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tech / sci.physics.relativity / Re: Lying piece of shit Pat Dolan cannot follow simple algebra

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

Pages:123456
Re: How to Write a Transformation--Part One

<2bfc2e06-bda5-48c6-a0fe-c27fec8b5968n@googlegroups.com>

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Tue, 10 May 2022 16:21 UTC

On Tuesday, May 10, 2022 at 6:52:20 AM UTC-7, Stan Fultoni wrote:
> On Monday, May 9, 2022 at 11:34:18 PM UTC-7, patdolan wrote:
> > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > when employing the LTs. .. prove that v = v'
>
> Your two assumptions are self-contradictory unless v=v'. In matrix notation you
> are assuming the linear relationships X'=LX and X=L'X', and denying that L' = L^-1.
Stan, we are doing logic here to confirm the consistency of the LTs.

We are using the method of proof by contradiction. We postulate two velocities, v and v', and assume they are not equal. Then we attempt to prove otherwise.

You can see above that rotchm was not able to accomplish this task. Are you?

I will now proclaim to this forum that none of its denizens, nay, no one alive or dead, can or could have ever completed the proof of which rotchm's (lackluster) effort is the most recent example of failure.

Re: How to Write a Transformation--Part One

<55954198-060d-4acd-bf89-403c99ce9838n@googlegroups.com>

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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Tue, 10 May 2022 17:36 UTC

On Tuesday, May 10, 2022 at 6:24:50 AM UTC-7, rotchm wrote:
> On Tuesday, May 10, 2022 at 2:34:18 AM UTC-4, patdolan wrote:
> > On Monday, May 9, 2022 at 7:43:12 PM UTC-7, rotchm wrote:
>
> > > To be clear, you are asking to show that from x' = (x-vt)g & t' = (t-xv/c²)g we can get x = (x'+vt')g ?
> > No, rotchm. Just as in the Galilean case, we are going to assume that there is a v for frame S and
> > another v' for frame S' that each observer O and O' must use when employing the LTs. Your
> > mission is to prove that v = v' just as you did in the Galilean Transformation case.
> That's the merest of child's play algebra.
> Starting with v and v', or not using v' and just algebra, are equivalent. The "same thing".
> In both cases, you are using the algebra to show that x = (x'+vt')g.
>
> Galilean case:
>
> x' = x - vt (1).
> t = t' (2)
>
> Put (2) in (1) to get
> x' = x - vt' . Rearrange, giving,
>
> x = x' + vt' . (3)
> This is the form you wanted. Its the same "v" in (1)&(3). The magnitude of the coeff of t in (1)&(3) are the same.
> No need to assume an extra v' . Just using the prescribed algebra sufficed.
>
> Same idea can be applied to the LT's to get
> x = (x'+vt')g

Wrong, rotchm. WRONG! Your method for the Galilean v does not work for the Lorentz v. It does not work in the case of x = (x'+vt')g because t is not interchangeable with t' in the LTs. The calculation of t' involves the term xv/c^2. This dooms your supposition and any attempt you or anyone else might try. Prove me wrong.
> But if you want to show that you can prove it via a longer way, with mistakes and typo's, then go on...if it makes you happy.

Re: How to Write a Transformation--Part One

<fe924362-b9c6-4191-9fb3-06f63d93a68bn@googlegroups.com>

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Subject: Re: How to Write a Transformation--Part One
From: rot...@gmail.com (rotchm)
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 by: rotchm - Tue, 10 May 2022 17:39 UTC

On Tuesday, May 10, 2022 at 12:21:57 PM UTC-4, patdolan wrote:
> On Tuesday, May 10, 2022 at 6:52:20 AM UTC-7, Stan Fultoni wrote:
> > On Monday, May 9, 2022 at 11:34:18 PM UTC-7, patdolan wrote:
> > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > when employing the LTs. .. prove that v = v'
> >
> > Your two assumptions are self-contradictory unless v=v'. In matrix notation you
> > are assuming the linear relationships X'=LX and X=L'X', and denying that L' = L^-1.
>
> You can see above that rotchm was not able to accomplish this task. Are you?

A lie. It is trivial to show. And others have shown it to you.
So you fail at math, and you lie. Now I know u r a dishonest crank.

> I will now proclaim to this forum that none of its denizens, nay, no one alive or dead, can or
could have ever completed the proof

It has already been shown to you.
If I'm in the mood, perhaps later I will show u. And contrary to your
proofs (as in ur OP), I will not have mistakes.

Re: How to Write a Transformation--Part One

<f6f95fde-cefb-4f6f-ac44-e2b28ffc47a9n@googlegroups.com>

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Subject: Re: How to Write a Transformation--Part One
From: rot...@gmail.com (rotchm)
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 by: rotchm - Tue, 10 May 2022 17:43 UTC

On Tuesday, May 10, 2022 at 1:36:17 PM UTC-4, patdolan wrote:
> On Tuesday, May 10, 2022 at 6:24:50 AM UTC-7, rotchm wrote:
> > On Tuesday, May 10, 2022 at 2:34:18 AM UTC-4, patdolan wrote:
> > > On Monday, May 9, 2022 at 7:43:12 PM UTC-7, rotchm wrote:
> >
> > > > To be clear, you are asking to show that from x' = (x-vt)g & t' = (t-xv/c²)g we can get x = (x'+vt')g ?
> > > No, rotchm. Just as in the Galilean case, we are going to assume that there is a v for frame S and
> > > another v' for frame S' that each observer O and O' must use when employing the LTs. Your
> > > mission is to prove that v = v' just as you did in the Galilean Transformation case.
> > That's the merest of child's play algebra.
> > Starting with v and v', or not using v' and just algebra, are equivalent. The "same thing".
> > In both cases, you are using the algebra to show that x = (x'+vt')g.
> >
> > Galilean case:
> >
> > x' = x - vt (1).
> > t = t' (2)
> >
> > Put (2) in (1) to get
> > x' = x - vt' . Rearrange, giving,
> >
> > x = x' + vt' . (3)
> > This is the form you wanted. Its the same "v" in (1)&(3). The magnitude of the coeff of t in (1)&(3) are the same.
> > No need to assume an extra v' . Just using the prescribed algebra sufficed.
> >
> > Same idea can be applied to the LT's to get
> > x = (x'+vt')g
> Wrong, rotchm. WRONG! Your method for the Galilean v does not work for the Lorentz v. It does not work in the case of x = (x'+vt')g

Yes it does. This shows that ur nil at math&algebra.

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 - Tue, 10 May 2022 18:08 UTC

On Tuesday, May 10, 2022 at 10:40:01 AM UTC-7, rotchm wrote:
> On Tuesday, May 10, 2022 at 12:21:57 PM UTC-4, patdolan wrote:
> > On Tuesday, May 10, 2022 at 6:52:20 AM UTC-7, Stan Fultoni wrote:
> > > On Monday, May 9, 2022 at 11:34:18 PM UTC-7, patdolan wrote:
> > > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > > when employing the LTs. .. prove that v = v'
> > >
> > > Your two assumptions are self-contradictory unless v=v'. In matrix notation you
> > > are assuming the linear relationships X'=LX and X=L'X', and denying that L' = L^-1.
> >
> > You can see above that rotchm was not able to accomplish this task. Are you?
> A lie. It is trivial to show. And others have shown it to you.
> So you fail at math, and you lie. Now I know u r a dishonest crank.
> > I will now proclaim to this forum that none of its denizens, nay, no one alive or dead, can or
> could have ever completed the proof
> It has already been shown to you.
> If I'm in the mood, perhaps later I will show u. And contrary to your
> proofs (as in ur OP), I will not have mistakes.

If you are in the mood! Now it is YOU who is the liar. You can't do it. rotchm, you little wimp, I would have put your environmental candy ass right through the plexiglass.

Re: How to Write a Transformation--Part One

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Subject: Re: How to Write a Transformation--Part One
From: maluwozn...@gmail.com (Maciej Wozniak)
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 by: Maciej Wozniak - Tue, 10 May 2022 18:11 UTC

On Tuesday, 10 May 2022 at 19:43:31 UTC+2, rotchm wrote:
> On Tuesday, May 10, 2022 at 1:36:17 PM UTC-4, patdolan wrote:
> > On Tuesday, May 10, 2022 at 6:24:50 AM UTC-7, rotchm wrote:
> > > On Tuesday, May 10, 2022 at 2:34:18 AM UTC-4, patdolan wrote:
> > > > On Monday, May 9, 2022 at 7:43:12 PM UTC-7, rotchm wrote:
> > >
> > > > > To be clear, you are asking to show that from x' = (x-vt)g & t' = (t-xv/c²)g we can get x = (x'+vt')g ?
> > > > No, rotchm. Just as in the Galilean case, we are going to assume that there is a v for frame S and
> > > > another v' for frame S' that each observer O and O' must use when employing the LTs. Your
> > > > mission is to prove that v = v' just as you did in the Galilean Transformation case.
> > > That's the merest of child's play algebra.
> > > Starting with v and v', or not using v' and just algebra, are equivalent. The "same thing".
> > > In both cases, you are using the algebra to show that x = (x'+vt')g..
> > >
> > > Galilean case:
> > >
> > > x' = x - vt (1).
> > > t = t' (2)
> > >
> > > Put (2) in (1) to get
> > > x' = x - vt' . Rearrange, giving,
> > >
> > > x = x' + vt' . (3)
> > > This is the form you wanted. Its the same "v" in (1)&(3). The magnitude of the coeff of t in (1)&(3) are the same.
> > > No need to assume an extra v' . Just using the prescribed algebra sufficed.
> > >
> > > Same idea can be applied to the LT's to get
> > > x = (x'+vt')g
> > Wrong, rotchm. WRONG! Your method for the Galilean v does not work for the Lorentz v. It does not work in the case of x = (x'+vt')g
> Yes it does. This shows that ur nil at math&algebra.

Speaking of math, it's always good to remind that your
bunch of fanatics had to announce its oldest part false, as
it didn't want to support the madness of your idiot guru.

Lying piece of shit Pat Dolan cannot follow simple algebra

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Subject: Lying piece of shit Pat Dolan cannot follow simple algebra
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Tue, 10 May 2022 18:15 UTC

On Tuesday, May 10, 2022 at 10:36:17 AM UTC-7, odious troll pat dolan wrote:
> Prove me wrong.

I already did, in another thread, lying piece of shit. You were unable to follow the simple algebra, this is why you are here, trolling again.

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 - Tue, 10 May 2022 18:39 UTC

On Tuesday, May 10, 2022 at 9:21:57 AM UTC-7, patdolan wrote:
> > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > when employing the LTs. .. prove that v = v'
> >
> > Your two assumptions are self-contradictory unless v=v'. In matrix notation you
> > are assuming the linear relationships X'=LX and X=L'X', and denying that L' = L^-1.
>
> We postulate two velocities, v and v', and assume they are not equal. Then we attempt
> to prove otherwise.

Right, that's what I said. In matrix notation, L is the Lorentz transformation with
parameter v, and L' is the Lorentz transformation with parameter v', and you are
assuming the linear relationships X'=LX and X=L'X', from which it follows trivially
that L' = L^-1.

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 - Tue, 10 May 2022 18:53 UTC

On Tuesday, May 10, 2022 at 11:15:35 AM UTC-7, Dono. wrote:
> On Tuesday, May 10, 2022 at 10:36:17 AM UTC-7, odious troll pat dolan wrote:
> > Prove me wrong.
>
> I already did, in another thread, lying piece of shit. You were unable to follow the simple algebra, this is why you are here, trolling again.
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

Go.

No one will ever arrive at the conclusion. Once the insolubility of the this problem is acknowledged by Dono, rotchm, Stan, Dirk, Paul(s), Pyth & Pro, Jan, Townes, Sylvia, etc. we will then seminar on what this means for the LTs in Part Two.

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 - Tue, 10 May 2022 18:53 UTC

On Tuesday, May 10, 2022 at 11:39:34 AM UTC-7, Stan Fultoni wrote:
> On Tuesday, May 10, 2022 at 9:21:57 AM UTC-7, patdolan wrote:
> > > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > > when employing the LTs. .. prove that v = v'
> > >
> > > Your two assumptions are self-contradictory unless v=v'. In matrix notation you
> > > are assuming the linear relationships X'=LX and X=L'X', and denying that L' = L^-1.
> >
> > We postulate two velocities, v and v', and assume they are not equal. Then we attempt
> > to prove otherwise.
> Right, that's what I said. In matrix notation, L is the Lorentz transformation with
> parameter v, and L' is the Lorentz transformation with parameter v', and you are
> assuming the linear relationships X'=LX and X=L'X', from which it follows trivially
> that L' = L^-1.

Not even wrong, Stan.

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 - Tue, 10 May 2022 19:01 UTC

On Tuesday, May 10, 2022 at 11:53:05 AM UTC-7, patdolan wrote:
> On Tuesday, May 10, 2022 at 11:15:35 AM UTC-7, Dono. wrote:
> > On Tuesday, May 10, 2022 at 10:36:17 AM UTC-7, odious troll pat dolan wrote:
> > > Prove me wrong.
> >
> > I already did, in another thread, lying piece of shit. You were unable to follow the simple algebra, this is why you are here, trolling again.
> 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
>
> Go.
>
> No one will ever arrive at the conclusion. Once the insolubility of the this problem is acknowledged by Dono, rotchm, Stan, Dirk, Paul(s), Pyth & Pro, Jan, Townes, Sylvia, etc. we will then seminar on what this means for the LTs in Part Two.

Pro Tip: strip away any prior meaning that you may have endowed the equations with. Treat the equations as pure schema; then derivate with the conclusion as your goal. Tom Roberts, can you do this?

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 - Tue, 10 May 2022 19:02 UTC

On Tuesday, May 10, 2022 at 11:53:44 AM UTC-7, patdolan wrote:
> > > We postulate two velocities, v and v', and assume they are not equal. Then we attempt
> > > to prove otherwise.
> >
> > Right, that's what I said. In matrix notation, L is the Lorentz transformation with
> > parameter v, and L' is the Lorentz transformation with parameter v', and you are
> > assuming the linear relationships X'=LX and X=L'X', from which it follows trivially
> > that L' = L^-1.
>
> Not even wrong, Stan.

If there's something about the simple explanation you don't understand, feel free
to ask for clarification.

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 - Tue, 10 May 2022 19:11 UTC

On Tuesday, May 10, 2022 at 12:02:14 PM UTC-7, Stan Fultoni wrote:
> On Tuesday, May 10, 2022 at 11:53:44 AM UTC-7, patdolan wrote:
> > > > We postulate two velocities, v and v', and assume they are not equal. Then we attempt
> > > > to prove otherwise.
> > >
> > > Right, that's what I said. In matrix notation, L is the Lorentz transformation with
> > > parameter v, and L' is the Lorentz transformation with parameter v', and you are
> > > assuming the linear relationships X'=LX and X=L'X', from which it follows trivially
> > > that L' = L^-1.
> >
> > Not even wrong, Stan.
> If there's something about the simple explanation you don't understand, feel free
> to ask for clarification.
Stan, yes, there is something that I don't understand: why are you trying to cast MY problem in terms of another problem of your own creation? Why don't you work out a solution to MY problem first. Then show us how the solution to YOUR problem is equivalent.

Please let me know if there is something about my suggestion that you don't understand.

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 - Tue, 10 May 2022 22:56 UTC

On Tuesday, May 10, 2022 at 12:11:13 PM UTC-7, patdolan wrote:
> > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > when employing the LTs. .. prove that v = v'. We postulate two velocities, v and v',
> > > and assume they are not equal. Then we attempt to prove otherwise.
> > >
> > In matrix notation, L is the Lorentz transformation with parameter v, and L' is the
> > Lorentz transformation with parameter v', and you are assuming the linear
> > relationships X'=LX and X=L'X', from which it follows trivially that L' = L^-1.
>
> Why are you trying to cast MY problem in terms of another problem of your own creation?

The above is the simple answer to _your_ question. You stipulate two systems of inertial
coordinates, S and S', which consist of t,x and t',x' respectively, and you have stipulated
that they are related by a Lorentz transformation. That settles the physics. Everything
from here on is just algebra. Letting X and X' denote column vectors with the components
t,x and t',x' respectively, and letting L denote the matrix with components
g -vg
-vg g
where g=1/sqrt(1-v^2), your stipulation is expressed as X' = LX. This signifies that
X = (L^-1)X'. There is no ambiguity about this. It is simple linear algebra... like saying
that if p=5q then q=(5^-1)p. As you can see, the components of the Lorentz transformation
L^-1 are
g vg
vg g
which are the same as the components of L, except that v is replaced with -v.

Do you have any remaining questions?

Re: How to Write a Transformation--Part One

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Date: Tue, 10 May 2022 16:46:37 -0700 (PDT)
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Subject: Re: How to Write a Transformation--Part One
From: patdo...@comcast.net (patdolan)
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 by: patdolan - Tue, 10 May 2022 23:46 UTC

On Tuesday, May 10, 2022 at 3:56:23 PM UTC-7, Stan Fultoni wrote:
> On Tuesday, May 10, 2022 at 12:11:13 PM UTC-7, patdolan wrote:
> > > > we are going to assume that there is a v for frame S and another v' for frame S' ...
> > > > when employing the LTs. .. prove that v = v'. We postulate two velocities, v and v',
> > > > and assume they are not equal. Then we attempt to prove otherwise.
> > > >
> > > In matrix notation, L is the Lorentz transformation with parameter v, and L' is the
> > > Lorentz transformation with parameter v', and you are assuming the linear
> > > relationships X'=LX and X=L'X', from which it follows trivially that L' = L^-1.
> >
> > Why are you trying to cast MY problem in terms of another problem of your own creation?
>
> The above is the simple answer to _your_ question. You stipulate two systems of inertial
> coordinates, S and S', which consist of t,x and t',x' respectively, and you have stipulated

Stan,
> that they are related by a Lorentz transformation. That settles the physics. Everything
> from here on is just algebra.
I stopped reading your post after the above sentence.

The physics to be settled is exactly what you are taking for granted; namely, that the Lorentz Transforms accurately reflect the physics. What is being questioned is that, in a universe where the components of velocity ( distance and elapsed time) are relative, how is it that the velocity between two observers is always and everywhere absolute?

It appears to me that you have not yet thought enough about LTs to actually understand what we are debating here. Please prove to me that you do; otherwise, please refrain from disturbing this thread with further pedantic displays and off-topic posts.

Letting X and X' denote column vectors with the components
> t,x and t',x' respectively, and letting L denote the matrix with components
> g -vg
> -vg g
> where g=1/sqrt(1-v^2), your stipulation is expressed as X' = LX. This signifies that
> X = (L^-1)X'. There is no ambiguity about this. It is simple linear algebra... like saying
> that if p=5q then q=(5^-1)p. As you can see, the components of the Lorentz transformation
> L^-1 are
> g vg
> vg g
> which are the same as the components of L, except that v is replaced with -v.
>
> Do you have any remaining questions?

Re: How to Write a Transformation--Part One

<38288de2-3e6f-41ca-9d2c-25bb34fa3f81n@googlegroups.com>

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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 - Wed, 11 May 2022 00:35 UTC

On Tuesday, May 10, 2022 at 4:46:39 PM UTC-7, patdolan wrote:
> You stipulate two systems of inertial coordinates, S and S', which consist
> of t,x and t',x' respectively, and you have stipulated that they are related by
> a Lorentz transformation. That settles the physics. Everything from here
> on is just algebra.
>
> The physics to be settled is exactly what you are taking for granted; namely,
> that the Lorentz Transforms accurately reflect the physics.

No, you stipulated that the two systems of coordinates, t,x and t',x', are
related by a Lorentz transformation, which signifies local Lorentz invariance
and all of special relativity. There's no ambiguity about this, once you
have specified that the inertial coordinate systems t,x and t',x' are related by
a Lorentz transformation.

> In a universe where the components of velocity ( distance and elapsed time)
> are relative, how is it that the velocity between two observers is always and
> everywhere absolute?

That doesn't make any sense, and is based on misunderstanding and false
premises: First, any given path at a given event has a specific velocity in
terms of a specified system of coordinates. Any given object has infinitely
many different velocities, in terms of infinitely many different systems of
coordinates. The velocity of a given object in terms of coordinate system x,t
is dx/dt. This is coordinate-dependent, just as are the components of intervals.

Now, if you want to talk about the "velocity between two objects" as a grown-up
you need to specify precisely what you are referring to. For example, you can
refer to the quantity v2 - v1 between two objects with velocity v1 and v2
respectively in terms of some specified system of coordinates S", but of
course this is coordinate dependent. So the premise of your question is
false.

Do you have any remaining questions?

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: rot...@gmail.com (rotchm)
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 by: rotchm - Wed, 11 May 2022 01:20 UTC

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.

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 - Wed, 11 May 2022 01:27 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.

Where is v = v' ???????????????

You spend all afternoon and early evening trying to prove v = v'. You fail. Then you post to tell me what it is I "really" want to prove. What a light weight. Can you even skate backwards?

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 - Wed, 11 May 2022 01:28 UTC

On Tuesday, May 10, 2022 at 5:35:31 PM UTC-7, Stan Fultoni wrote:
> On Tuesday, May 10, 2022 at 4:46:39 PM UTC-7, patdolan wrote:
> > You stipulate two systems of inertial coordinates, S and S', which consist
> > of t,x and t',x' respectively, and you have stipulated that they are related by
> > a Lorentz transformation. That settles the physics. Everything from here
> > on is just algebra.
> >
> > The physics to be settled is exactly what you are taking for granted; namely,
> > that the Lorentz Transforms accurately reflect the physics.
> No, you stipulated that the two systems of coordinates, t,x and t',x', are
> related by a Lorentz transformation, which signifies local Lorentz invariance
> and all of special relativity. There's no ambiguity about this, once you
> have specified that the inertial coordinate systems t,x and t',x' are related by
> a Lorentz transformation.
>
> > In a universe where the components of velocity ( distance and elapsed time)
> > are relative, how is it that the velocity between two observers is always and
> > everywhere absolute?
> That doesn't make any sense, and is based on misunderstanding and false
> premises: First, any given path at a given event has a specific velocity in
> terms of a specified system of coordinates. Any given object has infinitely
> many different velocities, in terms of infinitely many different systems of
> coordinates. The velocity of a given object in terms of coordinate system x,t
> is dx/dt. This is coordinate-dependent, just as are the components of intervals.
>
> Now, if you want to talk about the "velocity between two objects" as a grown-up
> you need to specify precisely what you are referring to. For example, you can
> refer to the quantity v2 - v1 between two objects with velocity v1 and v2
> respectively in terms of some specified system of coordinates S", but of
> course this is coordinate dependent. So the premise of your question is
> false.
> Do you have any remaining questions?

First of all Stan, are you actually Townes Olson? Or is Townes Olson actually Stan?

Second, your response is sophomoric and does not warrant a reply while we await rotchm's derivation of v = v'.

I will reply/educate you later. But for now I do not wish to provide rotchm and the rest of them any possible point on which to pivot away from what they claim is easy to derive. Look how cowardly rotchm tried to wiggle out of the trap. You, Stan, have also shown your inability to do so. You have not produced a single line of a proof for anything other than your own stilted question.

But I issue the challenge to you again. Strip the symbols of all physical content then try to perform the deduction just for fun. Even though you won't attempt it, I reproduce the assumptions and challenge conclusion below..

From these wellformed strings of arithmetic:

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 ]

derive this wellformed string of arithmetic:

v = v'

Go.

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 - Wed, 11 May 2022 01:37 UTC

On Tuesday, May 10, 2022 at 6:27:07 PM UTC-7, patdolan wrote:
> 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.
> Where is v = v' ???????????????
>
> You spend all afternoon and early evening trying to prove v = v'. You fail. Then you post to tell me what it is I "really" want to prove. What a light weight. Can you even skate backwards?

Do you not understand that you are assuming in your derivation the there is only v and no other possibility of another, different velocity in as measured in another frame? This is the issue under debate. You utterly incapable fellow, rotchm. Rec league?

Crank Pat Dolan keeps on trolling

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Subject: Crank Pat Dolan keeps on trolling
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Wed, 11 May 2022 01:43 UTC

On Tuesday, May 10, 2022 at 6:28:57 PM UTC-7, crank pat dolan trolled:
> From these wellformed strings of arithmetic:
> 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 ]
> derive this wellformed string of arithmetic:
>
> v = v'
>
> Go.

I did exactly that a few weeks ago, piece of shit. You couldn't follow the simple algebra.

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 - Wed, 11 May 2022 01:47 UTC

On Tuesday, May 10, 2022 at 6:43:47 PM UTC-7, Dono. wrote:
> On Tuesday, May 10, 2022 at 6:28:57 PM UTC-7, crank pat dolan trolled:
> > From these wellformed strings of arithmetic:
> > 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 ]
> > derive this wellformed string of arithmetic:
> >
> > v = v'
> >
> > Go.
> I did exactly that a few weeks ago, piece of shit. You couldn't follow the simple algebra.
Then link to it, chump.

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. - Wed, 11 May 2022 01:55 UTC

On Tuesday, May 10, 2022 at 6:47:04 PM UTC-7, crank pat dolan ate shit:
> On Tuesday, May 10, 2022 at 6:43:47 PM UTC-7, Dono. wrote:
> > On Tuesday, May 10, 2022 at 6:28:57 PM UTC-7, crank pat dolan trolled:
> > > From these wellformed strings of arithmetic:
> > > 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 ]
> > > derive this wellformed string of arithmetic:
> > >
> > > v = v'
> > >
> > > Go.
> > I did exactly that a few weeks ago, piece of shit. You couldn't follow the simple algebra.
> Then link to it, chump.

Chump is your mother's son, piece of shit.

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: rot...@gmail.com (rotchm)
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 by: rotchm - Wed, 11 May 2022 01:56 UTC

On Tuesday, May 10, 2022 at 9:27:07 PM UTC-4, patdolan wrote:

> Where is v = v' ???????????????

<sigh>. I repeat: "By comparison/inspection, your v' equals v.".
You can't do that part on your own? Do you really need me to show you how to do that last step?

> You spend all afternoon and early evening trying to prove v = v'.

Nope. Another lie on your part. I just got home (from a day on the lake, kayaking...).
Took me a few mins to type it in. I know you're jealous, but you need to admit that you are limited and that you are a loser. You will never get better if you just insult and ask for answers.

Re: Crank Pat Dolan keeps on trolling

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Date: Tue, 10 May 2022 18:58:24 -0700 (PDT)
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Subject: Re: Crank Pat Dolan keeps on trolling
From: eggy2001...@gmail.com (Dono.)
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 by: Dono. - Wed, 11 May 2022 01:58 UTC

On Tuesday, May 10, 2022 at 6:43:47 PM UTC-7, Dono. wrote:
> On Tuesday, May 10, 2022 at 6:28:57 PM UTC-7, crank pat dolan trolled:
> > From these wellformed strings of arithmetic:
> > 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 ]
> > derive this wellformed string of arithmetic:
> >
> > v = v'
> >
> > Go.
> I did exactly that a few weeks ago, piece of shit. You couldn't follow the simple algebra.

After elementary algebra, the 4 equations boil down to:

(v-v')*(t'^2-x'^2/c^2)=0

with the only reasonable conclusion v'=v

Eat shit, Pattycakes

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