SR IS DEFINITELY WRONG
Marcel Luttgens
In fact, like other SR experts, you forgot how the LT
have been derived.
In particular, time is counted from the instant at
which the origins of the two frames of reference S and S'
coincide. Iow, t=t'=0 when x=x'=0.
If a light pulse or an object starts from the coincident
origins at time t=t'=0, it will be at x=ct or x=Vt according
to S, and at x'=ct' or x'=V't' according to S'.
The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
Clearly, t' can be negative only if V>c, which is forbidden
by relativity.
Or the transformation t'=gamma(t-vx/c^2), applied to the event
x=a, t=0, gives t'= -gamma va/c^2, thus a negative time.
Consequently, the transformation is not generally applicable,
the Einsteinian space-time is physically wrong, and SRT itself
is definitely wrong.
No wonder that the classical LT, based on x=Vt, give false
results when x=a+Vt, which is the case of the event x=a, t=0.
In order to be applied to all possible cases, the LT must be
generalized to x'=gamma(x-vt) and t'=gamma(t-v(x-a)/c^2).
With x=a and t=0, x-a is of course 0, x'=gamma*a and
t'=0, a correct result.
The generalized LT can also be written
x'=gamma*a + gamma*t(V-v) and t'=gamma*t(1-Vv/c^2).
You wrote:
"An object is moving at the speed V in frame S.
It is at the position x = 1 at the time t = 0.
The object is thus moving according to the equation:
x = 1 + Vt.
Transformed to the frame S', this equation becomes:
x' = xo + ut'
where xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)
and u = (V-v)/(1 - vV/c^2)
So the position of the object in S' at t' = 0 is xo."
Your x0 is false. As a=1, it should be 1/sqrt(1-v^2/c^2).
Note also that the correct LT doesn't lead to silly infinities
any more.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> To Paul B. Andersen (paul.b....@hia.no)
>
> In fact, like other SR experts, you forgot how the LT
> have been derived.
> In particular, time is counted from the instant at
> which the origins of the two frames of reference S and S'
> coincide. Iow, t=t'=0 when x=x'=0.
> If a light pulse or an object starts from the coincident
> origins at time t=t'=0, it will be at x=ct or x=Vt according
> to S, and at x'=ct' or x'=V't' according to S'.
>
> The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> Clearly, t' can be negative only if V>c, which is forbidden
> by relativity.
Clearly, since t'=gamma*t(1-Vv/c^2), t' can be negative if t is negative.
Incredible idiot.
I have been patiently waiting for you. Welcome to the collection:
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html
Title: "SR IS DEFINITELY WRONG"
The rest of the stuff has been debunked a zillion times.
I'll leave it to Paul if he wants to play with it.
Dirk Vdm
Stewart Gordon
message news:243a1189.02031...@posting.google.com...
<snip>
> The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> Clearly, t' can be negative only if V>c, which is forbidden
> by relativity.
It's only "forbidden by relativity" if
(a) V is the velocity of a material object or a signal
(b) the supposed impossibility of FTL travel or communication is true
It doesn't forbid the general concept of a spacelike interval between two
events.
> Or the transformation t'=gamma(t-vx/c^2), applied to the event
> x=a, t=0, gives t'= -gamma va/c^2, thus a negative time.
<snip>
That's perfectly normal. It's called relativity of simultaneity.
Stewart.
--
My e-mail is valid but not my primary mailbox. Please keep replies on the
'group where everyone may benefit.
Dirk Van de moortel
"Stewart Gordon" <smjg...@yahoo.com> wrote in message news:a705b6$evo$1...@sun-cc204.lut.ac.uk...
> Marcel Luttgens <mlut...@club-internet.fr> strung together random words in
> message news:243a1189.02031...@posting.google.com...
> <snip>
> > The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> > written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> > Clearly, t' can be negative only if V>c, which is forbidden
> > by relativity.
>
> It's only "forbidden by relativity" if
> (a) V is the velocity of a material object or a signal
V *is* the velocity of an object as defined by our genius.
You snipped the definition. Here it is again:
| If a light pulse or an object starts from the coincident
| ^^^^^^^^^
| origins at time t=t'=0, it will be at x=ct or x=Vt according
| ^^^^^^
| to S, and at x'=ct' or x'=V't' according to S'.
| ^^^^^^^
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> > To Paul B. Andersen (paul.b....@hia.no)
> >
> > In fact, like other SR experts, you forgot how the LT
> > have been derived.
> > In particular, time is counted from the instant at
> > which the origins of the two frames of reference S and S'
> > coincide. Iow, t=t'=0 when x=x'=0.
> > If a light pulse or an object starts from the coincident
> > origins at time t=t'=0, it will be at x=ct or x=Vt according
> > to S, and at x'=ct' or x'=V't' according to S'.
> >
> > The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> > written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> > Clearly, t' can be negative only if V>c, which is forbidden
> > by relativity.
>
> Clearly, since t'=gamma*t(1-Vv/c^2), t' can be negative if t is negative.
> Incredible idiot.
You are the idiot. Of, course, mathematically, t' can be negative if t is
negative, but the LT has a *physical* meaning, that you systematically
ignore.
>
> I have been patiently waiting for you. Welcome to the collection:
> http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html
> Title: "SR IS DEFINITELY WRONG"
>
> The rest of the stuff has been debunked a zillion times.
> I'll leave it to Paul if he wants to play with it.
>
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<vtJk8.9858$DE4....@afrodite.telenet-ops.be>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> > > To Paul B. Andersen (paul.b....@hia.no)
> > >
> > > In fact, like other SR experts, you forgot how the LT
> > > have been derived.
> > > In particular, time is counted from the instant at
> > > which the origins of the two frames of reference S and S'
> > > coincide. Iow, t=t'=0 when x=x'=0.
> > > If a light pulse or an object starts from the coincident
> > > origins at time t=t'=0, it will be at x=ct or x=Vt according
> > > to S, and at x'=ct' or x'=V't' according to S'.
> > >
> > > The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> > > written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> > > Clearly, t' can be negative only if V>c, which is forbidden
> > > by relativity.
> >
> > Clearly, since t'=gamma*t(1-Vv/c^2), t' can be negative if t is negative.
> > Incredible idiot.
>
>
> You are the idiot. Of, course, mathematically, t' can be negative if t is
> negative, but the LT has a *physical* meaning, that you systematically
> ignore.
Ha yes, of course, I am the idiot.
So clearly, t' can be *physically negative* only if V>c, which is
forbidden by relativity.
Physical Negativity versus Mathematical Negativity.
Brilliant.
Dirk Vdm
Marcel Luttgens
> message news:243a1189.02031...@posting.google.com...
> <snip>
> > The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> > written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> > Clearly, t' can be negative only if V>c, which is forbidden
> > by relativity.
>
> It's only "forbidden by relativity" if
> (a) V is the velocity of a material object or a signal
> (b) the supposed impossibility of FTL travel or communication is true
>
> It doesn't forbid the general concept of a spacelike interval between two
> events.
>
> > Or the transformation t'=gamma(t-vx/c^2), applied to the event
> > x=a, t=0, gives t'= -gamma va/c^2, thus a negative time.
> <snip>
>
> That's perfectly normal. It's called relativity of simultaneity.
>
The correct transformation to use with the event x=a, t=0 is
t'=gamma(t-v(x-a)/c^2), not t'=gamma(t-vx/c^2).
If you disagree, explain the physical meaning of the solution t'= -gamma va/c^2,
instead of using unwarranted concepts.
> Stewart.
Marcel Luttgens
Stewart Gordon
news:243a1189.02031...@posting.google.com...
<snip>
> The correct transformation to use with the event x=a, t=0 is
Meaning?
> t'=gamma(t-v(x-a)/c^2), not t'=gamma(t-vx/c^2).
What's this for?
> If you disagree, explain the physical meaning of the solution t'= -gamma
va/c^2,
> instead of using unwarranted concepts.
It represents the fact that two spatially separated events that are
simultaneous in one inertial frame of reference, say x=0, t=0 and x=a, t=0,
happen at different times in other frames.
Marcel Luttgens
> Marcel Luttgens <mlut...@club-internet.fr> wrote in message
> news:243a1189.02031...@posting.google.com...
> <snip>
> > The correct transformation to use with the event x=a, t=0 is
>
> Meaning?
>
> > t'=gamma(t-v(x-a)/c^2), not t'=gamma(t-vx/c^2).
>
> What's this for?
>
> > If you disagree, explain the physical meaning of the solution t'= -gamma
> va/c^2,
> > instead of using unwarranted concepts.
>
> It represents the fact that two spatially separated events that are
> simultaneous in one inertial frame of reference, say x=0, t=0 and x=a, t=0,
> happen at different times in other frames.
>
> Stewart.
Sure, but could you present a physically credible scenario for the event
(x=a, t=0), using the two frames of reference S and S'?
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
Any event will do.
(x,t) = (a,0)
transforms to
(x',t') = ( gamma*a, -gamma*a*v / c^2 )
t' is negative, so for S' the event happens before the coincidence
of the two frames at (x,t) = (x',t') = (0,0)
Dirk Vdm
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
>
> In fact, like other SR experts, you forgot how the LT
> have been derived.
As the LT can be derived in an unlimited number of different
ways, that statement doesn't make much sense.
> In particular, time is counted from the instant at
> which the origins of the two frames of reference S and S'
> coincide.
It is of course an arbitrary choice which point on
the temporal axis you designate the value zero.
It has obviously no physical significance whatsoever.
You don't seriously propose that events with negative temporal
co-ordinate should be a problem, do you? :-)
The particular choice you are referring to is but
a convenient choice to make the LT simple.
(That is - the appearance of the LT obviously depend on
the relative orientations of the axes and the choice of
the origos in the involved frames of reference.)
> Iow, t=t'=0 when x=x'=0.
> If a light pulse or an object starts from the coincident
> origins at time t=t'=0, it will be at x=ct or x=Vt according
> to S, and at x'=ct' or x'=V't' according to S'.
So the equation x = ct (t>0) is the trajectory of the light pulse
in the K frame while x' = ct' (t>0) is the trajectory in the K' frame.
Of course no event on this particular trajectory can have negative
temporal co-ordinate in either of the frames; the trajectory
does not exist before the light was emitted.
So what?
Do you mean that if I emit a light pulse from the coinciding
origos tomorrow, then the LT is invalid today?
> The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
> written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
> Clearly, t' can be negative only if V>c, which is forbidden
> by relativity.
Mumbo-jumbo.
> Or the transformation t'=gamma(t-vx/c^2), applied to the event
> x=a, t=0, gives t'= -gamma va/c^2, thus a negative time.
> Consequently, the transformation is not generally applicable,
> the Einsteinian space-time is physically wrong, and SRT itself
> is definitely wrong.
The LT is a simple co-ordinate transform.
It transforms the co-ordinates of an event from one frame
of reference to another.
Why should it be a problem that the temporal co-ordinate
of an event is positive in one frame and negative in another?
Is it also a problem to you that you and I may disagree
about whether an event happened before or after midnight
according to our respective wrist watches? :-)
> No wonder that the classical LT, based on x=Vt, give false
> results when x=a+Vt, which is the case of the event x=a, t=0.
Why do you say that the LT is "based" on x = Vt?
It is not.
It is a consequence of the postulates of SR.
There are an infinite number of different ways you
from the postulates can derive the LT.
> In order to be applied to all possible cases, the LT must be
> generalized to x'=gamma(x-vt) and t'=gamma(t-v(x-a)/c^2).
> With x=a and t=0, x-a is of course 0, x'=gamma*a and
> t'=0, a correct result.
You are babbling.
The "classical LT" can be applied in all cases.
> The generalized LT can also be written
> x'=gamma*a + gamma*t(V-v) and t'=gamma*t(1-Vv/c^2).
Only if you define V = x/t, e.g. as the ratio between
the spatial and temporal co-ordinate of an arbitrary
event in K.
It should be blatantly obvious to any sane person
that there are no physical restrictions on this ratio.
It's an incredible stupid idea that the event
t = 0, x = 1 should in any way be "impossible"
just because the ratio V = x/t happens to be infinite.
Does New York disappear at midnight(0 hours)? :-)
> You wrote:
>
> "An object is moving at the speed V in frame S.
> It is at the position x = 1 at the time t = 0.
> The object is thus moving according to the equation:
> x = 1 + Vt.
>
> Transformed to the frame S', this equation becomes:
> x' = xo + ut'
> where xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)
> and u = (V-v)/(1 - vV/c^2)
>
> So the position of the object in S' at t' = 0 is xo."
And everything I wrote is correct.
> Your x0 is false. As a=1, it should be 1/sqrt(1-v^2/c^2).
You are wrong.
The co-ordinates of the event x = 1, t = 0 transforms
to x' = 1/sqrt(1-v^2/c^2), t' = - (v/c^2)/sqrt(1-v^2/c^2)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The event with the co-ordinates x' = xo, t' = 0 in the primed frame
is NOT the event that has the co-ordinates x = 1, t = 0 in the unprimed
frame.
Paul
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message > The correct transformation to use with the event x=a, t=0 is
> t'=gamma(t-v(x-a)/c^2), not t'=gamma(t-vx/c^2).
> If you disagree, explain the physical meaning of the solution t'= -gamma va/c^2,
> instead of using unwarranted concepts.
What a stupid question.
Do you really have a problem with the physical meaning of a point
in time just because it is designated by a negative number?
If I say: "two minutes before 0 hours (midnight)".
Do you really have to ask about the physical meaning
of that statement?
Paul
Marcel Luttgens
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> > To Paul B. Andersen (paul.b....@hia.no)
> >
> > In fact, like other SR experts, you forgot how the LT
> > have been derived.
>
> As the LT can be derived in an unlimited number of different
> ways, that statement doesn't make much sense.
>
> > In particular, time is counted from the instant at
> > which the origins of the two frames of reference S and S'
> > coincide.
>
> It is of course an arbitrary choice which point on
> the temporal axis you designate the value zero.
> It has obviously no physical significance whatsoever.
> You don't seriously propose that events with negative temporal
> co-ordinate should be a problem, do you? :-)
> > generalized to x'=gamma(x-vt) and t'=gamma(t-v(x-a)/c^2).
> > With x=a and t=0, x-a is of course 0, x'=gamma*a and
> > t'=0, a correct result.
>
> You are babbling.
> The "classical LT" can be applied in all cases.
>
False, as I will show below.
I snip some parts of your response, because they are irrelevant
if the "classical LT" cannot be applied in all cases.
> > The generalized LT can also be written
> > x'=gamma*a + gamma*t(V-v) and t'=gamma*t(1-Vv/c^2).
>
> Only if you define V = x/t, e.g. as the ratio between
> the spatial and temporal co-ordinate of an arbitrary
> event in K.
> It should be blatantly obvious to any sane person
> that there are no physical restrictions on this ratio.
>
> It's an incredible stupid idea that the event
> t = 0, x = 1 should in any way be "impossible"
> just because the ratio V = x/t happens to be infinite.
>
> Does New York disappear at midnight(0 hours)? :-)
Such event is only "possible" if x=a+Vt.
>
> > You wrote:
> >
> > "An object is moving at the speed V in frame S.
> > It is at the position x = 1 at the time t = 0.
> > The object is thus moving according to the equation:
> > x = 1 + Vt.
> >
> > Transformed to the frame S', this equation becomes:
> > x' = xo + ut'
> > where xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)
> > and u = (V-v)/(1 - vV/c^2)
> >
> > So the position of the object in S' at t' = 0 is xo."
>
> And everything I wrote is correct.
How did you obtain xo = sqrt(1-v^2/c^2)/(1 - vV/c^2) ?
~~~~~~~~~~~~~~~~~~
>
> > Your x0 is false. As a=1, it should be 1/sqrt(1-v^2/c^2).
>
> You are wrong.
> The co-ordinates of the event x = 1, t = 0 transforms
> to x' = 1/sqrt(1-v^2/c^2), t' = - (v/c^2)/sqrt(1-v^2/c^2)
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> The event with the co-ordinates x' = xo, t' = 0 in the primed frame
> is NOT the event that has the co-ordinates x = 1, t = 0 in the unprimed
> frame.
Wonderful!
You should ponder over the response of Dirk:
" (x,t) = (a,0)
transforms to
(x',t') = ( gamma*a, -gamma*a*v / c^2 )
t' is negative, so for S' the event happens before the coincidence
of the two frames at (x,t) = (x',t') = (0,0)."
He is right when he claims that for S', the event happens before
the coincidence of the two frames at (x,t) = (x',t') = (0,0).
There is no other "physical" explanation.
Let's call O the origin of S, and O' the origin of S'.
Either v, the velocity of S' relatively to S, is positive
(S' is going to the right) or v is negative (S' is going to the left).
The case of v=0 is not interesting here.
1) v is negative, t'= -gamma*a*v / c^2
Then O' is situated to the left of O, at a distance gamma*a*v^2 / c^2.
According to S', x' should be gamma*a + gamma*a*v^2/c^2 =
gamma*a (1 + v^2/c^2).
Or, the blind application of the LT gives x' = gamma*a, which is
contradictory.
2) v is positive, t'= -gamma*a*v / c^2
Then O' is situated to the right of O, at the same distance
gamma*a*v^2 / c^2. Hence, x' = gamma*a - gamma*a*v^2 / c^2 =
gamma*a (1 - v^2/c^2), another contradiction.
Iow, the classical LT cannot be used for the event (x,t) = (a,0).
Note that in both cases, O and O' will coincide when t'=0.
Only then, x=a and x'=gamma*a.
>
> Paul
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a7a6os$ece$1...@snipp.uninett.no>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> > > To Paul B. Andersen (paul.b....@hia.no)
> > >
> > > In fact, like other SR experts, you forgot how the LT
> > > have been derived.
> >
> > As the LT can be derived in an unlimited number of different
> > ways, that statement doesn't make much sense.
> >
> > > In particular, time is counted from the instant at
> > > which the origins of the two frames of reference S and S'
> > > coincide.
> >
> > It is of course an arbitrary choice which point on
> > the temporal axis you designate the value zero.
> > It has obviously no physical significance whatsoever.
> > You don't seriously propose that events with negative temporal
> > co-ordinate should be a problem, do you? :-)
> > > In order to be applied to all possible cases, the LT must be
> > > generalized to x'=gamma(x-vt) and t'=gamma(t-v(x-a)/c^2).
> > > With x=a and t=0, x-a is of course 0, x'=gamma*a and
> > > t'=0, a correct result.
> >
> > You are babbling.
> > The "classical LT" can be applied in all cases.
> >
>
> False, as I will show below.
No you won't.
You too apperently ;-)
>
> " (x,t) = (a,0)
> transforms to
> (x',t') = ( gamma*a, -gamma*a*v / c^2 )
>
> t' is negative, so for S' the event happens before the coincidence
> of the two frames at (x,t) = (x',t') = (0,0)."
>
> He is right when he claims that for S', the event happens before
> the coincidence of the two frames at (x,t) = (x',t') = (0,0).
>
> There is no other "physical" explanation.
>
> Let's call O the origin of S, and O' the origin of S'.
> Either v, the velocity of S' relatively to S, is positive
> (S' is going to the right) or v is negative (S' is going to the left).
> The case of v=0 is not interesting here.
>
> 1) v is negative, t'= -gamma*a*v / c^2
> Then O' is situated to the left of O, at a distance gamma*a*v^2 / c^2.
No, O' is on the right of O at a distance x' = gamma*a and
at time t' = -gamma*a*v / c^2 which is positive since v is negative.
Look at (x',t') = ( gamma*a, -gamma*a*v / c^2 )
> According to S', x' should be gamma*a + gamma*a*v^2/c^2 =
> gamma*a (1 + v^2/c^2).
> Or, the blind application of the LT gives x' = gamma*a, which is
> contradictory.
>
> 2) v is positive, t'= -gamma*a*v / c^2
> Then O' is situated to the right of O, at the same distance.
No, O' is on the right of O at a distance x' = gamma*a and
at time t' = -gamma*a*v / c^2 which is negative since v is positive
You should learn to look at equations before you use them
to play your stupid (and/or nasty) litle game.
Dirk Vdm
Dirk Van de moortel
>
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
> > 1) v is negative, t'= -gamma*a*v / c^2
> > Then O' is situated to the left of O, at a distance gamma*a*v^2 / c^2.
>
> No, O' is on the right of O at a distance x' = gamma*a and
> at time t' = -gamma*a*v / c^2 which is positive since v is negative.
> Look at (x',t') = ( gamma*a, -gamma*a*v / c^2 )
>
>
> > According to S', x' should be gamma*a + gamma*a*v^2/c^2 =
> > gamma*a (1 + v^2/c^2).
> > Or, the blind application of the LT gives x' = gamma*a, which is
> > contradictory.
> >
> > 2) v is positive, t'= -gamma*a*v / c^2
> > Then O' is situated to the right of O, at the same distance.
>
> No, O' is on the right of O at a distance x' = gamma*a and
> at time t' = -gamma*a*v / c^2 which is negative since v is positive
>
> You should learn to look at equations before you use them
> to play your stupid (and/or nasty) litle game.
>
> Dirk Vdm
Forget this. It was a fumble of my part ;-)
I Had been working like hell to recover a crashed disk all day.
But that's no excuse. I should have gone to bed.
Sorry.
Dirk Vdm
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...[snip]
Naughty Marcel. You apply the Galilean transformation *and* the
Lorentz transformation by adding the results together. Of course
you get a "contradiction" ;-)
The LT is supposed to *replace* the GT, not to complement it.
Nice try, as usual.
> Or, the blind application of the LT gives x' = gamma*a, which is
> contradictory.
>
> 2) v is positive, t'= -gamma*a*v / c^2
> Then O' is situated to the right of O, at the same distance
> gamma*a*v^2 / c^2. Hence, x' = gamma*a - gamma*a*v^2 / c^2 =
> gamma*a (1 - v^2/c^2), another contradiction.
Ditto.
>
> Iow, the classical LT cannot be used for the event (x,t) = (a,0).
Iow, the Luttgens cannot use the LT.
Just like Robert Winn, he has no clue about transformations,
coordinate systems, coordinates and events.
Dirk Vdm
Marcel Luttgens
Try to understand the scenario. Begin with the coicidence of O and O'
at t=0, t'=0. Then send S' to the left of S at a velocity v during a time
t'=gamma*a*v/c^2. Now O' is at a distance gamma*a*v^2/c^2 from O.
Put the clock of S' back 2t' and put the clock of S at t=0.
Then t'= -gamma*a*v/c^2, t=0, x'=gamma*a(1 + v2/c^2) instead of gamma*a and x=a.
Do the same with v positive.
If you disagree with my scenario, try to find another one instead of
just stupidly applying the false transformation.
In fact, the "classical" LT is only correct if v=0, when applied to the
event (a,0). I repeat, in such cases, the correct formula for time
is t'=gamma(t - (x-a)v/c^2).
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a7a6os$ece$1...@snipp.uninett.no>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
> > > To Paul B. Andersen (paul.b....@hia.no)
> > > You wrote:
> > >
> > > "An object is moving at the speed V in frame S.
> > > It is at the position x = 1 at the time t = 0.
> > > The object is thus moving according to the equation:
> > > x = 1 + Vt.
> > >
> > > Transformed to the frame S', this equation becomes:
> > > x' = xo + ut'
> > > where xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)
> > > and u = (V-v)/(1 - vV/c^2)
> > >
> > > So the position of the object in S' at t' = 0 is xo."
> >
> > And everything I wrote is correct.
>
>
> How did you obtain xo = sqrt(1-v^2/c^2)/(1 - vV/c^2) ?
> ~~~~~~~~~~~~~~~~~~
Straight forward application of the LT.
> > > Your x0 is false. As a=1, it should be 1/sqrt(1-v^2/c^2).
> >
> > You are wrong.
> > The co-ordinates of the event x = 1, t = 0 transforms
> > to x' = 1/sqrt(1-v^2/c^2), t' = - (v/c^2)/sqrt(1-v^2/c^2)
> > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> > The event with the co-ordinates x' = xo, t' = 0 in the primed frame
> > is NOT the event that has the co-ordinates x = 1, t = 0 in the unprimed
> > frame.
>
> Wonderful!
>
> You should ponder over the response of Dirk:
>
> " (x,t) = (a,0)
> transforms to
> (x',t') = ( gamma*a, -gamma*a*v / c^2 )
>
> t' is negative, so for S' the event happens before the coincidence
> of the two frames at (x,t) = (x',t') = (0,0)."
His answer is identical to mine (a = 1), so why should I ponder?
> He is right when he claims that for S', the event happens before
> the coincidence of the two frames at (x,t) = (x',t') = (0,0).
Of course. So what?
> There is no other "physical" explanation.
And what is it you want a "physical" explanation of?
> Let's call O the origin of S, and O' the origin of S'.
> Either v, the velocity of S' relatively to S, is positive
> (S' is going to the right) or v is negative (S' is going to the left).
> The case of v=0 is not interesting here.
>
> 1) v is negative, t'= -gamma*a*v / c^2
> Then O' is situated to the left of O, at a distance gamma*a*v^2 / c^2.
> According to S', x' should be gamma*a + gamma*a*v^2/c^2 =
> gamma*a (1 + v^2/c^2).
> Or, the blind application of the LT gives x' = gamma*a, which is
> contradictory.
I don't even understant what you are talking about.
But it is in fact quite simple.
In stead of using negative v, I will let S'
move to the left, and let v be a positive number.
As viewed from the S-frame:
===========================
t1' t2'
-|----------------|------> x' <- v
O' x1'
t1 t2
--|----------------|-------> x
O a
Since this is viewd from the S-frame, the co-ordinate clocks
t1 and t2 both reads zero.
1. An observer at O will when his clock t1 reads zero see:
- The adjacent point on the x'-axis is x' = 0.
- The adjacent t1' clock reads t1' = 0.
2. An observer at a will when his clock t1 reads zero see:
- The adjacent point on the x'-axis is x' = gamma*a
- The adjacent t2' clock reads t2' = gamma*a*v/c^2
Note that there are TWO events in this scenario,
E1: The origos are adjacent, x =0, t = 0 in S
x'=0, t'= 0 in S'.
E2: The body starts from x = a, t = 0 in S,
x' = gamma*a, t2' = gamma*a*v/c^2 in S'.
The reason why this can be drawn with one drawing as viewed
from S, is that these event are simultaneous in S.
As viewed from S', we have to use two different drawings,
one when E1 happens, another when E2 happens:
As viewed from the S-frame:
===========================
When E1 happens, e.g. t' = 0 :
------------------------------
t1 t2
--|----------------|----------> x v ->
O a
t1' t2'
--|----------------|------|---> x'
O' x2' x1'
1. An observer at O' will see O adjacent, t1 reading 0.
Note that this is identical to 1. above, both observers
are present at the event E1, and they agree about what
their own and the other clock show.
The point a has not yet reached x1'.
An observer adjacent to a will be at x2' = a/gamma.
He will see the t2 clock read -a*v/c^2
When E2 happens, e.g. t' = gamma*a*v/c^2
-----------------------------------------
t1 t2
--|----------------|----------> x v ->
O a
t1' t2'
--|------|----------------|---> x'
O' x3' x1'
2. An observer at x1' will see a adjacent and t2 = 0.
Note that this is identical to 2. above, both observers
are present at the event E2, and they agree about what
their own and the other clock show.
O has now passed O', it is at the point x3' = gamma*a*v^2/c^2
An observer at x3' will see t1 = a*v/c^2
What is contradictory about this?
>
> 2) v is positive, t'= -gamma*a*v / c^2
> Then O' is situated to the right of O, at the same distance
> gamma*a*v^2 / c^2. Hence, x' = gamma*a - gamma*a*v^2 / c^2 =
> gamma*a (1 - v^2/c^2), another contradiction.
Nonsense again.
You obviously don't know how to apply the LT.
> Iow, the classical LT cannot be used for the event (x,t) = (a,0).
Or more precisely put:
Marcel Luttgens cannot apply the LT for the event (x,t) = (a,0).
The problem is Marcel Luttgens, not the LT.
> Note that in both cases, O and O' will coincide when t'=0.
> Only then, x=a and x'=gamma*a.
As observed in S.
So what?
You obviously don't understand that you are confusing
two different events.
Namely:
x = a, t = 0, x'=gamma*a, t' = gamma*a*v/c^2
and:
x = a, t = -a*v/c^2, x'=a/gamma, t' = 0
(In the case S' is moving to the left)
Paul
Dirk Van de moortel
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
[snip]
> Try to understand the scenario. Begin with the coicidence of O and O'
> at t=0, t'=0. Then send S' to the left of S at a velocity v during a time
> t'=gamma*a*v/c^2. Now O' is at a distance gamma*a*v^2/c^2 from O.
> Put the clock of S' back 2t' and put the clock of S at t=0.
> Then t'= -gamma*a*v/c^2, t=0, x'=gamma*a(1 + v2/c^2) instead of gamma*a and x=a.
> Do the same with v positive.
>
> If you disagree with my scenario, try to find another one instead of
> just stupidly applying the false transformation.
Caught in the act of trolling.
What I wrote in my first reply was nonsense.
The fact that you haven't noticed once more proves that you
don't even read what is replied to you.
> In fact, the "classical" LT is only correct if v=0, when applied to the
> event (a,0). I repeat, in such cases, the correct formula for time
> is t'=gamma(t - (x-a)v/c^2).
Yeah, and since v=0, this implies t'=gamma*t.
Besides, event (x,t) = (a,0) implies
x = 0
which implies
x-a = 0,
so again
t'=gamma*t
in your "correct" formula.
Do at least pretend to read my correction and my reprise.
Dirk Vdm
Marcel Luttgens
Your approach is too muddled up.
Either S and S' are separated by a distance d when t'=-gamma*a*a/c^2
(an idea of Dirk Van de moortel, that he rejected the next day),
or they coincide.
As for me, the correct transformation t'=gamma(t-v(x-a)/c^2),
based on x= a + Vt, shows that t'=0, x=a and x'=gamma*a when t=0.
Note that a and gamma*a are the x,x'-coordinates of the object
moving at V. The coordinates of the origins O and O' are both
zero, Iow the two frames coincide when t=t'=0.
Could you be clearer?
Simply give:
1) the distance d, as seen by S and S'.
2) The transform for t' if the object is moving according to
x=a+vt.
3) Just to be complete, the way you applied the LT to obtain
xo = sqrt(1-v^2/c^2)/(1 - vV/c^2).
Thank you.
Marcel Luttgens
Marcel Luttgens
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a7fg6t$qk9$1...@snipp.uninett.no>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02032...@posting.google.com...
> > > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a7a6os$ece$1...@snipp.uninett.no>...
> > > > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02031...@posting.google.com...
In my last message, I wrote:
"Either S and S' are separated by a distance d when t'=-gamma*a*a/c^2
(an idea of Dirk Van de moortel, that he rejected the next day),
or they coincide.
As for me, the correct transformation t'=gamma(t-v(x-a)/c^2),
based on x= a + Vt, shows that t'=0, x=a and x'=gamma*a when t=0.
Note that a and gamma*a are the x,x'-coordinates of the object
moving at V. The coordinates of the origins O and O' are both
zero, Iow the two frames coincide when t=t'=0.
Could you be clearer?
Simply give:
1) the distance d, as seen by S and S'.
2) The transform for t' if the object is moving according to
x=a+Vt.
3) Just to be complete, the way you applied the LT to obtain
xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)."
Can you give the transform corresponding to x=a+Vt, or not?
My answer is x'=gamma(x-vt)
t'=gamma(t-v(x-a)/c^2)
If you disagree, you (and other SR experts!) should at least
be able to show your solution.
Marcel Luttgens
Paul B. Andersen
> In my last message, I wrote:
>
> "Either S and S' are separated by a distance d when t'=-gamma*a*a/c^2
> (an idea of Dirk Van de moortel, that he rejected the next day),
> or they coincide.
This make no sense, gamma*a*a/c^2 isn't a time, it's a squared time.
> As for me, the correct transformation t'=gamma(t-v(x-a)/c^2),
> based on x= a + Vt, shows that t'=0, x=a and x'=gamma*a when t=0.
> Note that a and gamma*a are the x,x'-coordinates of the object
> moving at V. The coordinates of the origins O and O' are both
> zero, Iow the two frames coincide when t=t'=0.
>
> Could you be clearer?
>
> Simply give:
>
> 1) the distance d, as seen by S and S'.
> 2) The transform for t' if the object is moving according to
> x=a+Vt.
> 3) Just to be complete, the way you applied the LT to obtain
> xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)."
>
> Can you give the transform corresponding to x=a+Vt, or not?
And what the heck is "the transform according to x=a+Vt"
supposed to mean?
We are talking about the Lorentz transform, are we not?
The equation x = a+Vt is the trajectory of an object in S.
That is, it is a locus; all events in S fulfilling the equation
are events on the object's world line.
Some examples of such events:
E1: x=a, t=0.
E2: x=0, t= -a/V
E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value
I can however transform this locus to S'.
All you have to do is to substitute x and t by their
transformed expression, e.g.:
x = gamma(x' + vt'), t = gamma(t' + x'v/c^2)
Doing this, we get:
The trajectory of the object in S': x' = xo + ut'
where xo = a*sqrt(1-v^2/c^2)/(1 - vV/c^2)
and u = (V-v)/(1 - vV/c^2)
The events E1, E2 and E3 above are obviously still on the locus.
Transformed to S', the co-ordinates are:
E1: x'=gamma*a, t' = -gamma*a*v/c^2
E2: x'=gamma*a*v/V, t' = -gamma*a/V
E3: x'=a*sqrt(1-v^2/c^2)/(1 - vV/c^2),t'=0
Insert these into the equation:
x' = a*sqrt(1-v^2/c^2)/(1 - vV/c^2) + ((V-v)/(1 - vV/c^2))t'
and you will see that they all fit.
This is simple and straight forward.
It is the world line of the object expressed in two
different frames of reference.
So what is your problem?
So to your question:
> 1) the distance d, as seen by S and S'.
Which distance are you talking about?
I infer from the above that you are talking about
the distance between S and S'. But S' is moving relative to S.
So the distance obviously depend ot the time.
> My answer is x'=gamma(x-vt)
> t'=gamma(t-v(x-a)/c^2)
Which is utterly nonsensical.
This is a _different_ co-ordinate transformation from the LT.
Why do you think the transformation should change just because
there is a moving object?
Would the LT split into a number of different transformations
if there are a number of different moving objects?
> If you disagree, you (and other SR experts!) should at least
> be able to show your solution.
Which I have done a number of times.
But you seem very confused.
Paul
Marcel Luttgens
> Marcel Luttgens wrote:
> > In my last message, I wrote:
> >
> > "Either S and S' are separated by a distance d when t'=-gamma*a*a/c^2
> > (an idea of Dirk Van de moortel, that he rejected the next day),
> > or they coincide.
>
> This make no sense, gamma*a*a/c^2 isn't a time, it's a squared time.
Wise guy, you know that I meant t'=gamma*a*v/c^2.
Now, what is the distance d at time t'?
And you are wrong, because you use a tranform that doesn't
apply.
You should realize that the "classical" LT is only valid
if x=Vt, not if x=a+Vt.
Indeed, according to the "classical" LT,
x'=gamma(x-vt)
t'=gamma(t-xv/c^2)
So,
x'/t' = (x-vt)/(t-xv/c^2) = (x/t - v)/(1-(x/t)v/c^2)
Or x'/t' and x/t are velocities. Let's call them V' and V.
We get V'=(V-v)/(1-Vv/c^2), the formula giving the relativistic
addition of velocities.
But what are those velocities? Well, V is the velocity of an
object starting from the coincident origin of S and S' at
time t=t'=0, and v is of course the velocity of S' wrt to S.
No wonder that for S, the object is moving at V, and will be
at a distance x=Vt after a time t.
But for S', the object is moving at V', and will be at a distance
x'=V't'.
As x=Vt, t'=gamma*t(1-Vv/c^2) and x'=V't'=gamma*t(V-v), which
is equivalent to t'=gamma(t-xv/c^2) and x'=gamma(x-vt).
With x=a+Vt, the object starts at a point P situated at a
distance a from the coincident origin, again at t=t'=0.
Then the LT is x'=gamma(x-vt), which is unchanged,
but t' becomes of course gamma(t-v(x-a)/c^2).
How don't SR experts realize the evidence? They must
be somewhat brainwashed after a century of Einsteinian
relativity!
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02040...@posting.google.com...
What utter nonsense.
How is it possible to believe that the validity of a co-ordinate
transform depend on how objects are moving?
This is simply too idiotic and nonsensical.
> Indeed, according to the "classical" LT,
> x'=gamma(x-vt)
> t'=gamma(t-xv/c^2)
> So,
> x'/t' = (x-vt)/(t-xv/c^2) = (x/t - v)/(1-(x/t)v/c^2)
> Or x'/t' and x/t are velocities. Let's call them V' and V.
This is nonsense. Utter nonsense.
x/t and x'/t' are not the velocities of anything.
> We get V'=(V-v)/(1-Vv/c^2), the formula giving the relativistic
> addition of velocities.
> But what are those velocities? Well, V is the velocity of an
> object starting from the coincident origin of S and S' at
> time t=t'=0, and v is of course the velocity of S' wrt to S.
> No wonder that for S, the object is moving at V, and will be
> at a distance x=Vt after a time t.
> But for S', the object is moving at V', and will be at a distance
> x'=V't'.
> As x=Vt, t'=gamma*t(1-Vv/c^2) and x'=V't'=gamma*t(V-v), which
> is equivalent to t'=gamma(t-xv/c^2) and x'=gamma(x-vt).
Mumbo jumbo.
> With x=a+Vt, the object starts at a point P situated at a
> distance a from the coincident origin, again at t=t'=0.
I will give you a hint: Relativity of simultaneity!
The object does NOT start when t'=0.
I have told you this a number of times, see my previous postings.
I won't bother to repeat the same thing over and over
since it obviously is to no avail anyway.
> Then the LT is x'=gamma(x-vt), which is unchanged,
> but t' becomes of course gamma(t-v(x-a)/c^2).
>
> How don't SR experts realize the evidence? They must
> be somewhat brainwashed after a century of Einsteinian
> relativity!
I understand what I am talking about, you obviously don't.
I won't bother to quarrel about these stupidities of yours any more.
You have got the correct answer. I see no point in repeating it.
Have a nice day anyway.
Paul
Marcel Luttgens
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02040...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a8ehf4$9fm$1...@snipp.uninett.no>...
> > > Marcel Luttgens wrote:
> > > > In my last message, I wrote:
> > > >
> > > > "Either S and S' are separated by a distance d when t'=-gamma*a*a/c^2
> > > > (an idea of Dirk Van de moortel, that he rejected the next day),
> > > > or they coincide.
> > >
> > > This make no sense, gamma*a*a/c^2 isn't a time, it's a squared time.
> >
> > Wise guy, you know that I meant t'=gamma*a*v/c^2.
> > Now, what is the distance d at time t'?
You can't answer! No wonder, such t' has no physical meaning!
So, for you, a formula transforming x=Vt gives correct results
when x=a+Vt!
> > Indeed, according to the "classical" LT,
> > x'=gamma(x-vt)
> > t'=gamma(t-xv/c^2)
> > So,
> > x'/t' = (x-vt)/(t-xv/c^2) = (x/t - v)/(1-(x/t)v/c^2)
> > Or x'/t' and x/t are velocities. Let's call them V' and V.
>
> This is nonsense. Utter nonsense.
> x/t and x'/t' are not the velocities of anything.
Then what do x/t and x'/t' represent, if not velocities?
> > We get V'=(V-v)/(1-Vv/c^2), the formula giving the relativistic
> > addition of velocities.
> > But what are those velocities? Well, V is the velocity of an
> > object starting from the coincident origin of S and S' at
> > time t=t'=0, and v is of course the velocity of S' wrt to S.
> > No wonder that for S, the object is moving at V, and will be
> > at a distance x=Vt after a time t.
> > But for S', the object is moving at V', and will be at a distance
> > x'=V't'.
> > As x=Vt, t'=gamma*t(1-Vv/c^2) and x'=V't'=gamma*t(V-v), which
> > is equivalent to t'=gamma(t-xv/c^2) and x'=gamma(x-vt).
>
> Mumbo jumbo.
Not at all. Let V=c (a light pulse is sent from the coincident
origin). Then V'=(c-v)/(1-v/c)=c, thus x/t and x'/t' are both
equal to c, which is correct.
How can you claim that x/t and x'/t' are not velocities?
>
> > With x=a+Vt, the object starts at a point P situated at a
> > distance a from the coincident origin, again at t=t'=0.
>
> I will give you a hint: Relativity of simultaneity!
> The object does NOT start when t'=0.
> I have told you this a number of times, see my previous postings.
>
> I won't bother to repeat the same thing over and over
> since it obviously is to no avail anyway.
Then where does he starts? Give a concrete example, just tell us
the distance separating the origins of S and S' when t'=-gamma*a*v/c^2.
Remember that t' is obtained by applying the transformation
t'=gamma(t-xv/c^2) to the event x=a, t=0.
>
> > Then the LT is x'=gamma(x-vt), which is unchanged,
> > but t' becomes of course gamma(t-v(x-a)/c^2).
> >
> > How don't SR experts realize the evidence? They must
> > be somewhat brainwashed after a century of Einsteinian
> > relativity!
>
> I understand what I am talking about, you obviously don't.
>
The opposite is true.
[snip]
>
> Have a nice day anyway.
>
Thanks, you too.
> Paul
Marcel Luttgens
Marcel Luttgens
Paul B. Andersen wrote:
"The equation x = a+Vt is the trajectory of an object in S.
That is, it is a locus; all events in S fulfilling the equation
are events on the object's world line.
Some examples of such events:
E1: x=a, t=0.
E2: x=0, t= -a/V
E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
Paul B. Andersen is a typical SR expert. He doesn't realize
that an equation must have a physical meaning.
For instance, the t coordinate of E2 is negative, which is
physically wrong.
Still worse with E3:
He says, v any value, but v cannot exceed c, nor can V
exceed c.
He doesn't even realize that his coordinates (x,t) are
physically nonsensical.
x can of course take any value. When x=0, we have
0=a/(1-vV/c^2). This is possible if a=0, but then, x=Vt,
or he is trying to "transform the locus x=a+Vt, not x=Vt,
to S'", so one must assume that a<>0.
Then we have 0=a+Vt, thus V = -a/t.
As t= a*v/(c^2*(1-vV/c^2)), V = (-a/av)*c^2*(1-vV/c^2)
Hence, Vv = -c^2+Vv, and c=0!
Iow, the S coordinates of E3 are only valid if c=0!
Paul B. Andersen is a true SR expert!
He then persists in his errors, claiming:
"I can however transform this locus to S'.
All you have to do is to substitute x and t by their
transformed expression, e.g.:
x = gamma(x' + vt'), t = gamma(t' + x'v/c^2)."
But he ignores that the "classical" LT apply only to cases
where x=Vt, not x=a+Vt!
Indeed, according to the "classical" LT,
x'=gamma(x-vt)
t'=gamma(t-xv/c^2)
So,
x'/t' = (x-vt)/(t-xv/c^2) = (x/t - v)/(1-(x/t)v/c^2)
To everybody, x'/t' and x/t correspond to velocities.
Let's call them V' and V.
We get V'=(V-v)/(1-Vv/c^2), the formula giving the relativistic
addition of velocities. This should prove to SR experts
that V' and V are indeed velocities.
But what are those velocities? Well, the logical explanation
is that V is the velocity of an object starting from the
coincident origin of S and S' at time t=t'=0, and v is of course
the velocity of S' wrt to S.
For S, the object is moving at V, and will be at a distance
x=Vt after a time t.
But for S', the object is moving at V', and will be at a distance
x'=V't'.
As x=Vt, t'=gamma*t(1-Vv/c^2) and x'=V't'=gamma*t(V-v), which
is equivalent to t'=gamma(t-xv/c^2) and x'=gamma(x-vt).
With x=a+Vt, the object starts at a point P situated at a
distance a from the coincident origin, again at t=t'=0.
Then the LT is x'=gamma(x-vt), which is unchanged,
but t' becomes of course gamma(t-v(x-a)/c^2).
But Paul B. Andersen claims that x/t and x'/t' are not
velocities. What are they in his opinion?
I am wondering why SR experts don't realize the evidence.
The only explanation I see is that they must be somewhat
brainwashed after a century of Einsteinian relativity!
Marcel Luttgens
Marcel Luttgens
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a8il1d$io9$1...@snipp.uninett.no>...
>
> Paul B. Andersen wrote:
>
> "The equation x = a+Vt is the trajectory of an object in S.
> That is, it is a locus; all events in S fulfilling the equation
> are events on the object's world line.
> Some examples of such events:
>
> E1: x=a, t=0.
> E2: x=0, t= -a/V
> E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
>
> Paul B. Andersen is a typical SR expert. He doesn't realize
> that an equation must have a physical meaning.
> For instance, the t coordinate of E2 is negative, which is
> physically wrong.
He should write t= -(-a)/V when a is negative, or t= -a/(-V)
when V is negative.
> Still worse with E3:
> He says, v any value, but v cannot exceed c, nor can V
> exceed c.
> He doesn't even realize that his coordinates (x,t) are
> physically nonsensical.
> x can of course take any value. When x=0, we have
> 0=a/(1-vV/c^2). This is possible if a=0, but then, x=Vt,
> or he is trying to "transform the locus x=a+Vt, not x=Vt,
> to S'", so one must assume that a<>0.
> Then we have 0=a+Vt, thus V = -a/t.
> As t= a*v/(c^2*(1-vV/c^2)), V = (-a/av)*c^2*(1-vV/c^2)
> Hence, Vv = -c^2+Vv, and c=0!
> Iow, the S coordinates of E3 are only valid if c=0!
> Paul B. Andersen is a true SR expert!
Anyhow, his E3 coordinates are awfully contradictory:
From x=a/(1-vV/c^2), we get (1-vV/c^2)=a/x
From t= a*v/(c^2*(1-vV/c^2)), we get (1-vV/c^2)=a*v/c^2*t
Thus, a/x=a*v/c^2*t, and x = c^2 * t/v.
Iow, his x-coordinate is in fact independent of V, contrary
to his basic relation x=a+Vt!
He should try to explain how x can be physically related to v,
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.0204...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a8il1d$io9$1...@snipp.uninett.no>...
>
> Paul B. Andersen wrote:
>
> "The equation x = a+Vt is the trajectory of an object in S.
> That is, it is a locus; all events in S fulfilling the equation
> are events on the object's world line.
> Some examples of such events:
>
> E1: x=a, t=0.
> E2: x=0, t= -a/V
> E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
>
> Paul B. Andersen is a typical SR expert. He doesn't realize
> that an equation must have a physical meaning.
> For instance, the t coordinate of E2 is negative, which is
> physically wrong.
The object is moving in the positive x-direction with
the speed V. At t = 0, it is at the point x = a.
Isn't it then blatantly obvious that the object passed through
x = 0 a time a/V EARLIER?
Do I really have to tell you that a/V before 0 is -a/V?
Why do you fail to understand something as simple as this?
One wouldn't expect that to be possible!
> Still worse with E3:
> He says, v any value, but v cannot exceed c, nor can V
> exceed c.
Wrong.
In the expression above v is but an arbitrary parameter
with dimension speed (length/time), which can have ANY
value. You simply get one pair of co-ordinates for each
value of v.
A very concrete example:
V = 0.1c, v = 2c.
x = 1.25*a, t = 2.5*a/c
This event is obviously on the locus: x = a + 0.1ct
> He doesn't even realize that his coordinates (x,t) are
> physically nonsensical.
> x can of course take any value. When x=0, we have
> 0=a/(1-vV/c^2). This is possible if a=0, but then, x=Vt,
> or he is trying to "transform the locus x=a+Vt, not x=Vt,
> to S'", so one must assume that a<>0.
> Then we have 0=a+Vt, thus V = -a/t.
> As t= a*v/(c^2*(1-vV/c^2)), V = (-a/av)*c^2*(1-vV/c^2)
> Hence, Vv = -c^2+Vv, and c=0!
> Iow, the S coordinates of E3 are only valid if c=0!
> Paul B. Andersen is a true SR expert!
This isn't even funny. Just incredible stupid.
Why don't you use the same logic on E1?
x can of course have any value.
When x = 0, we have 0 = a.
Hence a = 0!
Marcel Luttgens is a true mathematical genius!
> He then persists in his errors, claiming:
>
> "I can however transform this locus to S'.
> All you have to do is to substitute x and t by their
> transformed expression, e.g.:
> x = gamma(x' + vt'), t = gamma(t' + x'v/c^2)."
>
> But he ignores that the "classical" LT apply only to cases
> where x=Vt, not x=a+Vt!
I can only repeat my astonishment over the weird idea
that the co-ordinates of an object can only be transformed
if it moves in a specific way.
Paul
Marcel Luttgens
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.0204...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a8il1d$io9$1...@snipp.uninett.no>...
> >
> > Paul B. Andersen wrote:
> >
> > "The equation x = a+Vt is the trajectory of an object in S.
> > That is, it is a locus; all events in S fulfilling the equation
> > are events on the object's world line.
> > Some examples of such events:
> >
> > E1: x=a, t=0.
> > E2: x=0, t= -a/V
> > E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
> >
> > Paul B. Andersen is a typical SR expert. He doesn't realize
> > that an equation must have a physical meaning.
> > For instance, the t coordinate of E2 is negative, which is
> > physically wrong.
>
> The object is moving in the positive x-direction with
> the speed V. At t = 0, it is at the point x = a.
> Isn't it then blatantly obvious that the object passed through
> x = 0 a time a/V EARLIER?
> Do I really have to tell you that a/V before 0 is -a/V?
>
> Why do you fail to understand something as simple as this?
> One wouldn't expect that to be possible!
>
As the origins of S and S' concide at t=t'=0, where was S', in your
opinion, at a time a/V EARLIER? When and how did you synchronize
the clocks of S and S' before the coincidence of the origins?
> > Still worse with E3:
> > He says, v any value, but v cannot exceed c, nor can V
> > exceed c.
>
> Wrong.
> In the expression above v is but an arbitrary parameter
> with dimension speed (length/time), which can have ANY
> value. You simply get one pair of co-ordinates for each
> value of v.
> A very concrete example:
> V = 0.1c, v = 2c.
> x = 1.25*a, t = 2.5*a/c
> This event is obviously on the locus: x = a + 0.1ct
>
This is the origin of your misconception. The LT transform
distances and times between physical objects whose velocity
must be less than light velocity. They are more than pure
mathematical equations.
> > He doesn't even realize that his coordinates (x,t) are
> > physically nonsensical.
> > x can of course take any value. When x=0, we have
> > 0=a/(1-vV/c^2). This is possible if a=0, but then, x=Vt,
> > or he is trying to "transform the locus x=a+Vt, not x=Vt,
> > to S'", so one must assume that a<>0.
> > Then we have 0=a+Vt, thus V = -a/t.
> > As t= a*v/(c^2*(1-vV/c^2)), V = (-a/av)*c^2*(1-vV/c^2)
> > Hence, Vv = -c^2+Vv, and c=0!
> > Iow, the S coordinates of E3 are only valid if c=0!
> > Paul B. Andersen is a true SR expert!
>
> This isn't even funny. Just incredible stupid.
> Why don't you use the same logic on E1?
> x can of course have any value.
> When x = 0, we have 0 = a.
> Hence a = 0!
> Marcel Luttgens is a true mathematical genius!
>
Why couldn't a, unlike c, be 0?
> > He then persists in his errors, claiming:
> >
> > "I can however transform this locus to S'.
> > All you have to do is to substitute x and t by their
> > transformed expression, e.g.:
> > x = gamma(x' + vt'), t = gamma(t' + x'v/c^2)."
> >
> > But he ignores that the "classical" LT apply only to cases
> > where x=Vt, not x=a+Vt!
>
> I can only repeat my astonishment over the weird idea
> that the co-ordinates of an object can only be transformed
> if it moves in a specific way.
>
Very simple. When x=Vt, x'=V't', and x/t=V, x'/t'=V'.
From x = gamma(x' + vt'), t = gamma(t' + x'v/c^2), one gets
x/t = V = (V'+v)/(1+V'v/c^2), or V' = (V-v)/(1-Vv/c^2),
the expression for the relativistic addition of velocities.
But this this not possible with x=a+Vt.
The fact that the LT straightforwardly lead to such addition
is not a coincidence, as you seem to believe. They were obtained
from a thought experiment using the velocity c, with the
restriction x=ct, x'=ct'. And indeed, when V=c,
V' = (c-v)/(1-cv/c^2) = c. The LT imply that no object can
go faster than c, and also that the addition of velocities
is relativistic.
You will understand the physical meaning of the LT only if
you abandon the idea that v and V are but arbitrary parameters
with dimension speed (length/time), which can have ANY
value.
>
> Paul
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02040...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a8s8to$j2d$1...@snipp.uninett.no>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.0204...@posting.google.com...
> > > But he ignores that the "classical" LT apply only to cases
> > > where x=Vt, not x=a+Vt!
> >
> > I can only repeat my astonishment over the weird idea
> > that the co-ordinates of an object can only be transformed
> > if it moves in a specific way.
> Very simple. When x=Vt, x'=V't', and x/t=V, x'/t'=V'.
> From x = gamma(x' + vt'), t = gamma(t' + x'v/c^2), one gets
> x/t = V = (V'+v)/(1+V'v/c^2), or V' = (V-v)/(1-Vv/c^2),
> the expression for the relativistic addition of velocities.
> But this this not possible with x=a+Vt.
Of course it is. Nothing could be simpler.
I have shown you a number of times that the locus: x = a + Vt
transforms to: x' = xo + ut'
where xo = a*sqrt(1-v^2/c^2)/(1 - vV/c^2)
and u = (V-v)/(1 - vV/c^2)
See that u, Marcel?
Paul
Marcel Luttgens
And I have shown you that your demonstration is false.
But you should rather respond to my message.
>
> See that u, Marcel?
>
> Paul
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02041...@posting.google.com...
Read my previous postings again.
You will find all the answers including the correct solution there.
It is just too stupid to quarrel about basic issues like this.
Paul
Marcel Luttgens
I remind you that you wrote:
"The equation x = a+Vt is the trajectory of an object in S.
That is, it is a locus; all events in S fulfilling the equation
are events on the object's world line.
Some examples of such events:
E1: x=a, t=0.
E2: x=0, t= -a/V
E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
The problem is that you don't even realize how wrong you are.
Your example S3 for instance is self-contradictory, because
a is a constant in the relation x=a+Vt, or a=x-Vt.
Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
but a function of v,V and t!
Otoh, claiming that "v and V are but arbitrary parameters
with dimension speed (length/time), which can have ANY
value" is plain crackpottery!
Marcel Luttgens
Dirk Van de moortel
This is a special x on the locus x = a + Vt, it is the x for event E3
and it has a corresponding t given by t = xv/c^2 for this x of event E3.
> By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
> everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
> but a function of v,V and t!
This t is a specific t for event E3, namely t = a*v/(c^2*(1-vV/c^2)
Put it in the equation a=((c^2/v)-V)*t and you get
a = a
which is not a function of v, V and t.
>
> Otoh, claiming that "v and V are but arbitrary parameters
> with dimension speed (length/time), which can have ANY
> value" is plain crackpottery!
Confusing your ignorance with the correctness of a theory of which
you don't even remotely understand the very basic language: plain
simple elementary algebra and analytic geometry.
Enlighten us: are you fumbling or cheating or both?
Even if you are cheating, you thoroughly underestimate the intelligence
of the people you are talking to.
Idiot.
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02041...@posting.google.com...
> > > Read my previous postings again.
> > > You will find all the answers including the correct solution there.
> > > It is just too stupid to quarrel about basic issues like this.
> > >
> > > Paul
> >
> > I remind you that you wrote:
> >
> > "The equation x = a+Vt is the trajectory of an object in S.
> > That is, it is a locus; all events in S fulfilling the equation
> > are events on the object's world line.
> > Some examples of such events:
> >
> > E1: x=a, t=0.
> > E2: x=0, t= -a/V
> > E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
> >
> > The problem is that you don't even realize how wrong you are.
> > Your example S3 for instance is self-contradictory, because
> > a is a constant in the relation x=a+Vt, or a=x-Vt.
> > Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
>
> This is a special x on the locus x = a + Vt, it is the x for event E3
> and it has a corresponding t given by t = xv/c^2 for this x of event E3.
>
And yet, x=(c^2/v)*t, a physical nonsense, because c^2/v is a
velocity, and no velocity can physically exceed c.
> > By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
> > everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
> > but a function of v,V and t!
>
> This t is a specific t for event E3, namely t = a*v/(c^2*(1-vV/c^2)
> Put it in the equation a=((c^2/v)-V)*t and you get
> a = a
> which is not a function of v, V and t.
>
Of course, a = a, when you replace t by a*v/(c^2*(1-vV/c^2).
But the general relation a=((c^2/v)-V)*t shows that a is not a
constant, as it should be according to x=a+Vt.
> >
> > Otoh, claiming that "v and V are but arbitrary parameters
> > with dimension speed (length/time), which can have ANY
> > value" is plain crackpottery!
>
> Confusing your ignorance with the correctness of a theory of which
> you don't even remotely understand the very basic language: plain
> simple elementary algebra and analytic geometry.
> Enlighten us: are you fumbling or cheating or both?
> Even if you are cheating, you thoroughly underestimate the intelligence
> of the people you are talking to.
> Idiot.
>
> Dirk Vdm
The idiot is you. It is not your fault, you have been brainwashed, like
the majority of SRists.
Mahtematically, you can use the LT, but nevertheless obtain nonsensical
results. For instance, I can straightforwardly transform
x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), using my relations
x'=gamma*t*(V1-v),
t'=gamma*t(1-V1v/c^2),
where V1=c^2/v (from x=(c^2/v)t).
As V1-v = (c^2/v)-v = (c^2-v^2)/v = (c^2/v)*(1-v^2/c^2), you get
x' = gamma * a*v/(c^2*(1-vV/c^2)) * (c^2/v)*(1-v^2/c^2)
x' = a*sqrt(1-v^2/c^2)/(1-vV/c^2)
t' = gamma*t*(1-(c^2/v)*(v/c^2)) = 0
Those values of x' and t', obtained with my relations for x=V1t,
are identical to those obtained by Paul B. Andersen:
"The events E1, E2 and E3 above are obviously still on the locus.
Transformed to S', the co-ordinates are:
E1: x'=gamma*a, t' = -gamma*a*v/c^2
E2: x'=gamma*a*v/V, t' = -gamma*a/V
E3: x'=a*sqrt(1-v^2/c^2)/(1 - vV/c^2),t'=0"
Nevertheless, they are physically false, because V1 is greater than c.
Note that I did't use my transform for x=a+Vt.
Hopefully, you are intelligent enough to realize that SR cannot
be correct, because it is *blindly* based on the Lorentz transformation.
Marcel Luttgens
Dirk Van de moortel
I'm not going into this any more.
Bye
Dirk Vdm
Marcel Luttgens
Because you are unable to find a flaw in the above and accept that
SR is wrong. You are a desperate but typical case.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02041...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<2EZt8.15570$Ze....@afrodite.telenet-ops.be>...
> > "Marcel Luttgens" <mlut...@club-internet.fr> wrote in message news:243a1189.02041...@posting.google.com...
> > > Hopefully, you are intelligent enough to realize that SR cannot
> > > be correct, because it is *blindly* based on the Lorentz transformation.
> > >
> > > Marcel Luttgens
> >
> > I'm not going into this any more.
> > Bye
> >
> > Dirk Vdm
>
> Because you are unable to find a flaw in the above and accept that
> SR is wrong. You are a desperate but typical case.
>
> Marcel Luttgens
1) Suppose x = 1
then x^2 = 1
2) Suppose x = 2
then x^2 = 4
"HA," says Marcel, "that's impossible: if you suppose that x = 1
and then that x = 2, then you suppose that 1 = 2. That's clearly
impossible, therefore you are wrong."
Dirk Vdm
Bilge
Re: SR IS DEFINITELY WRONG to usenet:
>"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
>wrote in message news:<2EZt8.15570$Ze....@afrodite.telenet-ops.be>...
>>
>> Bye
>>
>
>SR is wrong. You are a desperate but typical case.
Since your lack of logic is the flaw, there's not much anyone but you
can do to fix the problem.
Marcel Luttgens
How clever! A typical response of a SR expert.
Marcel Luttgens
Marcel Luttgens
How clever! A typical response of a SR expert.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0204...@posting.google.com...
How clever! A typical response of a SR inpert.
Dirk Vdm
Paul B. Andersen
Marcel, your vendetta isn't against SR, but against reason and logic.
You are simply stating a lot of mathematical nonsense.
Anybody with the slightest knowledge of math will see
this, and find it at best funny.
But let's review some of it:
Paul B. Andersen wrote:
| "The equation x = a+Vt is the trajectory of an object in S.
| That is, it is a locus; all events in S fulfilling the equation
| are events on the object's world line.
| Some examples of such events:
|
| E1: x=a, t=0.
| E2: x=0, t= -a/V
| E3: x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2)), v any value."
Since the event E3 seems to confuse Marcel, I will comment
it a bit:
We can obviously express the x co-ordinate of the object
as a function of a parameter v with dimension m/s.
The v isn't the speed of anything, it is but a parameter.
The function f itself is dimensionless.
The function f can be any monotone function of v.
x = a*f(v)
We get a specific x for any value of the parameter v.
The t co-ordinate of the object when it is at the position x,
is obviously:
t = (x - a)/V = (a/V)*(f(v)-1)
I did of course have a reason for my particular choice of f,
but that's irrelevant in this context.
The simple point is that:
x = a*f(v) , t = (a/V)*(f(v)-1)
or
x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2))
very obviously are co-ordinate pairs of the object with
the trajectory x = a + Vt for any value of the parameter v.
Marcel Luttgens wrote:
| The problem is that you don't even realize how wrong you are.
| Your example S3 for instance is self-contradictory, because
| a is a constant in the relation x=a+Vt, or a=x-Vt.
| Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
The ratio between the co-ordinates for any co-ordinate pair
of the body is x/t = V*f(v)/f(v)-1)
or when f(v) = 1/(1-vV/c^2), x/t=(c^2/v)
So what?
Since x and t are expressed as functions of the parameter v,
why does it surprise you that their ratio also must be a function of v?
Marcel Luttgens continues:
| By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
| everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
| but a function of v,V and t!
It is amazing that Marcel doesn't realize how incredible ridiculous
this is without being explained why.
And it is even more amazing that Marcel doesn't realize it after
being told!
Since Marcel seems to be very confused by the complexity of
the expressions, let's redo his mathematical acrobatics
with simpler expressions:
Given an event on the body with trajectory x = a + Vt:
x = 5a, t = 4a/V
Thus x = 1.25V*t
By replacing x by 1.25V*t in the basic relation a=x-Vt,
everybody obtains a = 0.25V*t.
Marcel's conclusion?
"a is no more a constant but a function of V and t." !!!
It is of course a very obvious circularity, he gets the answer
he started with, namely t = 4a/V.
When Dirk Van de moortel pointed this out:
| This t is a specific t for event E3, namely t = a*v/(c^2*(1-vV/c^2)
| Put it in the equation a=((c^2/v)-V)*t and you get
| a = a
| which is not a function of v, V and t.
Marcel Luttgens didn't blush in embarrassment as he should have;
he still understands nothing and replies:
| Of course, a = a, when you replace t by a*v/(c^2*(1-vV/c^2).
| But the general relation a=((c^2/v)-V)*t shows that a is not a
| constant, as it should be according to x=a+Vt.
Not surprising that Dirk give him up!
So Marcel, when people won't bother to discuss with you,
it isn't because they cannot find flaws in your math.
It is because they get tired of pointing out your numerous flaws
over and over again, to no avail.
Your math is utter nonsense, Marcel.
And that has nothing whatsoever with SR to do.
Nothing seems to be so simple that you cannot screw it up.
Who would believe that it was possible to say so much
ridiculous about something as simple as a body moving
along the trajectory x = a + Vt?
I think you must have broken some kind of a record!
No, Marcel. I will not discuss this anymore.
If you still haven't got it, you never will.
And I am confident you never will.
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:a9som2$r7o$1...@snipp.uninett.no...
>
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0204...@posting.google.com...
> > ro...@radioactivex.lebesque-al.net (Bilge) wrote in message news:<slrnabu6k...@radioactivex.lebesque-al.net>...
> > > Marcel Luttgens said some stuff about
> > > Re: SR IS DEFINITELY WRONG to usenet:
> > > >"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> > > >wrote in message news:<2EZt8.15570$Ze....@afrodite.telenet-ops.be>...
> > > >>
> > > >> I'm not going into this any more.
> > > >> Bye
> > > >>
> > > >
> > > >Because you are unable to find a flaw in the above and accept that
> > > >SR is wrong. You are a desperate but typical case.
> > >
> > > Since your lack of logic is the flaw, there's not much anyone but you
> > > can do to fix the problem.
> >
> > How clever! A typical response of a SR expert.
> >
> > Marcel Luttgens
>
> Marcel, your vendetta isn't against SR, but against reason and logic.
> You are simply stating a lot of mathematical nonsense.
> Anybody with the slightest knowledge of math will see
> this, and find it at best funny.
Indeed, I remember one of his arguments against SR being just as
well applicable to Galilean Relativity ;-)
Can't trace it immediately though. Perhaps I'll do my best to find it
if he decides to insist on being right after all. But only perhaps.
Dirk Vdm
Marcel Luttgens
Not exactly irrelevant, you wanted t'=0.
>
> The simple point is that:
> x = a*f(v) , t = (a/V)*(f(v)-1)
> or
> x=a/(1-vV/c^2), t= a*v/(c^2*(1-vV/c^2))
> very obviously are co-ordinate pairs of the object with
> the trajectory x = a + Vt for any value of the parameter v.
You seem to have no clue about the physical signification of what
you call the parameter v. For you, and SR experts in general, the LT
are just mathematical relations. This is false, because the LT
transform the coordinates of a so-called event between two frames
of reference moving at v relatively to each other, and v cannot exceed
c. Just look at gamma, which becomes infinite if v=c.
Otoh, V is the velocity of light or of a physical object, hence V<=c
according to SR itself.
>
> Marcel Luttgens wrote:
> | The problem is that you don't even realize how wrong you are.
> | Your example S3 for instance is self-contradictory, because
> | a is a constant in the relation x=a+Vt, or a=x-Vt.
> | Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
>
> The ratio between the co-ordinates for any co-ordinate pair
> of the body is x/t = V*f(v)/f(v)-1)
> or when f(v) = 1/(1-vV/c^2), x/t=(c^2/v)
> So what?
> Since x and t are expressed as functions of the parameter v,
> why does it surprise you that their ratio also must be a function of v?
>
So what? As v is necessarily less than c (remember gamma), you get
V=c^2/v > c, which is physically nonsensical, because no material
object can move faster than c.
> Marcel Luttgens continues:
> | By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
> | everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
> | but a function of v,V and t!
>
> It is amazing that Marcel doesn't realize how incredible ridiculous
> this is without being explained why.
> And it is even more amazing that Marcel doesn't realize it after
> being told!
>
> Since Marcel seems to be very confused by the complexity of
> the expressions, let's redo his mathematical acrobatics
> with simpler expressions:
>
> Given an event on the body with trajectory x = a + Vt:
> x = 5a, t = 4a/V
> Thus x = 1.25V*t
> By replacing x by 1.25V*t in the basic relation a=x-Vt,
> everybody obtains a = 0.25V*t.
> Marcel's conclusion?
> "a is no more a constant but a function of V and t." !!!
>
I simply wanted to show, by way of joking, that your event E3
was as ridiculous as the above one.
Simply replace x by 5a and t by 4a/V in your relation x=a+Vt,
and you get a=a! (cf. the bright demonstration of Dirk).
Moreover, you begin with x=a+Vt, and you end with x=1.25V*t,
just like with E3, where you end up with x=(c^2/v)*t. Don't you
see the contradiction?
> It is of course a very obvious circularity, he gets the answer
> he started with, namely t = 4a/V.
>
> When Dirk Van de moortel pointed this out:
> | This t is a specific t for event E3, namely t = a*v/(c^2*(1-vV/c^2)
> | Put it in the equation a=((c^2/v)-V)*t and you get
> | a = a
> | which is not a function of v, V and t.
>
> Marcel Luttgens didn't blush in embarrassment as he should have;
> he still understands nothing and replies:
> | Of course, a = a, when you replace t by a*v/(c^2*(1-vV/c^2).
> | But the general relation a=((c^2/v)-V)*t shows that a is not a
> | constant, as it should be according to x=a+Vt.
>
> Not surprising that Dirk give him up!
>
>
> So Marcel, when people won't bother to discuss with you,
> it isn't because they cannot find flaws in your math.
> It is because they get tired of pointing out your numerous flaws
> over and over again, to no avail.
>
> Your math is utter nonsense, Marcel.
> And that has nothing whatsoever with SR to do.
> Nothing seems to be so simple that you cannot screw it up.
>
> Who would believe that it was possible to say so much
> ridiculous about something as simple as a body moving
> along the trajectory x = a + Vt?
>
Now you admit that V is the velocity of a body, but nevertheless
claim that V can be >c! Is your body immaterial?
> I think you must have broken some kind of a record!
Look at the beam in your eye.
>
> No, Marcel. I will not discuss this anymore.
> If you still haven't got it, you never will.
> And I am confident you never will.
>
> Paul
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02042...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a9som2$r7o$1...@snipp.uninett.no>...
> > Marcel Luttgens wrote:
> > | The problem is that you don't even realize how wrong you are.
> > | Your example S3 for instance is self-contradictory, because
> > | a is a constant in the relation x=a+Vt, or a=x-Vt.
> > | Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
> >
> > The ratio between the co-ordinates for any co-ordinate pair
> > of the body is x/t = V*f(v)/f(v)-1)
> > or when f(v) = 1/(1-vV/c^2), x/t=(c^2/v)
> > So what?
> > Since x and t are expressed as functions of the parameter v,
> > why does it surprise you that their ratio also must be a function of v?
> >
>
> So what? As v is necessarily less than c (remember gamma), you get
> V=c^2/v > c, which is physically nonsensical, because no material
> object can move faster than c.
V = c^2/v ???? :-)
> > Marcel Luttgens continues:
> > | By replacing x by (c^2/v)*t in the basic relation a=x-Vt,
> > | everybody obtains a=((c^2/v)-V)*t, Iow, a is no more a constant,
> > | but a function of v,V and t!
> >
> > It is amazing that Marcel doesn't realize how incredible ridiculous
> > this is without being explained why.
> > And it is even more amazing that Marcel doesn't realize it after
> > being told!
> >
> > Since Marcel seems to be very confused by the complexity of
> > the expressions, let's redo his mathematical acrobatics
> > with simpler expressions:
> >
> > Given an event on the body with trajectory x = a + Vt:
> > x = 5a, t = 4a/V
> > Thus x = 1.25V*t
> > By replacing x by 1.25V*t in the basic relation a=x-Vt,
> > everybody obtains a = 0.25V*t.
> > Marcel's conclusion?
> > "a is no more a constant but a function of V and t." !!!
> >
>
> I simply wanted to show, by way of joking, that your event E3
> was as ridiculous as the above one.
> Simply replace x by 5a and t by 4a/V in your relation x=a+Vt,
> and you get a=a! (cf. the bright demonstration of Dirk).
And you still haven't got it?
Amazing!
> Moreover, you begin with x=a+Vt, and you end with x=1.25V*t,
> just like with E3, where you end up with x=(c^2/v)*t. Don't you
> see the contradiction?
I see your confusion.
I thought I spelled this out so it shouldn't be possible to
miss the point. But you don't understand a thing!
If it is the time that confuses you, let's do it in two
spatial dimensions in stead:
y = a + kx
This is a straight line. You understand that much, don't you?
y = 5a, x = 4a/k is a point on this line, P(4a/k,5a)
Can you fail to understand that? Think not.
Now, Marcel. If we insert the co-ordinates of this point
into the equation for the line, we get:
5a = a + (4a/k)k = 5a
which can be simplified to:
a = a or 0 = 0 or 1 = 1 or 100 = 100 or anything = anything
e.g, we get an equality.
And why is that, Marcel?
We get an equality BECAUSE the point is on the line, of course.
How can something as obvious as this surprise you, Marcel?
Let's look at your ridiculous manipulations above:
y/x = 5k/4 or y = 1.25kx
What does this mean, Marcel?
It simply means that for the particular point P(4a/k,5a)
the value of the y co-ordinate is 1.25k times the value
of the x co-ordinate. It does bloody obvious NOT mean
that the ratio between the y and x co-ordinate of any other
point on the line is 1.25k. How can this escape you, Marcel?
Some examples of points on the line, and their y/x ratios:
P(x,y) y/x
-------- ------
P1(0,a) infinite
P2(-a/k,0) 0
P3(a/k,2a) 2k
P4(a,(1+k)a) 1+k
P5(4a/k,5a) 1.25k
You can get just about any ratio you want by selecting a point.
I will even draw it for you:
y /
^ X P3 P4 and P5 outside of the drawn area
| /
| /
|/
a X P1
/|
/ |
/ |
-X---------------------> x
P2 0
Look at this figure, Marcel. Isn't it obvious that the y/x
ratio is different for every point on the line?
But what do you do?
You take the ratio between the y and x co-ordinate
of one of the points and insert it into the equation for
the line:
For example: P5(4a/k,5a) y = 1.25kx
1.25kx = a + kx
a = 0.25kx
And your conclusion? a is no more a constant!
But what you REALLY have done is of course to find
the x co-ordinate of the point on the line that
has this y/x ratio: a = 0.25kx => x = 4a/k
This isn't a contradiction, Marcel.
It is a circularity.
You get out what you put in.
I didn't think I would ever use time to explain math at this
elementary level.
But now I have. Probably to no avail.
Your problem is that you do not realize your incompetence
in math. You are tumbling around with equations without
having the faintest idea of what you are doing.
Ignorance of own ignorance is the worst kind of ignorance.
Paul
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02042...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a9som2$r7o$1...@snipp.uninett.no>...
>
> > > Marcel Luttgens wrote:
> > > | The problem is that you don't even realize how wrong you are.
> > > | Your example S3 for instance is self-contradictory, because
> > > | a is a constant in the relation x=a+Vt, or a=x-Vt.
> > > | Or, from x=a/(1-vV/c^2) and x=a+Vt, you get x=(c^2/v)*t.
<snip>
It is not only the time that confuses YOU !
You systematically forget that the LT are mathematical relations,
that have a physical meaning!
In the relation x = a + Vt, V cannot exceed c, hence an infinity
of points situated on the straight line are physically nonsensical.
For instance, the point x=5a, t=4a/V (your point P5) is only
valid if 1.25*V <= c.
But you have been brainwashed by a century of Einsteinian relativity,
and you will never realize that the applicability of the LT is
limited by c, hence that SR -which uses them as pure mathematical
relations- is necessarily false.
> Paul
Marcel Luttgens
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0205...@posting.google.com...
And what the hell has the LT with this to to?
> In the relation x = a + Vt, V cannot exceed c, hence an infinity
> of points situated on the straight line are physically nonsensical.
> For instance, the point x=5a, t=4a/V (your point P5) is only
> valid if 1.25*V <= c.
Really?
So if I am at x = a = 1 lightsecond at t = 0, and go at 0.9c,
I will NOT be at x = 5 lightseconds at the time t = 4/0.9 = 4.44 seconds?
Where will I then be?
> But you have been brainwashed by a century of Einsteinian relativity,
> and you will never realize that the applicability of the LT is
> limited by c, hence that SR -which uses them as pure mathematical
> relations- is necessarily false.
>
> > Paul
>
> Marcel Luttgens
The math of a straight line is really confusing.
Isn't it? :-)
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:abeq83$7nc$1...@dolly.uninett.no...
>
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0205...@posting.google.com...
[snip]
> > In the relation x = a + Vt, V cannot exceed c, hence an infinity
> > of points situated on the straight line are physically nonsensical.
> > For instance, the point x=5a, t=4a/V (your point P5) is only
> > valid if 1.25*V <= c.
>
> Really?
> So if I am at x = a = 1 lightsecond at t = 0, and go at 0.9c,
> I will NOT be at x = 5 lightseconds at the time t = 4/0.9 = 4.44 seconds?
> Where will I then be?
In the brainwashing machine ;-)
> > But you have been brainwashed by a century of Einsteinian relativity,
> > and you will never realize that the applicability of the LT is
> > limited by c, hence that SR -which uses them as pure mathematical
> > relations- is necessarily false.
> >
> > > Paul
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0205...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<aa7ba4$s3d$1...@snipp.uninett.no>...
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02042...@posting.google.com...
> > > > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<a9som2$r7o$1...@snipp.uninett.no>...
>
> > Marcel Luttgens wrote:
> > It is not only the time that confuses YOU !
> > You systematically forget that the LT are mathematical relations,
> > that have a physical meaning!
>
> And what the hell has the LT with this to to?
You are not brainwashed, you have no brain at all.
Don't you see that all points must satisfy the relation x=a+Vt,
where x is the distance from the origin of a frame S, attained
after a time interval t by an object leaving with a velocity V,
at a time t=0, a point situated at a distance a from that origin?
Don't you see that the coordinates of your points are physically
nonsensical when V>c?
Don't you see that the LT directly apply to the cases where a=0,
Iow to the cases where the object (which can be a light pulse) leaves
the common origin of the frames S and S' at t=t'=0, at some velocity V?
Don't you see that the LT cannot physically apply when V>c?
>
> > In the relation x = a + Vt, V cannot exceed c, hence an infinity
> > of points situated on the straight line are physically nonsensical.
> > For instance, the point x=5a, t=4a/V (your point P5) is only
> > valid if 1.25*V <= c.
>
> Really?
> So if I am at x = a = 1 lightsecond at t = 0, and go at 0.9c,
> I will NOT be at x = 5 lightseconds at the time t = 4/0.9 = 4.44 seconds?
> Where will I then be?
>
Sorry, I meant that your point P5 is only valid if x/1.25t <=c.
Neddless to say that you chosed such point in order to obscure the
essential fact, i.e. that V must not exceed c.
I would not be surprised if a brainless SRist like you or your
buddy Van de moortel claimed that if he is at at x = a = 1 lightsecond
at t = 0, and go at 1.2c, he will be at x = 5 lightseconds at the
time t = 4/1.2 = 3.33 seconds!
> > But you have been brainwashed by a century of Einsteinian relativity,
> > and you will never realize that the applicability of the LT is
> > limited by c, hence that SR -which uses them as pure mathematical
> > relations- is necessarily false.
Do you still claim that the velocity of an object can exceed c, Iow
that the LT can be generally used?
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
[snip]
> Do you still claim that the velocity of an object can exceed c, Iow
> that the LT can be generally used?
>
> Marcel Luttgens
Suppose
x = a +Vt
After LT this becomes
x' = a' + V't'
with
{ a' = a / ( gamma(1-vV/c^2) )
{ V' = (V-v) / ( 1-vV/c^2 )
1) if V<c then
V' < c
This can describe a physical object.
2) if V = c then
V' = c
This can describe a light signal.
3) if V > c then
V' > c
This cannot describe a physical object or a light signal.
In every case the LT has been used.
Now what exactly is the problem?
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
>
> [snip]
>
> > Do you still claim that the velocity of an object can exceed c, Iow
> > that the LT can be generally used?
> >
> > Marcel Luttgens
>
> Suppose
> x = a +Vt
> After LT this becomes
> x' = a' + V't'
> with
> { a' = a / ( gamma(1-vV/c^2) )
> { V' = (V-v) / ( 1-vV/c^2 )
How did you derive a'?
What is the relation giving t'?
>
> 1) if V<c then
> V' < c
> This can describe a physical object.
>
> 2) if V = c then
> V' = c
> This can describe a light signal.
>
> 3) if V > c then
> V' > c
> This cannot describe a physical object or a light signal.
>
This is exactly the point, i.e. the LT are not pure mathematical
relations, because their applicability is limited to physical objects.
Their use is nonsensical if the velocity of such objects exceeds c.
So, after all, you have a brain :-)
Note that according to Paul, the LT can describe anything, which is
of course silly.
> In every case the LT has been used.
> Now what exactly is the problem?
>
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0205...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<Sk5D8.69438$Ze.1...@afrodite.telenet-ops.be>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> >
> > [snip]
> >
> > > Do you still claim that the velocity of an object can exceed c, Iow
> > > that the LT can be generally used?
> > >
> > > Marcel Luttgens
> >
> > Suppose
> > x = a +Vt
> > After LT this becomes
> > x' = a' + V't'
> > with
> > { a' = a / ( gamma(1-vV/c^2) )
> > { V' = (V-v) / ( 1-vV/c^2 )
>
> How did you derive a'?
Don't you know how to apply the LT?
Try it, it's simple algebra.
Start with x = a + Vt and apply the LT.
If you can't work it out, come back and I'll help you.
> What is the relation giving t'?
giving t' as a function of what?
>
> >
> > 1) if V<c then
> > V' < c
> > This can describe a physical object.
> >
> > 2) if V = c then
> > V' = c
> > This can describe a light signal.
> >
> > 3) if V > c then
> > V' > c
> > This cannot describe a physical object or a light signal.
> >
>
> This is exactly the point, i.e. the LT are not pure mathematical
> relations, because their applicability is limited to physical objects.
No, their applicability is limited to all events.
> Their use is nonsensical if the velocity of such objects exceeds c.
No it is not nonsensical. I just showed that the LT applied to an
equation like x = a + Vt transforms the equation to x' = a' + V't'
and that, independent of the chosen v or V, *both* equations
describe *together* either
- some physical object, or
- some light signal, or
- some non-physical object traveling faster than light
(like for instance a light spot on a distant object)
> So, after all, you have a brain :-)
> Note that according to Paul, the LT can describe anything, which is
> of course silly.
First of all it is not silly because it is true, like I just showed.
Second, the LT do not *describe* anything. They merely transform
descriptions or coordinates of events from one frame to another.
Dirk Vdm
Vertner Vergon
Vergon:
You are both on the wrong track. I've explained this elsewhere
and you both ignore it.
There are two velocities in the universe, that which is -- and the
observation of it.
V is the velocity that is -- (Newtonian). It goes to infinity.
v is the observation of V. It goes to c.
As V goes to infinity, v goes to c.
The relation is V (LT) = v
So superlumnal velocities exist. They just are not directly
observed. But they are *indirectly* observed.
The parameters that accompany v are the parameters of V.
In other words the parameters of V are observed in the company
of v -- as shown by the special theory of relativity.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.0205...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<Sk5D8.69438$Ze.1...@afrodite.telenet-ops.be>...
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > >
> > > [snip]
> > >
> > > > Do you still claim that the velocity of an object can exceed c, Iow
> > > > that the LT can be generally used?
> > > >
> > > > Marcel Luttgens
> > >
> > > Suppose
> > > x = a +Vt
> > > After LT this becomes
> > > x' = a' + V't'
> > > with
> > > { a' = a / ( gamma(1-vV/c^2) )
> > > { V' = (V-v) / ( 1-vV/c^2 )
> >
> > How did you derive a'?
>
> Don't you know how to apply the LT?
> Try it, it's simple algebra.
> Start with x = a + Vt and apply the LT.
> If you can't work it out, come back and I'll help you.
>
I don't need your help, but your solution is probably interesting.
Why don't you present it?
> > What is the relation giving t'?
>
> giving t' as a function of what?
>
As a function of t, v and V, or of t, v, x and a.
As a SRist, it should not be difficult for you to present both
solutions, thus showing your expertise.
>
> >
> > >
> > > 1) if V<c then
> > > V' < c
> > > This can describe a physical object.
> > >
> > > 2) if V = c then
> > > V' = c
> > > This can describe a light signal.
> > >
> > > 3) if V > c then
> > > V' > c
> > > This cannot describe a physical object or a light signal.
> > >
> >
> > This is exactly the point, i.e. the LT are not pure mathematical
> > relations, because their applicability is limited to physical objects.
>
> No, their applicability is limited to all events.
>
> > Their use is nonsensical if the velocity of such objects exceeds c.
>
> No it is not nonsensical. I just showed that the LT applied to an
> equation like x = a + Vt transforms the equation to x' = a' + V't'
> and that, independent of the chosen v or V, *both* equations
> describe *together* either
> - some physical object, or
> - some light signal, or
> - some non-physical object traveling faster than light
> (like for instance a light spot on a distant object)
Your "showing" is incomplete, as you have not given the relation for t'.
Otoh, do you have better examples of non-physical objects, because your
superluminal light spot is in fact a physical phenomenon.
>
> > So, after all, you have a brain :-)
> > Note that according to Paul, the LT can describe anything, which is
> > of course silly.
>
> First of all it is not silly because it is true, like I just showed.
You didn't show much.
> Second, the LT do not *describe* anything. They merely transform
> descriptions or coordinates of events from one frame to another.
>
Showing your time transform would be better than quibbling.
Of course, quibbling is easier.
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
[[ Warning: being in good mood, quibbling mode OFF ]]
I don't really find the derivation interesting, but let's see...
Take the set of events [coordinate pairs (x,t)] that satisfy the equation
x = a + Vt
Apply inverse Lorentz:
{ x = g(x'+vt')
{ t = g(t'+vx'/c^2)
Substitute x and t into the first equation to find what the
equation looks like in the frame S':
g(x'+vt') = a + Vg(t'+vx'/c^2)
Apply distributivity, commutativity and associativy
g(1-vV/c^2)x' = a + g(V-v)t'
Divide by g(1-vV/c^2)
x' = a/(g(1-vV/c^2)) + ((V-v)/(1-vV/c^2)) t'
to get final equation in the other frame:
x' = a' + V't'
where we define:
{ a' = a / ( g(1-vV/c^2) )
{ V' = (V-v) / ( 1-vV/c^2 )
The *result* is somewhat interesting:
|V| < c <==> |V'| < c
Physically moving objects are seen that way
in every frame
|V| = c <==> |V'| = c
Light signals are seen that way
in every frame
|V| > c <==> |V'| > c
Hypothetical superluminal objects or physical phenomena
are seen that way in every frame.
Do you really find the derivation interesting?
Sounds like you didn't have much algebra after all.
I must have overestimated your maths level. Sorry.
[[ Warning, quibbling mode *still* OFF, really ]]
>
> > > What is the relation giving t'?
> >
> > giving t' as a function of what?
> >
>
> As a function of t, v and V, or of t, v, x and a.
> As a SRist, it should not be difficult for you to present both
> solutions, thus showing your expertise.
For all events (x,t)/(x',t') in the whole of space-time we have
{ x' = g(x-vt)
{ t' = g(t-vx/c^2)
Therefore, for all the *special* events (x,t) that satisfy the relation
x = a + Vt
which, if V # 0, is equivalent with
t = (x-a)/V
we have
t' = g(t-v(a+Vt)/c^2)
or
t' = (g-vV/c^2)t - gav/c^2
which is a function of t, v, V, and a, valid for all the
events that satisfy the equations
x = a + Vt
So for each special event that satisfies x = a + Vt, you can
directly calculate the value of t' with the "transformation":
t' = (g-vV/c^2)t - gav/c^2
Of course this "transformation" is only valid for those special
events only. Certainly not for *all* events in space-time.
What is the issue?
Do you have a problem with this?
Or were you thinking of some other expression of t' that
causes trouble?
> > > >
> > > > 1) if V<c then
> > > > V' < c
> > > > This can describe a physical object.
> > > >
> > > > 2) if V = c then
> > > > V' = c
> > > > This can describe a light signal.
> > > >
> > > > 3) if V > c then
> > > > V' > c
> > > > This cannot describe a physical object or a light signal.
> > > >
> > >
> > > This is exactly the point, i.e. the LT are not pure mathematical
> > > relations, because their applicability is limited to physical objects.
> >
> > No, their applicability is limited to all events.
> >
> > > Their use is nonsensical if the velocity of such objects exceeds c.
> >
> > No it is not nonsensical. I just showed that the LT applied to an
> > equation like x = a + Vt transforms the equation to x' = a' + V't'
> > and that, independent of the chosen v or V, *both* equations
> > describe *together* either
> > - some physical object, or
> > - some light signal, or
> > - some non-physical object traveling faster than light
> > (like for instance a light spot on a distant object)
>
> Your "showing" is incomplete, as you have not given the relation for t'.
I have by now.
> Otoh, do you have better examples of non-physical objects, because your
> superluminal light spot is in fact a physical phenomenon.
Quite indeed, "Non-physical object" was extremely poor wording
of mine since we can safely assume that objects are physical by definition.
Like you suggest, I should have said "non-physical phenomenon".
So if you want other examples, you can think of the superluminal
lightspot, or you can think of hypothetical tachyons, or speeddragons
or whatevers that satisfy the equation x = a +Vt.
If you like, for V > c, you can imagine a set of explosions that
happen to take place at these places and times, with the condition
that they *appear* to move in a straight line and faster than light.
I look at it abstractly as a set of events that satisfy a certain
condition.
One could also contemplate the set { (0,1), (100, -0.4), (-5,6) }
which will be difficult to associate with some moving physical
object or phenomenon. But you can imagine 3 explosions
taking place at different places at different times.
>
> >
> > > So, after all, you have a brain :-)
> > > Note that according to Paul, the LT can describe anything, which is
> > > of course silly.
> >
> > First of all it is not silly because it is true, like I just showed.
>
> You didn't show much.
>
> > Second, the LT do not *describe* anything. They merely transform
> > descriptions or coordinates of events from one frame to another.
> >
>
> Showing your time transform would be better than quibbling.
I have shown, I hope...
> Of course, quibbling is easier.
[[ Quibbling mode still OFF ]]
Dirk Vdm
Paul B. Andersen
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<abeq83$7nc$1...@dolly.uninett.no>...
> > > In the relation x = a + Vt, V cannot exceed c, hence an infinity
> > > of points situated on the straight line are physically nonsensical.
> > > For instance, the point x=5a, t=4a/V (your point P5) is only
> > > valid if 1.25*V <= c.
> >
> > Really?
> > So if I am at x = a = 1 lightsecond at t = 0, and go at 0.9c,
> > I will NOT be at x = 5 lightseconds at the time t = 4/0.9 = 4.44 seconds?
> > Where will I then be?
> >
>
> Sorry, I meant that your point P5 is only valid if x/1.25t <=c.
Oh, no Marcel. That was not what you meant.
You insist that "an infinity of points situated on the straight line
[x = a + VT] are physically nonsensical."
You have now obviously realized that P5 above is not such a point.
Can you please give an example of such a physically nonsensical point(event)?
> Neddless to say that you chosed such point in order to obscure the
> essential fact, i.e. that V must not exceed c.
Can you please explain what you mean by that?
How should I have chosen the point NOT to obscure the obvious
fact that V can never exceed c?
It seems to me that you are trying to run away from the idiocies
you have said about the simple equation x = a + Vt,
and somehow make it appear that the issue is whether or not
V can exceed c. Which never was disputed.
Have you forgotten the big issue you made about the trivial
fact that the x/t ratio for different points on the line
x = a + vt are different? Do you remember that you insisted
that it was contradictory?
Do you want me to remind you? :-)
> I would not be surprised if a brainless SRist like you or your
> buddy Van de moortel claimed that if he is at at x = a = 1 lightsecond
> at t = 0, and go at 1.2c, he will be at x = 5 lightseconds at the
> time t = 4/1.2 = 3.33 seconds!
> Do you still claim that the velocity of an object can exceed c, Iow
> that the LT can be generally used?
Panic, Marcel? :-)
We both know I never claimed that.
Paul
Marcel Luttgens
How can you observe hypothetical objects or phenomena?
>
> >
> > > > What is the relation giving t'?
> > >
> > > giving t' as a function of what?
> > >
> >
> > As a function of t, v and V, or of t, v, x and a.
> > As a SRist, it should not be difficult for you to present both
> > solutions, thus showing your expertise.
>
> For all events (x,t)/(x',t') in the whole of space-time we have
> { x' = g(x-vt)
> { t' = g(t-vx/c^2)
>
> Therefore, for all the *special* events (x,t) that satisfy the relation
> x = a + Vt
> which, if V # 0, is equivalent with
> t = (x-a)/V
> we have
> t' = g(t-v(a+Vt)/c^2)
> or
> t' = (g-vV/c^2)t - gav/c^2
> which is a function of t, v, V, and a, valid for all the
> events that satisfy the equations
> x = a + Vt
>
> So for each special event that satisfies x = a + Vt, you can
> directly calculate the value of t' with the "transformation":
> t' = (g-vV/c^2)t - gav/c^2
You probably mean t' = gt(1-vV/c^2) - gav/c^2
> Of course this "transformation" is only valid for those special
> events only. Certainly not for *all* events in space-time.
>
> What is the issue?
> Do you have a problem with this?
> Or were you thinking of some other expression of t' that
> causes trouble?
Thank you for your derivations. You have been honest and courageous.
But you horrible relation
x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2)
can easily (with the help of a few Bols, try it!) be reduced to
x' = gt(V-v) + ga, which can straigthforwardy be obtained by replacing
the x in the LT x' = g(x-vt) by a+Vt!
Indeed, then, x' = g(a+Vt-vt) = gt(V-v) + ga.
The interpretation of this last relation is simple:
Seen from the frame S, the x-coordinate of the object after a time t
is of course a + Vt. For the observer moving at v wrt S, an object
leaving the common origin at t=t'=0, will be at x' = gt(V-v) after
a time interval t. But if the object starts from a distance a from the
common origin at t=t'=0, S' "sees" a as ga (that you can call a').
Hence, for S', the x'-coordinate of the point attained by the object
after a time interval t will be ga + gt(V-v). Very simple indeed.
But I don't agree with your transform t' = g(t-v(a+Vt)/c^2), that
you simply obtained by replacing x by a+Vt in the LT t' = g(t-vx/c^2).
Why? Simply because if the time interval t is 0, you get t' = -gav/c^2.
Or, after a time interval 0, the origins of the frames S and S' still
coincide at t=t'=0, Iow, t' must be zero, not -gav/c^2.
Hence, Imo, the correct time transform when x = a + Vt is
t' = g(t-v(x-a)/c^2). As x-a = Vt, t' = g(t-vVt/c^2) = gt(1-vV/c^2)!
Then, t'=t if t=0.
I'll add once more that the "classical" LT only make sense when x=Vt.
Iow, you can't use them for x=a+Vt.
What is the physical significance of transformations applied to
non-physical phenomena like blue fairies or speed-dragons?
> I look at it abstractly as a set of events that satisfy a certain
> condition.
>
> One could also contemplate the set { (0,1), (100, -0.4), (-5,6) }
> which will be difficult to associate with some moving physical
> object or phenomenon. But you can imagine 3 explosions
> taking place at different places at different times.
>
>
> Dirk Vdm
Marcel Luttgens
Paul B. Andersen
> Note that according to Paul, the LT can describe anything, which is
> of course silly.
Yes, that would indeed be silly. :-)
You seem to be very eager to make people forget the issue.
It was:
Can the equation for the trajectory of a body in S:
x = a + Vt
be transformed by the LT to S' moving with the speed v relative to S?
My answer is yes, obviously.
And the transformed equation is:
x' = xo + ut'
where xo = a*sqrt(1-v^2/c^2)/(1 - vV/c^2)
and u = (V-v)/(1 - vV/c^2)
Do you still refute this, Marcel?
Come on, Marcel.
Repeat some of your incredible naive, ridiculous, hilarious, idiotic
mathematical proofs for why the LT does not apply on this problem.
Someone may have missed them the first time.
And don't forget to include you especially funny arguments for why
there are an infinite number of physically nonsensical events
on the world-line x = a + Vt!
Paul
Dirk Van de moortel
I can't. Did I say I can?
Perhaps I was sloppy, but I don't recall and I don't see
myself saying it in my post...
You can think about hypotherical stuff though.
You can observe a light spot. The spot can be moving
faster than light.
Okay?
>
> >
> > >
> > > > > What is the relation giving t'?
> > > >
> > > > giving t' as a function of what?
> > > >
> > >
> > > As a function of t, v and V, or of t, v, x and a.
> > > As a SRist, it should not be difficult for you to present both
> > > solutions, thus showing your expertise.
> >
> > For all events (x,t)/(x',t') in the whole of space-time we have
> > { x' = g(x-vt)
> > { t' = g(t-vx/c^2)
> >
> > Therefore, for all the *special* events (x,t) that satisfy the relation
> > x = a + Vt
> > which, if V # 0, is equivalent with
> > t = (x-a)/V
> > we have
> > t' = g(t-v(a+Vt)/c^2)
> > or
> > t' = (g-vV/c^2)t - gav/c^2
> > which is a function of t, v, V, and a, valid for all the
> > events that satisfy the equations
> > x = a + Vt
> >
> > So for each special event that satisfies x = a + Vt, you can
> > directly calculate the value of t' with the "transformation":
> > t' = (g-vV/c^2)t - gav/c^2
>
> You probably mean t' = gt(1-vV/c^2) - gav/c^2
Yes, a typo.
>
> > Of course this "transformation" is only valid for those special
> > events only. Certainly not for *all* events in space-time.
> >
> > What is the issue?
> > Do you have a problem with this?
> > Or were you thinking of some other expression of t' that
> > causes trouble?
>
> Thank you for your derivations. You have been honest and courageous.
Okay, and you'll notice that I will do so again below :-)
Can I ask you to do the same and carefully read and
check every word in its context, and that you work through
it to the end before you start writing partial replies?
I hope you can do that.
And that I didn't make too many typo's.
> But you horrible relation
> x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2)
> can easily (with the help of a few Bols, try it!) be reduced to
> x' = gt(V-v) + ga, which can straigthforwardy be obtained by replacing
> the x in the LT x' = g(x-vt) by a+Vt!
Of course, since the monster was coming out of the LT and
x = a+Vt in the first place.
I was interested in x' as a function of t', iow in the equation
of motion according to S'.
> Indeed, then, x' = g(a+Vt-vt) = gt(V-v) + ga.
So you have a relation between the x' of one frame
and the t in another frame. Maybe that's interesting, maybe not...
But okay, let's look at what we can do with your relation...
> The interpretation of this last relation is simple:
> Seen from the frame S, the x-coordinate of the object after a time t
> is of course a + Vt.
Yes.
Okay, so we are considering an event R with
x_R = a+Vt
t_R = t
that takes place on the object. Let's say the object gives a Red flash.
> For the observer moving at v wrt S, an object
> leaving the common origin at t=t'=0, will be at x' = gt(V-v) after
> a time interval t.
Be very careful here: this is after a time interval t of the observer S.
The observer S' who uses x', thinks in terms of his own time t'.
This equation x' = gt(V-v) shows the relation between one observer's
distance x' with another observer's time t of an event on the moving
object.
Let's see what the coordinates of our event R are for observer S':
x'_R = g(x_R-vt_R)
= g(a+Vt-vt)
= gt(V-v)+ga
t'_R = g(t_R-vx_R/c^2)
= g(t-v(a+Vt)/c^2)
= g( t(1-vV/c^2) - av/c^2)
Remember, for S we had:
x_R = a+Vt
t_R = t
where t is the time of the Red flash according to S.
Keep all this in mind for later and let's go on...
> But if the object starts from a distance a from the
> common origin at t=t'=0, S' "sees" a as ga (that you can call a').
Let's call the event where the object starts G because it emits a
Green flash.
Since the equation of the object is x = a + Vt, and since the
distance is a, according to observer S the Green flash happens at
x_G = a
t_G = 0
So for observer S' the Green flash is happening at
x'_G = ga
t'_G = -gav/c^2
But look at the time of this event: it is t'_G = -gav/c^2,
it is not 0.
> Hence, for S', the x'-coordinate of the point attained by the object
> after a time interval t will be ga + gt(V-v). Very simple indeed.
Indeed, but this is about the Red flash event, and we already
know where and when it happens according to S': it happened at
x'_R = gt(V-v)+ga
t'_R = g( t(1-vV/c^2) - va/c^2)
x'_R can be *calculated* using the number t, but this is *happening*
at his time t'_R = g( t(1-vV/c^2) - va/c^2) which can also be *calculated*
using the number t.
> But I don't agree with your transform t' = g(t-v(a+Vt)/c^2), that
> you simply obtained by replacing x by a+Vt in the LT t' = g(t-vx/c^2).
The equation t' = g(t-v(a+Vt)/c^2) is exactly the same as
t'_R = g( t(1-vV/c^2) - va/c^2 )
Remember, it was the relation between the times t and t' for
all events happening on the object, including the Red flash
that took place at time t according to observer S.
> Why? Simply because if the time interval t is 0, you get t' = -gav/c^2.
Quite right, for the Green flash event.
> Or, after a time interval 0, the origins of the frames S and S' still
> coincide at t=t'=0, Iow, t' must be zero, not -gav/c^2.
No, why must it be zero?
Okay, the observers coincide at event C with
x_C = 0
t_C = 0
and
x'_C = 0
t'_C = 0
but the Green flash takes place at
x_G = a
t_G = 0
and
x'_G = ga
t'_G = -gav/c^2.
These C and G are entirely different events.
Two events that are seen to be simultaneous in one frame,
are not simultaneous in the other frame, unless they take
place at the same location.
> Hence, Imo, the correct time transform when x = a + Vt is
> t' = g(t-v(x-a)/c^2).
> As x-a = Vt, t' = g(t-vVt/c^2) = gt(1-vV/c^2)!
> Then, t'=t if t=0.
But this equation
t' = g(t-v(x-a)/c^2)
cannot be derived from x = a + Vt and the LT. You can
*only* arrive at it by forcing the start event G of the object
(Green flash) to be simultaneous with the Coincidence event
C of the frames, for *both* observers.
In fact, in order to obtain that equation, you have effectively
demanded that
0 = -gav/c^2.
I hope you realise that.
The event C does not happen on the object, so x = a + Vt
is not valid there (unless a = 0 of course).
> I'll add once more that the "classical" LT only make sense when x=Vt.
> Iow, you can't use them for x=a+Vt.
Because in *that* particular case your demand
0 = -gav/c^2
is absolutely legitimate because a = 0.
In this case the events C and G happen at the same place.
I'll repeat:
"Two events that are seen to be simultaneous in one frame,
are not simultaneous in the other frame, unless they take
place at the same location."
And we can add: "Not even if the observers coincide."
Phew...
I hope this helps?
You probably have to go through it again. I have done
so several times :-)
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
<snip>
> > But you horrible relation
> > x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2)
> > can easily (with the help of a few Bols, try it!) be reduced to
> > x' = gt(V-v) + ga, which can straigthforwardy be obtained by replacing
> > the x in the LT x' = g(x-vt) by a+Vt!
>
> Of course, since the monster was coming out of the LT and
> x = a+Vt in the first place.
> I was interested in x' as a function of t', iow in the equation
> of motion according to S'.
>
> > Indeed, then, x' = g(a+Vt-vt) = gt(V-v) + ga.
>
> So you have a relation between the x' of one frame
> and the t in another frame. Maybe that's interesting, maybe not...
> But okay, let's look at what we can do with your relation...
>
It was my relation since many months, now it is also yours.
Remember that you rejected it not so long ago.
> > The interpretation of this last relation is simple:
> > Seen from the frame S, the x-coordinate of the object after a time t
> > is of course a + Vt.
>
> Yes.
> Okay, so we are considering an event R with
> x_R = a+Vt
> t_R = t
> that takes place on the object. Let's say the object gives a Red flash.
>
> > For the observer moving at v wrt S, an object
> > leaving the common origin at t=t'=0, will be at x' = gt(V-v) after
> > a time interval t.
>
> Be very careful here: this is after a time interval t of the observer S.
Of course!
> The observer S' who uses x', thinks in terms of his own time t'.
Sure!
> This equation x' = gt(V-v) shows the relation between one observer's
> distance x' with another observer's time t of an event on the moving
> object.
> Let's see what the coordinates of our event R are for observer S':
> x'_R = g(x_R-vt_R)
> = g(a+Vt-vt)
> = gt(V-v)+ga
Yes, this is your (my) relation.
> t'_R = g(t_R-vx_R/c^2)
> = g(t-v(a+Vt)/c^2)
> = g( t(1-vV/c^2) - av/c^2)
> Remember, for S we had:
> x_R = a+Vt
> t_R = t
> where t is the time of the Red flash according to S.
> Keep all this in mind for later and let's go on...
False, because the object doesn't move from the common origin of
S and S' at a time t=t'=0, but starts from the distance a from that
origin, also at t=t'=0.
So, after a time interval t, the object is at a distance x=a+Vt from
the origin of S, after having travelled a distance Vt, not a+Vt.
Iow, you must replace the x in the LT t' = g(t-vx/c^2), not by
x = a+Vt, but by x-a = Vt
Then you obtain the correct time transformation t'_R = g(t-v(x-a)/c^2).
As x-a = Vt, the transformation can also be written under the form
t'_R = gt(1-vV/c^2), which doesn't include the term -gav/c^2.
Let's go back to your (my) relation x'_R = ga + gt(V-v), which can
also be written x'_R = a'+V't'.
Thus, a'+V't' = ga + gt(V-v).
As a'=ga (not a/( g(1-vV/c^2)), V't' = gt(V-v), and t' = gt(V-v)/V'.
By replacing V' by (V-v)/( 1-vV/c^2 ), we obtain t' = gt(1-vV/c^2).
This is another way to show that your relation g( t(1-vV/c^2) - av/c^2)
cannot be correct.
I agree, but your objection also applies to objects or light
pulses starting from a at t=t'=0. If you consider that it is
physically impossible to force the start to be simultaneous
with the frames coincidence, it is senseless to discuss scenarios
based on x = a+Vt, unless we consider them as thought experiments.
Then, it is valid to claim that t' = g(t-v(x-a)/c^2).
> In fact, in order to obtain that equation, you have effectively
> demanded that
> 0 = -gav/c^2.
> I hope you realise that.
> The event C does not happen on the object, so x = a + Vt
> is not valid there (unless a = 0 of course).
>
>
> > I'll add once more that the "classical" LT only make sense when x=Vt.
> > Iow, you can't use them for x=a+Vt.
>
> Because in *that* particular case your demand
> 0 = -gav/c^2
> is absolutely legitimate because a = 0.
> In this case the events C and G happen at the same place.
>
> I'll repeat:
>
> "Two events that are seen to be simultaneous in one frame,
> are not simultaneous in the other frame, unless they take
> place at the same location."
> And we can add: "Not even if the observers coincide."
>
> Phew...
> I hope this helps?
> You probably have to go through it again. I have done
> so several times :-)
>
See above, if x' = ga+gt(V-v), which you got by symplifying
your relation
x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2),
is correct, t' must be gt(1-vV/c^2), not g(t-v(a+Vt)/c^2).
As ga is straightforwardly obtained by replacing x by a+Vt
in x' = g(x-vt), I consider that ga is correct.
Anyhow, as the "classical" LT are only valid when x=Vt,
you can't use them for events like (x=a, t=0).
Replacing x by a, and t by 0, in t' = g(t-vx/c^2), gives
the false result t' = -gva/c^2, on which Einstein based his
relativity of simultaneity.
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
Your are talking about the Green flash (start of object), I am talking
about the Red one (time t for observer S).
And again you make the same error as before:
only event C (Coincidence of observers) has t=t'=0.
All other events that take place at t=0 for observer S,
do *not* take place at time t'=0 for observer S'.
Check the LT: they mix space and time: C is simultaneous
for both observers. G is not. That is the whole point.
Let's look at the pure events again (you snipped the
relevant stuff):
1) Event C: Coincidence and synchronization of observers S and S':
x_C = 0
t_C = 0
and
x'_C = 0
t'_C = 0
2) Event G: Green flash at start of object at time 0 for observer S:
x_G = a
t_G = 0
and
x'_G = ga
t'_G = -gav/c^2
3) Event R: Red flash of object at time t for observer S:
x_R = a+Vt
t_R = t
and
x'_R = gt(V-v)+ga
t'_R = g( t(1-vV/c^2) - va/c^2)
Comparing the times of events C and G you see that
t_C = t'_C = 0
and that
t_G = 0 # t'_G = -gav/c^2
Yet you insist that t_G = t'_G. It is not.
> So, after a time interval t, the object is at a distance x=a+Vt from
[snip]
I think I can snip the remainder because it is based on
the error that you made before, and that you just made
again.
You reject special relativity because you find impossible
things or contradictions. But you find them because you
make a fundamental mistake.
My whole idea was to make you realize that mistake.
I don't seem to have succeeded so far. Yet is so simple.
Again:
"Two events that are seen to be simultaneous in one frame,
are not simultaneous in the other frame, unless they take place
at the same location."
And we can add: "Not even if the observers coincide."
If you don't accept that, then there is nothing more to say
and you can not use the LT to show SR wrong. That is
mathematically impossible. Don't try it: it simply cannot be
done. The only way to show SR wrong, is by experiment.
Marcel, if you want to continue, first try to fully understand
the above. When you do understand, you must fully accept
it. Then we can go on and look at the rest (if you think it will
still stand).
Wat mij betreft, welwillend ondanks onze vroegere "ambras".
Gezondheid!
Dirk Vdm
Marcel Luttgens
No problem till now.
>
> 2) Event G: Green flash at start of object at time 0 for observer S:
> x_G = a
> t_G = 0
> and
> x'_G = ga
> t'_G = -gav/c^2
As for me, to time 0 for S corresponds time 0 for S'.
"a" is the coordinate of some point according to S, but S', which
is moving at v wrt S, "intellectually" sees the same point at a
distance ga at his time t'=0. I say "intellectually", because
S' didn't measure its distance to the point. Otoh, how can
you physically explain the negative time -gav/c^2? You can't,
you got it, simply by using the LT t' = g(t-vx/c^2). But
this is not allowed, because, as I showed many times, the LT
only apply when x = Vt and x' = V't'. There is no "a" in
x = Vt.
If you want to convince me, you have to physically explain
the origin of the negative time. If you tell me that no physical
explanation is needed, because it is inherent to the LT
themselves, I will not believe you.
Don't forget that the simplest physical derivation of the LT
is based on light pulses starting *from the common origin* of
the two frames S and S' when their time have been synchronized
to zero. Nowhere in such derivation has a negative time been spoken
about.
>
> 3) Event R: Red flash of object at time t for observer S:
> x_R = a+Vt
> t_R = t
> and
> x'_R = gt(V-v)+ga
> t'_R = g( t(1-vV/c^2) - va/c^2)
>
>
> Comparing the times of events C and G you see that
> t_C = t'_C = 0
> and that
> t_G = 0 # t'_G = -gav/c^2
>
> Yet you insist that t_G = t'_G. It is not.
>
"Extended" LT can be derived for the cases where x = a+Vt.
It is straightforward for x' = g(x-vt), because x is the x-coordinate
of the point attained by an object (or a light pulse) having leaved
the "a" point at t=t'=0. Thus, one is allowed to replace the x
in g(x-vt) by a+Vt.
Then, x' becomes g(a+Vt-vt), or x' = ga + gt(V-v).
If a=0, the "extended" LT reduces to the "classical" LT x' = gt(V-v),
where x = Vt. Otoh, if t=0, x' is simply ga for both LT, the classical
one and the "extended" one.
Now, we have to derive the "extended" LT for time.
The simplest way is to consider that x' = ga + gt(V-v) = a'+V't',
where a' = ga, and V' = (V-v)/(1-Vv/c^2).
Then, ga + gt(V-v) = ga + (V-v)/(1-Vv/c^2)*t'.
Hence, t' = gt(1-Vv/c^2), not g( t(1-vV/c^2) - va/c^2), which
is wrongly obtained by simply replacing the x in the "classical" LT
t' = g(t-vx/c^2) by a+Vt. This makes no sense, because the object
didn't travel during the whole distance x+Vt, but only during
the distance Vt.
Iow, the x in t' = g(t-vx/c^2) has to be replaced only by Vt.
The correct "extended" LT x' = ga + gt(V-v) and t' = gt(1-Vv/c^2)
are identical to the "classical" one if a=0. But if t=0, unlike with
the "classical" LT, one obtains the correct result t'=0!
No more inexplicable negative time t' with the "extended" LT!
Consequently, the concept of "relativity of simultaneity",
according to which "two events that are seen to be simultaneous
in one frame, are not simultaneous in the other frame, unless
they take place at the same location," is simply nonsensical!
> > So, after a time interval t, the object is at a distance x=a+Vt from
>
> [snip]
>
> I think I can snip the remainder because it is based on
> the error that you made before, and that you just made
> again.
Two things, first show the error, and second, explain
physically the origin of t' = -gva/c^2 when t = 0.
> You reject special relativity because you find impossible
> things or contradictions. But you find them because you
> make a fundamental mistake.
Which mistake?
> My whole idea was to make you realize that mistake.
> I don't seem to have succeeded so far. Yet is so simple.
>
Your problem is that you are too much convinced of the
validity of SR.
> Again:
> "Two events that are seen to be simultaneous in one frame,
> are not simultaneous in the other frame, unless they take place
> at the same location."
> And we can add: "Not even if the observers coincide."
>
> If you don't accept that, then there is nothing more to say
> and you can not use the LT to show SR wrong. That is
> mathematically impossible. Don't try it: it simply cannot be
> done. The only way to show SR wrong, is by experiment.
There is no mathematical error in the derivation of the
"classical" LT, but their applicability is restricted to
the scenario on which they are based.
>
> Marcel, if you want to continue, first try to fully understand
> the above. When you do understand, you must fully accept
> it. Then we can go on and look at the rest (if you think it will
> still stand).
Dirk, if you want to continue, ...
>
> Wat mij betreft, welwillend ondanks onze vroegere "ambras".
> Gezondheid!
Dank je, I reciprocate!
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<KvaF8.23$y54.5...@cacnews.cac.cpqcorp.net>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> > news:<IjVE8.81449$Ze.1...@afrodite.telenet-ops.be>...
> > > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > > > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
[snip]
"Intellectually" you say twice. That should be "Obsoletely
intellectually", according to the Galilean Relativity point of view,
which was wrong.
> Otoh, how can
> you physically explain the negative time -gav/c^2?
For S' the event G happens before event C. Nothing mysterious
about that. Notice the extremely small number: division by c^2.
You won't notice anything in every day life of it.
> You can't,
> you got it, simply by using the LT t' = g(t-vx/c^2). But
> this is not allowed, because, as I showed many times, the LT
> only apply when x = Vt and x' = V't'. There is no "a" in
> x = Vt.
I'l try to explain...
Notice that the pair of equations
{ x = Vt
{ x' = V't'
is merely a *consequence* of the LT.
The LT itself is *derived* by demanding that the pair of
equations for light signals
{ x = ct
{ x' = ct'
are satisfied together. Notice the same c in both equations.
The LT can easily be derived by demanding that the pair of
equations
{ x = p + ct
{ x' = p' + ct'
are satisfied together for any p you choose.
What matters, is that both equations have the same c.
The "p" does not matter.
Wait, read on...
The expression for V' (as a function of V,v and c) in the pair
{ x = Vt
{ x' = V't'
where
V' = (V-v) / ( 1-vV/c^2 )
*follows* from the LT, just like the expression of the a' (as a
function of a, V, v and c) in the pair
{ x = a + Vt
{ x' = a' + V't'
where
{ a' = a / ( g(1-vV/c^2) )
{ V' = (V-v) / ( 1-vV/c^2 )
*follows* from it. Note that the expressions of the V' are exactly
the same in both cases.
> If you want to convince me, you have to physically explain
> the origin of the negative time. If you tell me that no physical
> explanation is needed, because it is inherent to the LT
> themselves, I will not believe you.
It is not a question of believing. It follows from the constancy of light
speed. Either you believe *that*, or you don't. Do you or don't you?
Please answer?
> Don't forget that the simplest physical derivation of the LT
> is based on light pulses starting *from the common origin* of
> the two frames S and S' when their time have been synchronized
> to zero. Nowhere in such derivation has a negative time been spoken
> about.
A linear transformation in a 2 dimensional space is completely
determined by how 2 basis vectors are transformed. You can
take any pair of linearly independent vectors. If you know what
happens with them, the transformation is fixed. Usually one takes
a simple pair, like (1,0) and (0,1), but one can take any pair at
all. In this case one has taken a pair on the light lines
x = ct
x = -ct
and demand that they result in a pair of events on the transformed
light lines
x' = ct'
x' = -ct'
One can just as well take a pair on two other light lines
x = 10 + ct
x = -2023 - ct
and demand that they result in a pair of events on the transformed
light lines
x' = <some number> + ct'
x' = <some other number> -ct'
The resulting transformation is exactly the same:
{ x' = g(x-vt)
{ t' = g(t-vx/c^2)
and totally independent of the chosen numbers.
This is a very early theorem of basic linear algebra.
Does it ring a bell or didn't you get this at school?
>
> >
> > 3) Event R: Red flash of object at time t for observer S:
> > x_R = a+Vt
> > t_R = t
> > and
> > x'_R = gt(V-v)+ga
> > t'_R = g( t(1-vV/c^2) - va/c^2)
> >
> >
> > Comparing the times of events C and G you see that
> > t_C = t'_C = 0
> > and that
> > t_G = 0 # t'_G = -gav/c^2
> >
> > Yet you insist that t_G = t'_G. It is not.
> >
>
> "Extended" LT can be derived for the cases where x = a+Vt.
>
> It is straightforward for x' = g(x-vt), because x is the x-coordinate
> of the point attained by an object (or a light pulse) having leaved
> the "a" point at t=t'=0.
Sorry Marcel, here I must iterrupt because there is nothing new
and here is your error.
For the "a" point (event G) the time t' = -gav/c^2.
It simply is not 0. You want it to be 0, but it just is not.
It can not be if light speed is constant.
Like I just explained: the LT is what it is, and it is valid for
every event in spacetime. If you reject it for a#0, then you
also reject it for a=0, and then you reject the constancy of
light speed. It is that simple.
You don't have to elaborate and do calculations because
they are wrong.
I repeat, we have nothing more to say to each other unless
you realise that.
So I suggest you reply with either
- Okay, I accept, or
- Sorry, I do not accept.
In the first case we can look at other stuff.
Doe je best...
Dirk Vdm
Marcel Luttgens
I just wrote:
Now, we have to derive the "extended" LT for time.
The simplest way is to consider that x' = ga + gt(V-v) = a'+V't',
where a' = ga, and V' = (V-v)/(1-Vv/c^2).
Then, ga + gt(V-v) = ga + (V-v)/(1-Vv/c^2)*t'.
Hence, t' = gt(1-Vv/c^2), not g( t(1-vV/c^2) - va/c^2), which
is wrongly obtained by simply replacing the x in the "classical" LT
t' = g(t-vx/c^2) by a+Vt. This makes no sense, because the object
didn't travel during the whole distance x+Vt, but only during
the distance Vt.
Iow, the x in t' = g(t-vx/c^2) has to be replaced only by Vt.
The correct "extended" LT x' = ga + gt(V-v) and t' = gt(1-Vv/c^2)
are identical to the "classical" one if a=0. But if t=0, unlike with
the "classical" LT, one obtains the correct result t'=0!
No more inexplicable negative time t' with the "extended" LT!
Consequently, the concept of "relativity of simultaneity",
according to which "two events that are seen to be simultaneous
in one frame, are not simultaneous in the other frame, unless
they take place at the same location," is simply nonsensical!
I forgot one fundamental consideration:
I have demonstrated above that when x = a+Vt, the correct time LT
is t' = gt(1-Vv/c^2).
But this is also the case when x = Vt, because in Einstein's
scenario, x is the coordinate of the point attained after
a time interval t by a light pulse sent from the origin of
the frame S, when it coincided with the origin of the frame S',
at a time t = t' = 0. Hence, x cannot be considered as a
simple distance a from the origin of S, because it is the distance
Vt travelled by a light pulse or an object.
In order to avoid further misunderstandings, the "classical" LT should
be expressed as x' = gt(V-v) and t' = gt(1-Vv/c^2), where V = x/t.
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
Apparently you haven't seen my reply yet, so here's a quick
warm-up:
You have not demonstrated above that when x = a+Vt, the
correct time LT is t' = gt(1-Vv/c^2).
You have demonstrated that you do not accept the constancy
of light speed and the consequence, relativity of simultaneity.
You say that C and G are simultaneous in both frames. They
are not.
See my other reply for details and a little lesson on linear
algebra...
Tot later,
Dirk Vdm
Marcel Luttgens
What is a demonstration for you?
Here are the main steps:
1) I and *you* have shown that the position transform can be
obtained by replacing the x-coordinate in x' = g(x-vt) by
x = a+Vt, thus x' = g(a+Vt-vt), or x' = ga + gt(V-v).
Do you claim now that you made an error, and that x' is not
given by ga + gt(V-v)?
2) You agreed also that x' = a'+V't', where V' = (V-v)/(1-Vv/c^2).
3) Do you deny that a' = ga? If you don't, you must accept that
V't' = gt(V-v), hence that t' = gt(V-v) / V' = gt(1-Vv/c^2).
4) If you accept that t' = gt(1-Vv/c^2), you must also accept
that this transform can be obtained by replacing x in the
"classical" transform t' = g(t-xv/c^2) by x-a, i.e. by the
distance effectively travelled by the object with velocity V.
Indeed, then, t' = g(t-(x-a)v/c^2). As x = a+Vt, x-a = Vt,
and t' becomes g(t-Vtv/c^2), Iow, t' = gt(1-Vv/c^2).
QED
The origin of Einstein's blunder is now clear. He believed
(as the SRists still do) that the term x has the same physical
meaning in his "position" transform x' = g(x-vt) as in his
"time" transform t' = g(t-xv/c^2).
This is false, because in the "position" transform, x represents
the distance of some point from the origin of the frame S, but
in the "time" transform, x represents the distance travelled
by some object with a velocity V in a time interval t.
For instance, for an event x=a, t=0, the SRists simply
replace x by a, and t by 0 in both transforms, thus obtaining
x' = ga, and t' = -gav/c^2. From t' = -gav/c^2, Einstein invented
his so-called relativity of simultaneity.
As "a" represents the x-coordinate of some point, it is correct
to replace the x in the position transform by a, and obtain x' = ga,
but the distance a can only replace the x in the "time" transform
if it has been travelled by an object moving at V. Then,
a = Vt, and t' becomes gt(1-Vv/c^2). But as t = 0, t' is also
equal to 0, not to -gav/c^2!
Marcel Luttgens
Dirk Van de moortel
No error.
For all events on the object x = a + Vt we have
x' = ga + gt(V-v).
No problem. This is valid.
> 2) You agreed also that x' = a'+V't', where V' = (V-v)/(1-Vv/c^2).
Yes, it is the equation of motion for observer S' where we
have defined V' and a' as
{ a' = a / ( g(1-vV/c^2) )
{ V' = (V-v) / ( 1-vV/c^2 )
> 3) Do you deny that a' = ga?
By *all* means, yes!
In the context of x' = a' + V't' I have defined a' as
a' = a / ( g(1-vV/c^2) )
Now remember, for event G: the coordinates are
x_G = a
t_G = 0
and
x'_G = ga
t'_G = -gav/c^2
And *you* have defined a' as x'_G = ga.
I should have warned you at that stage. Sorry.
We are talking about two entirely different definitions of a'
My a' of point 2) is not your a' of point 3)
> If you don't, you must accept that
> V't' = gt(V-v), hence that t' = gt(V-v) / V' = gt(1-Vv/c^2).
Since I deny that a' = ga, I cannot accept this.
> 4) If you accept that t' = gt(1-Vv/c^2)
I don't:
t'_R = gt(1-vV/c^2) - gva/c^2
for the Red flash event, representing any event on the
object that has time t for observer S.
>, you must also accept
> that this transform can be obtained by replacing x in the
> "classical" transform t' = g(t-xv/c^2) by x-a, i.e. by the
> distance effectively travelled by the object with velocity V.
> Indeed, then, t' = g(t-(x-a)v/c^2). As x = a+Vt, x-a = Vt,
> and t' becomes g(t-Vtv/c^2), Iow, t' = gt(1-Vv/c^2).
> QED
No, you confused two definitions of a', so we can
safely snip the thing about Einstein's blunder.
Look at and reply to the other post please?
Dirk Vdm
Marcel Luttgens
Yes, but you a' is false, and your second error (t' = g(t-(a+Vt)v/c^2
instead of t' = g(t-(x-a)v/c^2) = gt(1-Vv/c^2), compensated the first one.
Remember that the relation you obtained with the wrong a' and t', i.e.
x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2)
reduces to the valid relation
x' = x'+V't' = ga + [(V-v)/( 1-vV/c^2 )] * gt(1-Vv/c^2) = ga + gt(V-v),
where ga is necessarily a'.
> Now remember, for event G: the coordinates are
> x_G = a
> t_G = 0
> and
> x'_G = ga
> t'_G = -gav/c^2
> And *you* have defined a' as x'_G = ga.
> I should have warned you at that stage. Sorry.
>
> We are talking about two entirely different definitions of a'
> My a' of point 2) is not your a' of point 3)
>
> > If you don't, you must accept that
> > V't' = gt(V-v), hence that t' = gt(V-v) / V' = gt(1-Vv/c^2).
>
> Since I deny that a' = ga, I cannot accept this.
>
See above.
> > 4) If you accept that t' = gt(1-Vv/c^2)
>
> I don't:
> t'_R = gt(1-vV/c^2) - gva/c^2
> for the Red flash event, representing any event on the
> object that has time t for observer S.
>
> >, you must also accept
> > that this transform can be obtained by replacing x in the
> > "classical" transform t' = g(t-xv/c^2) by x-a, i.e. by the
> > distance effectively travelled by the object with velocity V.
> > Indeed, then, t' = g(t-(x-a)v/c^2). As x = a+Vt, x-a = Vt,
> > and t' becomes g(t-Vtv/c^2), Iow, t' = gt(1-Vv/c^2).
> > QED
>
> No, you confused two definitions of a', so we can
> safely snip the thing about Einstein's blunder.
>
There is only one valid definition of a', which is a' = ga.
> Look at and reply to the other post please?
>
I did, and replied in <b45b8808.02051...@posting.google.com>
> Dirk Vdm
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:<5%KF8.84315$Ze.1...@afrodite.telenet-ops.be>...
> > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
[snip]
> > Look at and reply to the other post please?
> >
>
> I did, and replied in <b45b8808.02051...@posting.google.com>
I am referring to my answer to that post.
Look at kcsF8.83355$Ze.1...@afrodite.telenet-ops.be
If you first reply to that one, we might come back to
our definitions of a'.
See you...
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<5%KF8.84315$Ze.1...@afrodite.telenet-ops.be>...
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
>
> [snip]
>
> > > Look at and reply to the other post please?
> > >
> >
> > I did, and replied in <b45b8808.02051...@posting.google.com>
>
> I am referring to my answer to that post.
> Look at kcsF8.83355$Ze.1...@afrodite.telenet-ops.be
>
> If you first reply to that one, we might come back to
> our definitions of a'.
> See you...
>
> Dirk Vdm
Sorry, but I don't find that message.
Anyhow, if your definition of a' is false, SR is wrong, as I showed
in my last message b45b8808.02051...@posting.google.com :
Begin of message:
See above.
End of message.
I'll add:
Why is your a' false?
First of all, you get a' = ga, not a' = a / ( g(1-vV/c^2), when the x in
the position LT x' = g(x-vt) is replaced by x = a+Vt.
The second argument is a logical one.
The a in x = a+Vt is a constant, which has nothing to do with V.
As the frame S' is moving at v wrt the frame S, a' must be a function of v,
but there is absolutely no justification for it to be also a function of V.
Iow, your a' cannot be correct.
As a' = ga, one gets straightforwardly t' = gt(1-Vv/c^2), thus, if t = 0,
t' = 0, but not t' = -gav/c^2.
Consequently, there is no such thing as a "relativity of simultaneity",
and ST is wrong.
Marcel Luttgens
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<0nQF8.85184$Ze.1...@afrodite.telenet-ops.be>...
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<5%KF8.84315$Ze.1...@afrodite.telenet-ops.be>...
> > > > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > >
> > > [snip]
> > >
> > > > > Look at and reply to the other post please?
> > > > >
> > > >
> > > > I did, and replied in <b45b8808.02051...@posting.google.com>
> > >
> > > I am referring to my answer to that post.
> > > Look at kcsF8.83355$Ze.1...@afrodite.telenet-ops.be
> > >
> > > If you first reply to that one, we might come back to
> > > our definitions of a'.
> > > See you...
> > >
> > > Dirk Vdm
> >
> > Sorry, but I don't find that message.
>
> http://groups.google.com/groups?as_umsgid=kcsF8.83355$Ze.1...@afrodite.telenet-ops.be
> Just click the link or paste it into your explorer.
>
Thank you, I got it (see below).
> > Anyhow, if your definition of a' is false, SR is wrong, as I showed
> > in my last message b45b8808.02051...@posting.google.com :
>
> You did not because you swapped your definition of a'
> a' = ga
> with mine
> a' = a / ( g(1-vV/c^2) )
> upon which you agreed in post
> http://groups.google.com/groups?as_umsgid=T8rD8.72068%24Ze....@afrodite.telenet-ops.be
>
> You agreed on
> x = a + Vt
> getting LT transformed to
> x' = a' + V't'
> where we define (or abbreviate if you like)
> { a' = a / ( g(1-vV/c^2) )
> { V' = (V-v) / ( 1-vV/c^2 )
> I gave you the complete derivation and you even thanked me for it.
I thanked you because you gave it, but I never agreed with it.
>
> If you*really* feel you must insist on using
> a' = ga
> then if you like, I will use a different notation:
> x = a + Vt
> gets LT transformed to
> x' = Q + V't'
> where we define (or abbreviate if you like)
> { Q = a / ( g(1-vV/c^2) )
> { V' = (V-v) / ( 1-vV/c^2 )
>
> Now you are free to use a' = ga in
> x_G = a
> t_G = 0
> and
> x'_G = ga <=== your a'
> t'_G = -gav/c^2
>
> Think about it...
You can call Q your a' if you prefer, it is nevertheless false, as I
already explained to you:
"Your a' is false, and your second error t' = g(t-(a+Vt)v/c^2
instead of t' = g(t-(x-a)v/c^2) = gt(1-Vv/c^2), compensated the first one.
Remember that the relation you obtained with the wrong a' and t', i.e.
x' = a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2)
reduces to the valid relation
x' = x'+V't' = ga + [(V-v)/( 1-vV/c^2 )] * gt(1-Vv/c^2) = ga + gt(V-v),
where ga is necessarily a'.
Why is your a' false?
First of all, you get a' = ga, not a' = a / ( g(1-vV/c^2), when the x in
the position LT x' = g(x-vt) is replaced by x = a+Vt.
The second argument is a logical one.
The a in x = a+Vt is a constant, which has nothing to do with V.
As the frame S' is moving at v wrt the frame S, a' must be a function of v,
but there is absolutely no justification for it to be also a function of V.
Iow, your a' cannot be correct.
As a' = ga, one gets straightforwardly t' = gt(1-Vv/c^2), thus, if t = 0,
t' = 0, but not t' = -gav/c^2.
Consequently, there is no such thing as a "relativity of simultaneity",
and ST is wrong."
"The origin of Einstein's blunder is now clear. He believed
(as the SRists still do) that the term x has the same physical
meaning in his "position" transform x' = g(x-vt) as in his
"time" transform t' = g(t-xv/c^2).
This is false, because in the "position" transform, x represents
the distance of some point from the origin of the frame S, but
in the "time" transform, x represents the distance travelled
by some object with a velocity V in a time interval t.
For instance, for an event x=a, t=0, the SRists simply
replace x by a, and t by 0 in both transforms, thus obtaining
x' = ga, and t' = -gav/c^2. From t' = -gav/c^2, Einstein invented
his so-called relativity of simultaneity.
As "a" represents the x-coordinate of some point, it is correct
to replace the x in the position transform by a, and obtain x' = ga,
but the distance a can only replace the x in the "time" transform
if it has been travelled by an object moving at V. Then,
a = Vt, and t' becomes gt(1-Vv/c^2). But as t = 0, t' is also
equal to 0, not to -gav/c^2!"
The SRists should eventually think logically, and agree that the LT
t' = g(t-vx/c^2) can not be applied to the event (a,0), where
"a" is a constant.
>
> Dirk Vdm
Marcel Luttgens
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<kcsF8.83355$Ze.1...@afrodite.telenet-ops.be>...
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<KvaF8.23$y54.5...@cacnews.cac.cpqcorp.net>...
> > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
> news:<IjVE8.81449$Ze.1...@afrodite.telenet-ops.be>...
> > > > > "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02051...@posting.google.com...
> > > > > > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
>
> [snip]
>
> > > >
> > > > False, because the object doesn't move from the common origin of
> > > > S and S' at a time t=t'=0, but starts from the distance a from that
> > > > origin, also at t=t'=0.
> > >
> > > Your are talking about the Green flash (start of object), I am talking
> > > about the Red one (time t for observer S).
> > >
> > > And again you make the same error as before:
> > > only event C (Coincidence of observers) has t=t'=0.
> > > All other events that take place at t=0 for observer S,
> > > do *not* take place at time t'=0 for observer S'.
> > > Check the LT: they mix space and time: C is simultaneous
> > > for both observers. G is not. That is the whole point.
> > >
> > > Let's look at the pure events again (you snipped the
> > > relevant stuff):
> > >
> > > 1) Event C: Coincidence and synchronization of observers S and S':
> > > x_C = 0
> > > t_C = 0
> > > and
> > > x'_C = 0
> > > t'_C = 0
> >
> > No problem till now.
> >
> > >
> > > 2) Event G: Green flash at start of object at time 0 for observer S:
> > > x_G = a
> > > t_G = 0
> > > and
> > > x'_G = ga
> > > t'_G = -gav/c^2
> >
> > As for me, to time 0 for S corresponds time 0 for S'.
> > "a" is the coordinate of some point according to S, but S', which
> > is moving at v wrt S, "intellectually" sees the same point at a
> > distance ga at his time t'=0. I say "intellectually", because
> > S' didn't measure its distance to the point.
>
> "Intellectually" you say twice. That should be "Obsoletely
> intellectually", according to the Galilean Relativity point of view,
> which was wrong.
>
> > Otoh, how can
> > you physically explain the negative time -gav/c^2?
>
> For S' the event G happens before event C. Nothing mysterious
> about that. Notice the extremely small number: division by c^2.
> You won't notice anything in every day life of it.
>
Indeed, it is not mysterious, it is wrong, because for both frames,
S and S', the time is counted from the instant at which the two
clocks have been synchonized to zero. So, t' cannot be both
0 and -gav/c^2.
> > You can't,
> > you got it, simply by using the LT t' = g(t-vx/c^2). But
> > this is not allowed, because, as I showed many times, the LT
> > only apply when x = Vt and x' = V't'. There is no "a" in
> > x = Vt.
>
> I'l try to explain...
> Notice that the pair of equations
> { x = Vt
> { x' = V't'
> is merely a *consequence* of the LT.
I showed it a long time ago.
> The LT itself is *derived* by demanding that the pair of
> equations for light signals
> { x = ct
> { x' = ct'
> are satisfied together. Notice the same c in both equations.
Right.
> The LT can easily be derived by demanding that the pair of
> equations
> { x = p + ct
> { x' = p' + ct'
> are satisfied together for any p you choose.
I should say "extended" instead of "derived".
> What matters, is that both equations have the same c.
> The "p" does not matter.
> Wait, read on...
>
> The expression for V' (as a function of V,v and c) in the pair
> { x = Vt
> { x' = V't'
> where
> V' = (V-v) / ( 1-vV/c^2 )
> *follows* from the LT, just like the expression of the a' (as a
> function of a, V, v and c) in the pair
> { x = a + Vt
> { x' = a' + V't'
> where
> { a' = a / ( g(1-vV/c^2) )
> { V' = (V-v) / ( 1-vV/c^2 )
> *follows* from it. Note that the expressions of the V' are exactly
> the same in both cases.
>
And with your a', you get the "extended" transform t' = g(t-(a+Vt)v/c^2,
which is false, because the object moving at V didn't travel the whole
distance a+Vt. Can't you realize that t' transforms the time t during which
the object has actually travelled?
I give you once more the correct solution:
{ x = a + Vt
{ x' = a' + V't'
where
{ a' = ga
{ V' = (V-v) / ( 1-vV/c^2 )
{ t' = g(t-Vtv/c^2) = gt(1-Vv/c^2)
> > If you want to convince me, you have to physically explain
> > the origin of the negative time. If you tell me that no physical
> > explanation is needed, because it is inherent to the LT
> > themselves, I will not believe you.
>
> It is not a question of believing. It follows from the constancy of light
> speed. Either you believe *that*, or you don't. Do you or don't you?
> Please answer?
>
The negative time doesn't follow from the constancy of c, but from
a wrong interpretation of the time transform.
>
> > Don't forget that the simplest physical derivation of the LT
> > is based on light pulses starting *from the common origin* of
> > the two frames S and S' when their time have been synchronized
> > to zero. Nowhere in such derivation has a negative time been spoken
> > about.
>
You act like a typical SRist, trying to evade the real issue, which is
the non-applicability of the LT t' = g(t-xv/c^2) to events having the
coordinates (x=a, t=0), and the falseness of the solution t' = -gav/c^2.
>
> >
> > >
> > > 3) Event R: Red flash of object at time t for observer S:
> > > x_R = a+Vt
> > > t_R = t
> > > and
> > > x'_R = gt(V-v)+ga
> > > t'_R = g( t(1-vV/c^2) - va/c^2)
> > >
> > >
> > > Comparing the times of events C and G you see that
> > > t_C = t'_C = 0
> > > and that
> > > t_G = 0 # t'_G = -gav/c^2
> > >
> > > Yet you insist that t_G = t'_G. It is not.
> > >
> >
> > "Extended" LT can be derived for the cases where x = a+Vt.
>
> >
> > It is straightforward for x' = g(x-vt), because x is the x-coordinate
> > of the point attained by an object (or a light pulse) having leaved
> > the "a" point at t=t'=0.
>
> Sorry Marcel, here I must interrupt because there is nothing new
> and here is your error.
> For the "a" point (event G) the time t' = -gav/c^2.
> It simply is not 0. You want it to be 0, but it just is not.
> It can not be if light speed is constant.
You can repeat the same stuff ad nauseam, it will not make
your relation t'= g( t(1-vV/c^2) - va/c^2), obtained by
replacing the x in t' = g(t-xv/c^2) by x = a+Vt, right.
> Like I just explained: the LT is what it is, and it is valid for
> every event in spacetime.
No, they are neither valid for objects moving faster than light,
nor for events where a<>0 *and* t = 0.
> If you reject it for a#0, then you
> also reject it for a=0, and then you reject the constancy of
> light speed. It is that simple.
>
> You don't have to elaborate and do calculations because
> they are wrong.
> I repeat, we have nothing more to say to each other unless
> you realise that.
>
> So I suggest you reply with either
> - Okay, I accept, or
> - Sorry, I do not accept.
> In the first case we can look at other stuff.
Sorry, I cannot accept nonsense. Once again, the negative time
t' = -gav/c^2 doesn't follow from the constancy of c, but from
a wrong interpretation of the time transform.
>
> Doe je best...
>
> Dirk Vdm
Marcel Luttgens
Marcel Luttgens
message, and my response.
Marcel Luttgens
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<ugxG8.89928$Ze.1...@afrodite.telenet-ops.be>...
> Marcel, I'm starting a new thread because I don't get your last
> reply from my ISP. I have seen it in google groups.
> The only other option I have is to post through google, and I don't
> do that since they don't allow me to antispam my hotmail address.
>
> Let's recap and make sure we understand each other perfectly.
>
> We have the standard setup with observers S and S', coinciding
> at event C with x=x'=0 and t=t'=0.
>
> We have an object moving in S with equation
> x = a + Vt
> Applying the LT gives equation of motion according to S':
> x' = a' + V't'
> where we define
> { a' = a / ( g(1-vV/c^2) )
> { V' = (V-v) / ( 1-vV/c^2 )
>
> We look at 3 events and do nothing more than applying
> the Lorentz Transformation:
>
> 1) Event C: Coincidence and synchronization of observers S and S':
> { x_C = 0 { x'_C = 0
> { t_C = 0 { t'_C = 0
>
> 2) Event G: Green flash at start of object at time 0 for observer S:
> { x_G = a { x'_G = ga
> { t_G = 0 { t'_G = -gav/c^2
>
> 3) Event R: Red flash of object at time t for observer S:
> { x_R = a+Vt { x'_R = gt(V-v) + ga
> { t_R = t { t'_R = gt(1-vV/c^2) - gav/c^2
>
> The only thing I have done, is to apply the LT.
>
> Now, Marcel, what is wrong with this?
> What is your problem?
>
> Use my definitions and abbreviations, and when you introduce a new
> variable, use a name that I haven't used yet? Note that I already have
> defined a', as the thing that came out of the LT applied to x = a + Vt.
>
> And please, to make it somewhat easy to read, leave some white
> space between your text and your equations?
>
> Thanks...
>
> Dirk Vdm
Dirk, you will find the answers at
http://perso.wanadoo.fr/mluttgens/srfalse.htm
Our discussions have been helpful. Thanks!
Marcel Luttgens
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02052...@posting.google.com...
> > "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<acohut$f04$1...@dolly.uninett.no>...> > > news:AOtH8.93642$Ze.1...@afrodite.telenet-ops.be...
> > > > Don't come back anymore. You'll get the same bucket
> > > > of logic over you, and you'll have to "leave the discussion"
> > > > again.
> > >
> > > He will come back with a new idea.
> > >
> > > He has done this many times before:
> > > Got an idiotic idea, proved it mathematically in the ridiculous way
> > > you now know, and got buckets of logic over him. When he realizes
> > > that it is impossible to defend his point, he leave the discussion.
> > >
> > > Paul
> >
> > Do you claim that the t'-coordinate of some object
> > leaving with a velocity V a point P situated at a distance
> > "a" from the origin of the frame S can be obtained by
> > simply replacing the variable x in t' = g(t-xv/c^2)
> > by x = a+Vt, knowing that the object has only travelled
> > the distance x-a = Vt?
>
> Of course. Everyone claims that.
> It gives you what you call "the time transformation"
> t' as a function of t, a, v, V and c.
> It is a relation between the times t and t' and it is valid for
> all the events that satisfy the equation x = a + Vt.
> For t = 0, it gives t' = -gav/c^2 which by the way is
> independent of V.
>
> I think I have explained this before :-)
>
> Dirk Vdm
Did you ask Paul B. Andersen before pouring out nonsense
in his place?
In any case, I am responding to both of you:
First remember that in the "time" LT t' = g(t-xv/c^2),
the variable x corresponds, in the case of an object leaving
with a velocity V the common origin of the frames S and S'
at a time t = t' = 0, to the x-coordinate of the object after
a time interval t. Here, x = Vt.
If the object leaves a point P situated at a distance "a"
from the common origin, it will be situated at a distance a+Vt
from the common origin after the time t, but nevertheless,
it travelled only a distance Vt.
What you should realize is that the "time" LT transforms the
time interval, during which the object and the frame S' travelled
respectively the distances Vt and vt.
That the object started from the common origin, or from any
other point P, is irrelevant.
Iow, the variable x in the "time" LT doesn't represent
the x-coordinate of the object after the time interval t,
it represents the distance Vt travelled by the object
during that time interval.
Hence, the variable x should be replaced by Vt if the object
leaved the common origin, and by x-a if the object leaved
the point P.
When x = a+Vt, Vt = x-a, thus the correct time transform is
t' = g(t-(x-a)v/c^2), or t' = gt(1-Vv/c^2), but not
t' = g(t-(a+Vt)v/c^2), or gt(1-Vv/c^2) - gav/c^2.
As long as you don't realize this, you will make the same
mistakes, and claim for instance that
x' = a'+V't' =
a/( g(1-vV/c^2) + [(V-v)/( 1-vV/c^2 )] * g(t-v(a+Vt)/c^2).
Or, a' = a/( g(1-vV/c^2), as I already told you, is wholly
nonsensical, because the a in x = a+Vt is a constant, which
has nothing to do with V.
Consequently, a' cannot be a function of V.
The correct value of a' is obtained by replacing the variable x
in the "position" LT x' = g(x-vt) by x = a+Vt. This is licit,
because this LT doesn't transform a time coordinate, but a
position coordinate.
Thus, x' = g(a+Vt-vt) = ga + gt(V-v), where ga is only a
function of v and a, not of V.
By replacing a' by ga, V' by (V-v)/( 1-Vv/c^2) and
t' by gt(1-Vv/c^2) in the relation x' = a'+V't',
x' straightforwardly becomes ga + gt(V-v), which
is another proof that the correct t' is given by
gt(1-Vv/c^2), where V = (x-a)/t.
But the brainwashed SRists replace V by (a+Vt)/t in
t' = gt(1-Vv/c^2), thus obtaining
t' = gt(1-(a+Vt)v/tc^2) = g(t-(a+Vt)v/c^2) !
They don't even bother about the meaning of their funny
velocity (a+Vt)/t !
In fact, they are so eager to defend SR, that they
manipulate their own relations, don't think logically,
and act like true crackpots !
Note that I left the discussion, not because "it was
impossible to defend my point", but because you and Andersen
showed yourselves impervious to elementary logic.
I am just trying once more to present my arguments
and your errors, but I am not very hopeful!
Marcel Luttgens
Dirk Van de moortel
"Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02052...@posting.google.com...
Replied on thread "Marcel Luttgens x = a+Vt"
Dirk Vdm
Marcel Luttgens
> "Marcel Luttgens" <mutt...@wanadoo.fr> wrote in message news:b45b8808.02052...@posting.google.com...
>
> Replied on thread "Marcel Luttgens x = a+Vt"
>
> Dirk Vdm
Do you call that a reply? You didn't refute any argument!
Marcel Luttgens