On Wednesday, 19 October 2022 at 21:02:07 UTC+2, Paul B. Andersen wrote:
> 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!

And 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 Wednesday, October 19, 2022 at 3:30:31 PM UTC-7, Dono. wrote:
> On Wednesday, October 19, 2022 at 11:59:00 AM UTC-7, crank pat dolan dug himself deeper:
> > 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.
> Pattycakes,
>
> It is about the long list of imbecilities that led you to obtaining negative time.

Hey Dono, tell that to NASA: T-Minus Nine....T-Minus...T-Minus Eight....T-Minus Seven...

On 10/19/2022 9:51 AM, Ken Seto wrote:
> 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....?????

All you need is this observer to be able to measure x and t twice for
each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
that? How about (-6)/(-2) ? Can you do that yet?
>>
>> 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

It is implicit that a frame is chosen in which the observer is
stationary to get v1 and v2 relative to that frame. Your point is?
>>
>> And "absolute rest" is an assertion, so has no value in science.d V2
>
> How else you ae going to get v1 and v2...?

You choose a frame! Then you get v1 and v2 relative to that frame.
You can even choose a different frame, and get a different pair v1 and v2!

> And how can you claim that the closing speed is (v1+v2) ???

That's the definition of closing speed, Stupid Ken!

On Wednesday, October 19, 2022 at 12:02:07 PM UTC-7, Paul B. Andersen wrote:
> 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).

In K' clock B' and clock A' are synchro-Einsteined to zero by definition. Also, clock A is t=0 when clock A' t'=0 by definition. This is E1.

We will go round and round on this because of the built-in ambiguity of relativity. It is predictable that you will agree with some of my conclusions (as you already have) and vehemently disagree with others (as you already have) just as you do with the conclusions of Dirk, Dono, Python, Prokary, Sylvia, Tom Roberts, Legion, Jan, etc. Do you fools ever ask yourselves why you never disagree when it comes to Newtonian or Galilean calculations. It is because Einsteinianism is inherently marred by an illegitimate assumption at its very heart. After all the time I have spent tutoring you, Paul, you still claim that v' == v is the merest of trivialities. It was once claimed that heavy things fell faster than light things, and that absolute space and time were the also the merest of trivialities. I have demonstrated to this forum time and time again that v' != v, and frankly, my patience is beginning to wear thin. I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum. But I am willing to give this forum one final opportunity to raise itself out of the benighted state it finds itself in.

The good Doctor Hachel continues to pump chemo into the patient. Doctor, it is time to switch to embalming fluid.
>
> 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" rel="nofollow" target="_blank">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/

On 2022-10-20 08:03:55 +0000, patdolan said:
>
> [ … ]

> I have demonstrated to this forum time and time again that v' != v,
> and frankly, my patience is beginning to wear thin. I am considering
> never posting again...taking my genius and gratuitous instruction to a
> more accepting and appreciative forum.

It will be difficult, but somehow I think the group will manage to
survive such a disaster.

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

Den 20.10.2022 10:03, skrev patdolan:
> On Wednesday, October 19, 2022 at 12:02:07 PM UTC-7, Paul B. Andersen wrote:
>> 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?

I note that you have no comment to the obvious blunders you have made.
Unless you are a complete moron, there is no way you can fail to
understand that your calculations above are horribly wrong.

But you go on babbling, as if your errors are only differences
in opinion.

>
> We will go round and round on this because of the built-in ambiguity of relativity.

There is no ambiguity in SR.

> It is predictable that you will agree with some of my conclusions (as you already have) and vehemently disagree with others (as you already have) just as you do with the conclusions of Dirk, Dono, Python, Prokary, Sylvia, Tom Roberts, Legion, Jan, etc.

You were challenge to show that you could use the Lorentz transform.
You failed. What you call "your conclusions" are ERRORS.

What the Lorentz transform predicts is not a matter of opinion,
it is a matter of fact. None of the knowledgeable people you mention
will ever disagree about what the Lorentz transform predicts.

You, OTOH, have demonstrated that you can't use the Lorentz transform
without doing stupid errors.

You are obviously too mathematical illiterate to prove anything
about anything.

> Do you fools ever ask yourselves why you never disagree when it comes to Newtonian or Galilean calculations. It is because Einsteinianism is inherently marred by an illegitimate assumption at its very heart. After all the time I have spent tutoring you, Paul, you still claim that v' == v is the merest of trivialities. It was once claimed that heavy things fell faster than light things, and that absolute space and time were the also the merest of trivialities. I have demonstrated to this forum time and time again that v' != v, and frankly, my patience is beginning to wear thin. I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum. But I am willing to give this forum one final opportunity to raise itself out of the benighted state it finds itself in.
This is beyond what a sane mind can produce.

--
Paul

https://paulba.no/

On Thursday, October 20, 2022 at 1:03:57 AM UTC-7, patdolan wrote:
> I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum.

No, don't do that , Pattycakes. Your idiocies are fodder for daily entertainment, keep them coming.

Ken Seto wrote:

>> 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

in respect to the observed, fucking stoopid. *I_fuck_your_ass!*. Your
people died for you in Japonia. Two atomic bombs. That's NOT nothing.

On Thursday, October 20, 2022 at 6:02:01 AM UTC-7, Paul B. Andersen wrote:
> Den 20.10.2022 10:03, skrev patdolan:
> > On Wednesday, October 19, 2022 at 12:02:07 PM UTC-7, Paul B. Andersen wrote:
> >> 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?
> I note that you have no comment to the obvious blunders you have made.
> Unless you are a complete moron, there is no way you can fail to
> understand that your calculations above are horribly wrong.
>
> But you go on babbling, as if your errors are only differences
> in opinion.

Clocks synchronized in K' are not synchronous in K. Einstein discovered this nasty little surprise too late in the game and so had to invent the relativity of simultaneity to cover his shame. This may be the topic of my next lecture.

On a side note, Legion threw out the RoS as a Hail Mary--with no explanatory comment--as a last ditch effort to rescue v = v before he vanished into the woodwork.
> >
> > We will go round and round on this because of the built-in ambiguity of relativity.
> There is no ambiguity in SR.
> > It is predictable that you will agree with some of my conclusions (as you already have) and vehemently disagree with others (as you already have) just as you do with the conclusions of Dirk, Dono, Python, Prokary, Sylvia, Tom Roberts, Legion, Jan, etc.
> You were challenge to show that you could use the Lorentz transform.
> You failed. What you call "your conclusions" are ERRORS.
>
> What the Lorentz transform predicts is not a matter of opinion,
> it is a matter of fact. None of the knowledgeable people you mention
> will ever disagree about what the Lorentz transform predicts.
>
> You, OTOH, have demonstrated that you can't use the Lorentz transform
> without doing stupid errors.
>
> You are obviously too mathematical illiterate to prove anything
> about anything.
> > Do you fools ever ask yourselves why you never disagree when it comes to Newtonian or Galilean calculations. It is because Einsteinianism is inherently marred by an illegitimate assumption at its very heart. After all the time I have spent tutoring you, Paul, you still claim that v' == v is the merest of trivialities. It was once claimed that heavy things fell faster than light things, and that absolute space and time were the also the merest of trivialities. I have demonstrated to this forum time and time again that v' != v, and frankly, my patience is beginning to wear thin. I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum. But I am willing to give this forum one final opportunity to raise itself out of the benighted state it finds itself in.
> This is beyond what a sane mind can produce.
>
> --
> Paul
>
> https://paulba.no/

On Thursday, October 20, 2022 at 7:14:33 AM UTC-7, Dono. wrote:
> On Thursday, October 20, 2022 at 1:03:57 AM UTC-7, patdolan wrote:
> > I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum.
> No, don't do that , Pattycakes. Your idiocies are fodder for daily entertainment, keep them coming.
Then shape up Dono. Shape the hell up!

On Thursday, October 20, 2022 at 10:11:27 AM UTC-7, patdolan wrote:
> On Thursday, October 20, 2022 at 6:02:01 AM UTC-7, Paul B. Andersen wrote:
> > Den 20.10.2022 10:03, skrev patdolan:
> > > On Wednesday, October 19, 2022 at 12:02:07 PM UTC-7, Paul B. Andersen wrote:
> > >> 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" rel="nofollow" target="_blank">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?
> > I note that you have no comment to the obvious blunders you have made.
> > Unless you are a complete moron, there is no way you can fail to
> > understand that your calculations above are horribly wrong.
> >
> > But you go on babbling, as if your errors are only differences
> > in opinion.
> Clocks synchronized in K' are not synchronous in K. Einstein discovered this nasty little surprise too late in the game and so had to invent the relativity of simultaneity to cover his shame. This may be the topic of my next lecture.
>
> On a side note, Legion threw out the RoS as a Hail Mary--with no explanatory comment--as a last ditch effort to rescue v = v before he vanished into the woodwork.
** v' = v
> > >
> > > We will go round and round on this because of the built-in ambiguity of relativity.
> > There is no ambiguity in SR.
> > > It is predictable that you will agree with some of my conclusions (as you already have) and vehemently disagree with others (as you already have) just as you do with the conclusions of Dirk, Dono, Python, Prokary, Sylvia, Tom Roberts, Legion, Jan, etc.
> > You were challenge to show that you could use the Lorentz transform.
> > You failed. What you call "your conclusions" are ERRORS.
> >
> > What the Lorentz transform predicts is not a matter of opinion,
> > it is a matter of fact. None of the knowledgeable people you mention
> > will ever disagree about what the Lorentz transform predicts.
> >
> > You, OTOH, have demonstrated that you can't use the Lorentz transform
> > without doing stupid errors.
> >
> > You are obviously too mathematical illiterate to prove anything
> > about anything.
> > > Do you fools ever ask yourselves why you never disagree when it comes to Newtonian or Galilean calculations. It is because Einsteinianism is inherently marred by an illegitimate assumption at its very heart. After all the time I have spent tutoring you, Paul, you still claim that v' == v is the merest of trivialities. It was once claimed that heavy things fell faster than light things, and that absolute space and time were the also the merest of trivialities. I have demonstrated to this forum time and time again that v' != v, and frankly, my patience is beginning to wear thin. I am considering never posting again...taking my genius and gratuitous instruction to a more accepting and appreciative forum. But I am willing to give this forum one final opportunity to raise itself out of the benighted state it finds itself in.
> > This is beyond what a sane mind can produce.
> >
> > --
> > Paul
> >
> > https://paulba.no/

On Wednesday, October 19, 2022 at 11:08:22 PM UTC-4, Volney wrote:
> On 10/19/2022 9:51 AM, Ken Seto wrote:
> > 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....?????
> All you need is this observer to be able to measure x and t twice for
> each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
> that? How about (-6)/(-2) ? Can you do that yet?

Except that the distances x1 and x2 change are changing continuously and your coordinate system is moving along with you as you move.
> >>
> >> 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
> It is implicit that a frame is chosen in which the observer is
> stationary to get v1 and v2 relative to that frame. Your point is?
> >>
> >> And "absolute rest" is an assertion, so has no value in science.d V2
> >
> > How else you ae going to get v1 and v2...?
> You choose a frame! Then you get v1 and v2 relative to that frame.

But the x-distances between 1 and 2 are change continuously.

> You can even choose a different frame, and get a different pair v1 and v2!
> > And how can you claim that the closing speed is (v1+v2) ???
> That's the definition of closing speed, Stupid Ken!

On Thursday, October 20, 2022 at 11:13:41 AM UTC-4, Wilbert Sciacca wrote:
> Ken Seto wrote:
>
> >> 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
> in respect to the observed, fucking stoopid. *I_fuck_your_ass!*. Your
> people died for you in Japonia. Two atomic bombs. That's NOT nothing.

You are a fucking cunt....good bye.

On 10/21/2022 10:31 AM, Ken Seto wrote:
> On Wednesday, October 19, 2022 at 11:08:22 PM UTC-4, Volney wrote:
>> On 10/19/2022 9:51 AM, Ken Seto wrote:
>>> 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....?????
>> All you need is this observer to be able to measure x and t twice for
>> each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
>> that? How about (-6)/(-2) ? Can you do that yet?
>
> Except that the distances x1 and x2 change are changing continuously and your coordinate system is moving along with you as you move.

That's the point. In the new reference frame you get different values
for x1 and x2. Therefore different values for v1 and v2. But they are
events, each object is measured once at (x1,t1) and again at (x2,t2) and
since these are single values at instances of time so cannot "move". In
Galilean space, the closing speed v1+v2 will be the same for any
(inertial) reference frame.
>>>>
>>>> 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
>> It is implicit that a frame is chosen in which the observer is
>> stationary to get v1 and v2 relative to that frame. Your point is?
>>>>
>>>> And "absolute rest" is an assertion, so has no value in science.d V2
>>>
>>> How else you ae going to get v1 and v2...?
>> You choose a frame! Then you get v1 and v2 relative to that frame.
>
> But the x-distances between 1 and 2 are change continuously.

Why would they? (x1,t1) and (x2,t2) are events, and events don't move.
>
>> You can even choose a different frame, and get a different pair v1 and v2!
>>> And how can you claim that the closing speed is (v1+v2) ???
>> That's the definition of closing speed, Stupid Ken!

On Saturday, 22 October 2022 at 07:42:38 UTC+2, Volney wrote:
> On 10/21/2022 10:31 AM, Ken Seto wrote:
> > On Wednesday, October 19, 2022 at 11:08:22 PM UTC-4, Volney wrote:
> >> On 10/19/2022 9:51 AM, Ken Seto wrote:
> >>> 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....?????
> >> All you need is this observer to be able to measure x and t twice for
> >> each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
> >> that? How about (-6)/(-2) ? Can you do that yet?
> >
> > Except that the distances x1 and x2 change are changing continuously and your coordinate system is moving along with you as you move.
> That's the point. In the new reference frame you get different values
> for x1 and x2. Therefore different values for v1 and v2. But they are
> events, each object is measured once at (x1,t1) and again at (x2,t2) and
> since these are single values at instances of time so cannot "move". In
> Galilean space, the closing speed v1+v2 will be the same for any
> (inertial) reference frame.

For any (of 0) inertial frame.
In the meantime in the real world, however, forbidden
by your bunch of idiots GPS and TAI keep measuring
t'=t in forbidden by your bunch of idiots old seconds.

Volney wrote:
> On 10/21/2022 10:31 AM, Ken Seto wrote:
>> Except that the distances x1 and x2 change are changing continuously
>> and your coordinate system is moving along with you as you move.
>
> That's the point. In the new reference frame you get different values
> for x1 and x2. Therefore different values for v1 and v2. But they are
> events,
> each object is measured once at (x1,t1) and again at (x2,t2) and since

you two kiss ass. He cant remember Japan and the two atomic bombs, you play piano bombing gas pipelines in the middle of the sea, to kill the european people. What kind of shithole are you.

https://www.%70%72%65%73%73%74%76%2e%69%72/Detail/2022/10/21/691337/Russia-Nord-Stream-Sabotage-Peskov
A preliminary report released by Danish police stated that "powerful explosions" had caused the rupture of the gas pipeline.

Danish police said inspectors had confirmed that there had been extensive damage to Nord Stream 1 and 2 in the Danish exclusive economic zone.

Parts of the pipeline were either missing or buried in the seabed, the company which was inspecting the damage reported.

Swedish investigators reached a similar conclusion earlier this month, blaming the blasts on “gross sabotage.”

Peskov said on Friday that if the "truth" were to be revealed, it would "surprise" many Europeans who suspect Moscow of orchestrating the blasts.

He said Moscow is working "intensely" to be included in an international probe, which Russia has previously said is "tailored" against it.

Maciej Wozniak wrote:

> On Saturday, 22 October 2022 at 07:42:38 UTC+2, Volney wrote:
>> since these are single values at instances of time so cannot "move". In
>> Galilean space, the closing speed v1+v2 will be the same for any
>> (inertial) reference frame.
>
> For any (of 0) inertial frame.
> In the meantime in the real world, however, forbidden by your bunch of
> idiots GPS and TAI keep measuring t'=t in forbidden by your bunch of
> idiots old seconds.

you two terrorist states are illegaly blowing up undersea gas pipelines, killing people to increas your profit. You are a disgrace.

Top US diplomat welcomes Nord Stream sabotage as ‘tremendous opportunity’
https://www.%70%72%65%73%73%74%76%2e%69%72/Detail/2022/10/02/690210/Top-US-diplomat-welcomes-Nord-Stream-blasts-as-%E2%80%98tremendous-opportunity%E2%80%99

US Secretary of State Antony Blinken has expressed pleasure over the sabotage of the Nord Stream gas pipelines, hailing it as a “tremendous opportunity” to repel European Union states from Russian energy resources amid the upcoming winter cold.

Den 19.10.2022 20:00, skrev patdolan:
>
> Let's move on now, Paul, to calculating the coordinate relative velocity v' for clock B' as viewed from K.

The scenario can be found in:
https://paulba.no/pdf/Mutual_time_dilation.pdf

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

Too imprecise! Clock B' is moving at the speed v in K and it's
position in K depend on the time t.
But at a specific event, let's call it E₄, is t₄ = 0 and x₄ = -d/γ

The coordinates of E₄ in K' are:
t₄'= γ(t₄ - (v/c²)⋅x₄) = d⋅v/c²
x₄'= γ(x₄ - v⋅t₄) = -d (B' is stationary at x₂'= -d)

This means that clock B' shows d⋅v/c² when it is at x = -d/γ

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

Wrong sign.
The coordinates of E₂ (B' and A co-located)
are t₂ = d/γv and x₂ = 0 (A is stationary at x = 0)

>
> 3) But clock B' is time dilated by 1/γ when viewed from frame K.

The coordinates of E₂ in K' are:
t₂'= γ(t₂ - (v/c²)⋅x₂) = γ⋅d/γv = d/v
x₂'= γ(x₂ - v⋅t₂) = -d (B' is stationary at x₂'= -d)

This means that clock B' shows t₂'= d/v when it is at x = 0
whether you 'view' it from K, K' or the Moon.

The "when viewed from frame K" is nonsense.

> 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.

Wrong sign again. Clocks don't go backwards.

The proper time measured by B' between E₄ and E₂ is
ΔτB' = t₂'-t₄' = d/v-d⋅v/c² = d/(v⋅γ²)

But the "in frame K" is meaningless.
What clock B' shows at a particular event isn't frame dependent.

How can you fail to understand that a clock can't show different
time at a specific event depending on which frame you 'view'
it from? What a weird misconception!

Look at your watch NOW. It shows the time T.
If an observer passed by you at a speed v and
read your watch at the same NOW, would he view
that the clock read something different from T?

>
> 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.

This is nonsense.
In K the spatial interval between E₄ and E₂ is x₂-x₄ = 0-(-d/γ) = d/γ
In K the temporal interval between E₄ and E₂ is t₂-t₄ = d/γv- 0 = d/γv

The speed with which clock B' has moved in in K is:
(x₂-x₄)/(t₂-t₄) = (d/γ)/(d/γv) = v
(An obvious triviality)

In K' the spatial interval between E₄ and E₂ is x₂'-x₄' = 0
In K' the temporal interval between E₄ and E₂ is t₂'-t₄' = d/(v⋅γ²)

The speed width which clock B' has moved in in K' is:
(x₂'-x₄')/(t₂'-t₄') = 0/(d/(v⋅γ²)) = 0
(An obvious triviality since clock B' is stationary in K')

> 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!
>

This is nonsense.

You probably meant write:
v' = (B''s travelled distance in K)/(B''s proper time)
= (x₂-x₄)/(t₂'-t₄') = (d/γ)/(d/(v⋅γ²)) = v⋅γ

This can be interpreted as:
Observed by the moving clock B', the distance in K seems to
pass faster than the real speed should indicate.

This is because the moving clock B' appears to run slow by 1/γ.

Compare this to: (t₂-t₄)/(t₂'-t₄') = (d/γv)/(d/(v⋅γ²)) = γ

This is the same as my conclusion #3 (equation 5) in:
https://paulba.no/pdf/Mutual_time_dilation.pdf

Did you think this was something new of any significance? :-D

--
Paul

https://paulba.no/

On Sunday, 23 October 2022 at 12:45:10 UTC+2, Paul B. Andersen wrote:
> Den 19.10.2022 20:00, skrev patdolan:
> >
> > Let's move on now, Paul, to calculating the coordinate relative velocity v' for clock B' as viewed from K.
> The scenario can be found in:
> https://paulba.no/pdf/Mutual_time_dilation.pdf
>
> >
> > 1) We know from the previous calculations that clock B' is located -d/γ on the x-axis in K.
>
> Too imprecise! Clock B' is moving at the speed v in K and it's
> position in K depend on the time t.
> But at a specific event, let's call it E₄, is t₄ = 0 and x₄ = -d/γ
>
> The coordinates of E₄ in K' are:
> t₄'= γ(t₄ - (v/c²)⋅x₄) = d⋅v/c²
> x₄'= γ(x₄ - v⋅t₄) = -d (B' is stationary at x₂'= -d)
>
> This means that clock B' shows d⋅v/c² when it is at x = -d/γ
> >
> > 2) At velocity v it will take clock B' a total of -d/γv frame K seconds to reach A, when viewed from K.
> Wrong sign.
> The coordinates of E₂ (B' and A co-located)
> are t₂ = d/γv and x₂ = 0 (A is stationary at x = 0)
> >
> > 3) But clock B' is time dilated by 1/γ when viewed from frame K.
> The coordinates of E₂ in K' are:
> t₂'= γ(t₂ - (v/c²)⋅x₂) = γ⋅d/γv = d/v
> x₂'= γ(x₂ - v⋅t₂) = -d (B' is stationary at x₂'= -d)
>
> This means that clock B' shows t₂'= d/v when it is at x = 0
> whether you 'view' it from K, K' or the Moon.
>
> The "when viewed from frame K" is nonsense.
>
> > 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.
>
> Wrong sign again. Clocks don't go backwards.
>
> The proper time measured by B' between E₄ and E₂ is
> ΔτB' = t₂'-t₄' = d/v-d⋅v/c² = d/(v⋅γ²)
>
> But the "in frame K" is meaningless.
> What clock B' shows at a particular event isn't frame dependent.
>
> How can you fail to understand that a clock can't show different
> time at a specific event depending on which frame you 'view'
> it from? What a weird misconception!
>
> Look at your watch NOW. It shows the time T.
> If an observer passed by you at a speed v and
> read your watch at the same NOW, would he view
> that the clock read something different from T?
> >
> > 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.
> This is nonsense.
> In K the spatial interval between E₄ and E₂ is x₂-x₄ = 0-(-d/γ) = d/γ
> In K the temporal interval between E₄ and E₂ is t₂-t₄ = d/γv- 0 = d/γv
>
> The speed with which clock B' has moved in in K is:
> (x₂-x₄)/(t₂-t₄) = (d/γ)/(d/γv) = v
> (An obvious triviality)
>
> In K' the spatial interval between E₄ and E₂ is x₂'-x₄' = 0
> In K' the temporal interval between E₄ and E₂ is t₂'-t₄' = d/(v⋅γ²)
>
> The speed width which clock B' has moved in in K' is:
> (x₂'-x₄')/(t₂'-t₄') = 0/(d/(v⋅γ²)) = 0
> (An obvious triviality since clock B' is stationary in K')
> > 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!
> >
> This is nonsense.
>
> You probably meant write:
> v' = (B''s travelled distance in K)/(B''s proper time)
> = (x₂-x₄)/(t₂'-t₄') = (d/γ)/(d/(v⋅γ²)) = v⋅γ
>
> This can be interpreted as:
> Observed by the moving clock B', the distance in K seems to
> pass faster than the real speed should indicate.
>
> This is because the moving clock B' appears to run slow by 1/γ.
>
> Compare this to: (t₂-t₄)/(t₂'-t₄') = (d/γv)/(d/(v⋅γ²)) = γ
>
> This is the same as my conclusion #3 (equation 5) in:
> https://paulba.no/pdf/Mutual_time_dilation.pdf

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

On Saturday, October 22, 2022 at 1:42:38 AM UTC-4, Volney wrote:
> On 10/21/2022 10:31 AM, Ken Seto wrote:
> > On Wednesday, October 19, 2022 at 11:08:22 PM UTC-4, Volney wrote:
> >> On 10/19/2022 9:51 AM, Ken Seto wrote:
> >>> 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....?????
> >> All you need is this observer to be able to measure x and t twice for
> >> each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
> >> that? How about (-6)/(-2) ? Can you do that yet?
> >
> > Except that the distances x1 and x2 change are changing continuously and your coordinate system is moving along with you as you move.
> That's the point. In the new reference frame you get different values
> for x1 and x2. Therefore different values for v1 and v2. But they are
> events, each object is measured once at (x1,t1) and again at (x2,t2) and
> since these are single values at instances of time so cannot "move". In
> Galilean space, the closing speed v1+v2 will be the same for any
> (inertial) reference frame.

Let me try to understand your point:
1. you are at rest at the origin of the coordinate system you set up.
2. O1 and O2 are moving on the x-axis.....o1 is moving to the right od you and O2 is moving to the left of you.
3. you measure the velocity of O1 and get a value of Vi
4. you measure the velocity of O2 and get a value of V2.
5. you then conclude that the closing speed of O1 and O2 is simply (V1+V2).
6. I disagree because you assumed that you are in a state of absolute rest.
7 the closing speed between O1 and O2.must be determined by O1 and O2 individually. and the restlt is simply the relative speed between them.
> >>>>
> >>>> 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
> >> It is implicit that a frame is chosen in which the observer is
> >> stationary to get v1 and v2 relative to that frame. Your point is?
> >>>>
> >>>> And "absolute rest" is an assertion, so has no value in science.d V2
> >>>
> >>> How else you ae going to get v1 and v2...?
> >> You choose a frame! Then you get v1 and v2 relative to that frame.
> >
> > But the x-distances between 1 and 2 are change continuously.
> Why would they? (x1,t1) and (x2,t2) are events, and events don't move.
> >
> >> You can even choose a different frame, and get a different pair v1 and v2!
> >>> And how can you claim that the closing speed is (v1+v2) ???
> >> That's the definition of closing speed, Stupid Ken!

On 10/25/2022 11:12 AM, Ken Seto wrote:
> On Saturday, October 22, 2022 at 1:42:38 AM UTC-4, Volney wrote:
>> On 10/21/2022 10:31 AM, Ken Seto wrote:
>>> On Wednesday, October 19, 2022 at 11:08:22 PM UTC-4, Volney wrote:
>>>> On 10/19/2022 9:51 AM, Ken Seto wrote:
>>>>> On Saturday, October 15, 2022 at 9:54:47 PM UTC-4, Volney wrote:

>>>>>> Word Salad.
>>>>>>
>>>>>> What is this "initial observer" and why is it needed? d v2
>>>>>
>>>>> to get v1 and v2....?????
>>>> All you need is this observer to be able to measure x and t twice for
>>>> each of the two objects, then calculate v = (x2-x1)/(t2-t1). Can you do
>>>> that? How about (-6)/(-2) ? Can you do that yet?
>>>
>>> Except that the distances x1 and x2 change are changing continuously and your coordinate system is moving along with you as you move.
>> That's the point. In the new reference frame you get different values
>> for x1 and x2. Therefore different values for v1 and v2. But they are
>> events, each object is measured once at (x1,t1) and again at (x2,t2) and
>> since these are single values at instances of time so cannot "move". In
>> Galilean space, the closing speed v1+v2 will be the same for any
>> (inertial) reference frame.
>
> Let me try to understand your point:
> 1. you are at rest at the origin of the coordinate system you set up.
OK.
> 2. O1 and O2 are moving on the x-axis.....o1 is moving to the right od you and O2 is moving to the left of you.
OK.
> 3. you measure the velocity of O1 and get a value of Vi
V1, but OK.
> 4. you measure the velocity of O2 and get a value of V2.
OK.
> 5. you then conclude that the closing speed of O1 and O2 is simply (V1+V2).
OK.
> 6. I disagree because you assumed that you are in a state of absolute rest.
I made no such assumption, especially since there *is* no such thing as
absolute rest. Don't create fake straw men arguments just to knock down.
> 7 the closing speed between O1 and O2.must be determined by O1 and O2 individually.
Technically that's not allowed because closing speed is defined as the
total speed seen by a third party (myself in your example). No third
party observer here.

I suppose you *could* use O1's frame for the observer, which would then
be co-moving with O1, so O1's speed is 0 and O2 has a measurable speed
in O1's frame, so O2's speed and the closing speed are the same. Same
for an observer co-moving with O2.

> and the restlt is simply the relative speed between them.
If and only if my co-moving observer example is legit. But that does not
give an answer for the third party observer for the actual definition of
closing speed.