Why is the relation t' = gt nonsensical ?
Marcel Luttgens
asked me to show why the relation t' = gt, obtained with the LT
t' = g(t - xv/c^2) when x = 0, is nonsensical.
Here is a brief demonstration ( the full demonstration can be found at
http://perso.wanadoo.fr/mluttgens/LTfalse.htm ) :
In the LT
x' = g(x - vt) and
t' = g(t - xv/c^2), where g = 1/sqrt(1 - v^2/c^2),
x represents the position coordinate of some event (x,t).
By replacing x by Vt in those relations, one gets
equations that are valid for "events" taking place on objects
with coordinates (x=Vt, t) according to S :
(1) x' = gt(V - v) and
(2) t' = gt(1 - Vv/c^2).
These 2 equations are false:
According to S, since the object moves at V, its time t(o) is given by
(3) t(o) = t * sqrt(1 - V^2/c^2).
According to S', since the object moves at V' = (V-v)/(1-Vv/c^2)
relatively to S', its time t'(o) is given by
(4) t'(o) = t' * sqrt(1 - V'^2/c^2)
= t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).
If one considers that t(o) = t'(o), one gets
t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2) =
t * sqrt(1 - V^2/c^2), which reduces to
t' = gt(1 - Vv/c^2), which is identical to equation (2) !
Iow, the LT imply t(o) = t'(o). If it can be shown that t(o) is
different from t'(o), even in a specific case, the LT are necessarily
wrong. This is simple logic.
Let's consider the case where V = 0. Then, from relations (3) and (4),
one gets
t(o) = t * sqrt(1 - V^2/c^2) = t
t'(o) = t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2)
= t' * sqrt(1 - v^2/c^2)
= t'/g
For the LT to be correct, t'(o) must be equal to t(o), thus
t' = t / sqrt(1 - v^2/c^2) = gt.
And, indeed, when x = Vt = 0, the "time" LT t' = g(t - xv/c^2)
reduces to t' = gt.
As, according to S, the frame S' will be at a distance vt after
the time interval t, the clock of S' will read t' = t * sqrt(1 - v^2/c^2),
thus t = t' / sqrt(1 - v^2/c^2) = gt', and
t(o) = t = gt', whereas, as seen above,
t'(o) = t'/g.
Since, in the specific case where V = 0, t(o) is different from t'(o),
the LT are false.
By the way, SRists can, and probably will, claim that the t'
in the relation t'(o) = t'/g has not the same meaning as the t' in the
relation t' = t * sqrt(1 - v^2/c^2). In my opinion, such claim amounts to
sophistry, but if SRists are nevertheless right, they would illustrate the
incoherences of the LT and special relativity theory, because the same
symbol would represent different things. And, of course, an incoherent
theory is necessarily wrong.
QED
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> In the thread "The correct Lorentz transformation", Paul B. Andersen
> asked me to show why the relation t' = gt, obtained with the LT
> t' = g(t - xv/c^2) when x = 0, is nonsensical.
>
> Here is a brief demonstration ( the full demonstration can be found at
> http://perso.wanadoo.fr/mluttgens/LTfalse.htm ) :
>
This utterly clumsy and maliciously faulty attempt has been
thoroughly and conclusively debunked on my recent thread
http://groups.google.com/groups?&as_umsgid=Z29b9.92905$8o4....@afrodite.telenet-ops.be
where, after having been decisively defeated, Marcel ended
with a furious "Bye for GOOD."
Short memory, small brain, tiny penis?
This "Bye for GOOD" was preceded by two instances of the
weaker "Bye for good", so a burst of "BYE FOR GOOD"
must be imminent now.
Anyhow, let me repeat this:
Marcel, if there is anything you'd like me to explain about
events and coordinates, you are welcome to ask at any time.
If you somehow feel embarrassed by asking on this public
forum, feel free to go private. You've got my address.
Dirk Vdm
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
Splash!
Wrong "clock of S'".
The rest is thus wrong.
[..]
You failed again.
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:alodhf$i0n$1...@dolly.uninett.no...
Paul, allow me to clarify a bit?
Hope you don't mind? I just love this ;-)
No problem until now, since t" = t". See my thread
http://groups.google.com/groups?&as_umsgid=Z29b9.92905$8o4....@afrodite.telenet-ops.be
He copied from it.
> >
> > Let's consider the case where V = 0.
Here he is beginning to sneak, since he is now going to consider
events with V = 0, which means (Vx,t) = (0,t), i.o.w. only for
events taking place on clock S, where x=0.
I.o.w. he is going for a special case of what has already been
refuted. Watch him ;-)
> > Then, from relations (3) and (4),
> > one gets
> >
> > t(o) = t * sqrt(1 - V^2/c^2) = t
of course, since
t" = t(o) is the time coordinate, attributed by the object S" to
events taking place on itself. In this case t" is expressed as a
function of the time t, attributed by S on the observer S",
and we are talking about S" = S since V = 0.
So we have
t" = t"
or
t(o) = t'(o)
or
t = t'/g
He's going to claim that this is impossible.
> > t'(o) = t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2)
> > = t' * sqrt(1 - v^2/c^2)
> > = t'/g
of course, since
t" = t'(o) is the time coordinate, attributed by the object S" to
events taking place on itself. In this case t" is expressed as a
function of the time t', attributed by S' on the observer S",
and we are talking about S" = S since V = 0.
And we know that for events taking place on the clock S
itself that t' = g(t-vx) reduces to t'=gt, which is the same as
t = t'/g
Note that this equation is *only valid for events taking
place on the clock S* - This is rather important.
> >
> > For the LT to be correct, t'(o) must be equal to t(o), thus
> > t' = t / sqrt(1 - v^2/c^2) = gt.
Indeed, here he comes ;-)
> >
> > And, indeed, when x = Vt = 0, the "time" LT t' = g(t - xv/c^2)
> > reduces to t' = gt.
Yes, yes, watch him... ;-)
> >
> > As, according to S, the frame S' will be at a distance vt after
> > the time interval t, the clock of S' will read t' = t * sqrt(1 - v^2/c^2),
> > thus t = t' / sqrt(1 - v^2/c^2) = gt', and
>
> Splash!
> Wrong "clock of S'".
For those who might find this somewhat cryptic, specially
for our dearest friend Marcel, allow me to comment and
unsnip the rest?
Now Marcel suddenly switches to events taking place
on the clock S', since there we have x'=0 and hence
t=g(t'+vx/c^2) reduces to t = gt'.
Before we were talking about events taking place on the
clock S.
> >
> > t(o) = t = gt'
Yes, taking place on clock S'
>, whereas, as seen above,
> > t'(o) = t'/g.
Whereas above, we talked about events taking place on
clock S.
Dirty little sneak. Gotcha! ;-))
> >
> > Since, in the specific case where V = 0, t(o) is different from t'(o),
> > the LT are false.
QEDebunkandum - again.
So his entire post was a special case of my post
http://groups.google.com/groups?&as_umsgid=Z29b9.92905$8o4....@afrodite.telenet-ops.be
where halfway he took V = 0.
Naughty Marcel, caught with your pants down - again.
Next trial will be the case V = -0.253416673 * v?
Okay, I'll be ready.
Dirk Vdm
Stephen Speicher
[Big snips all around.]
> >
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> > >
> > > Since, in the specific case where V = 0, t(o) is different from t'(o),
> > > the LT are false.
>
> QEDebunkandum - again.
>
I usually like to discourage responding to cranks, but your
detailed analyses of Marcel's unending mistakes are so on the
money, and so entertaining, that I do encourage you to continue
for as long as you yourself can enjoy it.
It is fascinating to see how he comes back each time with another
incarnation of the same problem, all puffed up with his new
"proof" that the Lorentz transformation is wrong. With all of the
effort he expends on this, he actually could have built something
useful if he worked in a woodshop with his hands, rather than
misusing his mind. Delusion can run deep, and you do well in
exposing it.
--
Stephen
s...@compbio.caltech.edu
Printed using 100% recycled electrons.
-----------------------------------------------------------
Dirk Van de moortel
"Stephen Speicher" <s...@compbio.caltech.edu> wrote in message news:Pine.GSO.4.42.0209111615330.18900-100000@blinky...
> On Wed, 11 Sep 2002, Dirk Van de moortel wrote:
>
> [Big snips all around.]
>
> > >
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> > > >
> > > > Since, in the specific case where V = 0, t(o) is different from t'(o),
> > > > the LT are false.
> >
> > QEDebunkandum - again.
> >
>
> I usually like to discourage responding to cranks, but your
> detailed analyses of Marcel's unending mistakes are so on the
> money, and so entertaining, that I do encourage you to continue
> for as long as you yourself can enjoy it.
>
> It is fascinating to see how he comes back each time with another
> incarnation of the same problem, all puffed up with his new
> "proof" that the Lorentz transformation is wrong. With all of the
> effort he expends on this, he actually could have built something
> useful if he worked in a woodshop with his hands, rather than
> misusing his mind.
IIRC once I recommended starting collecting poodle droppings.
I think that was a mistake. He obviously can't resist eating them.
> Delusion can run deep, and you do well in
> exposing it.
I just hope I don't make any silly mistakes doing this.
If I do, and you or anyone notice(s), *please* jump on it and slap
me in the face?
If I make a real fumble, I will put it on my site.
Thanks Stephen, support warmly appreciated!
Dirk Vdm
Marcel Luttgens
And indeed, as expected, the "brightest" SRist just responded:
t = t'/g is only valid for events taking place on the clock S.
t = gt' is only valid for events taking place on clock S',
thus confirming that SR uses identical symbols, for instance t,
to represent different things.
Iow, SR is an incoherent theory if one accept the contorted views of
SRists.
But, in fact, SR is essentially false because Einstein's derivation
of the LT contains mathematical errors, as shown in
http://perso.wanadoo.fr/mluttgens/LTfalse.htm
>
> Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
[snip]
>
> And indeed, as expected, the "brightest" SRist just responded:
So, since you can't reply to me anymore ("Bye for GOOD"), you
reply to yourself. Sharp!
But what happened to "BYE FOR GOOD"?
> t = t'/g is only valid for events taking place on the clock S.
> t = gt' is only valid for events taking place on clock S',
>
> thus confirming that SR uses identical symbols, for instance t,
> to represent different things.
Just like x can represent a wide range of different things,
this must be a very difficult concept to grasp:
http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
> Iow, SR is an incoherent theory if one accept the contorted views of
> SRists.
> But, in fact, SR is essentially false because Einstein's derivation
> of the LT contains mathematical errors, as shown in
> http://perso.wanadoo.fr/mluttgens/LTfalse.htm
Containing *exactly* the same stuff as what I have
been debunking all over the place ;-)
Boy, do I love you!
Dirk Vdm
Paul B. Andersen
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:XWPf9.113379$8o4....@afrodite.telenet-ops.be...
>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:alodhf$i0n$1...@dolly.uninett.no...
>
> Paul, allow me to clarify a bit?
> Hope you don't mind? I just love this ;-)
By all means, do that.
I have not the patient of yours.
I responded only because he referred to me at the top.
But I pointed out only the most glaring error, and
I didn't bother to explain it in detail.
But when you do it so well, why should I? :-)
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
[big snipp]
> > As, according to S, the frame S' will be at a distance vt after
> > > the time interval t, the clock of S' will read t' = t * sqrt(1 - v^2/c^2),
> > > thus t = t' / sqrt(1 - v^2/c^2) = gt', and
> >
> > Splash!
> > Wrong "clock of S'".
>
> For those who might find this somewhat cryptic, specially
> for our dearest friend Marcel, allow me to comment and
> unsnip the rest?
>
> Now Marcel suddenly switches to events taking place
> on the clock S', since there we have x'=0 and hence
> t=g(t'+vx/c^2) reduces to t = gt'.
> Before we were talking about events taking place on the
> clock S.
Just a tiny comment on the expressions "the clock S" and
"the clock S'".
I understand that you mean a clock at the origo of S and S'
respectively. We can however imagine an infinite number
of stationary, synchronized clocks at every spatial point
in both frames, showing co-ordinate time.
That's why I said "wrong clock of S'".
The question is not what the clock at the origo of S' shows,
but what the co-ordinate clock of S' which is at the position
of the origo of S shows.
And that "clock of S'" shows t' = tg.
Sorry for nit-picking, but my point was only to explain my
somewhat cryptic remark: "wrong clock of S' ".
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:alq6vn$2ia$1...@dolly.uninett.no...
Yep, I usually write
"on the observer S", or
"on the origin S", or
"on the y-z-plane of S",
but
"on the clock S"
would do just as well.
I mean
"all events satisfying x = 0."
Ditto for the primed and doubly primed stuff.
> We can however imagine an infinite number
> of stationary, synchronized clocks at every spatial point
> in both frames, showing co-ordinate time.
Indeed, but in Special R, I prefer "the single clock at the origin"
view. An infinity of clocks makes me dizzy ;-) and talking about
the "times of these clocks as seen on other clocks" is needlessly
confusing. Specially for our good friend ML.
> That's why I said "wrong clock of S'".
Ha, but you wrote:
Wrong "clock of S'".
Mark the location of your opening quote, which made me
think you meant to say:
Wrong: "clock of S'"
but where you forget the colon ":".
*That* was the little detail that bothered me when I tried
to read your remark *through Marcel's eyes*.
> The question is not what the clock at the origo of S' shows,
> but what the co-ordinate clock of S' which is at the position
> of the origo of S shows.
> And that "clock of S'" shows t' = tg.
Yes, but again, I don't like this multi-clock view, so I
rather talk about "events taking place on the observer
at the origin" to avoid confusion.
>
> Sorry for nit-picking, but my point was only to explain my
> somewhat cryptic remark: "wrong clock of S' ".
Somewhat indeed ;-)
Cheers,
Dirk Vdm
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> But, in fact, SR is essentially false because Einstein's derivation
> of the LT contains mathematical errors, as shown in
> http://perso.wanadoo.fr/mluttgens/LTfalse.htm
Your fumbles are abundant.
The self-documenting ones are rare.
This is your third one:
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#SRSymbols
Title: "SR uses identical symbols to represent different things!"
Dirk VDm
Bilge
>Iow, SR is an incoherent theory if one accept the contorted views of
>SRists.
Only your version is incoherent. The version everyone uses works
fine. Where is your comparison to rotations, that I asked you to make?
Marcel Luttgens
_______________________________________
Here is a brief demonstration ( the full demonstration can be found at
http://perso.wanadoo.fr/mluttgens/LTfalse.htm ) :
In the LT
x' = g(x - vt) and
t' = g(t - xv/c^2), where g = 1/sqrt(1 - v^2/c^2),
x represents the position coordinate of some "event" (x,t).
By replacing x by Vt in those relations, one gets
equations that are valid for "events" taking place on objects
with coordinates (x=Vt, t) according to S :
(1) x' = gt(V - v) and
(2) t' = gt(1 - Vv/c^2).
These 2 equations are false:
According to S, since the object moves at V, its time t(o) is given by
(3) t(o) = t * sqrt(1 - V^2/c^2).
According to S', * since the object moves at V' = (V-v)/(1-Vv/c^2)
relatively to S' * , its time t'(o) is given by
According to SR experts,
t = t'/g is only valid for events taking place on the clock S.
t = gt' is only valid for events taking place on clock S',
meaning that SR uses identical symbols, in this case t,
to represent different things.
Or, logically, they should use two different symbols to represent t,
for instance t(events_S) = t'/g, and t(events_S') = gt'. But sophistry,
not logic, is their main strength.
In the above scenario, there are 3 frames: S, S' and S", the object's
frame, each having its own clock.
When the origins of the frames coincide, the 3 clocks are synchronized
to 0. Immediately after synchronization , the frame S' moves away at a
constant velocity v wrt the frame S, whereas the velocity V
of the object wrt S is zero. Hence, after a time t measured on the clock
of S, the object remains at the origin of S, and its clock reads t(o) = t,
according to the frame S considered at rest.
Iow, since the clock of S and that of the object are mutually at rest, they will
keep the same time, hence only the time t read on the clock of S needs to be
taken into consideration.
According to S, the clock of S' will read t' = t * sqrt(1 - v^2/c^2) = t/g,
after a time interval t.
Or, by applying the "time" LT, one gets t' = gt, instead of t' = t/g !
Iow, the direct application of SR (or LET) rules on one hand, and the use
of the LT on the other hand, lead to contradictory results.
According to SRists, there is no contradiction at all, since, as claimed
above,
" t = t'/g is only valid for events taking place on the clock S.
t = gt' is only valid for events taking place on clock S' ".
This is plain B.S., the origin of the contradiction is simply that in
the same scenario, S' is considered to be moving at v wrt S, which
leads to t = t'/g, whereas in order to derive the "time" LT, one has
to consider that S is moving at -v wrt S' (cf. relation (4) ), thus
obtaining t = gt'.
But SRists are too brainwashed to realize this evidence. Their contorted
thinking needs their deceitful concept of "events".
To everybody else, SR theory has just been shown, with the help of a simple
scenario, to be self-contradictory, hence false.
In fact, SR is essentially false because Einstein's derivation of
the LT contains mathematical errors, as shown in
http://perso.wanadoo.fr/mluttgens/LTfalse.htm
Of course, SRists will claim that his derivation is mathematically
correct, even if it leads, in practice, to blatant contradictions.
And they will not hesitate at resorting to sophisms, even in math,
to justify their point.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> _______________________________________
>
> Here is a brief demonstration ( the full demonstration can be found at
> http://perso.wanadoo.fr/mluttgens/LTfalse.htm ) :
>
[yawn, getting bored]
> In the LT
> x' = g(x - vt) and
> t' = g(t - xv/c^2), where g = 1/sqrt(1 - v^2/c^2),
> x represents the position coordinate of some "event" (x,t).
>
> By replacing x by Vt in those relations, one gets
> equations that are valid for "events" taking place on objects
> with coordinates (x=Vt, t) according to S :
You stole the above two lines:
http://groups.google.com/groups?&as_umsgid=Z29b9.92905$8o4....@afrodite.telenet-ops.be
Little cheat.
The rest is identical to what you farted before and has
already been debunked.
See also
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html#SRSymbols
and specially
http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
Dirk Vdm
Bilge
>Of course, SRists will claim that his derivation is mathematically
>correct, even if it leads, in practice, to blatant contradictions.
In other words, because you get the wrong result, you expect
rest of us to go along with your misunderstanding?
>And they will not hesitate at resorting to sophisms, even in math,
>to justify their point.
You'll be nominated for a fields medal any day now, too.
Dirk Van de moortel
"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message news:slrnao6c9...@radioactivex.lebesque-al.net...
Now he will reply that you, resorting to personal abuse, lose.
Dirk Vdm
Bilge
Re: Why is the relation t' = gt nonsensical ? to usenet:
Only because he doesn't count trolling as personal abuse, leaving him
way ahead in the abuse dept.
Marcel Luttgens
Try to debunk this last trolling !
Reminder:
S' moves at v wrt S, some object moves at V wrt S.
By replacing x by Vt in the "time" LT t' = g(t - xv/c^2), one gets
(1) t' = gt(1 - Vv/c^2).
According to S, since the object moves at V, its time t(o) is given by
t(o) = t * sqrt(1 - V^2/c^2), when the clock of S reads t.
This relation implies that S is considered at rest.
According to S', as the object moves at V' = (V-v)/(1-Vv/c^2)
relatively to S' , its time t'(o) is given by
t'(o) = t' * sqrt(1 - V'^2/c^2)
= t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).
This relation implies that S' is considered at rest.
If one considers that t(o) = t'(o), one gets
t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2) =
t * sqrt(1 - V^2/c^2), which reduces to
t' = gt(1 - Vv/c^2), or, by replacing V by x/t,
t' = g(t - xv/c^2).
This relation is identical to the Einsteinian "time" transformation,
which implies, as shown above, a jump from frame S to frame S',
meaning that it can't be correct.
Obtention of a correct relation:
In order to obtain a correct relation, one should use a single
rest frame of reference, not two, in the derivation.
This is possible by using the object's frame.
Then, according to the object,
S moves at -V, hence
t = t(o) * sqrt(1 - V^2/c^2),
when the object's clock reads t(o).
S' moves at -V', thus
t' = t(o) * sqrt(1 - V'^2/c^2), or
t' = t(o) * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2),
also when the object's clock reads t(o).
So, it is licit to replace t(o) by t/sqrt(1 - V^2/c^2) in this
last relation, thus obtaining, in a coherent way, the relation
t' = t * sqrt(1 - v^2/c^2) / (1 - Vv/c^2)
Let's note that when V = 0, the object's frame coincides with S, and
t' = t * sqrt(1 - v^2/c^2), which is correct, as S' moves at v wrt S.
But we must also note that t' is not generally the time of S'
when the clock of S reads t.
In fact, t' represents the time of the object according to S', as
a function of the time t of the object according to S.
As I claimed before, the time of S' is independant from any object
having left the common origin of the frames S and S' at t = t' = 0.
It only depends on v, the velocity of S' wrt S, and on t, according to
the relation t' = t * sqrt(1 - v^2/c^2).
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> > Dirk Van de moortel said some stuff about
> > Re: Why is the relation t' = gt nonsensical ? to usenet:
> > >
> > >"Bilge" <ro...@radioactivex.lebesque-al.net> wrote in message news:slrnao6c9...@radioactivex.lebesque-al.net...
> > >> Marcel Luttgens said some stuff about
> > >>
> > >> >Of course, SRists will claim that his derivation is mathematically
> > >> >correct, even if it leads, in practice, to blatant contradictions.
> > >>
> > >> In other words, because you get the wrong result, you expect
> > >> rest of us to go along with your misunderstanding?
> > >>
> > >> >And they will not hesitate at resorting to sophisms, even in math,
> > >> >to justify their point.
> > >>
> > >> You'll be nominated for a fields medal any day now, too.
> > >
> > >Now he will reply that you, resorting to personal abuse, lose.
> >
> >
> > Only because he doesn't count trolling as personal abuse, leaving him
> > way ahead in the abuse dept.
>
> Try to debunk this last trolling !
Again?
How many times have you said "Goodbye for GOOD" now???
Another copy/paste exercise?
>
> Reminder:
>
> S' moves at v wrt S, some object moves at V wrt S.
>
> By replacing x by Vt in the "time" LT t' = g(t - xv/c^2), one gets
> (1) t' = gt(1 - Vv/c^2).
Relation between times t and t' of events taking place on object S",
t measured by S, and t' measured by S'
>
> According to S, since the object moves at V, its time t(o) is given by
> t(o) = t * sqrt(1 - V^2/c^2), when the clock of S reads t.
>
You are looking at the object as a third reference frame here.
Call it S" and let it use the coordinates (x",t") for events in spacetime.
The LT wrt S are then:
{ x" = G(x-Vt)
{ t" = G(t-Vx/c^2)
with G = 1/sqrt(1-V^2/c^2)
and the inverse:
{ x = G(x"+Vt")
{ t = G(t"+Vx"/c^2)
On the object itself we have x" = 0, so we get
t = Gt"
or
t" = t/G
or in your lingo:
t(o) = t * sqrt(1 - V^2/c^2)
t" = t(o) is the time coordinate, attributed by the object S" to
events taking place on itself. In this case t" is expressed as a
function of the time t, attributed by S on the observer S".
> This relation implies that S is considered at rest.
What a daft thing to say ;-)
This relation implies what I said it says.
>
> According to S', as the object moves at V' = (V-v)/(1-Vv/c^2)
> relatively to S' , its time t'(o) is given by
>
> t'(o) = t' * sqrt(1 - V'^2/c^2)
> = t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).
>
Indeed, we also have the LT wrt S':
{ x" = G'(x'-V't')
{ t" = G'(t'-V'x'/c^2)
with G' = 1/sqrt(1-V'^2/c^2)
and V' = (V-v)/(1-Vv/c^2)
and
{ x' = G'(x"+V't")
{ t' = G'(t"+V'x"/c^2)
On the object itself we have x" = 0, so we get
t' = G't"
or
t" = t'/G'
or in your lingo:
t'(o) = t' * sqrt(1 - V'^2/c^2)
t" = t'(o) is the time coordinate, attributed by the object S" to
events taking place on itself. In this case t" is expressed as a
function of the time t', attributed by S' on the observer S".
> This relation implies that S' is considered at rest.
What a silly thing to say ;-)
This relation implies what I said it says.
>
> If one considers that t(o) = t'(o), one gets
>
> t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2) =
> t * sqrt(1 - V^2/c^2), which reduces to
> t' = gt(1 - Vv/c^2),
Of course you get this. You have been working with events
on the object S" all the time.
If you would get something else, you would have made an
error.
> or, by replacing V by x/t,
getting back where we started ;-)))
> t' = g(t - xv/c^2).
A MIRACLE!
We *are* back where we started!
A REAL MIRACLE!
>
> This relation is identical to the Einsteinian "time" transformation,
> which implies, as shown above, a jump from frame S to frame S',
> meaning that it can't be correct.
Meaning again that you still pretend to have no idea
what you are talking about.
Is this a stupidity contest with josX or with Spaceman or
something?
What do we do?
Do we continue?
Shall we have a look?
Okay, let's have a look.
After all, he promised to have something new :-))
>
> Obtention of a correct relation:
>
> In order to obtain a correct relation, one should use a single
> rest frame of reference, not two, in the derivation.
Now *this* is going to hurt.
>
> This is possible by using the object's frame.
>
> Then, according to the object,
>
> S moves at -V, hence
> t = t(o) * sqrt(1 - V^2/c^2),
> when the object's clock reads t(o).
This equation is valid for events taking place on S,
where x = 0.
>
> S' moves at -V', thus
> t' = t(o) * sqrt(1 - V'^2/c^2), or
> t' = t(o) * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2),
> also when the object's clock reads t(o).
This equation is valid for events taking place on S',
where x' = 0.
I sense some squashing in the make.
>
> So, it is licit to replace t(o) by t/sqrt(1 - V^2/c^2) in this
> last relation,
HUH? Licit?
Taking an equation valid for events with x = 0
t = t(o) * sqrt(1 - V^2/c^2),
and combining it with an equation valid for events with x'=0
t' = t(o) * sqrt(1 - V'^2/c^2)
?
Licit? You mean like "permissible"?
But yes, you are right, it *is* licit to consider events that
satisfy x=0 and x'=0 simultaneously.
Just take v = 0 and everything is just fine, right?
Of course v = 0 implies S = S' but what the hell, who
cares? After all, we are in the process of debunking a
complete century, right?
Okay: SQUASH!
> thus obtaining, in a coherent way, the relation
>
> t' = t * sqrt(1 - v^2/c^2) / (1 - Vv/c^2)
which with v = 0 licitly reduces to
t' = t
Another miracle!
>
> Let's note that when V = 0, the object's frame coincides with S, and
and with S', so they are all snugging cozily on top
of each other.
>
> t' = t * sqrt(1 - v^2/c^2), which is correct, as S' moves at v wrt S.
No, they don't, since v = 0.
Now we have
S = S' = S'
v = V = V' = 0
t = t' = t" for all events
x = x' = x" for all events
Licitly interesting!
>
> But we must also note that t' is not generally the time of S'
> when the clock of S reads t.
> In fact, t' represents the time of the object according to S', as
> a function of the time t of the object according to S.
Here is where we need to point to:
http://users.pandora.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
again.
>
> As I claimed before, the time of S' is independant from any object
> having left the common origin of the frames S and S' at t = t' = 0.
> It only depends on v, the velocity of S' wrt S, and on t, according to
> the relation t' = t * sqrt(1 - v^2/c^2).
Good dog.
When do we get a *real* challenge, Marcel?
Dirk Vdm
Bilge
>Try to debunk this last trolling !
I'm still waiting for you to do as I asked for rotations.
You seem to expect everyone else to accomodate you. You're
wrong. On more than one count.
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> > Reminder:
> >
> > S' moves at v wrt S, some object moves at V wrt S.
> >
> > By replacing x by Vt in the "time" LT t' = g(t - xv/c^2), one gets
> > (1) t' = gt(1 - Vv/c^2).
>
> Relation between times t and t' of events taking place on object S",
> t measured by S, and t' measured by S'
Conclusively, according to SRists, the variable t' in relation (1) is
different from the variable t' in the relation
(2) t' = t * sqrt(1 - v^2/c^2),
which gives the time of S' according to S.
This is a first clue that SR is incoherent.
Otoh,
t(o) = t * sqrt(1 - V^2/c^2),
t'(o) = t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).
As the validity of relation (2) is independant from the presence
of such moving objects, we also have
t = t' / sqrt(1 - v^2/c^2), thus
t(o) = t' * sqrt(1 - V^2/c^2) / sqrt(1 - v^2/c^2), which is
different from t'(o).
Hence, SR, which implies t(o) = t'(o), is necessarily false, except for
sophists or morons, who are implying that the relation (2) cannot be used
if some object with velocity V has left the common origin of S, S' and
the object's frame S", at t = t' = t" = 0.
Otoh, your argument according to which the relation
t = t(o) * sqrt(1 - V^2/c^2), valid for events with x = 0,
cannot be combined with the relation
t' = t(o) * sqrt(1 - V'^2/c^2), valid for events with x'=0,
unless x=0 and x'=0 are simultaneously satisfied,
is pure SRist crackpottery.
Indeed, both relations are physically correct, and in each of them,
t(o) represents the same time, so, mathematically,
t(o) = t / sqrt(1 - V^2/c^2) = t'/ sqrt(1 - V'^2/c^2), hence
t' = t * sqrt(1 - v^2/c^2) / (1 - Vv/c^2).
Your reasoning would make sense if the variable t(o) represented
two different times in those relations. This is not the case.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<2PEh9.9$OT3.2...@news.cpqcorp.net>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02091...@posting.google.com...
>
> > > Reminder:
> > >
> > > S' moves at v wrt S, some object moves at V wrt S.
> > >
> > > By replacing x by Vt in the "time" LT t' = g(t - xv/c^2), one gets
> > > (1) t' = gt(1 - Vv/c^2).
> >
> > Relation between times t and t' of events taking place on object S",
> > t measured by S, and t' measured by S'
(1) is valid for events taking place on object S", where x=Vt.
>
> Conclusively, according to SRists, the variable t' in relation (1) is
> different from the variable t' in the relation
> (2) t' = t * sqrt(1 - v^2/c^2),
> which gives the time of S' according to S
Wrong. It gives a relation between times t and t' of events taking
place on observer S' where x'=0, i.o.w. where x=vt.
> This is a first clue that SR is incoherent.
Wrong. It means that (1) and (2) can be only combined for events
taking place on object S" *and* on observer S' together,
i.o.w. for events satisfying x=vt and x=Vt together,
i.o.w. for events on observer S' while V=v, or
for event (x,t)=(x',t')=(x",t")=(0,0) which is trivial.
So this is not a clue that SR is incoherent.
[snipped at first error]
Dirk Vdm
Marcel Luttgens
Why can't you grasp the obvious ?
Premise: S' moves at v RELATIVELY to S, some object moves at V,
------- also RELATIVELY to S.
By replacing x by Vt in the "time" LT t' = g(t - xv/c^2), one gets
(1) t' = gt(1 - Vv/c^2).
According to S, since the object moves at V, its time t(o) is given by
t(o) = t * sqrt(1 - V^2/c^2), when the clock of S reads t.
According to S', as the object moves at V' = (V-v)/(1-Vv/c^2)
relatively to S' , its time t'(o) is given by
t'(o) = t' * sqrt(1 - V'^2/c^2)
= t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2).
If one considers that t(o) = t'(o), one gets
t' * sqrt(1 - v^2/c^2) * sqrt(1 - V^2/c^2) / (1 - Vv/c^2) =
t * sqrt(1 - V^2/c^2), which reduces to
t' = gt(1 - Vv/c^2).
Now, let's consider the case where V = 0.
Then, ACCORDING TO S, t(o) = t. Note that, since x = Vt = 0,
the object remains at the origin of S.
But, ACCORDING TO S', t'(o) = t' * sqrt(1 - v^2/c^2).
Otoh, since t(o) is considered to be equal to t'(o), one gets
t = t' * sqrt(1 - v^2/c^2), or
t' = t / sqrt(1 - v^2/c^2) = gt.
Note that one directly gets t' = gt by replacing x by 0 in
t' = g(t - xv/c^2).
Thus, the "time" LT implies that the frame S is moving at -v wrt the
frame S' (cf. t' = gt), in contradiction with the premise, according to which
the frame S' is moving at v wrt the frame S (then, of course, t' = t/g).
Iow, the "time" LT is self-contradictory.
As the "time" LT is false, so is SR itself.
>
> Dirk Vdm
Marcel Luttgens
Bilge
>wrote in message news:<8e4i9.122511$8o4....@afrodite.telenet-ops.be>...
>>
>> [snipped at first error]
>
>Why can't you grasp the obvious ?
to "debunking" rotations? You could always teem up with josX and make it
a collaborative fiasco.
Stephen Speicher
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<8e4i9.122511$8o4....@afrodite.telenet-ops.be>...
> >
> > So this is not a clue that SR is incoherent.
> > [snipped at first error]
>
> Why can't you grasp the obvious ?
>
Dirk is on vacation at the Alps for a week, and he asked that I
send you his regards when you show up. Regards from Dirk.
However, please do not let Dirk's absence hold you back. Just
assume that if he were here he would have pointed out the very
same error that you continuously make. So, you can respond with
the same evasion that you normally use when Dirk points out your
error to you. Consider this a surrogate therapy session for you,
while Dirk is away.
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Marcel Luttgens
> On 24 Sep 2002, Marcel Luttgens wrote:
>
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<8e4i9.122511$8o4....@afrodite.telenet-ops.be>...
> > >
> > > So this is not a clue that SR is incoherent.
> > > [snipped at first error]
> >
> > Why can't you grasp the obvious ?
> >
>
> Dirk is on vacation at the Alps for a week, and he asked that I
> send you his regards when you show up. Regards from Dirk.
>
> However, please do not let Dirk's absence hold you back. Just
> assume that if he were here he would have pointed out the very
> same error that you continuously make. So, you can respond with
> the same evasion that you normally use when Dirk points out your
> error to you. Consider this a surrogate therapy session for you,
> while Dirk is away.
Thank you. I wish Dirk a nice stay.
Perhaps will he find a way to explain why he gets
t' = gt when an object is resting at the origin of S,
forgetting that the object's frame coincides with S
and is thus, in the LT context, physically inseparable
from S, but t' = t/g when there is no such object.
You could give your own version in the meantime.
Marcel Luttgens
Stephen Speicher
Thanks for the offer, but following your illogic gives me a
headache.
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02092...@posting.google.com...
> Stephen Speicher <s...@speicher.com> wrote in message news:<Pine.LNX.4.33.020924...@localhost.localdomain>...
> > On 24 Sep 2002, Marcel Luttgens wrote:
> >
> > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<8e4i9.122511$8o4....@afrodite.telenet-ops.be>...
> > > >
> > > > So this is not a clue that SR is incoherent.
> > > > [snipped at first error]
> > >
> > > Why can't you grasp the obvious ?
> > >
> >
> > Dirk is on vacation at the Alps for a week, and he asked that I
> > send you his regards when you show up. Regards from Dirk.
Thanks Stephen.
Early return from Austria (usually called the pissoir of Europe)
after 3 days of looking at the rain, reading books, playing chess,
watching the weather forecasts on tv and waiting for supper.
Better luck next year in France :-)
> >
> > However, please do not let Dirk's absence hold you back. Just
> > assume that if he were here he would have pointed out the very
> > same error that you continuously make. So, you can respond with
> > the same evasion that you normally use when Dirk points out your
> > error to you. Consider this a surrogate therapy session for you,
> > while Dirk is away.
>
> Thank you. I wish Dirk a nice stay.
>
> Perhaps will he find a way to explain why he gets
> t' = gt when an object is resting at the origin of S,
t' = g(t-vx/c^2) reduces to t' = gt
<==> x=0 or v=0 (and g=1)
<==> Events taking place on (y-z-plane of) observer S
or no relative movement between S and S'
> forgetting that the object's frame coincides with S
> and is thus, in the LT context, physically inseparable
> from S, but t' = t/g when there is no such object.
t = g(t'+vx'/c^2) reduces to t' = t/g
<==> x'=0 or v=0 (and g=1)
<==> Events taking place on (y'-z'-plane of) observer S'
or no relative movement between S and S'
If we exclude the trivial case of v = 0 (squash!), there is only
one event where both t'=gt and t'=t/g are satisfied:
Following the arrows:
x = 0 and x' = 0
and thus, using Uncle x'=g(x-vt) and Aunt x=g(x'+vt'):
t = 0 and t' = 0
i.o.w.:
Trivial coincidence event of S and S': (x,t) = (x',t') = (0,0)
HTH.
Dirk Vdm