From: Dono. on
On Sep 26, 10:16 am, "Dirk Van de moortel"
<dirkvandemoor...(a)nospAm.hotmail.com> wrote:
>
> But note, Marcel, as always you fucked up where we take
> the meanings of the variables into account.
>
> In special relativity, the equation
> tA = t sqrt( 1-V^2 )
> says something about two events occuring at the same
> place in A's frame, whereas the equation
> tB = t sqrt (1-vB^2)
> says something about two events occuring at the same
> place in B's frame.
> So, still in special relativity the combination
> tB = tA sqrt( 1-v^2 ) / (1+ v V)
> is only meaningful if it says something about two events
> that are simultaneous in A's frame AND in B's frame.
> That's only possible if v = 0, in other words words if
> tB = tA sqrt( 1-0 ) / (1+0)
> i.o.w. if


Nooooo, you should have left the idiot in the dark.I gave him this
exact part as an exercise to solve, he never answered. I was planning
to keep him there forever (which may not be too long, given his
advanced Alzheimer)

From: Dirk Van de moortel on
Dirk Van de moortel <dirkvandemoortel(a)nospAm.hotmail.com> wrote in message
j0svm.174383$4f4.105151(a)newsfe11.ams2
> mluttgens <mluttgens(a)orange.fr> wrote in message
> 0e036cc7-d090-4958-8c55-6d872544f3d3(a)o35g2000vbi.googlegroups.com
>
> [snip]
>
>>>>>>>>>>>>>>> According to the CMBR observer (using c=1),
>>>>>>>>>>>>>>> vA = (v-vB)/(1-vB*v)
>>>>>>>>>>>>>>> tB = tCMBR * sqrt(1-vB^2)
>>>>>>>>>>>>>>> tA = tCMBR * sqrt(1-((v-vB)/(1-vB*v))^2)
>>>>>>>>>>>>>>> = tCMBR * sqrt(1-vB^2)*sqrt(1-v^2)/(1-vB*v)
>>>>>>>>>>>>>>> Hence, tA = tB * sqrt(1-v^2) / (1-vB*v).
>
> [snip]
>
>> How muddled up are the analyses of the twin paradox in terms.
>> of relativity theory! , see for instance
>> http://www.physicsforums.com/showthread.php?s=12dddfc2f78c05c783324da355c4780a&t=209625&page=2
>>
>> Anyhow, a simple analysis uses SR + acceleration.
>>
>> But the simplest one would use SR + cmbr, which
>> straightforwardly shows the absence of the paradox.
>> Note that the applicability of SR(cmbr) is wholly general:
>> When a clock A moves at V relative to the CMBR, and
>> a clock B moves at v relative to clock A, it can be shown
>> that tB = tA * sqrt(1-v^2)/(1+Vv)
>
> You obviously derived this formula as follows:
>
> You took the velocy of B w.r.t. CMBR as
> vB = (v+V) / (1+v V)
> so, using t for the CMBR-time you used
> tB = t sqrt (1-vB^2)
> = t sqrt( 1 - (v+V)^2 / (1+v V)^2 )
> = t sqrt( (1+v V)^2 - (v+V)^2 ) / (1+v V)
> = t sqrt( 1-v^2 ) sqrt ( 1-V^2 ) / (1+v V )
> and of course you also used
> tA = t sqrt( 1-V^2 )
> so you took these together to find
> tB = tA sqrt( 1-v^2 ) / (1+ v V).
> Good, the algebra is correct.
>
> But note, Marcel, as always you fucked up where we take
> the meanings of the variables into account.
>
> In special relativity, the equation
> tA = t sqrt( 1-V^2 )
> says something about two events occuring at the same
> place in A's frame, whereas the equation
> tB = t sqrt (1-vB^2)
> says something about two events occuring at the same
> place in B's frame.
> So, still in special relativity the combination
> tB = tA sqrt( 1-v^2 ) / (1+ v V)
> is only meaningful if it says something about two events
> that are simultaneous in A's frame AND in B's frame.

Typo.
That should be: ... about two events occuring at the same
place in both frames.

> That's only possible if v = 0, in other words words if
> tB = tA sqrt( 1-0 ) / (1+0)
> i.o.w. if
> tB = tA,
> but we already knew that, didn't we?

There's another way to have the events occur at the same
place in both frames, even without v = 0, namely if they
happen "on top of each other in time", i.o.w. when
tB = tA = 0,
in which case the equation says
0 = 0,
but we already knew that as well, didn't we, Marcel?

>
>> (see http://pagesperso-orange.fr/mluttgens/twinpdx1.htm
>> for more insight).
>
> Right. More insight in your failure to understand what
> the variables represent.
>
>> Let's note that V and v can have identical or opposite signs.
>> Let's also note that claiming that the CMBR can be moving
>> wrt an object is physically nonsensical.
>> So, in SR(cmbr), one is left with
>> tB = tA * sqrt(1-v^2)/(1+Vv)
>> tA = tB * (1+Vv)/sqrt(1-v^2)
>> Clearly, the time dilation effect is not reciprocal.
>
> Well, with the necessary condition that this can only happen when
> v = 0 ,

or when
tA = tB = 0 ,

> one is left with
> tB = tA
> tA = tB
> which looks quite reciprocal if you ask me.

specially when we turn it into
0 = 0
0 = 0.

>
> Quod Erat Debunkandum.

aka Quod Erat Deluttgandem

>
> Dirk Vdm

From: Dirk Van de moortel on
Dono. <sa_ge(a)comcast.net> wrote in message
cd418bd1-58a6-4a55-a1e4-ccaf5f7d8cf0(a)k13g2000prh.googlegroups.com
> On Sep 26, 10:16 am, "Dirk Van de moortel"
> <dirkvandemoor...(a)nospAm.hotmail.com> wrote:
>>
>> But note, Marcel, as always you fucked up where we take
>> the meanings of the variables into account.
>>
>> In special relativity, the equation
>> tA = t sqrt( 1-V^2 )
>> says something about two events occuring at the same
>> place in A's frame, whereas the equation
>> tB = t sqrt (1-vB^2)
>> says something about two events occuring at the same
>> place in B's frame.
>> So, still in special relativity the combination
>> tB = tA sqrt( 1-v^2 ) / (1+ v V)
>> is only meaningful if it says something about two events
>> that are simultaneous in A's frame AND in B's frame.
>> That's only possible if v = 0, in other words words if
>> tB = tA sqrt( 1-0 ) / (1+0)
>> i.o.w. if
>
>
> Nooooo, you should have left the idiot in the dark.I gave him this
> exact part as an exercise to solve, he never answered. I was planning
> to keep him there forever (which may not be too long, given his
> advanced Alzheimer)

Actually, there's still an exercise left for him.
Let's see if he finds it.

Dirk Vdm

From: Dono. on
On Sep 27, 2:32 am, "Dirk Van de moortel"
<dirkvandemoor...(a)nospAm.hotmail.com> wrote:
> Dono. <sa...(a)comcast.net> wrote in message
>
> cd418bd1-58a6-4a55-a1e4-ccaf5f7d8...(a)k13g2000prh.googlegroups.com
>
>
>
> > On Sep 26, 10:16 am, "Dirk Van de moortel"
> > <dirkvandemoor...(a)nospAm.hotmail.com> wrote:
>
> >> But note, Marcel, as always you fucked up where we take
> >> the meanings of the variables into account.
>
> >> In special relativity, the equation
> >> tA = t sqrt( 1-V^2 )
> >> says something about two events occuring at the same
> >> place in A's frame, whereas the equation
> >> tB = t sqrt (1-vB^2)
> >> says something about two events occuring at the same
> >> place in B's frame.
> >> So, still in special relativity the combination
> >> tB = tA sqrt( 1-v^2 ) / (1+ v V)
> >> is only meaningful if it says something about two events
> >> that are simultaneous in A's frame AND in B's frame.
> >> That's only possible if v = 0, in other words words if
> >> tB = tA sqrt( 1-0 ) / (1+0)
> >> i.o.w. if
>
> > Nooooo, you should have left the idiot in the dark.I gave him this
> > exact part as an exercise to solve, he never answered. I was planning
> > to keep him there forever (which may not be too long, given his
> > advanced Alzheimer)
>
> Actually, there's still an exercise left for him.
> Let's see if he finds it.
>
> Dirk Vdm



:-)
From: mluttgens on
On 26 sep, 19:16, "Dirk Van de moortel"
<dirkvandemoor...(a)nospAm.hotmail.com> wrote:
> mluttgens <mluttg...(a)orange.fr> wrote in message
>
>   0e036cc7-d090-4958-8c55-6d872544f...(a)o35g2000vbi.googlegroups.com
>
> [snip]
>
> >>>>>>>>>>>>>> According to the CMBR observer (using c=1),
> >>>>>>>>>>>>>> vA = (v-vB)/(1-vB*v)
> >>>>>>>>>>>>>> tB = tCMBR * sqrt(1-vB^2)
> >>>>>>>>>>>>>> tA = tCMBR * sqrt(1-((v-vB)/(1-vB*v))^2)
> >>>>>>>>>>>>>> = tCMBR * sqrt(1-vB^2)*sqrt(1-v^2)/(1-vB*v)
> >>>>>>>>>>>>>> Hence, tA = tB * sqrt(1-v^2) / (1-vB*v).
>
> [snip]
>
> > How muddled up are the analyses of the twin paradox in terms.
> > of relativity theory! , see for instance
> >http://www.physicsforums.com/showthread.php?s=12dddfc2f78c05c783324da....
>
> > Anyhow, a simple analysis uses SR + acceleration.
>
> > But the simplest one would use SR + cmbr, which
> > straightforwardly shows the absence of the paradox.
> > Note that the applicability of SR(cmbr) is wholly general:
> > When a clock A moves at V relative to the CMBR, and
> > a clock B moves at v relative to clock A, it can be shown
> > that tB = tA * sqrt(1-v^2)/(1+Vv)
>
> You obviously derived this formula as follows:
>
> You took the velocy of B w.r.t. CMBR as
>         vB = (v+V) / (1+v V)
> so, using t for the CMBR-time you used
>         tB = t sqrt (1-vB^2)
>              = t sqrt( 1 - (v+V)^2  / (1+v V)^2 )
>              = t sqrt( (1+v V)^2 - (v+V)^2 ) / (1+v V)
>              = t sqrt( 1-v^2 ) sqrt ( 1-V^2 ) / (1+v V )
> and of course you also used
>         tA = t sqrt( 1-V^2 )
> so you took these together to find
>         tB = tA sqrt( 1-v^2 ) / (1+ v V).
> Good, the algebra is correct.
>
> But note, Marcel, as always you fucked up where we take
> the meanings of the variables into account.
>
> In special relativity, the equation
>         tA = t sqrt( 1-V^2 )
> says something about two events occuring at the same
> place in A's frame, whereas the equation
>         tB = t sqrt (1-vB^2)
> says something about two events occuring at the same
> place in B's frame.
> So, still in special relativity the combination
>         tB = tA sqrt( 1-v^2 ) / (1+ v V)
> is only meaningful if it says something about two events
> that are simultaneous in A's frame AND in B's frame.
> That's only possible if v = 0, in other words words if
>         tB = tA sqrt( 1-0 ) / (1+0)
> i.o.w. if
>         tB = tA,
> but we already knew that, didn't we?
>
> > (seehttp://pagesperso-orange.fr/mluttgens/twinpdx1.htm
> > for more insight).
>
> Right. More insight in your failure to understand what
> the variables represent.
>
> > Let's note that V and v can have identical or opposite signs.
> > Let's also note that claiming that the CMBR can be moving
> > wrt an object is physically nonsensical.
> > So, in SR(cmbr), one is left with
> > tB = tA * sqrt(1-v^2)/(1+Vv)
> > tA = tB * (1+Vv)/sqrt(1-v^2)
> > Clearly, the time dilation effect is not reciprocal.
>
> Well, with the necessary condition that this can only happen when
>         v = 0 ,
> one is left with
>         tB = tA
>         tA = tB
> which looks quite reciprocal if you ask me.

Dirk, clock B moves at v relative to clock A in the scenario.
Whn v = 0, both clocks are at rest relativ to each other.
Then, of course, both clocks will show the same time.

Marcel Luttgens

>
> Quod Erat Debunkandum.
>
> Dirk Vdm