2AB/(t'A-tA) = c <=> A & B are at rest.
Androcles
Quote:
In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c
to be a universal constant, the velocity of light in empty space.
Unquote.
For those of you that love to nitpick, Einstein does not know the difference
between speed and velocity. The light ray reverses direction at B, so c is a
speed.
However, this is but a triviality, as is the phrase "In agreement with
experience", which is mere persuasion. Einstein is quite well versed in
persuasive rhetoric. What is more important to the mathematician is the
missing phrase "if and only if A and B are relatively at rest", for if A and
B are moving relatively one to the other with speed v, BA does not equal AB
since c is a finite and measurable speed.
This then begs the question, where does the 1/2 come from in
1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) ?
It appears to be an assumption derived from 2AB/(t'A-tA)=c, but that only
applies if A and B are at rest. So why not
3/4[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) or
pi/exp(1)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
tau(x',0,0,x'/(c-v))?
We can all make assumptions.
Homework for the relativist:
Calculate q in
q [tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v))
such that tau = t.
Hint: AB[1+v(tB-tA)] / (t'A-tA) = c.
Androcles
Eleaticus
"Androcles" <andr...@home.com> wrote:
> This then begs the question, where does the 1/2 come from in
> 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) ?
> It appears to be an assumption derived from 2AB/(t'A-tA)=c, but that only
> applies if A and B are at rest. So why not
> 3/4[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) or
> pi/exp(1)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
> tau(x',0,0,x'/(c-v))?
No assumption needed to explain the source of
the .5; the taus for light to the reflector
and light back to the moving source are supposed
to be equal, so the tau at the reflection must
be .5 the total tau-time.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus Oren C. Webster Thnk...@concentric.net ?
! "Anything and everything that requires or encourages systematic ?
! examination of premises, logic, and conclusions" ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Androcles
Eleaticus <Thnk...@concentric.net> wrote in message
news:mZV2tCSX...@concentric.net...
> In article <jihO2.1260$Rn5...@news.rdc1.pa.home.com>,
> "Androcles" <andr...@home.com> wrote:
>
> > This then begs the question, where does the 1/2 come from in
> > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) ?
> > It appears to be an assumption derived from 2AB/(t'A-tA)=c, but that
only
> > applies if A and B are at rest. So why not
> > 3/4[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v))
or
> > pi/exp(1)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
> > tau(x',0,0,x'/(c-v))?
>
>
> No assumption needed to explain the source of
> the .5; the taus for light to the reflector
> and light back to the moving source are supposed
> to be equal, so the tau at the reflection must
> be .5 the total tau-time.
>
>
> Eleaticus
and they are quite definitely unequal. I notice you say "supposed". Does
that mean the same as "assumed"?
What mathematician allows this assumption?
Moreover, dtau/dt < 1.0 for the ray to reach the reflector, and dtau/dt >
1.0 for the ray to return to the source.
Androcles.
Eleaticus
"Androcles" <andr...@home.com> wrote:
>
> Eleaticus <Thnk...@concentric.net> wrote in message
> news:mZV2tCSX...@concentric.net...
> > "Androcles" <andr...@home.com> wrote:
> > > This then begs the question, where does the 1/2 come from in
> > > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) ?
> > > It appears to be an assumption derived from 2AB/(t'A-tA)=c,...
> > No assumption needed to explain the source of
> > the .5; the taus for light to the reflector
> > and light back to the moving source are supposed
> > to be equal, so the tau at the reflection must
> > be .5 the total tau-time.
> > Eleaticus
> x'/(c+v) and x'/(c-v) are the times for the light to reach the reflector,
> and they are quite definitely unequal. I notice you say "supposed". Does
> that mean the same as "assumed"?
> What mathematician allows this assumption?
> Moreover, dtau/dt < 1.0 for the ray to reach the reflector, and dtau/dt >
> 1.0 for the ray to return to the source.
Those are the times, t; it is the times tau (t') that
are presumed equal, which is true also in Newtonian
mechanics when the light is emitted by a source
moving at rest wrt the moving system.
It is the farce of non-simultaneity that "informs"
the relating of unequal t values to equal t' (tau)
values.
In spite of the SR time dilation formula which
specifies for a given v and unprimed time duration
one and only one primed time duration, regardless
of location in either the primed or unprimed frame.
> Androcles.
Wayne Throop
::: This then begs the question, where does the 1/2 come from in
::: 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
::: tau(x',0,0,x'/(c-v)) ? It appears to be an assumption derived from
::: why not 3/4[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
::: tau(x',0,0,x'/(c-v)) or
::: pi/exp(1)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
::: tau(x',0,0,x'/(c-v))?
:: Eleaticus <Thnk...@concentric.net>
:: No assumption needed to explain the source of the .5; the taus for
:: light to the reflector and light back to the moving source are
:: supposed to be equal, so the tau at the reflection must be .5 the
:: total tau-time.
: "Androcles" <andr...@home.com>
: x'/(c+v) and x'/(c-v) are the times for the light to reach the
: reflector, and they are quite definitely unequal.
If by "times" you mean "the t-times", as opposed to the "tau-times"
which Thnky was talking about, you are correct but irrelevant.
Read what he wrote. The tau at reflection must be half the total tau.
The fact that the the t at the reflection is not half the total t
simply does not conflict.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Androcles
> In article <mnvO2.1308$Rn5...@news.rdc1.pa.home.com>,
> "Androcles" <andr...@home.com> wrote:
> >
> > Eleaticus <Thnk...@concentric.net> wrote in message
> > news:mZV2tCSX...@concentric.net...
>
> > > "Androcles" <andr...@home.com> wrote:
>
> > > > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
tau(x',0,0,x'/(c-v)) ?
>
>
> > > No assumption needed to explain the source of
> > > the .5; the taus for light to the reflector
> > > and light back to the moving source are supposed
> > > to be equal, so the tau at the reflection must
> > > be .5 the total tau-time.
>
>
> > x'/(c+v) and x'/(c-v) are the times for the light to reach the
reflector,
> > that mean the same as "assumed"?
> > What mathematician allows this assumption?
> > Moreover, dtau/dt < 1.0 for the ray to reach the reflector, and dtau/dt
>
> > 1.0 for the ray to return to the source.
>
>
> Those are the times, t; it is the times tau (t') that
> are presumed equal, which is true also in Newtonian
> mechanics when the light is emitted by a source
> moving at rest wrt the moving system.
In other words, you choose to use "presumed" in place of "assumed" or
"supposed".
Why do you not answer the question?
In Newtonian mechanics, the speed of a corpuscle of light is added the the
speed of the source, relative to the observer. There being no change since
the observer (reflector) is at rest with respect to the source, then it
holds true that x'/c = 2 x'/c divided by 2. However, the times "supposed"
are not x'/c, but x'/(c-v) and x'/(c+v).
Please account for this.
Androcles
Androcles
Wayne Throop <thr...@sheol.org> wrote in message
news:9234...@sheol.org...
> ::: "Androcles" <andr...@home.com>
> ::: This then begs the question, where does the 1/2 come from in
> ::: 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
> ::: tau(x',0,0,x'/(c-v)) ? It appears to be an assumption derived from
> ::: tau(x',0,0,x'/(c-v)) or
> ::: pi/exp(1)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
> ::: tau(x',0,0,x'/(c-v))?
>
> :: Eleaticus <Thnk...@concentric.net>
> :: No assumption needed to explain the source of the .5; the taus for
> :: light to the reflector and light back to the moving source are
> :: supposed to be equal, so the tau at the reflection must be .5 the
> :: total tau-time.
>
> : x'/(c+v) and x'/(c-v) are the times for the light to reach the
> : reflector, and they are quite definitely unequal.
>
> which Thnky was talking about, you are correct but irrelevant.
>
> Read what he wrote. The tau at reflection must be half the total tau.
> The fact that the the t at the reflection is not half the total t
> simply does not conflict.
>
> Wayne Throop thr...@sheol.org http://sheol.org/throopw
dain to comment, now you wish to ignore that and proceed with your own bias.
Quote:
So. Androcles. Tell us. What is the ground speed of the tennis ball?
Is it (c-v), as you insist for the scenario in section 3?
Unquote.
So. WayneTroop. Tell us. What is the closing velocity of the tennis ball
between the helicopters? Is it "not the speed of anything", as you insisted
earlier?
Androcles
Edward Schaefer
>
> Eleaticus <Thnk...@concentric.net> wrote in message
> news:O8e2tCSX...@concentric.net...
> > In article <mnvO2.1308$Rn5...@news.rdc1.pa.home.com>,
> > "Androcles" <andr...@home.com> wrote:
> > >
> > > Eleaticus <Thnk...@concentric.net> wrote in message
> > > news:mZV2tCSX...@concentric.net...
> >
> > > > "Androcles" <andr...@home.com> wrote:
> >
> > > > > 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]=
> tau(x',0,0,x'/(c-v)) ?
> >
> >
> > > > No assumption needed to explain the source of
> > > > the .5; the taus for light to the reflector
> > > > and light back to the moving source are supposed
> > > > to be equal, so the tau at the reflection must
> > > > be .5 the total tau-time.
> >
> >
> > > x'/(c+v) and x'/(c-v) are the times for the light to reach the
> reflector,
> > > that mean the same as "assumed"?
> > > What mathematician allows this assumption?
> > > Moreover, dtau/dt < 1.0 for the ray to reach the reflector, and dtau/dt
> >
> > > 1.0 for the ray to return to the source.
> >
> >
> > Those are the times, t; it is the times tau (t') that
> > are presumed equal, which is true also in Newtonian
> > mechanics when the light is emitted by a source
> > moving at rest wrt the moving system.
> > Eleaticus
> In other words, you choose to use "presumed" in place of "assumed" or
> "supposed".
> Why do you not answer the question?
> In Newtonian mechanics, the speed of a corpuscle of light is added the the
> speed of the source, relative to the observer. There being no change since
> the observer (reflector) is at rest with respect to the source, then it
> holds true that x'/c = 2 x'/c divided by 2. However, the times "supposed"
> are not x'/c, but x'/(c-v) and x'/(c+v).
> Please account for this.
It is required by the *postulate* that the speed of light is the same
in all inertial frames-of-reference. Your refusal to accept that
postulate does not affect the results of a mathematical exercise which
assumes it.
Eleaticus is giving you a perfectly decent answer. To respond with
"but in Newtonian Mechanics..." is a joke of an answer. Einstein is
*blatantly* disregarding Newtonain physics in that article, and makes
that clear at the beginning.
Right now, you are making Eleaticus look like a serious relativist. I
never thought that I would see that.
EMS
Wayne Throop
: In spite of the SR time dilation formula which specifies for a given v
: and unprimed time duration one and only one primed time duration,
: regardless of location in either the primed or unprimed frame.
Simple to prove Thnky wrong here. He says "regardless of location in
either the primed or unprimed frame". So let's do the derivation for a
location in the unprimed frame, and then for a location in the primed
frame. We'll do both derivations in both coordinates, to show clearly
that choice of coordinates doesn't affect things; only choice of which
clock we are considering for calculating dtau/dt.
Given: the primed frame moves at v in the unprimed frame.
Let's look at a clock at a location in the unprimed frame. That clock's position
is x=K, and we can without loss of generality choose K=0.
t' = (t-xv)/sqrt(1-v^2) ; transform (using geometric units)
t' = t/sqrt(1-v^2) ; substitute x=0
t'/t = 1/sqrt(1-v^2) ; tau is larger than t
But maybe that's just because we transformed from t to tau.
So let's do the transform from tau to t, and see what we get.
In k, that same clock's location is x'=K-v*t', and again without
loss of generality, we can choose K=0.
t = (t'+x'*v)/sqrt(1-v^2) ; transform (using geometric units)
t = (t'-t'*v^2)/sqrt(1-v^2) ; substitute xi=-v*tau
t = t'*(1-v^2)/sqrt(1-v^2) ; factor
t'/t = 1/sqrt(1-v^2) ; cancel and rearrange
Same result. Note, though, that in one case is't "because" of the factor
of gamma, and in the other is't "because" of the term -vt. That's the key
point of why relativity is both relative, and self-consistent.
But we've got two more derivations to go. Note in the above cases, we
were following the clock located in the unprimed frame. Now let's
follow the clock located in the primed frame. In x,t, the clock located
in the primed frame has position x=K+vt, and again, we can choose K=0
without loss of generality.
t' = (t-xv)/sqrt(1-v^2) ; transform (using geometric units)
t' = (t-t*v^2)/sqrt(1-v^2) ; substitute x=vt
t' = t*sqrt(1-v^2) ; factor and cancel
t'/t = sqrt(1-v^2) ; rearrange
And similarly for that same location, which in k is xi=0
t = (ta'+x'*v)/sqrt(1-v^2) ; transform (using geometric units)
t = t'/sqrt(1-v^2) ; substitute xi=0
t'/t = sqrt(1-v^2) ; rearrange
So. Swap which clock you're following, and you get the inverse result
for tau/t. The clock you follow is always falling behind.
Further, the entire argument that since tau/t is identical for all
locations in K means there is no relativity of simultaneity is fallacious.
Perhaps Thnky can't tell the difference between rate and setting, but I
think most people can cope with the fact that a clock might be running
at the right rate, but be showing the wrong time.
Androcles
Edward Schaefer <scha...@mitre.org> wrote in message
news:370BA4CF...@mitre.org...
in empty space postulate (false assumption #1), I have the "time from A to B
= time from B to A for light" postulate (false assumption #2), I have found
the "2AB/(t'A-tA)=c" postulate (false assumption #3, in the thread title,
which probably is derived from false assumption #2), but I cannot find a
"light speed same in all frames of reference" postulate. Where is it?
> Eleaticus is giving you a perfectly decent answer. To respond with
> "but in Newtonian Mechanics..." is a joke of an answer. Einstein is
> *blatantly* disregarding Newtonain physics in that article, and makes
> that clear at the beginning.
Then he is not permitted to use x'/(c+v) as a time, since x' is a fixed
distance, and the only way to derive a time from a distance is to divide by
the speed, as per Newtonian mechanics. c+v is a speed that is supposed to be
c, by Einstein's own postulate, which is the real joke. If the relativists
want to play mathematical games, I can too. Play by the rules, if you
please. The original question is: find q such that
q{tau0+tau2] = tau1.
> Right now, you are making Eleaticus look like a serious relativist. I
> never thought that I would see that.
Perhaps he is, but he is making assumptions which are not justified
mathematically.
Androcles
Wayne Throop
:: ball? Is it (c-v), as you insist for the scenario in section 3?
: Androcles <andr...@home.com>
: So. WayneTroop. Tell us. What is the closing velocity of the tennis
: ball between the helicopters? Is it "not the speed of anything", as
: you insisted earlier?
That's rather irrelevant.
I long ago noted that the velocity of light in x',y,z,t is (c-v).
But we were talking specifically about the speed of light in K. Over
and over I specifically said that (c-v) wasn't the speed of anything
*IN* *K* (or in k, for that matter).
K is the track-speed of the light. The ground-speed.
Androcles repeatedly claimed it was (c-v). Androcles was wrong.
So, again. Androcles. Tells us all. Do you still claim that the
ground speed (or track speed, or speed in K) of the tennis ball (or
light ray) must be (c-v)?
Eleaticus
"Androcles" <andr...@home.com> wrote:
> > Eleaticus is giving you a perfectly decent answer. To respond with
> > "but in Newtonian Mechanics..." is a joke of an answer. Einstein is
> > *blatantly* disregarding Newtonain physics in that article, and makes
> > that clear at the beginning.
> Then he is not permitted to use x'/(c+v) as a time, since x' is a fixed
> distance, and the only way to derive a time from a distance is to divide by
> the speed, as per Newtonian mechanics. c+v is a speed that is supposed to be
> c, by Einstein's own postulate, which is the real joke.
You're right about (c+v) or (c-v) being outlawed
velocity of light terms, but those are the effects
not of light speed differences, but changes in
whether the light's destination is moving toward
or away from the light. When the light starts
toward the mirror the mirror is only distance x'
away but the mirror is cheating and running away.
The light thus has to take t.1=x'/c to get to where
the mirror had been. But the mirror has made the
total light travel v*t.1 greater; this works out
to x'/(c-v):
t=(x'+vt)/c
ct-vt=x'
t(c-v)=x'
t=x'/(c-v).
In the other direction the light's destination is
being more cooperative, and is headed toward the
light. t=(x'-vt)/c, etc.
These two time interval formulas work in Newton
if one assumes the light is emitted by a source
at rest wrt the moving fram.
Einstein makes no such assumption or declaration
because by his light speed premise, it makes no
differece what the emitter's velocity might be.
Eleaticus
> : Thnk...@concentric.net (Eleaticus)
> : In spite of the SR time dilation formula which specifies for a given v
> : and unprimed time duration one and only one primed time duration,
> : regardless of location in either the primed or unprimed frame.
Throop:
Simple to prove Thnky wrong here. He says "regardless of location in
either the primed or unprimed frame". So let's do the derivation for a
location in the unprimed frame, and then for a location in the primed
frame. We'll do both derivations in both coordinates, to show clearly
that choice of coordinates doesn't affect things; only choice of which
clock we are considering for calculating dtau/dt.
Eleaticus:
Once again Throop says I'm wrong when I'm right.
This is a rare occasion when he illustrates his
point by thoroughly agreeing with me, and even
believing he proves it. I haven't checked his
'proofs' so who knows.
More of Throop:
Given: the primed frame moves at v in the unprimed frame.
Let's look at a clock at a location in the unprimed frame.
That clock's position
is x=K, and we can without loss of generality choose K=0.
t' = (t-xv)/sqrt(1-v^2) ; transform (using geometric units)
t' = t/sqrt(1-v^2) ; substitute x=0
t'/t = 1/sqrt(1-v^2) ; tau is larger than t
But maybe that's just because we transformed from t to tau.
So let's do the transform from tau to t, and see what we get.
In k, that same clock's location is x'=K-v*t', and again without
loss of generality, we can choose K=0.
t = (t'+x'*v)/sqrt(1-v^2) ; transform (using geometric units)
t = (t'-t'*v^2)/sqrt(1-v^2) ; substitute xi=-v*tau
t = t'*(1-v^2)/sqrt(1-v^2) ; factor
t'/t = 1/sqrt(1-v^2) ; cancel and rearrange
Same result. Note, though, that in one case is't "because" of the factor
of gamma, and in the other is't "because" of the term -vt. That's the key
point of why relativity is both relative, and self-consistent.
But we've got two more derivations to go. Note in the above cases, we
were following the clock located in the unprimed frame.
Eleaticus:
As the astute reader will notice, this ratios do not
depend on location in either frame. As he will show
now from the other frame's viewpoint, this holds true
there, too.
Interestingly, just a few months ago, Throop ranted
over me saying what he just said, and he swore up
and down that the interval formula was location
dependent in exactly the sense he now says contra
above. His argument was that since the ending form-
ula(s) were derived with an interim x, the end result
was dependent on that x - which could be if you had
to set x=0 to get some term to drop out.
Saying that from the opposite viewpoint
('moving' vs 'stationary') it also holds true that
the interval formula(s) do not depend on location
in either frame. Remember, 'viewpoint' is not location
here, but which system is being considered stationary,
we have,
Throop again:
Now let's
follow the clock located in the primed frame. In x,t, the clock located
in the primed frame has position x=K+vt, and again, we can choose K=0
without loss of generality.
t' = (t-xv)/sqrt(1-v^2) ; transform (using geometric units)
t' = (t-t*v^2)/sqrt(1-v^2) ; substitute x=vt
t' = t*sqrt(1-v^2) ; factor and cancel
t'/t = sqrt(1-v^2) ; rearrange
And similarly for that same location, which in k is xi=0
t = (ta'+x'*v)/sqrt(1-v^2) ; transform (using geometric units)
t = t'/sqrt(1-v^2) ; substitute xi=0
t'/t = sqrt(1-v^2) ; rearrange
So. Swap which clock you're following, and you get the inverse result
for tau/t. The clock you follow is always falling behind.
Eleaticus:
Oh, Jim Carrion suckophant Matthew Lybanon
is gonna be real mad at you. His first rant
against me was when I said what you have been
saying here. I pointed out that 'dilation' and
'contraction' are opposite effects but the
duration & interval formulas give the same
effect in general, depending on the viewpoint,
which system was being considered the stationary
one. [Just run the same four derivations for
x'. Remember that |v'| = |v| by Einstein's and
everyone else's assumption.]
Throop:
Further, the entire argument that since tau/t is identical for all
locations in K means there is no relativity of simultaneity is fallacious.
Perhaps Thnky can't tell the difference between rate and setting, but I
think most people can cope with the fact that a clock might be running
at the right rate, but be showing the wrong time.
Eleaticus:
Throop is just a little confused there.
Pick the viewpoint of your choice above, the
t'/t=1/gamma one, say. He just tried to tell
us his derivation proved that - for a given v -
the t frame can force a given t' on the other
system: t' = t/gamma. see? Just set your
clock to n and the other lock must read
n/gamma.
That's impossible; even Jim Carrion couldn't
come up with a bald-faced statement to that effect!
That leaves us with RATE and DURATION for possible
dommains for Throop's results, and if t'/t refers
to RATES then maintaining v for a given period
must produce a duration in the same ratio. If t'/t
refers to DURATION directly, then we're done.
Remember, there's only the three possible 'meanings'
for t'/t: READINGS (I can force your clock to say
anything I want) or RATE (the basis of duration),
and DURATION.
BTW, this is what he has to back off of:
"That clock's position
is x=K, and we can without loss of generality choose K=0."
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Eleaticus
Androcles
I recanted on that a long time ago too, Wayne. The speed in K is c. So your
comment is as irrelevant as mine. You just didn't read it, that's all.
However, the speed of light in k remains c+v and c-v as written, the
distance remains x' as written, and the time is still x'/(c-v) + x'/(c+v)
and is the time of the flight in k.
The distance across the court is x, the distance between the helicopters is
x' = x-vt, the ball is travels from baseline to baseline a distance x at
speed c, and it travels a distance x' between the helicopters at speeds c-v,
c+v, since the lead helicopter is moving away from the ball.
The supposition that this speed between the helicopters is to be called c so
that tau can be determined is mere supposition, not fact, and no postulate
has been given that the speed of light in all frames of reference. For the
ball bouncing between the helicopters, you would not claim that its speed is
measured as c as well as it being c on the ground, would you?
Androcles
Androcles
speed is c-v and the time is x'/(c-v), which is the interval tau1-tau0 for
the light to reach the mirror, and corresponds to t1-t0 = x/c in K, one on
one, isomorphic.
The interval tau2-tau1 is x'/(c+v) and corresponds to the light returning to
the position vt in K adjacent to the origin of k, at speed c. Thus the
return distance in K is x-vt, and tau2-tau1 in k corresponds to t2-t1 =
(x-vt)/c in K one on one, isomorphic.
In K we do not have 2AB/(t'A-tA), we do not even have 2AB. We do have 2AB in
k, but we do not have c. Instead, we have c+v and c-v given.
This contradicts the assumption 2AB/(t'A-tA) = c in k, and it contradicts
the same assumption in K, since the light doesn't reach A, the origin of K.
Androcles
Wayne Throop
: Once again Throop says I'm wrong when I'm right. This is a rare
: occasion when he illustrates his point by thoroughly agreeing with me,
: and even believing he proves it. I haven't checked his 'proofs' so
: who knows.
Here's what Thnky said. "the SR time dilation formula which specifies
for a given v and unprimed time duration one and only one primed time
duration, regardless of location in either the primed or unprimed frame"
That statement is false, and I provided the algebra to prove it false.
:: Further, the entire argument that since tau/t is identical for all
:: locations in K means there is no relativity of simultaneity is
:: fallacious.
: Throop is just a little confused there.
Not hardly.
: Pick the viewpoint of your choice above, the t'/t=1/gamma one, say.
: He just tried to tell us his derivation proved that - for a given v -
: the t frame can force a given t' on the other system: t' = t/gamma.
: see? Just set your clock to n and the other lock must read n/gamma.
I said no such thing. I said nothing at all about one sy stem
"forcing a value" on the other system. I said that at any event
at a location in the unprimed frame, that's what the clock settings
WILL BE; not that one "forces" the other.
Thnky's moronic misinterpretation is just as bad as if I'd said that
the mileage markers where I40 and I95 meet are x and y, then
x must "force" y.
Wayne Throop
: I recanted on that a long time ago too, Wayne. The speed in K is c.
: that's all.
Yet Androcles recently defended this in a recent posting.
:::: <9230...@sheol.org>
:::: The x' value is a distance between two objects coming in K at v,
:::: and x'/(c-v) is used as the time for light to reach one object from
:::: the other; Androcles concludes the light must move at (c-v) in K,
:::: which is obviously wrong.
::: "Androcles" <andr...@home.com>
::: <4R5N2.688$Rn5...@news.rdc1.pa.home.com>
::: Speed is DEFINED as distance divided by time.
: However, the speed of light in k remains c+v and c-v as written, the
: distance remains x' as written, and the time is still x'/(c-v) +
: x'/(c+v) and is the time of the flight in k.
Wrong, of course.
The distance in k is not x',
the time in k is not x'/(c+v)+x'/(c+v).
Nothing in the scenario whatsoever can validly lead to Androcles' conclusion.
You *can* construct a coordinate system in which what Androcles
says is true. That coordinate system is not k.
: The distance across the court is x, the distance between the
: helicopters is x' = x-vt, the ball is travels from baseline to
: baseline a distance x at speed c, and it travels a distance x' between
: the helicopters at speeds c-v, c+v, since the lead helicopter is
: moving away from the ball.
That's certainly not analogous to Einstein's use of those terms.
Androcles is still doing an excellent imitation of somebody
completely ignorant of the meaning and use of coordinate systems.
: The supposition that this speed between the helicopters is to be
: called c so that tau can be determined is mere supposition,
Not, however, something that I suppose. The helicopter example
was merely to show just how silly the "light has speed (c-v) in K".
The altered claim, that "light has speed (c-v) in k", is equally stupid,
since both claims directly contradict what Einstein stated as a
postulate. The whole point is to derive a linear transform in which the
speed of light is the same in both coordinates; Androcles claim
otherwise is about as lame-brained as anything known to mortals.
: For the ball bouncing between the helicopters, you would not claim
: that its speed is measured as c as well as it being c on the ground,
: would you?
If its speed were measured as c in one frame, it'd be measured as c in
all frames. But balls don't actually move at c, and for balls moving at
ordinary speed (eg: below supersonic) the galilean transform is a useful
approximation to what actually occurs.
Edward Schaefer
>
> Edward Schaefer wrote:
> >
> > Androcles wrote:
> > >
> > > In Newtonian mechanics, the speed of a corpuscle of light is
> > > added the the speed of the source, relative to the observer.
> > > There being no change since the observer (reflector) is at
> > > rest with respect to the source, then it holds true that
> > > x'/c = 2 x'/c divided by 2. However, the times "supposed"
> > > are not x'/c, but x'/(c-v) and x'/(c+v).
> > > Please account for this.
> >
> > It is required by the *postulate* that the speed of light is the same
> > in all inertial frames-of-reference. Your refusal to accept that
> > postulate does not affect the results of a mathematical exercise which
> > assumes it.
>
> in empty space postulate
That's it. My appologies for not including "in vacuo" in the above.
With me, its implied by a reference to c or the speed of light unless
explicitly stated otherwise.
> (false assumption #1)
That is your rejection of the postulate. Note that since Einstein
enters this gem into his paper as a postulate, you cannot state that
is is false. Instead, you are required as you continue to read the
paper to assume that it is true and note the consequences. The other
alternative is to stop reading the paper there since what follows
makes no sense without that postulate intact.
> I have the "time from A to B = time from B to A for light"
> postulate (false assumption #2),
No such postulate exists. Not in that form at any rate: You need to
state the frame-of-reference.
> I have found
> the "2AB/(t'A-tA)=c" postulate (false assumption #3, in the thread title,
> which probably is derived from false assumption #2), but I cannot find a
> "light speed same in all frames of reference" postulate. Where is it?
(I'm doing this from memory :-| )
Einstein states: "The rules of physics are the same in all inertial
frame of reference. [This is called] the Principle of Relativity.",
and that one of these rules is that "Light travels at a determined
velocity c anywhere in the universe". Put them together, and the
speed of light being c everywhere and in all inertial frames of
reference results.
"It will shown that these rules an only *apparently* in conflict".
Believe it or not, he actually does so.
EMS
Eleaticus
> : Thnk...@concentric.net (Eleaticus)
> : occasion when he illustrates his point by thoroughly agreeing with me,
> : and even believing he proves it. I haven't checked his 'proofs' so
> : who knows.
> for a given v and unprimed time duration one and only one primed time
> That statement is false, and I provided the algebra to prove it false.
the formula contains no x or x' term and is
independent of location.
> : Pick the viewpoint of your choice above, the t'/t=1/gamma one, say.
> : He just tried to tell us his derivation proved that - for a given v -
> : the t frame can force a given t' on the other system: t' = t/gamma.
> : see? Just set your clock to n and the other lock must read n/gamma.
> I said no such thing. I said nothing at all about one sy stem
> "forcing a value" on the other system.
Oh course you didn't, idiot child. You were too
unaware of what the consequences are of what you
say. If t'/t=F, then t'=tF, and all the umprimed
system has to do to force a particular t' on you
is set his t to whatever he wants. That's
if you were correct about t' and t there not
being (RATES or) DURATIONS. YOU insist that they
are merely settings with no relationship to
intervals/durations, and if so then choosing one's
t forces a particular t' on the other observer.
Settings/readings are completely arbitrary, like
daylight savings time, or Greenwich Standard Time,
etc.
It's YOUR idiocy, not mine, and it isn't
suprising that the idiocy's (e)spouse(r) should
be unaware of the implications of his idiocy.
> I said that at any event
> at a location in the unprimed frame, that's what the clock settings
> WILL BE; not that one "forces" the other.
You said t'/t=F; that is sayig there is a mathematical
relationship - a necessary, SR-scientific relationship -
between t' and t.
> Thnky's moronic misinterpretation is just as bad as if I'd said that
> the mileage markers where I40 and I95 meet are x and y, then
> x must "force" y.
I have never said that x/y=F(.).
But you're right, that would be a moronic assertion,
and is exactly what your're saying by asserting
t' and t are merely readings.
No? Then show us the mathematical statement for
the linear case of your x,y scenario. Here, let
me help: x/y=F; x=yF.
Androcles
> Androcles wrote:
> >
> > Edward Schaefer wrote:
> > >
> > > Androcles wrote:
> > > >
> > > > In Newtonian mechanics, the speed of a corpuscle of light is
> > > > added the the speed of the source, relative to the observer.
> > > > There being no change since the observer (reflector) is at
> > > > rest with respect to the source, then it holds true that
> > > > x'/c = 2 x'/c divided by 2. However, the times "supposed"
> > > > are not x'/c, but x'/(c-v) and x'/(c+v).
> > > > Please account for this.
> > >
> > > It is required by the *postulate* that the speed of light is the same
> > > in all inertial frames-of-reference. Your refusal to accept that
> > > postulate does not affect the results of a mathematical exercise which
> > > assumes it.
> >
> > Strange, but I cannot find that postulate. I have found the speed of
light
> > in empty space postulate
>
> That's it. My appologies for not including "in vacuo" in the above.
> With me, its implied by a reference to c or the speed of light unless
> explicitly stated otherwise.
referring to light in any medium.
> > (false assumption #1)
>
> That is your rejection of the postulate. Note that since Einstein
> enters this gem into his paper as a postulate, you cannot state that
> is is false. Instead, you are required as you continue to read the
> paper to assume that it is true and note the consequences. The other
> alternative is to stop reading the paper there since what follows
> makes no sense without that postulate intact.
Not so: "light is always propagated in empty space with a definite velocity
c which is independent of the state of the emitting body" says nothing
about it being constant in all frames of reference. All this postulate
states is that the speed of light in aether (if we temporarily assume such a
thing) is constant and c. It doesn't state anything about the observer's
motion or any other frame of reference. You cannot read into it anything
other than what it says, or you are making assumptions. If in reading the
paper through you come to the conclusion that it implies the speed of light
is constant in all frames of reference, that is a different story.
I do reject the postulate, but I do not do so out of self conviction and
inability to comprehend what has been written, as so many others might do,
or even on the basis that is defies "common sense". I can temporarily accept
it for the purpose of the discussion, but must eventually reject it on the
grounds that a contradiction is reached. I reject it on the basis that
light, like anything else, has a different velocity attached to it if I move
relative the the source, in accordance with the first postulate Einstein
gives, the "principle of relativity", otherwise know as Galilean Relativity.
If Einstein wishes to demonstrate that it remains constant for me regardless
of how I move, then I need to know why, if I move away from the source and
it remains constant, time does not change in a different manner if I move
toward the source and it remains constant. To answer that question, I need
to know where the 0.5 comes from in
0.5[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)).
> > I have the "time from A to B = time from B to A for light"
> > postulate (false assumption #2),
>
> No such postulate exists. Not in that form at any rate: You need to
> state the frame-of-reference.
Not so. Read Page 40 of Dover: "we establish by definition the 'time'
required by light to travel form A to B equals the 'time' it requires to
travel from B to A".
(Definition or postulate, its amounts to the same thing. )
Here, Einstein is setting us up for the k-frame and the bouncing light, but
he has not stated at this juncture that we need to say anything about the
frame of reference.
> > I have found
> > the "2AB/(t'A-tA)=c" postulate (false assumption #3, in the thread
title,
> > which probably is derived from false assumption #2), but I cannot find a
> > "light speed same in all frames of reference" postulate. Where is it?
>
> (I'm doing this from memory :-| )
On checking again, it is not derived from false assumption #2, it is a false
assumption #3 in its own right. He says "we further assume the quantity... "
(Page 40 of Dover)
> Einstein states: "The rules of physics are the same in all inertial
> frame of reference. [This is called] the Principle of Relativity.",
Which is correct,
> and that one of these rules is that "Light travels at a determined
> velocity c anywhere in the universe".
Not so. That is new postulate, "only apparently irreconcilable with the
former".
(page 38 of Dover)
>Put them together, and the speed of light being c everywhere and in all
inertial >frames of reference results.
If and only if *all* the assumptions are accepted. If any small detail is
found to be incorrect, the whole argument collapses and only the Principle
of Relativity remains, as employed by Galileo and Newton. So where does the
half come from?
In the k-frame, the speed of light is c-v, c+v, given because the distance
is x', the times are x'/(c-v) and x'/(c+v) and unequal, and speed is defined
as distance/time. That doesn't give
2AB/(t'A-tA) = c, we don't have c. What it does give (and this time I'll
give a definition that complies with "the rules of physics are the same in
all inertial frames of reference") is
AB/(tB-tA) = c-v, and BA/(t'A-tB) = c+v, BA = AB.
In the K-frame, we don't have 2AB, because the light returns to vt. What we
do have is AB/(tB-tA) = c and BD/(t'A-tB) = c, where D = vt.
> "It will shown that these rules an only *apparently* in conflict".
> Believe it or not, he actually does so.
Not so. You can believe it or not as you wish. I want proof to a
mathematical certainty, not assumptions, or I will not believe it. This is a
math game, and has nothing to do with physics. That means the 0.5 has to be
explained. It has not been shown, as you claim. It can be plucked out of a
hat, and is by Einstein to suit what he wants it to be. All I asked was that
the relativist discover the value of q such that
q[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,x'/(c-v)) gives
tau = t, or dtau/dt = 1.0.
(hint: q is equal to 0.5 if v = 0, and something else if v>0)
Androcles
Androcles
> : I recanted on that a long time ago too, Wayne. The speed in K is c.
> : So your comment is as irrelevant as mine. You just didn't read it,
> : that's all.
>
> Yet Androcles recently defended this in a recent posting.
> :::: <9230...@sheol.org>
> :::: The x' value is a distance between two objects coming in K at v,
> :::: and x'/(c-v) is used as the time for light to reach one object from
> :::: the other; Androcles concludes the light must move at (c-v) in K,
> :::: which is obviously wrong.
That is your statement, you love to rewrite your own garbage. Its not
relevant to anything I said, of course.
> ::: "Androcles" <andr...@home.com>
> ::: <4R5N2.688$Rn5...@news.rdc1.pa.home.com>
> ::: Speed is DEFINED as distance divided by time.
>
> : However, the speed of light in k remains c+v and c-v as written, the
> : distance remains x' as written, and the time is still x'/(c-v) +
> : x'/(c+v) and is the time of the flight in k.
>
> Wrong, of course.
> The distance in k is not x',
> the time in k is not x'/(c+v)+x'/(c+v).
Oh sure! What does Einstein say? "From the origin of k let a ray be emitted
at the time tau0 along the X-axis to x'..." which of course is not in k, and
not the distance the ray travels from the origin of k to x'.
You are merely attempting to make yourself look important, Wayne, when in
fact you know nothing. You dont want to discuss relativity, you just want to
disagree with anyone and everyone, and now you have disagreed with
Einstein. I see no purpose in continuing this discussion.
<remainder snipped>
Androcles
Eleaticus
> : "Androcles" <andr...@home.com>
> > ::: "Androcles" <andr...@home.com>
> > ::: <4R5N2.688$Rn5...@news.rdc1.pa.home.com>
> > ::: Speed is DEFINED as distance divided by time.
> And this is Androcles' response to that statement, which appears to
> defend the position that (c-v) is the velocity by definition.
What a farce. Wayne "non compos" Throoop putting down
ANDROnoncomposCLES as if the latter were saying
(c-v) was the speed of light, when the whole
point of A's piece(s) is his complaint about
(what he believes to be) Einstein's use of
(c-v) as the speed of light.
A typical Throoooopish distortion of what has been said.
Deliberate? Probably just addled.
Eleaticus
> : Thnk...@concentric.net (Eleaticus)
> : no x or x' term and is independent of location.
> dilation formula which specifies for a given v and unprimed time
> duration one and only one primed time duration, regardless of location
> in either the primed or unprimed frame".
> another answer. There was NOT "one and only one primed duration
> regardless of location in either the primed or unprimed frame".
Partular locations in this 1-dimensional context
are represented by particular values of x and x'.
Show us how the two formulas differ in result
depending on either x or x'.
Let gamma be .5. Let x and x' be whatever you wish:
t'/t = .5; wholly dependent on t', t, v and x'and x are not factors.
t'/t = 1/.5; wholly dependent on t', t, v and x'and x are not factors.
Come on now, Froopy. Show us how either of those
values/results changes if you change location in
either frame.
Not to mention that there are no locations on/in
the moving system that aren't locations on/in the
stationary system, and imagining a switch of sytem
viewpoint to be a switch of location is .. well,
it is Throopness.
Wayne Throop
: Usual nonsense from Wayne. Give us the date.
I already gave you a message ID and a quoted excerpt.
> :::: <9230...@sheol.org>
> :::: The x' value is a distance between two objects coming in K at v,
> :::: and x'/(c-v) is used as the time for light to reach one object from
> :::: the other; Androcles concludes the light must move at (c-v) in K,
> :::: which is obviously wrong.
: That is your statement,
Yes, that is my statement.
> ::: "Androcles" <andr...@home.com>
> ::: <4R5N2.688$Rn5...@news.rdc1.pa.home.com>
> ::: Speed is DEFINED as distance divided by time.
And this is Androcles' response to that statement, which appears to
defend the position that (c-v) is the velocity by definition.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Wayne Throop
: You proved it. Whichever viewpoint you select, the formula contains
: no x or x' term and is independent of location.
Wrong, of course. Look at what Thnky actually said: "the SR time
dilation formula which specifies for a given v and unprimed time
duration one and only one primed time duration, regardless of location
in either the primed or unprimed frame".
I chose one location. I got one answer. I chose another. I got
another answer. There was NOT "one and only one primed duration
regardless of location in either the primed or unprimed frame".
::: Pick the viewpoint of your choice above, the t'/t=1/gamma one, say.
::: He just tried to tell us his derivation proved that - for a given v
::: - the t frame can force a given t' on the other system: t' =
::: t/gamma. see? Just set your clock to n and the other lock must read
::: n/gamma.
:: I said no such thing. I said nothing at all about one sy stem
:: "forcing a value" on the other system.
: Oh course you didn't, idiot child.
Thank you for admitting your decipt you malicious moron.
: You were too unaware of what the consequences are of what you say. If
: t'/t=F, then t'=tF, and all the umprimed system has to do to force a
: particular t' on you is set his t to whatever he wants.
Nobody "forces" anything. Idiot.
: It's YOUR idiocy, not mine,
You are the moron who thinks mere calculation "forces" things to occur.
That description of a process can change the process.
Wayne Throop
: I want proof to a mathematical certainty, not assumptions,
Does Androcles even know what a mathematical proof is?
For example, the above suggests he thinks mathematical proofs
can proceed without any assumptions.
Androcles
> : "Androcles" <andr...@home.com>
> : I want proof to a mathematical certainty, not assumptions,
> : or I will not believe it.
>
> For example, the above suggests he thinks mathematical proofs
> can proceed without any assumptions.
have nothing of any importance to contribute to anyone. You are the idiot
that thinks x' is not a distance in k, even though Einstein says "from the
origin of system k let a ray be emitted to x'. Total idiotic nonsense from
the guy with the big red ball on his nose. When you have something of any
consequence to say, keep it to yourself, the rest of us know what a stupid
buffoon you are.
Androcles
Androcles
Wayne Throop <thr...@sheol.org> wrote in message
news:9236...@sheol.org...
> : "Androcles" <andr...@home.com>
>
> I already gave you a message ID and a quoted excerpt.
>
> > :::: <9230...@sheol.org>
> > :::: The x' value is a distance between two objects coming in K at v,
> > :::: and x'/(c-v) is used as the time for light to reach one object from
> > :::: the other; Androcles concludes the light must move at (c-v) in K,
> > :::: which is obviously wrong.
> : That is your statement,
>
> Yes, that is my statement.
>
> > ::: "Androcles" <andr...@home.com>
> > ::: <4R5N2.688$Rn5...@news.rdc1.pa.home.com>
> > ::: Speed is DEFINED as distance divided by time.
>
> And this is Androcles' response to that statement, which appears to
> defend the position that (c-v) is the velocity by definition.
>
Jan Bielawski
<
< Wayne Throop <thr...@sheol.org> wrote in message
< > : "Androcles" <andr...@home.com>
< >
< > : However, the speed of light in k remains c+v and c-v as written, the
< > : distance remains x' as written, and the time is still x'/(c-v) +
< > : x'/(c+v) and is the time of the flight in k.
< >
< > Wrong, of course.
< > The distance in k is not x',
< > the time in k is not x'/(c+v)+x'/(c+v).
<
< Oh sure! What does Einstein say? "From the origin of k let a ray be emitted
< at the time tau0 along the X-axis to x'..." which of course is not in k, and
< not the distance the ray travels from the origin of k to x'.
x' is in units of system K (since it's defined as x-vt) thus the
distance in k is *not* x'.
It's like this: the mirror is at rest in the system k and situated
x' K-units of length away from the origin of k.
--
Jan Bielawski )\._.,--....,'``.
Molecular Simulations Inc. /, _.. \ _\ ;`._ ,.
San Diego, CA fL `._.-(,_..'--(,_..'`-.;.'
j...@msi.com http://www.msi.com
-disclaimer-
unless stated otherwise, everything in the above message is personal opinion
and nothing in it is an official statement of molecular simulations inc.
Wayne Throop
: You are the idiot that thinks x' is not a distance in k, even though
: Einstein says "from the origin of system k let a ray be emitted to x'.
Which, of course, does not make x' the distance in k.
Androcles, of course, wants to sweep under the rug the fact that
Einstein said BEFORE his above snippet, that "we place x'=x-vt".
Hence, by definition, x' is a K-distance to the k-origin.
: When you have something of any consequence to say, keep it to
: yourself, the rest of us know what a stupid buffoon you are.
Hey. It took Androcles a month to figure out that the velocity
of light in K is c, and he STILL hasn't figured out that the
velocity of light in k is also c. Androcles' buffonery reaches
new and astounding heights of idiocy, unaspried to by mere mortals
such as myself.
Eleaticus
"Androcles" <andr...@home.com> wrote:
> > : However, the speed of light in k remains c+v and c-v as written, the
> > : distance remains x' as written, and the time is still x'/(c-v) +
> > : x'/(c+v) and is the time of the flight in k.
> > Wrong, of course.
> > The distance in k is not x',
> > the time in k is not x'/(c+v)+x'/(c+v).
> Oh sure! What does Einstein say? "From the origin of k let a ray be emitted
> at the time tau0 along the X-axis to x'..." which of course is not in k, and
> not the distance the ray travels from the origin of k to x'.
x' is the stationary constant (within a given
experiment) distance from the moving origin
to the moving mirror, as a stationary measurement.
'twas one of the many Einstein flubs to use
a primed variable to represent a stationary
system measurement. By defining x' as x-vt
x' stays a constant stationary length value
as the mirror's x-coordinate increases over
time: x'=(x+vt)-vt.
x' can't be an SR moving system value because
no gamma is involved. Well, gamma=1/1.
> You are merely attempting to make yourself look important, Wayne, when in
> fact you know nothing.
Actually, he knows every bit of professor and
textbook vomit he has ever swallowed. He
doesn't understand any of it, but he knows so
thoroughly the buzz phrases that constitute
ersatz understanding, he has no clue as to
his cluelessness.
You're trying to understand about as well as
he tries.
Edward Schaefer
>
> Edward Schaefer wrote:
> >
> > Einstein states: "The rules of physics are the same in all inertial
> > frame of reference. [This is called] the Principle of Relativity.",
> > Put them together, and the speed of light being c everywhere and in
> > all inertial frames of reference results.
>
> If and only if *all* the assumptions are accepted.
If you do not wish to accept the assumptions, then why are you arguing
the details?
EMS
Wayne Throop
: 'twas one of the many Einstein flubs to use a primed variable to
: represent a stationary system measurement.
Of course, it wasn't a "flub". In that paper, Einstein was using the
greek alphabet to represent coordinates of the moving system, and the
roman alphabet to represent coordinates of the stationary system. Since
x' uses the roman alplhabet, can you guess the system of which it is
a measure?
Wayne Throop
:: defend the position that (c-v) is the velocity by definition.
: Thnk...@concentric.net (Eleaticus)
: What a farce. Wayne "non compos" Throoop putting down
: ANDROnoncomposCLES as if the latter were saying (c-v) was the speed of
: light, when the whole point of A's piece(s) is his complaint about
: (what he believes to be) Einstein's use of (c-v) as the speed of
: light.
Right. Androcles thinks Einstein meant the speed of light in k
to be (c-v). Androcles is wrong about that. Tnky is just gibbering.
Wayne Throop
:: another answer. There was NOT "one and only one primed duration
:: regardless of location in either the primed or unprimed frame".
: Thnk...@concentric.net (Eleaticus)
: Partular locations in this 1-dimensional context are represented by
: particular values of x and x'.
Correct. If you choose a particular value of x, you get one result.
If you choose a particular value of x', you get the inverse.
This contradicts Thnky's claim that you can choose a position
in either frame, and get the same result.
: Come on now, Froopy. Show us how either of those values/results
: changes if you change location in either frame.
I already showed you. All you have to do is plug in the
values for x', or x, as the case may be.
Androcles
> Androcles wrote:
> >
> > Edward Schaefer wrote:
> > >
> > > Einstein states: "The rules of physics are the same in all inertial
> > > frame of reference. [This is called] the Principle of Relativity.",
> > > Put them together, and the speed of light being c everywhere and in
> > > all inertial frames of reference results.
> >
> > If and only if *all* the assumptions are accepted.
>
> the details?
>
Androcles
Jan Bielawski
< In article <m4bP2.1641$Rn5...@news.rdc1.pa.home.com>,
< "Androcles" <andr...@home.com> wrote:
<
< > > : However, the speed of light in k remains c+v and c-v as written, the
< > > : distance remains x' as written, and the time is still x'/(c-v) +
< > > : x'/(c+v) and is the time of the flight in k.
<
< > > Wrong, of course.
< > > The distance in k is not x',
< > > the time in k is not x'/(c+v)+x'/(c+v).
<
< > Oh sure! What does Einstein say? "From the origin of k let a ray be emitted
< > at the time tau0 along the X-axis to x'..." which of course is not in k, and
< > not the distance the ray travels from the origin of k to x'.
<
< x' is the stationary constant (within a given
< experiment) distance from the moving origin
< to the moving mirror, as a stationary measurement.
Right.
< 'twas one of the many Einstein flubs to use
< a primed variable to represent a stationary
< system measurement.
Now this is interesting. How can mere *notation* influence anything?
< By defining x' as x-vt
< x' stays a constant stationary length value
x' is also a new variable, defined as x-vt. It changes just like
x changes. In the paper the letter "x'" is also used to denote
a specific number: the mirror K-distance to k-origin. Such notational
recycling is very common in contexts similar to this one. As long as
one knows what "x'" denotes in each instance there is no problem.
< as the mirror's x-coordinate increases over
< time: x'=(x+vt)-vt.
x' = x-vt, not (x+vt)-vt.
Androcles
> In article <m4bP2.1641$Rn5...@news.rdc1.pa.home.com>,
> "Androcles" <andr...@home.com> wrote:
>
> > > : However, the speed of light in k remains c+v and c-v as written, the
> > > : distance remains x' as written, and the time is still x'/(c-v) +
> > > : x'/(c+v) and is the time of the flight in k.
>
> > > Wrong, of course.
> > > The distance in k is not x',
> > > the time in k is not x'/(c+v)+x'/(c+v).
>
> > Oh sure! What does Einstein say? "From the origin of k let a ray be
emitted
> > at the time tau0 along the X-axis to x'..." which of course is not in k,
and
> > not the distance the ray travels from the origin of k to x'.
>
> x' is the stationary constant (within a given
> experiment) distance from the moving origin
> to the moving mirror, as a stationary measurement.
>
> a primed variable to represent a stationary
> x' stays a constant stationary length value
> time: x'=(x+vt)-vt.
>
> no gamma is involved. Well, gamma=1/1.
>
> > You are merely attempting to make yourself look important, Wayne, when
in
> > fact you know nothing.
>
> Actually, he knows every bit of professor and
> textbook vomit he has ever swallowed. He
> doesn't understand any of it, but he knows so
> thoroughly the buzz phrases that constitute
> ersatz understanding, he has no clue as to
> his cluelessness.
>
> You're trying to understand about as well as
> he tries.
tolerate Wayne's venom. I think you are correct, he doesn't understand
anything, but he sure can spout it. If he were an educator, he'd be useful
cook. But alas, he never will be.
Androcles
Cat Lover
> : Thnk...@concentric.net (Eleaticus)
> : represent a stationary system measurement.
> greek alphabet to represent coordinates of the moving system, and the
> roman alphabet to represent coordinates of the stationary system. Since
> x' uses the roman alplhabet, can you guess the system of which it is
> a measure?
definition(s) of x', there was no gamma involved,
and if x' were moving system then the derivation of
the moving system x' formula would be invalid by
reduction to the absurd: x'=x-vt and x'=(x-vt)/gamma.
He should have just used some letter like L (length)
or D (distance) or some such.
The contents of that 1905 paper make Glird's assertion
that einstein made many dark-orifice changes at the
last second seem very likely.
Eleaticus
> < 'twas one of the many Einstein flubs to use
> < a primed variable to represent a stationary
> < system measurement.
> Now this is interesting. How can mere *notation* influence anything?
Now THAT is weird, jan.
When both tradition and his own usage in
that paper is to use primed symbols for
the moving system variables, and unprimed for the
stationary, he uses x' not only as a stationary
system value but as a CONSTANT, AND as a
moving system variable.
Hmm. Oh. I see. that was really smart of him. 8-(
> < By defining x' as x-vt
> < x' stays a constant stationary length value
>
> x' is also a new variable, defined as x-vt. It changes just like
> x changes. In the paper the letter "x'" is also used to denote
> a specific number: the mirror K-distance to k-origin. Such notational
> recycling is very common in contexts similar to this one. As long as
> one knows what "x'" denotes in each instance there is no problem.
In hundreds of articles on these ngs, it is
quite clear that most do not get it. And the
problem's source was entirely unnecessary and
uncalled for.
That would be true even if he had not treated
x' as a variable in the immediately subsequent
treatment of his taus in which the constant x'
was used.
> < as the mirror's x-coordinate increases over
> < time: x'=(x+vt)-vt.
> x' = x-vt, not (x+vt)-vt.
I quoted that in the article in which seem not to
have noticed I did so, then I showed how x' is a constant,
with x changing as x+vt over time. x'=(x+vt)-vt, where
I perhaps should have stated the even more obvious
idea that some x0=x at the time of emission, and
x'=(x0+vt)-vt.
> --
> Jan Bielawski )\._.,--....,'``.
> Molecular Simulations Inc. /, _.. \ _\ ;`._ ,.
> San Diego, CA fL `._.-(,_..'--(,_..'`-.;.'
> j...@msi.com http://www.msi.com
>
> -disclaimer-
> unless stated otherwise, everything in the above message is personal opinion
> and nothing in it is an official statement of molecular simulations inc.
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Cat Lover Sylvester-Heathcliff Acres ?
! Probably a non-profit land trust, plus a non-profit intentional ?
! community. Cat loving, anti-rat-race, lower-income folk. No ?
! smokes, other drugs. Explore www.ic.org for ideas. ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Wayne Throop
:: as the mirror's x-coordinate increases over time: x'=(x+vt)-vt.
: j...@iris8.msi.com (Jan Bielawski)
: x' = x-vt, not (x+vt)-vt.
Correct. But since the mirror's x-coordinate is (x0+vt), and of we
allow Thnky the "flub" of overloading x to mean x0 similarly to but
sillier than Einstein's overloading of x' to mean both the variable and
its constant value for the mirror, why then if you substitute the
mirror's x-coordinate (x+vt) into the definition of x'=x-vt you get
x'=(x+vt)-vt.
Which means that x' = x at t=0 and at no other time.
The point being, as Thnky says, the mirror's x coordinate changes
over time at exactly the correct rate to make the mirror's x' constant.
Wayne Throop
::: 'twas one of the many Einstein flubs to use a primed variable to
::: represent a stationary system measurement.
:: thr...@sheol.org (Wayne Throop)
:: Of course, it wasn't a "flub". In that paper, Einstein was using the
:: greek alphabet to represent coordinates of the moving system, and the
:: roman alphabet to represent coordinates of the stationary system.
:: Since x' uses the roman alplhabet, can you guess the system of which
:: it is a measure?
: Thnk...@concentric.net (Cat Lover)
: He should have just used some letter like L (length) or D (distance)
: or some such.
Only if Thnky tends to confuse greek with roman glyphs.
Jan Bielawski
< In article <F9xys...@msi.com>, j...@iris8.msi.com (Jan Bielawski) wrote:
<
< > < a primed variable to represent a stationary
< > < system measurement.
<
<
< Now THAT is weird, jan.
<
< When both tradition and his own usage in
< that paper is to use primed symbols for
< the moving system variables, and unprimed for the
< stationary, he uses x' not only as a stationary
< system value but as a CONSTANT, AND as a
< moving system variable.
In this particular derivation he uses Greek letters to denote the
"moving" system k inertial coordinates and the Latin ones for the
"stationary" K system.
The primed coordinate is a new label defined by x' = x-vt, so that
instead of using (x,y,z,t) we can use (x',y,z,t) instead. This
is easier to use because for a mirror at a fixed distance away from
the origin of k its x'-coordinate is constant. The *value*
of any such constant is also denoted by "x'" - a common practice.
Finally, performing a series of imaginary experiments with mirrors
at different distances from the k-origin yields the differential
equation that follows (actually, two such experiments would
suffice if one assumes linearity of tau ).
Edward Schaefer
> > Androcles wrote:
> > >
> > > Edward Schaefer wrote:
> > > >
> > > > Einstein states: "The rules of physics are the same in all inertial
> > > > frame of reference. [This is called] the Principle of Relativity.",
> > > > Put them together, and the speed of light being c everywhere and in
> > > > all inertial frames of reference results.
> > >
> > > If and only if *all* the assumptions are accepted.
> >
> > the details?
> >
> I want to know where the 1/2 comes from, of course.
> Androcles
Do you wish to accept the constancy of c in all inertial
frames-of-reference in this discussion? If so, I will tell you. If
not, then forget it.
EMS
Androcles
> > > Androcles wrote:
> > > >
> > > > Edward Schaefer wrote:
> > > > >
> > > > > Einstein states: "The rules of physics are the same in all
inertial
> > > > > frame of reference. [This is called] the Principle of
Relativity.",
> > > > > Put them together, and the speed of light being c everywhere and
in
> > > > > all inertial frames of reference results.
> > > >
> > > > If and only if *all* the assumptions are accepted.
> > >
> > > the details?
> > >
> > I want to know where the 1/2 comes from, of course.
> > Androcles
>
> Do you wish to accept the constancy of c in all inertial
> frames-of-reference in this discussion? If so, I will tell you. If
> not, then forget it.
No. There is no postulate to that effect, as I have already shown. I want
to know where the half comes from, but I do not accept Schaefer's postulate.
Androcles
Wayne Throop
:: frames-of-reference in this discussion?
: "Androcles" <andr...@home.com>
: There is no postulate to that effect, as I have already shown.
Of course not. It's an inevitable consequence of the two postulates
Einstein did give, though. The insight needed: the "law of propogation
of light" is a natural law. Therefore, it must be the same whether
refered to either of an arbitrary pair (or, in general, in all frames).
Glird
>:: Do you wish to accept the constancy of c in all inertial
>:: frames-of-reference in this discussion?
>: "Androcles" <andr...@home.com>
>: There is no postulate to that effect, as I have already shown.
>Of course not. It's an inevitable consequence of the two postulates
>Einstein did give, though. The insight needed: the "law of propogation
>of light" is a natural law. Therefore, it must be the same whether
>refered to either of an arbitrary pair (or, in general, in all frames).
>
It is a consequence of the third of Einstein's four Light Postulates:
"In agreement with experience we further assume the quantity
2AB/(tA' - tA) = c,
to be a universal constant -- the velocity of light in empty space."
In order for that to hold good in variably moving systems, their clocks
have to keep time at variable rates. That's one of the things Minkymath
blocks out of the minds of those taught to do the calculations his way.
glird
Androcles
Glird <gl...@aol.com> wrote in message
news:19990413125655...@ngol01.aol.com...
Yes, I agree with you on that point. Note that Einstein says velocity, not
speed as in his 2nd postulate, and in this case clearly it is speed that is
inferred, since the light reflects at B.
There is no agreement with experience here either. The speed of light is
finite, the principle of relativity holds true, and it takes longer for the
light to return to A from B than it did to reach B from A if A and B are
moving apart, and in the general case they will be in relative motion, IN
AGREEMENT WITH EXPERIENCE.
So the well-defined postulate should read:
[AB + v.(AB/c)]/(t'A-tA) = c, which reduces to 2AB/(t'A-tA) = c when v = 0.
But then, that is the thread title I chose:-)
Androcles.
Jan Bielawski
< In article <9239...@sheol.org>, thr...@sheol.org (Wayne Throop) writes:
<
< >:: Do you wish to accept the constancy of c in all inertial
< >:: frames-of-reference in this discussion?
< >: "Androcles" <andr...@home.com>
< >: There is no postulate to that effect, as I have already shown.
< >Of course not. It's an inevitable consequence of the two postulates
< >Einstein did give, though. The insight needed: the "law of propogation
< >of light" is a natural law. Therefore, it must be the same whether
< >refered to either of an arbitrary pair (or, in general, in all frames).
< >
< It is a consequence of the third of Einstein's four Light Postulates:
< "In agreement with experience we further assume the quantity
< 2AB/(tA' - tA) = c,
< to be a universal constant -- the velocity of light in empty space."
Calling it a "postulate" is a bit of a stretch as it does not refer
to any potential physical experimental result but is merely a
clock calibration convention. Of course the statement that such
a calibration is a *sensible thing to do* is in itself a conseqauence
of the second postulate.
< In order for that to hold good in variably moving systems, their clocks
< have to keep time at variable rates. That's one of the things Minkymath
< blocks out of the minds of those taught to do the calculations his way.
Clocks in relativity are only assumed to be identical in construction and
the meaning of their "rates" (as such) after separation is left undefined.
Only relative rates (relating two observers' clocks) are defined.
Wayne Throop
: So the well-defined postulate should read:
:
: [AB + v.(AB/c)]/(t'A-tA) = c, which reduces to 2AB/(t'A-tA) = c when v = 0.
Exactly. Einstein's explicitly said his version is for v=0. (that is,
"it is essential to have time defined by means of stationary clocks in
the stationary system"). So Einstein got it right, after all.
Androcles
> : "Androcles" <andr...@home.com>
> : So the well-defined postulate should read:
> :
> : [AB + v.(AB/c)]/(t'A-tA) = c, which reduces to 2AB/(t'A-tA) = c when v =
0.
>
> Exactly. Einstein's explicitly said his version is for v=0. (that is,
> "it is essential to have time defined by means of stationary clocks in
> the stationary system"). So Einstein got it right, after all.
>
As a mathematician, he'd make a good cook! :-)
Androcles
Wayne Throop
: As a mathematician, he'd make a good cook! :-)
But, for some reason, Androcles has yet to point out a
single flaw in Einstein's math. Not that he was a particularly
sophisticated mathematician in the first place, but still.
Oh, and :-)
Androcles
> "In agreement with experience we further assume the quantity
> 2AB/(tA' - tA) = c,
> to be a universal constant -- the velocity of light in empty space."
this case clearly it is speed that is inferred, since the light reflects at
B.
problemey
Androcleesey
Jim Carr
in message <9235...@sheol.org> ...
}
} :::: <9230...@sheol.org>
} :::: The x' value is a distance between two objects coming in K at v,
} :::: and x'/(c-v) is used as the time for light to reach one object from
} :::: the other; Androcles concludes the light must move at (c-v) in K,
} :::: which is obviously wrong.
In article <m4bP2.1641$Rn5...@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>That is your statement, you love to rewrite your own garbage.
It is his statement. It was not rewritten. Anyone who cares to
decide which half of your sentence is true and which half is false
just has to use the message ID given above and discover that
http://www.dejanews.com/getdoc.xp?AN=461876018
contains exactly those words in the footnote correctly describing
the calculation that was done.
>Its not relevant to anything I said, of course.
It is relevant to what you said earlier
} : "Androcles" <andr...@home.com>
} : However, the speed of light in k remains c+v and c-v as written, the
} : distance remains x' as written, and the time is still x'/(c-v) +
} : x'/(c+v) and is the time of the flight in k.
(This is in http://www.dejanews.com/getdoc.xp?AN=463888495)
}
} Wrong, of course.
} The distance in k is not x',
} the time in k is not x'/(c+v)+x'/(c+v).
and to the following reply.
In article <m4bP2.1641$Rn5...@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>Oh sure!
Yes, what Throop says is correct and what you said above was wrong.
>What does Einstein say? "From the origin of k let a ray be emitted
>at the time tau0 along the X-axis to x'..." which of course is not in k, and
>not the distance the ray travels from the origin of k to x'.
This establishes that what Throop said was correct. That is, the
value of x and t and x' are all measured in K, with x' clearly stated
to be on the X axis in K, while the value of tau and xi are measured
in k, with the distance in k being some value xi' measured along the
Xi axis. [The greek letters xi and Xi are pronounced "ksea" in Greek
and "ex-i" or "zai" by Americans; lower case is zeta with an extra
squiggle, while upper case is three horizontal lines, aka the Cascade
Hyperon symbol.] The constant x' is the distance between source and
mirror as measured in K, and the distance the light has to travel
from source to mirror as measured in K is equal to x' + vt.
This should be obvious if you read carefully and do not jump to the
conclusion that a primed variable is measured in the moving frame k.
>You are merely attempting to make yourself look important, Wayne, when in
>fact you know nothing.
He knows enough to correct your errors. Pay attention and you
can learn something.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Androcles
Jim Carr <j...@ibms48.scri.fsu.edu> wrote in message
news:7fgct0$enb$1...@news.fsu.edu...
> } : "Androcles" <andr...@home.com>
> >What does Einstein say? "From the origin of k let a ray be emitted
> >at the time tau0 along the X-axis to x'..." which of course is not in k,
and
> >not the distance the ray travels from the origin of k to x'.
>
> This establishes that what Throop said was correct. That is, the
> value of x and t and x' are all measured in K, with x' clearly stated
> to be on the X axis in K, while the value of tau and xi are measured
> in k, with the distance in k being some value xi' measured along the
> Xi axis. [The greek letters xi and Xi are pronounced "ksea" in Greek
> and "ex-i" or "zai" by Americans; lower case is zeta with an extra
> squiggle, while upper case is three horizontal lines, aka the Cascade
> Hyperon symbol.] The constant x' is the distance between source and
> mirror as measured in K, and the distance the light has to travel
> from source to mirror as measured in K is equal to x' + vt.
>
> This should be obvious if you read carefully and do not jump to the
> conclusion that a primed variable is measured in the moving frame k.
I get (0,0,0,t)
What is the K-coordinate of reflection?
I get (x,0,0,t+x/c)
What is the K-coordinate of reception?
I get (vt,0,0,t+x/c+x'/c)
What is the k-coordinate of emission?
I get (0,0,0,tau)
what is the k-coordinate of reflection?
I get (xi,0,0,tau+xi/c)
What is the k-coordinate of reception?
I get (0,0,0,tau+xi/c+xi/c)
How do we map the coordinates from K,(x,y,z,t) to k,(xi,eta,zeta,tau)?
I get
(0,0,0,t) |=> (0,0,0,tau)
(x,0,0,t+x/(c-v)) |=> (xi,0,0,tau+xi/c)
(vt,0,0,t+x/(c-v) + x'/(c+v)) => (0,0,0,tau+xi/c+xi/c)
Where am I wrong?
Androcles
Wayne Throop
: What is the K-coordinate of emission?
: [...]
Why is that relevant?
Hint: the function tau does not take K-coordinate parameters.
(but see below for K->k directly, rather than via x')
: What is the K-coordinate of reception?
: I get (vt,0,0,t+x/c+x'/c)
Wrong, of course. The return time is not x'/c.
Androcles is confusing himself trying to reuse "t" for multiple meanings.
: How do we map the coordinates from K,(x,y,z,t) to k,(xi,eta,zeta,tau)?
By taking the (x',y,z,t) to (xi,eta,zeta,tau) map that Einstein
derives, and substituting x-vt for x', just as Einstein does on
the page after he (quite correctly) states the three tau values
as
tau0 = tau(0,0,0,t)
tau1 = tau(x',0,0,t+x'/(c-v))
tau1 = tau(0,0,0,t+x'/(c-v)+x'/(c+v))
See there, on that page? It's just after Einstein says
"Substituting for x' it's value, we obtain". Seems quite clear, IMO.
: (0,0,0,t) |=> (0,0,0,tau)
: (x,0,0,t+x/(c-v)) |=> (xi,0,0,tau+xi/c)
: (vt,0,0,t+x/(c-v) + x'/(c+v)) => (0,0,0,tau+xi/c+xi/c)
: Where am I wrong?
Well, if you really want to know. Let's take the time of emission
to be 0 (with no loss of generality). This gets rid of some of
the ambiguity in Androcles' overloading of T. Now, let's cal
the x position of the mirror at time t=0 xM. GIven that, we can
now state the functions
(0,0,0,0) -> (0,0,0,tau0)
(xM+v*(xM/(c-v)),0,0,(xM/(c-v))) -> (xi1,0,0,tau1)
(xM+v*(xM/(c-v)+xM/(c+v)),0,0,(xM/(c-v)+xM/(c+v))) -> (0,0,0,tau2)
The above shows some of the reason why Einstein introduced x'
to simplify his job; the rest of it occurs when you try to
differentiate by chain rule.
So. Where is Androcles wrong? Well... right from the start,
straight through to the end, that's where Androcles is wrong.
Jim Carr
thr...@sheol.org (Wayne Throop) wrote:
}
} : Thnk...@concentric.net (Eleaticus)
} : represent a stationary system measurement.
}
} greek alphabet to represent coordinates of the moving system, and the
} roman alphabet to represent coordinates of the stationary system. Since
} x' uses the roman alplhabet, can you guess the system of which it is
} a measure?
In article <EYb3tCSX...@concentric.net>
Thnk...@concentric.net (Cat Lover) writes:
>
>No guessing involved, Throop. Besides the explicit
>definition(s) of x', there was no gamma involved,
>and if x' were moving system then the derivation of
>the moving system x' formula would be invalid by
>reduction to the absurd: x'=x-vt and x'=(x-vt)/gamma.
Since you did not have to guess to get the same answer everyone gets
and that Einstein used, it was all clear as originally written, as
Throop explained. Thank you for clarifying that there was no flub.
>He should have just used some letter like L (length)
>or D (distance) or some such.
He could have, and that is what I did in explaining this to Glird who
sometimes has a problem with subtle points like that, but as you clearly
point out, his paper is sufficiently clear that it is not necessary to
do so for the intended audience of the paper.
Jim Carr
} : Once again Throop says I'm wrong when I'm right. This is a rare
} : occasion when he illustrates his point by thoroughly agreeing with me,
} : and even believing he proves it. I haven't checked his 'proofs' so
} : who knows.
}
} Here's what Thnky said. "the SR time dilation formula which specifies
} for a given v and unprimed time duration one and only one primed time
} duration, regardless of location in either the primed or unprimed frame"
}
} That statement is false, and I provided the algebra to prove it false.
In article <Ff73tCSX...@concentric.net>
Thnk...@concentric.net (Eleaticus) writes:
>
>You proved it.
Very good. Like any good troll, Eleaticus has included a true
statement in amongst his erroneous ones.
} : Pick the viewpoint of your choice above, the t'/t=1/gamma one, say.
} : He just tried to tell us his derivation proved that - for a given v -
} : the t frame can force a given t' on the other system: t' = t/gamma.
} : see? Just set your clock to n and the other lock must read n/gamma.
}
} I said no such thing. I said nothing at all about one sy stem
} "forcing a value" on the other system.
>Oh course you didn't, ...
And another, thereby agreeing with Throop concerning what Eleaticus
wrote earlier.
Jim Carr
>a primed variable to represent a stationary
>system measurement.
You think that Einstein should have assumed that every reader
of his paper was mathematically illiterate and unable to follow
his clear statement that roman letters (primed or not) are used
for references to one frame and greek letters are used for the
other? It is really quite easy to follow a notational difference
as big as that.
>By defining x' as x-vt ....
he made it clear what coordinate system it was defined in.
Androcles
> : "Androcles" <andr...@home.com>
> : What is the K-coordinate of emission?
> : [...]
>
> Why is that relevant?
>
> Hint: the function tau does not take K-coordinate parameters.
then provides a hint.
He's right, of course. Tau does not take K-coordinate parameters. So?
> (but see below for K->k directly, rather than via x')
> : What is the K-coordinate of reception?
> : I get (vt,0,0,t+x/c+x'/c)
>
> Wrong, of course. The return time is not x'/c.
> Androcles is confusing himself trying to reuse "t" for multiple meanings.
Oh? Elaborate. Tell us, so that we may be less confused that we are. Please
educate us all, Oh Wise One.
> : How do we map the coordinates from K,(x,y,z,t) to k,(xi,eta,zeta,tau)?
>
> By taking the (x',y,z,t) to (xi,eta,zeta,tau) map that Einstein
> derives, and substituting x-vt for x', just as Einstein does on
> the page after he (quite correctly) states the three tau values
> as
> tau0 = tau(0,0,0,t)
> tau1 = tau(x',0,0,t+x'/(c-v))
> tau1 = tau(0,0,0,t+x'/(c-v)+x'/(c+v))
>
> See there, on that page? It's just after Einstein says
> "Substituting for x' it's value, we obtain". Seems quite clear, IMO.
Good. Glad you said "IMO". But what does it MEAN? When Einstein says "In
agreement with experience" (p 40, Dover and which we have no experience of)
what does THAT mean? Your opinion is only your prejudice, right?
> : (0,0,0,t) |=> (0,0,0,tau)
> : (x,0,0,t+x/(c-v)) |=> (xi,0,0,tau+xi/c)
> : (vt,0,0,t+x/(c-v) + x'/(c+v)) => (0,0,0,tau+xi/c+xi/c)
> : Where am I wrong?
>
> Well, if you really want to know. Let's take the time of emission
> to be 0 (with no loss of generality). This gets rid of some of
> the ambiguity in Androcles' overloading of T.
Not mine! I'll gladly accept (0,0,0,0) as the coordinate of the event of
emission!
Now, let's cal
> the x position of the mirror at time t=0 xM. GIven that, we can
> now state the functions
>
> (0,0,0,0) -> (0,0,0,tau0)
Stop right there. There is no need to use Throops overloading of tau
either.
Let's use
(0,0,0,0) -> (0,0,0,0), with the understanding that on the left is x,y,z,t
and on the right is xi,eta,zeta, tau.
Now, which frame is xM in?
I know I'm wrong, all I'm asking for is Froopy to show me where, instead of
telling me how wrong I always am.
Androcles
Wayne Throop
::: What is the K-coordinate of emission? [...]
:: Wayne Throop <thr...@sheol.org>
:: Why is that relevant?
:: Hint: the function tau does not take K-coordinate parameters.
: "Androcles" <andr...@home.com>
: So Mr Confused snips with [...] so we don't know what he's referring
: to, then provides a hint.
We know exactly what he's refering to. He's refering to
the K-coordinate of emission, and he's asking why that is
relevant in discussing Einstein's derivation, since in
Einstein's derivation, tau is not a function of K-coordinate.
: He's right, of course. Tau does not take K-coordinate parameters.
: So?
So why is discussing K-coordinates relevant in this context.
Just like I said the first time.
::: What is the K-coordinate of reception?
::: I get (vt,0,0,t+x/c+x'/c)
:: Wrong, of course. The return time is not x'/c. Androcles is
:: confusing himself trying to reuse "t" for multiple meanings.
: Oh? Elaborate. Tell us, so that we may be less confused that we are.
: Please educate us all, Oh Wise One.
It's hard to educate somebody as proud of their ignorance as Androcles.
The specific problem is that Androcles uses "t" to indicate the total time
from emission to the current event, even on opposite sides of the
reflection, and therefore invalidly substitutes x' for x-vt. True, x'
is *defined* as x-vt, but Androcles has overloaded the meaning of t, and
thus has lost track of the physical situation represented by the
expressions he is abusing.
:: See there, on that page? It's just after Einstein says "Substituting
:: for x' it's value, we obtain". Seems quite clear, IMO.
: Good. Glad you said "IMO". But what does it MEAN?
Well, let's see. The verb "substitute" means to put one thing in
the place of another. So we put x-vt in place of x' in the expressions
Einstein has got up to that point. The verb "obtain" means to gain
or attain by planning or effort, and in this case refers to the
expressions you get when you do the substitution just mentioned.
Are you getting any help for your reading comprehension problems, Androcles?
: When Einstein says "In agreement with experience" (p 40, Dover and
: which we have no experience of) what does THAT mean?
Hmmm. Apparently not. Well, the verb "agree" has a meaning
of "to be similar or to correspond". The noun "experience"
refers to observation and perception. So Einstein is starting
a sentence where he's refering to something which is similar to,
or corresponds with, observations and perception that people make.
Have you tried a dictionary for your problem?
:: (0,0,0,0) -> (0,0,0,tau0)
: Stop right there. There is no need to use Throops overloading of tau either.
What overloading of tau? I didn't even mention tau.
: Let's use (0,0,0,0) -> (0,0,0,0), with the understanding that on the
: left is x,y,z,t and on the right is xi,eta,zeta, tau.
Sure, no problem. For a linear traI know I'm wrong, all I'm asking for is Froopy to show me where, instead of
telling me how wrong I always am.nsform of the kind we're looking
for the origin always coincides, so we know tau0=0 if the emission
occurs at t=0 at the K-origin.
: Now, which frame is xM in?
Well, I said "Now, let's call the x position of the mirror at time t=0
xM". Since, as Androcles just pointed out, K uses x,y,z,t and k uses
xi,eta,zeta,tau, can you figure out "which frame" xM is "in"? Class?
Anybody? Let's not see all the same hands.
: I know I'm wrong, all I'm asking for is Froopy to show me where,
: instead of telling me how wrong I always am.
That was what I just got done doing. You are "wrong" in that you
are giving incorrect expressions for the K-coordinates of
reflectionan and detection. Of course, I could have made
a typo or mento and gotten it wrong, too. But Androcles'
versions are CLEARly wrong.
Eleaticus
some deleted garbage.
==========================================================
What follows is the "Jim Carr the Nut and Liar" faq.
The eligible content grows daily.
Lines/sections following "E:" are current notes
by Eleaticus.
Otherwise, lines without an initial set of attribution
indicators are by Jim Carr.
Lines with attribution indicators containing a j or c,
are by Jim Carr.
Lines with attribution indicators containing an e
are by Eleaticus.
***************************************************************
Message-ID: <6nb1hk$46g$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
e>x0' is not a constant you idiot!
You were the one who said it was in an "invariant equation",
an "invariant term", or other things to imply that it is a
scalar or otherwise time independent.
---------------------------------------------------------------
E: The content on the surface (and below) is a lie. I
have never said that x0' is a constant. I have always
said x0'=x0-vt, it being a transformed x0, whatever
x0 might be in a particular case.
E: You did note Jim said "in an" equation or term, right?
E: The concept of invariance - which the concept of a
scalar depends on but is not identical to - is about
the value of an expression/quantity AFTER some
operation(s): is the value the same? It has nothing to
do with what goes on IN the operations, nor the content/
appearance/details of the expression/quantity.
E: Jim presents himself as an expert on Special Relativity
and math. Do you really imagine such a person doesn't
know any of the above? We are left with the possibilities
only of his insanity or corruption, whether 'insanity' be
a function of drugs or illness or whatever.
--------------------------------------------------------------
After all, if it is a function of time, you need to write that
explicitly since that time dependence is also transformed. It
is wrong to say that x' - x0'(t) is a valid equation in frame K
since there is no "t" in that frame.
-----------------------------------------------------------------
E: As everyone knows that has ever dropped in on one
of these threads, the discussion is about the Galilean
tranformation, and t'=t.
E: Further, the transforms are of the nature: x'=x-vt,
which is the same - though more specific - as saying
x'=f(t). In this case, x0'=f(t), not x0'=f(t'), al-
though the two are in practice identical.
E: x' is also a function of t, so why not complain that
the equation should show x'(t) - x0'(t)? the answer
is: Jim Carr is a nutcase who responds to all of my
posts with his brain in 100% animus mode.
E: x'=x-vt=f(t,) and x0'=x0-vt=f(t,).
E: Later/elsewhere, Jim and his fellow crackpots insist
that (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0) is wrong, but
say (x'+vt-x0) is a correct expression. That expression
mixes notation, and is thus doubly wrong by his above
complaint, since there are terms there that fit neither
frame.
E: Jim presents himself as an expert on Special Relativity
and math. Do you really imagine such a person doesn't
know any of the above? We are left with the possibilities
only of his insanity or corruption, whether 'insanity' be
a function of drugs or illness or whatever.
***************************************************************
Message-ID: <6nb111$41a$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
[Subscribe to sci.physics.relativity]
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
j| Correct. Eleaticus is explicitly stating that x0' is a constant
j| when he asserts that this new equation is "invariant" (sic) and
j| this leads to his absurd conclusion.
E: That is absolutely a bald faced lie.
E: He made the same lie elsewhere, explaining his
'reasoning':
----------------------------------------------------------------
Only one problem: there is only one object since Eleaticus has
asserted earlier that x0 is a constant and this result proves
that (x'-x0') is invariant -- and thus that x0' is also a constant.
--------------------------------------------------------------------
E: The 'result' of which he speaks is:
(x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0),
showing invariance.
E: Obviously, I never asserted that x0 being a constant
proves (x'-x0') is invariant. If that is what he is
saying I said, that is just a bald-faced lie. If that
is HIS opinion, it is just nut-case idiocy.
E: If he is saying that HE believes x0' is a constant
that is just nut-case idiocy or crackpottery; x0'=x0-vt.
E: If he is saying that anything in what he says SHOWS
x0' is a constant that is just nut-case crackpottery;
x0'=x0-vt.
E: If he is saying that I believe x0' is a constant,
then he says so after I said x0'=x0-vt a few hundred
or a thousand times; that would be both lying and
crackpottery on his part.
E: He says similar things elsewhere in ways that lend
themselves to even more obvious debunking.
e>Perhaps thousands of times I insist that x0'
e>is not a constant, that x0'=x0-vt, ...
ROTFL. No, this is the first time. In the past you have claimed
it is an invariant, that is, a scalar.
E: That is just bald faced, unabated lying. My only claims
about invariance are about:
(x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0).
e> ... that x0'=x0-vt, ...
This means that x0' gets smaller when the ' frame is moving to
the left as it was in your example? But you pretty drawing says
it gets bigger. Better try again.
E: If the ' frame is moving to the left, then v < 0, and
x0' > x0, which is "it gets bigger". His comment is
just plain nut-cake idiocy.
E: Come to think about it - I'm editing this as I insert it
in a typical lying post of his to mews.admin - in that
post he says the faq contains bad arithmetic.
**************************************************************************
Message-ID: <6n94m9$dlm$1...@news.fsu.edu>
Newsgroups: sci.physics,sci.physics.relativity
e>The discussion was about the galilean transform
e>of a coordinate axis, x'=x-vt.
So why do you claim it applies to constants as well as coordinates,
E: Jim says there that no object or point has a
coordinate that doesn't change, that no coord-
inate is constant.
***********************************************************************
Message-ID: <6navsb$314$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
| e> Actually, I explicitly state that the
| e> traditional galilean transform equations
| e> are the correct galilean transform equations, ...
j| Of course you do; you just don't use them.
E: Contrast that with another section of the same
carrackpot article:
| e> Further, I use those traditional galilean
| e> transform equations over and over and over
| e> again.
j| ;-) Indeed you do.
j| One too many times, to be precise.
E: See? Not even the slightest compunction about
lying. Obviously, one or the other of his
statements must be false. Either I don't use
them XOR I "indeed ... do" use them.
e>What he means here is that some axis coordinates
e>cannot be tranformed when the axis is transformed.
Scalars are invariant under a transform.
E: It is (x-x0) that is invariant not some of
the various coordinates of a point across a
variety of two or more coordinate systems:
(x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0),
e>In particular, the coordinates of circle centers,
e>points of light emission, etc.
Those coordinates can change under a transformation of the variables
in the equation, as we have shown. It is your assertion that an
equation is "invariant", that is, that an equation is a number,
that is unsupportable.
E: From the eariest days of Einstein and before,
"invariant" in physics was about one central
question since it was the operational test of
the Principle of Relativity: is the equation
the same after transform as before. Do you
really imagine Jim Carr doesn't know that?
*********************************************************
Message-ID: <6n85ep$gne$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
What he did was use the notation of a constant for something
which is not constant. That is why he is unable to do any
problem that has numbers in it. He would either get the wrong
answer or expose the fact that he has done a _third_ transform
that has not been written explicitly. That is, since x0'=x0-vt,
(x'+vt') - (x0'+vt') = 0
==> (x'+vt') - (x0-vt'+vt') = 0
==> x' + vt' - x0 = 0
the equation we all say you get from the galilean transform,
one that clearly shows a lack of covariance by making the time
dependence of x0' explicit.
E: Notice that he mixes notation between frames. Above
here he complained because a form he provided mixed
notation. Since t'=t, x'=x-vt, x0'=x0-vt, we unmix
notation:
E: (x') - (vt' + x0) = 0
(x-vt) - (x0 + vt) = 0
(x-x0) = 0.
E: Which shows unmistakeably there is not time dependence
in that equation.
E: The time dependence of x0' is not in question, that's
just a lie by implication, where he has explicitly
asserted the lie elsewhere.
E: The time dependence of x0' is an irrelevancy; the only
question is whether the whole expression's value is
invariant, and the simple algebra shown above proves
that.
E: Jim presents himself as an expert on Special Relativity
and math. Do you really imagine such a person doesn't
know any of the above? We are left with the possibilities
only of his insanity or corruption, whether 'insanity' be
a function of drugs or illness or whatever.
E: You could just as easily have put in an oranges function
or alligators/acres function in the transforms and mixed
notation to make it look like the transformed equation
was oranges or alligator dependent. How old does the
typical child get before he/she understands that hiding
something doesn't make it non-existene? Even my cat knows
better. Crackpots don't apparently.
E: Let x'=x-vt+2A-3O), where A is alligators per acre and
O is the number of oranges in October in the nearest
orange grove to the stationary axis' origin.
Following the carrackpot, we have:
x' + vt' - 2A + 3O - x0 = 0,
which - according to his and others' crackpot logic -
makes the transformed equation both alligators/acre
and oranges dependent, even though simple reduction
of the equation shows us there are no such effects:
(x-vt+2A-3O) + vt - 2A + 3O - x0 = 0
x - x0 = 0.
-----------------------------------------------------------------------
... However, since he has x0' = x0 - vt, it
should be clear that x0' is not a constant, but rather a function of
t.
E: Thus proving in his own words he knows he is lying
everytime he says I say/believe/etc that x0' is
a constant. Unless he is insane enough to not know
how he continually misreperesents and/or says some-
thing delusive.
--------------------------------------------------------------------
Thus the error is one of notation hiding the time dependence, but
then applying it as if it really were a constant.
E: In other words, I hide the time dependence
by not using mixed-frame notation. His/their
(x'+vt'-x0) - as shown above and elsewhere
often - equals (x-vt +vt-x0)=(x-x0), which
proves there is no time dependence. True,
x' is time dependent, and so is x0', but as
shown, the relevant quantity in my thesis
- the only relevant quantity - (x-x0), trans-
forms as:
(x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0),
invariantly.
---------------------------------------------------------------------
t>To be consistent with his own notation, Thnky must Thnk that an
t>object (let's say, a rock) is motionless in both (t',x') coordinates and
t>in (t,x) coordinates. This is a difficult Thot to justify, since
t>(t',x') coordinates are in motion WRT (t,x) coordinates.
Correct.
E: It is hard to tell if that is a straight lie, or just
nut-cakery. Since I use the standard transform,
x'=x-vt, obviously only an extremely corrupt liar
(which includes Wayne Crackthroop and Jim Carr) could
imagine I ever said, hinted, or implied motionlessness
in both frames, and there has never been an explicit
hint, statement, or implication of such a silly idea.
I do suppose there may sometime or other been an extreme
sarcasm that an idiot could misconstrue, but I think not.
-----------------------------------------------------------
Eleaticus is explicitly stating that x0' is a constant
when he asserts that this new equation is "invariant" (sic) and
this leads to his absurd conclusion.
E: That is just bald faced lying. Or are the insane
exempt from the commission of lies by some decree
of power(s) that be? x' is not constant, so the
constancy of (x'-x0') is only possible if x0' is
not constant.
E: Jim presents himself as an expert on Special Relativity
and math. Do you really imagine such a person doesn't
know any of the above? We are left with the possibilities
only of his insanity or corruption, whether 'insanity' be
a function of drugs or illness or whatever.
******************************************************************
Message-ID: <6n19qg$fh6$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
e>An active transform is not a coordinate transform, ...
Right, it is a transform of the center (in the opposite direction)
done to effect the change of coordinates without a coordinate
transform. But if you know that, you would also know that you
cannot do both at the same time.
E: Transform of the center? Center of a circle?
He really is saying a circle center moves in
the opposite direction of the circle! Right?
********************************************************************
Message-ID: <6n1aho$g14$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
j...@ds16.scri.fsu.edu (Jim Carr) wrote:
|
|e> >E/!: His complaint is indeed that I transform the 'constant'
|e> > x0. And if you don't? What is the value of x0'? x0'=x0,
|e> > which is what he just lied about me having said he said.
j| Anyway, it was Eleaticus who claimed that x0' is a constant after
j| a Galilean transformation, even going so far as to claim that it
j| is invariant -- that is, the same as x0.
E: That is bald faced lying. There is absolutely no
way I have ever said, hinted, or implied that x0'
is constant/invariant. (x'-x0')=>(x-x0), and since
x'=x-vt, then obviously x0' must change accordingly.
That obvious logic makes his statement not just a
lie but extreme crackpottery.
j|Roberts is merely relying
j| on the Eleaticus FAQ that states this Eleaticus-error very clearly.
E: That is just plain lying.
E: Still about whether x0' is a constant:
e>I have always said that x0'=x0-vt, just as the
e>basic x'=x-vt tranmsform says, ...
e>This means using (x-x.c), for example, instead of
e>just x, so the transform of any such term in any
e>equation is, with x.c as the/a 'centroid' of im-
e>portance, such as a circle center or point of light
e>emission:
e> (x'-x.c') = [ (x-vt) - (x.c-vt) ] = (x-x.c),
You did not say that the equation in the K' frame is (x' -[x.c-vt]),
you said it was (x'-x.c') where x.c' and x.c are your notation for
constants, consistent with your claim that (x'-x.c') = (x-x.c) shows
E: What kind of extreme nut does it take to
see x.c'=(x.c-vt) and then mix notation
and say I did not say x.c'=[x.c-vt] in
effect?
E: And x.c' is not my notation for a constant. That's
just another version of one of his lies. And x.c
would only be a 'constant' for one particular circle,
but any equation using it is about an infinite number
of possible circles.
*******************************************************************
Message-ID: <6lmq47$8jj$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
Eleaticus wrote:
e} just what magic is it in equations of class A, B, C
e} (you fill in the expanded names) that would prevent
e} this simple algebra from working:
e} (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0),
The Lord Leto II <c72...@showme.missouri.edu> writes:
l>There is nothing wrong with the algebra -- those equations are perfectly
l>fine. That statement says that distance (in galilean relativity)
l>between two objects is not dependent upon a moving frame of reference.
Only one problem: there is only one object since Eleaticus has
asserted earlier that x0 is a constant and this result proves
that (x'-x0') is invariant -- and thus that x0' is also a constant.
E: The 'logic' is insane there. The algebra in question
contains x0'=x0-vt and Carr says x0' is a constant?
E: (x'-x0') is invariant [it always reduces to (x-x0)],
x' is not constant, so x0' is a constant?
That logic is, with C=constant, and V=non-constant:
C=(V+C).
What a monster of a nut-cake Carr is!
E: Jim presents himself as an expert on Special Relativity
and math. Do you really imagine such a person doesn't
know any of the above? We are left with the possibilities
only of his insanity or corruption, whether 'insanity' be
a function of drugs or illness or whatever.
-----------------------------------------------------------------
However, up above it is self-evident that x0' is a function of t.
At least it is to most of us.
E: After he argues against it for months, and after even
here he quotes me saying [ x0'=x0-vt ], "it is to
most of us"?!!?
**************************************************************
Message-ID: <6kalj0$7v4$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
e>(2) Apply the galilean transforms to all occurences
e>of values on the axes being transformed:
e> (x'-x_c')^2 + (y'-y_c')^2 + (z'-z_c')^2 = (ct')^2.
Here you assume that "values" are "variables", an interesting mistake.
E: Jim believes a coordinate axis has no place on it
where the place has a particular value. Right?
E: Jim believes that anything that doesn't change
is not a coordinate. Right?
E: Jim believes that an equation containing (x-x0),
where x0 is the x-coordinate of, say, a circle
center or point of emission, applies to only one
circle/etc ever? Otherwise, it is a variable, and
even he must allow the transform, IFF he is honest
and not insane.
e>(3) Which reduces - by collection of terms - to:
e>
e> (x-x_c)^2 + (y-y_c)^2 + (z-z_c)^2 = (ct)^2.
So you conclude that there is no Doppler effect in a Galilean world.
E: We're talking galilean transforms, and Jim
wants us to screw up the equation about the
expending light sphere, and make it differ
wrt every possible observer in the universe.
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Message-ID: <6naosf$qi5$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity,sci.physics
e>You obviously understand that the x0'=x0 stuff is
e>complete nonsense.
Just as you understand that your claim that x0' is not time
dependent is complete nonsense.
E: Just more of his fairly recent campaign to turn around
his claim that x0'<>x0-vt? It ALWAYS has been my claim
that x0 transforms just like any other x-coordinate,
explicitly and often: x0'=x0-vt. Is Carr 'just' insane?
Or 'just' so recklessly corrupt it LOOKS like insanity?
And thousands of times I have said otherwise.
E: Always saying x0'=x0-vt is always claiming that x0'
is time dependent, so even misunderstanding cannot be
an excuse for his lie here and elsewhere.
e>Now, just remember that the claim is about (x-x0)
e>and (x'-x0') and not just x, x', x0', or x0 individually.
Your original claim was about both.
E: Whatever that means. Both of the four things, x, x', x0',
and x0? Both of (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0)?
E: I have never claimed invariance/constancy for x' or x0';
I have always specified the standard transforms: x'=x-vt
amd x0'=x0-vt. It is just one more example of a carr lie.
********************************************************************
Message-ID: <6kd0fq$pte$1...@news.fsu.edu>
Newsgroups: sci.physics.relativity
e>(c) every point in the universe has transformed x-values
e>- as you admit above - at every instant. This is obvious; ...
Yes, but we have been trying to get you to recognize this
obvious point for weeks. I suspect that you do know it,
of course, just as I suspect that you know how to do the
chain rule correctly.
E: After he argues against it for months, "we have been
trying to get you to recognize this obvious point for
weeks." ?!??!
<g<carr
*********************************************************
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus Oren C. Webster Thnk...@concentric.net ?
! "Anything and everything that requires or encourages systematic ?
! examination of premises, logic, and conclusions" ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Eleaticus
Androcles
Froopy drew the diagram. Froopy's diagram shows the ray ends at vt in the
K-frame.
In K, in which the speed of light is c, with the understanding that the
chosen origin of the time coordinate is the event of emission, the
coordinate of emission is (0,0,0,0) because I don't want to overload t.
In k, this is (0,0,0,0).
So the coordinate in K (0,0,0,0) maps to the coordinate in k (0,0,0,0),
where again we do not want to overload tau.
That is
(0,0,0,0) => (0,0,0,0)
Ok so far?
For reflection, the light arrives at x in K, and because it moves at c in K,
the time to get there is x/c.
So
(x,0,0,x/c) => (xi,0,0,tau1), with tau1 as yet undiscovered.
For the event of reception, Froopy drew the diagram. Froopy's diagram shows
the ray ends at vt in the K-frame.
(vt, 0,0, x'/c+x/c) is the coordinate of reception in K, which is wrong, of
course,
What is the K-coordinate of reception?
> ::: I get (vt,0,0,t+x/c+x'/c)
>
> :: Wrong, of course. The return time is not x'/c. Androcles is
> :: confusing himself trying to reuse "t" for multiple meanings.
Froopy doesn't read what I said, and prefers to attempt to confuse in order
to make nasty remarks, because Froopy just loves to make nasty remarks,
because Froopy is just a nasty little man.
Androcles
Wayne Throop
: For the event of reception, Froopy drew the diagram. Froopy's diagram
: shows the ray ends at vt in the K-frame. (vt, 0,0, x'/c+x/c) is the
: coordinate of reception in K, which is wrong, of course,
Here we see AndyPandy's fundamental mistake, which he can't seem
to overcome. x'/c is not the K-coordinate duration between the time
of emission and reflection, x/c is not the K-coordinate duration between
the relfection and reception, and x'/c+x/c is not the K-coordinate
time of reception.
The K-coordinate time of reception is x'/(c-v)+x'/(c+v).
That's true whether you take lightspeed in k to be c or not.
That's simply the kinematics, refered ONLY to K.
Look folks, it is SIMPLY NOT MY FAULT that AndyPandy can't do
simple story problems involving algebra.
: Froopy just loves to make nasty remarks,
: because Froopy is just a nasty little man.
Not at all. I am merely reacting with appropriate disgust
to AndyPandy's pride in his own know-nothingism.
Jim Carr
news:371225CF...@mitre.org...
}
} Androcles wrote:
} >
} > Edward Schaefer <scha...@mitre.org> wrote in message
} > news:370E14C8...@mitre.org...
} > } Androcles wrote:
} > } > If and only if *all* the assumptions are accepted.
} > }
} > } If you do not wish to accept the assumptions, then why are you arguing
} > } the details?
} >
} > I want to know where the 1/2 comes from, of course.
}
} not, then forget it.
In article <%eyQ2.2366$Rn5....@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>No. There is no postulate to that effect, as I have already shown. I want
>to know where the half comes from, but I do not accept Schaefer's postulate.
It comes from the relationship defined in the first equation presented
in the paper you have been reading, an equation that appears on page
894 of Ann. d. Phys. 17, and on page 40 of the Dover translation/reprint.
You are directed to this equation in section I.1 by a sentence in the
section you are reading (I.3). Pretty clear to me.
Granted you have to do some simple algebra to rearrange it ...
Androcles
> : "Androcles" <andr...@home.com>
> : For the event of reception, Froopy drew the diagram. Froopy's diagram
> : shows the ray ends at vt in the K-frame. (vt, 0,0, x'/c+x/c) is the
> : coordinate of reception in K, which is wrong, of course,
>
> to overcome. x'/c is not the K-coordinate duration between the time
> of emission and reflection, x/c is not the K-coordinate duration between
> the relfection and reception, and x'/c+x/c is not the K-coordinate
> time of reception.
>
> The K-coordinate time of reception is x'/(c-v)+x'/(c+v).
in the K-frame is supposed to be c, and the distance it travels in the
K-frame to the reflector is x. So the time is x/c. As EMS pointed out to the
dumbo Froopy, the times x'/(c-v) and x'/(c+v) belong not to the K-frame, but
to the K'-frame. Froopy doesn't see the difference.
> That's true whether you take lightspeed in k to be c or not.
> That's simply the kinematics, refered ONLY to K.
>
> Look folks, it is SIMPLY NOT MY FAULT that AndyPandy can't do
> simple story problems involving algebra.
>
> : Froopy just loves to make nasty remarks,
> : because Froopy is just a nasty little man.
>
> Not at all. I am merely reacting with appropriate disgust
> to AndyPandy's pride in his own know-nothingism.
Right back at ya, Froopy. Eat your own words, as EMS showed you.
Androcles
Wayne Throop
: Utter nonsense. Pure, unadulterated bullshit from Froopy. The speed
: of light in the K-frame is supposed to be c, and the distance it
: travels in the K-frame to the reflector is x. So the time is x/c. As
: EMS pointed out to the dumbo Froopy, the times x'/(c-v) and x'/(c+v)
: belong not to the K-frame, but to the K'-frame. Froopy doesn't see
: the difference.
A lie. EMS never said x'/(c-v) and x'/(c+v) are not K-frame times.
: Eat your own words, as EMS showed you.
I'll repeat for the thinking impaired, such as Androcles.
EMS never said x'/(c-v) and x'/(c+v) are not K-frame times.
Androcles
> : "Androcles" <andr...@home.com>
> : Utter nonsense. Pure, unadulterated bullshit from Froopy. The speed
> : of light in the K-frame is supposed to be c, and the distance it
> : travels in the K-frame to the reflector is x. So the time is x/c. As
> : EMS pointed out to the dumbo Froopy, the times x'/(c-v) and x'/(c+v)
> : belong not to the K-frame, but to the K'-frame. Froopy doesn't see
> : the difference.
>
> A lie. EMS never said x'/(c-v) and x'/(c+v) are not K-frame times.
EMS said x'/(c-v) and x'/(c+v) were K'-frame times.
> : Eat your own words, as EMS showed you.
>
> I'll repeat for the thinking impaired, such as Androcles.
> EMS never said x'/(c-v) and x'/(c+v) are not K-frame times.
And I'll repeat for the thinking impaired Froopy.
Double negatives from Froopy again.
EMS said x'/(c-v) and x'/(c+v) were K'-frame times.
> : Eat your own words, as EMS showed you.
Moreover, Froopy raised the issue, not I. Nothing I said in recent posts
even remotely refers to x'/(c-v) and x'/(c+v), which is what Froopy is
changing the discussion to, to say I am wrong.
What I said was
What is the K-coordinate of reception?
> ::: I get (vt,0,0,t+x/c+x'/c)
>
To which Froopy replies:
> :: Wrong, of course.
So tell us, Oh Wise One, since I am wrong, what is the K (you understand K,
not k) coordinate of reception?
Are you getting any help for your reading comprehension problems, Froopy?
Androcles
Androcles
"In agreement with experience -[whose experience?]- we further assume the
quantity
2AB/(t'A-tA) =c.
Missing from this sentence is the caveat "if and only if A and B are
relatively at rest."
In the case where B is moving relatively to A, the statement is false. c is
finite, and the distance AB does not equal the distance BA. I can move B
from A at v=c/2, and the light takes much longer to reach B than it does to
return to A, even though it is moving at c in both directions. So where does
the 1/2 come from?
The sentence continues with
"to be a universal constant - the VELOCITY [my caps] of light in empty
space. A minor point,maybe, but speed should be used, not velocity, since
the light changes direction.
I'm still waiting for an honest answer from all those that find it pretty
clear, where does the 1/2 come from in
1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))]= tau(x',0,0,t+x'/(c-v)) ?
Androcles
Wayne Throop
:: EMS never said x'/(c-v) and x'/(c+v) are not K-frame times.
: "Androcles" <andr...@home.com>
: Double negatives from Froopy again.
Oh, that's right. AndyPandy can't understand simple English sentence
structure, as was demonstrated before. Here, I'll state it with a
single "does not exist". For expressons (x'/(c-v) and x'/(c+v)), there
exists no statement of EMS's which conflicts with the expression being
a K-frame time.
: EMS said x'/(c-v) and x'/(c+v) were K'-frame times.
Right. That's what he said. Note that what he said does not
conflict with those times being K-frame times.
As is dead obvious. Ask EMS.
: What I said was What is the K-coordinate of reception?
: I get (vt,0,0,t+x/c+x'/c)
: So tell us, Oh Wise One, since I am wrong, what is the K (you
: understand K, not k) coordinate of reception?
Still wrong. The K-coordinates of reception are
(v*(x'/(c-v)+x'/(c+v)),0,0,x'/(c-v)+x'/(c+v))
(for emission at t=0)
Note: Androcles doesn't seem to understand that he's got a VERY confused
use of "t"; it seems to be used in three different ways in his
expression, if we count use of x' as an implicit use of vt. He has one
symbol refering to three differing times. Trivially, his use of t in
both vt and t+[...] is at least poor form. But even if we overlook that
problem, and take the "t" in vt to be equal to the time coordinate of
the same tupple, the time he states is incorrect. You could use
x/c+x'/(c+v), where x has value x=x'+vt=x'+v(x'/(c-v)) (which is the
K-system x coordinate of the reflection). But that's not what AndyPandy
used. You can verify that x/c = x'/(c-v) from the above.
Jan Bielawski
<
< Wayne Throop <thr...@sheol.org> wrote in message
< > : "Androcles" <andr...@home.com>
< > : For the event of reception, Froopy drew the diagram. Froopy's diagram
< > : shows the ray ends at vt in the K-frame. (vt, 0,0, x'/c+x/c) is the
< > : coordinate of reception in K, which is wrong, of course,
< >
< > to overcome. x'/c is not the K-coordinate duration between the time
< > of emission and reflection, x/c is not the K-coordinate duration between
< > the relfection and reception, and x'/c+x/c is not the K-coordinate
< > time of reception.
< >
< > The K-coordinate time of reception is x'/(c-v)+x'/(c+v).
< in the K-frame is supposed to be c,
It is c.
< and the distance it travels in the
< K-frame to the reflector is x. So the time is x/c.
No, the distance mirror-origin is x as measured by K at the emission
instant (which occurs at t = 0 in your labelling scheme, it seems;
that's fine). By the time the light reaches the mirror it will have
travelled a distance greater than x. Let's say, the unknown flight
time origin->mirror is t+. Then the distance travelled by the light
will be:
x + v * t+
But the speed of light according to K is c, so:
distance/time = c
or:
( x + v * t+ ) / t+ = c
Solving for t+ :
t+ = x/(c-v)
In general, if the emission occurs at some time t other than 0, we'd
similarly get:
t+ = (x - vt)/(c-v)
(again, x denotes here the K-coordinate of the mirror, and vt amounts
to the K-coordinate of the moving k-origin.)
In the (x',y,z,t) coordinates the same formula looks like:
t+ = x'/(c-v)
Almost identical considerations for the light coming back to k-origin
(the emitter) yield the return flight time t-:
t- = x'/(c+v)
< As EMS pointed out to the
< dumbo Froopy, the times x'/(c-v) and x'/(c+v) belong not to the K-frame, but
< to the K'-frame. Froopy doesn't see the difference.
Strictly speaking, yes, but since the paper does not mention K' explicitly
at all and K' uses the same units as K, it seems entirely appropriate
to say "K" here.
Jim Carr
in message news:7fgct0$enb$1...@news.fsu.edu...
| >at the time tau0 along the X-axis to x'..." which of course is not in k,
| >and not the distance the ray travels from the origin of k to x'.
|
| This establishes that what Throop said was correct. That is, the
| value of x and t and x' are all measured in K, with x' clearly stated
| to be on the X axis in K, while the value of tau and xi are measured
| in k, with the distance in k being some value xi' measured along the
| Xi axis. [The greek letters xi and Xi are pronounced "ksea" in Greek
| and "ex-i" or "zai" by Americans; lower case is zeta with an extra
| squiggle, while upper case is three horizontal lines, aka the Cascade
| Hyperon symbol.] The constant x' is the distance between source and
| mirror as measured in K, and the distance the light has to travel
| from source to mirror as measured in K is equal to x' + vt.
|
| This should be obvious if you read carefully and do not jump to the
| conclusion that a primed variable is measured in the moving frame k.
In article <EiVS2.1116$xc5...@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>What is the K-coordinate of emission?
>I get (0,0,0,t)
Correct.
>What is the K-coordinate of reflection?
>I get (x,0,0,t+x/c)
Insufficient. You must also write that x = L+v*(x/c), where I
use L for the constant called x' by Einstein.
However, IMO it would be better to write L+v*t1 and t+t1 (where
c*t1 = L+v*t1) rather than x and t+x/c. This makes it clear that
the value of x' at this time, call it x'_1, is L.
Aside: Here I am really using delta-t_1 when I write t1, which is
different from the previous article in some places. The previous
article has some inconsistencies because of the non-zero start time
that requires terms like (t_1 - t_0) for the delta-time. One
should not do algebra as one types ....
>What is the K-coordinate of reception?
>I get (vt,0,0,t+x/c+x'/c)
This is wrong. Assuming you are using x' = L+v*(x'/c) for the length
of the return trip as measured in K, you need v*(x/c+x'/c) rather than
v*t for the x coordinate. Again, it would be better to write this
as v*(t1+t2) and t+t1+t2 -- where I am using t2 as (t_2 - t_1).
This makes it clear that the value of x' at this time, call it x'_2,
is zero, an important detail in the next part of the derivation.
>What is the k-coordinate of emission?
>I get (0,0,0,tau)
Correct, except the paper uses tau_0 for the tau value.
>what is the k-coordinate of reflection?
>I get (xi,0,0,tau+xi/c)
Note that we do not know the value of xi, which we could call L',
and the paper uses tau_1 for the tau value.
>What is the k-coordinate of reception?
>I get (0,0,0,tau+xi/c+xi/c)
Similarly, the paper uses tau_2 for the tau value. However, your
version does make the reason that tau_1 = (tau_2 + tau_0)/2 is
correctly used by Einstein in his next step very explicit.
>How do we map the coordinates from K,(x,y,z,t) to k,(xi,eta,zeta,tau)?
We have yet to figure this out.
>I get
>(0,0,0,t) |=> (0,0,0,tau)
>(x,0,0,t+x/(c-v)) |=> (xi,0,0,tau+xi/c)
>(vt,0,0,t+x/(c-v) + x'/(c+v)) => (0,0,0,tau+xi/c+xi/c)
>Where am I wrong?
Apart from not copying correctly from up above and the error in
the value of x_2, all this describes is the mapping that the
final transform will give you. It is not relevant to anything
in the derivation at this point. It is used a bit later.
If I were to quess at what you might be thinking at this point,
I would add that you could be erroneously thinking that there
is a one-to-one correspondence between the items in those lists
that describe the K and k coordinates of a point. If you are
thinking this, it is a huge error and you need to go back an
look at the entire concept of transformations again.
So now you must again read the paper and see what is to be
done next with the information you know in K. What you see
is that you now write tau as a linear function of x' and t,
carry out those steps in terms of what we know up above, and
find the function. Or just read my detailed description that
was written for Glird....
Jim Carr
In article <0Ko7tCSX...@concentric.net>
Thnk...@concentric.net (Eleaticus) writes:
>
>some deleted garbage.
If you don't read it in its entirety before deleting it, you will
never learn my true views and thus understand that your statements
are not based on what I actually say.
Eleaticus
> >some deleted garbage.
>
> If you don't read it in its entirety before deleting it, you will
> never learn my true views and thus understand that your statements
> are not based on what I actually say.
If you'd ever decide to not mix distortions,
etc into almost every one of your posts, you
might actually make a contribution to the
net-world.
For example, if you had been interested in being
informative and helpful in the world, I might have
learned you weren't using 'constant' and 'scalar'
in the normal basic way, but were using it to
mean invariant under rotations and reflections,
in which case I could have pointed out what an
idiot you were soon enough to avoid two years of
you making an idiot-ass out of yourself.
Jim Carr
: For the event of reception, Froopy drew the diagram. Froopy's diagram
: shows the ray ends at vt in the K-frame. (vt, 0,0, x'/c+x/c) is the
: coordinate of reception in K, which is wrong, of course,
In article <9248...@sheol.org>
thr...@sheol.org (Wayne Throop) writes:
>
>Here we see AndyPandy's fundamental mistake, ...
Hmmmm. It seems to me that you did not point out very clearly
(in the text I snipped) that the really major error is his
reference to an x coordinate and his failure to see that x=vt
means x'=0 is what is used as the argument for the tau function.
Your comments point out another error, but less fundamental
and possibly (as I posted) just poor notation.
: Froopy just loves to make nasty remarks,
: because Froopy is just a nasty little man.
>Not at all. I am merely reacting with appropriate disgust
>to AndyPandy's pride in his own know-nothingism.
You don't have to. It might actually distract him from what you
are saying when you point out specific errors.
Jim Carr
in message news:7fqrk7$76b$1...@news.fsu.edu...
|
| Edward Schaefer <scha...@mitre.org> wrote in message
| news:371225CF...@mitre.org...
| }
| } Androcles wrote:
| } > I want to know where the 1/2 comes from, of course.
| }
| } Do you wish to accept the constancy of c in all inertial
| } frames-of-reference in this discussion? If so, I will tell you. If
| } not, then forget it.
|
| In article <%eyQ2.2366$Rn5....@news.rdc1.pa.home.com>
| "Androcles" <andr...@home.com> writes:
| >No. There is no postulate to that effect, as I have already shown. I want
| >to know where the half comes from, but I do not accept Schaefer's postulate.
|
| It comes from the relationship defined in the first equation presented
| in the paper you have been reading, an equation that appears on page
| 894 of Ann. d. Phys. 17, and on page 40 of the Dover translation/reprint.
| You are directed to this equation in section I.1 by a sentence in the
| section you are reading (I.3). Pretty clear to me.
|
| Granted you have to do some simple algebra to rearrange it ...
In article <NuqU2.840$A46...@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>The sentence reads
>"In agreement with experience -[whose experience?]- we further assume the
>quantity 2AB/(t'A-tA) =c.
Exactly, and simple algebra then leads to the result you asked about,
thus verifying my observation above.
>Missing from this sentence is the caveat "if and only if A and B are
>relatively at rest."
Oops, sorry, my oversight. I forgot to remind you that you also
have to read more than one sentence to get the ideas expressed
in a scientific paper. Read the sentences that precede that one.
Androcles
> Jim Carr <j...@ibms48.scri.fsu.edu> wrote
> Oops, sorry, my oversight. I forgot to remind you that you also
> have to read more than one sentence to get the ideas expressed
> in a scientific paper. Read the sentences that precede that one.
>
Yes, I have read then again.
Missing from those sentences is the caveat "if and only if A and B are
relatively at rest."
Androcles
Jim Carr
in message news:7gd60j$qdi$1...@news.fsu.edu...
|
| Jim Carr <j...@ibms48.scri.fsu.edu> wrote
In article <5IFW2.68$zB3...@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>Sufficient because t is an arbitrary offset from the origin in the time
>coordinate, the light gets to x in K and travels at c in K, making its time
>coordinate in K x/c.
Nonsequitur. Of course t is an arbitrary offset. The problem
with what you wrote, and write again, is that x is not measured.
That is what made your answer incomplete, as opposed to wrong.
What you know is L (the length Einstein called x'), not x.
| >What is the K-coordinate of reception?
| >I get (vt,0,0,t+x/c+x'/c)
|
| This is wrong. Assuming you are using x' = L+v*(x'/c) for the length
| of the return trip as measured in K, you need v*(x/c+x'/c) rather than
| v*t for the x coordinate. Again, it would be better to write this
| as v*(t1+t2) and t+t1+t2 -- where I am using t2 as (t_2 - t_1).
|
| This makes it clear that the value of x' at this time, call it x'_2,
| is zero, an important detail in the next part of the derivation.
>Not wrong, but insufficient.
Wrong. The quantity "t" is already defined above and appears with
that same definition in the expression for the 4th coordinate.
You need to do what I said, as you do here.
>The x-coordinate would be better expressed as v(x/c +x'/c), where I used
> t = (x/c+x'/c), which then happens to be the correct t used as the offset
>for a repeat performance of the experiment.
However, you need to define x' explicitly, since x' is also not
something measured in the original formulation. See above for that.
>What do you mean by "the value of x' at this time, call it x'_2, is zero" ?
I mean the value of the variable (not the length) called x', the
argument that is taken by the function tau. This is different
from the (undefined) x' you use in your formula and what I have
called L in my exposition above (also see below).
>x' is the distance the light will travel and is never zero.
The character string "x'" has several meanings in that section of
the paper, and that is only one of them. As detailed in the long
exposition I wrote for Glird and suggested you read, I eventually
called that distance Einstein calls x' by the label L so there
would be no confusion with the coordinate x' = x - vt. [That is
actually x - v(t-t_0) at this point.] Neither of these is the
distance you call x'.
Using x' to mean a third distance, one that you do not define in
terms of L, only worsens the confusion Einstein created for the
naive reader when he used the same name for different things.
By x'_2 I mean the value x' has at time t_2, when x has the value x_2.
That value is clearly zero. I use L for the distance.
| >How do we map the coordinates from K,(x,y,z,t) to k,(xi,eta,zeta,tau)?
|
| We have yet to figure this out.
>Sure, but first we have to understand what we are discussing.
I think we are discussing not putting the cart before the horse. ;-)
Part of the mapping has been defined (that is what the tau function,
the function of x',y,z,t is doing there). Although not yet defined,
you could also define a function xi of x',y,z,t that does the rest
of this -- and that is what is worked with in the next phase of the
derivation after the tau function has been determined.
>I get
>(0,0,0,t) |=> (0,0,0,tau)
>(x,0,0,t+x/(c-v)) |=> (xi,0,0,tau+xi/c)
>(v.(x/(c-v) +x'/(c+v)),0,0,t+x/(c-v) + x'/(c+v)) => (0,0,0,tau+xi/c+xi/c)
>Where am I wrong?
Several places.
First, your second and third equations are wrong. They are not
what was obtained up above, mostly due to mixing notation even
worse than Einstein did.
Second, your " |=> " arrow hides the nature of the mapping, which
was made explicit when the functional dependence was defined in
Einstein's derivation. This is particularly dangerous.
Third, the mapping in Einstein's derivation is a function of
the coordinate x' rather than the 1st coordinate you use.
Wayne Throop
: j...@ibms48.scri.fsu.edu (Jim Carr)
: The character string "x'" has several meanings in that section of the
: paper, and that is only one of them.
Hrm? No, it's not even one of them. As near as I can tell, it's
not the distance light travels on any leg, in either coordinate system,
in any of the uses in the paper. Do you have a specific reference?
That's why Androcles' insistance that x/c+x'/c is the total time in K
is so fundamentally wrongheaded. That and the abuse of the term "x",
as Jim pointed out.
But mostly, x' is K-measure which denotes a point in k.
That's what it means as an argument to tau, and that's what
it's definition indicates: x'=x-vt. Obviously, it can be zero.
The value of the x' argument to tau which denotes the k-origin is zero.
The total time is L/(c-v)+L/(c+v), and that's exactly and precisely
because the speed of light is c, and the distance traveled by that
light is not L, but L+vt and L-vt.
Gerald L. O'Barr
In <9266...@sheol.org>
thr...@sheol.org (Wayne Throop) wrote:
. . . .
The total time is L/(c-v) + L/(c+v), and that's exactly and
precisely because the speed of light is c, and the distance
traveled by that light is not L, but L+vt and L-vt.
O'Barr comments:
The sun is shinning! What a beautiful day!!!
Wayne, you are the smartest man on this net. It is a
pleasure to respond to you. You are so perfectly correct,
the total time is exactly L/(c-v) + L/(c+v). And it is
perfect!!!! It is Newtonian perfection!!!!!! And it is
SR!!!!!!!!! This is SR in the ether.
Let there be an inertial frame A, with a train moving with
velocity v. Let the length of this moving train be L, as
measured in frame A, as the train is moving!!!!!! The time
for light, as measured in frame A, to go from the rear to
the front of this moving train, is L/(c-v). The time for
light, as measured in frame A, to go from the front to the
rear of the moving train, is L/(c+v). All of this is pure
Newtonian physics. In this Newtonian physics, the relative
velocity of the light past the train is exactly c +/- v.
The length of the moving train is exactly L. It is not any
perverted length. It is not just a perspective length! It
is a pure and simple L. Everything is pure and simple and
exactly what it is said to be. There is no SR physics what-
so ever on this level!!!!!!!!!!!!!!!!
And in the ether, this is, physically, all that is ever
there on the level of the physics. It is far simpler than
anything in SR. It is the simplest physics that has ever
been known: Simple Newtonian physics in Euclidian space!
And it is our reality. SR appears only as you want to enter
into the mathematical world of simplicity. But if you want
your simplicity in the physics, then one has to
scientifically chose the ether!!!
Sorry about all that, but that is the way it 'really' is!!!!!
--
Gerald L. O'Barr fl...@access1.net
Read Pete Brown's Aether FAQ at:
http://magna.com.au/~prfbrown/aeth_faq.htm
Read Jan 99 issue of Physics Today about the ether!
Wayne Throop
: respond to you. You are so perfectly correct, the total time is
: exactly L/(c-v) + L/(c+v). And it is perfect!!!! It is Newtonian
: perfection!!!!!! And it is SR!!!!!!!!! This is SR in the ether.
No, this is an arbitrary frame, not the ether.
Rent a clue sometime.
Gerald L. O'Barr
In <9267...@sheol.org>
thr...@sheol.org (Wayne Throop) wrote:
Ref: <EiVS2.1116$xc5...@news.rdc1.pa.home.com>
<7gd60j$qdi$1...@news.fsu.edu>
<5IFW2.68$zB3...@news.rdc1.pa.home.com>
<7hfjpd$19o$1...@news.fsu.edu> <9266...@sheol.org>
<373C74E3...@access1.net>
Gerald L. O'Barr <fl...@access1.net> (globarr) wrote:
: Wayne, you are the smartest man on this net. It is a
: pleasure to respond to you. You are so perfectly correct,
: the total time is exactly L/(c-v) + L/(c+v). And it is
: perfect!!!! It is Newtonian perfection!!!!!! And it is
: SR!!!!!!!!! This is SR in the ether.
Wasyne wrote:
No, this is an arbitrary frame, not the ether.
Rent a clue sometime.
O'Barr comments:
You are absolutely correct, that in SR, you were using an
arbitrary frame. But no matter how arbitrary you want to
make it, it is a correct arbitrary frame. It is correct in
its SR math, in its SR calculations, and in its SR results!
And this of course was the whole intent of the post, to show
that in this arbitrary but fully SR frame, the math was
Newtonian in every way!!!! Thus, SR math, even in its very
development, was and is Newtonian! As it still is today.
All you SR experts have to do is just re-examine what you are
doing!!!!! And this base math is simpler than what you end
up using. You could at any time go back to the original math
and still be exactly just as correct!!!! And this is the
ether approach!!!!!! The ether is based upon simple
Newtonian math, and Newtonian physics, and since all of
reality is contained in the one single ether frame, then the
ether makes all of reality this simple!
Thank you, Wayne, for being so smart!!!!!!
--
Gerald L. O'Barr fl...@access1.net
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
|
... Eleaticus again removes evidence of his authorship ...
| >some deleted garbage.
|
| If you don't read it in its entirety before deleting it, you will
| never learn my true views and thus understand that your statements
| are not based on what I actually say.
In article <LzYBtCSX...@concentric.net>
Thnk...@concentric.net (Eleaticus) writes:
>
>If you'd ever decide to not mix distortions,
>etc into almost every one of your posts, you
>might actually make a contribution to the
>net-world.
Nonsequitur. You will never understand what I wrote if you
filter it by deleting sections of it and the context that it
appeared in just so you can perpetuate your troll.
All of the things you complain were never explained were
explained from the very beginning.
Eleaticus
> | >some deleted garbage.
> |
> | If you don't read it in its entirety before deleting it, you will
> | never learn my true views and thus understand that your statements
> | are not based on what I actually say.
>
> In article <LzYBtCSX...@concentric.net>
> Thnk...@concentric.net (Eleaticus) writes:
> >
> >If you'd ever decide to not mix distortions,
> >etc into almost every one of your posts, you
> >might actually make a contribution to the
> >net-world.
>
> Nonsequitur. You will never understand what I wrote if you
> filter it by deleting sections of it and the context that it
> appeared in just so you can perpetuate your troll.
>
> All of the things you complain were never explained were
> explained from the very beginning.
You wouldn't want to directly quote me complaining
about things never explained to me, would you?
Didn't think so.
And you wrote that nonsense in response to me saying you
perhaps ought not mix distortions/etc into your
responses!
> James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
> http://www.scri.fsu.edu/~jac/ | desired to this or any address
> Supercomputer Computations Res. Inst. | that resolves to my account
> Florida State, Tallahassee FL 32306 | for any reason at any time.
Eleaticus
Wayne Throop
: And this of course was the whole intent of the post, to show that in
: this arbitrary but fully SR frame, the math was Newtonian in every
: way!!!!
Not in "every way"; the dynamics is not newtonian.
Kinematics using velocity differences and so on is the same as
in newton, of course. Who didn't know that? Rent a clue: it's old news.
: Thus, SR math, even in its very development, was and is Newtonian!
Nope. It's derived from a different symmetry altogether.
But even if it were, so what?
: All you SR experts have to do is just re-examine what you are
: doing!!!!! And this base math is simpler than what you end up using.
SR is simpler than LET, in all the examples MLuttgens has been working.
: You could at any time go back to the original math and still be
: exactly just as correct!!!!
Sure. As long as you are willing to do things the hard way,
by working th em in inconvenient frames, or using only a single frame,
and if you ignore dynamics, then you can "use newtonian math".
But really, it's not sensible to pretend that, just because you can
juggle with one hand tied behind your back, that that is the best
way to go about keeping balls up in the air.
: The ether is based upon simple Newtonian math, and Newtonian physics,
SR is based on simple Lorentzian math, and Lorentz-invariant physics.
It's much easier to use than "newton+gamma" theory.
: Thank you, Wayne, for being so smart!!!!!!
Then why do you always disagree with my actual positions?
Could it be you are simply being either hyporcitical or
gratuitously condescending? Hmmm?
Wayne Throop
"The ether is the Theory of The Future!"
Yep. And it always will be.
Jim Carr
|
| Oops, sorry, my oversight. I forgot to remind you that you also
| have to read more than one sentence to get the ideas expressed
| in a scientific paper. Read the sentences that precede that one.
In article <4v0Z2.1253$zB3....@news.rdc1.pa.home.com>
"Androcles" <andr...@home.com> writes:
>
>Yes, I have read then again.
>Missing from those sentences is the caveat "if and only if A and B are
>relatively at rest."
Then I guess you have poor judgement as to which sentences to read.
You need to start with the first sentence in section I, #1, and note
that because there is only one system of coordinates defined, the
statements "at point A" and "at point B" unambiguously refer to points
that are at rest with respect to one another in that coordinate system.
--
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
|
| :: x' is the distance the light will travel and is never zero.
| The character string "x'" has several meanings in that section of the
| paper, and that is only one of them.
In article <9266...@sheol.org>
thr...@sheol.org (Wayne Throop) writes:
>
>Hrm? No, it's not even one of them.
Without the context, I can't tell which x' is meant above. However,
you are quite right that I was unclear that I was talking about the
third meaning introduced by Androcles, one that appears in this
discussion but not in Einstein's paper. I probably meant to write
"not" and wrote "only" instead.
>That's why Androcles' insistance that x/c+x'/c is the total time in K
>is so fundamentally wrongheaded. That and the abuse of the term "x",
>as Jim pointed out.
It is not so much wrongheaded as adding to the confusion. There is
a distance that the light is measured to travel on each of those
two paths, and those distances are unequal and thus deserving of two
distinct names. However, using names already in use (twice) is a
very bad idea. Call them M and N or something. Anything but x and
x' -- and IMO it is better to use L for the distance called x' in
the original paper and reserve x' for x-vt.
>But mostly, x' is K-measure which denotes a point in k.
I would say it denotes a point. That point is "in" both K and k.
It is dangerous thinking to associate a point with only one of
the coordinate systems. In fact, this notation is associated with
an intermediate (artificial) coordinate system constructed by K,
so I would not associate it with the other coordinate system whose
measured values are denoted with greek letters.
>That's what it means as an argument to tau, and that's what
>it's definition indicates: x'=x-vt. Obviously, it can be zero.
Yes, but the other x' (L) is not zero, and the third x' (the
return path) also cannot be zero. It was the latter I was
talking about, but not clearly enough.
>The total time is L/(c-v)+L/(c+v), and that's exactly and precisely
>because the speed of light is c, and the distance traveled by that
>light is not L, but L+vt and L-vt.
Yes. The latter would be what is called "x'" in the text quoted above.
I thought I made that clear in that article, or the next one.
Wayne Throop
:: But mostly, x' is K-measure which denotes a point in k.
: j...@ibms48.scri.fsu.edu (Jim Carr)
: I would say it denotes a point. That point is "in" both K and k. It
: is dangerous thinking to associate a point with only one of the
: coordinate systems.
Let me explain why I disagree.
What x' denotes is actually a locus of events. Those events are
the same events in both coordinate systems. But the locus of events
denoted by constant spatial coordinates (ie, a point "in" a coordinate system)
is NOT the same between coordinate systems.
In my opinion, the vocabulary used needs to associate "a point"
(that is, a locus of events) with its rest coordinates. Otherwise,
you get ambiguity in saying that "x' denotes a point". Does that
mean it denotes the locus of events where x is x'? Well, you can
figure out in what Einstein said that it denotes a point where
x-vt is x', and hence, "a point at rest in k" (even though we don't
yet know the point's k-coordinates, we know they are constant).
Can you suggest an alternative phrasing to "a point in k"?
It seems a reasonable fit with the frequent usage of "an object in k",
so you'd have to come up with an alternative phrasing of that one, too.
One possibility is "a point-or-object *at* *rest* in k". But it
seems to me that what I said, in context, isn't enough different
from this to bother over. So; what alternatives are there?
: However, using names already in use (twice) is a very bad idea. Call
: them M and N or something. Anything but x and x' -- and IMO it is
: better to use L for the distance called x' in the original paper and
: reserve x' for x-vt.
Agreed one hundred percent. The issue about dummy names for the
parameters to tau is less clear, but using x' for all three things
(x-vt, first dummy parameter to tau, distance to the mirror at time zero)
is definitely prone to misinterpretation on a quick scan through,
and it can be hard to figure out what went wrong.
Jim Carr
In article <XCXFtCSX...@concentric.net>
>
>In article <7hktj0$njo$1...@news.fsu.edu>, j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
>
Eleaticus again feels it necessary to alter the contents of my
article, removing a statement pointing out past alterations of
its content.
| You will never understand what I wrote if you
| filter it by deleting sections of it and the context that it
| appeared in just so you can perpetuate your troll.
>You wouldn't want to directly quote me complaining
>about things never explained to me, would you?
Don't have to. You do an admirable job of it yourself.
There are only two ways to explain your repeated complaints that
certain statements are lies when in fact they are clearly correct
in their original context -- as has been shown in detail in the
past. Continuing complaints can only be evidence that you think
those explanations were inadequate or that you are just trolling.
You rhetorical question suggests it is the latter.
Thank you for clearing this up.
Eleaticus
>
> In article <XCXFtCSX...@concentric.net>
> Thnk...@concentric.net (Eleaticus) writes:
> >In article <7hktj0$njo$1...@news.fsu.edu>, j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
> Eleaticus again feels it necessary to alter the contents of my
> article, removing a statement pointing out past alterations of
> its content.
Jim may be taking the ultimate in designer drugs.
One that causes hallucinations/delusions of a
very specific kind.
> >You wouldn't want to directly quote me complaining
> >about things never explained to me, would you?
> Don't have to. You do an admirable job of it yourself.
How did I manage to foresee no quotes a-coming?
Must be psychic.
Jim Carr
:: But mostly, x' is K-measure which denotes a point in k.
j...@ibms48.scri.fsu.edu (Jim Carr)
| I would say it denotes a point. That point is "in" both K and k. It
| is dangerous thinking to associate a point with only one of the
| coordinate systems.
In article <9276...@sheol.org>
thr...@sheol.org (Wayne Throop) writes:
>
>Let me explain why I disagree.
>
>What x' denotes is actually a locus of events. Those events are
>the same events in both coordinate systems.
So you do not disagree on what I think is the important point.
>But the locus of events
>denoted by constant spatial coordinates (ie, a point "in" a coordinate system)
>is NOT the same between coordinate systems.
I would say that x' alone does not describe an event in the
coordinate system, since that also requires a time. In fact,
that might be the precise place where trouble arises: paying
attention to the time part of the spacetime coordinates.
>In my opinion, the vocabulary used needs to associate "a point"
>(that is, a locus of events) with its rest coordinates. Otherwise,
>you get ambiguity in saying that "x' denotes a point". Does that
>mean it denotes the locus of events where x is x'? Well, you can
>figure out in what Einstein said that it denotes a point where
>x-vt is x', and hence, "a point at rest in k" (even though we don't
>yet know the point's k-coordinates, we know they are constant).
Quite so. Sort of. I am convinced that a major part of Glird's
confusion is that even those words are not sharp enough. We have
a point at rest in k and we measure the coordinates of an event
(where that point is at some instant) as x and t in K; from those
we calculate x' (still in K) and note that x' is constant. For
*one* of those points in k, the constant value of x' is zero. For
another, it is something I took to calling L -- but that Einstein
called x' in his paper (to Glird's ongoing confusion).
However, I don't think you _need_ to associate the point with its
rest coordinates. I think there is a tendency to do so without
saying explicitly that this is what you are doing.
>Can you suggest an alternative phrasing to "a point in k"?
Yours. A point which is at rest in k.
However, once you start thinking of k and K as _spacetime_
coordinate systems, there is only "a point".
>It seems a reasonable fit with the frequent usage of "an object in k",
>so you'd have to come up with an alternative phrasing of that one, too.
I was pointing out that the object is in both k and K and that
the common view that the object is only "in" its rest frame can
lead to some confusion.
>One possibility is "a point-or-object *at* *rest* in k".
Or "object, which is at rest in k".
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
|
| In article <XCXFtCSX...@concentric.net>
| Thnk...@concentric.net (Eleaticus) writes:
| >
| >In article <7hktj0$njo$1...@news.fsu.edu>, j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
|
| Eleaticus again feels it necessary to alter the contents of my
| article, removing a statement pointing out past alterations of
| its content.
In article <dLQJtCSX...@concentric.net>
Thnk...@concentric.net (Eleaticus) writes:
>
>Jim may be taking the ultimate in designer drugs.
But Jim is not.
Note well that Eleaticus does not deny that he altered the content
of my article, he only pretends to by making a statement that has
no content whatsoever because of the use of "may" to avoid libel.