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Groups > sci.physics.relativity > #580849 > unrolled thread
| Started by | patdolan <patdolan@comcast.net> |
|---|---|
| First post | 2022-03-23 16:14 -0700 |
| Last post | 2022-03-30 12:34 -0700 |
| Articles | 20 on this page of 143 — 20 participants |
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v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-23 16:14 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-23 23:48 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-23 17:04 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 10:44 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-24 09:55 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 16:21 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-24 15:12 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 21:44 +0000
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 22:51 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-24 19:39 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 15:11 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-26 20:27 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-27 12:37 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-29 10:38 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-29 17:52 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-29 14:47 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-29 19:53 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-29 15:18 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-29 20:57 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-29 18:50 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-30 14:29 +0000
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-04-01 16:22 -0500
Re: v' = v in the Transforms Lorentz carl eto <carleto4157990662@gmail.com> - 2022-04-01 14:37 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-04-01 21:37 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-04-01 18:20 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-04-02 01:46 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-04-01 19:02 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-04-01 19:55 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-04-02 13:41 +0000
Cretin Pat Dolan admits he's the greatest imbecile "Dono." <eggy20011951@gmail.com> - 2022-04-02 07:20 -0700
Re: v' = v in the Transforms Lorentz Michael Moroney <moroney@world.std.spaamtrap.com> - 2022-04-02 00:35 -0400
Re: v' = v in the Transforms Lorentz Maciej Wozniak <maluwozniak@gmail.com> - 2022-03-29 22:15 -0700
Re: v' = v in the Transforms Lorentz RichD <r_delaney2001@yahoo.com> - 2022-03-27 14:33 -0700
Re: v' = v in the Transforms Lorentz whodat <whodaat@void.nowgre.com> - 2022-03-24 18:53 -0500
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 13:57 +0000
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-23 18:46 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-23 23:08 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-23 23:29 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-23 23:48 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 00:22 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 00:39 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 06:34 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 06:45 -0700
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 14:27 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 14:27 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 09:40 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 16:54 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 10:04 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 19:31 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 12:36 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 19:46 +0000
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 20:07 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 13:26 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 21:44 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 22:01 +0000
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 16:04 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 17:18 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 17:26 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 17:54 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 17:59 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 18:16 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 15:11 +0000
Re: v' = v in the Transforms Lorentz "Dono." <eggy20011951@gmail.com> - 2022-03-25 09:12 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 15:11 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-25 09:51 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-25 09:55 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 17:03 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-25 10:09 -0700
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 17:19 +0000
Cretin Pat Dolan perseveres "Dono." <eggy20011951@gmail.com> - 2022-03-25 10:39 -0700
Re: Cretin Pat Dolan perseveres patdolan <patdolan@comcast.net> - 2022-03-25 10:46 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 17:52 +0000
Re: Cretin Pat Dolan perseveres patdolan <patdolan@comcast.net> - 2022-03-25 10:59 -0700
Re: Cretin Pat Dolan perseveres patdolan <patdolan@comcast.net> - 2022-03-25 12:01 -0700
Re: Cretin Pat Dolan perseveres "Dono." <eggy20011951@gmail.com> - 2022-03-25 12:00 -0700
Village imbecile Pat Dolan inserts foot in mouth "Dono." <eggy20011951@gmail.com> - 2022-03-25 10:38 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-25 12:03 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-25 12:46 -0700
Re: v' = v in the Transforms Lorentz RichD <r_delaney2001@yahoo.com> - 2022-03-24 18:28 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 18:46 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 10:01 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 10:30 -0700
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 17:37 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 11:01 -0700
Cretin Pat Dolan perseveres "Dono." <eggy20011951@gmail.com> - 2022-03-24 11:24 -0700
Re: Cretin Pat Dolan perseveres patdolan <patdolan@comcast.net> - 2022-03-24 11:56 -0700
Re: Cretin Pat Dolan perseveres "Dono." <eggy20011951@gmail.com> - 2022-03-24 12:04 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 19:34 +0000
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 19:43 +0000
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-24 22:16 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-25 13:26 +0000
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-25 11:16 -0700
Re: Cretin Pat Dolan perseveres Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-03-25 14:58 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-26 00:52 +0000
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-25 21:58 -0700
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-27 23:56 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-28 13:09 +0000
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-29 11:48 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-29 19:25 +0000
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-30 11:07 -0700
Re: Cretin Pat Dolan perseveres Paul Alsing <pnalsing@gmail.com> - 2022-03-30 12:02 -0700
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-30 12:46 -0700
Re: Cretin Pat Dolan perseveres thor stoneman <consequently7990662@gmail.com> - 2022-03-30 13:23 -0700
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-30 21:31 -0700
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-30 22:10 -0700
Re: Cretin Pat Dolan perseveres The Starmaker <starmaker@ix.netcom.com> - 2022-03-31 10:40 -0700
Re: Cretin Pat Dolan perseveres carl eto <carleto4157990662@gmail.com> - 2022-04-01 13:34 -0700
Re: Cretin Pat Dolan perseveres Odd Bodkin <bodkinodd@gmail.com> - 2022-03-24 19:31 +0000
Re: v' = v in the Transforms Lorentz Michael Moroney <moroney@world.std.spaamtrap.com> - 2022-03-24 15:41 -0400
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 12:44 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 12:49 -0700
Re: v' = v in the Transforms Lorentz Maciej Wozniak <maluwozniak@gmail.com> - 2022-03-23 23:54 -0700
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 11:27 +0000
Re: v' = v in the Transforms Lorentz Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-03-26 00:45 +0100
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-25 17:01 -0700
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-26 10:48 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-26 10:38 +0000
Re: v' = v in the Transforms Lorentz Python <python@example.invalid> - 2022-03-26 16:25 +0100
Re: v' = v in the Transforms Lorentz Maciej Wozniak <maluwozniak@gmail.com> - 2022-03-26 08:37 -0700
Re: v' = v in the Transforms Lorentz Python <python@example.invalid> - 2022-03-26 16:41 +0100
Re: v' = v in the Transforms Lorentz Maciej Wozniak <maluwozniak@gmail.com> - 2022-03-26 10:51 -0700
Re: v' = v in the Transforms Lorentz Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-03-26 19:04 +0100
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-26 10:46 +0000
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-26 13:13 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-26 17:03 +0000
Re: v' = v in the Transforms Lorentz Odd Bodkin <bodkinodd@gmail.com> - 2022-03-26 17:12 +0000
Re: v' = v in the Transforms Lorentz Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2022-03-26 19:17 +0100
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-26 12:12 -0700
Re: v' = v in the Transforms Lorentz "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-03-24 13:43 +0100
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 13:53 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 14:01 +0000
Re: v' = v in the Transforms Lorentz Richard Hachel <r.hachel@tiscali.fr> - 2022-03-24 14:01 +0000
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 09:34 -0700
Re: v' = v in the Transforms Lorentz Dirk Van de moortel <dirkvandemoortel@notmail.com> - 2022-03-24 19:20 +0100
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 11:51 -0700
Re: v' = v in the Transforms Lorentz RichD <r_delaney2001@yahoo.com> - 2022-03-24 18:11 -0700
Re: v' = v in the Transforms Lorentz patdolan <patdolan@comcast.net> - 2022-03-24 18:16 -0700
Re: v' = v in the Transforms Lorentz Townes Olson <townesolson7@gmail.com> - 2022-03-24 18:25 -0700
Re: v' = v in the Transforms Lorentz RichD <r_delaney2001@yahoo.com> - 2022-03-24 18:46 -0700
Re: v' = v in the Transforms Lorentz Takabe Matsumura <tamu@frdesn.jp> - 2022-03-24 19:44 +0000
Re: v' = v in the Transforms Lorentz Takabe Matsumura <tamu@frdesn.jp> - 2022-03-24 20:00 +0000
Re: v' = v in the Transforms Lorentz Sylvia Else <sylvia@email.invalid> - 2022-03-25 11:21 +1100
Re: v' = v in the Transforms Lorentz carl eto <carleto4157990662@gmail.com> - 2022-03-30 12:34 -0700
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 00:39 -0700 |
| Message-ID | <7685ea6e-c544-493b-91cd-aa6441362f93n@googlegroups.com> |
| In reply to | #580864 |
On Thursday, March 24, 2022 at 12:22:14 AM UTC-7, Townes Olson wrote: > On Wednesday, March 23, 2022 at 11:48:23 PM UTC-7, patdolan wrote: > > > > > You're stipulating (with c=1) that x'=(x-vt)g and t'=(t-vx)g where g=1/sqrt(1-v^2). Multiplying through the equation for x' by v and adding it to the equation for t' gives vx'+t' = t(1-v^2)g = t/g, so we have t = (t'+vx')g. Likewise, multiplying through the equation for t' by v and adding it to the equation for x' gives vt'+x' = x(1-v^2)g = x/g, so we have x = (x'+vt')g. > > > > > > > You've ... demonstrated the fact that the transforms from S' to S can > > > > be derived from the transforms from S to S'. > > > > > > Right, with g=1/sqrt(1-v^2), the pair of equations x'=(x-vt)g and t'=(t-vx)g is algebraically equivalent to the pair of equations t=(t'+vx')g and x=(x'+vt')g, as shown above. > > > > Then it should be tautological and mere child's play for you to prime the v in the second > > pair of equations... > > You haven't defined what the phrase "prime the v" is supposed to mean. If you are defining the symbol v' to be identically equal to v, then indeed it is tautological that your symbol v' is identically equal to v. On the other hand, if you are defining the symbol v' to be your social security number, then it is tautologically your social security number. Tautologies of the kind you are engaged in have no cognitive significance. Remember, the dependence of g on v is fully accounted for in the grade school algebra that shows how the parameter v (defined as dx/dt of the spatial origin of S' in terms of S) appears in the transformation and its inverse. It follows per the above grade school algebra that dx'/dt' of the spatial origin of S in terms of S' is -v. See above. Townesy, let's forget all about Einstein, Relativity, velocity, time and distance. Let's consider four well constructed strings in algebra: x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] Now let's show that v' = v can be derived using the above four well constructed strings. Can you do it? Or do you decline?
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| From | Townes Olson <townesolson7@gmail.com> |
|---|---|
| Date | 2022-03-24 06:34 -0700 |
| Message-ID | <818825f4-156a-4367-aa61-06296592d689n@googlegroups.com> |
| In reply to | #580865 |
On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > Let's forget all about Relativity, velocity, time and distance. Right, your fundamental misunderstanding is much more elementary, and really has nothing to do with relativity in particular. > Let's consider four well constructed strings in algebra: > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > Now let's show that v' = v can be derived using the above four well constructed strings. That's a false proposition, because those equations imply only that v'= v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries reduce to the same kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable from arithmetic, and why you think v'=v is derivable from algebra. Your brain can't grasp the concept of non-linear functions.
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| From | Townes Olson <townesolson7@gmail.com> |
|---|---|
| Date | 2022-03-24 06:45 -0700 |
| Message-ID | <91ae80fa-1173-46f4-9d57-79da70e79cc5n@googlegroups.com> |
| In reply to | #580877 |
> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > > Let's forget all about Relativity, velocity, time and distance. > > Right, your fundamental misunderstanding is much more elementary, and really has nothing to do with relativity in particular. > > Let's consider four well constructed strings in algebra: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > Now let's show that v' = v can be derived using the above four well constructed strings. > That's a false proposition, because those equations imply only that v'= v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). That's actually the solution in terms of x' and t'. In terms of x and t it is v'=v or -v + 2xt(1-v^2)/(t^2-2txv+x^2). >You see, all your sophistries reduce to the same kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable from arithmetic, and why you think v'=v is derivable from algebra. Your brain can't grasp the concept of non-linear functions.
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| From | Richard Hachel <r.hachel@tiscali.fr> |
|---|---|
| Date | 2022-03-24 14:27 +0000 |
| Message-ID | <6NTkQMfaFEhNJ0PQ_BtCsK7vJd0@jntp> |
| In reply to | #580877 |
Le 24/03/2022 à 14:34, Townes Olson a écrit : > On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >> Let's forget all about Relativity, velocity, time and distance. > > Right, your fundamental misunderstanding is much more elementary, and really has > nothing to do with relativity in particular. > >> Let's consider four well constructed strings in algebra: >> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >> Now let's show that v' = v can be derived using the above four well constructed >> strings. > > That's a false proposition, because those equations imply only that v'= v or -v > - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries reduce to the same > kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b. > That's why you think 1=-1 is derivable from arithmetic, and why you think v'=v is > derivable from algebra. Your brain can't grasp the concept of non-linear > functions. There is a relation that gives x',y',z',To' in S', as a function of x,y,z,To in S. Ditto for cosµ and sinµ. Ditto also for the electromagnetic wavelengths (I gave all the correspondences) and on that, they are exactly the same as those of the physicists. So they can't be wrong. It is not possible to go wrong by going through two different paths. Let us take the case of a man who does not know how to read the signs, and who leaves Paris with a compass and an astrobal, and another leaves Paris with two totally different instruments. They must go to Rome. Arrived there, they are not sure, but, having gone through very different paths, they meet on the spot. It is then very likely that both are right: they are indeed in Rome. It would therefore be very surprising if all the equivalent calculations and equations given by Hachel and the scientists were wrong. That would be "weird". So they are definitely good. At least a priori. On the other hand, in some places (Langevin traveler, Tau Ceti traveler in an accelerated environment) there are very significant differences in the calculations. There is therefore, there, someone who is mistaken given the very significant discrepancies and the very different explanations (I have shown why it could not be me). But alone on a scale and all of them on the other pan, I can only bow to the number (and not to the reality of things). On the other hand, on the things that we have in common, and that I have accredited, I do not think it serious to contradict. The Lorentz transformations have a reciprocal transformation. To each event in R corresponds the same event in R (otherwise it is absurd), and vice versa. Here, it's very easy (the SR is very easy if you pay attention to the multitudes of little traps it conceals), you just have to change the direction of the displacement of the reference frame which slides on the other, and to transform v to -v. All the results are then concordant. If they don't match, it's because we made a small sign or calculation error. Otherwise, it's unstoppable. It works very well. Especially since nature is made like that, and we are talking about reality. R.H.
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| From | Richard Hachel <r.hachel@tiscali.fr> |
|---|---|
| Date | 2022-03-24 14:27 +0000 |
| Message-ID | <2EVwWGGfMmDpp_ztBXJWwvuOPag@jntp> |
| In reply to | #580877 |
Le 24/03/2022 à 14:34, Townes Olson a écrit : > On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >> Let's forget all about Relativity, velocity, time and distance. > > Right, your fundamental misunderstanding is much more elementary, and really has > nothing to do with relativity in particular. > >> Let's consider four well constructed strings in algebra: >> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >> Now let's show that v' = v can be derived using the above four well constructed >> strings. > > That's a false proposition, because those equations imply only that v'= v or -v > - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries reduce to the same > kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b. > That's why you think 1=-1 is derivable from arithmetic, and why you think v'=v is > derivable from algebra. Your brain can't grasp the concept of non-linear > functions. There is a relation that gives x',y',z',To' in S', as a function of x,y,z,To in S. Ditto for cosµ and sinµ. Ditto also for the electromagnetic wavelengths (I gave all the correspondences) and on that, they are exactly the same as those of the physicists. So they can't be wrong. It is not possible to go wrong by going through two different paths. Let us take the case of a man who does not know how to read the signs, and who leaves Paris with a compass and an astrobal, and another leaves Paris with two totally different instruments. They must go to Rome. Arrived there, they are not sure, but, having gone through very different paths, they meet on the spot. It is then very likely that both are right: they are indeed in Rome. It would therefore be very surprising if all the equivalent calculations and equations given by Hachel and the scientists were wrong. That would be "weird". So they are definitely good. At least a priori. On the other hand, in some places (Langevin traveler, Tau Ceti traveler in an accelerated environment) there are very significant differences in the calculations. There is therefore, there, someone who is mistaken given the very significant discrepancies and the very different explanations (I have shown why it could not be me). But alone on a scale and all of them on the other pan, I can only bow to the number (and not to the reality of things). On the other hand, on the things that we have in common, and that I have accredited, I do not think it serious to contradict. The Lorentz transformations have a reciprocal transformation. To each event in R corresponds the same event in R (otherwise it is absurd), and vice versa. Here, it's very easy (the SR is very easy if you pay attention to the multitudes of little traps it conceals), you just have to change the direction of the displacement of the reference frame which slides on the other, and to transform v to -v. All the results are then concordant. If they don't match, it's because we made a small sign or calculation error. Otherwise, it's unstoppable. It works very well. Especially since nature is made like that, and we are talking about reality. <http://news2.nemoweb.net/?DataID=2EVwWGGfMmDpp_ztBXJWwvuOPag@jntp> R.H. -- "Mais ne nous y trompons pas. Il n'y a pas que de la violence avec des armes. Il y a des situations de violence". Abbé Pierre.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 09:40 -0700 |
| Message-ID | <e29a811b-a2a4-41e2-aa76-b925aa604fc1n@googlegroups.com> |
| In reply to | #580877 |
On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: > On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > > Let's forget all about Relativity, velocity, time and distance. > > Right, your fundamental misunderstanding is much more elementary, and really has nothing to do with relativity in particular. > > Let's consider four well constructed strings in algebra: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > Now let's show that v' = v can be derived using the above four well constructed strings. > That's a false proposition, because those equations imply only that v'= v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries reduce to the same kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable from arithmetic, and why you think v'=v is derivable from algebra. Your brain can't grasp the concept of non-linear functions. Let me get this straight, Townes. You are saying that v' = v in my system of 4 equations is a false proposition. In other words, your testimony to this forum is that the equation v' = v can not be derived from the system of four equations. Correct?
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2022-03-24 16:54 +0000 |
| Message-ID | <t1i7ol$qu3$1@gioia.aioe.org> |
| In reply to | #580900 |
patdolan <patdolan@comcast.net> wrote: > On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >>> Let's forget all about Relativity, velocity, time and distance. >> >> Right, your fundamental misunderstanding is much more elementary, and >> really has nothing to do with relativity in particular. >>> Let's consider four well constructed strings in algebra: >>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >>> Now let's show that v' = v can be derived using the above four well constructed strings. >> That's a false proposition, because those equations imply only that v'= >> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >> reduce to the same kernel of idiocy, namely, your belief that the >> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >> from arithmetic, and why you think v'=v is derivable from algebra. Your >> brain can't grasp the concept of non-linear functions. > > Let me get this straight, Townes. You are saying that v' = v in my > system of 4 equations is a false proposition. In other words, your > testimony to this forum is that the equation > > v' = v > > can not be derived from the system of four equations. Correct? > No, that’s not true. You can do it the way I showed you in January, by using the transformation equations to track the location of the origin of one frame in the other frame. Or you can simply do algebra on the equations for x’ and t’ in terms of x, t, and v, to find equations for x and t in terms of x’, t’, and v, as both Paul and Townes showed you. Now you compare those to the equations of x and t in terms of x’, t’, and v’ as you wrote them. They are identical if and only if v=v’. It’s just algebra. -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 10:04 -0700 |
| Message-ID | <77d303f0-e093-47cb-aad5-98e5c9a49771n@googlegroups.com> |
| In reply to | #580903 |
On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: > patdolan <patd...@comcast.net> wrote: > > On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: > >> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > >>> Let's forget all about Relativity, velocity, time and distance. > >> > >> Right, your fundamental misunderstanding is much more elementary, and > >> really has nothing to do with relativity in particular. > >>> Let's consider four well constructed strings in algebra: > >>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > >>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > >>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > >>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > >>> Now let's show that v' = v can be derived using the above four well constructed strings. > >> That's a false proposition, because those equations imply only that v'= > >> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries > >> reduce to the same kernel of idiocy, namely, your belief that the > >> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable > >> from arithmetic, and why you think v'=v is derivable from algebra. Your > >> brain can't grasp the concept of non-linear functions. > > > > Let me get this straight, Townes. You are saying that v' = v in my > > system of 4 equations is a false proposition. In other words, your > > testimony to this forum is that the equation > > > > v' = v > > > > can not be derived from the system of four equations. Correct? > > > No, that’s not true. You can do it the way I showed you in January, by > using the transformation equations to track the location of the origin of > one frame in the other frame. > > Or you can simply do algebra on the equations for x’ and t’ in terms of x, > t, and v, to find equations for x and t in terms of x’, t’, and v, as both > Paul and Townes showed you. Now you compare those to the equations of x and > t in terms of x’, t’, and v’ as you wrote them. They are identical if and > only if v=v’. > > It’s just algebra. > -- > Odd Bodkin -- maker of fine toys, tools, tables If it's just algebra Bodkin, then why can't you, or Townes or Paul produce a theorem that reads "v' = v" from the four fundamental equations????
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2022-03-24 19:31 +0000 |
| Message-ID | <t1igtq$1gts$1@gioia.aioe.org> |
| In reply to | #580905 |
patdolan <patdolan@comcast.net> wrote: > On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: >> patdolan <patd...@comcast.net> wrote: >>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >>>>> Let's forget all about Relativity, velocity, time and distance. >>>> >>>> Right, your fundamental misunderstanding is much more elementary, and >>>> really has nothing to do with relativity in particular. >>>>> Let's consider four well constructed strings in algebra: >>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >>>>> Now let's show that v' = v can be derived using the above four well >>>>> constructed strings. >>>> That's a false proposition, because those equations imply only that v'= >>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >>>> reduce to the same kernel of idiocy, namely, your belief that the >>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >>>> from arithmetic, and why you think v'=v is derivable from algebra. Your >>>> brain can't grasp the concept of non-linear functions. >>> >>> Let me get this straight, Townes. You are saying that v' = v in my >>> system of 4 equations is a false proposition. In other words, your >>> testimony to this forum is that the equation >>> >>> v' = v >>> >>> can not be derived from the system of four equations. Correct? >>> >> No, that’s not true. You can do it the way I showed you in January, by >> using the transformation equations to track the location of the origin of >> one frame in the other frame. >> >> Or you can simply do algebra on the equations for x’ and t’ in terms of x, >> t, and v, to find equations for x and t in terms of x’, t’, and v, as both >> Paul and Townes showed you. Now you compare those to the equations of x and >> t in terms of x’, t’, and v’ as you wrote them. They are identical if and >> only if v=v’. >> >> It’s just algebra. >> -- >> Odd Bodkin -- maker of fine toys, tools, tables > > If it's just algebra Bodkin, then why can't you, or Townes or Paul > produce a theorem that reads "v' = v" from the four fundamental equations???? > Read what I just said. Paul and Townes *just* walked you through that algebra. Can you not even TRY to fill in some of the trivial steps yourself? Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz transformation equations. Write them out in full glory. Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) and t(x’,t’,v). You’ve done nothing but algebra. Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll see they are the same only if v=v’. -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 12:36 -0700 |
| Message-ID | <69f9bd07-e3b1-4220-bbd8-ff3db951af6bn@googlegroups.com> |
| In reply to | #580923 |
On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: > patdolan <patd...@comcast.net> wrote: > > On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: > >> patdolan <patd...@comcast.net> wrote: > >>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: > >>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > >>>>> Let's forget all about Relativity, velocity, time and distance. > >>>> > >>>> Right, your fundamental misunderstanding is much more elementary, and > >>>> really has nothing to do with relativity in particular. > >>>>> Let's consider four well constructed strings in algebra: > >>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > >>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > >>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > >>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > >>>>> Now let's show that v' = v can be derived using the above four well > >>>>> constructed strings. > >>>> That's a false proposition, because those equations imply only that v'= > >>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries > >>>> reduce to the same kernel of idiocy, namely, your belief that the > >>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable > >>>> from arithmetic, and why you think v'=v is derivable from algebra. Your > >>>> brain can't grasp the concept of non-linear functions. > >>> > >>> Let me get this straight, Townes. You are saying that v' = v in my > >>> system of 4 equations is a false proposition. In other words, your > >>> testimony to this forum is that the equation > >>> > >>> v' = v > >>> > >>> can not be derived from the system of four equations. Correct? > >>> > >> No, that’s not true. You can do it the way I showed you in January, by > >> using the transformation equations to track the location of the origin of > >> one frame in the other frame. > >> > >> Or you can simply do algebra on the equations for x’ and t’ in terms of x, > >> t, and v, to find equations for x and t in terms of x’, t’, and v, as both > >> Paul and Townes showed you. Now you compare those to the equations of x and > >> t in terms of x’, t’, and v’ as you wrote them. They are identical if and > >> only if v=v’. > >> > >> It’s just algebra. > >> -- > >> Odd Bodkin -- maker of fine toys, tools, tables > > > > If it's just algebra Bodkin, then why can't you, or Townes or Paul > > produce a theorem that reads "v' = v" from the four fundamental equations???? > > > Read what I just said. Paul and Townes *just* walked you through that > algebra. Can you not even TRY to fill in some of the trivial steps > yourself? > > Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz > transformation equations. Write them out in full glory. > > Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) > and t(x’,t’,v). You’ve done nothing but algebra. > > Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll > see they are the same only if v=v’. > -- > Odd Bodkin -- maker of fine toys, tools, tables Bodkin, I tell you that neither Townes nor Paul did that because nobody on earth can do that. Not even Dirk. Townes simply typed out a "v" then an apostrophe, then an equals sign, then another "v". That's all the effort and all the thought he put into it. Don't believe me? Watch, Townes will NEVER provide a derivation.
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2022-03-24 19:46 +0000 |
| Message-ID | <t1ihpp$1u52$1@gioia.aioe.org> |
| In reply to | #580926 |
patdolan <patdolan@comcast.net> wrote: > On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: >> patdolan <patd...@comcast.net> wrote: >>> On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: >>>> patdolan <patd...@comcast.net> wrote: >>>>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >>>>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >>>>>>> Let's forget all about Relativity, velocity, time and distance. >>>>>> >>>>>> Right, your fundamental misunderstanding is much more elementary, and >>>>>> really has nothing to do with relativity in particular. >>>>>>> Let's consider four well constructed strings in algebra: >>>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >>>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >>>>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >>>>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >>>>>>> Now let's show that v' = v can be derived using the above four well >>>>>>> constructed strings. >>>>>> That's a false proposition, because those equations imply only that v'= >>>>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >>>>>> reduce to the same kernel of idiocy, namely, your belief that the >>>>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >>>>>> from arithmetic, and why you think v'=v is derivable from algebra. Your >>>>>> brain can't grasp the concept of non-linear functions. >>>>> >>>>> Let me get this straight, Townes. You are saying that v' = v in my >>>>> system of 4 equations is a false proposition. In other words, your >>>>> testimony to this forum is that the equation >>>>> >>>>> v' = v >>>>> >>>>> can not be derived from the system of four equations. Correct? >>>>> >>>> No, that’s not true. You can do it the way I showed you in January, by >>>> using the transformation equations to track the location of the origin of >>>> one frame in the other frame. >>>> >>>> Or you can simply do algebra on the equations for x’ and t’ in terms of x, >>>> t, and v, to find equations for x and t in terms of x’, t’, and v, as both >>>> Paul and Townes showed you. Now you compare those to the equations of x and >>>> t in terms of x’, t’, and v’ as you wrote them. They are identical if and >>>> only if v=v’. >>>> >>>> It’s just algebra. >>>> -- >>>> Odd Bodkin -- maker of fine toys, tools, tables >>> >>> If it's just algebra Bodkin, then why can't you, or Townes or Paul >>> produce a theorem that reads "v' = v" from the four fundamental equations???? >>> >> Read what I just said. Paul and Townes *just* walked you through that >> algebra. Can you not even TRY to fill in some of the trivial steps >> yourself? >> >> Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz >> transformation equations. Write them out in full glory. >> >> Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) >> and t(x’,t’,v). You’ve done nothing but algebra. >> >> Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll >> see they are the same only if v=v’. >> -- >> Odd Bodkin -- maker of fine toys, tools, tables > > Bodkin, I tell you that neither Townes nor Paul did that because nobody > on earth can do that. Not even Dirk. Townes simply typed out a "v" then > an apostrophe, then an equals sign, then another "v". That's all the > effort and all the thought he put into it. Don't believe me? Watch, > Townes will NEVER provide a derivation. > You are saying, Pat, that no, you cannot do 5th grade algebra. And furthermore, you are saying NO ONE can do 5th grade algebra, until they show you how to do 5th grade algebra on your demand. That’s really what you want to say? -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2022-03-24 20:07 +0000 |
| Message-ID | <t1ij2t$h5f$1@gioia.aioe.org> |
| In reply to | #580931 |
Odd Bodkin <bodkinodd@gmail.com> wrote: > patdolan <patdolan@comcast.net> wrote: >> On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: >>> patdolan <patd...@comcast.net> wrote: >>>> On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: >>>>> patdolan <patd...@comcast.net> wrote: >>>>>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >>>>>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >>>>>>>> Let's forget all about Relativity, velocity, time and distance. >>>>>>> >>>>>>> Right, your fundamental misunderstanding is much more elementary, and >>>>>>> really has nothing to do with relativity in particular. >>>>>>>> Let's consider four well constructed strings in algebra: >>>>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >>>>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >>>>>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >>>>>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >>>>>>>> Now let's show that v' = v can be derived using the above four well >>>>>>>> constructed strings. >>>>>>> That's a false proposition, because those equations imply only that v'= >>>>>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >>>>>>> reduce to the same kernel of idiocy, namely, your belief that the >>>>>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >>>>>>> from arithmetic, and why you think v'=v is derivable from algebra. Your >>>>>>> brain can't grasp the concept of non-linear functions. >>>>>> >>>>>> Let me get this straight, Townes. You are saying that v' = v in my >>>>>> system of 4 equations is a false proposition. In other words, your >>>>>> testimony to this forum is that the equation >>>>>> >>>>>> v' = v >>>>>> >>>>>> can not be derived from the system of four equations. Correct? >>>>>> >>>>> No, that’s not true. You can do it the way I showed you in January, by >>>>> using the transformation equations to track the location of the origin of >>>>> one frame in the other frame. >>>>> >>>>> Or you can simply do algebra on the equations for x’ and t’ in terms of x, >>>>> t, and v, to find equations for x and t in terms of x’, t’, and v, as both >>>>> Paul and Townes showed you. Now you compare those to the equations of x and >>>>> t in terms of x’, t’, and v’ as you wrote them. They are identical if and >>>>> only if v=v’. >>>>> >>>>> It’s just algebra. >>>>> -- >>>>> Odd Bodkin -- maker of fine toys, tools, tables >>>> >>>> If it's just algebra Bodkin, then why can't you, or Townes or Paul >>>> produce a theorem that reads "v' = v" from the four fundamental equations???? >>>> >>> Read what I just said. Paul and Townes *just* walked you through that >>> algebra. Can you not even TRY to fill in some of the trivial steps >>> yourself? >>> >>> Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz >>> transformation equations. Write them out in full glory. >>> >>> Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) >>> and t(x’,t’,v). You’ve done nothing but algebra. >>> >>> Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll >>> see they are the same only if v=v’. >>> -- >>> Odd Bodkin -- maker of fine toys, tools, tables >> >> Bodkin, I tell you that neither Townes nor Paul did that because nobody >> on earth can do that. Not even Dirk. Townes simply typed out a "v" then >> an apostrophe, then an equals sign, then another "v". That's all the >> effort and all the thought he put into it. Don't believe me? Watch, >> Townes will NEVER provide a derivation. >> > > You are saying, Pat, that no, you cannot do 5th grade algebra. And > furthermore, you are saying NO ONE can do 5th grade algebra, until they > show you how to do 5th grade algebra on your demand. > > That’s really what you want to say? > By the way, Pat, that 5th grade algebra that you say NO ONE can do, is in fact in Chapter 4 of that book you said you’d work through. -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 13:26 -0700 |
| Message-ID | <772e0712-f03e-4608-9523-10de74664ff1n@googlegroups.com> |
| In reply to | #580931 |
On Thursday, March 24, 2022 at 12:46:04 PM UTC-7, bodk...@gmail.com wrote: > patdolan <patd...@comcast.net> wrote: > > On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: > >> patdolan <patd...@comcast.net> wrote: > >>> On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: > >>>> patdolan <patd...@comcast.net> wrote: > >>>>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: > >>>>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: > >>>>>>> Let's forget all about Relativity, velocity, time and distance. > >>>>>> > >>>>>> Right, your fundamental misunderstanding is much more elementary, and > >>>>>> really has nothing to do with relativity in particular. > >>>>>>> Let's consider four well constructed strings in algebra: > >>>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > >>>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > >>>>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > >>>>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > >>>>>>> Now let's show that v' = v can be derived using the above four well > >>>>>>> constructed strings. > >>>>>> That's a false proposition, because those equations imply only that v'= > >>>>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries > >>>>>> reduce to the same kernel of idiocy, namely, your belief that the > >>>>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable > >>>>>> from arithmetic, and why you think v'=v is derivable from algebra. Your > >>>>>> brain can't grasp the concept of non-linear functions. > >>>>> > >>>>> Let me get this straight, Townes. You are saying that v' = v in my > >>>>> system of 4 equations is a false proposition. In other words, your > >>>>> testimony to this forum is that the equation > >>>>> > >>>>> v' = v > >>>>> > >>>>> can not be derived from the system of four equations. Correct? > >>>>> > >>>> No, that’s not true. You can do it the way I showed you in January, by > >>>> using the transformation equations to track the location of the origin of > >>>> one frame in the other frame. > >>>> > >>>> Or you can simply do algebra on the equations for x’ and t’ in terms of x, > >>>> t, and v, to find equations for x and t in terms of x’, t’, and v, as both > >>>> Paul and Townes showed you. Now you compare those to the equations of x and > >>>> t in terms of x’, t’, and v’ as you wrote them. They are identical if and > >>>> only if v=v’. > >>>> > >>>> It’s just algebra. > >>>> -- > >>>> Odd Bodkin -- maker of fine toys, tools, tables > >>> > >>> If it's just algebra Bodkin, then why can't you, or Townes or Paul > >>> produce a theorem that reads "v' = v" from the four fundamental equations???? > >>> > >> Read what I just said. Paul and Townes *just* walked you through that > >> algebra. Can you not even TRY to fill in some of the trivial steps > >> yourself? > >> > >> Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz > >> transformation equations. Write them out in full glory. > >> > >> Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) > >> and t(x’,t’,v). You’ve done nothing but algebra. > >> > >> Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll > >> see they are the same only if v=v’. > >> -- > >> Odd Bodkin -- maker of fine toys, tools, tables > > > > Bodkin, I tell you that neither Townes nor Paul did that because nobody > > on earth can do that. Not even Dirk. Townes simply typed out a "v" then > > an apostrophe, then an equals sign, then another "v". That's all the > > effort and all the thought he put into it. Don't believe me? Watch, > > Townes will NEVER provide a derivation. > > > You are saying, Pat, that no, you cannot do 5th grade algebra. And > furthermore, you are saying NO ONE can do 5th grade algebra, until they > show you how to do 5th grade algebra on your demand. > > That’s really what you want to say? > -- > Odd Bodkin -- maker of fine toys, tools, tables I'll say this, Bodkin: There is no pathway, comprised of validly deduced equations, that leads from these four premises of well-formed algebraic strings: x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] to this well-formed conclusion string: v' = v PROVE ME WRONG!!! Respectfully, TMWBTLTs
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| From | Odd Bodkin <bodkinodd@gmail.com> |
|---|---|
| Date | 2022-03-24 21:44 +0000 |
| Message-ID | <t1iood$17oh$2@gioia.aioe.org> |
| In reply to | #580937 |
patdolan <patdolan@comcast.net> wrote: > On Thursday, March 24, 2022 at 12:46:04 PM UTC-7, bodk...@gmail.com wrote: >> patdolan <patd...@comcast.net> wrote: >>> On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: >>>> patdolan <patd...@comcast.net> wrote: >>>>> On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: >>>>>> patdolan <patd...@comcast.net> wrote: >>>>>>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >>>>>>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >>>>>>>>> Let's forget all about Relativity, velocity, time and distance. >>>>>>>> >>>>>>>> Right, your fundamental misunderstanding is much more elementary, and >>>>>>>> really has nothing to do with relativity in particular. >>>>>>>>> Let's consider four well constructed strings in algebra: >>>>>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >>>>>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >>>>>>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >>>>>>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >>>>>>>>> Now let's show that v' = v can be derived using the above four well >>>>>>>>> constructed strings. >>>>>>>> That's a false proposition, because those equations imply only that v'= >>>>>>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >>>>>>>> reduce to the same kernel of idiocy, namely, your belief that the >>>>>>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >>>>>>>> from arithmetic, and why you think v'=v is derivable from algebra. Your >>>>>>>> brain can't grasp the concept of non-linear functions. >>>>>>> >>>>>>> Let me get this straight, Townes. You are saying that v' = v in my >>>>>>> system of 4 equations is a false proposition. In other words, your >>>>>>> testimony to this forum is that the equation >>>>>>> >>>>>>> v' = v >>>>>>> >>>>>>> can not be derived from the system of four equations. Correct? >>>>>>> >>>>>> No, that’s not true. You can do it the way I showed you in January, by >>>>>> using the transformation equations to track the location of the origin of >>>>>> one frame in the other frame. >>>>>> >>>>>> Or you can simply do algebra on the equations for x’ and t’ in terms of x, >>>>>> t, and v, to find equations for x and t in terms of x’, t’, and v, as both >>>>>> Paul and Townes showed you. Now you compare those to the equations of x and >>>>>> t in terms of x’, t’, and v’ as you wrote them. They are identical if and >>>>>> only if v=v’. >>>>>> >>>>>> It’s just algebra. >>>>>> -- >>>>>> Odd Bodkin -- maker of fine toys, tools, tables >>>>> >>>>> If it's just algebra Bodkin, then why can't you, or Townes or Paul >>>>> produce a theorem that reads "v' = v" from the four fundamental equations???? >>>>> >>>> Read what I just said. Paul and Townes *just* walked you through that >>>> algebra. Can you not even TRY to fill in some of the trivial steps >>>> yourself? >>>> >>>> Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz >>>> transformation equations. Write them out in full glory. >>>> >>>> Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) >>>> and t(x’,t’,v). You’ve done nothing but algebra. >>>> >>>> Now compare these equations with your x(x’,t’,v’) and t(x’,t’,v’). You’ll >>>> see they are the same only if v=v’. >>>> -- >>>> Odd Bodkin -- maker of fine toys, tools, tables >>> >>> Bodkin, I tell you that neither Townes nor Paul did that because nobody >>> on earth can do that. Not even Dirk. Townes simply typed out a "v" then >>> an apostrophe, then an equals sign, then another "v". That's all the >>> effort and all the thought he put into it. Don't believe me? Watch, >>> Townes will NEVER provide a derivation. >>> >> You are saying, Pat, that no, you cannot do 5th grade algebra. And >> furthermore, you are saying NO ONE can do 5th grade algebra, until they >> show you how to do 5th grade algebra on your demand. >> >> That’s really what you want to say? >> -- >> Odd Bodkin -- maker of fine toys, tools, tables > I'll say this, Bodkin: > > There is no pathway, comprised of validly deduced equations, that leads > from these four premises of well-formed algebraic strings: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > to this well-formed conclusion string: > > v' = v > > PROVE ME WRONG!!! > > Respectfully, > > TMWBTLTs > Do you, or do you not, know how to do the 5th grade algebraic steps outlined to you by others here? Can you, or can you not, find those algebraic steps in the cited chapter of the cited book in your possession? -- Odd Bodkin -- maker of fine toys, tools, tables
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| From | Richard Hachel <r.hachel@tiscali.fr> |
|---|---|
| Date | 2022-03-24 22:01 +0000 |
| Message-ID | <-2L6ROTAXlV7SYg10shfVKTpkA8@jntp> |
| In reply to | #580937 |
Le 24/03/2022 à 21:26, patdolan a écrit : > On Thursday, March 24, 2022 at 12:46:04 PM UTC-7, bodk...@gmail.com wrote: >> patdolan <patd...@comcast.net> wrote: >> > On Thursday, March 24, 2022 at 12:31:10 PM UTC-7, bodk...@gmail.com wrote: >> >> patdolan <patd...@comcast.net> wrote: >> >>> On Thursday, March 24, 2022 at 9:54:50 AM UTC-7, bodk...@gmail.com wrote: >> >>>> patdolan <patd...@comcast.net> wrote: >> >>>>> On Thursday, March 24, 2022 at 6:34:16 AM UTC-7, Townes Olson wrote: >> >>>>>> On Thursday, March 24, 2022 at 12:39:32 AM UTC-7, patdolan wrote: >> >>>>>>> Let's forget all about Relativity, velocity, time and distance. >> >>>>>> >> >>>>>> Right, your fundamental misunderstanding is much more elementary, and >> >>>>>> really has nothing to do with relativity in particular. >> >>>>>>> Let's consider four well constructed strings in algebra: >> >>>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] >> >>>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] >> >>>>>>> x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] >> >>>>>>> t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] >> >>>>>>> Now let's show that v' = v can be derived using the above four well >> >>>>>>> constructed strings. >> >>>>>> That's a false proposition, because those equations imply only that v'= >> >>>>>> v or -v - 2xt(1-v^2)/(t^2+2txv+x^2). You see, all your sophistries >> >>>>>> reduce to the same kernel of idiocy, namely, your belief that the >> >>>>>> relation f(a)=f(b) implies a=b. That's why you think 1=-1 is derivable >> >>>>>> from arithmetic, and why you think v'=v is derivable from algebra. Your >> >>>>>> brain can't grasp the concept of non-linear functions. >> >>>>> >> >>>>> Let me get this straight, Townes. You are saying that v' = v in my >> >>>>> system of 4 equations is a false proposition. In other words, your >> >>>>> testimony to this forum is that the equation >> >>>>> >> >>>>> v' = v >> >>>>> >> >>>>> can not be derived from the system of four equations. Correct? >> >>>>> >> >>>> No, that’s not true. You can do it the way I showed you in January, by >> >>>> using the transformation equations to track the location of the origin of >> >>>> one frame in the other frame. >> >>>> >> >>>> Or you can simply do algebra on the equations for x’ and t’ in terms of >> x, >> >>>> t, and v, to find equations for x and t in terms of x’, t’, and v, as >> both >> >>>> Paul and Townes showed you. Now you compare those to the equations of x and >> >> >>>> t in terms of x’, t’, and v’ as you wrote them. They are identical if >> and >> >>>> only if v=v’. >> >>>> >> >>>> It’s just algebra. >> >>>> -- >> >>>> Odd Bodkin -- maker of fine toys, tools, tables >> >>> >> >>> If it's just algebra Bodkin, then why can't you, or Townes or Paul >> >>> produce a theorem that reads "v' = v" from the four fundamental equations? ? >> ? ? >> >>> >> >> Read what I just said. Paul and Townes *just* walked you through that >> >> algebra. Can you not even TRY to fill in some of the trivial steps >> >> yourself? >> >> >> >> Start with x’(x,t,v) and t’(x,v,t). Those are shorthand for the Lorentz >> >> transformation equations. Write them out in full glory. >> >> >> >> Now do the algebra to solve for x and t, so that you now have x(x’,t’,v) >> >> and t(x’,t’,v). You’ve done nothing but algebra. >> >> >> >> Now compare these equations with your x(x’,t’,v’) and >> t(x’,t’,v’). You’ll >> >> see they are the same only if v=v’. >> >> -- >> >> Odd Bodkin -- maker of fine toys, tools, tables >> > >> > Bodkin, I tell you that neither Townes nor Paul did that because nobody >> > on earth can do that. Not even Dirk. Townes simply typed out a "v" then >> > an apostrophe, then an equals sign, then another "v". That's all the >> > effort and all the thought he put into it. Don't believe me? Watch, >> > Townes will NEVER provide a derivation. >> > >> You are saying, Pat, that no, you cannot do 5th grade algebra. And >> furthermore, you are saying NO ONE can do 5th grade algebra, until they >> show you how to do 5th grade algebra on your demand. >> >> That’s really what you want to say? >> -- >> Odd Bodkin -- maker of fine toys, tools, tables > I'll say this, Bodkin: > > There is no pathway, comprised of validly deduced equations, that leads from > these four premises of well-formed algebraic strings: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > to this well-formed conclusion string: > > v' = v > > PROVE ME WRONG!!! > > Respectfully, > > TMWBTLTs Vo=0.8c x=12 y=9 z=0 To=-15 (local time) t=0 (watch O proper time) x'=40 y'=9 z'=0 To'=-41 (local time) t'=0 (watch O' proper time) x' = (x-v.To)/sqrt(1-v²/c²) To' = (To-v.x/c²)/sqrt(1-v²/c²) x = (x'-v.To')/sqrt(1-v²/c²) To = (To'-v.x'/c²)/sqrt(1-v²/c²) These equations, I found them myself starting from the principle of invariance of the observable transverse speed of light (ie Vo=c). They are correct. Henri Poincaré was not mistaken. Do the mathematical verification, and you will see that the transformations from S to S' are the same as from S' to S. It suffices to set v=-v because the relativistic observable velocities are reciprocal, since there is no privileged universal reference frame. Attention, I understand your objection, simply, what it is necessary to understand that in the reciprocal of the transformation, -v is already transformed into +v. You don't have to do it again, otherwise you have for example x=x'+(-v)To'/sqrt(1-v²/c²) and the calculation obviously becomes wrong. R.H.
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| From | Townes Olson <townesolson7@gmail.com> |
|---|---|
| Date | 2022-03-24 16:04 -0700 |
| Message-ID | <66fa3597-9412-41e8-9fb1-4a115326ee20n@googlegroups.com> |
| In reply to | #580937 |
On Thursday, March 24, 2022 at 1:26:03 PM UTC-7, patdolan wrote: > There is no pathway, comprised of validly deduced equations, that leads from these > four premises of well-formed algebraic strings: > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > to this well-formed conclusion string: v' = v Not true. To avoid confusion, let’s use the symbol u instead of v’. You’ve been shown explicitly (and agreed) that the first pair of equations x’=(x-vt)g[v] and t’=(t-vx)g[v] is equivalent to the inverted pair x=(x’+vt’)g[v] and t=(t’+vx’)g[v], which have the same form as the next two of your equations, x=(x’+ut’)g[u] and t=(t’+ux’)g[u]. So, the two expressions for x give the relation (x’+vt’)/sqrt(1-v^2) = (x’+ut’)/sqrt(1-u^2) Now, presumably you do not dispute that this equation is satisfied if u=v, so your question is whether it’s possible for this equation to also be satisfied for some value of u that is not equal to v. By simple grade school algebra (I gave you the explicit solution previously), you can see that the answer is yes, and it is also possible for there to be a value of u other than v that satisfies the corresponding relation given by setting the two expressions for t equal to each other (t’+vx’)/sqrt(1-v^2) = (t’+ux’)/sqrt(1-u^2) However, depending on whether t is greater than or less than x (spacelike or timelike), only one of them gives equality, and the other gives negative equality (i.e., same magnitude, opposite sign). So, for any u other than v, you can’t make both of these relations true at the same time, although you can make their squares both equal. To see this, just square both sides, clear the fractions, bring everything over to one side, and factor. One fact is (u-v), and the other factor gives the other value of u that makes the squares of these expressions equal.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 17:18 -0700 |
| Message-ID | <5ce78f8c-7ec2-4fc7-b83a-146e28dc4127n@googlegroups.com> |
| In reply to | #580943 |
On Thursday, March 24, 2022 at 4:04:17 PM UTC-7, Townes Olson wrote: > On Thursday, March 24, 2022 at 1:26:03 PM UTC-7, patdolan wrote: > > There is no pathway, comprised of validly deduced equations, that leads from these > > four premises of well-formed algebraic strings: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > to this well-formed conclusion string: v' = v > Not true. To avoid confusion, let’s use the symbol u instead of v’. You’ve been shown explicitly (and agreed) that the first pair of equations x’=(x-vt)g[v] and t’=(t-vx)g[v] is equivalent to the inverted pair x=(x’+vt’)g[v] and t=(t’+vx’)g[v], which have the same form as the next two of your equations, x=(x’+ut’)g[u] and t=(t’+ux’)g[u]. So, the two expressions for x give the relation > > (x’+vt’)/sqrt(1-v^2) = (x’+ut’)/sqrt(1-u^2) > > Now, presumably you do not dispute that this equation is satisfied if u=v, so your question is whether it’s possible for this equation to also be satisfied for some value of u that is not equal to v. By simple grade school algebra (I gave you the explicit solution previously), you can see that the answer is yes, and it is also possible for there to be a value of u other than v that satisfies the corresponding relation given by setting the two expressions for t equal to each other > > (t’+vx’)/sqrt(1-v^2) = (t’+ux’)/sqrt(1-u^2) > > However, depending on whether t is greater than or less than x (spacelike or timelike), only one of them gives equality, and the other gives negative equality (i.e., same magnitude, opposite sign). So, for any u other than v, you can’t make both of these relations true at the same time, although you can make their squares both equal. To see this, just square both sides, clear the fractions, bring everything over to one side, and factor. One fact is (u-v), and the other factor gives the other value of u that makes the squares of these expressions equal.
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 17:26 -0700 |
| Message-ID | <585b70ba-b5e6-47d3-ac3b-bb0b9d6c7931n@googlegroups.com> |
| In reply to | #580943 |
On Thursday, March 24, 2022 at 4:04:17 PM UTC-7, Townes Olson wrote: > On Thursday, March 24, 2022 at 1:26:03 PM UTC-7, patdolan wrote: > > There is no pathway, comprised of validly deduced equations, that leads from these > > four premises of well-formed algebraic strings: > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > to this well-formed conclusion string: v' = v > Not true. To avoid confusion, let’s use the symbol u instead of v’. You’ve been shown explicitly (and agreed) that the first pair of equations x’=(x-vt)g[v] and t’=(t-vx)g[v] is equivalent to the inverted pair x=(x’+vt’)g[v] and t=(t’+vx’)g[v], which have the same form as the next two of your equations, x=(x’+ut’)g[u] and t=(t’+ux’)g[u]. So, the two expressions for x give the relation > > (x’+vt’)/sqrt(1-v^2) = (x’+ut’)/sqrt(1-u^2) > > Now, presumably you do not dispute that this equation is satisfied if u=v, so your question is whether it’s possible for this equation to also be satisfied for some value of u that is not equal to v. This is a wonderful admission by you Townes. By simple grade school algebra (I gave you the explicit solution previously), you can see that the answer is yes, and it is also possible for there to be a value of u other than v that satisfies the corresponding relation given by setting the two expressions for t equal to each other > > (t’+vx’)/sqrt(1-v^2) = (t’+ux’)/sqrt(1-u^2) > > However, depending on whether t is greater than or less than x (spacelike or timelike), only one of them gives equality, and the other gives negative equality (i.e., same magnitude, opposite sign). So, for any u other than v, you can’t make both of these relations true at the same time, although you can make their squares both equal. To see this, just square both sides, clear the fractions, bring everything over to one side, and factor. One fact is (u-v), and the other factor gives the other value of u that makes the squares of these expressions equal. This is baloney Townes. I also note that you have not attempted to derive a damned thing, because like Bodkin, you can't. But I can! I can prove that your u and v CAN NEVER EQUAL EACH OTHER. NEVER! This means that the LTs rely on the false assumption that v = v'. What do you say to this Townes? And how does it feel to have mathematical sword of Damocles over your sorry head?
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| From | Townes Olson <townesolson7@gmail.com> |
|---|---|
| Date | 2022-03-24 17:54 -0700 |
| Message-ID | <4c29b330-b504-44f5-af8f-2c889b95a8c0n@googlegroups.com> |
| In reply to | #580949 |
On Thursday, March 24, 2022 at 5:26:36 PM UTC-7, patdolan wrote: > > > There is no pathway, comprised of validly deduced equations, that leads from these > > > four premises of well-formed algebraic strings: > > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > > to this well-formed conclusion string: v' = v > > Not true. To avoid confusion, let’s use the symbol u instead of v’. You’ve been shown explicitly (and agreed) that the first pair of equations x’=(x-vt)g[v] and t’=(t-vx)g[v] is equivalent to the inverted pair x=(x’+vt’)g[v] and t=(t’+vx’)g[v], which have the same form as the next two of your equations, x=(x’+ut’)g[u] and t=(t’+ux’)g[u]. So, the two expressions for x give the relation > > > > (x’+vt’)/sqrt(1-v^2) = (x’+ut’)/sqrt(1-u^2) > > > > Now, presumably you do not dispute that this equation is satisfied if u=v... > > I can prove that your u and v CAN NEVER EQUAL EACH OTHER. With u=v both sides of that equation (and the corresponding t equation) are explicitly identical, so to claim that those relations are not satisfied is insane. The only sane question you could possibly be asking is whether it’s possible for this equation to *also* be satisfied for some value of u that is not equal to v. Now, by simple grade school algebra (I gave you the explicit solution previously), you can see that the answer is yes, i.e., the above equality has two solutions, one with u=v and one with the other value of u that I gave you previously. And it is also possible for there to be a value of u other than v that satisfies the corresponding relation given by setting the two expressions for t equal to each other (t’+vx’)/sqrt(1-v^2) = (t’+ux’)/sqrt(1-u^2) However, depending on whether t is greater than or less than x, only one of them gives equality, and the other gives negative equality (i.e., same magnitude, opposite sign). So, for any u other than v, you can’t make both of these relations true at the same time. Do you understand this? > This means that the LTs rely on the false assumption that v = v'. Goodness no. Remember, you're not even talking about relativity or space or time or velocity any more, you are talking purely about your struggles with grade school algebra, and you have stipulated the relationship between x,t and x',t', so the derivation of the Lorentz transformation is not even in dispute. You have stipulated it in this discussion. Remember? You're essentially just trying to understand why the inverse of a system of linear equations is unique. That is pure grade school algebra. For example, given X'=X-K, we have the inverse X=X'-K. Your claim is that the K in the first equation cannot be the same as the K in the second equation. Your claim is obviously insane. Understand?
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| From | patdolan <patdolan@comcast.net> |
|---|---|
| Date | 2022-03-24 17:59 -0700 |
| Message-ID | <65267c29-7805-4a8c-8a44-5a901c3d7b4en@googlegroups.com> |
| In reply to | #580954 |
On Thursday, March 24, 2022 at 5:54:17 PM UTC-7, Townes Olson wrote: > On Thursday, March 24, 2022 at 5:26:36 PM UTC-7, patdolan wrote: > > > > There is no pathway, comprised of validly deduced equations, that leads from these > > > > four premises of well-formed algebraic strings: > > > > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ] > > > > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ] > > > > x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ] > > > > t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ] > > > > to this well-formed conclusion string: v' = v > > > Not true. To avoid confusion, let’s use the symbol u instead of v’. You’ve been shown explicitly (and agreed) that the first pair of equations x’=(x-vt)g[v] and t’=(t-vx)g[v] is equivalent to the inverted pair x=(x’+vt’)g[v] and t=(t’+vx’)g[v], which have the same form as the next two of your equations, x=(x’+ut’)g[u] and t=(t’+ux’)g[u]. So, the two expressions for x give the relation > > > > > > (x’+vt’)/sqrt(1-v^2) = (x’+ut’)/sqrt(1-u^2) > > > > > > Now, presumably you do not dispute that this equation is satisfied if u=v... > > > > I can prove that your u and v CAN NEVER EQUAL EACH OTHER. > With u=v both sides of that equation (and the corresponding t equation) are explicitly identical, so to claim that those relations are not satisfied is insane. The only sane question you could possibly be asking is whether it’s possible for this equation to *also* be satisfied for some value of u that is not equal to v. > > Now, by simple grade school algebra (I gave you the explicit solution previously), you can see that the answer is yes, i.e., the above equality has two solutions, one with u=v and one with the other value of u that I gave you previously. And it is also possible for there to be a value of u other than v that satisfies the corresponding relation given by setting the two expressions for t equal to each other > (t’+vx’)/sqrt(1-v^2) = (t’+ux’)/sqrt(1-u^2) > However, depending on whether t is greater than or less than x, only one of them gives equality, and the other gives negative equality (i.e., same magnitude, opposite sign). So, for any u other than v, you can’t make both of these relations true at the same time. Do you understand this? > > This means that the LTs rely on the false assumption that v = v'. > Goodness no. Remember, you're not even talking about relativity or space or time or velocity any more, you are talking purely about your struggles with grade school algebra, and you have stipulated the relationship between x,t and x',t', so the derivation of the Lorentz transformation is not even in dispute. You have stipulated it in this discussion. Remember? You're essentially just trying to understand why the inverse of a system of linear equations is unique. > > That is pure grade school algebra. For example, given X'=X-K, we have the inverse X=X'-K. Your claim is that the K in the first equation cannot be the same as the K in the second equation. Your claim is obviously insane. Understand? Townes, I'm tired of reading your word walls. You never derive. You only type more words. Useless words. I am going to destroy your words....tomorrow. Or maybe later tonight. Go to sleep tonight secure in the knowledge that your LTs will be dead before the sun sets tomorrow.
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