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tech / sci.physics.relativity / Re: How to Write a Transformation--Part One

SubjectAuthor
* How to Write a Transformation--Part Onepatdolan
+* Cretin Pat Dolan fails basic algebraDono.
|+* Re: Cretin Pat Dolan fails basic algebrapatdolan
||`* Re: Cretin Pat Dolan fails basic algebraDono.
|| `* Re: Cretin Pat Dolan fails basic algebrapatdolan
||  `- Re: Cretin Pat Dolan fails basic algebraDono.
|`* Re: Cretin Pat Dolan fails basic algebraRichard Hertz
| `- Re: Cretin Pat Dolan fails basic algebra and fellow crank Dick HertzDono.
+- Re: How to Write a Transformation--Part Onerotchm
+* Re: How to Write a Transformation--Part Onerotchm
|`* Re: How to Write a Transformation--Part Onepatdolan
| `* Re: How to Write a Transformation--Part Onerotchm
|  `* Re: How to Write a Transformation--Part Onepatdolan
|   `- Re: How to Write a Transformation--Part Onerotchm
+* Re: How to Write a Transformation--Part OneStan Fultoni
|+- Re: How to Write a Transformation--Part Onepatdolan
|`* Re: How to Write a Transformation--Part Onerotchm
| `* Re: How to Write a Transformation--Part Onepatdolan
|  `* Re: How to Write a Transformation--Part Onepatdolan
|   +* Re: How to Write a Transformation--Part Onerotchm
|   |+* Re: How to Write a Transformation--Part Onepatdolan
|   ||+* Re: How to Write a Transformation--Part Onerotchm
|   |||`* Re: How to Write a Transformation--Part Onepatdolan
|   ||| +* Re: How to Write a Transformation--Part Onerotchm
|   ||| |`- Re: How to Write a Transformation--Part OneMaciej Wozniak
|   ||| `* Lying piece of shit Pat Dolan cannot follow simple algebraDono.
|   |||  `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||   `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    +* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |`* Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    | `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |  `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    |   +* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraDono.
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraDono.
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    |   |+* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   ||+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraPython
|   |||    |   ||`- Re: Lying piece of shit Pat Dolan cannot follow simple algebraMaciej Wozniak
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebraDono.
|   |||    |   |+- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    |   |`- Re: Lying piece of shit Pat Dolan cannot follow simple algebraDono.
|   |||    |   `- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||    `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     +* Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |`* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     | +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     | `* Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |  `* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||     |   +- Re: Lying piece of shit Pat Dolan cannot follow simple algebraStan Fultoni
|   |||     |   `- Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     +* Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||     |`* Re: Lying piece of shit Pat Dolan cannot follow simple algebrapatdolan
|   |||     | `- Re: Lying piece of shit Pat Dolan cannot follow simple algebrarotchm
|   |||     `* Re: Lying piece of shit Pat Dolan cannot follow simple algebraPaul B. Andersen
|   |||      `* Re: Lying piece of shit Pat Dolan cannot follow simple algebraRichard Hachel
|   |||       `* Re: Lying piece of shit Pat Dolan cannot follow simple algebraPaul B. Andersen
|   |||        `- Re: Lying piece of shit Pat Dolan cannot follow simple algebraRichard Hachel
|   ||`* Re: How to Write a Transformation--Part OneStan Fultoni
|   || `* Re: How to Write a Transformation--Part Onepatdolan
|   ||  +* Re: How to Write a Transformation--Part Onerotchm
|   ||  |`- Re: How to Write a Transformation--Part Onepatdolan
|   ||  `* Re: How to Write a Transformation--Part OneStan Fultoni
|   ||   `* Re: How to Write a Transformation--Part Onepatdolan
|   ||    `* Re: How to Write a Transformation--Part OneStan Fultoni
|   ||     `* Re: How to Write a Transformation--Part Onepatdolan
|   ||      `* Re: How to Write a Transformation--Part OneStan Fultoni
|   ||       `* Re: How to Write a Transformation--Part Onepatdolan
|   ||        `* Re: How to Write a Transformation--Part OneStan Fultoni
|   ||         `* Re: How to Write a Transformation--Part Onepatdolan
|   ||          +* Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingrotchm
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingPaul Alsing
|   ||          |+- Re: Crank Pat Dolan keeps on trollingrotchm
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |+- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          |+- Re: Crank Pat Dolan keeps on trollingpatdolan
|   ||          |`- Re: Crank Pat Dolan keeps on trollingDono.
|   ||          +- Re: How to Write a Transformation--Part OneStan Fultoni
|   ||          +- Re: How to Write a Transformation--Part Onepatdolan
|   ||          +- Re: How to Write a Transformation--Part OneStan Fultoni
|   ||          +- Re: How to Write a Transformation--Part Onepatdolan
|   ||          +- Re: How to Write a Transformation--Part OneStan Fultoni
|   ||          +* Re: How to Write a Transformation--Part OneTom Roberts
|   ||          +- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   ||          +- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   ||          +- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   ||          +- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   ||          +- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   ||          `- Re: Crank Pat Dolan keeps on trollingStan Fultoni
|   |`- Re: How to Write a Transformation--Part Onepatdolan
|   `- Cretin Pat Dolan keeps trollingDono.
`* Re: How to Write a Transformation--Part Onepatdolan

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Re: Crank Pat Dolan keeps on trolling

<3c2b326e-4d7f-4daf-807d-171f57589746n@googlegroups.com>

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Subject: Re: Crank Pat Dolan keeps on trolling
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Fri, 13 May 2022 02:30 UTC

On Thursday, May 12, 2022 at 6:34:02 PM UTC-7, patdolan wrote:
> ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]

Again, you are writing the relations for a particle at rest in x,t, which means
dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
dx'/dt', which you have found is equal to dx'/dt'. Magnificent.

> ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]

Again, for the interval you are describing, dx=0, so your equation is
dx' dt/dt' = 0, which obviously does not follow rationally from anything..
> ∆x∆x'^2/c^2∆t'^2 = ∆x [6]

Again, for the interval you are describing, dx=0, so this equation is
0 = 0. Magnificent.

> ∆x'^2 = c^2∆t'^2 [7]

Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Fri, 13 May 2022 12:30 UTC

On Thursday, May 12, 2022 at 7:30:17 PM UTC-7, Stan Fultoni wrote:
> On Thursday, May 12, 2022 at 6:34:02 PM UTC-7, patdolan wrote:
> > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> Again, you are writing the relations for a particle at rest in x,t, which means
> dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> Again, for the interval you are describing, dx=0, so your equation is
> dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> Again, for the interval you are describing, dx=0, so this equation is
> 0 = 0. Magnificent.
> > ∆x'^2 = c^2∆t'^2 [7]
> Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.

Stan, you BSer. Have a look:

https://www.khanacademy.org/science/physics/special-relativity/einstein-velocity-addition/v/einstein-velocity-addition-formula-derivation

Let me know if there is anything in this video that you don't understand or is unclear to a novice.

Re: Crank Pat Dolan keeps on trolling

<28f322fc-ae40-453f-8032-beb9cc793509n@googlegroups.com>

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Date: Fri, 13 May 2022 06:26:34 -0700 (PDT)
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Subject: Re: Crank Pat Dolan keeps on trolling
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Fri, 13 May 2022 13:26 UTC

On Friday, May 13, 2022 at 5:30:56 AM UTC-7, patdolan wrote:
> > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > Again, you are writing the relations for a particle at rest in x,t, which means
> > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > Again, for the interval you are describing, dx=0, so your equation is
> > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > Again, for the interval you are describing, dx=0, so this equation is
> > 0 = 0. Magnificent.
> > > ∆x'^2 = c^2∆t'^2 [7]
> > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
>
> Have a look [at a youtube video]. Let me know if there is anything in this
> video that you don't understand or is unclear to a novice.

Again, your absurd claim that the Lorentz transformation implies v=c has
been thoroughly debunked (see above). Your mistake was not realizing that
the interval you are referring to has dx'/dt' = -v and dx/dt = 0. If you need
more help understanding this, just ask. (Exhibiting infantile echoalia is
not helping you learn.)

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Fri, 13 May 2022 16:17 UTC

On Friday, May 13, 2022 at 6:26:36 AM UTC-7, Stan Fultoni wrote:
> On Friday, May 13, 2022 at 5:30:56 AM UTC-7, patdolan wrote:
> > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > Again, for the interval you are describing, dx=0, so your equation is
> > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > Again, for the interval you are describing, dx=0, so this equation is
> > > 0 = 0. Magnificent.
> > > > ∆x'^2 = c^2∆t'^2 [7]
> > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
> >
> > Have a look [at a youtube video]. Let me know if there is anything in this
> > video that you don't understand or is unclear to a novice.
> Again, your absurd claim that the Lorentz transformation implies v=c has
> been thoroughly debunked (see above). Your mistake was not realizing that
> the interval you are referring to has dx'/dt' = -v and dx/dt = 0. If you need
> more help understanding this, just ask. (Exhibiting infantile echoalia is
> not helping you learn.)
Slippery Stan, why are you responding to me again instead of Khan's video? Instruct this forum how Khan's derivation is different from mine.

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Fri, 13 May 2022 17:21 UTC

On Thursday, May 12, 2022 at 6:34:02 PM UTC-7, patdolan wrote:
> On Thursday, May 12, 2022 at 6:21:23 PM UTC-7, rotchm wrote:
> > On Thursday, May 12, 2022 at 1:46:44 PM UTC-4, patdolan wrote:
> >
> > > Well rotchm? Well Stan? What does either of you have to say
> > Well, I already answered you. You agreed that I was right in that you are wrong. That is all that matters in this thread.
> > Now you are asking you question. For that , you must start a new thread.. And I suggest you correct the many errors you made before you repost your question in a new thread. Yes of course I saw what you wrote and you have many many errors. This shows that you are very confused and that you are nil at math. Hence, you are no longer worthy of our time. You are just a loser seeking attention. You should try a new hobby like collect rocks or try to catch flies with your mouth. You will have more success with that then you have with basic algebra.
> x' = g(v)(x - vt) [A]
> t' = g(v)(t - vx/c^2) [B]
>
> ∆x'/∆t' = g(v)(∆x-v∆t)/g(v)(∆t-v∆x/c^2) [2]
>
> ∆x'/∆t' = (∆x-v∆t)/(∆t-v∆x/c^2) [3]

This is ok
> ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
>

This is wrong since, like the idiot you are, you are taking v' to be equal to ∆x'/∆t'

Face it, Pattycakes, you are inept.

> v' = c [9]

No. Idiot.

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Fri, 13 May 2022 18:21 UTC

On Friday, May 13, 2022 at 10:21:40 AM UTC-7, Dono. wrote:
> On Thursday, May 12, 2022 at 6:34:02 PM UTC-7, patdolan wrote:
> > On Thursday, May 12, 2022 at 6:21:23 PM UTC-7, rotchm wrote:
> > > On Thursday, May 12, 2022 at 1:46:44 PM UTC-4, patdolan wrote:
> > >
> > > > Well rotchm? Well Stan? What does either of you have to say
> > > Well, I already answered you. You agreed that I was right in that you are wrong. That is all that matters in this thread.
> > > Now you are asking you question. For that , you must start a new thread. And I suggest you correct the many errors you made before you repost your question in a new thread. Yes of course I saw what you wrote and you have many many errors. This shows that you are very confused and that you are nil at math. Hence, you are no longer worthy of our time. You are just a loser seeking attention. You should try a new hobby like collect rocks or try to catch flies with your mouth. You will have more success with that then you have with basic algebra.
> > x' = g(v)(x - vt) [A]
> > t' = g(v)(t - vx/c^2) [B]
> >
> > ∆x'/∆t' = g(v)(∆x-v∆t)/g(v)(∆t-v∆x/c^2) [2]
> >
> > ∆x'/∆t' = (∆x-v∆t)/(∆t-v∆x/c^2) [3]
> This is ok
> > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> >
> This is wrong since, like the idiot you are, you are taking v' to be equal to ∆x'/∆t'
>
> Face it, Pattycakes, you are inept.
>
>
>
> > v' = c [9]
>
> No. Idiot.
Dono, if you were going to represent v' in coordinate space and time, how would YOU do it? Just plain v ? If so, then how would YOU represent v ?

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Fri, 13 May 2022 18:46 UTC

On Friday, May 13, 2022 at 9:17:49 AM UTC-7, patdolan wrote:
> > > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > > Again, for the interval you are describing, dx=0, so your equation is
> > > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > > Again, for the interval you are describing, dx=0, so this equation is
> > > > 0 = 0. Magnificent.
> > > > > ∆x'^2 = c^2∆t'^2 [7]
> > > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
>
> ...how [is youtube] derivation [of velocity addition formula] different from mine?

You have not presented a derivation of the velocity addition formula, you have presented
a childishly absurd pseudo-derivation of the ridiculous claim that the Lorentz transformation
implies v=c. Your fallacious pseudo-reasoning was debunked above. If you see anything
that you think is wrong or unclear in the debunking, go ahead and point it out. Keep in
mind that for the interval you are considering, ∆x'/∆t' = -v and ∆x = 0.

Re: Crank Pat Dolan keeps on trolling

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Subject: Re: Crank Pat Dolan keeps on trolling
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Fri, 13 May 2022 18:56 UTC

On Friday, May 13, 2022 at 11:21:59 AM UTC-7, patdolan wrote:
> On Friday, May 13, 2022 at 10:21:40 AM UTC-7, Dono. wrote:
> > On Thursday, May 12, 2022 at 6:34:02 PM UTC-7, patdolan wrote:
> > > On Thursday, May 12, 2022 at 6:21:23 PM UTC-7, rotchm wrote:
> > > > On Thursday, May 12, 2022 at 1:46:44 PM UTC-4, patdolan wrote:
> > > >
> > > > > Well rotchm? Well Stan? What does either of you have to say
> > > > Well, I already answered you. You agreed that I was right in that you are wrong. That is all that matters in this thread.
> > > > Now you are asking you question. For that , you must start a new thread. And I suggest you correct the many errors you made before you repost your question in a new thread. Yes of course I saw what you wrote and you have many many errors. This shows that you are very confused and that you are nil at math. Hence, you are no longer worthy of our time. You are just a loser seeking attention. You should try a new hobby like collect rocks or try to catch flies with your mouth. You will have more success with that then you have with basic algebra.
> > > x' = g(v)(x - vt) [A]
> > > t' = g(v)(t - vx/c^2) [B]
> > >
> > > ∆x'/∆t' = g(v)(∆x-v∆t)/g(v)(∆t-v∆x/c^2) [2]
> > >
> > > ∆x'/∆t' = (∆x-v∆t)/(∆t-v∆x/c^2) [3]
> > This is ok
> > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > >
> > This is wrong since, like the idiot you are, you are taking v' to be equal to ∆x'/∆t'
> >
> > Face it, Pattycakes, you are inept.
> >
> >
> >
> > > v' = c [9]
> >
> > No. Idiot.
> Dono, if you were going to represent v' in coordinate space and time, how would YOU do it? Just plain v ? If so, then how would YOU represent v ?

v.v' are the relative speeds between the frames S and S'. They have nothing to do with ∆x/∆t and ∆x'/∆t' . You being a pathological imbecile will never understand this despite this being explained to you multiple times.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sat, 14 May 2022 16:17 UTC

On Monday, May 9, 2022 at 11:02:26 AM UTC-7, patdolan wrote:
> Galilean Relativity is a text book example of a logically coherent piece of mathematics and physical science. Let's take a look at the Galilean Transformations
>
> x' = x - vt (1)
> t = t' (2)
>
> from which we derive
>
> x = x' + vt' (3)
> or
> x = x' + vt (4) because t = t'
>
> We now prove the logical coherence of the Galilean transforms by showing that the magnitudes of the v in eqn 3 and eqn 4 are identical. Spoiler alert--this will not be the case for the Lorentz Transformations.
>
> Step One: Let's assume the v's in eqn's 1 and 3 are different. Let's call them v and v' respectively. We now rewrite eqns 1 and 3 as
>
> x' = x - vt (5)
> x = x' + v't (6)
>
> Step Two: Solve for v
>
> v = [ -x' + x ]/t (7)
>
> Step Three: Solve for v'
>
> v' = [ x - x' ]/t (8)
>
> v' = -v
>
> Step Four: acknowledge that ||v|| = ||-v|| therefore there is nothing ambivalent or equivocal about v in the Galilean Transformations. Observers in either frame acknowledge the equivalence of the magnitude of v. Easy peasy.
>
> Stay tuned for Part Two....

*********************
THREAD DIGEST
*********************

1) Dolan introduced the possibility that the v in the Galilean Transforms many not have the same magnitude for observers O and O' in two different Galilean frames of references. Dolan postulated to velocities, v and v', and that v != v'. Dolan then proved by mathematical deduction that in fact v == v' in all cases.

2) Dolan then went on to introduce the same possibility in the case of the Lorentz Transforms. But his proposal was in the negative. Dolan proposed what will one day be known as the Dolan Conjecture, namely, that v == v' it is an undecidable proposition in the case of the Lorentz Transforms.

3) At this point rotchm and Stan leaped into the fray as the protagonists for the other point of view. They advance all sorts of specious arguments purporting to prove that v == v'. But none of these arguments were deductive in nature and so were discarded out of hand. The Dolan Conjecture remained intact.

It is also to be noted that Dono participated in the case against the Dolan Conjecture. Dono's efforts characteristically amounted to lumbering onto the thread, raising his tail and spraying the place down with that god-awful aroma we have all come to know so well.

4) The wily Dolan was not yet done with protagonists rotchm & Stan. Dolan adroitly pivoted to accepting the validity of v == v' in the case of the LTs. He then used this premise to validly deduce a reductio ad absurdum, thereby destroying the mathematical validity of the LTs and their illegitimate spawn, Special Relativity.

5) By this time rotchm had had enough and quit the field. But Stan, Van der mootally wounded, fought on. Stan substituted "0" into the v' variable, not realizing that all he had accomplished was to show that c = 0. When Stan continued to protest in ignorance, he was directed to a Khan Academy video demonstrating the same derivation Dolan used. This was enough for protagonist Stan and he too promptly quit the field also.

6) The Dolan Conjecture assumed its place among the vitally important questions of post-modern Physics.

Imbecile Pat Dolan perseveres

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Subject: Imbecile Pat Dolan perseveres
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Sat, 14 May 2022 16:46 UTC

On Saturday, May 14, 2022 at 9:17:54 AM UTC-7, imbecile pat dolan wrote:

> > x' = x - vt (5)
> > x = x' + v't (6)
> >
> > Step Two: Solve for v
> >
> > v = [ -x' + x ]/t (7)
> >
ok
> > Step Three: Solve for v'
> >
> > v' = [ x - x' ]/t (8)
> >
ok
> > v' = -v

Wrong. Imbecile. you can't do basic algebra.

> > Stay tuned for Part Two....
>
> *********************
> THREAD DIGEST
> *********************
>
> 1) Dolan introduced imbecilities

Yep
>
> 2) Dolan then went on to introduce more imbecilities

Yep
>


> 6) The Dolan Conjecture assumed its place among the immortal fumbles.

Yep

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Sat, 14 May 2022 16:55 UTC

On Saturday, May 14, 2022 at 9:17:54 AM UTC-7, patdolan wrote:
> > > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > > Again, for the interval you are describing, dx=0, so your equation is
> > > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > > Again, for the interval you are describing, dx=0, so this equation is
> > > > 0 = 0. Magnificent.
> > > > > ∆x'^2 = c^2∆t'^2 [7]
> > > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
> > >
> > > ...how [is youtube] derivation [of velocity addition formula] different from mine?
> >
> > You have not presented a derivation of the velocity addition formula, you have presented
> > a childishly absurd pseudo-derivation of the ridiculous claim that the Lorentz transformation
> > implies v=c. Your fallacious pseudo-reasoning was debunked above. If you see anything
> > that you think is wrong or unclear in the debunking, go ahead and point it out. Keep in
> > mind that for the interval you are considering, ∆x'/∆t' = -v and ∆x = 0.
>
> You substituted "0" into the v' variable...

Nope, you have agreed that what you called v' is actually v, which is the relative speed between the two systems of coordinates x,t and x',t'. This means that an interval along the path of an object at rest in x',t' has ∆x'=0 and ∆x/∆t = v, and an interval along the path of an object at rest in x,t has ∆x = 0 and ∆x'/∆t' = -v.

> a video demonstrating the same derivation Dolan used.

Nope. You referred to a (truly abysmal) video on deriving the velocity addition formula, which has nothing to do with your absurdly fallacious pseudo-reasoning in this thread. If you want me to explain the derivation of the velocity addition formula, just ask. In the mean time, the thorough debunking of your fallacious pseudo-reasoning is shown above. Changing the subject to derivations of the velocity addition formula will not change the debunking of your fallacies, although it could conceivably help you learn a little about the subject so that you would understand the debunking.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sat, 14 May 2022 17:07 UTC

On Saturday, May 14, 2022 at 9:55:10 AM UTC-7, Stan Fultoni wrote:
> On Saturday, May 14, 2022 at 9:17:54 AM UTC-7, patdolan wrote:
> > > > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > > > Again, for the interval you are describing, dx=0, so your equation is
> > > > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > > > Again, for the interval you are describing, dx=0, so this equation is
> > > > > 0 = 0. Magnificent.
> > > > > > ∆x'^2 = c^2∆t'^2 [7]
> > > > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh..
> > > >
> > > > ...how [is youtube] derivation [of velocity addition formula] different from mine?
> > >
> > > You have not presented a derivation of the velocity addition formula, you have presented
> > > a childishly absurd pseudo-derivation of the ridiculous claim that the Lorentz transformation
> > > implies v=c. Your fallacious pseudo-reasoning was debunked above. If you see anything
> > > that you think is wrong or unclear in the debunking, go ahead and point it out. Keep in
> > > mind that for the interval you are considering, ∆x'/∆t' = -v and ∆x = 0.
> >
> > You substituted "0" into the v' variable...
>
> Nope, you have agreed that what you called v' is actually v, which is the relative speed between the two systems of coordinates x,t and x',t'. This means that an interval along the path of an object at rest in x',t' has ∆x'=0 and ∆x/∆t = v, and an interval along the path of an object at rest in x,t has ∆x = 0 and ∆x'/∆t' = -v.
>
> > a video demonstrating the same derivation Dolan used.
>
> Nope. You referred to a (truly abysmal) video on deriving the velocity addition formula, which has nothing to do with your absurdly fallacious pseudo-reasoning in this thread. If you want me to explain the derivation of the velocity addition formula, just ask. In the mean time, the thorough debunking of your fallacious pseudo-reasoning is shown above. Changing the subject to derivations of the velocity addition formula will not change the debunking of your fallacies, although it could conceivably help you learn a little about the subject so that you would understand the debunking.

A last gasp from Stan. Poor fella...listen how he mumbles incoherently...it sounds like he believes himself to be a greater educator Khan, the man who brought higher education to the masses and watched by millions of people around the world. I wonder how many listen to Stan.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Sat, 14 May 2022 17:20 UTC

On Saturday, May 14, 2022 at 10:07:03 AM UTC-7, patdolan wrote:
> On Saturday, May 14, 2022 at 9:55:10 AM UTC-7, Stan Fultoni wrote:
> > On Saturday, May 14, 2022 at 9:17:54 AM UTC-7, patdolan wrote:
> > > > > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > > > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > > > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > > > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > > > > Again, for the interval you are describing, dx=0, so your equation is
> > > > > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > > > > Again, for the interval you are describing, dx=0, so this equation is
> > > > > > 0 = 0. Magnificent.
> > > > > > > ∆x'^2 = c^2∆t'^2 [7]
> > > > > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
> > > > >
> > > > > ...how [is youtube] derivation [of velocity addition formula] different from mine?
> > > >
> > > > You have not presented a derivation of the velocity addition formula, you have presented
> > > > a childishly absurd pseudo-derivation of the ridiculous claim that the Lorentz transformation
> > > > implies v=c. Your fallacious pseudo-reasoning was debunked above. If you see anything
> > > > that you think is wrong or unclear in the debunking, go ahead and point it out. Keep in
> > > > mind that for the interval you are considering, ∆x'/∆t' = -v and ∆x = 0.
> > >
> > > You substituted "0" into the v' variable...
> >
> > Nope, you have agreed that what you called v' is actually v, which is the relative speed between the two systems of coordinates x,t and x',t'. This means that an interval along the path of an object at rest in x',t' has ∆x'=0 and ∆x/∆t = v, and an interval along the path of an object at rest in x,t has ∆x = 0 and ∆x'/∆t' = -v.
> >
> > > a video demonstrating the same derivation Dolan used.
> >
> > Nope. You referred to a (truly abysmal) video on deriving the velocity addition formula, which has nothing to do with your absurdly fallacious pseudo-reasoning in this thread. If you want me to explain the derivation of the velocity addition formula, just ask. In the mean time, the thorough debunking of your fallacious pseudo-reasoning is shown above. Changing the subject to derivations of the velocity addition formula will not change the debunking of your fallacies, although it could conceivably help you learn a little about the subject so that you would understand the debunking.
>
> A last...

So, you still have no substantive response at all to the thorough debunking of your fallacious pseudo-reasoning, and are still too proud to ask for help understanding the velocity addition derivation. Until you can bring yourself to actually engage your mind in trying to understand the subject, you can't expect to make any progress. Example of valid reasoning:

For a particle moving at speed u in terms of x',t' we have dx'/dt' = u, and we wish to find the speed of that particle in terms of x,t. From the relations dx = (dx'+vdt')g, dt = (dt'+vdx')g, we have dx/dt = (dx'+vdt')/(dt'+vdx'), and dividing the numerator and denominator by dt', this gives dx/dt = (u+v)/(1+uv).

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sat, 14 May 2022 18:02 UTC

On Saturday, May 14, 2022 at 10:20:20 AM UTC-7, Stan Fultoni wrote:
> On Saturday, May 14, 2022 at 10:07:03 AM UTC-7, patdolan wrote:
> > On Saturday, May 14, 2022 at 9:55:10 AM UTC-7, Stan Fultoni wrote:
> > > On Saturday, May 14, 2022 at 9:17:54 AM UTC-7, patdolan wrote:
> > > > > > > > ∆x'/∆t' = (∆x+[∆x'∆t/∆t'])/(∆t+∆x∆x'/∆t'c^2) [4]
> > > > > > > Again, you are writing the relations for a particle at rest in x,t, which means
> > > > > > > dx/dt = 0 and d'x/dt' = -v, so the right side of the above equation is simply
> > > > > > > dx'/dt', which you have found is equal to dx'/dt'. Magnificent.
> > > > > > > > ∆x'∆t/∆t' + ∆x∆x'^2/c^2∆t'^2 = ∆x + ∆x ∆'t/∆t' [5]
> > > > > > > Again, for the interval you are describing, dx=0, so your equation is
> > > > > > > dx' dt/dt' = 0, which obviously does not follow rationally from anything.
> > > > > > > > ∆x∆x'^2/c^2∆t'^2 = ∆x [6]
> > > > > > > Again, for the interval you are describing, dx=0, so this equation is
> > > > > > > 0 = 0. Magnificent.
> > > > > > > > ∆x'^2 = c^2∆t'^2 [7]
> > > > > > > Nope, the relation 0 = 0 does not imply that v = c. (Duh)^Duh.
> > > > > >
> > > > > > ...how [is youtube] derivation [of velocity addition formula] different from mine?
> > > > >
> > > > > You have not presented a derivation of the velocity addition formula, you have presented
> > > > > a childishly absurd pseudo-derivation of the ridiculous claim that the Lorentz transformation
> > > > > implies v=c. Your fallacious pseudo-reasoning was debunked above. If you see anything
> > > > > that you think is wrong or unclear in the debunking, go ahead and point it out. Keep in
> > > > > mind that for the interval you are considering, ∆x'/∆t' = -v and ∆x = 0.
> > > >
> > > > You substituted "0" into the v' variable...
> > >
> > > Nope, you have agreed that what you called v' is actually v, which is the relative speed between the two systems of coordinates x,t and x',t'. This means that an interval along the path of an object at rest in x',t' has ∆x'=0 and ∆x/∆t = v, and an interval along the path of an object at rest in x,t has ∆x = 0 and ∆x'/∆t' = -v.
> > >
> > > > a video demonstrating the same derivation Dolan used.
> > >
> > > Nope. You referred to a (truly abysmal) video on deriving the velocity addition formula, which has nothing to do with your absurdly fallacious pseudo-reasoning in this thread. If you want me to explain the derivation of the velocity addition formula, just ask. In the mean time, the thorough debunking of your fallacious pseudo-reasoning is shown above. Changing the subject to derivations of the velocity addition formula will not change the debunking of your fallacies, although it could conceivably help you learn a little about the subject so that you would understand the debunking.
> >
> > A last...
>
> So, you still have no substantive response at all to the thorough debunking of your fallacious pseudo-reasoning, and are still too proud to ask for help understanding the velocity addition derivation. Until you can bring yourself to actually engage your mind in trying to understand the subject, you can't expect to make any progress. Example of valid reasoning:
>
> For a particle moving at speed u in terms of x',t' we have dx'/dt' = u, and we wish to find the speed of that particle in terms of x,t. From the relations dx = (dx'+vdt')g, dt = (dt'+vdx')g, we have dx/dt = (dx'+vdt')/(dt'+vdx'), and dividing the numerator and denominator by dt', this gives dx/dt = (u+v)/(1+uv).

It is you who are about to get the lesson Stan.

In your desperation to free yourself from my trap you have apparently hallucinated a 3rd object of interest for which we need to work out the coordinate velocity in the two FoRs of interest.

Now read this carefully Stan: T. H. E. R. E. I.S. N.O. T. H. I .R.D. O. B. J. E. C. T.

Got that Stan? No third object in this scenario. So no Einstein velocity addition formula. But let's go with the EVAF for a moment. In the absence of a third party/object u = 0. So we are left with, according to your own reasoning, dx/dt = (0+v)/(1+0) = v.

So, Stan, we are back to considering just the v between the origins of the two FoRs. We now reiterate the original question:

"Can it be deductively demonstrated that the magnitude of the v in the LTs is equivalent for all pairs of FoRs?"

The answer is the Dolan Conjecture: Assuming a velocity v and a coordinate velocity v' for all pairs of FoRs, it is undecidable that v = v'

Furthermore we have the corollary: It is demonstrated that contradictory results are produced by assuming v == v' for all pairs of FoRs

Please let me know if there is something about this situation which remains unclear to you; or when you can bring yourself to actually engage your mind in trying to understand the subject.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Sat, 14 May 2022 18:31 UTC

On Saturday, May 14, 2022 at 11:02:41 AM UTC-7, patdolan wrote:
> > For a particle moving at speed u in terms of x',t' we have dx'/dt' = u, and we wish to find the speed of that particle in terms of x,t. From the relations dx = (dx'+vdt')g, dt = (dt'+vdx')g, we have dx/dt = (dx'+vdt')/(dt'+vdx'), and dividing the numerator and denominator by dt', this gives dx/dt = (u+v)/(1+uv).

> You have apparently hallucinated a 3rd object of interest for which we need to
> work out the coordinate velocity in the two FoRs of interest.

There is no hallucination, this is simply the derivation of the velocity composition formula, which involves three mutual speeds, namely the speed u of an object in terms of S', the speed v of S' in terms of S, and the speed (call it w) of the object in terms of S. As shown above, these three speeds are related by w = (u+v)/(1+uv).

> No third object in this scenario.

Which scenario? Are you talking about your fallacious idiocy, or are you talking about the derivation of the velocity composition formula? We've already dubunked your idiocy, and you seem to have shifted to the velocity composition, which involves 3 entities with 3 pairwise mutual speeds, as explained above.

> So, we are back to considering just the v between the origins of the two FoRs.

Well, you are the one who brought up the derivation of the velocity composition formula, even after I told you it wasn't relevant to your idiocy.

> "Can it be deductively demonstrated that the magnitude of the v in the LTs is equivalent
> for all pairs of FoRs?"

But you already conceded this point, remember? In a truly impressive display of intellectual integrity (I'm sincere about that) you announced that you had been persuaded that v'=v. Then you resorted (again) to your idiocy about deriving v=c, which has also been thoroughly debunked. Now you go back to your previous idiocy (forfeiting your newly won respect for displaying your intellectual integrity... didn't last long!) which you already admitted was idiocy. What gives?

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sat, 14 May 2022 18:50 UTC

On Saturday, May 14, 2022 at 11:31:50 AM UTC-7, Stan Fultoni wrote:
> On Saturday, May 14, 2022 at 11:02:41 AM UTC-7, patdolan wrote:
> > > For a particle moving at speed u in terms of x',t' we have dx'/dt' = u, and we wish to find the speed of that particle in terms of x,t. From the relations dx = (dx'+vdt')g, dt = (dt'+vdx')g, we have dx/dt = (dx'+vdt')/(dt'+vdx'), and dividing the numerator and denominator by dt', this gives dx/dt = (u+v)/(1+uv).
> > You have apparently hallucinated a 3rd object of interest for which we need to
> > work out the coordinate velocity in the two FoRs of interest.
> There is no hallucination, this is simply the derivation of the velocity composition formula, which involves three mutual speeds, namely the speed u of an object in terms of S', the speed v of S' in terms of S, and the speed (call it w) of the object in terms of S. As shown above, these three speeds are related by w = (u+v)/(1+uv).
> > No third object in this scenario.
> Which scenario? Are you talking about your fallacious idiocy, or are you talking about the derivation of the velocity composition formula? We've already dubunked your idiocy, and you seem to have shifted to the velocity composition, which involves 3 entities with 3 pairwise mutual speeds, as explained above.
>
> > So, we are back to considering just the v between the origins of the two FoRs.
>
> Well, you are the one who brought up the derivation of the velocity composition formula, even after I told you it wasn't relevant to your idiocy.
> > "Can it be deductively demonstrated that the magnitude of the v in the LTs is equivalent
> > for all pairs of FoRs?"
> But you already conceded this point, remember? In a truly impressive display of intellectual integrity (I'm sincere about that) you announced that you had been persuaded that v'=v. Then you resorted (again) to your idiocy about deriving v=c, which has also been thoroughly debunked. Now you go back to your previous idiocy (forfeiting your newly won respect for displaying your intellectual integrity... didn't last long!) which you already admitted was idiocy. What gives?

Stan, You obviously want to substitute your own problem for my problem. You are recalcitrant and ineluctable in this regard. Given your situation right now, I don't blame you. At least the vanquished rotchm had the honor to surrender his sword in defeat. And Dono scuttled back to his burrow.

What gives is that a better man set you up. Remember that the next time you want lecture relativity from your high throne. You believed in realtivity before you ever studied it, didn't you Stan. I seem to recall that I did too. Perhaps an even more profound line of inquiry is why you and I have chosen such different...

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Sat, 14 May 2022 19:02 UTC

On Saturday, May 14, 2022 at 11:50:14 AM UTC-7, patdolan wrote:
> > > > For a particle moving at speed u in terms of x',t' we have dx'/dt' = u, and we wish to find the speed of that particle in terms of x,t. From the relations dx = (dx'+vdt')g, dt = (dt'+vdx')g, we have dx/dt = (dx'+vdt')/(dt'+vdx'), and dividing the numerator and denominator by dt', this gives dx/dt = (u+v)/(1+uv).
> > > You have apparently hallucinated a 3rd object of interest for which we need to
> > > work out the coordinate velocity in the two FoRs of interest.
> > There is no hallucination, this is simply the derivation of the velocity composition formula, which involves three mutual speeds, namely the speed u of an object in terms of S', the speed v of S' in terms of S, and the speed (call it w) of the object in terms of S. As shown above, these three speeds are related by w = (u+v)/(1+uv).
> > > No third object in this scenario.
> > Which scenario? Are you talking about your fallacious idiocy, or are you talking about the derivation of the velocity composition formula? We've already dubunked your idiocy, and you seem to have shifted to the velocity composition, which involves 3 entities with 3 pairwise mutual speeds, as explained above.
> >
> > > So, we are back to considering just the v between the origins of the two FoRs.
> >
> > Well, you are the one who brought up the derivation of the velocity composition formula, even after I told you it wasn't relevant to your idiocy.
> > > "Can it be deductively demonstrated that the magnitude of the v in the LTs is equivalent
> > > for all pairs of FoRs?"
> > But you already conceded this point, remember? In a truly impressive display of intellectual integrity (I'm sincere about that) you announced that you had been persuaded that v'=v. Then you resorted (again) to your idiocy about deriving v=c, which has also been thoroughly debunked. Now you go back to your previous idiocy (forfeiting your newly won respect for displaying your intellectual integrity... didn't last long!) which you already admitted was idiocy. What gives?
>
> You obviously want to substitute your own problem for my problem.

Wait, you posed three different questions here, and I've thoroughly explained each one. First you claimed v' was not v, and that was debunked. Then you claimed v=c, and that was debunked. Then you claimed the velocity composition formula only involves one speed, and that was debunked. Now you're back to claiming v' is not v. Wow. Isn't it a little early in the day to be drunk already (or still)?

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: rot...@gmail.com (rotchm)
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 by: rotchm - Sun, 15 May 2022 00:52 UTC

On Saturday, May 14, 2022 at 12:17:54 PM UTC-4, patdolan wrote:

> 2) Dolan then went on to introduce the same possibility in the case of the Lorentz Transforms.
> But his proposal was in the negative. Dolan proposed what will one day be known as the Dolan
> Conjecture, namely, that v == v' it is an undecidable proposition in the case of the Lorentz Transforms.

And you were proven that v does equal v'.

> 3) At this point rotchm and Stan ...arguments purporting to prove that v == v'. But none of these arguments
> were deductive in nature and so were discarded out of hand.

A lie on your part.
You said such a proof could not be done. I gave you the proof.
You saw the proof and you acknowledged that I was right, that v = v'.
And no where did you argue that my proof was wrong. You simply agreed that it was correct.
Do you remember that you agreed that was the case?
If not, if your memory is failing you, Google kept the record for you.

> The Dolan Conjecture remained intact.

Not according to your own claims. Contradict yourself.
You are either a grade-a idiot, or a troll.

> 4) The wily Dolan was not yet done with protagonists rotchm & Stan. Dolan adroitly pivoted to
> accepting the validity of v == v' in the case of the LTs.

Now you contradict yourself from just above.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sun, 15 May 2022 02:30 UTC

On Saturday, May 14, 2022 at 5:52:59 PM UTC-7, rotchm wrote:
> On Saturday, May 14, 2022 at 12:17:54 PM UTC-4, patdolan wrote:
>
>
> > 2) Dolan then went on to introduce the same possibility in the case of the Lorentz Transforms.
> > But his proposal was in the negative. Dolan proposed what will one day be known as the Dolan
> > Conjecture, namely, that v == v' it is an undecidable proposition in the case of the Lorentz Transforms.
> And you were proven that v does equal v'.
>
> > 3) At this point rotchm and Stan ...arguments purporting to prove that v == v'. But none of these arguments
> > were deductive in nature and so were discarded out of hand.
> A lie on your part.
> You said such a proof could not be done. I gave you the proof.
> You saw the proof and you acknowledged that I was right, that v = v'.
> And no where did you argue that my proof was wrong. You simply agreed that it was correct.
> Do you remember that you agreed that was the case?
> If not, if your memory is failing you, Google kept the record for you.
> > The Dolan Conjecture remained intact.
> Not according to your own claims. Contradict yourself.
> You are either a grade-a idiot, or a troll.
> > 4) The wily Dolan was not yet done with protagonists rotchm & Stan. Dolan adroitly pivoted to
> > accepting the validity of v == v' in the case of the LTs.
> Now you contradict yourself from just above.
rotchm, give me your proof again, so that I may consider it anew.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: rot...@gmail.com (rotchm)
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 by: rotchm - Sun, 15 May 2022 02:49 UTC

On Saturday, May 14, 2022 at 10:30:31 PM UTC-4, patdolan wrote:

> rotchm, give me your proof again, so that I may consider it anew.

Are you saying that you do not have the skills to search this thread?
If you are lacking in such skills, then you have no business being here.

You are now obviously just a loser trolling.
Spam reported.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sun, 15 May 2022 03:09 UTC

On Saturday, May 14, 2022 at 7:49:28 PM UTC-7, rotchm wrote:
> On Saturday, May 14, 2022 at 10:30:31 PM UTC-4, patdolan wrote:
>
> > rotchm, give me your proof again, so that I may consider it anew.
> Are you saying that you do not have the skills to search this thread?
> If you are lacking in such skills, then you have no business being here.
>
> You are now obviously just a loser trolling.
> Spam reported.

You might as well report your wimpy bicep while you are reporting spam. I'll bet that bicep is about as soft and gooey as a loaf of spam too.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: rot...@gmail.com (rotchm)
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 by: rotchm - Sun, 15 May 2022 13:20 UTC

On Saturday, May 14, 2022 at 11:09:51 PM UTC-4, patdolan wrote:

> You might as well report your pythonic bicep while you are reporting spam.

In case you haven't noticed, your reply is off-topic.
Spam reported.

Re: Lying piece of shit Pat Dolan cannot follow simple algebra

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Date: Sun, 15 May 2022 16:46:37 -0700 (PDT)
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Subject: Re: Lying piece of shit Pat Dolan cannot follow simple algebra
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Sun, 15 May 2022 23:46 UTC

On Tuesday, May 10, 2022 at 6:21:00 PM UTC-7, rotchm wrote:
> On Tuesday, May 10, 2022 at 2:53:05 PM UTC-4, patdolan wrote:
>
> > Here are the assumptions for this proof:
> >
> > x' = g( v )[ x - vt ]
> > t' = g( v )[ t - vx/c^2 ]
> > x = g( v' )[ x' + v't' ]
> > t = g( v' )[ t' + v'x'/c^2 ]
> >
> > Here is the conclusion that is being asked for:
> > v' = v
> > No one will ever arrive at the conclusion.
> You are either trolling or a retard. And if you are trolling that you are a retard. A conclusion is you are a retard.
>
> x' = (x - vt )g
> t' = (t - vx/c²)g
>
> Expanding gives
>
> x' = xg - v(tg) [1]
> t' = (tg) - vxg/c² [2]
>
> Isolating tg in [2] gives
>
> (tg) = t' + vxg/c² [3]
>
> Replacing this (tg) into [1], expanding & simplifying gives
>
> x' = xg - vt' - xgv²/c². Rewrite as
>
> x' +vt' = xg - xgv²/c² = (factor out xg) = xg(1-v²/c²) = xg/g² = x/g. Rewrite as
>
> (x' +vt')g = x
>
> This is what you set out to prove (and you failed, even in the Galilean case, you failed).
> Note that I didn't even need your last two eqs, nor did I need to invoke a v'.
> By comparison/inspection, your v' equals v.
> This is simple highschool algebra, which you failed to do. Are you a retard or something?
> Seriously, you don't belong in this NG.

rotchm, is this the proof you proffer? If so then you are being just too pigheaded to understand the problem we are addressing here. STOP READING RIGHT HERE. STAND UP. LIFT ONE FOOT OFF THE FLOOR. RUB YOUR STOMACH AND PAT YOUR DOLAN...UM...YOUR HEAD AT THE SAME TIME. DO TEN JUMPING JACKS. TRY AND CURL 8LBS ONCE OR TWICE. NOW SIT BACK DOWN AND RESUME READING.

What we are doing in this exercise just for fun is assuming that a relativity doubter has claimed that the Lorentz transforms are incorrectly written and that the same velocity value v cannot be use in the following transformations:

S -> S' transforms
x' = g(v)[ x - vt ]
t' = g(v)[ t-vx/c^2 ]

S' -> S transforms
x = g(v)[ x' + vt' ]
x = g(v)[ t' + vx'/c^2 ]

The doubter claims that one velocity value v must be used in the S -> S' transforms while another velocity value v' must be used in the S' -> S transforms. Now how do we go about proving to this doubter that there is only one v for use in both sets of transforms? The best way is to assume the doubter's point by postulating two velocities v and v' for use in the transforms S -> S' and S -> S' then **derive** the fact that v = v' perforce.

If we assume the doubter's point of view, our starting point now becomes

S -> S' transforms
x' = g(v)[ x - vt ]
t' = g(v)[ t-vx/c^2 ]

S' -> S transforms
x = g(v')[ x' + v't' ]
x = g(v')[ t' + v'x'/c^2 ]

Do you understand what is being asked of you in this problem rotchm (and Stan)? Good. Now from this new starting point can you derive the string v = v' ? What about you Tom Roberts? Paul B. Anderson? Dirk is incapable of this sort of meta-algebraic reasoning. Bodkin would try to abort the problem before the first line of derivation began. Dono will fake a proof, showing only the last two lines of his supposed proof.

Re: Lying piece of shit Pat Dolan cannot follow simple algebra

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Subject: Re: Lying piece of shit Pat Dolan cannot follow simple algebra
From: fultonis...@gmail.com (Stan Fultoni)
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 by: Stan Fultoni - Mon, 16 May 2022 00:10 UTC

On Sunday, May 15, 2022 at 4:46:38 PM UTC-7, patdolan wrote:
> Can you derive the string v = v' ?

The trivial derivation has been presented to you many times, in many different forms. Again, it's grade school algebra to show that the first two equations imply
x = g( v )[ x' + vt' ]
t = g( v )[ t' + vx'/c^2 ]
so your second two equations are true for arbitrary t and x if and only if g(v)=g(v') and vg(v)=v'g(v'). From the first of these we have v' = +-v, and from the second we have that their signs are the same, so v'=v. What part of this do you think is wrong or unclear?

Re: Lying piece of shit Pat Dolan cannot follow simple algebra

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Subject: Re: Lying piece of shit Pat Dolan cannot follow simple algebra
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Mon, 16 May 2022 00:37 UTC

On Sunday, May 15, 2022 at 5:10:17 PM UTC-7, Stan Fultoni wrote:
> On Sunday, May 15, 2022 at 4:46:38 PM UTC-7, patdolan wrote:
> > Can you derive the string v = v' ?
>
> The trivial derivation has been presented to you many times, in many different forms. Again, it's grade school algebra to show that the first two equations imply
> x = g( v )[ x' + vt' ]
> t = g( v )[ t' + vx'/c^2 ]
> so your second two equations are true for arbitrary t and x if and only if g(v)=g(v') and vg(v)=v'g(v'). From the first of these we have v' = +-v, and from the second we have that their signs are the same, so v'=v. What part of this do you think is wrong or unclear?
Stan, you are hopeless. Nothing penetrates your thick skull. You probably don't even know that your argument is v = v' if and only if v = v'.

What is it about proofing that you just can't understand?

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