On Sunday, October 16, 2022 at 9:15:01 AM UTC-7, Tom Roberts wrote:
> 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

How is "the apparent superluminal, velocity" of certain jets not "illusory"?

It's said "these apparently superluminal jets are just illusory",
then how are they apparent?

"Superluminal motion of up to ~9.6c has been observed along
the (inner) jet of this quasar."

-- https://en.wikipedia.org/wiki/Superluminal_motion

It seems if the image is "illusory" the theory's "imaginary".

Volney wrote:

> On 10/15/2022 10:18 AM, Ken Seto wrote:
>> 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?

watch this nazi khazar bitch, blathering about a nonexistent "garden",
built on cheap energy stolen from Russia on *fake_funny_money*, since
before the terrorist traitor agents, gorbacheov and lech walensa.

Watch shitty EU imperialist speaking about invading civilized countries
https://%62%69%74%63%68%75%74%65.com/video/bTvRyJn67FEq/

Little Persian birdy kills mighty s300 in US-occupied Ukraine, lol
https://%62%69%74%63%68%75%74%65.com/video/sdSlD13l1VX6/

Exterminating ~160 Pentagon terrorists in Zaporozhye, lol
https://%62%69%74%63%68%75%74%65.com/video/F2gWEMyDCwdf/

VACCINE KILLS MILLIONS AS khazar PFIZER NOW ADMITS GUILT....
https://www.bitchute.com/video/iWvORr7GSDaX/

On Sunday, October 16, 2022 at 3:27:06 AM UTC-7, 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
>
> 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.

Agreed. By definition IN FRAME K' clock B' is displaced -d x'-axis length units from Event 1. BUT when the x-axis is viewed from Frame K' we count dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.

Your turn.
>
> 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, October 16, 2022 at 3:27:06 AM UTC-7, 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
>
> 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'.

Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1. BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.

Your turn.
>
> ----
>
> 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, October 16, 2022 at 3:27:06 AM UTC-7, 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
>
> 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'.

Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1. BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT ON THE X-AXIS OPPOSITE CLOCK B' and Event 1.

Your turn.
>
> ----
>
> 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/

Den 17.10.2022 19:20, skrev patdolan:
> On Sunday, October 16, 2022 at 3:27:06 AM UTC-7, Paul B. Andersen wrote:
>> 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'.
>
> Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1.

Your use of "Event 1" can only mean that your statement is:
"IN FRAME K' at t' = 0 we count -d x'-axis length units
on the x'-axis between clock B' and x' = 0."

In other less convoluted words:
The position of clock B' in K' at t' = 0 is x' = -d.

Note that this is an event with coordinates (t',x') = (0,-d)

> BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.

This is a very ambiguous statement which doesn't really make sense,
but let's try to decipher it.
You are trying to define an event, but which event?

"THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1."
Must mean that we are talking about the position of clock B'
in frame K.
Your reference to "Event 1" must mean at the time is t = 0.

So the event is:
The position of clock B' in K at t = 0.
You claim that the coordinates of this event in K are (t,x) = (0,-dγ)

This i wrong.

An exercise:
Use the Lorentz transform to find
the correct position of clock B' in K at t = 0.

Show your calculations!

>>
>> ----
>>
>> 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 Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> Den 17.10.2022 19:20, skrev patdolan:
> > On Sunday, October 16, 2022 at 3:27:06 AM UTC-7, Paul B. Andersen wrote:
> >> 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'.
> >
> > Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1.
> Your use of "Event 1" can only mean that your statement is:
> "IN FRAME K' at t' = 0 we count -d x'-axis length units
> on the x'-axis between clock B' and x' = 0."
>
> In other less convoluted words:
> The position of clock B' in K' at t' = 0 is x' = -d.
>
> Note that this is an event with coordinates (t',x') = (0,-d)
>
> > BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.
>
>
> This is a very ambiguous statement which doesn't really make sense,
> but let's try to decipher it.
> You are trying to define an event, but which event?
>
> "THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1."
> Must mean that we are talking about the position of clock B'
> in frame K.
> Your reference to "Event 1" must mean at the time is t = 0.
>
> So the event is:
> The position of clock B' in K at t = 0.
> You claim that the coordinates of this event in K are (t,x) = (0,-dγ)
>
> This i wrong.
>
> An exercise:
> Use the Lorentz transform to find
> the correct position of clock B' in K at t = 0.
>
> Show your calculations!

Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.

1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.

2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.

3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.

4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A has traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).

> >>
> >> ----
> >>
> >> 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 Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> Den 17.10.2022 19:20, skrev patdolan:
> > On Sunday, October 16, 2022 at 3:27:06 AM UTC-7, Paul B. Andersen wrote:
> >> 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'.
> >
> > Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1.
> Your use of "Event 1" can only mean that your statement is:
> "IN FRAME K' at t' = 0 we count -d x'-axis length units
> on the x'-axis between clock B' and x' = 0."
>
> In other less convoluted words:
> The position of clock B' in K' at t' = 0 is x' = -d.
>
> Note that this is an event with coordinates (t',x') = (0,-d)
>
> > BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.
>
>
> This is a very ambiguous statement which doesn't really make sense,
> but let's try to decipher it.
> You are trying to define an event, but which event?
>
> "THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1."
> Must mean that we are talking about the position of clock B'
> in frame K.
> Your reference to "Event 1" must mean at the time is t = 0.
>
> So the event is:
> The position of clock B' in K at t = 0.
> You claim that the coordinates of this event in K are (t,x) = (0,-dγ)
>
> This i wrong.
>
> An exercise:
> Use the Lorentz transform to find
> the correct position of clock B' in K at t = 0.
>
> Show your calculations!

Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.

1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.

2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.

3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units--albeit they are contracted by 1/γ--because the proper length interval AC is -dγ x-axis length units by definition.

4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is aligned with clock B'. And we have just demonstrated that when viewed from Frame K' clock A has traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).
> >>
> >> ----
> >>
> >> 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 Monday, October 17, 2022 at 8:51:14 PM UTC-7, patdolan wrote:
> And we have just demonstrated that Pat Dolan is acretin

Yup. You have absolutely no clue what "v" means.

On Monday, 17 October 2022 at 22:03:28 UTC+2, Paul B. Andersen wrote:
> Den 17.10.2022 19:20, skrev patdolan:
> > On Sunday, October 16, 2022 at 3:27:06 AM UTC-7, Paul B. Andersen wrote:
> >> 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'.
> >
> > Agreed. By definition IN FRAME K' we count -d x'-axis length units on the x'-axis between clock B' and Event 1.
> Your use of "Event 1" can only mean that your statement is:
> "IN FRAME K' at t' = 0 we count -d x'-axis length units
> on the x'-axis between clock B' and x' = 0."
>
> In other less convoluted words:
> The position of clock B' in K' at t' = 0 is x' = -d.
>
> Note that this is an event with coordinates (t',x') = (0,-d)
>
> > BUT when we look at the x-axis from our position in Frame K' we count -dγ x-axis length units on the x-axis between THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1.
>
>
> This is a very ambiguous statement which doesn't really make sense,
> but let's try to decipher it.
> You are trying to define an event, but which event?
>
> "THE POINT OPPOSITE CLOCK B' ON THE X-AXIS and Event 1."
> Must mean that we are talking about the position of clock B'
> in frame K.
> Your reference to "Event 1" must mean at the time is t = 0.
>
> So the event is:
> The position of clock B' in K at t = 0.
> You claim that the coordinates of this event in K are (t,x) = (0,-dγ)
>
> This i wrong.
>
> An exercise:
> Use the Lorentz transform to find
> the correct position of clock B' in K at t = 0.

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 Monday, October 17, 2022 at 10:17:52 PM UTC-7, Dono. wrote:
> On Monday, October 17, 2022 at 8:51:14 PM UTC-7, patdolan wrote:
> > And we have just demonstrated that Pat Dolan is acretin
>
> Yup. You have absolutely no clue what "v" means.
Wrong Dono. I am the only one to ever ask "If there is a coordinate and proper length, and a coordinate and proper time, then is there also a coordinate and proper relative velocity?" I am the first to have awaken to woke relativity! My epitaph:

v' = -v x gamma^2

On Monday, October 17, 2022 at 10:55:59 PM UTC-7, patdolan wrote:
> On Monday, October 17, 2022 at 10:17:52 PM UTC-7, Dono. wrote:
> > On Monday, October 17, 2022 at 8:51:14 PM UTC-7, patdolan wrote:
> > > And we have just demonstrated that Pat Dolan is acretin
> >
> > Yup. You have absolutely no clue what "v" means.
> Wrong Dono. I am the only one to ever ask "If there is a coordinate and proper length, and a coordinate and proper time, then is there also a coordinate and proper relative velocity?" I am the first to have awaken to woke relativity! My epitaph:
>
> v' = -v x gamma^2

Hey, that's 2.5 ordinates, ....

As a "coordinate system" it's a contraction of sorts, in contraction and relaxation,
that in it's planar expression, 2 ordinates then this "closed term the peripheral shift
of the image under parallax", is that, after parallax, is, peripheral parallax.

"What is down", left and right, .... The usual stereoptic vision is very coordinative.

Reading the clock is of its image, ....

So, you pick it up and run it out in the coordinate system then put it back,
then for what must have transpired how it's the least flex that's not break.
(Or most as it were.)

This is for flex and break that acceleration is a flex, while, flux to zero is a break.
(Invariance theory.)

There is not symmetry-breaking for as that usual terms are "asymmetry-breaks",
not at all symmetry-breaks but impulse in terms. This is that impulse is only
a unit-al function so it's possible to say that without being wrong. [While of course
it's mathematically general.] Attaching a system of constants and ratio to it
is about usual in terms.

Let's all come down from acceleration instead of flying off to it,
and get together. That's how things started. Of course that's
just the integral instead of the derivative, while of course the
derivative has great gradient. It's just about that simple.
(Or at least it is.)

So, when you say "I've written my coordinates", no, Patty,
write your system of coordinates, then "... here is the system
of coordinates".

Then it would look more rulial.

I can't but imagine in all prior art there's somebody who's written just that.

Den 18.10.2022 05:35, skrev patdolan:
> On Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
>>
>> An exercise:

https://paulba.no/pdf/Mutual_time_dilation.pdf
Look at fig. 1

>> Use the Lorentz transform to find

THIS:
>> the correct position of clock B' in K at t = 0.
===================================================

>>
>> Show your calculations!

>
> Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.
>
> 1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.
>
> 2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.
>
> 3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.
>
> 4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A h".as traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).

You haven't answered the question,
and you haven't used the Lorentz transform,
which is:

γ = 1/√(1−v²/c²)

t' = γ(t - (v/c²)⋅x)
x' = γ(x - v⋅t)

t = γ(t' + (v/c²)⋅x')
x = γ(x' + v⋅t')

You pride yourself of your "Lorentzian virtuosity".
SO PLEASE DEMONSTRATE IT by using the equations above to find
the position of clock B' in K at t = 0.
=======================================

I predict you can't do it.
Prove me wrong!

>
>
>>>>
>>>> ----
>>>>
>>>> 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/

Le 18/10/2022 à 10:27, "Paul B. Andersen" a écrit :

> You haven't answered the question,
> and you haven't used the Lorentz transform,
> which is:
>
> γ = 1/√(1−v²/c²)
>
> t' = γ(t - (v/c²)⋅x)
> x' = γ(x - v⋅t)
>
> t = γ(t' + (v/c²)⋅x')
> x = γ(x' + v⋅t')
>

Paul, my dear, my honey, did you warn him that in the Poincaré-Lorentz
transformations, the notion of time is negative?

I'm not sure even the greatest pundits of relativity have understood this.

Do you understand Paul?

I'm telling you this so that you remain human, and that you don't take
pride in having understood something, which ultimately isn't worth so
much.

<http://news2.nemoweb.net/jntp?r57q26p4R_facYshV7tS_WQg32s@jntp/Data.Media:1>

R.H.

On Monday, October 17, 2022 at 10:55:59 PM UTC-7, patdolan wrote:
> On Monday, October 17, 2022 at 10:17:52 PM UTC-7, Dono. wrote:
> > On Monday, October 17, 2022 at 8:51:14 PM UTC-7, patdolan wrote:
> > > And we have just demonstrated that Pat Dolan is acretin
> >
> > Yup. You have absolutely no clue what "v" means.
> I am the only one to ever ask "If there is a coordinate and proper length, and a coordinate and proper time, then is there also a coordinate and proper relative velocity?" I am the first to have awaken to woke relativity! My epitaph:
>
> v' = -v x gamma^2

Your epitaph: "Here lies an imbecile"

On Tuesday, October 18, 2022 at 1:28:49 AM UTC-7, Paul B. Andersen wrote:
> Den 18.10.2022 05:35, skrev patdolan:
> > On Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> >>
> >> An exercise:
>
> https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> Look at fig. 1
> >> Use the Lorentz transform to find
> THIS:
> >> the correct position of clock B' in K at t = 0.
> ==================================================> >>
> >> Show your calculations!
>
> >
> > Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.
> >
> > 1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.
> >
> > 2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.
> >
> > 3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.
> >
> > 4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A h".as traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).
>
> You haven't answered the question,
> and you haven't used the Lorentz transform,
> which is:
>
> γ = 1/√(1−v²/c²)
>
> t' = γ(t - (v/c²)⋅x)
> x' = γ(x - v⋅t)
>
> t = γ(t' + (v/c²)⋅x')
> x = γ(x' + v⋅t')
>
> You pride yourself of your "Lorentzian virtuosity".
> SO PLEASE DEMONSTRATE IT by using the equations above to find
> the position of clock B' in K at t = 0.
> =======================================
>
> I predict you can't do it.
> Prove me wrong!
> >
> >
> >>>>
> >>>> ----
> >>>>
> >>>> 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/

-d = x' = γ(x - v⋅t)
so at t = t' = 0
-d/γ = x
therefore clock B' is aligned with the position -d/γ on the x-axis in K

t' = γ(t - (v/c²)⋅x)
0 = γ(t - (v/c²)⋅-d/γ)
0 = γt + dv/c²
t = dv/γc²
so when viewed from K, clock B' reads dv/γc² seconds while clock A and clock A' read 0 seconds.

Next I will perform the above calculations using the coordinate relative velocity v' which are just as valid Lorentz Transformations as those using the proper relative velocity v, which was used in the above two calculations.

On Tuesday, October 18, 2022 at 7:52:21 PM UTC-7, patdolan wrote:
> On Tuesday, October 18, 2022 at 1:28:49 AM UTC-7, Paul B. Andersen wrote:
> > Den 18.10.2022 05:35, skrev patdolan:
> > > On Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> > >>
> > >> An exercise:
> >
> > https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> > Look at fig. 1
> > >> Use the Lorentz transform to find
> > THIS:
> > >> the correct position of clock B' in K at t = 0.
> > ===================================================
> > >>
> > >> Show your calculations!
> >
> > >
> > > Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.
> > >
> > > 1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.
> > >
> > > 2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.
> > >
> > > 3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.
> > >
> > > 4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A h".as traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).
> >
> > You haven't answered the question,
> > and you haven't used the Lorentz transform,
> > which is:
> >
> > γ = 1/√(1−v²/c²)
> >
> > t' = γ(t - (v/c²)⋅x)
> > x' = γ(x - v⋅t)
> >
> > t = γ(t' + (v/c²)⋅x')
> > x = γ(x' + v⋅t')
> >
> > You pride yourself of your "Lorentzian virtuosity".
> > SO PLEASE DEMONSTRATE IT by using the equations above to find
> > the position of clock B' in K at t = 0.
> > =======================================
> >
> > I predict you can't do it.
> > Prove me wrong!
> > >
> > >
> > >>>>
> > >>>> ----
> > >>>>
> > >>>> 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/
> -d = x' = γ(x - v⋅t)
> so at t = t' = 0
> -d/γ = x
> therefore clock B' is aligned with the position -d/γ on the x-axis in K
> t' = γ(t - (v/c²)⋅x)
> 0 = γ(t - (v/c²)⋅-d/γ)
> 0 = γt + dv/c²
> t = dv/γc²
> so when viewed from K, clock B' reads dv/γc² seconds while clock A and clock A' read 0 seconds.
>
> Next I will perform the above calculations using the coordinate relative velocity v' which are just as valid Lorentz Transformations as those using the proper relative velocity v, which was used in the above two calculations.
t = -dv/γc²
so when viewed from K, clock B' reads dv/γc² seconds behind clock A and clock A', which both read 0 seconds.

On Saturday, October 15, 2022 at 9:54:47 PM UTC-4, Volney wrote:
> 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? d v2

to get v1 and v2....?????
>
> The observer only to measure a position and time twice, for each of the
> two objects to get the two speeds.

You assumed that the observer is not moving
>
> And "absolute rest" is an assertion, so has no value in science.d V2

How else you ae going to get v1 and v2...?
And how can you claim that the closing speed is (v1+v2) ???

On Tuesday, October 18, 2022 at 7:52:21 PM UTC-7, cretin pat dolan wrote:

> 0 = γt + dv/c²
> t = dv/γc²

hahahahaha

On Tuesday, October 18, 2022 at 8:12:43 PM UTC-7, patdolan wrote:
> On Tuesday, October 18, 2022 at 7:52:21 PM UTC-7, patdolan wrote:
> > On Tuesday, October 18, 2022 at 1:28:49 AM UTC-7, Paul B. Andersen wrote:
> > > Den 18.10.2022 05:35, skrev patdolan:
> > > > On Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> > > >>
> > > >> An exercise:
> > >
> > > https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > Look at fig. 1
> > > >> Use the Lorentz transform to find
> > > THIS:
> > > >> the correct position of clock B' in K at t = 0.
> > > ===================================================
> > > >>
> > > >> Show your calculations!
> > >
> > > >
> > > > Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.
> > > >
> > > > 1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.
> > > >
> > > > 2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.
> > > >
> > > > 3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.
> > > >
> > > > 4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A h".as traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).
> > >
> > > You haven't answered the question,
> > > and you haven't used the Lorentz transform,
> > > which is:
> > >
> > > γ = 1/√(1−v²/c²)
> > >
> > > t' = γ(t - (v/c²)⋅x)
> > > x' = γ(x - v⋅t)
> > >
> > > t = γ(t' + (v/c²)⋅x')
> > > x = γ(x' + v⋅t')
> > >
> > > You pride yourself of your "Lorentzian virtuosity".
> > > SO PLEASE DEMONSTRATE IT by using the equations above to find
> > > the position of clock B' in K at t = 0.
> > > =======================================
> > >
> > > I predict you can't do it.
> > > Prove me wrong!
> > > >
> > > >
> > > >>>>
> > > >>>> ----
> > > >>>>
> > > >>>> 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/
> > -d = x' = γ(x - v⋅t)
> > so at t = t' = 0
> > -d/γ = x
> > therefore clock B' is aligned with the position -d/γ on the x-axis in K
> > t' = γ(t - (v/c²)⋅x)
> > 0 = γ(t - (v/c²)⋅-d/γ)
> > 0 = γt + dv/c²
> > t = dv/γc²
> > so when viewed from K, clock B' reads dv/γc² seconds while clock A and clock A' read 0 seconds.
> >
> > Next I will perform the above calculations using the coordinate relative velocity v' which are just as valid Lorentz Transformations as those using the proper relative velocity v, which was used in the above two calculations.
> t = -dv/γc²
> so when viewed from K, clock B' reads dv/γc² seconds behind clock A and clock A', which both read 0 seconds.

Let's move on now, Paul, to calculating the coordinate relative velocity v' for clock B' as viewed from K.

1) We know from the previous calculations that clock B' is located -d/γ on the x-axis in K.

2) At velocity v it will take clock B' a total of -d/γv frame K seconds to reach A, when viewed from K.

3) But clock B' is time dilated by 1/γ when viewed from frame K. So in frame K clock B' will show an elapsed time of -d/(γ^2) when traveling from -d/γ on the x-axis to clock A.

4) Also recall that in frame K we will still count -d x'-axis length units between clock B' and clock A. They are contracted, but there are still -d of them between clock B' and clock A.

5) We now form the coordinate relative velocity v' by taking the ratio of coordinate distance to coordinate elapsed time

v' = -d/ -d/(γ^2) = γ^2

Again we find--using your own terms and conditions, Paul--that in frame K the coordinate relative velocity of clock B' is v' = γ^2

Prefect, right down to the sign convention. The ring of truth!

On Tuesday, October 18, 2022 at 8:12:43 PM UTC-7, patdolan wrote:
> On Tuesday, October 18, 2022 at 7:52:21 PM UTC-7, patdolan wrote:
> > On Tuesday, October 18, 2022 at 1:28:49 AM UTC-7, Paul B. Andersen wrote:
> > > Den 18.10.2022 05:35, skrev patdolan:
> > > > On Monday, October 17, 2022 at 1:03:28 PM UTC-7, Paul B. Andersen wrote:
> > > >>
> > > >> An exercise:
> > >
> > > https://paulba.no/pdf/Mutual_time_dilation.pdf" rel="nofollow" target="_blank">https://paulba.no/pdf/Mutual_time_dilation.pdf
> > > Look at fig. 1
> > > >> Use the Lorentz transform to find
> > > THIS:
> > > >> the correct position of clock B' in K at t = 0.
> > > ===================================================
> > > >>
> > > >> Show your calculations!
> > >
> > > >
> > > > Gladly. I am always looking for opportunities to show off my Lorentzian virtuosity.
> > > >
> > > > 1) Pick a point C on the x-axis such that it is displaced the proper distance -dγ x-axis length units from clock A. Call this distance the length interval AC.
> > > >
> > > > 2) When viewed in Frame K' the x-axis length interval AC will be contracted by 1/γ. Therefore in Frame K' the length interval AC will be -dγ/γ or -d x'-axis length units long. So when viewed from Frame K' point C on the x-axis will exactly align with clock B' on the x'-axis at t1 = t'1 = 0.
> > > >
> > > > 3) Even though AC measures only -d x'-axis length units in Frame K', from Frame K' we still count -dγ x-axis length units because the proper length interval AC is -dγ x-axis length units by definition.
> > > >
> > > > 4) Now we are off to the races. By your own calculation, in Frame K' clock A reads d/vγ when it is opposite clock B'. And we have just demonstrated that when viewed from Frame K' clock A h".as traveled -dγ x-axis length units when it is aligned with clock B'. Soooo the coordinate relative velocity v' of clock A when watched from Frame K' is v' = -dγ/( d/vγ ) or v' = -vγ^2. Perfect right down to the minus sign ( that's for Starmaker ).
> > >
> > > You haven't answered the question,
> > > and you haven't used the Lorentz transform,
> > > which is:
> > >
> > > γ = 1/√(1−v²/c²)
> > >
> > > t' = γ(t - (v/c²)⋅x)
> > > x' = γ(x - v⋅t)
> > >
> > > t = γ(t' + (v/c²)⋅x')
> > > x = γ(x' + v⋅t')
> > >
> > > You pride yourself of your "Lorentzian virtuosity".
> > > SO PLEASE DEMONSTRATE IT by using the equations above to find
> > > the position of clock B' in K at t = 0.
> > > =======================================
> > >
> > > I predict you can't do it.
> > > Prove me wrong!
> > > >
> > > >
> > > >>>>
> > > >>>> ----
> > > >>>>
> > > >>>> 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/
> > -d = x' = γ(x - v⋅t)
> > so at t = t' = 0
> > -d/γ = x
> > therefore clock B' is aligned with the position -d/γ on the x-axis in K
> > t' = γ(t - (v/c²)⋅x)
> > 0 = γ(t - (v/c²)⋅-d/γ)
> > 0 = γt + dv/c²
> > t = dv/γc²
> > so when viewed from K, clock B' reads dv/γc² seconds while clock A and clock A' read 0 seconds.
> >
> > Next I will perform the above calculations using the coordinate relative velocity v' which are just as valid Lorentz Transformations as those using the proper relative velocity v, which was used in the above two calculations.
> t = -dv/γc²
> so when viewed from K, clock B' reads dv/γc² seconds behind clock A and clock A', which both read 0 seconds.

Let's move on now, Paul, to calculating the coordinate relative velocity v' for clock B' as viewed from K.

1) We know from the previous calculations that clock B' is located -d/γ on the x-axis in K.

2) At velocity v it will take clock B' a total of -d/γv frame K seconds to reach A, when viewed from K.

3) But clock B' is time dilated by 1/γ when viewed from frame K. So in frame K clock B' will show an elapsed time of -d/v(γ^2) when traveling from -d/γ on the x-axis to clock A.

4) Also recall that in frame K we will still count -d x'-axis length units between clock B' and clock A. They are contracted, but there are still -d of them between clock B' and clock A.

5) We now form the coordinate relative velocity v' by taking the ratio of coordinate distance to coordinate elapsed time

v' = -d/ -d/v(γ^2) = vγ^2

Again we find--using your own terms and conditions, Paul--that in frame K the coordinate relative velocity of clock B' is v' = vγ^2

Prefect, right down to the sign convention. The ring of truth!

On Wednesday, October 19, 2022 at 11:05:11 AM UTC-7, crank pat dolan wrote:
> > > 0 = γt + dv/c²
> > > t = dv/γc²

Pattycakes,

Like your crank buddy Richard Hertz, you are at your most entertaining when you attempt to do calculations. Keep it up, dumbestfuck!

On Wednesday, October 19, 2022 at 11:42:45 AM UTC-7, Dono. wrote:
> On Wednesday, October 19, 2022 at 11:05:11 AM UTC-7, crank pat dolan wrote:
>
> > > > 0 = γt + dv/c²
> > > > t = dv/γc²
> Pattycakes,
>
> Like your crank buddy Richard Hertz, you are at your most entertaining when you attempt to do calculations. Keep it up, dumbestfuck!

Dono, you underhanded fiend. I corrected the dropped minus sign in the very next post. If my calculations are good enough for the tacitly accepting and dejected Paul B. Andersen, then who are you to critique them.

Den 19.10.2022 05:12, skrev patdolan:
> On Tuesday, October 18, 2022 at 7:52:21 PM UTC-7, patdolan wrote:
>> On Tuesday, October 18, 2022 at 1:28:49 AM UTC-7, Paul B. Andersen wrote:
>>>
>>> https://paulba.no/pdf/Mutual_time_dilation.pdf
>>>
>>> You haven't answered the question,
>>> and you haven't used the Lorentz transform,
>>> which is:
>>>
>>> γ = 1/√(1−v²/c²)
>>>
>>> t' = γ(t - (v/c²)⋅x)
>>> x' = γ(x - v⋅t)
>>>
>>> t = γ(t' + (v/c²)⋅x')
>>> x = γ(x' + v⋅t')
>>>
>>> You pride yourself of your "Lorentzian virtuosity".
>>> SO PLEASE DEMONSTRATE IT by using the equations above to find
>>> the position of clock B' in K at t = 0.
>>> =======================================
>>>
>>> I predict you can't do it.
>>> Prove me wrong!

Note that the position of clock B' in K at t = 0 is an EVENT.
In the following I will call this 'the event'.

Note that the temporal coordinate of 'the event' is t = 0.

>> -d = x' = γ(x - v⋅t)
>> so at t = t' = 0
>> -d/γ = x

You have here correctly used the coordinates of 'the event'
x' = -d and t = 0 (temporal coordinate in K', spatial coordinate in K)

and have calculated the coordinates of 'the event' in K to be:
t = 0, x = -d/γ. (t = 0 is given in the definition of 'the event')

This is correct. So you could do it!

BUT:
You claim that the coordinates of 'the event' in K' are:
t'= 0 and x' = -d

This is wrong, because:
t' = γ(t - (v/c²)⋅x) = γ(0 + (v/c²)⋅d/γ) = (v/c²)⋅d

So the coordinates of 'the event' in K' are:
t' = (v/c²)⋅d x' = -d

>> therefore clock B' is aligned with the position -d/γ on the x-axis in K
>> t' = γ(t - (v/c²)⋅x)
>> 0 = γ(t - (v/c²)⋅-d/γ)
>> 0 = γt + dv/c²
>> t = dv/γc²
>> so when viewed from K, clock B' reads dv/γc² seconds while clock A and clock A' read 0 seconds.

Good grief, you have no idea what you are doing!
We are talking about the coordinates of ONE event
in K and K'.

The coordinates of 'the event' in K are t = 0, x = -d/γ
(as YOU correctly found).

You _guess_ that the temporal coordinate of 'the event'
in K' is t' = 0, and transform the temporal coordinate
back to frame K and find: t = (v/c²)⋅d

So you are claiming that the temporal coordinate of the event
"the position of clock B' in K at t = 0" in K is t = (v/c²)⋅d
^^^^^^
How is it possible to do such a giant idiocy without noticing
that it is just that?

>>
>> Next I will perform the above calculations using the coordinate relative velocity v' which are just as valid Lorentz Transformations as those using the proper relative velocity v, which was used in the above two calculations.
> t = -dv/γc²
> so when viewed from K, clock B' reads dv/γc² seconds behind clock A and clock A', which both read 0 seconds.

Mindless babble!

Let's calculate the velocity v of K' relative to K
and the 'coordinate relative velocity' v' of K relative to K'.

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

The coordinates of event E₁ (clock A and clock A' are adjacent) are:
In K: t₁ = 0, x₁ = 0
In K': t₁'= 0, x₁'= 0

The coordinates of event E₂ (clock A and clock B' are adjacent) are:
In K : t₂ = d/vγ x₂ = 0
In K': t₂' = C x₂' = -d

The coordinates of event E₃ (clock B and clock A' are adjacent) are:
In K : t₃ = d/v x₃ = d
In K': t₃'= d/vγ x₃'= 0

Clock A' at x' = 0 (origin of K') moves from x = 0 to x = d in K
at the velocity v = (x₃-x₁)/(t₃-t₁) = d/(d/v) = v

Clock A at x = 0 (origin of K) moves from x' = 0 to x' = -d in K'
at the velocity v' = (x₂'-x₁')/(t₂'-t₁') = -d/(d/v) = -v

|v| = |-v|

The relative speed of K' in K is equal to the relative speed of K in K'.

WHICH IS AN OBVIOUS TRIVIALITY!

--
Paul

https://paulba.no/

Den 19.10.2022 20:05, skrev patdolan:
>
> Let's move on now, Paul, to calculating the coordinate relative velocity v' for clock B' as viewed from K.

Done!

--
Paul

https://paulba.no/