"What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be exactly correct. There is only v. All observes will always and only use v when calculating the coordinate values x' and t'. This from Hyperphysics:

Transforms from S to S'
x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]

and going back the other way

Transforms to S' to S
x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2

Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A great reasoner and philosopher once wrote that if v = v in both cases then it should be provable.

I dare any denizen of this forum to prove that the v used in the Transforms to S' equals/is identical to, the v used in the Transforms to S.

I don't think even Townes Olson has a clue how to do this. He may not even understand what I am typing about. And Bodkin? Fuhgedaboudit.

With deepest sympathy for a lost cause,

TMWBTLT

patdolan <patdolan@comcast.net> wrote:
> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you
> would be exactly correct. There is only v. All observes will always and
> only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
>
> Transforms from S to S'
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> and going back the other way
>
> Transforms to S' to S
> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>
> Conspicuous by its absence from the Transforms Lorentz is the quantity
> v'. A great reasoner and philosopher once wrote that if v = v in both
> cases then it should be provable.

Which I did for you already, using the Lorentz transform.
Why would you need this repeated?

>
> I dare any denizen of this forum to prove that the v used in the
> Transforms to S' equals/is identical to, the v used in the Transforms to S.
>
> I don't think even Townes Olson has a clue how to do this. He may not
> even understand what I am typing about. And Bodkin? Fuhgedaboudit.
>
> With deepest sympathy for a lost cause,
>
> TMWBTLT
>
>
>
>

--
Odd Bodkin — Maker of fine toys, tools, tables

On Wednesday, March 23, 2022 at 4:48:57 PM UTC-7, bodk...@gmail.com wrote:
> patdolan <patd...@comcast.net> wrote:
> > "What!" you howl, "There is no v' in the Transforms Lorentz!" And you
> > would be exactly correct. There is only v. All observes will always and
> > only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
> >
> > Transforms from S to S'
> > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> >
> > and going back the other way
> >
> > Transforms to S' to S
> > x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> > t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
> >
> > http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
> >
> > Conspicuous by its absence from the Transforms Lorentz is the quantity
> > v'. A great reasoner and philosopher once wrote that if v = v in both
> > cases then it should be provable.
> Which I did for you already, using the Lorentz transform.
> Why would you need this repeated?

Yes, why repeat? Give us a link to your proof.
> >
> > I dare any denizen of this forum to prove that the v used in the
> > Transforms to S' equals/is identical to, the v used in the Transforms to S.
> >
> > I don't think even Townes Olson has a clue how to do this. He may not
> > even understand what I am typing about. And Bodkin? Fuhgedaboudit.
> >
> > With deepest sympathy for a lost cause,
> >
> > TMWBTLT
> >
> >
> >
> >
> --
> Odd Bodkin — Maker of fine toys, tools, tables

On Wednesday, March 23, 2022 at 4:14:47 PM UTC-7, patdolan wrote:
> Transforms from S to S'
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> I dare any denizen of this forum to prove that the v used in the Transforms to S'
> equals/is identical to, the v used in the Transforms to S.

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.

On Wednesday, March 23, 2022 at 6:46:54 PM UTC-7, Townes Olson wrote:
> On Wednesday, March 23, 2022 at 4:14:47 PM UTC-7, patdolan wrote:
> > Transforms from S to S'
> > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> > I dare any denizen of this forum to prove that the v used in the Transforms to S'
> > equals/is identical to, the v used in the Transforms to S.
> 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.

First of all, Townes, you forgot to prime your x in your first derivation. Secondly, all you've done is demonstrated the fact that the transforms from S' to S can be derived from the transforms from S to S'. Lesson One, Townes: if you are going to prove anything about v then you will have to expand or divide out your g because it contains v.

Townes, I'm going to help you out for your second try. Let's re-write the Transforms Lorentz for S to S' and S' to S using the variables v and v' accordingly, like this:

Transforms from S to S'
x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]

Transforms to S' to S
x = ( x' + v't' )/sqrt[ 1 - (v'/c)^2 ]
t = ( t' + v'x'/c^2 )/sqrt[ 1 - (v'/c)^2 ]

Now Townes, and this is very important, prove that v = v'. Do you understand? Let me know if I can help you.

On Wednesday, March 23, 2022 at 11:08:25 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 forgot to prime your x..

Right, the result of inverting the transformation is t=(t'+vx')g and 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'.

It isn't a "transform", it is a "transformation" (there is a difference), and each pair of equations constitutes a single transformation, not plural. And yes, the transformation from one system of coordinates to the other is algebraically invertible, and the parameter v characterizes the relationship between those two systems. Remember, 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. See above.

> If you are going to prove anything about v then you will have to expand or
> divide out your g because it contains v.

No, the dependence of g on v is fully accounted for in the grade school algebra that shows how the parameter v appears in the transformation and it inverse. See above.

On Wednesday, March 23, 2022 at 11:29:49 PM UTC-7, Townes Olson wrote:
> On Wednesday, March 23, 2022 at 11:08:25 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 forgot to prime your x..
>
> Right, the result of inverting the transformation is t=(t'+vx')g and 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'.
> It isn't a "transform", it is a "transformation" (there is a difference), and each pair of equations constitutes a single transformation, not plural.. And yes, the transformation from one system of coordinates to the other is algebraically invertible, and the parameter v characterizes the relationship between those two systems. Remember, 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. See above.

Then it should be tautological and mere child's play for you to prime the v in the second pair of equations then prove that v' = v. Do you intend to do this? Or do you decline?
>
> > If you are going to prove anything about v then you will have to expand or
> > divide out your g because it contains v.
> No, the dependence of g on v is fully accounted for in the grade school algebra that shows how the parameter v appears in the transformation and it inverse. See above.

On Thursday, 24 March 2022 at 07:29:49 UTC+1, Townes Olson wrote:

> It isn't a "transform", it is a "transformation" (there is a difference), and each pair of equations constitutes a single transformation, not plural.. And yes, the transformation from one system of coordinates to the other is algebraically invertible, and the parameter v characterizes the relationship between those two systems. Remember, 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. See above.

In the meantime in the real world, however, forbidden by
your insane religion TAI keep measuring t'=t, just like
all serious clocks always did. Remember.

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.

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?

patdolan <patdolan@comcast.net> wrote:
> On Wednesday, March 23, 2022 at 4:48:57 PM UTC-7, bodk...@gmail.com wrote:
>> patdolan <patd...@comcast.net> wrote:
>>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you
>>> would be exactly correct. There is only v. All observes will always and
>>> only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
>>>
>>> Transforms from S to S'
>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> and going back the other way
>>>
>>> Transforms to S' to S
>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>>
>>> Conspicuous by its absence from the Transforms Lorentz is the quantity
>>> v'. A great reasoner and philosopher once wrote that if v = v in both
>>> cases then it should be provable.
>> Which I did for you already, using the Lorentz transform.
>> Why would you need this repeated?
>
> Yes, why repeat? Give us a link to your proof.

Who is “us”?

If you can’t stay sober enough to pay attention to conversations you
initiate here, whose problem is that?

>>>
>>> I dare any denizen of this forum to prove that the v used in the
>>> Transforms to S' equals/is identical to, the v used in the Transforms to S.
>>>
>>> I don't think even Townes Olson has a clue how to do this. He may not
>>> even understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>>
>>> With deepest sympathy for a lost cause,
>>>
>>> TMWBTLT
>>>
>>>
>>>
>>>
>> --
>> Odd Bodkin — Maker of fine toys, tools, tables
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Le 24/03/2022 à 00:14, patdolan a écrit :
> Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A
> great reasoner and philosopher once wrote that if v = v in both cases then it
> should be provable.
>
> I dare any denizen of this forum to prove that the v used in the Transforms to
> S' equals/is identical to, the v used in the Transforms to S.
>
> I don't think even Townes Olson has a clue how to do this. He may not even
> understand what I am typing about. And Bodkin? Fuhgedaboudit.
>
> With deepest sympathy for a lost cause,

Yes. It's the same v.

v'=v

Covariance notion.

"Il n'y a pas de référentiel absolu".

Si un référentiel S' se déplace à 0.8c par rapport à un autre, alors
l'autre se déplace à la même vitesse.

On peut d'ailleurs résoudre ça par l'absurde.

Pourquoi l'autre se déplacerait-il, par exemple à 0.2c vis à vis du
premier? ? ?

Donc la question ne se pose pas.

Ou alors il faut retirer le principe de covariance.

Mais je n'en vois pas, aujourd'hui, l'intérêt.

Par contre, j'apprécie le fait que vous cherchiez à rendre cette
théorie plus cohérente, car il est évident que des cailloux dans la
chaussure s'y trouvent.

Mais pas dans la notion de réciprocité des référentiels.

"Les effets de la physique sont réciproques par permutation de
référentiel"

Autre exemple : deux montres battent alors réciproquement plus vite
(selon le même coefficient) que celle qu'elles observent.

R.H.

Den 24.03.2022 00:14, skrev patdolan:
> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be exactly correct. There is only v. All observes will always and only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
>
> Transforms from S to S'
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> and going back the other way
>
> Transforms to S' to S
> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>
> Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A great reasoner and philosopher once wrote that if v = v in both cases then it should be provable.
>
> I dare any denizen of this forum to prove that the v used in the Transforms to S' equals/is identical to, the v used in the Transforms to S.
>
> I don't think even Townes Olson has a clue how to do this. He may not even understand what I am typing about. And Bodkin? Fuhgedaboudit.
>
> With deepest sympathy for a lost cause,
>
> TMWBTLT
>

Forget SR, forget transformations.

x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]

This is two equations where x', t', c and v are constants,
x and t are unknowns to be determined.
We solve for x and t, and get:

x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]

This is two equations where x, t, c and v are constants,
x' and t' are unknowns to be determined.
We solve for x' and t', and get:

x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]

The two sets of two equations are the same!
When one set is known, the other is only
a rearranging of x',t', x, t, c and v.
So of course x',t', x, t, c and v are
all the same in both sets of equations.

--
Paul

https://paulba.no/

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.

> 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.

Le 24/03/2022 à 13:43, "Paul B. Andersen" a écrit :
>
>
> Den 24.03.2022 00:14, skrev patdolan:
>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be
>> exactly correct. There is only v. All observes will always and only use v when
>> calculating the coordinate values x' and t'. This from Hyperphysics:
>>
>> Transforms from S to S'
>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>
>> and going back the other way
>>
>> Transforms to S' to S
>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>
>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>
>> Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A
>> great reasoner and philosopher once wrote that if v = v in both cases then it
>> should be provable.
>>
>> I dare any denizen of this forum to prove that the v used in the Transforms to
>> S' equals/is identical to, the v used in the Transforms to S.
>>
>> I don't think even Townes Olson has a clue how to do this. He may not even
>> understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>
>> With deepest sympathy for a lost cause,
>>
>> TMWBTLT
>>
>
> Forget SR, forget transformations.
>
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> This is two equations where x', t', c and v are constants,
> x and t are unknowns to be determined.
> We solve for x and t, and get:
>
> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> This is two equations where x, t, c and v are constants,
> x' and t' are unknowns to be determined.
> We solve for x' and t', and get:
>
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>
> The two sets of two equations are the same!
> When one set is known, the other is only
> a rearranging of x',t', x, t, c and v.
> So of course x',t', x, t, c and v are
> all the same in both sets of equations.

Oui, évidemment.

Il suffit de changer v en -v, et au lien que S défile dans S' de gauche
à droite, il défile de droite à gauche.

Ca revient au même.

Les correspondances seront les mêmes.

R.H.

Le 24/03/2022 à 14:53, Richard Hachel a écrit :
> Le 24/03/2022 à 13:43, "Paul B. Andersen" a écrit :
>>
>>
>> Den 24.03.2022 00:14, skrev patdolan:
>>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be
>>> exactly correct. There is only v. All observes will always and only use v when
>>> calculating the coordinate values x' and t'. This from Hyperphysics:
>>>
>>> Transforms from S to S'
>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> and going back the other way
>>>
>>> Transforms to S' to S
>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>>
>>> Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A
>>> great reasoner and philosopher once wrote that if v = v in both cases then it should
>>> be provable.
>>>
>>> I dare any denizen of this forum to prove that the v used in the Transforms to
>>> S' equals/is identical to, the v used in the Transforms to S.
>>>
>>> I don't think even Townes Olson has a clue how to do this. He may not even
>>> understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>>
>>> With deepest sympathy for a lost cause,
>>>
>>> TMWBTLT
>>>
>>
>> Forget SR, forget transformations.
>>
>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>
>> This is two equations where x', t', c and v are constants,
>> x and t are unknowns to be determined.
>> We solve for x and t, and get:
>>
>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>
>> This is two equations where x, t, c and v are constants,
>> x' and t' are unknowns to be determined.
>> We solve for x' and t', and get:
>>
>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>
>> The two sets of two equations are the same!
>> When one set is known, the other is only
>> a rearranging of x',t', x, t, c and v.
>> So of course x',t', x, t, c and v are
>> all the same in both sets of equations.
>
> Oui, évidemment.
>
> Il suffit de changer v en -v, et au lien que S défile dans S' de gauche à
> droite, il défile de droite à gauche.
>
> Ca revient au même.
>
> Les correspondances seront les mêmes.
>
> R.H.

EXAMPLE :

1. Calculate the coordinates and angles of E in R'.

2. Then re-calculate the coordinates and angles of E in R.

<http://news2.nemoweb.net/jntp?WJVs61et92rRXAe9qN2yBs6U514@jntp/Data.Media:1>

R.H.

Le 24/03/2022 à 15:01, Richard Hachel a écrit :
> Le 24/03/2022 à 14:53, Richard Hachel a écrit :
>> Le 24/03/2022 à 13:43, "Paul B. Andersen" a écrit :
>>>
>>>
>>> Den 24.03.2022 00:14, skrev patdolan:
>>>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be
>>>> exactly correct. There is only v. All observes will always and only use v when
>>>> calculating the coordinate values x' and t'. This from Hyperphysics:
>>>>
>>>> Transforms from S to S'
>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>
>>>> and going back the other way
>>>>
>>>> Transforms to S' to S
>>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>
>>>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>>>
>>>> Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A
>>>> great reasoner and philosopher once wrote that if v = v in both cases then it should
>>>> be provable.
>>>>
>>>> I dare any denizen of this forum to prove that the v used in the Transforms to
>>>> S' equals/is identical to, the v used in the Transforms to S.
>>>>
>>>> I don't think even Townes Olson has a clue how to do this. He may not even
>>>> understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>>>
>>>> With deepest sympathy for a lost cause,
>>>>
>>>> TMWBTLT
>>>>
>>>
>>> Forget SR, forget transformations.
>>>
>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> This is two equations where x', t', c and v are constants,
>>> x and t are unknowns to be determined.
>>> We solve for x and t, and get:
>>>
>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> This is two equations where x, t, c and v are constants,
>>> x' and t' are unknowns to be determined.
>>> We solve for x' and t', and get:
>>>
>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>
>>> The two sets of two equations are the same!
>>> When one set is known, the other is only
>>> a rearranging of x',t', x, t, c and v.
>>> So of course x',t', x, t, c and v are
>>> all the same in both sets of equations.
>>
>> Oui, évidemment.
>>
>> Il suffit de changer v en -v, et au lien que S défile dans S' de gauche à
>> droite, il défile de droite à gauche.
>>
>> Ca revient au même.
>>
>> Les correspondances seront les mêmes.
>>
>> R.H.
>
> EXAMPLE :
>
> 1. Calculate the coordinates and angles of E in R'.
>
> 2. Then re-calculate the coordinates and angles of E in R.
>
> <http://news2.nemoweb.net/jntp?WJVs61et92rRXAe9qN2yBs6U514@jntp/Data.Media:1>
>
> R.H.

<http://news2.nemoweb.net/?DataID=GlPCRs-_FpCvevdDh3FSuhWXX5U@jntp>

R.H.

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.

On 3/24/2022 5:44 AM, Odd Bodkin wrote:
> patdolan <patdolan@comcast.net> wrote:
>> On Wednesday, March 23, 2022 at 4:48:57 PM UTC-7, bodk...@gmail.com wrote:
>>> patdolan <patd...@comcast.net> wrote:
>>>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you
>>>> would be exactly correct. There is only v. All observes will always and
>>>> only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
>>>>
>>>> Transforms from S to S'
>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>
>>>> and going back the other way
>>>>
>>>> Transforms to S' to S
>>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>
>>>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>>>
>>>> Conspicuous by its absence from the Transforms Lorentz is the quantity
>>>> v'. A great reasoner and philosopher once wrote that if v = v in both
>>>> cases then it should be provable.
>>> Which I did for you already, using the Lorentz transform.
>>> Why would you need this repeated?
>>
>> Yes, why repeat? Give us a link to your proof.
>

> Who is “us”?

Standard English usage. Get a grip already.

> If you can’t stay sober enough to pay attention to conversations you
> initiate here, whose problem is that?

That's a question that could easily be asked about you. Without
defending dolan you lack common courtesy that is expected of adults
om settings like this one.

>>>>
>>>> I dare any denizen of this forum to prove that the v used in the
>>>> Transforms to S' equals/is identical to, the v used in the Transforms to S.
>>>>
>>>> I don't think even Townes Olson has a clue how to do this. He may not
>>>> even understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>>>
>>>> With deepest sympathy for a lost cause,
>>>>
>>>> TMWBTLT
>>>>
>>>>
>>>>
>>>>
>>> --
>>> Odd Bodkin — Maker of fine toys, tools, tables
>>
>
>
>

whodat <whodaat@void.nowgre.com> wrote:
> On 3/24/2022 5:44 AM, Odd Bodkin wrote:
>> patdolan <patdolan@comcast.net> wrote:
>>> On Wednesday, March 23, 2022 at 4:48:57 PM UTC-7, bodk...@gmail.com wrote:
>>>> patdolan <patd...@comcast.net> wrote:
>>>>> "What!" you howl, "There is no v' in the Transforms Lorentz!" And you
>>>>> would be exactly correct. There is only v. All observes will always and
>>>>> only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
>>>>>
>>>>> Transforms from S to S'
>>>>> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
>>>>> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>>
>>>>> and going back the other way
>>>>>
>>>>> Transforms to S' to S
>>>>> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
>>>>> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
>>>>>
>>>>> http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
>>>>>
>>>>> Conspicuous by its absence from the Transforms Lorentz is the quantity
>>>>> v'. A great reasoner and philosopher once wrote that if v = v in both
>>>>> cases then it should be provable.
>>>> Which I did for you already, using the Lorentz transform.
>>>> Why would you need this repeated?
>>>
>>> Yes, why repeat? Give us a link to your proof.
>>
>
>
>> Who is “us”?
>
> Standard English usage. Get a grip already.

I didn’t know he spoke for you too. I know Pat is too lazy to use his
Usenet reader to find his own conversations where he last asked this
question. I’m a little surprised you’re in the same boat.

>
>> If you can’t stay sober enough to pay attention to conversations you
>> initiate here, whose problem is that?
>
> That's a question that could easily be asked about you. Without
> defending dolan you lack common courtesy that is expected of adults
> om settings like this one.
>

People have history here. Pat has a long one. You think this should be
overlooked? Have you overlooked mine?

>
>
>>>>>
>>>>> I dare any denizen of this forum to prove that the v used in the
>>>>> Transforms to S' equals/is identical to, the v used in the Transforms to S.
>>>>>
>>>>> I don't think even Townes Olson has a clue how to do this. He may not
>>>>> even understand what I am typing about. And Bodkin? Fuhgedaboudit.
>>>>>
>>>>> With deepest sympathy for a lost cause,
>>>>>
>>>>> TMWBTLT
>>>>>
>>>>>
>>>>>
>>>>>
>>>> --
>>>> Odd Bodkin — Maker of fine toys, tools, tables
>>>
>>
>>
>>
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Thursday, March 24, 2022 at 5:43:15 AM UTC-7, Paul B. Andersen wrote:
> Den 24.03.2022 00:14, skrev patdolan:
> > "What!" you howl, "There is no v' in the Transforms Lorentz!" And you would be exactly correct. There is only v. All observes will always and only use v when calculating the coordinate values x' and t'. This from Hyperphysics:
> >
> > Transforms from S to S'
> > x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> > t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> >
> > and going back the other way
> >
> > Transforms to S' to S
> > x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> > t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
> >
> > http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2
> >
> > Conspicuous by its absence from the Transforms Lorentz is the quantity v'. A great reasoner and philosopher once wrote that if v = v in both cases then it should be provable.
> >
> > I dare any denizen of this forum to prove that the v used in the Transforms to S' equals/is identical to, the v used in the Transforms to S.
> >
> > I don't think even Townes Olson has a clue how to do this. He may not even understand what I am typing about. And Bodkin? Fuhgedaboudit.
> >
> > With deepest sympathy for a lost cause,
> >
> > TMWBTLT
> >
> Forget SR, forget transformations.
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> This is two equations where x', t', c and v are constants,
> x and t are unknowns to be determined.
> We solve for x and t, and get:
> x = ( x' + vt' )/sqrt[ 1 - (v/c)^2 ]
> t = ( t' + vx'/c^2 )/sqrt[ 1 - (v/c)^2 ]
> This is two equations where x, t, c and v are constants,
> x' and t' are unknowns to be determined.
> We solve for x' and t', and get:
> x' = ( x - vt )/sqrt[ 1 - (v/c)^2 ]
> t' = ( t - vx/c^2 )/sqrt[ 1 - (v/c)^2 ]
> The two sets of two equations are the same!
> When one set is known, the other is only
> a rearranging of x',t', x, t, c and v.
> So of course x',t', x, t, c and v are
> all the same in both sets of equations.
>
> --
> Paul
>
> https://paulba.no/

Yours, Paul, is an answer to a question I didn't ask.

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?

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

On Thursday, March 24, 2022 at 9:40:15 AM UTC-7, patdolan 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?

An by the way Clownes this sentence of yours

"all your sophistries reduce to the same kernel of idiocy, namely, your belief that the relation f(a)=f(b) implies a=b."

is in fact the definition of a one to one function.