid. lol.
>
> Fucking idiot......in order to get v1 and v2 the observer had to be in a
> state of absolute rest. No such observer available. Gee you are so
> fucking stupid.

wrt the two, my japanese friend. Why not?

On 10/15/2022 10:18 AM, Ken Seto wrote:
> On Saturday, October 1, 2022 at 1:02:56 PM UTC-4, tjrob137 wrote:
>> On 9/30/22 10:17 PM, rotchm wrote:

>>> Its "v1 ± v2" depending of sign conventions.
>> Only when the two objects are moving along a single axis in the inertial
>> frame used. If that isn't so, the calculation is MUCH more complicated.
>>
> Since there is no initial observer in a gravitational field then there is no initial observer who is qualified to measure the closing speed between two objects. The only observer who is qualified is the observer who is in a state of absolute rest.....but no such observer available. Therefore closing speed between two objects is not measurable.

Word Salad.

What is this "initial observer" and why is it needed?

The observer only to measure a position and time twice, for each of the
two objects to get the two speeds.

And "absolute rest" is an assertion, so has no value in science.

Den 15.10.2022 14:51, skrev patdolan:
> On Saturday, October 15, 2022 at 2:45:58 AM UTC-7, Paul B. Andersen wrote:
>> Den 15.10.2022 05:47, skrev patdolan:
>>>> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
>>>>> https://paulba.no/pdf/Mutual_time_dilation.pdf
>>>>>

>>>
>>> At E1 clock A is dγ away from clock B' as viewed from Frame K'.

>> Event E1 is that clock A and clock A' are adjacent.(colocated)
>> So your statement above is equivalent to:
>> "At E1 clock A' is dγ away from clock B' as viewed from Frame K'."
>>

>>> This follows from Paul's statements

>> Which were:
>> In frame K the distance between the stationary clocks A and B is d.
>> In frame K' the distance between the stationary clocks A' and B' is d.
>> and
>>>
>>> In frame K the distance between the moving clocks A' and B' is d/γ.
>>> In frame K' the distance between the moving clocks A and B is d/γ.

>> 'nuff said!
>>
>> --
>> Paul
>>
>> https://paulba.no/

> But I am finding the distance between clocks A and B' at E1 as viewed from Frame K'. Your drawing ignores this distance. That distance is -dγ.

First a word about your statement:
"At E1 clock A is dγ away from clock B' as viewed from Frame K'."

An 'event' is a point in time and space, and the clocks
that are present at the event E1 are A and A' (co-located).
The coordinates of the event are:
In K : t = 0, x = 0.
In K': t' = 0, x' = 0.

It doesn't make sense to talk about the position of clock B'
at the event E1, because B' isn't present at E1.

So I have interpreted your statement as:
"At t' = 0 clock A is dγ away from clock B' as viewed from Frame K'."

Which is wrong, because:

https://paulba.no/pdf/Mutual_time_dilation.pdf

Look at fig 1.
It shows the situation at t = t'= 0, when clock A and A' are co-located.
In K' A is at x' = 0, and B' is at x' = -d BY DEFINITION.

So:
At t' = 0 clock A is d away from clock B' as viewed from Frame K'.

----

You seem to have problems with applying the Lorentz transform.
So here is an exercise:

What is the position of clock B in K' at t' = 0?

--
Paul

https://paulba.no/

On Sunday, 16 October 2022 at 12:27:06 UTC+2, Paul B. Andersen wrote:
> Den 15.10.2022 14:51, skrev patdolan:
> > On Saturday, October 15, 2022 at 2:45:58 AM UTC-7, Paul B. Andersen wrote:
> >> Den 15.10.2022 05:47, skrev patdolan:
> >>>> On Friday, October 14, 2022 at 5:40:19 AM UTC-7, Paul B. Andersen wrote:
> >>>>> https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> >>>>>
>
> >>>
> >>> At E1 clock A is dγ away from clock B' as viewed from Frame K'.
>
> >> Event E1 is that clock A and clock A' are adjacent.(colocated)
> >> So your statement above is equivalent to:
> >> "At E1 clock A' is dγ away from clock B' as viewed from Frame K'."
> >>
>
> >>> This follows from Paul's statements
>
> >> Which were:
> >> In frame K the distance between the stationary clocks A and B is d.
> >> In frame K' the distance between the stationary clocks A' and B' is d.
> >> and
> >>>
> >>> In frame K the distance between the moving clocks A' and B' is d/γ.
> >>> In frame K' the distance between the moving clocks A and B is d/γ.
>
> >> 'nuff said!
> >>
> >> --
> >> Paul
> >>
> >> https://paulba.no/
>
> > But I am finding the distance between clocks A and B' at E1 as viewed from Frame K'. Your drawing ignores this distance. That distance is -dγ..
> First a word about your statement:
> "At E1 clock A is dγ away from clock B' as viewed from Frame K'."
> An 'event' is a point in time and space, and the clocks
> that are present at the event E1 are A and A' (co-located).
> The coordinates of the event are:
> In K : t = 0, x = 0.
> In K': t' = 0, x' = 0.
>
> It doesn't make sense to talk about the position of clock B'
> at the event E1, because B' isn't present at E1.
>
> So I have interpreted your statement as:
> "At t' = 0 clock A is dγ away from clock B' as viewed from Frame K'."
>
> Which is wrong, because:
>
> https://paulba.no/pdf/Mutual_time_dilation.pdf

In the meantime in the real world, of course,
forbidden by your bunch of idiots GPS and TAI
keep measuring t'=t in forbidden by your
bunch of idiots old seconds.

On 2022-10-16 10:55:31 +0000, Maciej Wozniak said:

> On Sunday, 16 October 2022 at 12:27:06 UTC+2, Paul B. Andersen wrote:
>> Den 15.10.2022 14:51, skrev patdolan:> > On Saturday, October 15, 2022
>> at 2:45:58 AM UTC-7, Paul B. Andersen wrote:> >> Den 15.10.2022 05:47,
>> skrev patdolan:> >>>> On Friday, October 14, 2022 at 5:40:19 AM UTC-7,
>> Paul B. Andersen wrote:> >>>>>
>> https://paulba.no/pdf/Mutual_time_dilation.pdf " rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf> >>>>>>> >>>
>>>>> At E1 clock A is dγ away from clock B' as viewed from Frame K'.>> >>
>>>>> Event E1 is that clock A and clock A' are adjacent.(colocated)> >> So
>>>>> your statement above is equivalent to:> >> "At E1 clock A' is dγ away
>>>>> from clock B' as viewed from Frame K'."> >>>> >>> This follows from
>>>>> Paul's statements>> >> Which were:> >> In frame K the distance between
>>>>> the stationary clocks A and B is d.> >> In frame K' the distance
>>>>> between the stationary clocks A' and B' is d.> >> and> >>>> >>> In
>>>>> frame K the distance between the moving clocks A' and B' is d/γ.> >>>
>>>>> In frame K' the distance between the moving clocks A and B is d/γ.>> >>
>>>>> 'nuff said!> >>> >> --> >> Paul> >>> >> https://paulba.no/>> > But I am
>>>>> finding the distance between clocks A and B' at E1 as viewed from Frame
>>>>> K'. Your drawing ignores this distance. That distance is -dγ.
>> First a word about your statement:
>> "At E1 clock A is dγ away from clock B' as viewed from Frame K'."
>> An 'event' is a point in time and space, and the clocks> that are
>> present at the event E1 are A and A' (co-located).> The coordinates of
>> the event are:> In K : t = 0, x = 0.> In K': t' = 0, x' = 0.>> It
>> doesn't make sense to talk about the position of clock B'> at the event
>> E1, because B' isn't present at E1.>> So I have interpreted your
>> statement as:> "At t' = 0 clock A is dγ away from clock B' as viewed
>> from Frame K'.">> Which is wrong, because:>>
>> https://paulba.no/pdf/Mutual_time_dilation.pdf
> In the meantime in the real world, of course,
> forbidden by your bunch of idiots GPS and TAI
> keep measuring t'=t in forbidden by yourbunch of idiots old seconds.

515.

--
Athel -- French and British, living mainly in England until 1987.

On 10/9/22 12:42 PM, Paul Alsing wrote:
> On Sunday, October 9, 2022 at 9:03:47 AM UTC-7, tjrob137 wrote:
>> Any (locally) inertial frame can be used to determine the
>> closing speed between two objects, including either one of their
>> rest frames (as long as the object is moving inertially).
>>
>> [Note that the value of the closing speed depends on which
>> (locally) inertial frame is used.]
>>
>> Closing speed is the rate of change of the distance between the two
>> objects, as measured in the specified (locally) inertial frame. If
>> one uses the rest frame of either object, the closing speed is
>> equal to the relative speed of the other object.
>
> I do not follow. 2 vehicles are approaching each other along the same
> axis. Let's say each is moving at .7c relative to my position in my
> chair on my balcony looking down on that axis.

Your confusion is due to thinking speed is "relative to my position",
when in fact speed is ALWAYS relative to some set of coordinates. By
convention, unless otherwise specified it is presumed that the
coordinates are (locally) inertial.

Moreover, the value of the closing speed for a given pair of objects
DEPENDS ON WHICH INERTIAL FRAME IS USED TO MEASURE IT.

> I can measure the rate
> of change of the distance between them and come up with a closing
> speed of 1.4c... but neither one of these guys can make any kind of
> measurement of the other that will ever exceed c...

Yes, but each of them can measure the other's speed relative to their
own inertial frame. As each is at rest in their own frame, the speed of
the other relative to their frame is equal to the closing speed in that
frame (here 0.94 c).

> their
> measurements are limited by the LT...

Their measurements are limited by the physics of how the world behaves,
not by any abstract theory. But the theory models the physical world
very well, and neither of them ever measures a speed or velocity greater
than c.

> isn't that correct? Isn't that
> just basic relativity? Wouldn't this also be true even for much lower
> speeds? Even at, say, 50 mph each, their closing speed would be 100
> mph but technically you could still apply the LT and find that from
> the frame of either car their perceived relative speed would be less
> than 100 mph by some incredibly small amount? Like 99.99999999 mph or
> some such?

Yes. For such small speeds, Galilean relativity holds to good accuracy,
and in Galilean relativity, the closing speed for a given pair of
objects is the same in all inertial frames. It is only for speeds that
are an appreciable fraction of c that closing speed depends
significantly on which frame is used.

Tom Roberts