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How many inconsistencies are sufficient?

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Richard

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Sep 15, 2003, 8:47:15 AM9/15/03
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Here's another one for the SRist crackpots to crack:

Two events occur along the x axis, they are simultaneous in frame K, and
separated by the non-zero space interval x. According to the crackpots
these events will not be simultaneous wrt any frame K' that is in motion
wrt K along the x axis. Now since it is maintained that the order of
events doesn't change between frames, then it follows that any event
that occurs other than the given two events, must, according to K, occur
either before or after those events. Thus according to the math, any
other event, besides these two, must also occur either before or after
these two events wrt K'. Thus there are no events in the universe
occurring during the interval dt' wrt K'.

K' must then conclude that time ceases to flow in the universe over that
interval, i.e. that the two events were in fact simultaneous in his
frame as well.

SR crackpots, work on that one will you;)

Richard Perry

http://www.cswnet.com/~rper
Electromagnetism: First Principles

Dirk Van de moortel

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Sep 15, 2003, 8:54:22 AM9/15/03
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"Richard" <no_mail...@yahoo.com> wrote in message news:3F65B4D...@yahoo.com...

>
> Here's another one for the SRist crackpots to crack:
>
> Two events occur along the x axis, they are simultaneous in frame K, and
> separated by the non-zero space interval x. According to the crackpots
> these events will not be simultaneous wrt any frame K' that is in motion
> wrt K along the x axis. Now since it is maintained that the order of
> events doesn't change between frames,

This is "maintained" only for timelike separated events
where the squared interval
(c*Dt)^2 - Dx^2 > 0
The pair you have chosen is (0,0) and (0,x),
(c*Dt)^2 - Dx^2 = -x^2 < 0
They are spacelike separated.

Try to have a serious go at
http://www.physicsguy.com/ftl/html/FTL_part1.html
http://www.physicsguy.com/ftl/html/FTL_part2.html
and let's get beyond section 1.1 of part 1 this time.
If you have a problem, just ask a question.

[snipped on first error]

Dirk Vdm


Richard

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Sep 15, 2003, 1:24:41 PM9/15/03
to

Let's put it to practice Dirk.

In the train experiment lightning flashes simultaneously at the ends of
the railway carriage wrt the embankment observer. According to the
observer on the train, who is positioned at its center, the light from
the leading end of the carriage traverses a smaller space interval than
the light from the trailing end. Thus, believing that light also
propagates wrt him at c, he calculates that the leading flash occurred
before the trailing flash. So far this is straight from the book.

Now if the interval that the carriage observer measures is dt', then we
can, theoretically allow a second leading flash to occur at some time
less than dt' after the first, and the carriage observer will will
intercept that light before he intercepts the trailing flash, thus he
has an event occurring in the midst of the initial interval. The
sequence, according to K' is

1-Forward flash
2-Second forward flash
3-Rearward flash

Keep in mind that the first and third of these is simultaneous wrt the
track observer.

Wrt K the order of 2 and 3 is thus reversed wrt the same events as
perceived from frame K'. Now lets suppose that the second forward flash
consists of green light, while the rearward flash consists of red light.
On board the train, in the train observers hand and also at rest wrt the
carriage, is a device that trips one of two relays depending upon the
frequency of detected light. Upon first detection the corresponding
relay trips and locks out the other.

These observers will not agree on which relay trips, simply because they
don't agree on which flash arrived at the detector first. Moreover the
creed that according to SR, "events do not reverse when transforming
between frames" has also been shown to be false, as Dirk also correctly
noted. Now it probably didn't occur to him however, that events must in
fact maintain their order through transformations, as the experiment
proves beyond doubt. Thus SR has once again been shown to be
inconsistent.

Richard Perry

http://www.cswnet.com/~rper

Richard

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Sep 15, 2003, 1:37:44 PM9/15/03
to

*Argument corrected below []

[track observer, the light from


the leading end of the carriage traverses a smaller space interval than

the light from the trailing end before being intercepted by the carriage
observer, thus the leading flash intercepts the carriage observer first.

The carriage observer OTOH, believing that light also [end correction]]

> propagates wrt him at c, calculates that the leading flash occurred


> before the trailing flash. So far this is straight from the book.
>

> Now if the interval that the carriage observer measures is dt' between these flashes, then we


> can, theoretically allow a second leading flash to occur at some time
> less than dt' after the first, and the carriage observer will

> intercept that light before he intercepts the trailing flash, thus he
> has an event occurring in the midst of the initial interval. The
> sequence, according to K' is
>
> 1-Forward flash
> 2-Second forward flash
> 3-Rearward flash
>
> Keep in mind that the first and third of these is simultaneous wrt the
> track observer.
>
> Wrt K the order of 2 and 3 is

Rearward flash/Second forward flash, and thus events 2 and 3
are reversed wrt the two frames. Now lets suppose that the second

Dirk Van de moortel

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Sep 15, 2003, 1:47:50 PM9/15/03
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"Richard" <no_mail...@yahoo.com> wrote in message news:3F65F5D9...@yahoo.com...

Of course they agree.
They do not agree on the order of *emitting* the green and red
since these events are - remember? - spacelike related:
(c*Dt)^2 - Dx^2 < 0
and of course
(c*Dt')^2 - Dx'^2 < 0
O.t.o.h. the events of *arrival* of the green and red take place
on the worldline of the observer on the train where the
coordinate-differences of these events satisfy
(c*Dt)^2 - Dx^2 > 0
and of course
(c*Dt')^2 - Dx'^2 > 0
These arrival events are timelike related, remember?
So for these events all observers agree on the order.
Draw a little diagram and you will immediately see.

> Moreover the
> creed that according to SR, "events do not reverse when transforming
> between frames" has also been shown to be false, as Dirk also correctly
> noted. Now it probably didn't occur to him however, that events must in
> fact maintain their order through transformations, as the experiment
> proves beyond doubt. Thus SR has once again been shown to be
> inconsistent.

Once again you have shown:
- that you haven't looked at my answer
- that you do not to know what you are talking about,
So in order for you not to make a fool of yourself, I will
repeat my kind offer:
try to have a serious go at


http://www.physicsguy.com/ftl/html/FTL_part1.html
http://www.physicsguy.com/ftl/html/FTL_part2.html
and let's get beyond section 1.1 of part 1 this time.
If you have a problem, just ask a question.

Dirk Vdm


Dirk Van de moortel

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Sep 15, 2003, 1:48:56 PM9/15/03
to

"Richard" <no_mail...@yahoo.com> wrote in message news:3F65F8E8...@yahoo.com...
>
>
> *Argument corrected below []

Correction was irrelevant.

Dirk Vdm


AndroclesInEngland

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Sep 15, 2003, 1:54:30 PM9/15/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3f65b6f7$1...@usenet01.boi.hp.com...
I took a quick glance at it, there was a picture of a light beam purporting
to be in a train. So I snipped on first error, see
http://www.androc1es.pwp.blueyonder.co.uk/Gardner.htm
for why it is nonsense.

Dirk Van de moortel

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Sep 15, 2003, 1:59:40 PM9/15/03
to

Paul B. Andersen

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Sep 15, 2003, 5:37:20 PM9/15/03
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"Richard" <no_mail...@yahoo.com> skrev i melding news:3F65F8E8...@yahoo.com...

From whence have you got this silly idea?
It is of course utterly irrelevant what different observers
have to say about the sequence of _emission_ of the flashes.
The question is - what are the sequence of the three events:
light from flash 1, 2 and 3 hits the detector.
Since the detector is present at all these events,
they are sparated by time like intervals.
ALL observers will agree on their sequence.

> > Moreover the
> > creed that according to SR, "events do not reverse when transforming
> > between frames" has also been shown to be false, as Dirk also correctly
> > noted.

You are babbling.
Different observers may dissagree on the sequence of events
separated by space like intervals.
Different observers will always agree on the sequence of events
separated by time like intervals.

This is the "creed according to SR". It is not falsified.
But YOU obviously didn't know the difference.

In your scenario, the events "emission of the flashes" are of the former class,
while the events "flashes hit detector" are of the latter class.

> > Now it probably didn't occur to him however, that events must in
> > fact maintain their order through transformations, as the experiment
> > proves beyond doubt. Thus SR has once again been shown to be
> > inconsistent.

You ARE confused, are you not? :-)

I bet you even didn't realize that "light hits detector in the middle
of the train" is a different event from "light is emitted from end of train".

Paul


Old Man

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Sep 15, 2003, 5:51:04 PM9/15/03
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Richard <no_mail...@yahoo.com> wrote in message
news:3F65B4D...@yahoo.com...
>
> Here's another one for the SRist crackpots to crack:
>
> Two events occur along the x axis, they are simultaneous in frame K, and
> separated by the non-zero space interval x. According to the crackpots
> these events will not be simultaneous wrt any frame K' that is in motion
> wrt K along the x axis. Now since it is maintained that the order of
> events doesn't change between frames, then ....

The time order is maintained only for causally related events. Two
simultaneous events that are separated by a finite distance are not
causally related. Causally related events have delta_x < c*delta_t
[Old Man]

> Richard Perry

Eric Prebys

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Sep 15, 2003, 5:53:02 PM9/15/03
to
Richard wrote:
> Here's another one for the SRist crackpots to crack:
>
> Two events occur along the x axis, they are simultaneous in frame K, and
> separated by the non-zero space interval x. According to the crackpots
> these events will not be simultaneous wrt any frame K' that is in motion
> wrt K along the x axis. Now since it is maintained that the order of
> events doesn't change between frames,

Order is only maintained for events with *timelike* separation. Order
of events with spacelike separation can change, which is why
under the current theory they can have no causal connection.

This topic is covered pretty early in any relativity text I've
ever read. Perhaps you should consider actually studying something
before you comment on it.

-Eric

Mark Palenik

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Sep 15, 2003, 7:30:37 PM9/15/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message
news:r1n9b.64$H31...@news-binary.blueyonder.co.uk...
>

This page says

"In addition to apparent changes in length, there are also apparent changes
in time. Astronauts on each ship will find that clocks on the other ship are
running faster. A simple thought experiment shows this must indeed be the
case. Imagine you are looking out the porthole of one spaceship into the
porthole of another ship. The two ships are passing each other with uniform
speed close to that of light. As they pass, a beam of light on the other
ship is sent from its ceiling at the bow to its floor at midships. That is
as the nose of the ship is passing your porthole, the light begins at the
ceiling, and arrives at the floor as the middle of the ship, which has now
moved directly opposite your porthole. There it strikes a mirror and is
reflected back to the ceiling again, now at the tail. You will see this as a
vertical line. If you had sufficiently accurate instruments (of course no
such instruments exist), you could clock the time it takes this light beam
to traverse the vertical path. By dividing the length of the path by the
time, you obtain the speed of light."

This argument is unbelievably erronious and shows a strong lack of
visualization capabilities on the part of the writer.

Picture this - two ships passing eachother. One shines a beam of light from
the cealing to the floor. The astronaut in the other ship observes it.

He sees the light beam start at the cealing at point a along the x axis.
The point on the floor which it is about to hit is also at point a. But,
since the ship is moving, before the light beam hits the floor, everything
has moved along the x axis to point b. So, not only has the light traveled
along the y axis (from the cealing to the floor), but also along the x axis
from point a to point b. Therefore, the astraunaut in the other ship sees
its total distance as sqr(y^2 + (b-a)^2). Of course, b-a = vt, and we know
that the total distance has to be ct, where t is the amount of time it takes
for the light to travel from the cealing to the floor and c is the speed of
light. So, solving for y, we get :

y^2 + (vt)^2 = (ct)^2
y^2 = (ct)^2 - (vt)^2
and
y = sqr((ct)^2 - (vt)^2)

we can factor out a t, to get

y = t*sqr(c^2 - v^2)

The astronaut in the ship that has sent the beam, however, likewise, sees
the beam as traveling from the cealing to the floor, but not being aware of
any change in his x position, sees the total distance as y. Since light is
traveling at c, ct' must then equal y, where t' is the amount of time he
thinks it takes the light to get from the cealing to the floor.

so, we have one astronaut who says y = t*sqr(c^2-v^2) and another who says y
= ct'. So therefore:

t*sqr(c^2 - v^2) = ct'

dividing both sides by c, we get:

t*sqr(c^2 - v^2)/c = t'

Factoring the c into the square root, we get:

t*sqr(1-v^2/c^2) = t'

Therefore, the astronaut who is moving should experience a time slow down
proportional to sqr(1-v^2/c^2), and that is what our "stationary" astronaut
will report. He will say that time has slowed down for the moving astronaut
in the t' frame. Of course, for the astronaut in the "moving" frame, things
will be reversed, and he will say the same thing about our astronaut.

This may seem somewhat strange, but it makes perfect sense in the context of
minkowski spacetime.


Dirk Van de moortel

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Sep 16, 2003, 3:07:04 AM9/16/03
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"Mark Palenik" <markp...@wideopenwest.com> wrote in message news:u26dnX10Gpg...@wideopenwest.com...

That's what we all have been trying to tell "Androcles" so
many times. It did not help.
You'll find out that whatever you say, it will not help either:
he has made up his mind and there is nothing anyone can do
about it. Just check the google searches I provided.
Have a look at this one:
http://groups.google.com/groups?&threadm=rfM17.2362$z7.5...@afrodite.telenet-ops.be
and understand why everything will always bounce.
Otherwise, enjoy ;-)

Dirk Vdm


AndroclesInEngland

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Sep 16, 2003, 6:56:57 PM9/16/03
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"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bk5bf6$phi$1...@dolly.uninett.no...
I bet you didn't even realize that a V-shaped light path in one frame of reference can be seen as a vertical light path in the other, AND they can be reversed, did you? Well, they can. You ARE confused, are you not? But you shouldn't be, should you? After all, you successfully proved to me that
t = b(t+vx/c2),  and not t = b(t-vx/c2), didn't you? Or was that just confusion, Paul? How simple can it be, Paul? Richard is right, SR is inconsistent.

AndroclesInEngland

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Sep 16, 2003, 7:22:08 PM9/16/03
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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:g6n9b.21896$Ur2.1...@phobos.telenet-ops.be...
Thanks for the publicity, Drik. I should hire you as my agent, you are
obviously infatuated with me :)


Mark Palenik

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Sep 16, 2003, 7:33:39 PM9/16/03
to
Since I'm operating this computer from somewhere other than the place with a
wideopenwest connection, I am unable to respond to e-mails sent to
MarkP...@WideOpenWest.com

I got an e-mail from Richard, or Androcles, or whoever, and will respond to
it here, since I can't from my mail box.

----- Original Message -----
From: "Androcles" <jp006...@blueyonder.co.uk>
To: "Mark Palenik" <markp...@wideopenwest.com>
Sent: Tuesday, September 16, 2003 6:05 PM
Subject: Re: How many inconsistencies are sufficient?


>
> You haven't read all of it, have you? Try again. Just because you know how
> to handle simple algebra doesn't mean you are capable of simple logic. It
> doesn't take much to realize that a V-shaped path in one frame of
reference
> can be seen as a vertical line in the other, AND vice -versa, and when we
> look at the vice-versa, we conclude that the clock on the other ship runs
> FASTER. Perfect sense, right?
> http://www.androc1es.pwp.blueyonder.co.uk/
>

If we create a vertical beam of light in the "moving" reference frame, it
will appear as a sloped line in the "stationary" one. If we create a sloped
beam of light in then the moving frame, it will appear as a sloped beam of
light with a smaller slope (closer to horizontal) in the stationary frame.
To create something that appears to be a horizontal line we would have to
send a beam of light out backwards with respect to the moving observer (or
rather, at an angle toward the back and floor of the ship). In this case,
the horizontal velocity frame of the beam of light is no longer that of the
moving frame, instead it is in the frame of the stationary observer. Doing
the same calculations again, we can see that this means that the "moving"
observer sees the stationary observer's time as running slower as well,
which is exactly what SR states - that each observer sees the other's clock
as running slower that his.

AndroclesInEngland

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Sep 16, 2003, 7:45:34 PM9/16/03
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"Mark Palenik" <markp...@wideopenwest.com> wrote in message
news:u26dnX10Gpg...@wideopenwest.com...
[snip]

> Picture this - two ships passing eachother. One shines a beam of light
from
> the cealing to the floor. The astronaut in the other ship observes it.

Picture this - oh, I did -see the GIF at
http://www.androc1es.pwp.blueyonder.co.uk/gardner.htm

He sees the light beam start at the cealing (ceiling) at point 'a' along
the x axis.
The point on the floor which it is about to hit is not also at point b, but
at midships, because I tilted the flashlight toward the back of the ship.


But, since the ship is moving, before the light beam hits the floor,

everything has moved along the x axis to point b (on the 'moving' ship), but
is directly opposite the porthole on the 'stationary ship'. So, not only
has the light traveled along the y axis (from the cealing (ceiling) to the
floor), but also along the x axis from point a to point b, but rmains at
point 'a' for the other astronaut. Therefore, the astraunaut (astronaut) in
the other ship sees its total distance as y, whereas it is really the
hypotenuse in the moving ship. Of course, b-a = vt, and we know that the


total distance has to be ct, where t is the amount of time it takes

for the light to travel from the cealing (ceiling) to the floor and c is the


speed of light. So, solving for y, we get :

y^2 = (ct)^2
and
y = y
but the real light path on the 'moving' ship is sqrt(y^2 + c^2t^2)


The astronaut in the ship that has sent the beam, however, likewise, sees

the beam as traveling from the cealing (ceiling) to the floor, but being
fully aware of a change in its x position, sees the total distance as
sqrt(y^2 + c^2t^2). Since light is traveling at c, ct' must then not be
equal to y, where t' is the amount of time he thinks it takes the light to
get from the cealing (ceiling) to the floor.
so, we have one astronaut who says y = t*sqr(c^2 (PLUS) v^2) and another who


says y = ct'. So therefore:

t*sqr(c^2 + v^2) = ct'


dividing both sides by c, we get:

t*sqr(c^2 + v^2)/c = t'

Factoring the c into the square root, we get:

t*sqr(1+v^2/c^2) = t'

Therefore, the astronaut (spelling correct this time) who is moving should
experience a time speed up
proportional to sqr(1+v^2/c^2), and that is what our "stationary" astronaut
will report. He will say that time has sped up for the moving astronaut in


the t' frame. Of course, for the astronaut in the "moving" frame, things
will be reversed, and he will say the same thing about our astronaut.
This may seem somewhat strange, but it makes perfect sense in the context of

minkowski spacetime, and that is the 'perfect' sense of a relativist, or
nonsense.

Mark Palenik

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Sep 16, 2003, 9:30:14 PM9/16/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message
news:ygN9b.801$z11...@news-binary.blueyonder.co.uk...


I made a response to that in this thread. See my other message. Basically,
the time dilation that we are measuring is that of the frame of the
horizontal component of the light's velocity. So, shining the light
backwards will cause the "moving" astronaut to see a horizontal component of
the light's velocity equal to that of the stationary astronaut's horizontal
velocity. The stationary astronaut will see it as zero, or in other words
as being his own horizontal velocity. So, from the frame where the
horizontal component is non-zero, the frame of the moving astronaut, time
appears to be running slower for the stationary astronaut, which is exactly
what relativity predicts. Each astronaut should see the other's time as
running slower, or to put it another way, each astronaut sees his own time
as running faster than the other's.

Or, as I put it in my other response:

tadchem

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Sep 16, 2003, 10:41:29 PM9/16/03
to

"Mark Palenik" <markp...@wideopenwest.com> wrote in message
news:AIOcnSsKv7p...@wideopenwest.com...

> Since I'm operating this computer from somewhere other than the place with
a
> wideopenwest connection, I am unable to respond to e-mails sent to
> MarkP...@WideOpenWest.com
>
> I got an e-mail from Richard, or Androcles, or whoever, and will respond
to
> it here, since I can't from my mail box.

Yes, you can.
Log on to:
http://webmail.wideopenwest.com/cgi-bin/webmail.cgi?cmd=abc

Any ISP worth $9.95 / month will let you work your mail remotely.


Tom Davidson
Richmond, VA


Jim Greenfield

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Sep 17, 2003, 2:15:57 AM9/17/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:<%yM9b.403$z11...@news-binary.blueyonder.co.uk>...

> "Paul B. Andersen" <paul.b....@hia.no> wrote in message
> news:bk5bf6$phi$1...@dolly.uninett.no...
> > > > >
> > > > > > Two events occur along the x axis, they are simultaneous in
> frame K, and
> > > > > > separated by the non-zero space interval x. According to the
> crackpots
> > > > > > these events will not be simultaneous wrt any frame K' that is
> in motion
> > > > > > wrt K along the x axis. Now since it is maintained that the
> order of
> > > > > > events doesn't change between frames,
> > > > >
> > > > > This is "maintained" only for timelike separated events
> > > > > where the squared interval
> > > > > (c*Dt)^2 - Dx^2 > 0
> > > > > The pair you have chosen is (0,0) and (0,x),
> > > > > (c*Dt)^2 - Dx^2 = -x^2 < 0
> > > > > They are spacelike separated.
> > > > >
> > > > > Try to have a serious go at
> > > > > http://www.physicsguy.com/ftl/html/FTL part1.html
> > > > > http://www.physicsguy.com/ftl/html/FTL part2.html
> > have to say about the sequence of emission of the flashes.

Why confuse the issue?
Observer on platform with train stationary sees light path from
ceiling of carriage heading towards center of earth. When train is
moving, path becomes diagonal to said observer, returning to a
different position with RT him and the railway line. ERGO: light beam
now is not directed to earth center, and light IS SOURCE DEPENDENT

Jim G

Starblade Darksquall

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Sep 17, 2003, 4:03:52 AM9/17/03
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Richard <no_mail...@yahoo.com> wrote in message news:<3F65B4D...@yahoo.com>...

You need to understand the concept of light cones. If one event exerts
a causal influence on another in one reference frame then it does so
in all reference frames. But it can only do so if the two events are
within eachother's light cones.

However, if two events are outside eachother's light cones, then
neither can truly be said to occur either before, after, or at the
same time as eachother.

The concepts of Past, Future, Elsewhere, and Present (meaning the two
events occur at the same time and the same space as the other) are all
preserved.

(...Starblade Riven Darksquall...)

AndroclesInEngland

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Sep 17, 2003, 5:42:22 AM9/17/03
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"Mark Palenik" <markp...@wideopenwest.com> wrote in message
news:V-Wcne8pwJK...@wideopenwest.com...
[snip as read]

> I made a response to that in this thread. See my other message.
Basically,
> the time dilation that we are measuring is that of the frame of the
> horizontal component of the light's velocity. So, shining the light
> backwards

Not backwards, just a small tilt from the vertical, through an angle of
magnitude arctan(v/c).

> will cause the "moving" astronaut to see a horizontal component of
> the light's velocity equal to that of the stationary astronaut's
horizontal
> velocity.

That statement isn't clear. How does the stationary astronaut have a
horizontal velocity and still be stationary?

>The stationary astronaut will see it as zero, or in other words
> as being his own horizontal velocity. So, from the frame where the
> horizontal component is non-zero, the frame of the moving astronaut, time
> appears to be running slower

No no... stop right there. This is really quite simple. If the astronaut
carrying the light (call him the shiner) shines it vertically, it has a
shorter path in his view than the astronaut (call him the watcher) watching
him go by does, because the watcher sees a V- shaped path. From this you can
easily deduce time dilation, right? Piece of cake, not rocket science. Just
simple algebra. However, if the shiner's light is tilted backwards, he sees
the longer V-shaped path and the watcher sees the shorter, vertical path.
Again with the simple algebra, and we have time contraction instead of time
dilation. Think about it.

> for the stationary astronaut, which is exactly
> what relativity predicts.

Yes, because relativity can only deal with the special case that suits
relativity, and cannot cope with simple logic.

Each astronaut should see the other's time as
> running slower, or to put it another way, each astronaut sees his own time
> as running faster than the other's.

Nope, each astronaut sees the other's clock as running FASTER, or to put it
another way, each astronaut sees his own time as running slower than the
other's, once the flashlight is tilted from the vertical toward the rear of
the ship. Of course what you are trying to do is apply the vector addition
of velocities while keeping the magnitude the same, and apply it only to a
special case. In the general case, time dilation is an absurdity.

>
> Or, as I put it in my other response:
>
> If we create a vertical beam of light in the "moving" reference frame, it
> will appear as a sloped line in the "stationary" one. If we create a
sloped
> beam of light in then the moving frame, it will appear as a sloped beam of
> light

No, it wil appear as a VERTICAL beam if we choose the angle as arctan(v/c).
Look, this is really simple, just look at the GIF I created at
http://www.androc1es.pwp.blueyonder.co.uk/gardner.htm where you can actually
see it happening.


with a smaller slope (closer to horizontal) in the stationary frame.
> To create something that appears to be a horizontal line

Nobody is trying to do that...I think you are confused, sorry.

> we would have to
> send a beam of light out backwards with respect to the moving observer (or
> rather, at an angle toward the back and floor of the ship). In this case,
> the horizontal velocity frame of the beam of light is no longer that of
the
> moving frame, instead it is in the frame of the stationary observer.


Doing
> the same calculations again, we can see that this means that the "moving"
> observer sees the stationary observer's time as running slower as well,
> which is exactly what SR states - that each observer sees the other's
clock
> as running slower that his.

I know what SR states, it really is very simple. Unfortunately, you can't
keep on repeating the same mantra and claim to have proved anything, since I
have quite clearly shown you a disproof, a contradiction. Proof by
contradiction is perfectly valid to any mathematician.
Androcles


AndroclesInEngland

unread,
Sep 17, 2003, 6:05:03 AM9/17/03
to

"Jim Greenfield" <greenf...@hotmail.com> wrote in message
news:3c4afb26.03091...@posting.google.com...
My post was to Paul, who shouldn't be confused by it, though I expect he
is:)


> Observer on platform with train stationary sees light path from
> ceiling of carriage heading towards center of earth. When train is
> moving, path becomes diagonal to said observer, returning to a
> different position with RT him and the railway line. ERGO: light beam
> now is not directed to earth center, and light IS SOURCE DEPENDENT

Hmmm... suppose you are bouncing a ball on the train, vertically.
Well, let's make it an airplane instead, and go a little faster. Or even an
orbiting shuttle, perhaps, to make the point. Does the ball still travel
toward the Earth's centre? I think it does, for both the astronaut and the
ground based observer. The ball will be in orbit also as it moves toward the
earth and away again, but will still return to the astronaut's hand. But we
on the ground will still determine the motion of the ball to be toward the
Earth's centre, even though it has a horizontal component. Actually, if the
astronaut throws the ball exceeding slowly, so that it takes a full orbit of
about 90 minutes to complete a single cycle from hand to floor and back
again, it will still have moved toward the center of the Earth and back
again, wouldn't it?

What you are really trying to say is that the vector addition of velocities
applies to light as it does to everything else. The relativists accept that,
but insist (without proof) that the magnitude of light's velocity (the
speed) remains constant. But then, they are a weird bunch, aren't they?
Androcles


AndroclesInEngland

unread,
Sep 17, 2003, 6:16:11 AM9/17/03
to

"Mark Palenik" <markp...@wideopenwest.com> wrote in message
news:AIOcnSsKv7p...@wideopenwest.com...
Not trying to do that here, but that is essentially the way Einstein set up
his thought experiment (the light travels along the x-axis only and is
reflected back to the origin) in his paper, "On the Electrodynamics of
Moving Bodies". The flaw in that I've given in
http://www.androc1es.pwp.blueyonder.co.uk/Fundamental.htm,
but I realize that some students may not be fully prepared for a full
mathematical treatment.


we would have to
> send a beam of light out backwards with respect to the moving observer (or
> rather, at an angle toward the back and floor of the ship). In this case,
> the horizontal velocity frame of the beam of light is no longer that of
the
> moving frame, instead it is in the frame of the stationary observer.
Doing
> the same calculations again, we can see that this means that the "moving"
> observer sees the stationary observer's time as running slower as well,
> which is exactly what SR states - that each observer sees the other's
clock
> as running slower that his.

No, doing the calculations properly shows that each observer sees the
other's clock running FASTER. Maybe you are doing the same calculations
again, instead of the calculations that map the new situation :)


Paul B. Andersen

unread,
Sep 17, 2003, 6:23:45 AM9/17/03
to
With no reference to the ongoing discussion,
"AndroclesInEngland" wrote:

> I bet you didn't even realize that a V-shaped light path in one frame of
> reference can be seen as a vertical light path in the other, AND they can
> be reversed, did you? Well, they can. You ARE confused, are you not?
> But you shouldn't be, should you?

How much did you bet?
Send me your Visa number so we can settle it, please.

> After all, you successfully proved to me that

> tau = beta(t+vx/c^2), and not tau = beta(t-vx/c^2), didn't you?

Right, I didn't.

I did however show you that the Lorentz transform may be
either: tau = (t+vx/c^2)/sqrt(1-v^2/c^2),
or: tau = (t-vx/c^2)/sqrt(1-v^2/c^2),
depending on how the frames of reference are defined.
The former is when the "Greek frame" is moving in the negative x-direction,
while the latter is when the "Greek frame" is moving in the positive x-direction.

But I am sure this is obvious to an intelligent guy like yourself.
So did you have a point?

BTW, Androcles,
has sqrt(1) two answers, 1 and -1, in England as well as in Florida?

Paul


Mathew Orman

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Sep 17, 2003, 6:47:36 AM9/17/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message
news:%%V9b.5633$z11....@news-binary.blueyonder.co.uk...

Mathematical fallacy modeling nonexistent property!

Particle move in space and time is the ratio of motions between this
particle and the reference motion represented by
instruments like clocks.
There is all the evidence that space exist.
There is no connection between space and time that means that one has no
physical influence on the other.
The definition of space does not exist if there are no objects of matter.
The definition of time does not exist if there is no motion of matter in
space.
Space has no shape or dimension only the object of matter in space have
dimensions and shapes.
Bending is a process describing the change of geometrical shape of a
material object.
Bending of space is impossible because ones again the space have no shape or
dimension.
All statements above where proven true in physical experiments.

My definition of time states that time is only a ration of motion of objects
of matter.
My definition of time is based on physical evidence and only defines the
practical scientific aspect of it that can be measured with physical
instruments and can be proven in a physical experiment.
The geometric time definition only exists if the geometry of the matter is
changing and does not state that
time is a ratio of motion.
Feather more there is the only one way of measuring what is called the time
and that is the way of measuring the ratio of motion.
All clocking devices which represent the reference motion are based on
periodic motion of matter (particles or waves).

The property of matter that cannot be measured or detected has only a
philosophical or metaphysical application.

I do not tolerate fallacy!
Physics are about properties of reality!
If you say that there exist a property of space that you call space-time
than if you do not provide physical evidence of that property.
Than I will immediately call you theory FALSE or Sci-Fi.

Sincerely,
ps.

My theory of relativity:
Everything in the universe is relative!


My definition of time:
Time is a ratio of motions and is always measured as such,
ratio between measured and reference motions.
The reference motions are periodic (pendulums clocks) or linear (predictable
displacements).
Example:
I walk 5m while clocks pendulum swings 3 times(periodic reference).
I walk 3m while 1kg weight falls down from the height of 1m at see level at
20deg centigrade etc.(predictable displacement reference)
Time is a method of measuring or synchronizing motions.
Anything else is Sci-Fi!

The has not been a single experiment proving that static fields of any know
kind can propagate.
Propagation of any static field can only be done by moving its source.
The source of gravity always contains matter in forms of atoms it exits with
the matter and it can be controlled by changing the densities of the matter.
Changes in gravity detected on the earth are the results of planetary
motions in the universe not by
propagation of the gravity field.
The precision clocks (reference periodic motion counters) are based on
counting periodic motion of the electrons in specially chosen
atoms. Which means that they are affected by gravity and gravity forces
differ with the altitude change above the earth.
Saying that time goes slower is correct since in my definition time is a
ratio of motions not a constant derive from c.
The space and matter in its various form exists, but the definition of time
only exists if there is matter in motion.
Every atom in the universe has it own gravity field.

Well, if you build your theory on definition used in existing theories,
than that means that you are attempting to modify an or improve the
existing ones.
The existing theories your are creating your vocabulary based up on, have
faults and lack of experimental
prove, so your modified theories will carry on the unreliable and unproven
elements.
The new theories should be based on prime definition and proven in physical
experiments.
Your asking to agree with your statements may result in endless waste of
time.

So let's present new and solid theories instead and have discussions about
them!

When you try to measure time than you will notice
that all you are doing is measuring a ratio of motions!
No motion, no time can be measured.

Assumption of nonexistent property.
Clocks do not keep time.
They count repetitive motion.
There is no definition of "proper time".

Simple, it is repetitive motion.
And time is a ratio of motion,
so to measure the repetitive motion of the first clock we use the count of
more grainer
motion.
The same applies to distance between the marks on given rule,
one uses the more grainer ruler and so on.
And this metrology logic is several thousand years old.

Now since we know what clock do, maybe now you can define the "proper
time"?

There is a hierahical existance of definitions and that is:
Definition of space exists only if there are objects of matter.
Definition of time exists if there is motion of matter.
Definition of space distance exists if there is a ratio of size of objects
of matter.
Ans so on...

No,
motion passes in ratio of time.
Time is a method and not a property of space.
We measure time by comparing reference motion to measured motion.
Examples:
I walk 3m (measured motion) while clock swings his pendulum 2 times
(reference motion).
I walk 1.2m (measured motion) while weight of 1kg drops down from height
of 1m at sea level and 20deg centigrade temperature (reference motion)..
For practical reasons humans agreed to use period motion as a reference
motion.
We use planetary periodic motion and its subdivisions (day, hour,sec etc..)
as the reference motion
for measuring and or synchronizing motions.


As an engineer I describe your time in the following way:
Time is a reference count for synchronizing the human motion.
Clocks are the mechanisms of distributing the reference count.
Distributed reference count allow people and machines to move synchronously.
Finally clocks do not measure time(time is the method) but count periodic
motion.
"What time it is?" people and machines are asking when they choose to find
if and when their motion is
out of phase.
Connecting time with the space has no useful practical porpoise.
Spacetime is a word invented by GTR inventers and
despite wishful thinking, it does not alter the motion of the matter.


Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com


Paul B. Andersen

unread,
Sep 17, 2003, 6:56:48 AM9/17/03
to

"Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...

What is the point in quoting a lot of text which you don't
comment or make any references to whatsoever?

> Why confuse the issue?

And in order to not confuse the issue,
you introduce an entirely different scenario?

> Observer on platform with train stationary sees light path from
> ceiling of carriage heading towards center of earth. When train is
> moving, path becomes diagonal to said observer, returning to a
> different position with RT him and the railway line. ERGO: light beam
> now is not directed to earth center, and light IS SOURCE DEPENDENT

Why are you shouting trivialities?
Of course the direction of the velcoity of light depend
on the velocity of the source.
Or in other words:
The direction of the velocity of light is frame dependent.
So what?
Did you have a point?

Paul


Mathew Orman

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Sep 17, 2003, 7:06:25 AM9/17/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message
news:IvW9b.5845$z11....@news-binary.blueyonder.co.uk...

Except clocks do no measure time.
Time is a ratio of motion so you need clock and the measured motion
to calculate that
and in physical reality your thought experiment fails because
clocks have no ability to register when and why it's own periodic rate
changes.

You are a mathematical time fallacy worshiper!

//----------------------------------------------------------
Quick definitions (worshiper)
noun: someone who admires too much to recognize faults
noun: a person who has religious faith
See worship
//----------------------------------------------------------

Paul Cardinale

unread,
Sep 17, 2003, 10:16:41 AM9/17/03
to
Richard <no_mail...@yahoo.com> wrote in message news:<3F65F5D9...@yahoo.com>...

> Dirk Van de moortel wrote:
> >
> > "Richard" <no_mail...@yahoo.com> wrote in message news:3F65B4D...@yahoo.com...
> > >
> > > Here's another one for the SRist crackpots to crack:
> > >
> > > Two events occur along the x axis, they are simultaneous in frame K, and
> > > separated by the non-zero space interval x. According to the crackpots
> > > these events will not be simultaneous wrt any frame K' that is in motion
> > > wrt K along the x axis. Now since it is maintained that the order of
> > > events doesn't change between frames,
> >
> > This is "maintained" only for timelike separated events
> > where the squared interval
> > (c*Dt)^2 - Dx^2 > 0
> > The pair you have chosen is (0,0) and (0,x),
> > (c*Dt)^2 - Dx^2 = -x^2 < 0
> > They are spacelike separated.
> >
> > Try to have a serious go at
> > http://www.physicsguy.com/ftl/html/FTL_part1.html
> > http://www.physicsguy.com/ftl/html/FTL_part2.html

> > and let's get beyond section 1.1 of part 1 this time.
> > If you have a problem, just ask a question.
> >
> > [snipped on first error]
> >
> > Dirk Vdm
>
> Let's put it to practice Dirk.
>
> In the train experiment lightning flashes simultaneously at the ends of
> the railway carriage wrt the embankment observer. According to the
> observer on the train, who is positioned at its center, the light from

> the leading end of the carriage traverses a smaller space interval than
> the light from the trailing end. Thus, believing that light also
> propagates wrt him at c, he calculates that the leading flash occurred

> before the trailing flash. So far this is straight from the book.
>
> Now if the interval that the carriage observer measures is dt', then we

> can, theoretically allow a second leading flash to occur at some time
> less than dt' after the first, and the carriage observer will will

> intercept that light before he intercepts the trailing flash, thus he
> has an event occurring in the midst of the initial interval. The
> sequence, according to K' is
>
> 1-Forward flash
> 2-Second forward flash
> 3-Rearward flash
>
> Keep in mind that the first and third of these is simultaneous wrt the
> track observer.
>
> Wrt K the order of 2 and 3 is thus reversed wrt the same events as
> perceived from frame K'. Now lets suppose that the second forward flash

> consists of green light, while the rearward flash consists of red light.
> On board the train, in the train observers hand and also at rest wrt the
> carriage, is a device that trips one of two relays depending upon the
> frequency of detected light. Upon first detection the corresponding
> relay trips and locks out the other.
>
> These observers will not agree on which relay trips, simply because they
> don't agree on which flash arrived at the detector first. Moreover the

> creed that according to SR, "events do not reverse when transforming
> between frames" has also been shown to be false, as Dirk also correctly
> noted. Now it probably didn't occur to him however, that events must in

> fact maintain their order through transformations, as the experiment
> proves beyond doubt. Thus SR has once again been shown to be
> inconsistent.
>
> Richard Perry
>

If you were smarter than a drooling babboon, you might be able to
learn enough math to apply the LT and see that there are no
inconsistencies.
But you're not smarter than a drooling babboon and you can't apply
simple algebra, as you have demonstrated many times.

Paul Cardinale

Dirk Van de moortel

unread,
Sep 17, 2003, 10:25:05 AM9/17/03
to

"Paul Cardinale" <pcard...@volcanomail.com> wrote in message news:64050551.03091...@posting.google.com...

> Richard <no_mail...@yahoo.com> wrote in message news:<3F65F5D9...@yahoo.com>...

[snip]

> > proves beyond doubt. Thus SR has once again been shown to be
> > inconsistent.
> >
> > Richard Perry
> >
>
> If you were smarter than a drooling babboon, you might be able to
> learn enough math to apply the LT and see that there are no
> inconsistencies.
> But you're not smarter than a drooling babboon and you can't apply
> simple algebra, as you have demonstrated many times.

I think his algebra is more or less okay, but
1) He thinks that
t = t' gamma'
is a Lorentz transformation.
Remember:
http://groups.google.com/groups?&threadm=3F426001...@yahoo.com
Explanations don't help.

2) He does not know what the transformation actually does.
He has no idea what the x, t, x' and t' represent.

In short, he is Fun ;-)

Dirk Vdm


AndroclesInEngland

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Sep 17, 2003, 11:37:02 AM9/17/03
to

"Paul B. Andersen" <paul.b....@hia.no> wrote in message
news:bk9co9$r33$1...@dolly.uninett.no...
Yes, it does, Paul. 1 times 1 = 1, and (-1) times (-1) = 1, Paul, at least
on my computer, anyway. Dunno about yours...maybe you should try it and let
me know what answer you get by multiplying -1 by -1. I 'd be very interested
in knowing if there was an answer different to 1. Who knows, you might be
able to teach me something.


Mathew Orman

unread,
Sep 17, 2003, 12:21:59 PM9/17/03
to

"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03091...@posting.google.com...

Except clocks do no measure time.

Dirk Van de moortel

unread,
Sep 17, 2003, 12:53:01 PM9/17/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:zc%9b.8523$z11....@news-binary.blueyonder.co.uk...

So what do we say:
sqrt(1) = 1 or sqrt(1) = -1
or do we say
sqrt(1) = 1 and sqrt(1) = -1
?

Dirk Vdm


Paul Cardinale

unread,
Sep 17, 2003, 2:35:21 PM9/17/03
to
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<3f686f48$1...@usenet01.boi.hp.com>...

> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message news:64050551.03091...@posting.google.com...
> > Richard <no_mail...@yahoo.com> wrote in message news:<3F65F5D9...@yahoo.com>...
>
> [snip]
>
> > > proves beyond doubt. Thus SR has once again been shown to be
> > > inconsistent.
> > >
> > > Richard Perry
> > >
> >
> > If you were smarter than a drooling babboon, you might be able to
> > learn enough math to apply the LT and see that there are no
> > inconsistencies.
> > But you're not smarter than a drooling babboon and you can't apply
> > simple algebra, as you have demonstrated many times.
>
> I think his algebra is more or less okay, but

Yes, but he can't apply it (which is what I wrote) because he can't
comprehend how to connect the letters to anything.

> 1) He thinks that

Now your just being silly.

Paul Cardinale

AndroclesInEngland

unread,
Sep 17, 2003, 2:58:35 PM9/17/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:Nj0ab.24988$vn5.1...@phobos.telenet-ops.be...
Idiocy definitely does become you, Drik.
Surely you realize that you should say sqrt(1) = 1 xor sqrt(1) = -1:)
But then, you wouldn't be able to work that out for yourself, would you,
since you've never beeen able to work out ANYTHING for yourself.
Please publish my reply to Paul on your page, I can use the publicity, even
adverse publicity. Of course, it only makes YOU look like a fool, doesn't
it?
Oops... did I say 'look like'? :)

Androcles


Mathew Orman

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Sep 17, 2003, 3:08:58 PM9/17/03
to

"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03091...@posting.google.com...
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:<3f686f48$1...@usenet01.boi.hp.com>...
> > "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03091...@posting.google.com...
> > > Richard <no_mail...@yahoo.com> wrote in message
news:<3F65F5D9...@yahoo.com>...
> >
> > [snip]
> >
> > > > proves beyond doubt. Thus SR has once again been shown to be
> > > > inconsistent.
> > > >
> > > > Richard Perry
> > > >
> > >
> > > If you were smarter than a drooling babboon, you might be able to
> > > learn enough math to apply the LT and see that there are no
> > > inconsistencies.
> > > But you're not smarter than a drooling babboon and you can't apply
> > > simple algebra, as you have demonstrated many times.
> >
> > I think his algebra is more or less okay, but
>
> Yes, but he can't apply it (which is what I wrote) because he can't
> comprehend how to connect the letters to anything.
>
> > 1) He thinks that
>
> Now your just being silly.
>
> Paul Cardinale

And you cannot comprehend that one cannot create and entity or property of
matter
with mathematical formula!

Math is a modeling tool and not a tool for creating physical objects or
physical properties.
But you cannot comprehend that because you and Dirk are blind clock
worshipers.
Bragging about your mathematical correctness that has nothing to do with the
reality.
Modeling fantastic worlds using math is a trivial task.
And modeling physical phenomena is not because it requires to test the model
against the
experimental results.
And that is where you always fail!

I bet that you cannot model a simple capacitor!

Dirk Van de moortel

unread,
Sep 17, 2003, 3:08:28 PM9/17/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:v92ab.71$6l6...@news-binary.blueyonder.co.uk...

>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:Nj0ab.24988$vn5.1...@phobos.telenet-ops.be...
> >

[snip]

> >
> > So what do we say:
> > sqrt(1) = 1 or sqrt(1) = -1
> > or do we say
> > sqrt(1) = 1 and sqrt(1) = -1
> > ?
> Idiocy definitely does become you, Drik.
> Surely you realize that you should say sqrt(1) = 1 xor sqrt(1) = -1:)

Under which circumstances would you write sqrt(1) = 1
and under which would you write sqrt(1) = -1?

Dirk Vdm


John Anderson

unread,
Sep 17, 2003, 8:36:37 PM9/17/03
to

Richard wrote:

> Here's another one for the SRist crackpots to crack:
>
> Two events occur along the x axis, they are simultaneous in frame K, and
> separated by the non-zero space interval x. According to the crackpots
> these events will not be simultaneous wrt any frame K' that is in motion
> wrt K along the x axis. Now since it is maintained that the order of
> events doesn't change between frames,

You are an idiot. The temporal ordering of spacelike separated eventsIS
manifestly frame-dependent in SR.

SR doesn't have to explain the fact that you don't understand what it
predicts.

John Anderson

Richard

unread,
Sep 17, 2003, 11:11:34 PM9/17/03
to

AndroclesInEngland wrote:
>
> "Mark Palenik" <markp...@wideopenwest.com> wrote in message
> news:V-Wcne8pwJK...@wideopenwest.com...

<snipped same old debate>

>
> Doing
> > the same calculations again, we can see that this means that the "moving"
> > observer sees the stationary observer's time as running slower as well,
> > which is exactly what SR states - that each observer sees the other's
> clock
> > as running slower that his.
> I know what SR states, it really is very simple. Unfortunately, you can't
> keep on repeating the same mantra and claim to have proved anything, since I
> have quite clearly shown you a disproof, a contradiction. Proof by
> contradiction is perfectly valid to any mathematician.
> Androcles

Yes Androcles, your argument is clear and correct, and has been posted
innumerable
times already, to no avail. The SRists hold that reciprocity constitutes
no contradictions, LOL, perhaps no internal contradictions, but there is
no correspondence to reality as you well know:)

Now despite the fact that Dirk caught my error in the original post,
there
is yet another argument that seems to me to be quite sound, and that
permanently disproves SR without appealing to any of the contradictions
that have thus far been presented. That argument is this:

Given an observer at rest on a non-rotating planet, two land vehicles
accelerate away from the origin of rest frame along the ground (away
from the rest frame observer) in opposite directions, equally, achieving
equal very
high speeds wrt the rest frame in very short equal time intervals. The
rest frame is frame K, the vehicle moving to the right is accelerated
into frame K', and the vehicle moving to the left is accelerated into
frame K''. The vehicles continue in this uniform motion (in frames K'
and K'' respectively) wrt rest frame K for a time interval that is
orders
of magnitude larger than the acceleration interval until finally they
decelerate simultaneously and equally wrt K back into the rest
frame K, i.e. they are now at rest wrt the original frame and thus wrt
the ground, but some large distance apart.

Now here is where SR must get off the logic train, because regardless of
what any other observer perceives pertaining to this sequence, it
remains that the distance of each of these vehicles to the the origin of
K will be equal when measured by any arbitrary observer, and this must
be true even when these distances (ground-'lengths' are 1/gamma or even
'otherwise' contracted wrt them. The lengths will still be equal wrt the
arbitrary observer. Thus, for instance, from the frame of reference of
an
observer that was at rest wrt frame K' throughout the sequence, the
final distances of the two ships
from the original observer at the origin of K must be equal. What this
requires is that the wrt an observer in frame K' , K'' was moving
exactly twice the speed of K. IOW he must find that W = v + w.

According to SR two observers will reciprocally agree on their speed wrt
each other, thus v (the speed of K wrt K'' according to both observers
in these two frames) is equal to w, since w is the speed of K wrt K' as
measured by both of these observers). That is, K measures his speed to
be v wrt K' and w wrt K'', thus from his frame the speed of K' wrt K''
must be 2v wrt each of these observers, and thus v+w. IOW W=v+w, and
normal velocity addition necessarily applies.

Now this doesn't' preclude the physical contraction of objects in
motion, nor does it preclude a physical change in a clock's ticking
rate,
but these would necessarily be non-reciprocal alterations, and thus
would require some absolute frame wrt to reference them, a frame such as
that of a large gravitating body for instance, a body that extends a
composite inverse squared electromagnetic field through which those
clocks and meter sticks are moving, and thus providing something other
than pixies to explain the changes to these malleable
measuring devices. OTOH we lose constancy of light speed along the x
axis wrt those moving observers, but the two-way propagational speed can
still be c wrt them with no contradictions. Now take into consideration
that only the two-way
trip can actually be measured, and we are still in conformity with the
premise that "light speed will always be 'measured' to be c wrt all
frames in motion.

However none of this will hold in the transition area when moving from
one field into the next, such as when an observer is situated perfectly
between two neighboring stars. Instead he will have to superpose the
alterations to his clocks and meter sticks that are caused be each field
respectively. OTOH this is what must be done anyway, and this anywhere
at
all, but when an experiment is conducted in a predominant local field,
the effects of neighboring fields will be negligible. Moreover one might
expect that the contraction of a meter stick will depend upon its
orientation, as well as upon its speed, wrt the local field, which is
exactly the case when considering the MMX where contraction is a
function of angle wrt horizontal motion.

Richard Perry

http://www.cswnet.com/~rper

AndroclesInEngland

unread,
Sep 18, 2003, 3:27:18 AM9/18/03
to

"Richard" <no_mail...@yahoo.com> wrote in message
news:3F692266...@yahoo.com...

>
> Yes Androcles, your argument is clear and correct, and has been posted
> innumerable
> times already, to no avail.

Thank you, Richard. It is always good to find someone with the power to
reason.


The SRists hold that reciprocity constitutes
> no contradictions, LOL, perhaps no internal contradictions,

Oh, but there are. One of my favourites (p 40, Dover) is
"In agreement with experience (who's experience? Certainly not mine...) we
ASSUME (my capitals) the quantity 2AB/(t'A -tA) =c to be a universal
constant...".
Beautiful, isn't it? The persuasive rhetoric that sold relativity in the
first place, encapsulated right there. It does look good, because even a
10-year old can understand it. The distance the light travels from A to B
and back again divided by the time it takes to do so gives the speed. Of
course, Einstein fails to mention what happens if A and B are in relative
motion. If we do consider that, then the distance AB is NOT the same as the
distance BA when the light makes the return trip. Why anyone would make such
a stupid assumption is beyond me.

Here's another (p 45, Dover) "But the ray moves relatively to the initial
point in k, when measure in the stationary system, with the velocity c-v, so
that..."

There you have it. Einstein's own statement that the velocity of light is
(c-v) MEASURED in his 'stationary' system. If that isn't an internal
contradiction, I don't know what an internal contradiction is.


but there is
> no correspondence to reality as you well know:)

Yep, absolutely none at all.

Very nice, Richard, but while it is fun to invent these scenarios ( I used
to do it) you will quickly find out that you are only providing cannon
fodder to the relativists, who will jeer at you, as you are doing at them.
You see, you are attacking their religion, and that will only serve to raise
their hackles. You only need to read a few posts where people are called
idiots and crackpots because they find relativity to be an impossible
theory. The relativists will always say it is you that doesn't understand
it.
Once one is actually aware of the flaw in relativity, as you are, it is
quite easy to invent literally hundreds of "what if" experiments such as the
one you've given, and I advise against it, it is futile.
What I would suggest you do is move on.
Here are some things to ponder.
1) What is a photon?
1a) Is the speed of light in a medium such as air or water constant wrt the
medium?
1b) Does a magnetic field constitute a medium?
1c) Does an electric filed constitute a medium?
1d) Does a gravitational field constitute a medium?
2) What emits light?
Is it
a) Hot stars?
b) Hot light bulb filaments?
c) Hot molecules?
d) Electrons changing quantum levels?
(if you answered d, how fast does the electron throw off the light?)
3) If you pass a current through a wire, it produces a magnetic field.
Let there be a point P at a distance d from the wire. From the moment you
throw the switch, how long will it take for the field to be detected at P?
What is the speed of a magnetic field?
4) MMX disproves the aether, so what does MMX actually show?
5) If the velocity of light in empty space is source dependent, what do we
actually see when a star moves in an elliptical orbit because it has a
companion?
Hints:
a: A stick in water appears bent, but the appearance is an illusion.
b: What illusions can we expect from 3) above?
c: http://www.astro.uiuc.edu/stardial/variables/mira.html
And of course my own site, I've been studying this for many years...
http://www.androc1es.pwp.blueyonder.co.uk/
Good seeing,
Androcles


AndroclesInEngland

unread,
Sep 18, 2003, 4:17:29 AM9/18/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:Mi2ab.25307$FN3.1...@phobos.telenet-ops.be...
Oops, you snipped with saying why. I'll put it back for you...

But then, you wouldn't be able to work that out for yourself, would you,
since you've never beeen able to work out ANYTHING for yourself.
Please publish my reply to Paul on your page, I can use the publicity, even
adverse publicity. Of course, it only makes YOU look like a fool, doesn't
it?
Oops... did I say 'look like'? :)
Now to answer your question, so that you can learn very basic math that
most children over the age of 12 already know.
-1 times -1 = 1
1 times 1 = 1
that means that the square roots of 1 are -1 and 1.
i.e sqrt(1) = 1 , sqrt(1) = -1.

Now for something a little more advanced, probably far too difficult for you
to comprehend.
Did you know that 1 has three cube roots? Or that in the general case, the
nth root of x has n roots?
To find them, find just one root, place a circle of radius nth-root(x)
centred on the origin in the complex plane, (that's the Cartesian plane with
'i' on the vertical axis and x on the horizontal), and divide the circle
into n segments, like an orange. The points on the circle will then be ALL
the n-roots of x.
Example:
In the case of 1, find one cube root of 1 (it's 1) and thus the radius of
the circle is 1. Divide the circle into 3 segments, so that you have an
equilateral triangle circumscribed (not circumcized, you dumb cluck) by the
circle with apex at (1, 0i). The points of the triangle will be the cube
root of 1. Multiply any one of them by itself and by itself again and the
answer will be (1,0i).
thus the three cube roots of 1 are
1, 0i
-0.5, 0.866i
-0.5, -0.866i
What's 'i', you ask? It's one of the square roots of -1, of course. The
other one is -i.
Should I go on? Nah, it's beyond you...


Dirk Van de moortel

unread,
Sep 18, 2003, 5:39:37 AM9/18/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:uSdab.3017$sj....@news-binary.blueyonder.co.uk...

I will repeat a simple question: under which circumstances


would you write sqrt(1) = 1 and under which would you
write sqrt(1) = -1?

When you see sqrt(9) in a text, how do you know that
they mean 3 or -3?
Another simple question: do you know that the expression


sqrt(1) = 1 xor sqrt(1) = -1

implies


sqrt(1) = 1 or sqrt(1) = -1

?

Dirk Vdm


Robert J. Kolker

unread,
Sep 18, 2003, 5:43:35 AM9/18/03
to

Dirk Van de moortel wrote:

> Another simple question: do you know that the expression
> sqrt(1) = 1 xor sqrt(1) = -1
> implies
> sqrt(1) = 1 or sqrt(1) = -1
> ?
>

Could it be that sqrt is not a one valued function and has to be spit up
into two proper one valued functions? This is exactly what is done on
the complex number domain. A two sheaf Reimann surface is constructed so
that sqrt behaves itself.

Bob Kolker

Dirk Van de moortel

unread,
Sep 18, 2003, 5:59:07 AM9/18/03
to

"Robert J. Kolker" <bobk...@attbi.com> wrote in message news:bkbuoc$rdl2j$1...@ID-76471.news.uni-berlin.de...

In the complex context one could write 'sqrt'(1) = {-1,1}
But this is not the complex context. This is elementary
real algebra (for engineers if you like).
By the way, a function is always singlevalued, by definition.
When in the complex context one defines 'sqrt'(1) = {-1,1},
the 'sqrt'-function value of 1 is the *set* {-1,1}, which is a
single value.
Just ask yourself this: when *you* see an equation that
contains the string sqrt(9), do you have to wonder whether


they mean 3 or -3?

Dirk Vdm


Dirk Van de moortel

unread,
Sep 18, 2003, 6:38:18 AM9/18/03
to

"Richard" <no_mail...@yahoo.com> wrote in message news:3F692266...@yahoo.com...

[snip]

we have
velocity(K,K') = v
velocity(K',K") = W (is unknown)
velocity(K,K") = w
So standard addition of velocities(K,K',K") gives
w = (v+W)/(1+vW)
Solve for W:
W = (w-v)/(1-vw)

The other way around:
velocity(K',K) = -v
velocity(K",K') = X (is unknown)
velocity(K",K) = -w
So standard addition of velocities(K",K',K) gives
-w = (X-v)/(1-Xv)
Solve for X:
X = (v-w)/(1-vw)

Taken together:
X = -W (reciprocal)

In your special case you have w = -v, so
W = (-v-v)/(1+v^2) = -2v/(1+v^2).
X = (v+w)/(1+vw) = 2v/(1+v^2)
so of course
W = -X

Dirk Vdm


AndroclesInEngland

unread,
Sep 18, 2003, 7:33:19 AM9/18/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3f697de6$1...@usenet01.boi.hp.com...
Any math text I've ever read expects the reader to know that the roots of 9
are 3 and -3. Possibly you haven't read any math texts.. try it sometime.

> Another simple question: do you know that the expression
> sqrt(1) = 1 xor sqrt(1) = -1
> implies
> sqrt(1) = 1 or sqrt(1) = -1
> ?
>
> Dirk Vdm

Wrong again, Ding. Look at a truth table for xor.
A B Xor
0 0 0
0 1 1
1 0 1
1 1 0
Three rows correspond to OR, one does not. Can you find the odd one out?
Hint: 1 OR 1 = 1, 1 XOR 1 = 0. To phrase it in English, A exclusive or B is
A or B but not both. Does 'but' confuse you?
Here's a simple application
_______
o----------/ A B \---------0----o
_______

Light switches A and B, shown in the up position, conduct electricity to the
bulb, 0. They do when both are in the down position, too. But when on is up
and the other is down, the bulb is not light. This is actually NOT XOR, but
can be made into XOR by crossing the wires over at one light switch.
This is the configuration for OR.
__/ A____
o--------| |--------0-------o
|___/ B___|
Here, the light switch remains on even with both switches are down.

And I see you've been snipping again.. here, let me put it back in for you.

Dirk Van de moortel

unread,
Sep 18, 2003, 7:44:15 AM9/18/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:4Kgab.5658$sj....@news-binary.blueyonder.co.uk...

(A xor B)
<==> (A and B') or (A' and B)
<==> (A or A') and (A or B) and (B' or A') and (B' or B)
<==> (A or B) and (B' or A')
==> (A or B)
QED

Or if you prefer truth tables:
A B Or


0 0 0
0 1 1
1 0 1

1 1 1

A B Xor
0 0 0
0 1 1
1 0 1

1 1 1

A B =>
0 0 1
0 1 1
1 0 0
1 1 1

(A xor B ) => (A or B)
0 0 0 1 0 0 0


0 1 1 1 0 1 1

1 1 0 1 1 1 0
1 0 1 1 1 1 1
QED

Unless you use another kind of logic of course.

Dirk Vdm


Dirk Van de moortel

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Sep 18, 2003, 9:47:02 AM9/18/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:4Kgab.5658$sj....@news-binary.blueyonder.co.uk...

>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:3f697de6$1...@usenet01.boi.hp.com...
> >

[snip]

> > I will repeat a simple question: under which circumstances
> > would you write sqrt(1) = 1 and under which would you
> > write sqrt(1) = -1?
> > When you see sqrt(9) in a text, how do you know that
> > they mean 3 or -3?

[hadn't seen this hidden part
- added empty line to separate]

> Any math text I've ever read expects the reader to know that the roots of 9
> are 3 and -3. Possibly you haven't read any math texts.. try it sometime.

Yes, but I was asking about sqrt(9).
What is it? 3 or -3 or both?

Dirk Vdm


pst...@ix.netcom.com

unread,
Sep 18, 2003, 10:03:21 AM9/18/03
to

By definition, mathematically it can be both. If it were to
represent a physical quantity it can only be 3... You cannot have a
physical quantity of -3! I can have 3 rocks, but I cannot have -3
rocks...

Paul Stowe

Dirk Van de moortel

unread,
Sep 18, 2003, 10:20:26 AM9/18/03
to

<pst...@ix.netcom.com> wrote in message news:t5ejmv8bipr0ll5o9...@4ax.com...

> On Thu, 18 Sep 2003 15:47:02 +0200, "Dirk Van de moortel"
> <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
> >
> >"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:4Kgab.5658$sj....@news-binary.blueyonder.co.uk...
> >>
> >> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> >> in message news:3f697de6$1...@usenet01.boi.hp.com...
> >> >
> >
> >[snip]
> >
> >> > I will repeat a simple question: under which circumstances
> >> > would you write sqrt(1) = 1 and under which would you
> >> > write sqrt(1) = -1?
> >> > When you see sqrt(9) in a text, how do you know that
> >> > they mean 3 or -3?
> >
> >[hadn't seen this hidden part
> > - added empty line to separate]
> >
> >> Any math text I've ever read expects the reader to know that the roots of 9
> >> are 3 and -3. Possibly you haven't read any math texts.. try it sometime.
> >
> > Yes, but I was asking about sqrt(9). What is it? 3 or -3 or both?
>
> By definition, mathematically it can be both.

By definition it is 3.
Open a textbook on physics, engineering, calculus,
statistics, chemistry, whatever... when you see sqrt(9),
do you stop and ask yourself
"Halt, what would they mean here, 3 or -3" ?
Do you?
No. You know they mean 3 because sqrt is defined
that way: it gives the positive root of a postive number.

> If it were to
> represent a physical quantity it can only be 3... You cannot have a
> physical quantity of -3! I can have 3 rocks, but I cannot have -3
> rocks...

And you cannot have a velocity of -3 m/s?
And your height cannot be -3 m?
And events with time -3 s cannot exist?
Where have you been?
James Driscoll Spaceman has a follower.
Congratulations.

Dirk Vdm


pst...@ix.netcom.com

unread,
Sep 18, 2003, 10:39:37 AM9/18/03
to
On Thu, 18 Sep 2003 16:20:26 +0200, "Dirk Van de moortel"
<dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:

>
><pst...@ix.netcom.com> wrote in message news:t5ejmv8bipr0ll5o9...@4ax.com...
>> On Thu, 18 Sep 2003 15:47:02 +0200, "Dirk Van de moortel"
>> <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:

[Snip...]

>>> Yes, but I was asking about sqrt(9). What is it? 3 or -3 or both?
>>
>> By definition, mathematically it can be both.
>
> By definition it is 3.
> Open a textbook on physics, engineering, calculus,
> statistics, chemistry, whatever... when you see sqrt(9),
> do you stop and ask yourself

Have reading comprehension problem Dirk old boy? I said,
mathematically...

(-3)(-3) = ?

> "Halt, what would they mean here, 3 or -3" ?
> Do you?
> No. You know they mean 3 because sqrt is defined
> that way: it gives the positive root of a postive number.
>
>> If it were to represent a physical quantity it can only
> be 3... You cannot have a physical quantity of -3! I
> can have 3 rocks, but I cannot have -3 rocks...
>
> And you cannot have a velocity of -3 m/s?

Direction is arbitrary Dirk. Is it -3 m/s or, perhaps,
3 m/s? Well, than solely depends on HOW 'I' define my
coordinate system. You most certainly cannot have -3 m/s
speed!

> And your height cannot be -3 m?

Nope.

> And events with time -3 s cannot exist?

Nope. 'Did' exist yes, does exist, no! You cannot have,
physically 3 s since that does not 'yet' exist!

> Where have you been? James Driscoll Spaceman has a follower.
> Congratulations.

Well facts are facts dirk...

Paul Stowe

Robert J. Kolker

unread,
Sep 18, 2003, 10:58:26 AM9/18/03
to

Dirk Van de moortel wrote:

> Yes, but I was asking about sqrt(9).
> What is it? 3 or -3 or both?

Strictly speaking a function has either no value or one value for an
argument. Sqrt is generally extended to apply to a Reimann surface so it
is plus on one part of the Reimann surface and negative on the other. So
there strict answer is either one or the other, but not both. Generally
the positive value is taken to be the value of sqrt of lamda x(x)^(1/2

Bob Kolker


Dirk Van de moortel

unread,
Sep 18, 2003, 10:58:41 AM9/18/03
to

<pst...@ix.netcom.com> wrote in message news:qtfjmvgks8cg78sv8...@4ax.com...

> On Thu, 18 Sep 2003 16:20:26 +0200, "Dirk Van de moortel"
> <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
> >
> ><pst...@ix.netcom.com> wrote in message news:t5ejmv8bipr0ll5o9...@4ax.com...
> >> On Thu, 18 Sep 2003 15:47:02 +0200, "Dirk Van de moortel"
> >> <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote:
>
> [Snip...]
>
> >>> Yes, but I was asking about sqrt(9). What is it? 3 or -3 or both?
> >>
> >> By definition, mathematically it can be both.
> >
> > By definition it is 3.
> > Open a textbook on physics, engineering, calculus,
> > statistics, chemistry, whatever... when you see sqrt(9),
> > do you stop and ask yourself
>
> Have reading comprehension problem Dirk old boy? I said,
> mathematically...

Mathematically, sqrt(9) = 3.

>
> (-3)(-3) = ?

9

>
> > "Halt, what would they mean here, 3 or -3" ?
> > Do you?
> > No. You know they mean 3 because sqrt is defined
> > that way: it gives the positive root of a postive number.

Well, *do* you stop and ask


"Halt, what would they mean here, 3 or -3" ?

?
In a textbook on physics, engineering, calculus,
statistics, chemistry, whatever?
Have you seen the words "calculus" and "statistics"?
Reading comprehension problem?

> >
> >> If it were to represent a physical quantity it can only
> > be 3... You cannot have a physical quantity of -3! I
> > can have 3 rocks, but I cannot have -3 rocks...
> >
> > And you cannot have a velocity of -3 m/s?
>
> Direction is arbitrary Dirk. Is it -3 m/s or, perhaps,
> 3 m/s? Well, than solely depends on HOW 'I' define my
> coordinate system. You most certainly cannot have -3 m/s
> speed!

That is why I said velocity.
Reading comprehension problem?

>
> > And your height cannot be -3 m?
>
> Nope.

Here is was careless.
If height is defined as length, you cannot have a
height of -3.
If height is defined as position, you can have
a height of -3. I was referring to this.

>
> > And events with time -3 s cannot exist?
>
> Nope. 'Did' exist yes, does exist, no! You cannot have,
> physically 3 s since that does not 'yet' exist!

And events with time -3 s cannot have existed?

>
> > Where have you been? James Driscoll Spaceman has a follower.
> > Congratulations.
>
> Well facts are facts dirk...

And word games are word games Paul.

Androcles' argument and entire "world view"
http://groups.google.com/groups?&threadm=l4677.22938$EP6.5...@news1.rdc2.pa.home.com
is based on the conviction that in a standard text book
the string
sqrt(9)
can be replaced by
3
or by
-3.
If you want to have a say, first work your way through
the entire thread and laugh your ass off.

Dirk Vdm


Dirk Van de moortel

unread,
Sep 18, 2003, 11:09:08 AM9/18/03
to

"Robert J. Kolker" <bobk...@attbi.com> wrote in message news:bkch6n$5u6o$1...@ID-76471.news.uni-berlin.de...

>
>
> Dirk Van de moortel wrote:
>
> > Yes, but I was asking about sqrt(9).
> > What is it? 3 or -3 or both?
>
> Strictly speaking a function has either no value or one value for an
> argument.

Strictly speaking a function is a subset of the product
set of two sets with the additional property that if (a,b)
and (a,c) are elements of the function, then b = c.
So it makes sense to introduce the abbreviation f(a):
b = c = f(a).

> Sqrt is generally extended to apply to a Reimann surface so it
> is plus on one part of the Reimann surface and negative on the other. So
> there strict answer is either one or the other, but not both. Generally
> the positive value is taken to be the value of sqrt of lamda x(x)^(1/2

Open a textbook on physics, engineering, calculus,


statistics, chemistry, whatever... when you see sqrt(9),
do you stop and ask yourself

"Halt, what would they mean here, 3 or -3" ?
Do you?
No. You know they mean 3 because sqrt is defined

that way: it gives the positive root of a positive number.

I'm well aware of the extensions, but this is the
context were are talking about :-)

Dirk Vdm


Paul Cardinale

unread,
Sep 18, 2003, 11:25:29 AM9/18/03
to
"Mathew Orman" <or...@nospam.com> wrote in message news:<bkabf6$5fj$1...@news.onet.pl>...

Orman is no ordinary crackpot. He is actively engaged in criminal
fraud. Unforntunately, when it comes to snake-oil peddlers, the U.S.
Gov't. closes it's eye's. So orman will probably never get the free
stay in Uncle Sam's Iron Hotel that he deserves.

Paul Cardinale

AndroclesInEngland

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Sep 18, 2003, 11:30:50 AM9/18/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3f69...@usenet01.boi.hp.com...
Correct.

> A B Xor
> 0 0 0
> 0 1 1
> 1 0 1
> 1 1 1

Wrong. You table is identical to the 'Or', but the light switch diagrams
below are clearly different even to a moron - but obviously not to a
sub-moron.
But you snipped again, didn't you?

> A B =>


> 0 0 1
> 0 1 1
> 1 0 0
> 1 1 1
>
> (A xor B ) => (A or B)
> 0 0 0 1 0 0 0
> 0 1 1 1 0 1 1
> 1 1 0 1 1 1 0
> 1 0 1 1 1 1 1
> QED
>
> Unless you use another kind of logic of course.
>
> Dirk Vdm

Of course I do, and so does everyone else (except relativists, of course).


AndroclesInEngland

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Sep 18, 2003, 11:35:57 AM9/18/03
to

"Robert J. Kolker" <bobk...@attbi.com> wrote in message
news:bkch6n$5u6o$1...@ID-76471.news.uni-berlin.de...
>
>
You can't help Dinky, Bob.
He thinks that 'xor' is identical to 'or'.
Androcles


Dirk Van de moortel

unread,
Sep 18, 2003, 12:15:34 PM9/18/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:Kckab.7745$sj....@news-binary.blueyonder.co.uk...

Indeed, that was an obvious typo.
You could have checked that I used the correct subsitution
in the last part.
So here we go again:

(A xor B)
<==> (A and B') or (A' and B)
<==> (A or A') and (A or B) and (B' or A') and (B' or B)
<==> (A or B) and (B' or A')
==> (A or B)
QED

Or if you prefer truth tables:
A B Or
0 0 0
0 1 1
1 0 1
1 1 1

A B Xor
0 0 0
0 1 1
1 0 1
1 1 0

A B =>

Dirk Van de moortel

unread,
Sep 18, 2003, 12:17:45 PM9/18/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:xhkab.7780$sj....@news-binary.blueyonder.co.uk...

I said xor *implies* or.

(A xor B)
<==> (A and B') or (A' and B)
<==> (A or A') and (A or B) and (B' or A') and (B' or B)
<==> (A or B) and (B' or A')
==> (A or B)
QED

Or if you prefer truth tables:
A B Or

0 0 0
0 1 1
1 0 1

1 1 1

A B Xor
0 0 0
0 1 1
1 0 1

1 1 0 (this was a typo)

A B =>
0 0 1
0 1 1


1 0 0
1 1 1

(A xor B ) => (A or B)

0 0 0 1 0 0 0


0 1 1 1 0 1 1

Mathew Orman

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Sep 18, 2003, 1:43:13 PM9/18/03
to

Funny capacitor :)

Richard

unread,
Sep 18, 2003, 2:09:36 PM9/18/03
to

Dirk Van de moortel wrote:
>

When K' comes to a stop, virtually instantly, K retains his position wrt
K, while K'' instantly moves further away. What does the deceleration of
K' into rest frame do other than alter his perception of the positions
in time? Now here is the real contradiction Dirk, because we cannot
maintain that one frame maintains its position in space and time, while
another changes in position, because we could just as well have let K''
assume the role of K, by introducing yet another frame K''' moving twice
the speed of K'' wrt K'. IOW the speed of K'' wrt K' is arbitrary, and
thus we cannot single out one particular speed of K'' wrt K' that will
not cause a change in spacetime position through the deceleration of K'
back into rest frame. Thus K'' will not instantly move further away from
K' upon the deceleration of K' any more than K does. Thus simultaneity
must hold wrt frames. IOW

x' = (x-vt)g
t' = t g

The ticking rates of clocks in these frames is constant, i.e. the two
way trip need not be accounted for to derive the ticking rates as it is
in the lorentz transformation, and it is just

dt = dt'g

As you can see the relationship of x and x' intervals remains unaltered.

Try it. The two way speed of a light beam moving along the x axis will
still be c wrt all observers, but the one way speed will be c only wrt a
particular frame. Moreover simultaneity will hold wrt translations
between frames.

Here's another thought or two:

Suppose we have a laser clock oriented parallel to the Earths surface.
As the clock is raised from ground level to orbital level (we are
assuming no angular acceleration of the clock) the ticking rate of the
clock increases. Since L remains constant, because the clock is at rest
wrt the ground based clock, we can only account for the increased
ticking rate in terms of a higher light speed than that measured at
ground level. Thus we can deduce that only in free space, far distant
from any local fields (objects), will light propagate at its maximum c.
Near the surface of the Earth light speed must be somewhat less than c.
We can compare an altitude with a given speed, namely the speed that the
clock would be moving had it fallen from infinity to its present
altitude, and I've already shown that this approach does in fact give
exactly the gravitational time dilation as prescribed by Einstein. Thus
time dilation occurs in this case without doubt as a function of motion
and position wrt a local field, the gravitational field of the Earth. It
is not a reach then to assume that velocity dependent time dilation is
also due to motion wrt that same field, as it is really logically
impossible to provide for exactly the same effect being caused by two
separate and independent causes.

Richard Perry

http://www.cswnet.com/~rper

Mathew Orman

unread,
Sep 18, 2003, 2:19:32 PM9/18/03
to

Speculation: Tell the truth about Einstein and the cash cow dies (Einstein
loses credibility). Engage in half truths about Einstein for decades, get
Time Magazine to elevate Einstein to the status of Person of the Century and
the supreme con works. All you have to do to make the con work is suppress
unfavorable information about Einstein He was a plagiarist yet somehow this
information never made it to the American public. How many of you ever heard
that Einstein was a blatant, brazen plagiarist? Why havent you heard it? The
physics community buried it. Then the physics community makes billions of
dollars from grants, awards, power, prestige, stipends, money, travel, book
deals, honoraria, etc over the next several decades. Advertising people
would readily agree, that Person of the Century is worth billions of dollars
to the physics community over the next 25 to 50 years. This is similar to
artificially inflating a stock price with false stories and then cashing in
at the lofty price. Substitute getting research grants approved instead of
cashing in at a lofty price, and you have the physics community.
It should be obvious that there is absolutely no incentive for the physics
community to tell the truth about Einstein except that they might get
caught. They have a tremendous conflict of interest between telling the
truth about Einstein or promoting physics. It appears that telling the truth
about Einstein is not high on the list of priorities of the physics
community. One of the ways that the physics community has legal problems is
with respect to the phrase passive fraud. I define passive fraud as the
willingness to leave on the record false information and then benefit from
that false information. In other words, the physics community may not have
to do anything proactive. All they have to do is permit false and misleading
information to go unchallenged and then benefit from it.
The physics community may face the following charges: 1)Conspiracy, for
failing to get out the truth about Einstein, 2)Fraud, for the placement (or
covert support) of false information on the record and then benefitting from
it, 3)Perjury, for false statements in research grants, 4)Racketeering, For
acting like organized criminals. 5)Obstruction of justice as the scientists
try to destroy documents. The justice department should squeeze
undergraduates first, then graduates, then adjunct professors, then full
professors, then Department Heads, then the chancellors of the universities,
and then Nobel Laureates. The level of criminality that the physics
community has engaged in is perhaps fourth or fifth behind organized crime
as the leading class of criminals in our society in terms of total felonies
committed and magnitude of the money involved.
One of the biggest problems facing the physics community is that they are
incredibly arrogant and view themselves as being intellectually superior.
Consider these passages from Gleicks book, CHAOS, Making a New Science:
These scientists had experience with brilliance and with eccentricity. p. 2,
I understand youre real smart, Agnew said to Feigenbaum, If youre so smart,
why dont you just solve laser fusion? p.2, To a physicist, creating laser
fusion was a legitimate problem; puzzling out the spin and color and flavor
of small particles was a legitimate problem; dating the origin of the
universe was a legitimate problem. Understanding clouds was a job for a
meteorologist. Like other physicists, Feigenbaum used an understated,
tough-guy vocabulary to rate such problems. p.3, The most passionate
advocates of the new science go so far as to say that Twentieth century
science will be remembered for just three things: relativity, quantum
mechanics, and chaos. p.6
What the physics community is going to realize is that, painfully, their
very intelligence is their greatest enemy; a Mafia underling with an IQ of
80, might not realize he was in the middle of a conspiracy. Is a brilliant
physicist unaware of what it meant if Einstein was a plagiarist or that the
physics community had falsified the data surrounding the eclipse of 1919 and
1922? Were they completely unaware that falsehoods with respect to Einstein
would result in an increase in funding to the physics community? The physics
community has two choices: They may argue that they are incredible
ignoramuses and have no idea of illegal activities. The basic problem with
this approach is that the phyics community has gone to great pains to tell
us how brilliant they are. Now that they are caught conducting illegal
activity, they have to play dumb. Somehow, a defense along the lines of:
When Im a physicist, Im brilliant, but when Im a crook, Im stupid, has poor
prospects of success. This is not a recommended defense for the physics
community. So why are we funding stupid people? Or, they knew what was going
on and deliberately covered it up. In which case, they are crooks.

Why are we funding crooks?

Richard Moody Jr.


Check the hard evidece at : www.ultra-faster-than-light.com

Sincerely,

Mathew Orman


Dirk Van de moortel

unread,
Sep 18, 2003, 2:30:19 PM9/18/03
to

"Richard" <no_mail...@yahoo.com> wrote in message news:3F69F4E0...@yahoo.com...

>
> Dirk Van de moortel wrote:
> >
> > "Richard" <no_mail...@yahoo.com> wrote in message news:3F692266...@yahoo.com...

[snip]

> > > According to SR

K always retains his position wrt K,
so I suppose you want to say
K' always retains his position wrt K.
That probably was a typo. Okay.

> while K'' instantly moves further away.

Yes.

> What does the deceleration of
> K' into rest frame do other than alter his perception of the positions
> in time? Now here is the real contradiction Dirk, because we cannot
> maintain that one frame maintains its position in space and time, while
> another changes in position, because we could just as well have let K''
> assume the role of K, by introducing yet another frame K''' moving twice
> the speed of K'' wrt K'. IOW the speed of K'' wrt K' is arbitrary, and
> thus we cannot single out one particular speed of K'' wrt K' that will
> not cause a change in spacetime position through the deceleration of K'
> back into rest frame. Thus K'' will not instantly move further away from
> K' upon the deceleration of K' any more than K does. Thus simultaneity
> must hold wrt frames.

I don't understand a single word of this, so let's
look at your IOW:

> IOW
>
> x' = (x-vt)g
> t' = t g

The second equation is valid for x = 0 only, so
the first equation says
x' = -vgt
and thus
x' = -vt'
which is simply the equation of motion of the spatial
origin (x=0) of the K-frame, expressed in the K'-frame.
(just like x = vt is the equation of motion of the spatial
origin (x'=0) of the K'-frame, expressed in the K-frame).
Do you realize this?
Do you understand this?
Are these other words for whatever it was that you
were trying to explain?

Dirk Vdm


Uncle Al

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Sep 18, 2003, 2:56:30 PM9/18/03
to
Mathew Orman wrote:
>
> Speculation: Tell the truth about Einstein and the cash cow dies (Einstein
> loses credibility).

http://w0rli.home.att.net/youare.swf
http://www.mazepath.com/uncleal/sunshine.jpg

http://arXiv.org/abs/hep-th/0307140
GR structure, especially Part 4/p. 7
<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/Volume4/2001-4will/index.html>
Experimental constraints on General Relativity.
<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/Volume6/2003-1ashby/index.html>
http://www.eftaylor.com/pub/projecta.pdf
Relativity in the GPS system

[snip]

Learn how to write a subject heading, asshole.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

AndroclesInEngland

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Sep 18, 2003, 6:43:49 PM9/18/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:GSkab.26657$s36.1...@phobos.telenet-ops.be...
Nope. Obvious typos are aloud < (obvious typo, I should have said
'allowed') in written Englihs <obvious typo, but what you've written are
called stupid errors.

> You could have checked that I used the correct subsitution < obvious typo,
allowed - I understand it means substitution)
> in the last part.

Nothing to check. Your conclusion is wrong, do your own homework.


> So here we go again:
>
> (A xor B)
> <==> (A and B') or (A' and B)
> <==> (A or A') and (A or B) and (B' or A') and (B' or B)
> <==> (A or B) and (B' or A')
> ==> (A or B)
> QED

Utter gibberish.
Use a Venn diagram, you should spot your own stupid error.

Paul Cardinale

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Sep 18, 2003, 6:57:32 PM9/18/03
to
"Mathew Orman" <or...@nospam.com> wrote in message news:<bkcsug$je6$1...@news.onet.pl>...

[snip crap]

You are the only person fascinated by your own vomit.

Paul Cardinale

Jim Greenfield

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Sep 19, 2003, 12:02:19 AM9/19/03
to
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bk9em7$rnm$1...@dolly.uninett.no>...
> "Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...
>
> What is the point in quoting a lot of text which you don't
> comment or make any references to whatsoever?
>
> > Why confuse the issue?
>
> And in order to not confuse the issue,
> you introduce an entirely different scenario?
>
> > Observer on platform with train stationary sees light path from
> > ceiling of carriage heading towards center of earth. When train is
> > moving, path becomes diagonal to said observer, returning to a
> > different position with RT him and the railway line. ERGO: light beam
> > now is not directed to earth center, and light IS SOURCE DEPENDENT
>
> Why are you shouting trivialities?
> Of course the direction of the velcoity of light depend
> on the velocity of the source.
> Or in other words:
> The direction of the velocity of light is frame dependent.
> So what?
> Did you have a point?
>
> Paul

Far from shouting Paul- just remarking that velocity IS directional,
and if the train's motion alters the light's motion IN ANY CASE, then
light is defacto source dependent both in Speed AND direction.
I can't show you the math on this detector in the middle scenario, but
I am confident that an analysis would show that a photon travelling
toward the rear is treated differently than one coming from the guards
van: train velocity is subtracted in one case from c, but not added
in the other.

Jim G

Jim Greenfield

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Sep 19, 2003, 1:15:49 AM9/19/03
to
John Anderson <and...@attglobal.net> wrote in message news:<3F68FE15...@attglobal.net>...
> Richard wrote:
>
> > Here's another one for the SRist crackpots to crack:
> >
> > Two events occur along the x axis, they are simultaneous in frame K, and
> > separated by the non-zero space interval x. According to the crackpots
> > these events will not be simultaneous wrt any frame K' that is in motion
> > wrt K along the x axis. Now since it is maintained that the order of
> > events doesn't change between frames,
>
> You are an idiot. The temporal ordering of spacelike separated eventsIS
> manifestly frame-dependent in SR.
>
> SR doesn't have to explain the fact that you don't understand what it
> predicts.
>
> John Anderson

John,
Take the case of a submarine travelling on the surface, and one below
it. At speed, the surface vessel LTs using c in air. The one below
would use velocity of c in water---- so for the identical boats
travelling at the same speed, we get a different size. Of course we
could fill the top one with glass and do the measurements with the
light speed in that medium....or take it to a higher altitude (less
atmospheric pressure, etc etc!
Ergo: neither boat "really" shrinks, and errors caused by illusions
are just that- errors. ie the observer is MISTAKEN! It is not a point
of view situation.
Velocity doesn't shrink anything, and any amount of (+) and (-)
arbitrarily and changing and choosing 'reference frames' will not make
it so.
Jim G

Mathew Orman

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Sep 19, 2003, 2:51:05 AM9/19/03
to

"Paul Cardinale" <pcard...@volcanomail.com> wrote in message
news:64050551.03091...@posting.google.com...

http://www.mazepath.com/uncleal/sunshine.jpg

AndroclesInEngland

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Sep 19, 2003, 3:29:16 AM9/19/03
to

"Jim Greenfield" <greenf...@hotmail.com> wrote in message
news:3c4afb26.03091...@posting.google.com...

That's a somewhat strange statement, Jim. If, as you say, the velocity of
light is source dependent, why would it be treated differently depending on
direction? It would seem to me that you want it both ways, and I can't
agree with that. Why on earth would it be added to one and not be added to
the other?


Dirk Van de moortel

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Sep 19, 2003, 3:55:16 AM9/19/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:9Iqab.41$Z_...@news-binary.blueyonder.co.uk...

>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:GSkab.26657$s36.1...@phobos.telenet-ops.be...
> >

[snip]

> Nothing to check. Your conclusion is wrong, do your own homework.
> > So here we go again:
> >
> > (A xor B)
> > <==> (A and B') or (A' and B)
> > <==> (A or A') and (A or B) and (B' or A') and (B' or B)
> > <==> (A or B) and (B' or A')
> > ==> (A or B)
> > QED
> Utter gibberish.
> Use a Venn diagram, you should spot your own stupid error.

My Venn diagram says that (A xor B) is a subset of (A or B)
http://users.pandora.be/vdmoortel/dirk/Stuff/AxorB.gif
QED (3)

Well, I must say that you really did your best:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
Title: "Logic is utter gibberish"

Dirk Vdm


AndroclesInEngland

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Sep 19, 2003, 6:41:49 AM9/19/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3f6ab6f8$1...@usenet01.boi.hp.com...
What ARE you babbling about? From your own diagram, the intersection that is
white is different from the intersection that is blue, and therefore doesn't
imply they are the same. Kindly change the title to "Dinky van der Mumble
doesn't use or understand logic"


Dirk Van de moortel

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Sep 19, 2003, 7:06:33 AM9/19/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:O3Bab.10577$Z_5....@news-binary.blueyonder.co.uk...

The white part in the top figures is (A and B), the intersection.
The blue part in the top figure is (A xor B), the exclusive union
The blue part in the bottom figure is (A or B), the union.

Dirk Vdm


Randy Poe

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Sep 19, 2003, 7:39:17 AM9/19/03
to
On Fri, 19 Sep 2003 11:41:49 +0100, "AndroclesInEngland"
<jp006...@blueyonder.co.uk> wrote:
>> My Venn diagram says that (A xor B) is a subset of (A or B)
>> http://users.pandora.be/vdmoortel/dirk/Stuff/AxorB.gif
>> QED (3)
>>
>> Well, I must say that you really did your best:
>> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
>> Title: "Logic is utter gibberish"
>What ARE you babbling about? From your own diagram, the intersection that is
>white is different from the intersection that is blue, and therefore doesn't
>imply they are the same. Kindly change the title to "Dinky van der Mumble
>doesn't use or understand logic"
>

You're defending an indefensible position and embarrassing yourself.

Ask yourself what the elements of A xor B are. They are things which
are in A, or in B, but not both. Do you really think that's the area
in white? Hint: No, it's the area in blue.

Now what are the elements of A or B? It's the union of A and B, or the
entire blue area in the bottom diagram. Clearly the blue above is a
subset of the blue below.

I've noticed in this thread you keep trying to pretend Dirk is saying
something different than what he is. Here you try to pretend "A xor B"
is the intersection of A and B. Do you really want to defend that?
Before I saw you take the statement "A xor B IMPLIES A or B" and say
"Dirk claims these two statements are equivalent".

p => q is not the same claim as "p is equivalent to q". I can say "A
is a dog implies A is a mammal" but that doesn't mean mammal and dog
are equivalent. Just look at the verbal DESCRIPTION of these two sets:
A or B contains everything which is in A, or in B.
A xor B contains everything which is in A or in B but not both.

Clearly to create the second set you start with the things in the
first set ("everything which is in A or in B") and eliminate some of
them ("but not both"). When you take a set and eliminate some
elements, you get what we like to call a "subset".

- Randy

AndroclesInEngland

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Sep 19, 2003, 11:38:21 AM9/19/03
to

"Randy Poe" <rpo...@yahoo.com> wrote in message
news:7bqlmv0j8bqkqrlvb...@4ax.com...
I'm not arguing about it being a subset, the problem is Drinks insistence
that (A xor B) IMPLIES (A or B). He often makes errors, then brushes them
off as typos.


Paul Cardinale

unread,
Sep 19, 2003, 12:31:13 PM9/19/03
to
"Mathew Orman" <or...@nospam.com> wrote in message news:<bke8vi$cu2$1...@news.onet.pl>...

> "Paul Cardinale" <pcard...@volcanomail.com> wrote in message
> news:64050551.03091...@posting.google.com...
> > "Mathew Orman" <or...@nospam.com> wrote in message
> news:<bkcsug$je6$1...@news.onet.pl>...
> >
> > [snip crap]
> >
> > You are the only person fascinated by your own vomit.
> >
> > Paul Cardinale
>
> http://www.mazepath.com/uncleal/sunshine.jpg
>
That picture of you taking a meal demonstrates why your vomit and your
excrement are the same.

Dirk Van de moortel

unread,
Sep 19, 2003, 12:49:58 PM9/19/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:QpFab.2935$562...@news-binary.blueyonder.co.uk...

An implication in logic is isomorphic with a subset in set theory:

A subset of B
<==> for all x: IF x in A THEN x in B
i.o.w.
for all x: x in A ==> x in B
i.o.w.
for all x: x in A IMPLIES x in B

(P xor Q) IMPLIES (P or Q) and there is nothing
you can do about it.

What does it take before you ask yourself that there
might just be a small possibility that *you* are flatly
wrong?

Dirk Vdm


Paul B. Andersen

unread,
Sep 19, 2003, 2:42:54 PM9/19/03
to

"Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:<bk9em7$rnm$1...@dolly.uninett.no>...
> > "Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...
> >
> > What is the point in quoting a lot of text which you don't
> > comment or make any references to whatsoever?
> >
> > > Why confuse the issue?
> >
> > And in order to not confuse the issue,
> > you introduce an entirely different scenario?
> >
> > > Observer on platform with train stationary sees light path from
> > > ceiling of carriage heading towards center of earth. When train is
> > > moving, path becomes diagonal to said observer, returning to a
> > > different position with RT him and the railway line. ERGO: light beam
> > > now is not directed to earth center, and light IS SOURCE DEPENDENT
> >
> > Why are you shouting trivialities?
> > Of course the direction of the velcoity of light depend
> > on the velocity of the source.
> > Or in other words:
> > The direction of the velocity of light is frame dependent.
> > So what?
> > Did you have a point?
> >
> > Paul
>
> Far from shouting Paul- just remarking that velocity IS directional,
> and if the train's motion alters the light's motion IN ANY CASE, then
> light is defacto source dependent both in Speed AND direction.

So:
If the velocity of light transform in such a way that its
speed is invariant but the direction changes,
then the speed changes IN ANY CASE.

Shouting helps!

> I can't show you the math on this detector in the middle scenario, but
> I am confident that an analysis would show that a photon travelling
> toward the rear is treated differently than one coming from the guards
> van: train velocity is subtracted in one case from c, but not added
> in the other.

"I can't show it, but I am confident that it is"
is a real convincing argument.
I am not sure for what, though.

Paul


ste...@nomail.com

unread,
Sep 19, 2003, 2:48:15 PM9/19/03
to
In sci.physics.relativity AndroclesInEngland <jp006...@blueyonder.co.uk> wrote:

: "Randy Poe" <rpo...@yahoo.com> wrote in message
: news:7bqlmv0j8bqkqrlvb...@4ax.com...
:>
:> I've noticed in this thread you keep trying to pretend Dirk is saying


:> something different than what he is. Here you try to pretend "A xor B"
:> is the intersection of A and B. Do you really want to defend that?
:> Before I saw you take the statement "A xor B IMPLIES A or B" and say
:> "Dirk claims these two statements are equivalent".

:>
<snip>
:>
: I'm not arguing about it being a subset, the problem is Drinks insistence


: that (A xor B) IMPLIES (A or B). He often makes errors, then brushes them
: off as typos.

Dirk is correct. (A xor B) IMPLIES (A or B). In order to show him
wrong, you would need to demonstrate a case where (A xor B) is true
but (A or B) is false. This is simple logic.

Stephen

Dirk Van de moortel

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Sep 19, 2003, 2:55:30 PM9/19/03
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"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bkfio8$jku$1...@dolly.uninett.no...

>
> "Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...

[snip]

> > Far from shouting Paul- just remarking that velocity IS directional,
> > and if the train's motion alters the light's motion IN ANY CASE, then
> > light is defacto source dependent both in Speed AND direction.
>
> So:
> If the velocity of light transform in such a way that its
> speed is invariant but the direction changes,
> then the speed changes IN ANY CASE.

Of course, and exactly how much it changes
would dramaticaly depend on the material the
CASE IS MADE OFF .

>
> Shouting helps!

hm yes, sometimes it does ;-)

>
> > I can't show you the math on this detector in the middle scenario, but
> > I am confident that an analysis would show that a photon travelling
> > toward the rear is treated differently than one coming from the guards
> > van: train velocity is subtracted in one case from c, but not added
> > in the other.
>
> "I can't show it, but I am confident that it is"
> is a real convincing argument.
> I am not sure for what, though.

He's working on the mathematical analysis. Just wait...
but don't expect any equations.

Dirk Vdm


Dirk Van de moortel

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Sep 19, 2003, 3:00:32 PM9/19/03
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<ste...@nomail.com> wrote in message news:bkfj1f$647$3...@msunews.cl.msu.edu...

It's so simple to verify on the Venn diagram he wanted
http://users.pandora.be/vdmoortel/dirk/Stuff/AxorB.gif
If something is sitting in the blue area of the top figure, THEN
it sits in the blue area of the bottom figure.
So now he's got 3 'proofs' and he still doesn't get it.
This is not simple logic, this is baby logic. Sheesh.

Dirk Vdm


Randy Poe

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Sep 20, 2003, 1:14:38 AM9/20/03
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On Fri, 19 Sep 2003 16:38:21 +0100, "AndroclesInEngland"
<jp006...@blueyonder.co.uk> wrote:
>I'm not arguing about it being a subset,

Good, because you sure seemed to be in this exchange:


>> >> My Venn diagram says that (A xor B) is a subset of (A or B)

>> >What ARE you babbling about? From your own diagram, the intersection that


>> >white is different from the intersection that is blue, and therefore doesn't
>> >imply they are the same

> the problem is Drinks insistence


>that (A xor B) IMPLIES (A or B).

Um, that's the same thing as saying that the set that satisfies "A xor
B" is a subset of the set that satisfies "A or B". Belonging to the
first set implies belonging to the second set if and only if the first
set is a subset of the second.

> He often makes errors, then brushes them
>off as typos.

This is neither, as it's correct.

You may be confused as to what "implies" means.

X is a dog implies that X is a mammal. The set of dogs is a subset of
the set of mammals.

- Randy

Jim Greenfield

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Sep 20, 2003, 4:53:14 AM9/20/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:<hfyab.7352$Z_5....@news-binary.blueyonder.co.uk>...

On the contrary, it looks to me as if that is the way light is treated
by Fitzgerald/Lorentz, and I think that is Wrong. The contraction is
brought about by the Arbitrary use of +, - in formula involving
velocity (speed/direction)
To me (c+v) along direction of travel is equivalent to (c+v) towards
source, and that arbitrarily changing 'frame of reference' causes
illusion of contraction.

Jim G

Jim Greenfield

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Sep 20, 2003, 5:03:14 AM9/20/03
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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:<CiIab.28538$dy5.1...@phobos.telenet-ops.be>...

Actually, I think the math is up to you: with it you can show me why a
surface submarine is calculated by LT to be different from identical
one submerged, taken same position and velocity of both- shouldn't be
too hard......

Jim G

AndroclesInEngland

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Sep 20, 2003, 6:34:20 AM9/20/03
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"Randy Poe" <rpo...@yahoo.com> wrote in message
news:i9onmvcdhq9srvglt...@4ax.com...

You may be confused as to the meaning of parentheses.
X is a dog or Y is a dog implies (X or Y) is a mammal - true.
X is a dog and Y is a dog implies (X and Y) is a mammal - true.
(X is a dog) xor (Y is a teacup) implies (X and (not Y)) is a mammal - true.
(X is a dog) xor (Y is a dog) implies (X and (not Y)) implies not Y is a
mammal - false.
The white intersection of Dinky's Venn isn't a subset of the blue
intersection, nor the blue a subset of the white. Having found a single case
where his proposition is false, his entire proposition is false.
Hence (A xor B) does NOT imply [all of] (A or B), anymore than his assertion
that sqrt(x) has only one possible answer does.

Paul B. Andersen

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Sep 20, 2003, 8:38:51 AM9/20/03
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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> skrev i melding
news:CiIab.28538$dy5.1...@phobos.telenet-ops.be...

>
> "Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bkfio8$jku$1...@dolly.uninett.no...
> >
> > "Jim Greenfield" <greenf...@hotmail.com> skrev i melding news:3c4afb26.03091...@posting.google.com...
>
> [snip]
>
> > > Far from shouting Paul- just remarking that velocity IS directional,
> > > and if the train's motion alters the light's motion IN ANY CASE, then
> > > light is defacto source dependent both in Speed AND direction.
> >
> > So:
> > If the velocity of light transform in such a way that its
> > speed is invariant but the direction changes,
> > then the speed changes IN ANY CASE.
>
> Of course, and exactly how much it changes
> would dramaticaly depend on the material the
> CASE IS MADE OFF .

Or rather of what material the case is filled with.
Let's say the case is Jims head.
Would the speed of light be c?

Paul


Dirk Van de moortel

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Sep 20, 2003, 8:43:07 AM9/20/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:bDXab.169$FP....@news-binary.blueyonder.co.uk...

You perfectly know that I said:
The white part in the top figures is (A and B), the intersection.
The blue part in the top figure is (A xor B), the exclusive union
The blue part in the bottom figure is (A or B), the union.

http://users.pandora.be/vdmoortel/dirk/Stuff/AxorB.gif
I will add this:
I was not talking about the intersection (the with part).
The blue part in the top figure is (A xor B), the exclusive union,
is a subset of the blue part in the bottom figure is (A or B), the
union.

I also said:
An implication in logic is isomorphic with a subset in set theory:

A subset of B
<==> for all x: IF x in A THEN x in B
i.o.w.
for all x: x in A ==> x in B
i.o.w.
for all x: x in A IMPLIES x in B

(P xor Q) IMPLIES (P or Q) and there is nothing
you can do about it.

I gave you 3 proofs: formal logic, truth tables, and
a Venn diagram.

Have you even *looked* at this?
(A xor B ) => (A or B)
0 0 0 1 0 0 0


0 1 1 1 0 1 1

1 1 0 1 1 1 0
1 0 1 1 1 1 1
You see all ones under the implication arrow "=>".
Can you tell me where the error is?

You got this:


(A xor B)
<==> (A and B') or (A' and B)
<==> (A or A') and (A or B) and (B' or A') and (B' or B)
<==> (A or B) and (B' or A')
==> (A or B)

Can you tell me where the error is?


> Having found a single case
> where his proposition is false, his entire proposition is false.
> Hence (A xor B) does NOT imply [all of] (A or B), anymore than his assertion
> that sqrt(x) has only one possible answer does.

So, you still maintain that sqrt(9) can be -3?
So, when you take an engineering book and you
see a square root, you don't know what they mean?
You see sqrt(9) in a textbook and you ask yourself
whether they could mean -3?
You still have not answered this simple question.

Androcles, what does it take before you realize that
you *might* be in the process of *learning* something
from someone?

Dirk Vdm


Dirk Van de moortel

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Sep 20, 2003, 8:49:12 AM9/20/03
to

"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:bkhhpn$4h2$1...@dolly.uninett.no...

Hm, indeed.
In Jim's case, the material being *that* dense, light
speed in it would be experimentally indistinguishable
from c I guess?

Dirk Vdm


AndroclesInEngland

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Sep 20, 2003, 8:55:09 AM9/20/03
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"Jim Greenfield" <greenf...@hotmail.com> wrote in message
news:3c4afb26.0309...@posting.google.com...
What contraction and what illusion of contraction? I've never seen anything
actually contract, or produce the illusion of contraction (except the quite
normal experience of cooling something down, or letting the air out of a
balloon, of course, and I don't think you mean that. As you appear to be
referring to the Lorentz contraction, that's a figment of the relativist's
imagination, not a physical event.
Look, suppose you are moving toward a star that has been emitting light in
your direction for eons. Isn't the speed of that light c + v, where v is
your velocity with respect to the star? Suppose you turn around and head
away from it. Isn't the speed of light c-v now? And if you stop dead, so
that you are at rest relative to the star, isn't the speed of light from the
star now c? What is it about your motion that automatically changes the
speed of light from a star when it is (say) ten light years away from you?
What about if the star is in orbit around another, so that it alternately
moves toward you and away from you. Does that make its speed c, or not?
Of course, if it were not, some newer faster photons would catch up with
slower, older photons, and the star would appear from more than one position
at the same time when it arrived, wouldn't it? And it would also appear to
vary in brightness, too. So that can't be right.... or can it? See
http://www.astro.uiuc.edu/stardial/variables/mira.html
Androcles


AndroclesInEngland

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Sep 20, 2003, 11:22:05 AM9/20/03
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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:vXXab.29620$PE5.1...@phobos.telenet-ops.be...
[snip]
> Dirk Vdm
Does it occur to you, dirtbag, that lying to make your pathetic jokes is
extremely unethical? Nowhere have I ever said "Logic is utter gibberish",
but you, scumbag, have chosen to maliciously quote me as saying that. You
are a pathetic liar, not to be believed by anyone, and the only reply you
can expect if you answer one of my posts again is this one.
Androcles.


ste...@nomail.com

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Sep 20, 2003, 2:26:21 PM9/20/03
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In sci.physics.relativity AndroclesInEngland <jp006...@blueyonder.co.uk> wrote:

: "Randy Poe" <rpo...@yahoo.com> wrote in message
:> You may be confused as to what "implies" means.


:>
:> X is a dog implies that X is a mammal. The set of dogs is a subset of
:> the set of mammals.

: You may be confused as to the meaning of parentheses.
: X is a dog or Y is a dog implies (X or Y) is a mammal - true.
: X is a dog and Y is a dog implies (X and Y) is a mammal - true.
: (X is a dog) xor (Y is a teacup) implies (X and (not Y)) is a mammal - true.
: (X is a dog) xor (Y is a dog) implies (X and (not Y)) implies not Y is a
: mammal - false.
: The white intersection of Dinky's Venn isn't a subset of the blue
: intersection, nor the blue a subset of the white. Having found a single case
: where his proposition is false, his entire proposition is false.
: Hence (A xor B) does NOT imply [all of] (A or B), anymore than his assertion
: that sqrt(x) has only one possible answer does.

Where did this [all of] come from? You cannot just insert words
in the middle of an argument and claim you are correct.
"Implies" is well defined in logic, and you were simply wrong
that (A xor B) Implies (A or B) is false. Now you dishonestly
insert [all of], which is not even well defined, and pretend
this is what the argument was about all the time.

Stephen

Dirk Van de moortel

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Sep 20, 2003, 2:47:27 PM9/20/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:1r0bb.13$Ph...@news-binary.blueyonder.co.uk...

Where is the lie?
I gave you logic:


| > (A xor B)
| > <==> (A and B') or (A' and B)
| > <==> (A or A') and (A or B) and (B' or A') and (B' or B)
| > <==> (A or B) and (B' or A')
| > ==> (A or B)

| > QED
And you replied with:
| Utter gibberish.

Evidence in
http://groups.google.com/groups?&threadm=9Iqab.41$Z_...@news-binary.blueyonder.co.uk

I gave you 2 other proofs, and you said:
- Utter gibberish.
- Use a Venn diagram, you should spot your own stupid error.
- Nothing to check. Your conclusion is wrong,
do your own homework
- What ARE you babbling about?
- He often makes errors, then brushes them off as typos
- his entire proposition is false.

What do you think that means?

Dirk Vdm


Dirk Van de moortel

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Sep 20, 2003, 2:55:45 PM9/20/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:1r0bb.13$Ph...@news-binary.blueyonder.co.uk...

... and in case you wonder about the phrase "Logic is utter
gibberish"... this is the title of your fumble. Not a quote.
Compare with the other fumbles. If it was a quote, I would
have used double quotes around it in the title.

Dirk Vdm


me...@cars3.uchicago.edu

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Sep 20, 2003, 5:54:54 PM9/20/03
to
In article <bki64d$2li8$1...@msunews.cl.msu.edu>, ste...@nomail.com writes:
>In sci.physics.relativity AndroclesInEngland <jp006...@blueyonder.co.uk> wrote:
>
>: "Randy Poe" <rpo...@yahoo.com> wrote in message
>:> You may be confused as to what "implies" means.

>:>
>:> X is a dog implies that X is a mammal. The set of dogs is a subset of
>:> the set of mammals.
>
>: You may be confused as to the meaning of parentheses.
>: X is a dog or Y is a dog implies (X or Y) is a mammal - true.
>: X is a dog and Y is a dog implies (X and Y) is a mammal - true.
>: (X is a dog) xor (Y is a teacup) implies (X and (not Y)) is a mammal - true.
>: (X is a dog) xor (Y is a dog) implies (X and (not Y)) implies not Y is a
>: mammal - false.
>: The white intersection of Dinky's Venn isn't a subset of the blue
>: intersection, nor the blue a subset of the white. Having found a single case
>: where his proposition is false, his entire proposition is false.
>: Hence (A xor B) does NOT imply [all of] (A or B), anymore than his assertion
>: that sqrt(x) has only one possible answer does.
>
>Where did this [all of] come from? You cannot just insert words
>in the middle of an argument and claim you are correct.
>"Implies" is well defined in logic, and you were simply wrong
>that (A xor B) Implies (A or B) is false.

No, it is most certainly true. The opposite isn't.

Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"

AndroclesInEngland

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Sep 20, 2003, 5:49:29 PM9/20/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:3h1bb.30101$vN3.1...@phobos.telenet-ops.be...

You claimed I said "Logic is utter gibberish". I did not say that. Nor do
you have a record of me saying that. I have a record of YOU saying that,
falsely accusing me of the statement. Therefore you are an unethical liar,
not to be trusted.
Androcles


AndroclesInEngland

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Sep 20, 2003, 5:57:10 PM9/20/03
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:Ro1bb.30114$sc6.1...@phobos.telenet-ops.be...

You lie again. See
news:nNUab.29288$n05.1...@phobos.telenet-ops.be...
Quoting what you said by copying and pasting.
open quote
Androcles:
(A xor B) does not imply (A or B)
Logic is utter gibberish!
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
close quote.
I see no quotes. I do see it indented after my name. You have falsely
attributed that statement to me. You are a scoundrel.


Randy Poe

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Sep 20, 2003, 6:47:50 PM9/20/03
to
On Sat, 20 Sep 2003 11:34:20 +0100, "AndroclesInEngland"
<jp006...@blueyonder.co.uk> wrote:

>Hence (A xor B) does NOT imply [all of] (A or B),

"imply all of"? What does that mean? (A xor B) means that either
property A is true or property B is true, but not both. For example,
let A = "X is a dog" and B = "X > 50 kg". If X is an adult elephant,
(A xor B) is true. If X is a 75 kg St. Bernard, (A xor B) is false.
Clear?

OK, now the statement (A or B) is true if either A is true, or B is
true, or both. (A or B) is true for the elephant. (A or B) is true for
the St. Bernard. That's not a problem. The claim is that
(A xor B) => (A or B), and what that means is if I choose an X such
that (A xor B) is true, then (A or B) is true.

Symbolically, this is written with a double arrow to the right =>
Nobody would read this as "implies all of". There's no such phrase. If
the second statement implies the first, we can write a double arrow to
the left <=. If both implications hold, we write a double-headed
arrow: <=> This can be read as "the two statements are equivalent" or
"the first statement is true if and only if the second statement is
true". That seems to be what you think "implies" means, when you
mutate it into "implies all of".

For the elephant, (A xor B) is true. If I say that this implies (A or
B) is true, it means the elephant satisfies (A or B). What the hell
would it mean to say that the elephant satisfies "all of" (A or B).

> anymore than his assertion
>that sqrt(x) has only one possible answer does.

sqrt(x) is the symbol which is conventionally taken to stand for the
non-negative root when x >=0, and i times a nonnegative value when x <
0. It stands for only one of the two possible square roots of x. The
reason it is useful is that it is useful to have a "square root
function", and functions are single-valued.

- Randy

AndroclesInEngland

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Sep 20, 2003, 7:03:16 PM9/20/03
to

<ste...@nomail.com> wrote in message
news:bki64d$2li8$1...@msunews.cl.msu.edu...
If you really want to be pedantic, X is a dog implies that X is a mammal
isn't a valid statement at all. Missing is "All dogs are mammals". Then:
For All dogs: dogs are mammals. There exists a dog X. Therefore X is a
mammal.
How's this?
1) Dinky logic

> > > (A xor B)
> > > <==> (A and B') or (A' and B)
> > > <==> (A or A') and (A or B) and (B' or A') and (B' or B)
> > > <==> (A or B) and (B' or A')
> > > ==> (A or B)

1) My copy of Dinky logic, with NOT reversed throughout
(A xor B)
<==> (A' and B) or (A and B')
<==> (A' or A) and (A or B) and (B' or A') and (B or B')
<==> (A' or B') and (B or A)
==> (A' or B')
That what you want?


Dirk Van de moortel

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Sep 20, 2003, 7:18:10 PM9/20/03
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"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:535bb.126$tw...@news-binary.blueyonder.co.uk...

P and Q ==> P
P and Q ==> Q

P and Q and R ==> P
P and Q and R ==> Q
P and Q and R ==> R

Want me to continue?

;-)

Dirk Vdm


Dirk Van de moortel

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Sep 20, 2003, 7:20:30 PM9/20/03
to

"AndroclesInEngland" <jp006...@blueyonder.co.uk> wrote in message news:s34bb.21$Tf...@news-binary.blueyonder.co.uk...

Where is the lie?
I gave you logic:


| > (A xor B)
| > <==> (A and B') or (A' and B)
| > <==> (A or A') and (A or B) and (B' or A') and (B' or B)
| > <==> (A or B) and (B' or A')
| > ==> (A or B)

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